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THEORY and PROBLEMS

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MODERNALGEBRA

hy FRANK AYRBS.JR.

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iSCHAUM PUBLISHING CO,

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SCHAVM'S OUTLIISE OF

THEORY AI^D PROBLEMS

OF

MODERX ALGEBRA

BY

FRANK AYRES, JR., Ph.D.

Formerly Professor and Head,

Department of Mathematics

Dickinson College

SCHAVM PlTBUSHmO CO.257 Park Avenue South, New York 10010

/

/

Copyright (311965^ By The

SCHAUM PUBLISHING COMPANY

All rights reserved. This book or anypart thereof may not be reproduced in

any form without written permission

from the publishers.

PRINTED IN THE UNITED STATES OF AMERICA

Preface

This study of algebraic systems is designed to be used either as a supplement tocurrent texts or as a text for a course in modern abstract algebra at the junior-seniorlevel. As such, it is intended to provide a solid foundation for future study of avariety of systems rather than to be a study in depth of any one or more.

The basic ingredients of algebraic systems - sets of elements, relations, operations,mappings -are treated in the first two chapters. The pattern established here,

(i) a simple and concise presentation of each topic,

(ii) a wide variety of familiar examples,(iii) proofs of most theorems included among the solved problems,(iv) a carefully selected set of supplementary exercises,

is followed throughout the book.

Beginning in Chapter 3 with the Peano postulates for the natural numbers, theseveral number systems of elementary algebra are constructed in turn and theirsalient properties deduced. This not only introduces the reader to a detailed andrigorous development of these number systems but also provides him with muchneeded practice for the deduction of properties of the abstract systems which follow.

The first abstract algebraic system — the Group — is considered in Chapter 9.

Coaets of a subgroup, invariant subgroups and their quotient groups are investigated,and the chapter ends with the Jordan-Holder Theorem for finite groups.

Chapters 10-11 are concerned with Rings, Integral Domains and Fields. Poly-n<nnials over rings and fields are then considered in Chapter 12 along with a certainamount of ekimentary theory of equations. Throughout these chapters, considerableattention is given to finite rings.

Vector spaces are introduced in Chapter 13. The algebra of linear transformationson a vector space of finite dimension leads naturally to the algebra of matrices(Chapter 14). Matrices are then used to solve systems of linear equations and, thus,provide simpler solutions of a number of problems connected with vector spaces.Matrix polynomials are considered in Chapter 15 as an example of a non-commutativepolynomial ring. The characteristic polynomial of a square matrix over a field isthen defined. The characteristic roots and associated invariant vectors of real sym-metric matrices are used to reduce the equations of conies and quadric surfaces tostandard form. Linear algebras are formally defined in Chapter 16 and otherexamples briefly considered.

In the final chapter. Boolean algebras are introduced and the important applica-tion to simple electric circuits indicated.

The author wishes to take this opportunity to express his appreciation to thestaff of the Schaum Publishing Company, especially to Jeffrey Albert and AlanHopenwasser, for their unfailing cooperation.

Frank Ayres, Jr.

Carlisle, Peniâ.

June, 1966

CONTENTS

Page

Chapter 1 SETSSets. Equal sets. Subsets of a set. Universal sets. Intersection and unionof sets. Venn diagrams. Operations with sets. The product set. Mappings.One-to-one mappings. One-to-one mapping of a set onto itself.

Chapter 2 RELATIONS AND OPERATIONS 15Relations. Properties of binary relations. Equivalence relations. Equivalencesets. Ordering in sets. Operations. Types of binary operations. Well-defined

operations. Isomorphisms. Permutations. Transpositions. Algebraic systems.

Chapter 3 THE NATURAL NUMBERS 30The Peano postulates. Addition. Multiplication. Mathematical induction.

The order relations. Multiples and powers. Isomorphic sets.

Chapter i THE INTEGERS 38Introduction. Binary relation ~ . Addition and multiplication. The positive

integers. Zero and the negative integers. The integers. Order relations.

Subtraction. Absolute value. Other properties of integers. Multiples andpowers.

Chapter 5 SOME PROPERTIES OF INTEGERS 49Divisors. Primes. Greatest common divisor. Relatively prime integers. Primefactors. Congruences. The algebra of residue classes. Linear congruences.Positional notation for integers.

Chapter 6 THE RATIONAL NUMBERS 60The rational numbers. Addition and multiplication. Subtraction and divi-

sion. Replacement. Order relations. Reduction to lowest terms. Decimalrepresentation.

Chapter 7 THE REAL NUMBERS 65Introduction. Dedekind cuts. Positive cuts. Multiplicative inverses. Additiveinverses. Multiplication. Subtraction and division. Order relations. Proper-ties of the real numbers. Summary.

Chapter 8 THE COMPLEX NUMBERS 75The system of complex numbers. Addition and multiplication. Properties ofcomplex numbers. Subtraction and division. Trigonometric representation.Roots. Primitive roots of unity.

Chapter 9 GROUPS 82Groups. Simple properties of groups. Subgroups. Cyclic groups. Permuta-tion groups. Homomorphisms. Isomorphisms. Cosets. Invariant subgroups.Quotient groups. Product of subgroups. Composition series.

CONTENTS

Page

Chapter 10 RINGS 101

Rings. Properties of rings. Subrings. Types of rings. Characteristic. Divi-

sors of zero. Homomorphisms and isomorphisms. Ideals. Principal ideals.

Prime and maximal ideals. Quotient rings. Euclidean rings.

Chapter // INTEGRAL DOMAINS, DIVISION RINGS, FIELDS 114

Integral domains. Unit, associate, divisor. Subdomains. Ordered integral

domains. Division algorithm. Unique factorization. Division rings. Fields.

Chapter 12 POLYNOMIALS 124

Introduction. Polynomial forms. Monic polynomials. Division. Commutative

polynomial rings with unity. Substitution process. The polynomial domain

f{x]. Prime polynomials. The polynomial domain C[x\. Greatest commondivisor. Properties of the polynomial domain y[a;].

Chapter 13 VECTOR SPACES 143

Introduction. Vector spaces. Subspace of a vector space. Linear dependence.

Bases of a vector space. Vector spaces over R. Linear transformations. The

algebra of linear transformations.

Chapter 14 MATRICES 164

Introduction. Square matrices. Total matrix algebra. A matrix of order

m X w. Solutions of a system of linear equations. Elementary transformations

on a matrix. Upper triangular, lower triangular and diagonal matrices. Acanonical form. Elementary column transformations. Elementary matrices.

Inverses of elementary matrices. The inverse of a non-singular matrix.

Minimum polynomial of a square matrix. Systems of linear equations. De-

terminant of a square matrix. Properties of determinants. Evaluation of

determinants.

Chapter 15 MATRIX POLYNOMIALS 198

Matrices with polynomial elements. Elementary transformations. Normal

form. Polynomials with matrix coefficients. Division algorithm. The charac-

teristic roots and vectors of a matrix. Similar matrices. Real symmetric

matrices. Orthogonal matrices. Conies and quadric surfaces.

Chapter 16 LINEAR ALGEBRAS . . . .

.

Linear algebra. An isomorphism.

Chapter 17

219

BOOLEAN ALGEBRAS 222

Boolean algebra. Boolean functions. Normal forms. Changing the form of a

Boolean function. Order relation in a Boolean algebra. Algebra of electrical

networks. Simplification of networks.

INDEX 239

INDEX OF SYMBOLS 245

chapter 1

Sets

SETS

Any collection of objects as (a) the points of a given line segment, (b) the lines througha given point in ordinary space, (c) the natural numbers less than ten, (d) the five Jonesboys and their dog, (e) the pages of this book will be called a set or class. Theindividual points, lines, numbers, boys and dog, pages will be called elements of therespective sets. Generally, sets will be denoted by capital letters and arbitrary elementsof sets will be denoted by small letters.

Let A be a given set and let p and q denote certain objects. When p is an element ofA, we shall indicate this fact by writing p G A; when both p and g are elements of A,we shall write p,qGA instead of p G A and q G A; when q is not an element of A, weshall write g S A.

Although in much of our study of sets we will not be concerned with the type ofelements, sets of numbers will naturally appear in many of the examples and problems.For convenience, we shall now reserve

A^ to denote the set of all natural numbers/ to denote the set of all integers

Q to denote the set of all rational numbersR to denote the set of all real numbers

Example 1:

(a) 1 G A^ and 205 G N since 1 and 205 are natural numbers; |, -5 € AT since 4and —5 are not natural numbers.

(6) The symbol S indicates membership and may be translated as "in", "is in","are in", "be in" according to context. Thus, "Let r e Q" may be' read as"Let r be in Q" and "For any p,g £ I" may be read as "For any p and qin /". We shall at times write nÔei instead of n ^ 0, n G I; alsop¥=0,q G I instead of p,q G I with p ¥- 0.

The sets to be introduced here will always be well-defined, that is, it will always bepossible to determine whether any given object does or does not belong to the particularset. The sets of the first paragraph were defined by means of precise statements in wordsAt times, a set will be given in tabular form by exhibiting its elements between a pair ofbraces; for example,

A = {a} is the set consisting of the single element a.

B = {a, b] is the set consisting of the two elements a and b.

C - {1, 2, 3, 4} is the set of natural numbers less than 5.

K = (2, 4, 6, ... } is the set of all even natural numbers.L = {..., -15, -10, -5, 0, 5, 10, 15, ... } is the set of all integers

having 5 as a factor.

The sets C, K, and L above may also be defined as follows:

C = {x: xGN, x<5}K = {x: X GN, xis even}

L = {x: xGl, a; is divisible by 5}

Here each set consists of all objects x satisfying the conditions following the colon.

See Problem 1.

2 SETS [CHAP. 1

EQUAL SETSWhen two sets A and B consist of the same elements, they are called equal and we

shall write A = B. To indicate that A and B are not equal, we shall write A¥' B.

Example 2: (i) When A = {Mary, Helen, John} and B = {Helen, John, Mary}, then A = B.

Note that a variation in the order in which the elements of a set are tabulated

is immaterial.

(ii) When A - {2, 3, 4} and B = {3, 2, 3, 2, 4}, then A = B since each element of

A is in B and each element of 5 is in A. Note that a set is not changed by

repeating one or more of its elements,

(iii) When A = {1,2} and B = {1,2,3,4}, then A^ B since 3 is an element of

B but not of A.

SUBSETS OF A SETLet S be a given set. Any set A, each of whose elements is also an element of S, is

said to be contained in S and called a subset of S.

Example 3: The sets A = {2}, B = {1,2,3}, and C = {4,5} are subsets of S = {1,2,3,4,5}.

Also, D - {1,2,3,4,5} = S is a subset of S.

The set E = {1, 2, 6} is not a subset of S since 6 e £; but 6 g S.

Let A be a subset of S. If A ^ S, we shall call A a proper subset of S and write

Acs (to be read "A is a proper subset of S" or "A is properly contained in S"). More

often and in particular when the possibility A = S is not excluded, we shall write A C S

(to be read "A is a subset of S" or "A is contained in S"). Of all the subsets of a given

set S, only S Itself is improper, that is, is not a proper subset of S.

Example 4: For the sets of Example 3 we may write A c S, B C S, C c S, D C S, E ^ S. The

precise statements, of course, are AcS, BcS,CcS, D = S,E^S.

Note carefully that €. connects an element and a set, while C and C connect two

sets. Thus, 2 G S and {2} C S are correct statements, while 2 C S and {2} G Sare incorrect.

Let A be a proper subset of S. Now S consists of the elements of A together with

certain elements not in A. The totality of these latter elements, i.e. those not in A, con-

stitute another proper subset of S called the complement of the subset A in S.

Example 5: For the set S = {1,2,3,4,5} of Example 3, the complement of A = {2} in S is

F = {1, 3, 4, 5}. Also, B - {1, 2, 3} and C = {4, 5} are complementary subsets in S.

Our discussion of complementary subsets of a given set implies that these subsets be

proper. The reason is simply that, thus far, we have been depending upon intuition

regarding sets; that is, we have tacitly assumed that every set must have at least one

element. In order to remove this restriction (also to provide a complement for the improper

subset S in S), we introduce the empty or null set as the set having no elements. There

follow readily^j^ ^ .^ ^ ^^^^^^ ^^ ^^^^^ ^^^ g

(ii) is a proper subset of every set S ^ 0.

Example 6: The subsets of S - {a,b,c} are 0, {a}, {6}, {c}, {a,b}, {a,c}, {b,c), and {a,b,e}.

The pairs of complementary subsets are:

{a,b,e} and {0,6} and {c}

{a,c} and {5} {b,c} and {a}

There is an even number of subsets and, hence, an odd number of proper subsets of

a set of 3 elements. Is this true for a set of 303 elements? of 303,000 elements?

UNIVERSAL SETSIf U ¥^0 is a given set whose subsets are under consideration, the given set will

often be referred to as a universal set.

CHAP. 1] SETS

Example 7: Consider the equation

(a; + l)(2a;-3)(3a! + 4)(a;2-2)(a;2 + l) =

whose solution set, that is, the set whose elements are the roots of the equation, is

S = {—1, 3/2, —4/3, yfi, —V2, », —»} provided the universal set is the set of

all complex numbers. However, if the universal set is R, the solution set is

A = {-1, 3/2, -4/3, V2, -V2 }. What is the solution set if the universal set is Q?

is/? is AT?

If, on the contrary, we are given two sets A — {1,2,3} and B — {4,5,6,7}, andnothing more, we have little knowledge of the universal set C/ of which they are subsets.

For example, V might be {1,2,3, . . .,7}, [x: x gN, x^ 1000}, N, I, .... Nevertheless,

when dealing with a number of sets A,B,C, . . ., we shall always think of them as subsets

of some universal set U not necessarily explicitly defined. With respect to this universal set,

the complements of the subsets A, B, C, . . . will be denoted by A', B', C, . . . respectively.

INTERSECTION AND UNION OF SETS

Let A and B be given sets. The set of all elements which belong to both A and B is

called the intersection of A and B. It will be denoted by A n B (read either as "the inter-

section of A and B" or as "A cap B"). Thus,

AnB = {x: x&A and x&B)The set of all elements which belong to A alone or to B alone or to both A and B is

called the union of A and B. It will be denoted hy A\J B (read either as "the union ofA and B" or as "A cup B"). Thus,

A\J B = {x: x & A alone or a; £ S alone or x & A CiB}

More often, however, we shall write

AKJ B = {x: x&A or x&B)The two are equivalent since every element of A n B is an element of A.

Example 8: Let A = {1,2,3,4} and B = {2,3,6,8,10}; then A \J B = {1,2,3,4,5,8,10} andA nB = {2,3}. o 1 T, i_, « .See also Problems 2-4.

Two sets A and B will be called disjoint if they have no element in common, that is, if

A n B = 0. In Example 6, any two of the sets {a}, {6}, {c} are disjoint; also the sets{a, 6} and {c}, the sets {a,c) and {b), and the sets {6,c} and {a} are disjoint.

VENN DIAGRAMSThe complement, intersection, and union of sets may be pictured by means of Venn

diagrams. In the diagrams below the universal set U is represented by points (not indi-

cated) in the interior of a rectangle, and any of its non-empty subsets by points in theinterior of closed curves. (To avoid confusion, we shall agree that no element of U is

represented by a point on the boundary of any of these curves.) In Fig. l-l(a), the subsetsA and B of U satisfy A C B; in Fig. 1-1(6), A n B = 0; in Fig. l-l(c), A and B have atleast one element in common so that A n B ^ 0.

'^(o)

SETS [CHAP. 1

Suppose now the interior of U, except for the interior of A, in the diagrams above beshaded. In each case, the shaded area will represent the complementary set A' of A in U.

The union AuB and the intersection A n B of the sets A and B of Fig. l-l(c) arerepresented by the shaded area in Fig. l-2(a) and (b) respectively. In Fig. l-2(a), theunshaded area represents {A U B)', the complement of A U 5 in U; in Fig. 1-2(6), theunshaded area represents {Ar\B)'. From these diagrams, as also from the definitions of

U and n, it is clear that A U 5 = B U A and Ar\B = B r\ A.See Problems 5-7.

(̂6)

Fig. 1-2

OPERATIONS WITH SETS

In addition to complementation, union, and intersection, which we shall call operations

with sets, we define:

The difference A — B, in that order, of two sets A and B is the set of all elements of Awhich do not belong to B, i.e.,

A-B = {x: xGA, x^B)In Fig. 1-3, A- B i& represented by the shaded area

and B — A by the cross-hatched area. There follow

A-B = AnB' = B'-A'A-B ^ ^ if and only if A C BA-B = B-A if and only if A = BA- B = A if and only if A n B =

Example 9: Prove:

ie) A

(a) A-

(a) A - B = A n B' = B' - A'; (6) A-B- B = A if and only if A n B = 0.

Fig. 1-3

if and only if A C B;

= {x: xGA, x€B} = {« : « G A and a; e B'} = AnB'= {« : « « A', a; G B'} = B'-A'

(b) Suppose A-B = 0. Then, by (o), A n B' = 0, i.e., A and B' are disjoint.

Now B and B' are disjoint; hence, since B U B' = U, we have A C B.

Conversely, suppose A C B. Then A n B' — and A — B = 0.

(c) Suppose A—B = A. Then A n B' = A, i.e., A c B'. Hence, by (6),

A n (B')' = A n B =

Conversely, suppose A n B = 0. Then Aand A-B = A.

-B' - 0, A cB', A n B' = A

In Problems 5-7, Venn diagrams have been used to illustrate a number of properties

of operations with sets. Conversely, further possible properties may be read out of these

diagrams. For example. Fig. 1-3 suggests

(A-B) U (B-A) = (A U B) - (A n B)

It must be understood, however, that while any theorem or property can be illustrated by

a Venn diagram, no theorem can be proved by the use of one.

CHAP. 1] SETS

Example 10: Prove (A - B) U (B - A) = (AuB) - (AnB).

The proof consists in showing that every element of (A — B) U {B — A) is anelement of (AuB) — (AnB) and, conversely, every element of (AuB) — (AnB)is an element of (A — B) U (B — A). Each step follows from a previous definition

and it will be left for the reader to substantiate these steps.

Let X e (A-B) u (B- A); then x & A ~ B or x e B - A.then xGA but x S B; if x G B - A, then xG B but x g A.xeAuB but xÂnB. Hence, x G (AuB) - (AnB) and

(A-B)u(B-A) c (AuB) -(AnB)

If a; e A - B,

In either case.

Conversely, let a; G (A uB) - (A nB); then a; G A U B but a; € A n B. Noweither a; G A but a; S B, i.e., a; G A - B, or xG B but x & A, i.e., x e B - A.Hence, a; G (A - B) U (B - A) and (AuB) -(AnB) C (A-B)U(B-A).

Finally, (A - B) u (B - A) c (A u B) - (A n B) and(AuB)~(AnB) c (A-B)U(B-A) imply (A - B) U (B - A) = (AuB) - (AnB)

For future reference we list below the more important laws governing operations withsets. Here the sets A,B,C are subsets of U the universal set.

LAWS OF OPERATIONS WITH SETS

(1.1) (A'Y = A

(1.2) 0' = U (1.2') U' =

(1.3) A-A = 0, A-0 ^ A, A - B = A n B'

(1.4) A u = A (1.4') A n t; = A(1.5) AuU = U (15') A n =0(1.6) A u A = A (1.6') An A = A(1.7) A u A' = U (1.7') An A' =

Associative Laws

(1.8) (AuB)uC = Au(BuC) (1.8') (AnB)nC = An (BnC)

Commutative Laws(1.9) AuB = B u A (1.9') AnB = B n A

Distributive Laws

(1.10) Au(BnC) = (AuB)n(AuC) (i.io') A n (BuC) = (AnB) u (AnC)

De Morgan's Laws(1.11) (AuB)' = A'nB' (1.11') (AnB)' = A' u B'

(1.12) A - (BuC) = (A - B) n (A - C) (1.12') A -(BnC) = (A-B)U(A-C)

See Problems 8-16.

THE PRODUCT SETLet A ^ {a,b} and B = {b,c,d}. The set of distinct ordered pairs

C = {{a, b), (a, c), (a, d), (&, b), (b, c), {b, d)}

in which the first component of each pair is an element of A while the second is an elementof B, is called the product set C = AxB (in that order) of the given sets. Thus, if Aand B are arbitrary sets, we define

AxB = {{x,y): x G A, y G B}

SETS [CHAP. 1

Example 11: Identify the elements of X^ = {1,2,3}

as the coordinates of points on the a;-axis

(see Fig. 1-4), thought of as a numberscale, and the elements of F = {1, 2, 3, 4}as the coordinates of points on the y-axis,

thought of as a number scale. Then the

elements of X x Y are the rectangular

coordinates of the 12 points shown.

Similarly, when X = Y — N, the set

X X Y are the coordinates of all points

in the first quadrant having integral

coordinates.

s-

2

1-

• • •C1.4) (2,4) (S,4)

(1,3) (2,3) X3, 3)

• • •(1,2) (2.2) (3,2)

(1.1) (2,1) (8,1)

Fig. 1-4

MAPPINGSConsider the set H = {hi, hi, ha, . . ., ha} of all houses on a certain block of Main Street

and the set C = [Ci, d, Cz, . . ., C39] of all children living in this block. We shall be con-

cerned here with the natural association of each child of C with the house of H in which

the child lives. Let us assume that this results in associating Ci with hi, d with h^, Cz with

hi, C4 with As, C5 with hi, . . ., C39 with hz. Such an association of or correspondence between

the elements of C and H is called a mapping of C into H. The unique element of Hassociated with any element of C is called the image of that element (of C) in the mapping.

Now there are two possibilities for this mapping: (1) every element of H is an image,

that is, in each house there lives at least one child; (2) at least one element of H is not an

image, that is, in at least one house there live no children. In the case (1), we shall call

the correspondence a mapping of C onto H. Thus, the use of 'onto' instead of 'into' calls

attention to the fact that in the mapping every element of H is an image. In the case (2),

we shall call the correspondence a mapping of C into, but not onto, H. Whenever we write

"a is a mapping of A into B" the possibility that a may, in fact, be a mapping of A onto Bis not excluded. Only when it is necessarj' to distinguish between cases will we write either

"a is a mapping of A onto B" or "a is a mapping of A into, but not onto, B".

A particular mapping a of one set into another may be defined in various ways. For

example, the mapping of C into H above may be defined by listing the ordered pairs

a = {{ci,h2), {C2,hi), {cz,h2), {Ci,hi), {C5,hg), . . ., {ca9,ha)}

It is now clear that « is simply a certain subset of the product setCxH oi C and H. Hence,

we define

A mapping of a set A into a set B is a subset of A x B in which each element of Aoccurs once and only once as the first component in the elements of the subset.

In any mapping a of A into B, the set A is called the domain and the set B is called the

co-domain of a. If the mapping is "onto", B is also called the range of a; otherwise, the

range of « is the proper subset of B consisting of the images of all elements of A.

A mapping of a set A into a set B may also be displayed by the use of -* to connect

associated elements.

Example 12: Let A = {a,b,c} and B - {1,2}. Then

a: a^l, 6^2, c-»2

is a mapping of A onto B (every element of B is an image) while

/? : 1 -» a, 2 -* 6

is a mapping of B into, but not onto, A (not every element of A is an image).

CHAP. 1] SETS

In the mapping a, A is the domain and B is both the co-domain and the range.

In the mapping p, B is the domain, A is the co-domain, and C = {a, 6} C A is the

range.

When the number of elements involved is small, Venn diagrams may be used

to advantage. Fig. 1-5 below displays the mappings a and p of this example.

.A

'n

Fig. 1-5

A third way of denoting a mapping is discussed in

Example 13: Consider the mapping a oi N into itself, that is, of N into N,

a: 1 -> 3, 2 ^ 5, 3 -* 7, 4 ^ 9, . . .

or, more compactly, a : m-»2m + 1, n & N

Such a mapping will frequently be defined by

a : la = 3, 2a = 5, 3a = 7, 4a = 9, ...

or, more compactly, by a : na = 2n + 1, n G N

Here N is the domain (also the co-domain) but not the range of the mapping. Therange is the proper subset M oi N given by

M = {x: X = 2n + l, n&N)

or M = {x: x€.N, X \& odd}

Mappings of a set X into a set Y, especially when X and Y are sets of numbers, arebetter known to the reader as functions. For instance, defining X = N and Y = M inExample 13 and using / instead of a, the mapping (function) may be expressed in func-tional notation as

(i) y = fix) = 2a; -h 1

We say here that y is defined as a function of x. It is customary nowadays to distinguishbetween "function" and "function of". Thus, in the example, we would define the func-tion / by

/ = {{x,y): y = 2x + l,xGX}

or / = {(x, 2a; + 1) : x G X}

that is, as the particular subset of X x Y, and consider (i) as the "rule" by which thissubset is determined. Throughout much of this book we shall use the term mapping ratherthan function and, thus, find little use for the functional notation.

Let a be a mapping of A into B and j8 be a mapping of B into C. Now the effect of ais to map aGA into aa G B and the effect of B is to map aa G B into {aa)p G C. Thenet result of applying a followed by |8 is a mapping of A into C which we define by

a^ : a(a/3) = iaa)p, aGAWe shall call a^ the product of the mappings a and ^ in that order. Unfortunately, thenotation has not been standardized so that a/3 is sometimes used to denote the effect of themapping /3 followed by the mapping «. Note also that we have used the term producttwice in this chapter with meanings quite different from the familiar product, say, of twointegers. This is unavoidable unless we keep inventing new names.

SETS [CHAP. 1

Example 14: Let A = {a,b,c}, B = {d,e}, C = {f,g,h,i} and

a '. aa =^ d, ba == e, Ca = e

P : d/3 = f, ep = h

Then a/3 : a(al3) = (aa)p = d/} = f, 6(a/3) = efi = h, c(aji) = h

Fig. 1-6

ONE-TO-ONE MAPPINGSA mapping a-^ a' of a set A into a set B is called a one-to-one mapping of A into B

if the images of distinct elements of A are distinct elements of B; if, in addition, every

element of B is an image, the mapping is called a one-to-one mapping of A onto B. In the

latter case, it is clear that the mapping a-^ a' induces a mapping a' -» a of B onto A.

The two mappings are usually combined into a <-> a' and called a one-to-one correspondence

between A and B.

Example 15: (a) The mapping a of Example 14 is not a one-to-one mapping of A into B (the dis-

tinct elements h and c of A have the same image).

(6) The mapping /8 of Example 14 is a one-to-one mapping of B into, but not onto,

C (fif € C is not an image).

(c) When A = {a, b, c, d} and B = {p, q, r, s},

(i) ffi : o «-> p, 6 <^ g, c <^ r, d <-> s

and (ii) ag : a <-» r, 6 <^ p, c <-> g, d «-» s

are examples of one-to-one mappings of A onto B.

Two sets A and B are said to have the same number of elements if and only if a

one-to-one mapping of A onto B exists. A set A is said to have n elements if there exists

a one-to-one mapping of A onto the subset S = {1,2, 3, . . .,n} of N. In this case, A is

called a finite set.

The mapping « : na = 2n, n G Nof N onto the proper subset M =^ {x: x G N, x is even} of A^ is both one-to-one and onto.

Now N is an infinite set; in fact, we may define an infinite set as one for which there exists

a one-to-one correspondence between it and one of its proper subsets.

An infinite set is called countable or denumerable if there exists a one-to-one corre-

spondence between it and the set A'^ of all natural numbers.

ONE-TO-ONE MAPPING OF A SET ONTO ITSELF

Let a: x-^x + 1, jS: x <^Sx, y:a;<^2a;-5, S: x (^x-1

be one-to-one mappings of R onto itself. Since for any x GR

CHAP. 1] SETS 9

while xl3a = (3x)a = Sx + 1

we see that (i) a/? ¥- â

However, iCyS = (2a; -5)8 - 2x-6

and Xa{yS) = {x + l)y8 = 2(x + 1) - 6 = 2a; - 4

while xay = (a; + l)y = 2a; — 3

and a;(ay)8 = (2a; - 3) - 1 = 2x - 4

Thus, (ii) a(y8) = {ay)8

Now xa8 = (a; + 1)8 = x

and a;Sa = {x-l)a = x

that is, a followed by 8 (also, 8 followed by a) maps each x GR into itself. Denote by J,the identity mapping,

J : X <^ X

Then (iii) „s = 8a = JJ

that is, 8 undoes whatever a does (also, a undoes whatever 8 does). In view of (iii), 8 is

called the inverse mapping of a and we write 8 = «"'; also, « is the inverse of 8 and wewrite a = 8~K

See Problem 18.

In Problem 19, we prove

Theorem I, If « is a one-to-one mapping of a set S onto a set T, then a has a uniqueinverse, and conversely.


Theorem II. If « is a one-to-one mapping of a set S onto a set T and /? is a one-to-onemapping of T onto a set U, then (a/S)"' = /J-i • «->.

Solved Problems

1. Exhibit in tabular form: (a) A = {a: a G N, 2 < a <6}, (b) B = {p: p G N, p < 10,pis odd}, (c) C = {a;: a; G/, 2a;2-)-a;-6 = 0}.

(a) Here A consists of all natural numbers (a e N) between 2 and 6; thus, A = {3, 4, 5}.

(6) B consists of the odd natural numbers less than 10; thus, B = {1,3,5,7,9}.

(c) The elements of C are the integral roots of 2x2 + x - 6 = (2a; - 3){a; + 2) = 0; thus C = {-2}.

10 SETS [CHAP. 1

2. Let A = {a,h,c,d}, B - {a,c,g}, C - {c,g,m,n,p}. Find:

AuB = {a,b,c,d,g}, AuC = {a,b,c,d,g,m,n,p}, BuC = {a,c,g,m,n,p};

AnB = {a,c}, AnC = {c], BnC = {c,g}; A n (BuC) = {a,c};

{AnB) U C = {a,c,g,m,n,p}, (AuB) nC = {c,g}, (AnB) U (AnC) = A n (BuC).

3. Consider the subsets K = {2,4,6,8}, L = {1,2,3,4), M = {3,4,5,6,8} of f7 ={1,2,3, ...,10}. (a) Exhibit K',L',M' in tabular form. (6) Show that {KkjL)' =K' n L'.

(a) K' = {1,3,5,7,9,10}, L' = {5,6,7,8,9,10}, M' = {1,2,7,9,10}.

(6) KU L = {1, 2, 3, 4, 6, 8} so that (K U L)' = {5, 7, 9, 10}. Then K' n L' = (5, 7, 9, 10} = (K U L)'.

4. For the sets of Problem 2, show: (a) (A U B) U C = A U (B U C), (6) (A n B) n C =A n (B n C).

(o) Since A \J B = {a,h,c,d,g} and C = {cgyVi.n.p}, we have {A U B) U C = {a,b,e,d,g,m,n,p}.

Since A = {a, b, c, d} and B u C = {a,e, g, m, n, p}, we have A u {B u C) = {a, b, c, d, g, m, n, p} =(A u B) u C.

(b) Since AnB = {a,c}, we have (A n B) n C = {c}. Since BnC = {o,g), we have

A n (B n C) = {c} = (A n B) n C.

5. In Fig. l-l(c), let C = A n B, D = A n B', E = B r\ A' and F = {A\J B)'. Verify:

(a) (A U B)' = A' n B', (6) (A n B)' = A' U B'.

(a) A' n B' = (B u F) n (D u F) = F = (A U B)'

(6) A' U B' =: (EuF)\j(DkjF) = (EuF)uD = C = {AnB)'

6. Use the Venn diagram of Fig. 1-7 to verify:

(a) E -- (AnB) nC (c) A U B n C is ambiguous

(6) A U B U C = (A U B) U C = A U (B U C) (d) A' D C = G U L

(o) A n B = D U £? and C - E U F u G u L; then

(A n B) n C = JS;

(6) AoBuC = EuFuGuDuHuJuK. NowAuB = EuFuGuDuHuJ

and C = D u H u J u Kso that

(AuB)uC = EuFuGuDuHuJvK pjg ^.^

= A uB U C

Also, BuC = EuGuDuHuJuK and A=£uii'uZ)uH

so that Au(BuC) =EuFuGuDliHuJuK = AliBuC

(c) A U B n C could be interpreted either as (A U B) n C or as A U (B n C). Now (A U B) n C =

D u H uJ, while A u (B n C) = A u (D u J) = A u J. Thus, A u B n C is ambiguous.

(d) A' = GuJuKuL and C = £ U F U G U L; hence, A' n C = G u L.

7. Let A and B be subsets of U. Use Venn diagrams to illustrate: A n B' = A if and

only if A n B = 0.

Suppose A n B = and refer to Fig. l-l(b). Now A C B'; hence A n B' = A.

Suppose A n B ^ and refer to Fig. l-l(c). Now A (^ B'; hence A n B' ¥- A.

Thus, A n B' = A if and only if A n B = 0.

8. Prove: (A U B) U C = A U (B U C).

Let a; G (A u B) u C. Then a; G A U B or x&C, so that a; e A or x&B or x & C. When

a e A, then a; € A U (B U C); when a; e B or a; G C, then a; G B U C and hence a; G A U (B U C).

Thus (A u B) u C c A U (B U C).

CHAP. 1] SETS 11

Let a; e A u (B u C). Then x&A or a; e B u C, so that x&A or x e. B or x&C. Whenx&A or x&B, then x e A U B and hence a; € (A U B) U C; when sc £ C, then x e (A U B) U C.

Thus A u (B u C) c (A u B) u C.

Now (AuB) u C c A u (BuC) and A u (BuC) c (AuB) u C imply (AuB) u C = A u (BuC)

as required. Thus, A U B U C is unambiguous.

9. Prove: (A n fi) n C = A n (B n C).

Let a; G (A n B) n C. Then x e A n B and x G C, so that a; G A and x G B and x G C. Since

X G B and « G C, then x G B n C; since x e. A and x G B n C, then x G A n (B n C). Thus

(A n B) n C c A n (B n C).

Let X G A n (B n C). Then x G A and x G B n C, so that x€. A and x e^ B and x G C. Since

X G A and x&B, then x G A n B; since x G A n B and x€:C, then x G (A n B) n C. Thus

A n (B n C) c (A n B) n C and (A n B) n C = A n (B n C) as required. Thus, A n B n C is unam-

biguous.

10. Prove: A n (B U C) = (A n B) U (A n C).

Let xe.A(^(BuC). Then xGA and x G B U C (xGB or xG C), so that a: G A and xGBor X G A and x G C. When xGA and x G B, then x G A n B and so x G (A n B) U (A n C);

similarly, when x&A and x&C, then x G A n C and so x G (A n B) U (A nC). Thus A n (BuC) C(AnB) u (AnC).

Let X G (A n B) U (A n C), so that xGAnB or xGAnC. When x G A n B, then x&Aand X G B so that x G A and x G B U C; similarly, when x G A n C, then x G A and x G C so that

xe.A and x G B u C. Thus x G A n (B u C) and (A n B) U (A n C) C A n (B u C). Finally

A n (B U C) = (A n B) u (A n C) as required.

11. Prove: (A U B)' = A' n B'.

Let X G (A U B)'. Now x € A U B, so that x € A and x€B. Then x G A' and x G B', that is,

X G A' n B'; hence (A u B)' c A' n B'.

Let X G A' n B'. Now x G A' and x G B', so that x S A and x g B. Then x € A U B, so thatxG(AuB)'; hence A' D B' Q (AuB)'. Thus (A U B)' = A' D B' as required.

12. Prove: (A n J?) U C = (A U C) n (B U C).

Cu (AnB) = (CuA)n (CUB) by (1.10), Page 5.

Then (A n B) u C = (A u C) n (B u C) by (1.9), Page 5.

13. Prove: A - (B U C) = (A - B) n (A - C).

Let X G A - (B U C). Now x G A and x « B U C, that is, x G A but x € B and x € C. ThenX G A - B and xeA-C, so that x G (A - B) n (A - C) and A - (B U C) C (A - B) n (A - C).

Let X G (A - B) n (A - C). Now x G A - B and x G A - C, that is, x G A but x € B and x S C.Then x G A but x^B U C, so that x G A - (B U C) and (A - B) n (A - C) C A - (B U C). ThusA - (B u C) = (A - B) n (A - C) as required.

14. Prove: (A U B) n B' = A if and only if A n B = 0.

Using (1.10') and (1.7'), Page 5, we find

(A u B) n B' = (An B') u (B n B') = AnB'

We are then to prove: A n B' — A if and only if A n B = 0.

(a) Suppose A n B = (2). Then A C B' and A n B' = A.

(6) Suppose A n B' = A. Then A c B' and A n B = 0.

Thus, (A u B) n B' = A if (by (a)) and only if (by (6)) A n B = 0.

12 SETS [CHAP. 1

15. Prove: XcY if and only if Y' c X'.

(i) Suppose X qY. Let y' e Y'. Then y' € X since y' g F; hence, 2/' G Z' and Y' c X'.

(ii) Conversely, suppose Y' C X^'. Now, by (i), (X')' c {Y')'; hence, A' c F as required.

(A - B) U (B - A)

16. Prove the identity {A-B)U (B-A) = {A u B) - {A n B) of Example 10 using theidentity A - B - A n B' of Example 9.

We have

(AnB') u (BnA')

[(A n B') u B] n [(A n B') u A']

[(A u B) n {B' u B)] n [(A u A') n (B' u A')]

[(AuB) n 17] n [i/n (B'uA')]

(A u B) n (B' u A')

(AuB) n (A'uB')

(A u B) n (A n B)'

(A u B) - (A n B)

by (1.10), Page 5

by (1.10)

by (1.7)

by (1.4')

by (1.9)

by (1.11')

17. Show that any two line segments have the same number of points.

Let the line segments be AB and A'B' of Fig. 1-8. We are

to show that it is always possible to establish a one-to-one

correspondence between the points of the two segments. De-

note the intersection of AB' and BA' by P. On AB take anypoint C and denote the intersection of CP and A'B' by C.The mapping

C ^ Cis the required correspondence, since each point of AB has a

unique image on A'B' and each point of A'B' is the image of

a unique point on AB. Fig. 1-8

18. Prove: (a) x-»x + 2 is a mapping of N into, but not onto, A^. (b) x-*3x — 2 is a

one-to-one mapping of Q onto Q, (c) x-* x^-3x^~x is a mapping of R onto R but is

not one-to-one.

(a) Clearly a; -I- 2 S A^ when x G N. The mapping is not onto since 2 is not an image.

(6) Clearly 3a; - 2 G Q when x G Q. Also, each r G Q is the image of x = (r + 2)/3 G Q.

(c) Clearly x^-3x^-xS R when x G R. Also, when reR, x^-3x^-x = r always has a real

root X whose image is r. When r = —3, x^ — Sx^ — x = r has 3 real roots x = —1,1,3. Since

each has r = —3 as image, the mapping is not one-to-one.

19. Prove: If a is a one-to-one mapping of a set S onto a set T, then a has a unique inverse

and conversely.

Suppose a is a one-to-one mapping of S onto T; then for any s & S, we have

s^sa = t G T

Since t is unique, it follows that a induces a one-to-one mapping

p : tp ^ s

Now s(oy8) = (sa)/8 = tp — s; hence, afi = J and ;S is an inverse of a. Suppose this inverse is not

unique; in particular, suppose ;8 and y are inverses of a. Since

ap = Pa - S aild «Y = Y« = (5

it follows that

CHAP. 1] SETS 13

Pay = p{ay) = j3 • J ^ p

and /Say = (/3a)y — J) • y — y

Thus p = y; the inverse of a is unique.

Conversely, let the mapping a of S into T have a unique inverse a""'. Suppose for Sj, S2GS,with Si # S2, we have Sja = S2«- Then (sia)a~^ = {s2a)a~^, so that Siia'a"^) = S2(a:'a~i) and

Si = 82, a contradiction. Thus a is a one-to-one mapping. Now, for any tGT, we have (ta^^)a =t{a~^ • a) = t' J — t; hence, t is the image of s = ta"^ S S and the mapping is onto.

20. Prove: If a is a one-to-one mapping of a set S onto a set T and jS is a one-to-one

mapping of T onto a set U, then ia/3)~^ = j8~i'a""^

Since (a/8)(;8î • a"') = a(/8 • ;8~i)(»~i = a'a~i = ^, p^^ ' a^^ is an inverse of ap. ByProblem 19 such an inverse is unique; hence, (a/8)~i = p~^'a~^.

Supplementary Problems

21. Exhibit each of the following in tabular form:

(a) the set of negative integers greater than —6,

(b) the set of integers between —3 and 4,

(c) the set of integers whose squares are less than 20,

(d) the set of all positive factors of 18,

(e) the set of all common positive factors of 16 and 24,

(/) {p: pGN, p2 < 10}

(g) {b: b e N, 3 ^ b ^ 8}

(h) {x: a; G /, 3a;2 + 7a; + 2 = 0}

(i) {x: a; e Q, 2a;2 + 5a; + 3 = 0}

Partial answer: (a) {-5, -4, -3, -2, -1}, (d) {1, 2, 3, 6, 9, 18}, (/) {1, 2, 3}, {h) {-2}

22. Verify: (a) {a; : a; 6 AT, a; < 1} = 0, (6) {x : x G /, 6a;2 + 5a; - 4 = 0} = 0.

23. Exhibit the 15 proper subsets oi S = {a, b, c, d}.

24. Show that the number of proper subsets of S — {a^, 02. • • • . <»«} is 2» — 1.

25. Using the sets of Problem 2, verify:

(a) (A u B) u C = A u (B u C), (b) (A n B) n C = A n (B n C), (c) (Au B) n C ¥- A u (B n C).

26. Using the sets of Problem 3, verify: (a) (K')' = K, (6) (K n L)' - K' U L', (c) (K U L U M)' =K' n L' n M', id) K n(Lu M) = (K n L)u (Kn M).

27. Let "n|m" mean "n is a factor of m". Given A = {a; : a; G iV, 3

|a;} and B = {a; : x G AT, 5

|a;},

list 4 elements of each of the sets A', B', A U B, A r\ B, A U B', A n B', A' U B' where A' and B'are the respective complements of A and B in AT.

28. Prove those laws of (1.8)-(1.12'), Page 5, which were not treated in Problems 8-13.

14 SETS [CHAP. 1

29. Let A and B be subsets of a universal set V . Prove:

(a) A U B = A n B if and only if A = B,

(5) A n B = A if and only if A c B,

(c) (A n B') U (A' n B) = A U B if and only if A n B = 0.

30. Given n{l]) = 692, n(A) = 300, n(B) = 230, n{0 = 370, n(A n B) = 150, w(A n C) = 180,

«(B n C) = 90, n(A n B' n C) = 10 where w(S) is the number of distinct elements in the set S, find:

(a) ri(AnBnC) = 40 (c) n(A'nB'nC') = 172

(6) m(A' n B n C) = 30 (d) %((A n B) u (A n C) u (B n C)) = 340

31. Given the mappings a : « ^ n^ + 1 and /3 : n -> 3w + 2 of Af into "N, find: «« = «* + 2w2 + 2, /3/8, a/3 =3n2 + 5, and â.

32. Which of the following mappings of / into /:

(a) a; ^ a; + 2, (6) x -> Sx, (c) a; ^ a;^, (d) a; ^ 4 — x, (e) a; -> x^, (/) a; ^ a;^ — a;

are (i) mappings of / onto /, (ii) one-to-one mappings of / onto /? Ans. (i), (ii); (a), (d)

33. Same as Problem 32 with / replaced by Q. Âns. (i), (ii); (a), (6),(d)

34. Same as Problem 32 with / replaced by R. Ans. (i),(ii); (a), (6), (d), (e)

35. (a) If E is the set of all even positive integers, show that xâ; + l, a;Gj& is not a mapping of Eonto the set F of all odd positive integers.

(6) If E* is the set consisting of zero and all even positive integers (i.e., the non-negative integers),

show that X ^ x + 1, xG E* is a mapping of E* onto F.

36. Given the one-to-one mappings

J): 1^ = 1, 2J = 2, 3JJ = 3, 4J = 4

la =2, 2a = 3, 3a = 4, 4a = 1

ip =4, 2j8 = 1, 3)3 = 2, 4/3=3

ly = 3, 2y = 4, 3y = 1, 4y = 2

18 =1, 25 = 4, 3S = 3, 48 = 2

of S = {1, 2, 3, 4} onto itself, verify:

(a) a/8 = j8a = ^, hence p = a-»; (6) ay = ya = P; (c) aS ¥= ha; (d) a- = aa - y; (e) y2 = ^,

hence y-i = y; (/) a* = JJ, hence <^ = aî; (ff) (a^)-! = (a-i)^.

chapter 2

Relations and Operations

RELATIONS

Consider the set P = {a,b,c, . . .,t} of all persons living on a certain block of MainStreet. We shall be concerned in this section with statements such as "a is the brother

of p", "c is the father of g", . . ., called relations on (or in) the set P. Similarly, "is

parallel to", "is perpendicular to", "makes an angle of 45° with", . . . , are relations onthe set L of all lines in a plane.

Suppose in the set P above that the only fathers are c,d,g and that

c is the father of a, g, m, p, q

d is the father of /

g is the father of h, n

Then, with % meaning "is the father of", we may write

c%a, c%g, c%m, c%p, c%q, d%f, g%h, g%nNow c%a may be thought of as determining an ordered pair, either (a, c) or (c, a), of the

product set PxP. Although both will be found in use, we shall always associate

c%a with the ordered pair (a, c)

With this understanding, "3^ determines on P the set of ordered pairs

(a,c), {g,c), (m.c), {p,c), (q^rc), (/,d), (h,g), (n,g)

As in the case of the term function in Chapter 1, we define this subset of P x P to be therelation %. Thus,

A relation 'tZ? on a set S (more precisely, a binary relation on S since it will be arelation between pairs of elements of S) is a subset of S x S.

Example 1 : (a) Let S = {2, 3, 5, 6} and let <^ mean "divides".

Since 2 ^ 2, 2 '5^ 6, 3 ^ 3, 3 "S^ 6, 5 -i^ 5, 6 '5^ 6, we have

% = {(2,2), (6,2), (3,3), (6,3), (5,5), (6,6)}

(6) Let S = {1,2,3, . . .,20} and '3f mean "is three times".

Then 3 ^ 1, 6 '3i 2, 9 '^^ 3, 12 '2^ 4, 15 '?{ 5, 18 gj 6 and

% := {(1,3), (2,6), (3,9), (4,12), (5,15), (6,18)}

(c) Consider % = {(x, y) : 2x-y = 6, xeR}. Geometrically, each («, y) & % is

a point on the graph of the equation 2x — y = 6. Thus, while the choice

c%a means (a, c) G % rather than (c, a) &%may have appeared strange at the time, it is now seen to be in keeping with theidea that any equation y = f(x) is merely a special type of binary relation.

PROPERTIES OF BINARY RELATIONSA relation 9^ on a set S is called reflexive \i a%a for every a e S.

15

16 RELATIONS AND OPERATIONS [CHAP. 2

Example 2: (a) Let T be the set of all triangles in a plane and % mean "is congruent to". Nowany triangle t G T is congruent to itself; thus, t%t for every ( G T, and % is

reflexive.

(6) For the set T let % mean "has twice the area of". Clearly, t ^ t and % is notreflexive.

A relation "J^ on a set S is called symmetric if whenever a%b then b%a.

Example 3: (a) Let P be the set of all persons living on a block of Main Street and % mean"has the same surname as". When x G P has the same surname as y & P,

then y has the same surname as x; thus, x%y implies y%x and % is symmetric.

(6) For the same set P, let % mean "is the brother of" and suppose x%y. Nowy may be the brother or sister of x; thus, xliy does not necessarily imply yl^xand % is not symmetric.

A relation '5^ on a set S is called transitive if whenever a%b and b%c then a%c.

Example 4: (a) Let S be the set of all lines in a plane and % mean "is parallel to". Clearly,

if line a is parallel to line 6 and if b is parallel to line c, then a is parallel to c

and % is transitive.

(6) For the same set S, let % mean "is perpendicular to". Now if line a is per-

pendicular to line 6 and if b is perpendicular to line c, then a is parallel to c.

Thus, 1{ is not transitive.

EQUIVALENCE RELATIONS

A relation <5ft on a set S is called an equivalence relation on S when % is (i) reflexive,

(ii) symmetric, and (iii) transitive.

Example 5: The relation "=" on the set R is undoubtedly the most familiar equivalence relation.

Example 6: Is the relation "has the same surname as" on the set P of Example 3 an equivalence

relation?

Here we must check the validity of each of the following statements involving

arbitrary x,y,z 6 P:

(i) X has the same surname as x.

(ii) If X has the same surname as y, then y has the same surname as x.

(iii) If X has the same surname as y and if y has the same surname as z, then x

has the same surname as z.

Since each of these is valid, "has the same surname as" is (i) reflexive, (ii) sym-

metric, (iii) transitive and, hence, is an equivalence relation on P.

Example 7: It follows from Example 3(6) that "is the brother of" is not symmetric and, hence,

is not an equivalence relation on P. ^ -,»»,.. .»

See Problems 1-3.

EQUIVALENCE SETS

Let S be a set and "J^ be an equivalence relation on S. If a e S, the elements y G Ssatisfying y%a constitute a subset, [a], of S, called an equivalence set or equivalence class.

Thus, formally,^ ,

[a] = {y: yes, y%a}

(Note the use of brackets here to denote equivalence classes.)

Example 8: Consider the set T of all triangles in a plane and the equivalence relation (see Prob-

lem 1) "is congruent to". When a,b & T we shall mean by [a] the set or class of

all triangles of T congruent to the triangle a, and by [6] the set or class of all tri-

angles of T congruent to the triangle 6. We note, in passing, that triangle a is

included in [a] and that if triangle c is included in both [a] and [b] then [a] and [6]

are merely two other ways of indicating the class [c].

CHAP. 2] RELATIONS AND OPERATIONS 17

A set {A,B,C, . . .} of non-empty subsets of a set S will be called a partition of Sprovided (i) Au B U C U • • • = S and (ii) the intersection of every pair of distinct

subsets is the empty set. The principal result of this section is

Theorem I. An equivalence relation "SJ on a set S effects a partition of S and, conversely,

a partition of S defines an equivalence relation on S.

Example 9: Two integers will be said to have the same parity if both are even or both are odd.

The relation "has the same parity as" on / is an equivalence relation. (Prove this.)

The relation establishes two subsets of /:

A = {x: X e I, X is even} and B = {« : a; € /, « is odd}

Now every element of / will be found either in A or in B but never in both.

Hence, A U B = I and A n B = 0, and the relation effects a partition of /.

Example 10: Consider the subsets A={3,6,9 24}, B = {1,4,7, .. .,26} and C={2,5,8 23}of S=: {1,2,3 25}. Clearly, A u B u C = S and A nB = A n C = Bn C = 0,so that {A,B,C} is a partition of S. The equivalence relation which yields this par-tition is "has the same remainder when divided by 3 as".

In proving Theorem I, (see Problem 6), use will be made of the following propertiesof equivalence sets:

(1) ae[a]

(2) If 6 G [a], then [b] = [a].

(3) If [a] n [6] ^ 0, then [a] = [b].

The first of these follows immediately from the reflexive property a%a of an equivalencerelation. For proofs of the others, see Problems 4-5.

12

*'•

3<

2<

ORDERING IN SETS

Consider the subset A = {2, 1, 3, 12, 4} of N. In writ-ing this set we have purposely failed to follow a naturalinclination to give it as A = {1, 2, 3, 4, 12} so as to pointout that the latter version results from the use of thebinary relation (^) defined on N. This ordering of theelements of A (also, of N) is said to be total, since forevery a,b G A {m,n G N) either a<b, a-b, or a>b{m<n, m = n, m>n). On the other hand, the binaryrelation

( |), (see Problem 27, Chapter 1) effects only a

partial ordering on A, i.e., 2|4 but 2/3. These order-ings of A can best be illustrated by means of diagrams.Fig. 2-1 shows the ordering of A effected by (^). Webegin at the lowest point of the diagram and follow thearrows to obtain

1^2^3=^4^12It is to be expected that the diagram for a totally ordered set is always a straight line.

Fig. 2-2 shows the partial ordering of A effected by the relation( |

).

See also Problem 7.

A set S will be said to be partially ordered (the possibility of a total ordering is notexcluded) by a binary relation % if for arbitrary a,b,c G S,

(i) % is reflexive, i.e., a%a(ii') % is anti-symmetric, i.e., a%b and b%a if and only if a = 6

(iii) % is transitive, i.e., a%b and b%c implies a%c.

1*

FiK.M Fig.^2


It will be left for the reader to check that these properties are satisfied by each of

the relations (^) and( |

) on A and also to verify the properties contain a redundancy in

that (ii') implies (i). The redundancy has been introduced to make perfectly clear the

essential difference between the relations of this and the previous section.

Let S be a partially ordered set with respect to "5?.. Then:

(1) every subset of S is also partially ordered with respect to % while some subsets maybe totally ordered. For example, in Fig. 2-2 the subset {1,2,3} is partially ordered

while the subset {1,2,4} is totally ordered by the relation (|).

(2) the element a e S is called a first element of S if a%x for every x G S.

(3) the element g G S is called a last element of S if x%g for every x G S.

(The first (last) element of an ordered set, assuming there is one, is unique.)

(4) the element a G S is called a minimal elem,ent of S if x%a implies x = a for every

xGS.

(5) the element g G S is called a maximal element of S if g%x implies g = x for every

xGS.Example 11: (a) In the orderings of A of Fig. 2-1 and 2-2 the first element is 1 and the last ele-

ment is 12. Also, 1 is a minimal element and 12 is a maximal element.

(6) In Fig. 2-3, S = {a, b, e, d} has a first element a but no last element. Here, a is

a minimal element while c and d are maximal elements.

(c) In Fig. 2-4, S - {a, b, c, d, e) has a last element e but no first element. Here,

a and 6 are minimal elements while e is a maximal element.

Fig. 2-3

An ordered set S having the property that each of its non-empty subsets has a first

element, is said to be well-ordered. For example, consider the sets N and Q each ordered

by the relation (^). Clearly, N is well-ordered but, since the subset {x: xGQ, x>2}of Q has no first element, Q is not well-ordered. Is / well-ordered by the relation (=^)?

Is A = {1,2,3,4,12} well-ordered by the relation (|)?

Let S be well-ordered by the relation %. Then for arbitrary a,b gS, the subset

{a,b} of S has a first element and so either a%b or b%a. We have proved

Theorem II. Every well-ordered set is totally ordered.

OPERATIONS

Let Q* = {x: xGQ, x>0}. For every a,6eQ+, we have

a + b,b + a, a'b, b'a, a-r-b, bâ G Q+

Addition, multiplication and division are examples of binary operations on Q+. (Note

that such operations are simply mappings of Q+ xQ+ into Q+.) For example, addition

associates with each pair a,b G Q+ an element a + bGQ*. Now a + b = b + a but, in

general, a^b ¥- bâ; hence, to insure a unique image it is necessary to think of these

operations as defined on ordered pairs of elements. Thus

o a b e d e

\

a b e d e

b hs

d e a

c c d a b

d d e a ^K ®\

e e a b e ^d

CHAP. 2] RELATIONS AND OPERATIONS 19

A binary operation " o " on a non-empty set S is a mapping which associates with

each ordered pair (a, b) of elements of S a uniquely defined element aob of S. In brief,

a binary operation on a set S is a mapping of S x S into S.

Example 12: (a) Addition is a binary operation on the set of even

natural numbers (the sum of two even natural

numbers is an even natural number) but is not

a binary operation on the set of odd natural

numbers (the sum of two odd natural numbers

is an even natural number).

(6) Neither addition nor multiplication are binary

operations on S = {0, 1, 2, 3, 4} since, for ex-

ample, 2 + 3 = 5 « S and 2 • 3 = 6 € S.

(c) The adjoining Table 2-1, defining a certain bi-

nary operation ° on the set A — {a, b, e, d, e} is Table 2-1

to be read as follows: For every ordered pair

(x,y) of A X A, we find x°y as the entry common to the row labeled x and the

column labeled y. For example, the encircled element is d ° e (not e°d).

The fact that o is a binary operation on a set S is frequently indicated by the equivalent

statement: The set S is closed with respect to the operation o . Example 12(a) may then

be expressed: The set of even natural numbers is closed with respect to addition; the set

of odd natural numbers is not closed with respect to addition.

TYPES OF BINARY OPERATIONS

A binary operation o on a set S is called commutative whenever xoy = yox for all

x,yGS.Example 13: (a) Addition and multiplication are commutative binary operations while division

is not a commutative binary operation on Q + .

(6) The operation o on A of Table 2-1 is commutative. This may be checked readily

by noting that (i) each row {b,e,d,e,a in the second row, for example) and the

same numbered column (6, c, d, e, a in the second column) read exactly the sameor that (ii) the elements of S are symmetrically placed with respect to the prin-

cipal diagonal (dotted line) extending from the upper left to the lower right of

the table.

A binary operation o on a set S is called associative whenever {xoy)oz — xo(yoz)

for all x,y,z €: S.

Example 14: (a) Addition and multiplication are associative binary operations on Q^.

(5) The operation o on A of Table 2-1 is associative. We find, for instance,

{boe)°d = dod = b and bo{cod) = boa = b; {do e)od = c°d = a andd°(eod) = d°e — a; .... Completing the proof here becomes exceedingly

tedious but it is suggested that the reader check a few other random choices.

(c) Let ° be a binary operation on R defined by

ao6 = + 26 for all a.fcefi

Since (aob)°e - (a + 2b)°c - a + 26 + 2c

while a o (6 o c) = a o (6 + 2c) = a + 2(6 + 2c) = a + 26 + 4c

the operation is not associative.

A set <S is said to have an identity {unit or neutral) element with respect to a binary

operation o on S if there exists an element u&S with the property uox = xou — x for

every x GS.Example 15 : (o) An identity element of Q with respect to addition is since + x = x + = x

for every x G Q; an identity element of Q with respect to multiplication is 1

since !• x — x'l = x for every x & Q.

(6) N has no identity element with respect to addition, but 1 is an identity element

with respect to multiplication.


(c) An identity element of the set A of Example 12(c) with respect to o is a. Notethat there is only one.

In Problem 8, we prove:

Theorem III. The identity element, if one exists, of a set S with respect to a binaryoperation on S is unique.

Consider a set S having the identity element u with respect to a binary operation o

.

An element y S S is called an inverse of x G S provided xoy = yox = u.

Example 16: (a) The inverse with respect to addition, or additive inverse of a; e / is -x sinceX + (-a;) = 0, the additive identity element of /. In general, x&I does nothave a multiplicative inverse.

(6) In Example 12(c), the inverses of a, b, c, d, e are respectively a, e, d, c, h.

It is not difficult to prove

Theorem IV. Let o be a binary operation on a set S. The inverse with respect to o ofX G S, if one exists, is unique.

Finally, let S be a set on which two binary operations a and o are defined. Theoperation a is said to be left distributive with respect to o if

aD(6oc) = (ai=ib)o(aoc) for all a,&,c e S (a)

and is said to be right distributive with respect to o if

(6oc)aa = (6na)o(cna) forall a,6,cGS (6)

When both (o) and (6) hold, we say simply that n is distributive with respect to o . Notethat the right members of (a) and (6) are equal whenever a is commutative.

Example 17: (a) For the set of all integers, multiplication (a = •) is distributive with respect to

addition (° = +) since x-{y + z) = x-y + x'z for all x,y,z G /.

(h) For the set of all integers, let ° be ordinary addition and a be defined by

x'='y = x^'y — x^y for all x,y G /

Since a a (6 + c) = a'-h + a^c = (a a 6) + (a a c)

a is left distributive with respect to +. Since

(6 + c) n a = ah^ + 2abc + ac^ # (6 a o) + (c a a) = 62a + eâ

=1 is not right distributive with respect to +.

WELL-DEFINED OPERATIONSLet S = {a,b,c, . . .} be a set on which a binary operation o is defined and let the

relation % partition S into a set E = {[a],[b],[c], . . .} of equivalence classes. Let abinary operation © on £7 be defined by

[a] © [6] = [ao6] for every [a],[b]GE

Now it is not immediately clear that, for arbitrary p,q G [a] and r,s G [b], we have

[por] = [qos] = [ao&] (c)

We shall say that © is well-defined on E, that is,

[p]®[r] = [q]®[s] = [a]®[b]if and only if (c) holds.

Example 18: The relation "has the same remainder when divided by 9 as" partitions N into

nine equivalence classes [1], [2], [3], . . ., [9]. If o is interpreted as addition on N,it is easy to show that ffi as defined above is well-defined. For example, whenx.yeN, 9a; + 2G[2] and 9y + 5 G [5]; then [2] ffi [5] = [(9a; + 2) + (9a/ + 6)] =[9(x4-j/) + 7] = [7] = [2 + 5], etc.

CHAP. 21 RELATIONS AND OPERATIONS

ISOMORPHISMS

Throughout this section we shall be using two sets:

A = {1,2,3,4} and B = {p,q,r,s}

Now that ordering relations have been introduced, there will be a tendency

the familiar ordering used in displaying the elements of each set. Weorder to warn the reader against giving to any set properties which

stated. In (1) below we consider A and B as arbitrary sets of four e

nothing more; for instance, we might have used {*,+,$,%} as A or B; in

ordering relations on A and B but not the ones mentioned above; in (3)

operations on the unordered sets A and B; in (4) we define binary c

ordered sets of (2).

(1) The mapping «: l<^p, 2<^q, S<^r, 4<^s

is one of twenty-four establishing a 1-1 correspondence between A ^nd B.

(2) Let A be ordered by the relation "5^ =( |

)

and B be ordered by the relation %' as in-

dicated in the diagram of Fig. 2-5. Since

the diagram for A is as shown in Fig. 2-6,

it is clear that the mapping

y8: l<^r, 2«-»s, S*-*^, 4<-»p

is a 1-1 correspondence between A and Bwhich preserves the order relations, that

is, for u,v G A and x,y G B with u*^ x

and V <^y then

u%' V implies x%yand conversely.

to note here

point this out in

are not explicitly

ements each and

(2) we introduce

we define binary

perations on the

Fig. 2-5

(3) On the unordered sets A and B, define the respective binary

operation tables

o 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

and

Table 2-2

g r

r s

s p

P <i

Table 2-3

It may be readily verified that the mapping

y : l<-»s, 2<^p, 3<-»r, 4<^q'

is a 1-1 correspondence between A and B which preserves the

wheneverw E. A <r^ X G B and v G A «-» y G B

(to be read "w in A corresponds to a; in 5 and v in A corresponds

wov <r^ xi=iy

(4) On the ordered sets A and B of (2), define the respective binary o

with operation tables

21

Fig. 2-6

operations o and '=> with

B

q r s

s p

p q

q r

r s

o])erations, that is,

tp y in B"), then

derations o and a


Bo 1 2 3 4

1 1 2 3 4

2 2 4 1 3

3 3 1 4 2

4 4 3 2 1

a P Q r 8

P r s P Q

9 8 P 9 r

r V <? r 8

s Q r s P

and

Table 2-4 Table 2-5

It may be readily verified that the mapping

/8: l*-*r, 2'(^s, 3*^q, 4*^pis a 1-1 correspondence between A and B which preserves both the order relations

and the operations.

By an algebraic system S we shall mean a set S together with any relations andoperations defined on S. In each of the cases (l)-(4) above, we are concerned then with acertain correspondence between the sets of the two systems. In each case we shall saythat the particular mapping is an isomorphism of A onto B or that the systems A and Bare isomorphic under the mapping in accordance with:

Two systems S and T are called isomorphic provided

(i) there exists a 1-1 correspondence between the sets S and T, and

(ii) any relations and operations defined on the sets are preserved in the correspondence.

Let us now look more closely at, say, the two systems A and B of (3). The binaryoperation o ia both associative and commutative; also, with respect to this operation,

A has 1 as identity element and every element of A has an inverse. One might suspect

then that Table 2-2 is something more than the vacuous exercise of constructing a squarearray in which no element occurs twice in the same row or column. Considering the

elements of A as digits rather than abstract symbols, it is easy to verify that the binaryoperation o may be defined as: for every x,y G A, xoy is the remainder when X'y is

divided by 5. (For example, 2-4 = 8 = l'5 + 3 and 2 o 4 = 3.) Moreover, the system Bis merely a disguised or coded version of A, the particular code being the 1-1 correspond-

ence y.

We shall make use of isomorphisms between algebraic systems in two ways:

(a) having discovered certain properties of one system (for example, those of A listed

above) we may without further ado restate them as properties of any other systemisomorphic with it.

(b) whenever more convenient, we may replace one system by any other isomorphic withit. Examples of this will be met with in Chapters 4 and 6.

PERMUTATIONSLet S = {1, 2,3, . . .,n} and consider the set S„ of the «! permutations of these n

symbols. (No significance is to be given the fact that they are natural numbers.) Thedefinition of the product of mappings in Chapter 1 leads naturally to the definition of a"permutation operation" o on the elements of Sn. First, however, we shall introduce

more useful notations for permutations.

Let ii, t2, ia, . . .,in be some arrangement of the elements of S. We now introduce a

two-line notation for the permutation

12 3 ... n

ii 12 is ... in

CHAP. 2] RELATIONS AND OPERATIONS

which is simply a variation of the notation for the mapping

a : la = ii, 2a = {2, Sa = ia, . . ., na — in

Similarly, if ji, jz, is, . . .,}n is another arrangement of the elements of ^, we write

12 3 ... n

Jl 32 J3 ... 3ny/? =

By the product a 0/3 we shall mean that a and /S are to be performed in

a rearrangement of any arrangement of the elements of S is simply

of these elements. Thus, for every a, /? e Sn, a o /? g Sn and o is a binary

"hat order. Nowanoiher arrangement

operation on S„.

Example 19: Let12 3 4 5

2 3 4 5 1

,12 3 4 5^ = U 3 2 5 4 ^ ^""^ ^

three of the 5! permutations in the set S5 of all permutations

Since the order of the columns of any permutation is imîaterial

2 3 4 5 1

3 2 5 4 1

be

write /8 as ;8 =Then

12 3 4 5

12 4 5 3

j)n S = {1,2,3,4,5}.

, we may re-

in which the upper line of p is the lower line of a.

a° p12 3 4 5

2 3 4 5 1

2 3 4 5 1

3 2 5 4 1

Similarly, rewriting a:as(),we find j3 o a =( ^

° is not commutative. 2 4 3 15

Writing y as (^ ^ ^ * ^

) , we findV4 2 3 5 1/

(a° p) ° y = 12 3 4 5

3 2 5 4 1

It is left for the reader to obtain fi°y =

3 2 5 4 1

4 2 3 5 1

2 3 4 5 1

4 2 3 5 1

a°{j3° y). Thus, ° is associative in this example. It is nowassociative on S^ and also on S„.

The identity permutation is JJ = 12 3 4 5

,12 3 4 5

Finally, interchanging the two lines of a, we have

2 3 4 5 1

12 3 4 5

12 3 4 5

5 12 3 4

since a° a i = a ^° a = ^. Moreover, it is evident that eve^y element of S5 hasan inverse.

Another notation for permutations will now be introduced. The iêrmutation

12 3 4 5'

.23451,cycle (12345) is

5, and 5 by 1.

of Example 19 can be written in cyclic notation as (12345) where the

interpreted to mean: 1 is replaced by 2, 2 is replaced by 3, 3 by 4, 4 pyThe permutation

/I 2 3 4 5

^ ~ VI 2 4 5 3;

can be written as (345) where the cycle (345) is interpreted to mean: 1 s^nd 2, the missing

23

2 3 4 5

2 5 4 1

2 3 4 5

4 3 15 Thus,

^12 3 4 5

^,4 2 3 5 1

and sjiow that (a° p)°y =

to show that ° iseasy

clearly


symbols, are unchanged while 3 is replaced by 4, 4 by 5, and 5 by 3. The permutation pcan be written as (23)(45). The interpretation is clear: 1 is unchanged; 2 is replaced by 3

and 3 by 2; 4 is replaced by 5 and 5 by 4. We shall call (23)(45) a product of cycles. Notethat these cycles are disjoint, i.e., have no symbols in common. Thus, in cyclic notation

we shall expect a permutation on n symbols to consist of a single cycle or the product of

two or more mutually disjoint cycles. Now (23) and (45) are themselves permutations of

S = {1,2,3,4,5} and, hence, j8 = (23)o(45) but we shall continue to use juxtaposition to

indicate the product of disjoint cycles. The reader will check that aop = (135) andpoa — (124). In this notation the identity permutation S will be denoted by (1).

See Problem 11.

TRANSPOSITIONS

A permutation such as (12), (25), . . . which involves interchanges of only two of the

n symbols of S — {1,2,3, .. .,n} is called a transposition. Any permutation can be

expressed, but not uniquely, as a product of transpositions.

Example 20: Express each of the following permutations

(a) (23), (6) (135), (c) (2345), (d) (12345)

on S = (1, 2, 3, 4, 5} as products of transpositions.

(a) (23) = (12) ° (23) o (13) = (12) ° (13) ° (12)

(6) (135) = (13) ° (15) = (15) o (35) = (15) ° (13) o (15) o (13)

(c) (2345) = (23) ° (24) o (25) = (25) ° (34) o (35)

(d) (12345) = (12) o (13) o (14) o (15)

The example above illustrates

Theorem V. Let a permutation a on n symbols be expressed as the product of r trans-

positions and also as a product of s transpositions. Then r and s are either

both even or both odd.p^^ ^ p^^^^^ ^^^ p^^^j^^ ^2

A permutation will be called even {odd) if it can be expressed as a product of an even

(odd) number of transpositions. In Problem 13, we prove

Theorem VI. Of the n\ permutations of n symbols, half are even and half are odd.

Example 20 also illustrates

Theorem VII. A cycle of m symbols can be written as a product of m - 1 transpositions.

ALGEBRAIC SYSTEMSMuch of the remainder of this book will be devoted to the study of various algebraic

systems. Such systems may be studied in either of two ways:

(a) we begin with a set of elements (for example, the natural numbers or a set isomorphic

to it), define the binary operations addition and multiplication, and derive the familiar

laws governing operations with these numbers.

(6) we begin with a set S of elements (not identified); define a binary operation o ; lay

down certain postulates, for example, (i) o is associative, (ii) there exists in S an

identity element with respect to o,

(iii) there exists in S an inverse with respect to o

of each element of S; and establish a number of theorems which follow.

We shall use both procedures here. In the next chapter we shall follow (a) in the study

of the natural numbers.


Solved Problems

1. Show that "is congruent to" on the set T of all triangles in a plane

relation.

(i) "a is congruent to a for all a G T" is valid.

(ii) "If a is congruent to h, then h is congruent to a" is valid.

(iii) "If a is congruent to 6 and h is congruent to c, then a is congruent to c'

Thus "is congruent to" is an equivalence relation on T.

2. Show that " < " on / is not an equivalence relation.

(i) "a < a for all a S /" is not valid.

(ii) "If a < b, then 6 < a" is not valid.

(iii) "If a < b and b < c, then a < c" is valid.

Thus "<" on / is not an equivalence relation. (Note that (i) or (ii) is sufficient

3. Let % be an equivalence relation and assume c%a and c%b. Prove

Since c%a, then a%c (by the symmetric property). Since a%c and e%btransitive property).

4. Prove: If 6 e [a], then [b] = [a].

Denote by % the equivalence relation defining [a]. By definition, b G [a]

X G [6] implies x%b. Then x%a for every x G [6] (by the transitive property)

A repetition of the argument using al^b (which follows by the symmetric property

(whenever 3/G[a]) yields [a] c [6]. Thus [6] = [a], as required.

5. Prove: If [a] n [b] ^ 0, then [a] =

Suppose [a] n [b] = {r, s, ...}. Then

(by the transitive property of =).

[6].

[r] - [a] and [r] = [b] (by Problem 4), and [o] = [b]

Prove: An equivalence relation 9( on a set S effects a partition of ^ and, conversely,

a partition of S defines an equivalence relation on S.

Let % be an equivalence relation on S and define for each p G S,

Tp = [p] = {x: x&S, x'Kp}

Since p G [p], it is clear that S is the union of all the distinct subsets r„

since.

by Problem 5. Thus, {r„, T^,, T^, . . .} is the partition of S effected by %.

Conversely, let {r,,, T^, T^, . . .} be any partition of S. On S define the binary relation % by<^lear that % is both

there exist subsets

TjH Tf, ^ and

so Tj = T^. Since p,r & Tj, then p%r and % is transitive. This completes th^ proof that % is anequivalence relation.

p%q if and only if there is a Tj in the partition such that p, q G Tj. It is

reflexive and symmetric. Suppose p%q and q'Jir; then by the definition of '2?

Tj and T^ (not necessarily distinct) for which p, q G Tj and q,r e T^. Now

25

is an equivalence

is valid.

.)

a%b.

then a'J(_b (by the

implies b%a andand [6] C [a].

of %) and y%a

induced by %.


7. Diagram the partial ordering of (a) the set of subsets of S = (a, b, c} effected by the

binary relation ( C ), (6) the set B = {2, 4, 5, 8, 15, 45, 60} effected by the binary relation( |

).

{a, b, e}

{a.b} {h.c}

Fig. 2-7 Fig. 2-8

These figures need no elaboration once it is understood a miniinum number of line segments is to

be used. In Pig. 2-7, for example, is not joined directly to {a, b, c} since £ {a, b, c} is indicated

by the path Q {a} C {a,b} C {a,b,c}. Likewise, in Fig. 2-8, segments joining 2 to 8 and 5 to 45

are unnecessary.

Prove: The identity element, if one exists, with respect to a binary operation o on a

set S is unique.

Assume the contrary, that is, assume qi and 92 *<> be identity elements of S. Since gj is an

identity element we have gi ° 92 — Q2. "while since ^2 is an identity element we have gi ° 92 — 9i-

Thus 91 — ?! ° 92 = 12' the identity element is unique.

9. Show that multiplication is a binary operation on S = {1, —l,i, —i} where i — y/—i.

This can be done most easily by forming the adjoining table and

noting that each entry is a unique element of S.

The order in which the elements of S are placed as labels on the

rows and columns is immaterial; there is some gain, however, in using

the same order for both.

The reader may show easily that multiplication on S is both asso-

ciative and commutative, that 1 is the identity element and that the

inverses of 1, —1, i, —i are respectively 1, —1, —i, i.

1 -1 t —I

1 1 -1 —i

-1 -1 1 -i i

i i —i -1 1

—i —i i -1

Table 2-6

10. Determine the properties of the binary operations o and <=> defined on S = [a, b, c, d}

by the given tables:

a b c d

a a b c d

b b c d a

e e d a b

d d a b c

a b c d

a d a c b

b a e b d

c b d a c

d c b d a

Table 2-7 Table 2-8


theThe binary operation ° defined by Table 2-7 is commutative (check that

metrically placed with respect to the principal diagonal) and associative (again, a

identity element a (the column labels are also the elements of the first column and

also the elements of the first row). The inverses of o, 6, e, d are respectively a, d,

c°e = d°b = a).

The binary operation <=> defined by Table 2-8 is neither commutative (a'=>e

associative (a => (6 a c) = a a 5 = o, (o a 6) a c = a n c = c). There is no identity

no element of 5 has an inverse.

Since o a (d o c) = a ¥' d = (o a d) o (a a c) and (d o c) a o ¥ (d = o) <> (c a o)

nor right distributive with respect to o; since d o (c = 6) = c ¥' a = (d o c) a (d o

^)

tative, o is neither left nor right distributive with respect to >=>.

11. (o) Write the permutations (23) and (13)(245) on 5 symbols in two line notation

(b) Express the products (23) o (13)(245) and (13)(245)o(23) in eye'

(c) Express in cyclic notation the inverses of (23) and of (13)(245).

(a) (23)/I 2 3 4 5\

- Vl 3 2 4 5/and (13)(245) -il

2 3 4

3 4 15

27

entries are sym-

jihore). There is anthe row labels are

c,b {a°a = b°d =

e, c a a = 6) nor

element and, hence,

=> is neither left

and o is commu-

ic notation.

(6) (23) o (13)(245)

and

(13)(246) ° (23)

/l 2 3 4 5\ /l 3 2 4 5\ _ /l 2 3 4 5\~

Vl 3 2 4 5/ Vs 1 4 5 2/ Vs 1 4 5 2/

/l 2 3 4 5'N /3 4 1 5 2\ _ /l 2 3 4 5\-

V3 4 1 5 2;/ ° \2 4 1 5 3/~

1,2 4 1 6 3/

(c) The inverse of (23) - (l | | J 5) = (j 3 2 4 5) = ^^S)

The inverse of (13)(245) is (12345) = (35124) ^ (13)(254),

of12. Prove: Let a permutation a on « symbols be expressed as the product

and also as the product of s > r transpositions. Then r and s are either

both odd.

Using the distinct symbols Xi, x^, Xg, . . ., x„, we form the product

A = («i - K2)(«i - »3)(«i - «4) • • • (*i - «n)

(«2 - "'sX^B -Xi) • (X2 - »n)

(«n-l-«n)

A transposition {u, v), where m < v, on A has the following effect: (1) any ftJctor

neither «„ nor x„ is unchanged, (2) the single factor «„ — x„ changes sign, (3) the

each of which contains either «„ or x„ but not both, can be grouped into pairs

where u<v<w, («„ — «„)(«„, — x„) where tt < w < 1;, and («„ — x„)(a!„ — x„) whernare all unchanged. Thus, the effect of the transposition on A is to change its sigii

Now the effect of a on A is to produce (—1)''A or (—1)'A according as a is written

of r or 8 transpositions. Since (—1)''A = (—1)''A, we have A = (—1)'-^. so that s

r and s are either both even or both odd.

13. Prove: Of the n! permutations on n symbols, half are even and half

Denote the even permutations by Pi, P2, P3, ...,?« ''^^ ^he odd permutationsLet t be any transposition. Now t o p^, t ° P2, t ° p^, . . ., t°p„ are permutationsare distinct since Pi, Pa, P3, . . . , p„ are distinct and they are odd; thus, u — v.

t°q3. ..., t°g„ are distinct and even; thus, v — u. Hence u = v = -^nl

(13452)

(12453)

r transpositions

both even or

which involves

remaining factors,

w <u<v, which

as the product— r is even. Thus,

are odd.

1 n symbols. TheyAlso, to 91, to 92.



14. Which of the following are equivalence relations?

(o) "Is similar to" for the set T of all triangles in a plane.

(6) "Has the same radius as" for the set of all circles in a plane.

(c) "Is the square of" for the set N.

(d) "Has the same number of vertices as" for the set of all polygons in a plane.

(e) "C" for the set of sets S = {A,B,C, ...}.

(J)"^" for the set R.

Ans. (a), (h), (<£).

15. (a) Show that "is a factor of" on N is reflexive and transitive but is not symmetric.

(6) Show that "costs within one dollar of" for men's shoes is reflexive and symmetric but nottransitive.

(c) Give an example of a relation which is symmetric and transitive but not reflexive.

(d) Conclude from (a), (6), (c) that no two of the properties reflexive, symmetric, transitive of abinary relation implies the other.

16. Diagram the partial ordering of

(a) A = {1,2,3,6}

(6) B = {1,2,3,5,30,60} andeffected on each by the relation

(

|

(c) C = {1,3,5,15,30,45}

17. Let S = {a, 6, c, d, e, /} be ordered by the relation % as shown in

Fig. 2-9. (a) List all pairs x,y ^ S for which x'^y. (b) List all sub-

sets of three elements each of which are totally ordered.

18. Verify:

(o) the ordered set of subsets of S in Problem 7(a) has as first element (also, as minimal element)

and S as last element (also, as maximal element).

(b) the ordered set B of Problem 7(6) has neither a first nor last element. What are its minimal andmaximal elements?

(c) the subset C = {2, 4, 5, 15, 60} of B of Problem 7(6) has a last element but no first element.

What are its minimal and maximal elements?

19. Show:(a) multiplication is a binary operation on S = {1, —1} but not on T — {1, 2},

(6) addition is a binary relation on S — {x : x B I, x < 0} but multiplication is not.

20. Let S = {A,B,C,D} where A = 0, B = {a, 6}, C = {a,c}, D = {a,b,c}. Construct tables to show

that U is a binary relation on S but n is not.

21. For the binary relations ° and a defined on S = {a,b,c,d,e} by Tables 2-9 and 2-10, assume

associativity and investigate for all other properties.

o a b e d e

a a d a d e

b d b b d e

e a b c d e

d d d d d e

e e e e e e

a 6 c d e

a a c c a a

6 c c c 6 b

c e c c c c

d a 6 c d d

e a 6 c d e

Table 2-9 Table 2-10

22. Let S = {A,B,C,D} where A ^ 0, B = {a}, C = {a,b}, D = {a.b.e}.

(o) Construct tables to show that U and n are binary operations on S.

(6) Assume associativity for each operation and investigate all other properties.

CHAP. 21 RELATIONS AND OPERATIONS 29

23. For the binary operation on S = {a, b, e, d, e, f, g, h}

defined by Table 2-11, assume associativity and in-

vestigate all other properties.

24. Show that ° defined in Problem 23 is a binary opera-

tion on the subsets Sp = {a}, Si = {a, c}, Sj = {a, e},

S3 = {a,f}, S4 = {a,g}, S5 = {a, h}, Sg = {a, b, c,d},

S7 = {a, c, e, /}, Sg = {o, c, g, h} but not on the sub-

sets r, = {a, b} and T2 = {a, f, g) of S.

25. Prove Theorem IV. Hint: Assume y and z to be

inverses of x and consider z°(xoy) = (z°x)°y.

a h e d e f 9 h

a a b e d e f 9 h

b b c d a h 9 e f

c c d a b f e h 9

d d a b c 9 h f e

e e 9 f h a e b d

f f h e 9 e a d b

9 9 f h e d b a e

h h e 9 f b d e a

Table 2-Jl

26. (a) Show that the set N of all natural numbers under addition and the set M = {2x: x& N} under

addition are isomorphic. Hint: Use nGN *^ 2n^M.

(b) Is the set N under addition isomorphic to the set P = {2x — 1 : x S N} und4r addition?

(c) Is the set M of (a) isomorphic to the set P of (6)?

27. Let A and B be sets with respective operations ° and <=! Suppose A and B are isbmorphic and show:

(a) if the associative (commutative) law holds in A it also holds in B.

(6) if A has an identity element u, then its correspondent u' is the identity element in B.

(c) if each element in A has an inverse with respect to °, the same is true of the elements of B with

respect to <=>

.

Hint: In (a), let aGA*^ a'eB, b<r^b', c<rê'. Then

a°{b°c) <-» o' a (6' a c'), (a°b)°c <r^ (a' a 6') ^ c'

and o o (6 o c) = (a°b)oc implies a' a (6' a c') = (a' n 6') a c'

28. Express each of the following permutations on 8 symbols as a product of disjoint cycles and as a

product of transpositions (minimum number).

(a)12345678

(6)1234567834567812 (c)

123456783416827 6^23415678/

(d) (2468)° (348) (e) (15)(2468) ° (37)(15468) (/) (135) o (3456) o (4678)

Partial answer, (a) (1234) = (12)(13)(14)

(c) (13)(246)(58) = (13)(24)(26)(58)

(d) (28)(346) = (28)(34)(36)

(/) (1637845) = (16)(13)(17)(18)(14)(15)

Note: For convenience, ° has been suppressed in indicating the products of transpjositions.

29. Show that the cycles (1357) and (2468) of Problem 28(6) are commutative,

covering this.

!State the theorem

30. Write in cyclic notation the 6 permutations on SPi, P2> Ps, • • • > Pe, and form a table of products Pj o

Pj.

{1,2,3}, denote them in some order by

31. Form the operation (product) table for the set S = {(1), (1234), (1432), (13)(24)} of permutations on

four symbols. Using Table 2-3, show that S is isomorphic to B.

chapter 3

The Natural Numbers

THE PEANO POSTULATES

Thus far we have assumed those properties of the number systems necessary to

provide examples and exercises in the earlier chapters. In this chapter we propose to

develop the system of natural numbers assuming only a few of its simpler properties.

These simple properties, known as the Peano Postulates (Axioms) after the Italian

mathematician who in 1899 inaugurated the program, may be stated as follows:

Let there exist a non-empty set A^ such that

Postulate I : leNPostulate II : For each n&N there exists a unique n* G N, called the successor of n.

Postulate III : For each » G N we have n* ¥-1.

Postulate IV : If in,n G N and m* — n*, then m = n.

Postulate V : Any subset K of N having the properties

(a) IGK(6) k* gK whenever kGK

is equal to N.

First, we shall check to see that these are in fact well-known properties of the natural

numbers. Postulates I and II need no elaboration; III states that there is a first natural

number 1; IV states that distinct natural numbers m and « have distinct successors m +

1

and n + 1; V states essentially that any natural number can be reached by beginning with 1

and counting consecutive successors.

It will be noted that, in the definitions of addition and multiplication on N which

follow, nothing beyond these postulates is used.

ADDITION ON NAddition on N is defined by

(i) TC + 1 = «*, for every n GN(ii) n + m* — (n + m)* whenever « + m is defined.

It can be shown that addition is then subject to the following laws:

For all in,n,p G N,

Ai. Closure Law n +mGNA2. Commutative Law n + m = m + n

A3. Associative Law m + {n + p) — (m + n) +pA4. Cancellation Law If m + p = n + p, then m = n.

30

CHAP. 3] THE NATURAL NUMBERS 31

MULTIPLICATION ON NMultiplication on N is defined by

(iii) n • 1 = «

(iv) n • TO* = n-m + n whenever « • to is defined.

It can be shown that multiplication is then subject to the following laws:

For all m,n,p G N,

Ml. Closure Law wm G NM2. Commutative Law m-n = m • to

Ms. Associative Law m-(rfp) = (to • w) • p

M4. Cancellation Law If rri'p = n-p, then m = n.

Addition and multiplication are subject to the Distributive Laws:

For all m,n,p G N,

Di. m- (n + p) — m-n -\- m-pD2. (n + p)'TO = 'rfm + p'Tti

MATHEMATICAL INDUCTION

Consider the proposition

P{m) : TO* ¥= m, for every m GNWe shall now show how this proposition may be established using only the Postulates

I-V. DefineK = {k: kGN, P(k) is true}

Now IGN (Postulate I)

and 1* ^ 1 (Postulate III)

Thus P(l) is true and 1 G X

Next, let k be any element of K; then

(a) P{k) : k*^k

is true. Now if (k*)* = k* it follows by Postulate IV that k* = k, a contradiction of (a).

HenceP(k*) : (fc*)* # k*

is true and so k* G N. Now K has the two properties stated in Postulate V; thus, K — Nand the proposition is valid for every mGN.

In establishing the validity of the above proposition, we have at the same time

established the following

Principle of Mathematical Induction

A proposition P(to) is true for all mGN provided:

P(l) is true

and, for each kGN, P{k) is true implies P{k*) is true

The several laws A1-A4, M1-M4, Di-D2 can be established by mathematical induction.

Ai is established in Example 1, As in Problem 1, A2 in Problems 2 and 3, and D2 in

Problem 5.

32 THE NATURAL NUMBERS [CHAP. 3

Example 1: Prove the Closure Law: n +mGN for all m,nGN.

We are to prove that n + m is defined (is a natural number) by (i) and (ii)

for all m,w G N. Suppose n to be some fixed natural number and consider the

propositionP{m) : -re + m G AT, for every m €. N

Now P(l) : n+ieN

is true since n + 1 = w* (by (i)) and w* e iV (by Postulate II). Suppose next that

for some k G N(a) P{k) : n + ke N is true

It then follows that P{k*) : n + k*eN

is true since n + A;* = (n + fc)* (by (ii)) and (n + k)*&N whenever n+keN(by Postulate II). Thus, by induction, P(m) is true for all m 6 2V and, since n was

any natural number, the Closure Law for Addition is established.

In view of the Closure Laws Ai and Mi, addition and multiplication are (see Chapter 2)

binary operations on N. The laws As and Ma suggest as definitions for the sum and

product of three elements ai, ai, ag e N,

(v) tti + tta + tta - (ai + az) -\- a& - Oi + (a^ + as)

(vi) ai • tta • as = (ai • a2) -as = ai • (a2 • as)

Note that for a sum or product of three natural numbers, parentheses may be inserted

at will. The sum of four natural numbers is considered in Problem 4. The general case

is left as an exercise.


Theorem I. Every element n^ 1 of iV is the successor of some other element of "N.

THE ORDER RELATIONS

For each m,nGN, we define "<" by

(vii) m<n if and only if there exists some p GN such that m + p = n.

In Problem 8 it is shown that the relation < is transitive but neither reflexive nor

symmetric. By Theorem I,^ , „1 < n for all n^-l

and, by (i) and (vii), n<n* for all nGN

For each m,nGN, we define ">" by

(viii) m> n if and only if n<m

There follows

The Trichotomy Law: For any m,n G N one and only one of the following

(a) m = n, (6) m < n, (c) m> n

is true. For a proof, see Problem 10.

Further consequences of the order relations are Theorems II and II':

Theorem II. If m,nGN and m<n, then for each p GN,

(a) m + p < n + p (b) m-p < n-p

and, conversely, if (a) or (6) hold for some pGN, then m<n.


Theorem II'. If m,nS N and m> n, then for each p €. N,

(a) m + p > n + p

(b) ni'p > n-p

and, conversely, if (a) and (ft) hold for some p & N, then m>n.

Since Theorem U' is merely Theorem II with m and n interchanged, it is clear that

the proof of any part of Theorem II (see Problem 11) establishes the corresponding partof Theorem IP.

The relations "less than or equal to" (=^) and "equal to or greater than" (^) aredefined as follows:

For m,n G N

,

m^n ii either m <n or m = n

m — n if either m = n or m> nholds.

Let A be any subset of N (i.e., A c N). An element p of A is called the least elementof A provided p ^ a for every a G A. Notice that in the language of sets, p is the first

element of A with respect to the ordering ^. In Problem 12, we prove

Theorem III. The set N is well-ordered.

MULTIPLES AND POWERSLet a G S, on which binary operations + and • have been defined, and define

la = a a^ = a

and {k + l)a = ka -\-

a

a^+i - a^-a

whenever ka and a'^, for k G N, are defined.

Example 2: Since la = a and a* = a, we have

2o = (l + l)a = la + lo = a + a and a^ = a' + i = a^'a = a-a

3o = (2 + l)a = 2a + la — a + a + a and a^ = a^+i = o2«a = a«o»a

etc.

It must be understood here that in Example 2 the + in 1 + 1 and the + in a + aare to be presumed quite different, since the first denotes addition on iV and the secondon S. (It might be helpful to denote the operations on S by © and O). In particular,fea = a + a+---+aisa multiple of a, and can be written as fc • a only if k G S.

Using the induction principle, the following properties may be established for all

a,b G S and all m,nG N:

(ix) ma + na = (m + n)a (ix)' a"" -a" - a'"^"

(x) m{na) — {m-n)a (x)' (a")™ = a""-"

and, when + and • are commutative on S,

(xi) na-Vnh = n{a + 6) (xi)' a" • 6" = (a6)"

ISOMORPHIC SETSIt should be evident by now that the set {1, 1*, (1*)*, . . .}, together with the opera-

tions and relations defined on it developed here differs from the familiar set {1,2,3, .. .}with operations and relations in vogue at the present time only in the symbols used.Had a Roman written this chapter he would, of course, have reached the same conclusionwith his system {I, II, III, ...}. We say simply that the three are isomorphic.


Solved Problems

1. Prove the Associative Law A3 : m + (n + p) — (m + n) + p for all m,n,p G N.

Let m and n be fixed natural numbers and consider the proposition

P(p) : m + (n + p) = (m + n) + p for all p € AT

We first check the validity of P(l): w + (m + 1) = (wi + n) + 1. By (i) and (ii), Page 30,

w + (re + 1) = m + n* = (m + n)* = (m + n) + 1

and P(l) is true.

Next, suppose that for some fc G AT,

P(k) : m + (re + fc) = (m + n) + fe

is true. We need to show that this assures

P(k*) : m + (n + k*) = (m + n) + k*

is true. By (ii), m + (n + k*) - m + (n + k)* - [m + (n + k)]*

and (m + n) + k* = \(m + n) + k\*

Then, whenever P(k) is true,

m+(n + k)* = [m + (n + k)]* = [(m + n) + k]* = (m + n) + k*

and P(k*) is true. Thus, P(p) is true for all p G iST and, since m and n were any natural numbers,

A3 follows.

2. Prove P{n): n + 1 = 1 + n for all n e iV.

Clearly P(l) : 1 + 1 = 1 + 1 is true. Next, suppose that for some ke. N,

P(k) : fc + 1 = 1 + A;

is true. We are to show that this assures

P(k*) : k* + 1 = 1 + k*

is true. Using in turn the definition of fe*, A3, the assumption that P(k) is true, and the definition

of A;*, we have1 + fc* = 1 + (fc + 1) = (1 + fc) + 1 = (fc + 1) + 1 = fc* + 1

Thus P(k*) is true and P(w) is established.

3. Prove the Commutative Law A2: m + n = n + m for all m,nGN.

Let re be a fixed but arbitrary natural number and consider

P(m) : m + n = n + m for all m G iV

By Problem 2, P(l) is true. Suppose that for some fc G AT,

P(fc) : fc + Tt = re + fc

is true. Now

fc* + re = (fc + 1) + re = fc + (1 + re) = fe + (re + 1) = fc + re* = (fc + re)* = (re + fc)* = re + fe*

Thus P(fc*) : k* +n = n + k*

is true and Aj follows.

The reader will check carefully that in obtaining the sequence of equalities above, only definitions,

postulates, such laws of addition as have been proved and, of course, the critical assumption P(fc) is

true for some arbitrary fc G iV have been used. Both here and in later proofs, it will be understood

that when the evidence supporting a step in a proof is not cited, the reader is expected to supply it.


4. (a) Let Oi, 02, as, uaGN and define 01 + 02 + 03 + 04 = (oi + 02 + 03) + 04. Show that

in oi + 02 + Os + 04 we may insert parentheses at will.

Using (v), we have aj + 02 + as + 04 = (oj + 02 + 03) + 04 = (oj + ag) + 03 + 04 = (o, + aj) +(as + a4) = aj + 02 + (03 + a4) = a, + (02 + 03 + 04), etc.

(6) For &,ai,02,a3 G iV, prove 6 '(01 + 02 + 03) = b-oi + 6*02 + 6*03.

6 • (oi + 02 + 03) = 6 • [(oi + 02) + as] = 6 • (aj + 02) + 6 • as = 6 • Oj + 6 • a2 + 6 • as

5. Prove the Distributive Law D2 : {n + p)-m = n"m + p-m for all m,n,p E N.

Let n and p be fixed and consider

P{m) : {n + p) "m = n'm + p'm for all mG N

Using Ai and (iii), we find

P(l) : (n + p)'l = n + p = n'l + p>l

is true. Suppose that for some fc e iV,

P(k) : (n + p)-k = n'k + vk

is true. Then (n + p)'k* = {n + p) • k + {n + p) = wk + p'k + n + p

= n'k + {p'k + n) + p = n'k+(n + p'k) + p

— (n-k + n) + (p'k + p) = » • fc* + p • fc*

Thus, P{k*): (n + p)'k* - tfk* + p-k*

is true and Dj is established.

6. Prove: Every element n ^ 1 of N is the successor of some other element of N.

First, we note that Postulate III excludes 1 as a successor. Denote by K the set consisting of

the element 1 and all elements of N which are successors, i.e.,

K = {k: k G N, k = 1 or k = m* for some mS N}

Now every k & K has a unique successor k* E N (Postulate II) and, since k* is a successor, we havefe* G K. Then K = N (Postulate V). Hence, for any n G N v/e have either % = 1 or n = m* for

some m. G N.

7. Prove: m + n ¥= m for all m,n G N.

Let n be fixed and consider P(m): 'm + n ¥= m for all m G N. By Postulate III, P(l): 1 + w # 1

is true. Suppose that for some k G N,

(a) P{k) : k + n ¥- k

is true. Now (fe + n)* ¥= k* since, by Postulate IV, {k + n)* — k* implies k + n — k, a contradiction

of (o). ThusPik*) : k* + n ¥- k*

is true and the theorem is established.

8. Show that < is transitive but neither reflexive nor symmetric.

Let m,n,p & N and suppose that m < n and n < p. By (vii) there exist r,s & N such that

m + r — n and n + s = p. Then

n + s = {'m + r) + 8 = m + (r + s) = p

Thus, m < p and < is transitive.


Let neN. Now w < m is false since, if it were true, there would exist some kGN such that

n + k = n, contrary to the result in Problem 7. Thus, < is not reflexive.

Finally, let m.nS N and suppose m < n and n < m. Since < is transitive, it follows that

m <m., contrary to the result in the paragraph immediately above. Thus, < is not symmetric.

9. Prove: 1 — n, for every nG.N.

When M = 1 the equality holds; otherwise, by Problem 6, n = m* = m + 1, for some m e AT,

and the inequality holds.

10. Prove the Trichotomy Law: For any m.nGN, one and only one of

(a) ^, = 11 (b) in<n {c) m>nis true.

Let m be any element of N and construct the subsets

Ni = {m}, N^ = {x: xGN, x<m}, N3 = {x: x e N, x > m}

We are to show that {N^.N^.N^} is a partition of N relative to {=,<,>}.

(1) Suppose m = 1; then N^ = {1}, iVg = (Problem 9) and N3 = {x: x e N, x > 1}. Clearly

iVi U ATg U ATj = N. Thus, to complete the proof for this case, there remains only to check

Ni n N2 - Ni n N3 = N2 n Ns = 0.

(2) Suppose m¥'l. Since 1 G N^, it follows that 1 € NÛ N^V N^. Now select any w # 1 G

iVi U iV2 U N3. There are three cases to be considered:

(i) n G iVj. Here, n - m and so n* G N3.

(ii) neNi so that n + p = m for some pSN. It p - 1, then w* = m G ATj; if p # 1 so

that p = 1 + q for some 9 G N, then n* + q = m and so n* G N^.

(iii) n € N^. Here n* > n> m and so n* G iVg.

Thus, for every ne N,

n G Ni U iVj U N3 implies n* G Nj U N2 u N3

Since 1 G iVj U N2 U N3 we conclude that N = NiV NÛ N3.

Now m € iVg, since m < w; hence, iV, n iVj = 0. Similarly, m > m and so N^n Ng = 0.

Suppose N^n Ng = {p} for some p G AT. Then p < w and p > w or, what is the same, p < mand m < p. Since < is transitive, we have p < p, a contradiction. Thus, we must conclude

that Nzn N3 — and the proof is now complete for this case.

11. Prove: If m,n€.N and m<n, then for each pSN, m + p <7i + p and conversely.

Since m < n, there exists some kGN such that m + k — n. Then

n + p = (m+k) + p - m + k + p = m + p + k = (m + p) + k

and so m + p < n + p

For the converse, assume m + p < n + p. Now either m = n, m < n, or m > n. If w = n, then

m + p = n + p; if m > m, then w + p > n + p (Theorem II'). Since these contradict the hypothesis,

we conclude that m < n.

12. Prove: The set N is well-ordered.

Consider any subset S'f^0 of N. We are to prove that S has a least element. This is certainly

true if 1 G S. Suppose 1 € S; then 1 < s for every s G S. Denote by K the set

K - {k: keN, k£ 8 for each s G S}

Since 1 G X^, we know that K ¥' 0. Moreover K ¥= N; hence, there must exist an r G K such that

r* € K. Now this r G S since, otherwise, r < s and so r* =^ 8 for every « G S. But then r* G K, a

contradiction of our assumption concerning r. Thus, S has a least element. Now S was any non-

empty subset of JV; hence, every non-empty subset of N has a least element and N is well-ordered.



13. Prove by induction: 1 • n = w for every tiG N.

14. Prove by induction: (o) Mi, (6) M2, (c) M3.

Hint: Use the result of Problem 13 and Dg in (6).

15. Prove: (a) Dj by following Problem 5, (6) Di by using M2.

16. Prove: (a) (m + n*)* = m* + n*

(6) (m • «*)* =»!•» + m*

(c) (m* • w*)* = m* + m'n + n*

where m,n& N.

17. Prove: (a) (tn + n) ' (p + q) — (m • p + «i • 5) + (m • p + n • q)

(6) m' {n + p) • q = (m • n) • g + m • (p • g)

(c) wi* + n* = (wi + n)* + 1

(d) w* • n* = (w • n)* + m + n

18. Let m,n,p,q& N and define m'wp'g = (m'wp)'?. (a) Show that in m-n'p'q we mayinsert parentheses at will. (6) Prove: m,' {n-\-p + q) = m'n + m-p + m'q.

19. Identify S - {x : x e N, n < x < n* for ail n e N).

20. If m,n,p,qGN and if m<M and p < q, prove: (a) m + p < n + q, (5) wp < wg

21. Let m,nG N. Prove: (a) If m = w, then k* •m > n for every k&N. (h) It k* •m = n for somekG N, then w < w.

22. Prove A4 and M4 using the Trichotomy Law and Theorems II, Page 32, and II', Page 33.

23. For all m G N define m^ = m and mP+' = -mP'rn provided ?n» is defined. When m, n, p, qr G iV,

prove: (a) mP "mi = mP+i, (6) (»np) 2m • n if m # «.

25. Prove, by induction, for all n G N:

(a) H-2 + 3+---+M = ^n(n + 1)

(6) 12 + 22 + 32 + • + n* = ^n(n + l)(2re + 1)

(C) 13 + 23 + 33 + • + w3 = ^w2(m + 1)2

26. For ffli, 02,03, .. .,a„ G iV define oj + 02 + 03+ • • • + o^ = (oj + 02 + 03+ • • • + Ofc_i) + a^ forfc = 3, 4, 5, . .

. , M. Prove:

(o) «! + 02 + O3 + • • • + a„ = (Oi + 02 + 03 + • • • + «r) + ("r+l + ar+2 + Or+S + " " " + aj(6) In any sum of n natural numbers, parentheses may be inserted at will.

27. Prove each of the following alternate forms of the Induction Principle:

(a) With each n G AT let there be associated a proposition P(n). Then P(«) is true for every n G AT

provided:

(i) P(l) is true.

(ii) For each mG N the assumption P{k) is true for all k < m implies P(m) is true.

(6) Let 6 be some fixed natural number, and with each natural number w — 6 let there be associateda proposition P(n). Then P(n) is true for all values of n provided:

(i) P(b) is true.

(ii) For each m > b the assumption that P(k) is true for all kG N such that 6 - fc < mimplies P(m) is true.

chapter 4

The Integers

INTRODUCTIONThe system of natural numbers has an obvious defect in that, given m,s GN, the

equation m + x = s may or may not have a solution. For example, m + x = m has no

solution (see Problem 7, Chapter 3) while m + x = m* has the solution x = 1. Everyone

knows that this state of affairs is remedied by adjoining to the natural numbers (then

called positive integers) the additional numbers zero and the negative integers to form

the set / of all integers.

In this chapter it will be shown how the system of integers can be constructed from

the system of natural numbers. For this purpose, we form the product set

L = N xN = {(s, m): sGN,mGN}

Now we shall not say that (s, m) is a solution of m + x = s. However, let it be perfectly

clear, we shall proceed as if this were the case. Notice that if (s,m) were a solution of

m + x - s, then {s,m) would also be a solution of m* + x = s* which, in turn, would

have (s*,TO*) as solution. This observation motivates the partition of L into equivalence

classes such that {s,m) and (s*,m*) are members of the same class.

BINARY RELATION ~

Let the binary relation "~", read "wave", be defined on all {s,m),(t,n) G L by

(s,to) ~ {t,n) if and only if 8 + n = t + m

Example 1: (a) (5,2) ~ (9,6) since 5 + 6 = 9 + 2

(6) (5,2)'/' (8,4) since 5 + 4 ¥= 8 + 2

(c) (r, r) ~ («, s) since r + s - s + r

id) (r*,r) ~ {s*,s) since r* + « - s*+r

(e) (r*,8*) ~ (r,8) since r* + s = r + s*

whenever r,8 G N.

Now ~ is an equivalence relation (see Problem 1) and thus partitions L into a set of

equivalence classes S = {[s,m],[t,n], . . .} where

[s,m] = {(a,b): ia,b) G L, {a,b) ~ (s.m)}

We recall from Chapter 2 that (s,w) e [s,m] and that, if (c,d) G [s,m], then [c,d] =

[s,m]. Thus[s,m]^[t,n] if and only if (s,w) -(«,«)

It will be our purpose now to show that the set S of equivalence classes of L relative

to ~ is, except for the symbols used, the familiar set / of all integers.

38

CHAP. 4] THE INTEGERS 39

ADDITION AND MULTIPLICATION ON gAddition and multiplication on g will be defined respectively by

(i) \s, m] + [t, n] = [{s + 1), {m + n)]

(ii) [s,m]'[t,n] = [{s-t + m'n), {s-n + m-t)]

for all [s,TO],[<,n] G g.

An examination of the right members of (i) and (ii) shows that the Closure Laws

Ai. x + y G g for all x,y G gand

Ml. x'y&S for all x,y G gare valid.


Theorem I. The equivalence class to which the sum (product) of two elements, oneselected from each of two equivalence classes of J, belongs is independentof the particular elements selected.

Example 2: If {a,b),{c,d) e [s,m] and {e,f),(g,h)G[t,n], we have not only

[a,b] = [c,d] - [s.m] and [e,f] = [g,h] = [t,n]

but, by Theorem I, also

[a,b] + [e,f] = [e,d\ + [g,h] = [s.m] + [t.n]

and [a,b]-[e,f] = [c,d\-[g,h] = {s.m] • [t,n]

Theorem I may also be stated as follows: Addition and multiplication on g arewell-defined.

By using the commutative and associate laws for addition and multiplication on N, it

is not difficult to show that addition and multiplication on g obey these same laws. Theassociative law for addition and one. of the distributive laws are proved in Problems 4 and 5.

THE POSITIVE INTEGERSLet r G N. From 1 + r = r* it follows that r is a solution of 1 + x = r*. Consider

now the mapping[»*, l]*^n, n gN (i)

For this mapping, we find

[r*,l] + [s*,l] = [(r*+8*),il + l)] = [(r + s)M]^r + s

and

[rM]-[sM] = [(r*-s* + l-l), (r*-H-s*-l)] = [(r-s)*,!] ^ r-s

Thus, (1) is an isomorphism of the subset {[n*, l]:nGN} of ^ onto N.

Suppose now that [s,m] = [r*,l]. Then (s,m) ~ (r*,l), 8 = r + m, and s>m. Thissuggests defining the set /+ of positive integers by

/+ = {[s,m]: [s,m]G g, s> m}

In view of the isomorphism (1) the set /+ may be replaced by the set N whenever thelatter is found more convenient.

40 THE INTEGERS [CHAP. 4

ZERO AND THE NEGATIVE INTEGERS

Let r,s G N. Now [r, r] = [s, s] for any choice of r and s, and [r, r] = [s, t] if and

only if t = s. We now define the integer zero, 0, to correspond to the equivalence class

[r,r], r G N. Its familiar properties are

[s, m] + [r, r] = [s, m] and [s, m] [r, r] = [r, r]

proved in Problems 2(6) and 2(c). The first of these leads to the designation of zero as

the identity element for addition.

Finally, we define the set /~ of negative integers by

/- = {[s,m]: [s,TO]e^, s<m}

It follows now that for each integer [a, b], a^b, there exists a unique integer [b,a] such

that (see Problem 2(d)) ^^^^

[a,b] + {b,a] = [r,r] ^ (2)

We denote [b,a] by - [a, b] and call it the negative of [a, b]. The relation {2) suggests the

designation of [b,a] or -[a,b] as the additive inverse of [a,b].

THE INTEGERS

Let p,qGN. By the Trichotomy Law for natural numbers, there are three possibilities:

(a) p = q, whence [p, q] = [q, p] ^ 0.

(6) p<q, so that p + a = q for some a G N; then p + a* = q + 1 and [q,p] = [a*,l] <-» a.

(c) p> q, so that p - q + a for some aGN and [p, q] <^ a.

Suppose [p,q] <-^ nGN. Since [q,p] = -[p,q], we introduce the symbol -n to

denote the negative of nGN and write [q,p] ^ -n. Thus, each equivalence class of îs now mapped onto a unique element of / = (0, ±1, ±2, . .

.}. That S and / are isomorphic

follows readily once the familiar properties of the minus sign have been established. In

proving most of the basic properties of integers, however, we shall find it expedient to

use the corresponding equivalence classes of ^.

Example 3: Let o, 6 G /. Show that (-a) • 6 = -(a • 5). Let a «• [s, m] so that -a^ [m, a] and

let 6 «-> [t,n]. Then

(-a) • 6 <-> [m, s] • [t, n] = [{m' t + s'v), (m-n + s' t)]

while o • 6 <^ [s, m] • [t, w] = [(s • t + m • n), (s • b + m • t)]

Now -{a -b) «• [(s-n + wt), (s't + m-n)] <-» [{-a) • b]

and so (-«) • 6 = -(o • 6)

See Problems 6-7.

ORDER RELATIONS

For a,bGl, let a <^ [s,m] and b ^ [t,n]. The order relations "<" and ">" for

integers are defined byab if and only if (s + n)>{t + m)

In Problem 8, we prove the Trichotomy Law: For any a,b Gl, one and only one of

(o) a^b, (b) a<b, (c) a>bis true.


When a,b,cE I, we have:

(1) a + cb.

(2) If c> 0, then a'C<b-c if and only if a<b.

(2') If c> 0, then a-Ob-c if and only if a>b.

(3) If c < 0, then a-c <b'C if and only if a> b.

(3') If c < 0, then a-Ob'C if and only if a<b.

For proofs of (1') and (3), see Problems 9-10.

The Cancellation Law for multiplication on /,

M4. If 2 ^ and if X'z = yz, then x=^y

may now be established.

As an immediate consequence, we have

Theorem 11. If a,b G I and if a-b = 0, then either a = or b = 0.

For a proof see Problem 11.

The order relations permit the customary listing of the integers

. . ., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...

and their representation as equally spaced points on a line. Then "a < b" means "a lies

to the left of b", and "a > b" means "a lies to the right of b".

T T— I1 1 1 1 —I,-10 12 3 4 6

Fig. 4-1

From the above listing of the integers, we note

Theorem III. There exists no nGl+ such that < n < 1.

This theorem (see Problem 12 for a proof) is a consequence of the fact that the set /+ ofpositive integers (being isomorphic to N) is well-ordered.

SUBTRACTION "-"

Subtraction, "-", on / is defined by a-b = a + {-b). Subtraction is clearly a binaryoperation on /. It is, however, neither commutative nor associative although multiplicationis distributive with respect to subtraction.

Example 4: Prove: a-{b-e) ¥- {a-V)-e for a,6,cG/ and c ^ 0.

Let a.<r^\8,m\, 6<->[t,n], and c<^[m,p]. Then

h- e = b + (-c) <^ [{t + p), (n + m)]

-(b - c) <-» [(to + u), (t + p)]

and a-{b-c) = a+ (-(6 - c)) <^ [{s + n + u), (m + t + p)]

while a-b - a+ (-6) <^ [(s + to), (m + t)]

and (a-b)- c = (a+b) + (-c) <^ [(s + to + p), (w + t + u)]

Thus, when c ^ 0, a — {b- c) ¥= {a- b) - c.


ABSOLUTE VALUE \a\

The absolute value, "\a\", of an integer a is defined by

f a when a —a =

-a when a <Thus, except when a — 0, |a| G /+.

The following laws

(1) -\a\ ^ a ^ \a\ (2) |a-6| = |a|-|6|

(3) |a| - |6| ^ \a + b\ (3') |a + b| ^ |a| + |&|

(4) |a|-|b| ^ \a-b\ (4') |a-6| ^ |a| + 16|

are evidently true when at least one of a, b is 0. They may be established for all a,b G I

by considering the separate cases as in Problems 14 and 15.

ADDITION AND MULTIPLICATION ON /

The operations of addition and multiplication on / satisfy the laws A1-A4, M1-M4, and

D1-D2 of Chapter 3 (when stated for integers) with the single modification

M4. Cancellation Law: If wp = n-p and if pÔG/, then m = n for all

m,n G /.

We list below two properties of / which N lacked

As. There exists an identity element, G /, relative to addition, such that n + =+ n = n for every n G I.

Ae. For each n G / there exists an additive inverse, -n G /, such that n + (—n) =

(-») +n = 0.

and a common property of N and /

M5. There exists an identity element, 1 G /, relative to multiplication, such that

1 • n = n • 1 = n for every nG I.

By Theorem III, Chapter 2, the identity elements in As and Ms are unique; by TheoremIV, Chapter 2, the additive inverses in As are unique.

OTHER PROPERTIES OF INTEGERS

Certain properties of the integers have been established using the equivalence classes

of ^. However, once the basic laws have been established, all other properties may be

obtained using the elements of / themselves.

Examples: Prove: For all a,6,c6/,

(a) o • = • o = (b) a{-b) - -(06) (c) 0(6 - c) = 06 - oc

(a) + = (As)

Then o'o + O = a' a = o(a + 0) = o'o + a'O (Dj)

and = o'O (A4)

Now, by M2, • o = o • = 0, as required. However, for reasons which

will not be apparent until a later chapter, we will prove

O'o = o'O

without appealing to the commutative law of multiplication. We have


o-o + O - a- a = (a + 0)a = a'o + O'o (D2)

hence, = 0*o

and 0*0 = a •

(6) = o • = a[6 + i-b)] = o • 6 + a{-b) (Dj)

thus, o(—6) is an additive inverse of o • 6. But —(a • 6) is also an additive in-

verse of o • 6; hence,a(-b) - -(a • b) (Theorem IV, Chapter 2)

(c) a{b -e) = a[b + (-c)] = ab + a{-e) (Dj)

= ab + (-ac) ((b) above)

= ab — ae

Note: In (c) we have replaced a • 6 and —(o • c) by the familiar 06 and —ac

respectively.

MULTIPLES AND POWERSThe section with this title in Chapter 3 (Page 33) may be repeated here after replacing

N by /+.4

It is to be noted that for the case S = /, we may now identify ka with k-a. Moreover,

since / contains the additive inverse of each of its elements, we have (ix)-(xi), but not

(ix)'-(xi)', for all m,n Gl. This is largely trivial, of course, since when a,b,m,nG I

the properties (ix) and (xi) are then the distributive laws.

Solved Problems

1. Show that ~ on L is an equivalence relation.

Let (s, m), {t, n), (u, p) e L. We have

(a) («, m) ~ (s, m) since s + m = a -\- m; ~is reflexive.

(6) If {s,m) ~ (t,n), then (t,n) ~ (s, »i) since each requires s + n = t + m; ~is symmetric.

(c) If (s, to) ~ (t, n) and (t, n) ~ (u, p), then 8 + w = t + w, t + p = u + n, and 8 + n+t + p =t + m + u + n. Using A4 of Chapter 3, Page 30, the latter equality can be replaced by

8 + p — m + u; then (s, w) ~ (u, p) and ~ is transitive.

Thus, ~, being reflexive, symmetric, and transitive, is an equivalence relation.

2. When s.m.p.r G N, prove:

(a) [{r + p), p] = [r*, 1] (c) [s, m] • [r, r] = [r, r] (e) [s, m] • [r*, r] = [s, m]

(6) [s, m] + [r, r] = [s, m] (d) [s, m] + [m, s] = [r, r]

(o) ((r + p), p) ~ (r*, 1) since r + p + 1 = r* + p.

Hence, [(r + p), p] = [r*, 1] as required.

(6) [s, w] + [r, r] = [(» + r), (m + r)]. Now ((« + r), (to + r)) ~ (s, to) since (s + r) + to — s + (to + r).

Hence, [(« + r), (to + r)] = [s, to] + [r, r] = [«,to].

(c) [«, to] • [r, r] = [(g • r + TO • r), (s • r + TO • r)] = [r, r] since s'r + TO'r + r = s-r + m'r + r.

(d) [s, to] + [to, «] = [(s + to), (s + to)] = [r, r]

(e) [8,to] • [r*,r] = [(s 'r* + wr), (s -r + m'r*)] = [s, to] since 8'r* + TO»r + TO = 8 + 8'r +nfr* = 8*r* + wi* r*.


3. Prove: The equivalence class to which the sum (product) of two elements, one selected

from each of two equivalence classes of S' belongs is independent of the elements

selected.

Let [a, 5] = [«,w] and [e,d] = [t,n]. Then (a,h) ~ (s,m) and (e.d) ~ (t,«) so that

o + nt = 8 + 6 and c + « = t + d. We shall prove:

(a) [a, 6] + [c, d] = [s,m] + [t,n], the first part of the theorem;

(6) a'C+h'd + 8'n + m't = o«d+6'c + 8't + »fn, a necessary lemma;

(c) [o, 6] • [e,d] = [s.m] • [t,w], the second part of the theorem.

(a) Since a + m+c + n= s + 6 + t+d,

(a + c) + (m + m) = («+<) + (6 + d)

((a + c), (ft + d)) ~ ((«+t), (m + w))

[(a + c), (6 + d)] = [{8 + t)Am + n)]

and [a, 6] + [c, d] = [«, w] + [t, w]

(b) We begin with the evident equality

(a + m) ' (c + t) + (8 + 6) • (d + w) + (c + n) • (o + s) + (d+t) • (6 + m)

= (s + 6) • (c + t) + (a + m)' (d + n) + (d + «) • (a + s) + (c + m) • (6 + m)

which reduces readily to

2(a'c + 6'd + 8'n + m't) + (o-t + m-c + 8'd+b'n) + (8'C + wa+&'t + wd)= 2(a •d+6'c + 8't + m'w) + (o't + m-c + s«d+6'7i) + (8'C + Ti'a+6't + wd)

and, using the Cancellation Laws of Chapter 3, to the required identity.

(c) From (6), we have

(a • c + 6 • d) + (s • n + «i • t) = (s • t + m • w) + (a • d + 6 • c)

Then ((a • c + 6 • d), (a • d + 6 • c)) ~ ((s • t + m • w), (s • w +w t))

[(a • c + 6 • d), (o • d + 6 • c)l = [(8 • t +w m), (s • w + m • f)]

and so [a,h]-[c,d] = [s,m] • [t,n]

4. Prove the Associative Law for addition:

{[s,m] + [t,n]) + [u,p] = [s,m] + {[t,n] + [u,p])

for all [s, m], [t, n], [u, p] G S-

We find

([s,m] + [t,n]) + [u,p] = [(8 + t),{m + n)] + [u,p] = [{s + t + u), (m + n + p)]

while [s,m] + {[t,n] + [u,p]) = [s,m] + [(t + u), (n + p)] = [(s + 1 + u), Cm + n + p)]

and the law follows.

5. Prove the Distributive Law D2:

{[s,m] + [t,n])'[u,p] = [s,m]-[u,p] + [t,n]-[u,p]

for all [s,m],[t,n],[u,p] e ^.

We have

([a, m] + [t, n]) • [u, p] = [(s + t), (m + n)] • [u, p]

= [((s + t)'M + (m + n)'p), ((s + t)-p + (m. + n)'u)]

= [(s'U + t-u + m-p + n-p), (s ' p + t • p + m • u + 71' u)]

= [(^(s'U + rn'p) + (t'u + ri'p)), ((s • p + m • m) + (t'p + wu))]

= [(s'u + m-p), (s'p + m'u)] + [(t'u + n' p), (t- p + n'u)]

= [s,m.] • [u,p] + [t,n] ' [u,p]


6. (a) Show that a + (-a) = for every a € 7.

Let a<^[8,m\; then —a«-»[w, s],

a + (-o) <-> [s,w] + [m,s] = [(« + m), (m + s)] = [r,r] *^

and a + (—o) = 0.

(&) If a; + a = ft for a,b Gl, show that a; = b + (-a).

When a; = b + (-o), x + o = (b + (-o)) + a= h + ((-a) + a) = b; thus, a; = b + (-a) is a

solution of the equation x + a = b. Suppose there is a second solution y. Then j/ + o = b =

x + a and, by A4, y = x. Thus the solution is unique.

7. When a.bGl, prove: (1) (-a) + (-5) = -(a + ft), (2) (-a)-(-6) = a- 6.

Let a <-» [s,m] and b <^ [t,n]; then -o *^ [m,8] and -b <-* [w, t].

(1) (-0) + (-b) «• [m,8] + [w.t] = [(w + m), (s+t)]

and a + b ^ [s.m] + [t,n] = [(s + «), (m + n)]

Then -{a+b) <^ [(m + w), (a + 1)] <^ (-0) + (-b)

and (-0) + (-b) = -(a+b)

(2) (-a) • (-b) «^ [mi.g] '[nit] = [(m»»i + s • t), (»»• t + s 'n)]

and a' b ** [s, m] • [t, n] = [(« • t + w n), (s • w + m • t)]

Now [(j» • w + « • t), (w t + « • m)] = [(8 • t + m • w), (s • w + m • t)]

and (-0) • (-b) = a • b

8. Prove the Trichotomy Law: For any a,b G I one and only one of

(a) a = b, (b) a< b, (c) a > 6

is true.

Let a«-» [8, to] and b <-» [«,ri]; then by the Trichotomy Law of Chapter 3, Page 32, one and

only one of (a) B + n = t +m and a = b, (b) 8 + n < * + to and o < b, {c) s + n > t + m and

a > b is true.

9. When a,b,cGl, prove: a + c > b + c if and only if a > 6.

Take o <^ [8,m], b <^ [t,7t], and e <-» [u,p]. Suppose first that

a + e > b + c or ([s.m] + [u,p]) > ([t,n] + [u,p])

Now this implies [{s + u), (w + p)] > [{t + u), (n. + p)]

which, in turn, implies (s + u) + {n + p) > {t + u) + {m + p)

Then, by Theorem II', Chapter 3, Page 33, (a + n) > (t + to) or [«, m] > [t, n] and a > b, as required.

Suppose next that o > b or [s,m] > [t,n]; then {s + n) > {t + m). Now to compare

a + c ^ [{8 + u),{m + p)] and b + c *^ [{t + u), {n + p)]

we compare [(s + u), (to + p)] and [(« + u), (n + p)]

or {s + u) + {n + p) and (t + m) + (to + p)

or (s + n) + {u + p) and {t + m) + {u + p)


Since (s + m) > (t + m), it follows by Theorem 11', Chapter 3, Page 33, that

(s + n) + (m + p) > (t + m) + (u + p)

Then (s + u) + (n + p) > (t + u) + (m + p)

[{8 + u),{m + p)] > Ht + u).{n + p)]

and a + e > b + c

as required.

10. When a,b,c G I, prove: If c < 0, then a • c < 6 • c if and only if a> b.

Take a <-> [«, m], 6 <-» [t,n], and c «-* [m, p] in which u < p since c < 0.

(o) Suppose a • c < 6 • c; then

[(s • M + m • p), (s • p + m • m)] < [(t • M + w • p), (t • p + w • m)]

and (« • M + m • p) + (( • p + w m) < (t • m + m • p) + (s • p + m • m)

Since m < p, there exists some k G. N such that u + k — p. When this replacement for p is

made in the inequality immediately above, we have

{s'u + m'u + m'k + t'u+t'k + n-u) < {t'u + n'u + ri'k + s'u + s-k + m'u)

whence wfc + *'fc < wfc + 8'fc

Then (m+t)- k < {n + s) • k

m + t < 71 + 8

s + n > t + m

and o > 6

(b) Suppose a > b. By simply reversing the steps in (a), we obtain a- c < b ' c, as required.

11. Prove: If a,b G I and wb = 0, then either a-0 or 6 = 0.

Suppose o ?^ 0; then o • 6 = = a • and by M4, 6 = 0. Similarly, if 6 # then a = 0.

12. Prove: There exists no nG I-^ such that < « < 1.

Suppose the contrary and let m G 7+ be the least such integer. From < m < 1, we have by

the use of (2), Page 41, < m^ < m < 1. Now < m^ < 1 and m^ < m contradict the assumption

that m is the least, and the theorem is established.

13. Prove: When a,b Gl, then o [s,w] and 6 «-»[«,«]. Then

a-b = a + {-b) ^ [s.m] + [n,t] = [(s + w), (w + t)]

If a < 6, then s + n < m+t and a—b < 0. Conversely, if o — 6 < 0, then s + n < m+t and

a<b.

14. Prove: \a + b\ ^ \a\ + \b\ for all a.bGl.

Suppose a>0 and 6 > 0; then |a + 6| = + 6 = |o| + |6|.

Suppose a<0 and 6 < 0; then |o + 6| = -(0 + 6) = -a + (-b) = loj + |6|.


Suppose a > and 6 < so that \a\ = o and |5| = -6. Now, either a + 6 = and |a+ 6| =

< |a| + |6| or

a+b < and |a+6| = -(o + 6) = -a + (-b) = -|a| + |6| < \a\ + \b\

or a + b > and \a + h\ = a + b = a - {-b) = \a\ - \b\ < \a\ + \b\

The case a < and ft > is left as an exercise.

15. Prove: |a'6| = |a|'lb| for all a,bGl.

Suppose a > and b > 0; then lo| = a and \b\ = b. Then \a'b] = a-b = \a\'\b\

.

Suppose a < and 5 < 0; then ]o| = -a and |6| = -6. Now a-b > 0; hence, |a'6| = a'b =

(-o)-(-5) = lal-l^l-

Suppose a>0 and 6 < 0; then |a| = o and |b| = -6. Since a-b < 0, |o»6| = -(a-b) =

a-(-b) = \a\-\b\.

The case a < and b > is left as an exercise.

16. Prove: If a and b are integers such that a-b = 1 then a and b are either both 1 or

both -1.

First we note that neither o nor b can be zero. Now |a- b| = |o| • |*1 = 1 and, by Problem 12,

\a\ s= 1 and \b\ ^ 1. If \a\ > 1 (also, if \b\ > 1), \a\ • \b\ ¥- 1. Hence \a\ = |b| = 1 and, in view of

Problem 7(b), the theorem follows.


17. Prove: When r,8 G N,

(a) (r,r) ~ (8,8) ~ (1,1) (d) (r*,r) -f- (r,r*)

(b) (r*,r) ~ (8*, 8) ~ (2,1) (e) (r*,r) -f- {s,g*)

(c) (r,r*) ~ (8,8*) ~ (1,2) (/) (r* • 8* + 1, r* + 8*) ~ ((r-s)*, l)

18. State and prove: (a) the Associative Law for multiplication, (b) the Commutative Law for addition,

(c) the Commutative Law for multiplication, (d) the Cancellation Law for addition on ^.

19. Prove: [r*, r] <^ 1 and [r, r*] *^ -1

20. If ae I, prove: (a) a • = • o = 0, (b) (-1) • a = -a, (c) -0 = 0.

21. If a, be I, prove: (o) - (-o) = +o, (b) {-a)(-b) = a'b, (c) (-a) + b = - (a + (-b)).

22. When b € / + , show that a - b < o + b for all a G /.

23. When a,bGl, prove (1), (2), (2') and (3'), Page 41, of the order relations.

24. When a,b,cGl, prove a'{b — c) = o'b — o*c.


25. Prove: If a,be7 and a<b, then there exists some cGl'*' such that a+c = b.

Hint. For a and 6 as represented in Problem 7, take c <-> [(t + m), (n + s)].

26. Prove: When a,h,c,d&I,

(a) —a>-b if a < b.

(6) a + c < b + d if a<6 and c < d.

(c) If a < (6 + c), then a—b<c.(d) a— b = c — d if and only if a + d = b + c.

27. Prove that the order relations are well-deflned.

28. Prove the Cancellation Law for multiplication.

29. Define sums and products of n > 2 elements of I and show that in such sums and products parentheses

may be inserted at will.

30. Prove: (a) m? > for all integers m ¥= 0.

(5) m3 > for all integers m > 0.

(c) w? < for all integers m < 0.

31. Prove without using equivalence classes (see Example 5):

(a) —(—a) = a

(6) (-a)(-6) = ab

(c) (b-c) = (6 + a) - (c + a)

(d) 0.(6 — c) = ab — ac

(e) (a + b){c + d) = (ac + ad) + (be + 6d)

(/) (a + b)(c - d) = (ac + be) - (ad + bd)

(g) (a - b)(c -d) = (ac + bd) -(ad+ be)

chapter 5

Some Properties of Integers

DIVISORS

An integer a ^ is called a divisor (factor) of an integer b (written "a\b") if there

exists an integer c such that b - ac. When a|6 we shall also say that b is an integral

multiple of a.

Example 1 : (a) 2{6 since 6 = 2*3

(6) -31 15 since 15 = (-3)(-5)

(c) aI0, for all aG I, since = a •

In order to show that the restriction o t^ is necessary, suppose 1b. If b ¥- 0, we

must have b = • c for some c £ /, which is impossible; while if 6 = 0, we would have

= • c, which is true for every c G I.

When b,c;x,y El, the integer bx + cy is called a linear combination of b and c. In

Problem 1, we prove

Theorem I. If a I 6 and a I c then a I (bx + cy) for all x,y G I.

See also Problems 2, 3.

PRIMES

Since a-1 — (—a)(— 1) = a for every aGl, it follows that ±1 and ±a are divisors

of a. An integer p¥= 0,±1 is called a prime if and only if its only divisors are ±1 and ±p.

Example 2: (a) The integers 2 and —5 are primes, while 6 — 2*3 and —39 = 3(—13) are not

primes.

(6) The first 10 positive primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

It is clear that —p is a prime if and only if p is a prime. Hereafter, we shall restrict

our attention mainly to positive primes. In Problem 4, we prove:

The number of positive primes is infinite.

When a = be with \b\ > 1 and \c\ > 1, we call a composite. Thus every integer

a ^ 0, ±1 is either a prime or a composite.

GREATEST COMMON DIVISOR

When a\b and a\c, we call a a common divisor of b and c. When, in addition, every

common divisor of b and c is also a divisor of a, we call a a greatest common divisor

{highest common factor) of b and c.

Example 3: (a) ±1, ±2, ±3, ±4, ±6, ±12 are common divisors of 24 and 60.

(6) ±12 are greatest common divisors of 24 and 60.

(c) The greatest common divisors of 6 = and c ^ are ±c.

49

50 SOME PROPERTIES OF INTEGERS [CHAP. 5

Suppose c and d are two different greatest common divisors of o ^ and h ¥'Q. Then

cI

d and d\c; hence, by Problem 3, c and d differ only in sign. As a matter of convenience,

we shall hereafter limit our attention to the positive greatest common divisor of two

integers a and 6 and use either d or (a, b) to denote it. Thus, d is truly the largest (greatest)

integer which divides both a and 6.

Example 4: A familiar procedure for finding (210,510) consists in expressing each integer as a

product of its prime factors, i.e., 210 = 2 • 3 • 5 • 7, 510 = 2 • 3 • 5 • 17, and forming

the product 2 • 3 • 5 = 30 of their common factors.

In Example 4 we have tacitly assumed (a) that every two non-zero integers have a

positive greatest common divisor and (6) that any integer a > 1 has a unique factoriza-

tion, except for the order of the factors, as a product of positive primes. Of course, in

(6) it must be understood that when a is itself a prime, "a product of positive primes"

consists of a single prime. We shall prove titese propositions later. At the moment, wewish to exhibit another procedure for finding the greatest common divisor of two non-zero

integers. We begin with:

The Division Algorithm. For any two non-zero integers a and h, there exist unique

integers q and r, called respectively quotient and remainder, such that

a ^ hq + r, ^ r < |6| (1)

For a proof, see Problem 5.

Example 5: (a) 780 = -48(-16) + 12 (c) 826 = 25-33 + 1

(6) -2805 = 119(-24) + 51 (d) 758 = 242(3) 4- 32

From {!) it follows that 6|a and (a,h) = 6 if and only if r = 0. When r#0 it is

easy to show that a common divisor of a and h also divides r and a common divisor of

b and r also divides o. Then (a,h)\{b,r) and {b,r)\(a,b) so that, by Problem 3,

(a, h) = (5, r). Now either r|6 (see Examples 5(a) and (5c)) or r)^h (see Examples 5(b) and

5(d)). In the latter case, we use the division algorithm to obtain

b = rqi + ri, < ri < r (2)

Again, either ri|r and (a, b) = ri or, using the division algorithm,

r = riqi -\- r^, < rz < ri {S)

and (a, b) = (6, r) = (r, n) = (n, Vi).

Since the remainders ri, ri, . . ., assuming the process to continue, constitute a set of

decreasing non-negative integers there must eventually be one which is zero. Suppose

the process terminates with

{Jc) rk-3 = rfc-z'Qk-i + rk-i (i<rk-i<rk-%

{k + 1) rk-2 = rk-i-qk + rk < n < n-i

(fe-t-2) rk-i = rk'Qfc+i +

Then (a, ft) = (b,r) = (r,ri) = ••• = (rk-2,rfc-i) = (rk-,,rk) = r^.

Example 6: (a) In Example 5(6), 51/119. Proceeding as in (2), we find 119 = 51(2) + 17. Now17

I

51; hence, (-2805, 119) = 17.

(6) In Example 5(d), 32/242. From the sequence

758 = 242(3) + 32

242 = 32(7) + 18

32 = 18(1) + 14

18 = 14(1) + 4

14 = 4(3) + 2

4 = 2(2)

we conclude (758, 242)

CHAP. 5] SOME PROPERTIES OF INTEGERS 51

Now solving (J) for r = a — bq = a + (—q)b = rriia + nib ; and

substituting in (2), n — b — rqi — b — (mia + nib)qi

= —viiqia + (1 — niqi)b — TO20 + njo

substituting in {S), ri = r — Viq^ = {rriia + nib) — {m%a + 1126)92

= (wii — q2.m2)a + (rii — q%ni)b = mâ + n^b

and continuing, we obtain finally

ric = TOk + ia + «)c+i6Thus, we have

Theorem II. When d = (a, b), there exist m,n G I such that d = {a, b) = ma + nb.

Example 7: Find (726, 276) and express it in the form of Theorem II.

From We obtain

726 = 275-2 + 176 11 = 77 - 22 • 3 = 77 - (99 - 77) • 3

275 = 176*1 + 99 = 77 • 4 - 99 • 3

176 = 99 • 1 + 77 - (176 - 99) • 4 - 99 • 3

99 = 77'1 + 22 = 176 '4 - 99 '7

77 = 22 • 3 + 11 = 176 • 4 - (275 - 176) • 7

22 = 11 • 2 = 176 • 11 - 275 • 7

= (726 - 275 • 2) • 11 - 275 • 7

Thus, m = 11 and n = —29.= 11 '726 + (-29) "275

Note 1. The procedure for obtaining m and n here is an alternate to that used in

obtaining Theorem II.

Note 2. In (a, b) = m.a + nb, the integers m and n are not unique; in fact, (a, b) ={m + kb)a + (n — ka)b for every k G N.

See Problem 6.

The importance of Theorem II is indicated in

Example 8: Prove: If o|c, if 6

|c, and if (o, 6) = d, then 06

1 ed.

Since o|c and 6

|e, there exist integers s and t such that c = as = bt. By

Theorem II there exist m,re G / such that d — ma + nb. Then

cd = cma + enb = btma + asnb = ab(tm + sn)and ab

|ed.

A second consequence of the Division Algorithm is

Theorem III. Any non-empty set K of integers which is closed under the binary opera-tions addition and subtraction is either {0} or consists of all multiples of

its least positive element.

An outline of the proof when K ¥- {0} follows. Suppose K contains the integer o # 0.

Since K is closed with respect to addition and subtraction, we have:

(i) a-a = OGK(ii) - a = -aGK(iii) K contains at least one positive integer.

(iv) K contains a least positive integer, say, e.

(v) By induction on n, K contains all positive multiples ne of e (show this),

(vi) K contains all integral multiples me of e.

(vii) If b G K, then b = q-e + r where — r < e; hence, r = and so every elementof K is an integral multiple of e.


RELATIVELY PRIME INTEGERS

For given a,b Gl, suppose there exist m,nGl such that am + bn = 1. Now every

common factor of a and 6 is a factor of the right member 1; hence, (a, 6) = 1. Twointegers a and b for which (a, 6) = 1 are said to be relatively prime.

gee Problem 7


Theorem IV. If (a,s) = (b,s) = 1, then (ab.s) = 1.

PRIME FACTORS


Theorem V. If p is a prime and if p |ab, where a,b e I, then p\a or p\b.

By repeated use of Theorem V there follows

Theorem V. If p is a prime and if p is a divisor of the product a-b-c- .. .-t of n

integers, then p is a divisor of at least one of these integers.


The Unique Factorization Theorem. Every integer a > 1 has a unique factorization,

except for order,

(a) a = pi • P2 • P3 •. . .

• Pn

into a product of positive primes.

Evidently, if (a) gives the factorization of a, then

-a = -(pi • Pa • P3 • ... • Pn)

Moreover, since the p's of (a) are not necessarily distinct, we may write

a ^z p"' • p"^ • p"3 •. . . • p"'

where each aj ^ 1 and the primes pi, P2, ps, . . . , Ps are distinct.

Example 9: Express each of 2,241,756 and 8,566,074 as a product of positive primes and obtain

their greatest common divisor.

2,241,756 = 22 • 3* • 11 • 17 • 37 and 8,566,074 = 2 • 3* • 112 • 19 • 23

Their greatest common divisor is 2 • 3* • 11.

CONGRUENCESLet TO be a positive integer. The relation "congruent modulo m", {= (mod w)), is

defined on all a,b G/ by a = b(mod m) if and only if m\{a-b).

Example 10: (a) 89 = 25(mod 4) since 4|(89 - 25) = 64 (e) 24 # 3(mod 5) since 5/21

(6) 89 = l(mod 4) since 4 | 88 (/) 243 ^ 167(mod 7) since 7/76

(c) 25 = l(mod 4) since 4|24 (g) Any integer a is congruent

modulo m to the remainder(d) 153 = -7(mod 8) since 8 1

160 obtained by dividing a by m.

An alternate definition, often more useful than the original, is a = b(mod m) if and

only if a and b have the same remainder when divided by to.

As immediate consequences of these definitions, we have:

Theorem VI. If a = 6(mod to) then, for any nGl, mn + a s 6(mod to) and conversely.


Theorem VII. If a = 6(mod m), then, for all x e /, a + x = b + a;(mod m) and ax =6x(niod m).

Theorem VIII. If o = 6(mod m) and c = e(mod m), then a + c = b + e(mod m), a-c =b-e{modm), ac ^ beimodm). ^^ Problem 11.

Theorem IX. Let (c, m)-d and write m = mid. If ca = c6(mod m), then a = b{niod TOi)

and conversely.


As a special case of Theorem IX, we have

Theorem X. Let (c,m) = 1. If ca = c6(modw), then a = 6(modTO) and conversely.

The relation = (mod m) on / is an equivalence relation and separates the integers intom equivalence classes, [0], [1], [2], . . ., [m- 1], called residue classes modulo m, where

[r] = {a: a G /, a = r(modm)}

Example 11: The residue classes modulo 4 are:

[0] = { . . . , -16, -12, -8,^-4, 0, 4, 8, 12, 16, ...

}

[1] = { . .. , -15, -11, -7, -3, 1, 5, 9, 13, 17, ...

}

[2] = { . . . , -14, -10, -6, -2, 2, 6, 10, 14, 18, ...

}

[3] = {. . ., -13, -9, -5, -1, 3, 7, 11, 15, 19, . . .}

We will denote the set of all residue classes modulo m by Il{m). For example, 7/(4) ={[0]. [1]. [2],j3]} and Il(m) = {[0], [1], [2], [3], . . ., [w-1]}. Of course, [3] € 7/(4) =[3] e 7/(m) if and only if m = 4. Two basic properties of the residue classes modulo m are:

If a and b are elements of the same residue class [s], then a = &(mod m).

If [s] and [t] are distinct residue classes with a £ [s] and h G [t], then a # 6(mod m).

THE ALGEBRA OF RESIDUE CLASSESLet "0" (addition) and "O" (multiplication) be defined on the elements of 7/(m) as

follows:

[a] © [6] = [a + b]

[a] O [6] = [a • 6]

for every [a], [b] G Il(m).

Since © and O on 7/(m) are defined respectively in terms of + and • on 7, it followsreadily that © and O satisfy the laws At-A4, M1-M4, and D1-D2 as modified in Chapter 4,Page 42.

Example 12: The addition and multiplication tables for //(4) are:

and

© 1 2 3

1 2 3

1 1 2 3

2 2 3 1

3 3 1 2

1 2 3

1 1 2 3

2 2 2

3 3 2 1

Table 5-1 Table 5-2

where, for convenience, [0], [1], [2], [3] have been replaced by 0,1,2,3.


LINEAR CONGRUENCESConsider the linear congruence

(b) ax = 6(mod m)

in which a, b, m are fixed integers with to > 0. By a solution of the congruence we shall

mean an integer x = xi for which m\{axi — b). Now if Xi is a solution of (6) so that

TOI

(axi — 6) then, for any k G I, m\ {a{xi + km) - b) and Xi + km is another solution.

Thus, if Xi is a solution so also is every other element of the residue class [xi] modulo to.

If then the linear congruence (6) has solutions, they consist of all the elements of one or

more of the residue classes of //(to).

Example 13: (a) The congruence 2x = 3(mod 4) has no solution since none of 2*02 • 2 - 3, 2 • 3 - 3 has 4 as a divisor.

3, 2 • 1 - 3,

(6) The congruence 3a; = 2(mod 4) has 6 as a solution and, hence, all elements of

[2] G 7/(4) as solutions. There are no others.

(c) The congruence 6a; = 2(mod 4) has 1 and 3 as solutions. Since 3 # l(mod 4),

we shall call 1 and 3 incongruent solutions of the congruence. Of course, all ele-

ments of [1], [3] e 7/(4) are solutions. There are no others.

Returning to (&), suppose (a, to) = 1 = sa + tm. Then b - bsa + btm and xi = bs is

a solution. Now assume X2 # a;i(mod to) to be another solution. Since axi = 6(mod to)

and ax2 = b{mod to), it follows from the transitive property of = (mod to) that axi =

aa;2(mod to). Then to|a{xi - X2) and xi = a;2(mod to) contrary to our assumption. Thus,

(b) has just one incongruent solution, say Xi, and the residue class [xi] G I/{m), also called

a congruence class, includes all solutions.

Next, suppose that (a, to) = d = sa + tm, d>l. Since a = aid and m-mid, it

follows that if (5) has a solution x = Xi then axi-b = mq = m,idq and so d\b. Con-

versely, suppose that d = (a, to) is a divisor of b and write b = bid. By Theorem IX,

Page 53, any solution of (b) is a solution of

(c) ttix = bi(mod TOi)

and any solution of (c) is a solution of (b). Now (ai,TOi) = 1 so that (c) has a single

incongruent solution and, hence, (6) has solutions. We have proved the first part of

Theorem XI. The congruence ax = 6(mod to) has a solution if and only if d = (a, to)

is a divisor of b. When d\b, the congruence has exactly d incongruent

solutions (d congruence classes of solutions).

To complete the proof, consider the subset

S = {xi, xi + mi, xi + 2mi, Xi + Smi, . . ., xi + (d-l)mi]

of [xi], the totality of solutions of ais; = 6i(mod TOi). We shall now show that no two

distinct elements of S are congruent modulo to (thus, (b) has at least d incongruent solu-

tions) while each element of [xi] - S is congruent modulo to to some element of S (thus,

(6) has at most d incongruent solutions).

Let Xi + STOi and Xi + tmi be distinct elements of S. Now if Xi + smi = Xi + tTOi(mod to)

then TOI

(s - t)mi; hence, d\(s — t) and s = t, a contradiction of the assumption s ¥- 1.

Thus, the elements of S are incongruent modulo to. Next, consider any element of

[a;i] - S, say Xi + {qd + r)mi where q^l and ^ r < d. Now Xi + {qd + r)mi =xi + rTOi + qm = Xi + rTOi(mod to) and Xi + rmi G S. Thus, the congruence (b), with

(a, to) = d and d\b, has exactly d incongruent solutions. „ 1.1 , a^ ' '

I ' See Problem 14.


POSITIONAL NOTATION FOR INTEGERS

It is well known to the reader that

827,016 = 8-10^ + 2-104 + 7-103 + o-lO^ + 1-10 + 6

What is not so well known is that this representation is an application of the congruence

properties of integers. For, suppose a is a positive integer. By the division algorithm,

a = 10 - go + ro, ^ n < 10. If Qo = 0, we write a = n; if go > 0, then qo = 10 - gi + u,

0^n< 10. Now if qi = 0, then a = 10 - ri + r2 and we write a — r\n; if qx > 0, then

qi = 10 • Q2 + rz, — r2< 10. Again, if 92 = 0, then a = 10^ - rz + 10 - ri + n and wewrite a = r2riro; if 92 > 0, we repeat the process. That it must end eventually and wehave

a = 10*-n + 10»~i-r,-i + ••• + 10-ri + n = nn-i •rin

follows from the fact that the g's constitute a set of decreasing non-negative integers.

Note that in this representation the symbols n used are from the set (0,1,2,3, . . .,9} of

remainders modulo 10. (Why is the representation unique?)

In the paragraph above, we chose the particular integer 10, called the base, since this

led to our system of representation. However, the process is independent of the base andany other positive integer may be used. Thus, if 4 is taken as base, any positive integer

will be represented by a sequence of the symbols 0, 1, 2, 3. For example, the integer

(base 10) 155 = 4"- 2 + 4^- 1 + 4-2 + 3 = 2123 (base 4).

Now addition and multiplication are carried out in much the same fashion, regardless

of the base; however, new tables for each operation must be memorized. These tables

for base 4 are:

+ 1 2 3

1 2 3

1 1 2 3 10

2 2 3 10 11

3 3 10 11 12

and

• 1 2 3

1 1 2 3

2 2 10 12

3 3 12 21

Table 5-3 Table 5-4

See Problem 15.

Solved Problems

1. Prove: If a|6 and a

|c, then a

|

(bx + cy) where x,y & I.

Since a\b and a

\e, there exist integers s, t such that b = as and e — at. Then bx + cy

asx + aty — a(sx + ty) and a|

(bx + cy).

2. Prove: If a|6 and b ¥- 0, then |6| — \a\.

Since o| 6, we have 6 = ac for some c S I. Then |6| = \a\ - \c\ with

follows that |o|-|c| s: |o|^ that is, |6| - |a|.

1. Since |c| s= 1, it


3. Prove: If a|6 and b

\a, then b = a or b = -a.

Since a\b implies a y^ and 6

|a implies b ¥- 0, write b = ae and a = bd, where c, d S /. Now

a-b = {bd)(ac) = abed and, by the Cancellation Law, 1 = cd. Then by Problem 16, Chapter 4,

c = 1 or — 1 and 6 = ac = a or —a.

4. Prove: The number of positive primes is infinite.

Suppose the contrary, i.e., suppose there are exactly n positive primes Pi,P2'P3> •••.Pn> written

in order of magnitude. Now form a = Pi • P2 ' Pa '• • • " Pn and consider the integer a. + 1. Since

no one of the p's is a divisor of a + 1, it follows that a + 1 is either a prime > p„ or has a prime

> p„ as factor, contrary to the assumption that p„ is the largest. Thus, there is no largest positive

prime and their number is infinite.

5. Prove the Division Algorithm: For any two non-zero integers a and b, there exist

unique integers q and r such that

a = bq + r, Q^r <\b\

Define S = {a-bx: x & I). If 6 < 0, i.e., 6 ^ -1, then b-\a\^ -\a\^ a and o - 6 • |a| ^ 0.

If 6 > 0, i.e., 6^1, then 6 • (- |a|) ^-\a\â and a- 6(- la|) ^ 0. Thus, S contains non-negative

integers; denote by r the smallest of these (r ^ 0) and suppose r = a-bq. Now if r ^ |6|, then

r-l&l s= and r-|6t = a-bq-\b\ = a - (9 + 1)6 < r or a-(q-\)b < r, contrary to the

choice of r as the smallest non-negative integer S S. Hence, a < \b\.

Suppose we should find another pair q' and r' such that

a = bq' + r', ^ r' < \b\

Now bq' + r' = bq + r or b(q' - q) = r-r' implies b\(r-r') and, since \r-r'\ < \b\, then

r-r' = 0; also g' - g = since b ¥- 0. Thus, r' = r, q' = q, and q and r are unique.

6. Find (389,167) and express it in the form 389m + 167n.

From We find

389 = 167'2 + 55 1 = 55 - 2 '27

167 = 55-3 + 2 = 55-82 - 167-27

55 = 2-27 + 1 = 389-82 - 167-191

2 = 1-2

Thus, (389,167) = 1 = 82 • 389 + (-191)(167).

7. Prove: If c\ab and if (a, c) = 1, then c\b.

From 1 = mo + ne, we have b = mab + neb. Since c is a divisor of mab + neb, it is a divisor

of b and c|6 as required.

8. Prove: If (a,s) = {b,s) = 1, then {ab,s) = 1.

Suppose the contrary, i.e., suppose (06, s) = d > 1 and let d = (ab,s) = mab + ns. Now

dI

a6 and dI

s. Since (a, s) = 1, it follows that d](a; hence, by Problem 7, d!

ft. But this contradicts

(6,s) = 1; thus, (ab,s) = 1.

9. Prove: If p is a prime and if p \ ab, where a,b Gl, then p\a or p |

b.

If p I a we have the theorem. Suppose then that p / a. By definition, the only divisors of p are

±1 and ±p; then {p,a) = 1 = mp + na for some w,nG7 by Theorem II. Now 6 = mpb + nab

and, since p |

{mpb + nab), p |5 as required.


10. Prove: Every integer a > 1 has a unique factorization (except for order)

a = Pi'Pz'Pa' . . .• Pn

into a product of positive primes.

If a is a prime, the representation in accordance with the theorem is immediate. Suppose thenthat a is composite and consider the set S = {x : a; > 1, x

|a}. The least element s of S has no

positive factors except 1 and s; hence s is a prime, say p,, and

« = Pi • 6i, 6i > 1

Now either 6, is a prime, say p^, and a = Pi • Pa or 6i, being composite, has a prime factor P2 and

a - Pi' P2' ^2' 62 > 1

A repetition of the argument leads to a = Pi P2 * Ps or

o = p, • Pa • Pa • 63, 63 > 1and so on.

Now the elements of the set B = {6,, 62. *3. • •• } have the property bj > 62 > ^s > • • ; hence,B has a least element, say b„, which is a prime p„ and

a = pi • P2 • Ps • . . .• p„

as required.

To prove uniqueness, suppose we have two representations

1 = Pi • P2 • Ps •• • - • Pn = 9l • 92 • 93 • • • • • 9m

Now Qi is a divisor of Pi • P2 • Pa '- • • * Pn! hence, by Theorem V, q, is a divisor of some one of the

p's, say pj. Then 9, = pj, since both are positive primes, and by M4 of Chapter IV,

P2 • Ps •• • • • Pn = 92 • 93 •

• •• • 9m

After repeating the argument a sufficient number of times, we find that m = n and the factorizationis unique.

11. Find the least positive integers modulo 5 to which 19, 288, 19 • 288 and 193 . 288^ arecongruent.

We find

19 = 5-3 + 4; hence 19 = 4(mod 5).

288 = 5 • 57 + 3; hence 288 = 3(mod 5).

19-288 = 5(- • •) + 12; hence 19 - 288 = 2(mod 5).

193 - 2882 = 5(- • •) + 43 - 32 = 5(- • •) + 576; hence 193 . 2882 = l(mod 5).

12. Prove: Let (c, m) = d and write m = rriid. If ca = c6(mod m), then a = 6(mod mi)and conversely.

mWrite e = Cid so that (c^.ttii) = 1. If m\e(a-b), that is, if mjd

| Cid(a- 6), theni\ci{a-b) and, since (c„m,) = 1, mi\ia-b} and a = 6(mod Wi).

For the converse, suppose a = 6(mod m,). Since Wi|(a - 6), it follows that mi |

c^â - b) andWid

I

Cjd(a - 6). Thus, m\c{a— b) and ca = c6(mod m).

13. Show that, when a,b,p > G /, (a + b)" = a" + 6''(mod p).

By the binomial theorem, (o + 6)p - op + p( • • • ) + 6» and the theorem is immediate.

14. Find the least positive incongruent solutions of:

(a) 13a; = 9(mod 25) (c) 259x = 5(mod 11) (e) 222a: = 12{mod 18)

(6) 207a; = 6(mod 18) (d) 7a; s 5(mod 256)

(o) Since (13, 25) = 1, the congruence has, by Theorem XI, a single incongruent solution.


Solution I. If xi is the solution, then it is clear that a;, is an integer whose unit's digit is

either 3 or 8; thus Xi G {3,8,13,18,23}. Testing each of these in turn, we find Xi = 18.

Solution II. By the greatest common divisor process we find (13, 25) = 1 = —1 • 25 + 2 • 13.

Then 9 - - 9 • 25 + 18 • 13 and 18 is the required solution.

(6) Since 207 = 18 • 11 + 9, 207 = 9(mod 18), 207a; = 9a;(mod 18) and, by transitivity, the given

congruence is equivalent to 9a; = 6(mod 18). By Theorem IX this congruence may be reduced to

3a; = 2(mod 6). Now (3, 6) = 3 and 3/2; hence, there is no solution.

(c) Since 259 = 11 • 23 + 6, 259 = 6(mod 11) and the given congruence is equivalent to 6x = 5(mod 11).

This congruence has a single incongruent solution which by inspection is found to be 10.

(d) Using the greatest common divisor process, we find (256, 7) = 1 = 2 • 256 + 7(-73); thus,

5 = 10 • 256 + 7(-365). Now -365 = 147(mod 256) and the required solution is 147.

(e) Since 222 = 18 • 12 + 6, the given congruence is equivalent to 6a; = 12(mod 18). Since

(6, 18) = 6 and 6 1 12, there are exactly 6 incongruent solutions. As shown in the proof of

Theorem XI, these 6 solutions aro the first 6 positive integers in the set of all solutions of

X = 2(mod 3), that is, the first 6 positive integers in [2] G //(mod 3). They are then

2,5,8,11,14,17.

15. Write 141 and 152 with base 4. Form their sum and product, and check each result.

141 — 43.2 + 42-0 + 4*3 + 1; the representation is 2031.

162 = 43 • 2 + 42 • 1 + 4 • 2 + 0; the representation is 2120.

Sum.

1 + = 1; 3 + 2 = 11, we write 1 and carry 1; 1 + 1 + = 2; 2 + 2 = 10

Thus, the sum is 10211, base 4, and 293, base 10.

Product.

Multiply by 0:0000

Multiply by 2: 2 • 1 = 2; 2 • 3 = 12, write 2 and carry 1; etc. 10122

Multiply by 1:2031

Multiply by 2: 1^122

11032320

The product is 11032320, base 4, and 21432, base 10.


16. Show that the relation ( |) is reflexive and transitive but not symmetric.

17. Prove: If a\b, then -a|6, a\-b, and -a\-b.

18. List all the positive primes (a) < 50, (b) < 200.

Ans. (o) 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

19. Prove: If a = b • q + r, where a,b,q,r e I, then any common divisor of a and 6 also divides r

while any common divisor of b and r also divides a.

20. Find the greatest common divisor of each pair of integers and express it in the form of Theorem II:

(a) 237, 81 Ans. 3 = 13 • 237 + (-38) • 81

(b) 616, 427 Ans. 7 = -9 • 616 + 13 • 427

(c) 936, 666 Ans. 18 = 5 • 936 + (-7) • 666

(d) 1137, 419 Ans. 1 = 206 • 1137 + (-559) • 419

21. Prove: If s # 0, then (sa, sb) = \s\ • (a, b).


22. Prove: (a) If a [a, b\s and (o, 6) = 1, then ab|s.

(b) If m = dm, and if m\am^, then d

\a.

23. Prove: If p, a prime, is a divisor of a • 5 • c, then p |a or p |

6 or p 1c.

24. The integer e = [a, i>] is called the least common multiple oi the positive integers o and h when(1) o

I

e and 6|e, (2) if a\x and 6

|x then e

|a;.

25. Find: (a) [3.7], (6) [3,12], (c) [22,715]. Am«. (a) 21, (6) 12, (c) 1430

26. (a) Write the integers a - 19,500 and 6 = 54,450 as products of positive primes.

(6) Find d = (a,b) and e = {a,h\.

(c) Verify d • e = a • 6

(d) Prove the relation in (c) when a and 6 are any positive integers.

Ans. (b) 2'3'52; 22 • 3^ • 5^ • 112 • 13

27. Prove: If m > 1, m K a, m jfb, then jnj(a-5) implies a- mq^ = r = b-mq^, < r < m, andconversely.

28. Find all solutions of:

(a) 4x = 3(mod 7) (e) 153a; = 6(mod 12)

(&) 9x = ll(mod 26) (/) a; + 1 = 3(mod 7)

(c) 3a; + 1 s 4(mod 5) {g) 8x = 6(mod 422)

(d) 8x = 6(mod 14) {h) 363a; = 345(mod 624)

Ans. (a) [6], (6) [7], (c) [1], (d) [6], [13], (e) [2], [6], [10], (/) [2], (g) [159], [370], {h) [123], [331], [539]

29. Prove: Theorems V, VI, VII, VIII.

30. Prove: If a = 6(mod w) and c = 6(mod m), then a = c(mod m). See Examples 10(o), (6), (c).Page 52.

31. (a) Prove: If a + x = 6 + a;(mod m), then a = 6(mod w).

(6) Give a single numerical example to disprove: If ax = bx(mod m), then ax = 6(mod m).(c) Modify the false statement in (6) to obtain a true one.

32. (a) Interpret a = 6(mod 0).

(6) Show that every a; G / is a solution of ax = 6(mod 1).

33. (o) Construct addition and multiplication tables for //(5).

(6) Use the multiplication table to obtain 32 = 4(mod 5), 3* = l(mod 5), 38 = l(mod 5).

(c) Obtain 3256 = i(mod 6), 35" = 4(mod 5), 3i«24 = x(^^ 5)

34. Construct addition and multiplication tables for 7/(2), 7/(6), 7/(7), 7/(9).

35. Prove: If [s] e 7/(w) and if 0,6 G[s], then a = 6(mod m).

36. Prove: If [s], \t\ G 7/(m) and if o G [s] and 6 G [«], then a = 6(mod m) if and only if [s] = \t\.

37. Express 212 using in turn the base (a) 2, (6) 3, (c) 4, (d) 7, and (e) 9.

Ans. (a) 11010100, (6) 21212, (c) 3110, (d) 422, (e) 255

38. Express 89 and 111 with various bases, form the sum and product, and check.

39. Prove the first part of the Unique Factorization Theorem using the induction principle stated inProblem 27, Chapter 3, Page 37.

chapter 6

The Rational Numbers

THE RATIONAL NUMBERSThe system of integers has an obvious defect in that, given integers m 7^ and s, the

equation mx — s may or may not have a solution. For example, Zx — & has the solution

x = 2 but 4a; = 6 has no solution. This defect is remedied by adjoining to the integers

additional numbers (common fractions) to form the system Q of rational numbers. The

construction here is, in the main, that used in Chapter 4.

We begin with the set of ordered pairs

K = /x(/-{0)) = {(s,ot): sG/, toG/-{0}}

and define the binary relation ~ on all (s, m), (t, n) G K by

{s,m) ~ {t,n) if and only if sn = mt

(Note carefully that may appear as first component but never as second component in

any (s,m).)

Now ~ is an equivalence relation (prove it) and thus partitions K into a set of

equivalence classes ^ , ^ , r^ i

S = {[s,m],[t,n], ...}

where [s, m] = { (a, b) : (a, b) G K, (a, b) ~ (s, m) }

We shall call the equivalence classes of S rational numbers and in the following sections

indicate that S is isomorphic to the system Q as we know it.

ADDITION AND MULTIPLICATION

Addition and Multiplication on S will be defined respectively by

(i) [s,m] + [t,n] = [{sn + mt),mn]

and(ii) [s, m] • [t, n] = [st, mn]

These operations, being defined in terms of well-defined operations on integers, are

(see Problem 1) themselves well-defined.

We now define two special rational numbers

zero: [0,m] <-» one: [m,m] «^ 1

and the inverses

(additive): -[s,m] = [-s,m] for each [s,m]e^

(multiplicative): [s,m]-' = [m,s] for each [s, m] e ^ when s v^ 0.

By paralleling the procedures in Chapter 4, it is easily shown that addition and

multiplication obey the laws Ai-Ae, M1-M5, D1-D2 as stated for integers.

A property of S' l>ut not of /, is:

Mb: For every x -r^ G ^ there exists a multiplicative inverse x'^ G S such

that X'X~^ = x'^'x = 1.

60

CHAP. 6] THE RATIONAL NUMBERS 61

By Theorem IV, Chapter 2, the inverses defined in Me are unique.


Theorem I. If x and y are non-zero elements of ^ then {x-y)~^ = y-^'x~^.

SUBTRACTION AND DIVISION

Subtraction and division are defined on ^ by

(iii) x-y = X + {-y) for all x,y G ând

(iv) x-i-y = x-y-^ for all x G ^, 2/ ^ E ^respectively.

These operations are neither associative nor commutative (prove this). However, ason I, multiplication is distributive with respect to subtraction.

REPLACEMENTThe mapping [t,l]& (j ^ tGl

is an isomorphism of a certain subset of ^ onto the set of integers. We may then, when-ever more convenient, replace the subset ^* = {[t,l] : [t,l] e ^} by /. To complete theidentification of ^ with Q, we have merely to replace

and, in particular, [s, m] by s/m. * ^ V ^'V

ORDER RELATIONSAn element xGQ, i.e., x -^ [s, m] G (j, is called positive if and only if s • m > 0.

The subset of all positive elements of Q will be denoted by Q+ and the correspondingsubset of ^ by ^+. Similarly, [s,m] is called negative if and only if s-m<0. Thesubset of all negative elements of Q will be denoted by Q" and the corresponding subsetof ^ by ^ . Since, by the Trichotomy Law of Chapter 4, either s-m>0, s-m<0 orS'tn = 0, it follows that each element of ^ is either positive, negative or zero.

The order relations < and > on Q are defined as follows:

For each x,y G Q,X <y if and only if x — y < Q

x>y if and only if x - y >

These relations are transitive but neither reflexive nor symmetric.

Q also satisfies

The Trichotomy Law: If x,y G Q, one and only one of

holds.(«) * = 2/ (b) x<y {c) x>y

REDUCTION TO LOWEST TERMSConsider any arbitrary [s, m] G S with s ^ 0. Let the (positive) greatest common

divisor of s and to be d and write s = dsu to = dmu Since (s, to) ~ (si, toi), it follows that[s,m] == [s,,TOi], i.e., s/m = sjmi. Thus, any rational number ¥- can be writtenuniquely in the form alb where a and b are relatively prime integers. Whenever s/m hasbeen replaced by a/b, we shall say that s/m has been reduced to lowest terms. Hereafter,any arbitrary rational number introduced in any discussion is to be assumed reduced tolowest terms.

62 THE RATIONAL NUMBERS fCHAP. 6

In Problem 3 we prove:

Theorem II. If x and y are positive rationals with x <y, then 1/x > \ly.

In Problems 4 and 5, we prove:

The Density Property. If x and y, with x <y, are two rational numbers, there exists a

rational number z such that x <z <y.and

The Archimedean Property. If x and y are positive rational numbers, there exists a posi-

tive integer p such that vx > y\

DECIMAL REPRESENTATION

Consider the positive rational number a/b in which 6 > 1. Now

a = qob + ro 0^n<b

lOro ^ qib + n ^ r-i < 6

Since ro<b and, hence, qib + ri = lOro < 10b, it follows that qi < 10. If n = 0, then

r^ = Blb^ a .^ qob+^b, and f= 9o + 9i/10. We write a/b = go.gi and call qo.qi the

decimal representation of a/b. If ri ^ 0, we have

lOn = q2b + r2 ^ rg < b

in which q2<10. If r2 = 0, then n = ||6 so that r, = ^b+^.b and the decimal

representation of a/b is 90.9192; if rz^n, the decimal representation of a/b is the repeat-

ing decimal 90.91929292 ; if r2 ^ O.ri, we repeat the process.

Now the distinct remainders n, r,, r2, . . . are elements of the set {0, 1, 2, 3, . .. ,

b - 1}

of residues modulo b so that, in the extreme case, n must be identical with some one of

ro,ri,r2, ....n-i, say Vc, and the decimal representation of a/b is the repeating decimal

90.919293 .... 9b-i9c + i9c+2 . . . 9b-i9c + i9c + 2 .... 9b-i • • • •

Thus, every rational number can be expressed as either a terminating or a repeating decimal.

Example 1: (a) 5/4 = 1.25

(6) 3/8 - 0.375

(c) For 11/6, we find^^ ^ 1.6 + 5; 9o = L *-o

= 5

10 • 5 = 8 • 6 + 2; gj = 8, n = 2

10-2 = 3 • 6 + 2; 92 = 3, rg = 2 = rj

and 11/6 = 1.833333

(d) For 25/7, we find

25 = 3 • 7 + 4; 9o = 3, ro = 4

10 • 4 = 5 • 7 + 5; gj = 5, rj = 5

10-5 = 7*7 + 1; 92 = 7, r2 = 1

10 • 1 = 1 • 7 + 3; 93 = 1. »"3 = 3

10 • 3 = 4 • 7 + 2; 94 = 4, r4 = 2

10 • 2 = 2 • 7 + 6; 95 = 2, rg = 6

10-6 = 8-7 + 4; 96 = 8, rg = 4 = Tq

and 25/7 = 3.571428 571428....

CHAP. 6] THE RATIONAL NUMBERS 63

Conversely, it is clear that every terminating decimal is a rational number. Forexample, 0.17 = 17/100 and 0.175 = 175/1000 = 7/40.


Theorem III. Every repeating decimal is a rational number.

The proof makes use of two preliminary theorems:

(i) Every repeating decimal may be written as the sum of an infinite geometric pro-gression.

(ii) The sum of an infinite geometric progression whose common ratio r satisfies \r\ < 1is a finite number.

A discussion of these theorems can be found in any College Algebra book.

Solved Problems

1. Show that addition and multiplication on ^ are well-defined.

Let [a,b] = [s,m] and [e,d] = [t,n]. Then (a.b) ~ (s,m) and (c,d) -~ (t,n), so that am = bsand en = dt. Now

[a, ft] + [c, d] = [(ad + be), bd] = [(ad + be)mn, bd • mn]

= [(am • dn + en' bm), bd • mn]

— [(bs 'dn + dt' bm), bd • mn]

— [bd(sn + tm,), bd • mn]

= [sn+tm,mn] = [s,m] + [t,n]and addition is well-defined.

Also [a, b] • [c, d] = [ae, bd] = [ae • mn, bd • mn]

= [am ' en, bd • mn] = [6g • dt, bd • mn]

= [bd ' St, bd ' mn] = [st, mn]

= [s,m] ' [t,n]

and multiplication is well-defined.

1.^-12. Prove: If x,y are non-zero rational numbers then (x-y)-^ = y~^'x

Let X ^ [s,m] and y <^ [t,n], so that a;-i <^ [m,s] and j/-» «-> [n,t]. Then x-y ^ [s,m] • [t,n] =[st,mn] and (x'y)'^ <-> [mn,st] = [n,t]'[m,s] <-» y~^ • x^^.

3. Prove: If x and y are positive rationals with x <y, then 1/x > 1/y.

Let X *^ [s,m] and y *-* [t,n]; then sm > 0, tn > 0, and sn < mt. Now, for 1/x = x'^ ** [m,s]and l/y *^ [n,t], the inequality mt > sn implies \lx>\ly, as required.

4. Prove: If x and y, with x<y, are two rational numbers, there exists a rationalnumber z such that x <z <y.

Since x < y, we have

2x = x + x < x + y and x + y < y + y - 2y

Then 2x < x + y < 2y

and, multiplying by ^, x < ^(x + y) < y. Thus, i(x + y) meets the requirement for z.

64 THE RATIONAL NUMBERS [CHAP. 6

5. Prove: If x and y are positive rational numbers, there exists a positive integer p

such that px> y. .. j ,

Let x<^[s,m] and y *^ [t,n\, where s, m, t, n are positive integers. Now px > y if and only

if psn > mt. Since sn ^ 1 and 2sm > 1, the inequality is certainly satisfied if we take p - 2mt.

6. Prove: Every repeating decimal represents a rational number.

Consider the repeating decimal

x.yzdefdef.... = x.yz + O.OOdef + O.OOOOOde/ +

Now x.yz is a rational fraction since it is a terminating decimal, while O.OOde/ + O.OOOOOde/ + • ••

is an infinite geometric progression with first term a = O.OOdef, common ratio r = 0.001, and sum

c - _?_ = M2M = _M^, a rational fraction^ ~' l-r 0.999 99900

'

Thus the repeating decimal, being the sum of two rational numbers, is a rational number.

7. Express (a) H with base 4, (b) i with base 5.

(a) 27/32 = 3(1) + 3/32 =3(i) + Ki)^ + 1/32 = 3(i) + 1(1)^ + 2(1)'

The required representation is 0.812.

(6) 1/3 = l(i) + T% = Ki) + 3(i)2 + 1/75

= l(i) +W + l(i)^ + 2/375

= Ki) + 3(i)='+ Ki)' + 3(i)* + i/is'^s

The required representation is 0.131313


8. Verify: (a) [s,m] + [0,n] = [s,m] (c) [s,m] + [s.m] = [O.n] = [s,m] + [s,-m]

(b) [s,m]'[0,n] = [0,n\ {d) [s.m] ' [m.,s] = [n,n]

9. Restate the laws A,-Ae, M,-M„ Di-D^ of Chapter 4 for rational numbers and prove them.

10. Prove: (a) ^+ is closed with respect to addition and multiplication.

(6) If [s,m] e ^ +, so also does [s,m\-^.

11. Prove: (o) ^- is closed with respect to addition but not with respect to multiplication.

(6) If [s,m] e ^-, so also does [s,w]-».

12. Prove: If x,y e Q and x-y = 0, then a; = or 3/ = 0.

13. Prove: If x,y G Q, then (a) -(ac + y) = -x-y and (6) -(-«) = x.

14. Prove: The Trichotomy Law.

15. If x.y.z G Q, prove:

(a) « + 2 < J/ + 2 if and ""^y i* « < 2/;

(6) when z > 0, xz < yz if and only if x < y;

(c) when z < 0, xz < j/2 if and only if x > y.

16. If w.x.J/.z G Q with xz¥'0 in (a) and (b), and xj/z t^ in (c), prove:

(a) (w -^ x) ± (!/-^ z) = (wz ± xy) -^ xz

(6) (w -^ x) • (j/ H- z) = wj/ -^ xz

(c) (w^x) ^ (y^z) = wz -i- xy

17. Prove: If a,bGQ+ and a < b, then a^ < a5 < b^. What is the corresponding inequality if

a,beQ-2

chapter 7

The Real NumbersINTRODUCTION

Chapters 4 and 6 began with the observation that the system A' of numbers previouslystudied had an obvious defect. This defect was remedied by enlarging the system X. Indoing so, we defined on the set of ordered pairs of elements of X an equivalence relation,etc., etc. In this way we developed from N the systems / and Q satisfying N c I C Q.For what is to follow, it is important to realize that each of the new systems / and Q hasa single simple characteristic, namely,

/ is the smallest set in which, for arbitrary m,s&N, the equation m + x - salways has a solution.

Q is the smallest set in which, for arbitrary integers m^Q and s, the equationmx = s always has a solution.

Now the situation here is not that the system Q has a single defect; rather, it is thereare many defects and these so diverse that the procedure of the previous chapters will notremedy all of them. We mention only two:

(1) The equation x^ = S has no solution in Q. For, suppose the contrary, and assumethat the rational a/b, reduced to lowest terms, be such that {a/bf = 3. Sincea^ := 3b\ it follows that 3

|

a^ and, by Theorem V, Chapter 5, that 3|a. Write a = 3a,-

then 3af = b^ so that 3i

b' and, hence, 3 1 b. But this contradicts the assumptionthat a/b was expressed in lowest terms.

(2) The circumference c of a circle of diameter d G Q is not an element of Q, i.e., inc = 7rd, n^Q. Moreover, n^ ^ Q so that r is not a solution of x^ = q for any q EQ.(In fact, TT satisfies no equation of the form ax" + bx"'^ + + sx + t = witha,b, .. .,s,teQ.)

The method, introduced in the next section, of extending the rational numbers to thereal numbers is due to the German mathematician R. Dedekind. In an effort to motivatethe definition of the basic concept of a Dedekind cut or, more simply, a cut of the rationalnumbers, we shall first discuss it in non-algebraic terms.

Consider the rational scale of Fig. 7-1, that is,

a line L m which the non-zero elements of Q areattached to points at proper (scaled) distances fromthe origin which bears the label 0. For convenience,call each point of L to which a rational number is

attached a rational point. (Not every point of Lis a rational point. For, let P be an intersection ofthe circle with center at and radius 2 units withthe line parallel to and 1 unit above L. Then let theperpendicular to L through P meet L in T; by (1)

above, T is not a rational point.) Suppose the line Lto be cut into two pieces at some one of its points.There are two possibilities:

— L

Fig. 7-1

65

66 THE REAL NUMBERS [CHAP. 7

(a) The point at which L is cut is not a rational point. Then every rational point of L

is on one but not both of the pieces.

(6) The point at which L is cut is a rational point. Then, with the exception of this

point, every other rational point is on one but not both of the pieces. Let us agree

to place the exceptional point always on the right-hand piece.

In either case, then, the effect of cutting L at one of its points is to determine two

non-empty proper subsets of Q. Since these subsets are disjoint while their union is Q,

either defines the other and we may limit our attention to the left-hand subset. This

left-hand subset is called a cut and we now proceed to define it algebraically, that is,

without reference to any line.

DEDEKIND CUTS

By a cut C in Q we shall mean a non-empty proper subset of Q having the additional

properties:

(i) if c G C and a G Q with a<c, then a G C.

(ii) for every c&C there exists h &C such that b > c.

The gist of these properties is that a cut has neither a least (first) element nor a

greatest (last) element. There is, however, a sharp difference in the reasons for this

state of affairs. A cut C has no least element because, if c G C, every rational number

a < c is an element of C. On the other hand, while there always exist elements of C

which are greater than any selected element c G C, there also exist rational numbers

greater than c which do not belong to C, i.e., are greater than every element of C.

Example 1: Let r be an arbitrary rational number. Show that C(r) = {a : a e Q, a < r} is a cut.

Since Q has neither a first nor last element, it follows that there exists r^^Q

such that ri < r (thus, C(r) ^ 0) and rj G Q such that r.2,> r (thus, Or) ¥^ Q).

Hence, C(r) is a non-empty proper subset of Q. Let c G C(r), that is, c < r. Now

for any a G Q such that a < c, we have a < c < r; thus, a G C{r) as required in (i).

By the Density Property of Q there exists d G Q such that c < d < r; then d> c

and d G C(r), as required in (ii). Thus, C(r) is a cut.

The cut defined in Example 1 will be called a rational cut or, more precisely, the cut

at the rational number r. For an example of a non-rational cut, see Problem 1.

When C is a cut, we shall denote by C the complement of C in Q. For example, if

C = C{r) of Example 1, then C = C'(r) = {o' : a'GQ.a'^ r}. Thus, the complement

of a rational cut is a proper subset of Q having a least but no greatest element. Clearly,

the complement of the non-rational cut of Problem 1 has no greatest element; m Problem 2,

we show that it has no least element.


Theorem I. If C is a cut and r G Q, then

(a) Z) - {r + a: a GC) is acut and (b) D' = {r + a':

a' G C'}

It is now easy to prove

Theorem II. If C is a cut and r G Q+, then

(a) E = [ra: a G C} is a cut and (b) E' = {ra' : a' G C'}


Theorem III. If C is a cut and r G Q +, there exists bGC such that r + bGC.

CHAP. 7] THE REAL NUMBERS 67

POSITIVE CUTS

Denote by X the set of all cuts of the rational numbers and by '3^+ the set of all cuts(called positive cuts) which contain one or more elements of Q^. Let the remaining cuts

in -^ be partitioned into the cut 0, i.e., = C(0) = {a: o G Q}, and the set %- of all

cuts containing some but not all of the elements of Q". For example, C{2) GX* whileC(-5) G X~- We shall, for the present, restrict our attention solely to the cuts of X'*' forwhich it is easy to prove

Theorem IV. If CGX^ and r>lGO +, there exists cGC such that re G C.

Each C GX+ consists of all elements of Q~, 0, and see (Problem 5) an infinitude of

elements of Q+. For each C GX+, define C = (a: a EC, a>0} and denote the set of

all C's by Ji. For example, if C = C(3) then C(3) = {a : a G Q, < a < 3} and C may bewritten as C = C(3) = Q- U {0} U C(3). Note that each CGX^ defines a unique C ^ Jiand, conversely, each C ^ ^ defines a unique C G '^+. Let us agree on the convention:when Ci G '^+, then d ^ Q' U {0} U d-

We define addition (+) and multiplication (•) on X* as follows:

C1 + C2 = Q- u {0} u (C, + C2)

for each Cud GX^Cx'd - Q-U{0}U(C,-C2)

where (i)

and

{Cx + C2 = K + c,: c^GC,, c^&C)

ICi'G = [cr<^2- C,GC„ c^GC,}

It is easy to see that both Ci + d and Ci'C2 are elements oi X^- Moreover, since

C1 + C2 = {a: aGCi + d, a>0}

Ci'Ci == {a: aGCi'Ci, a>0)

it follows that J( is closed under both addition and multiplication as defined in (i).

Example 2: Verify: (a) C(3) + C(7) = C(10), (6) C(3) C(7) = C(21).

Denote by C(3) and C(^) respectively the subsets of all positive rational numbersof C(3) and C(7). We need then only verify

C(3) + C(7) = C(10) and C(3) • C(7) = a21)

(a) Let c, G C(3) and eg e C(7). Since < Ci < 3 and < C2 < 7, we have< c, + C2 < 10. Then Ci + eê C(10) and C(3) + C(7) c C(10). Next, suppose

C3 G C(10). Then, since < Cj < 10,

< JqC3 < 3 and < ^Cg < 7

that is, YJC3 G C(3) and ^Cg G C(7). Now "s = 3^63 + ^Cg; hence,

C(10) C C(3) + a?). Thus, C(3) + C(7) = aiO) as required.

(6) For cj and Cj as in (a), we have < Cj • C2 < 21. Then Ci • Cg G C(21) and

C(3) • C(7) C C(21). Now suppose C3 G C(21) so that < C3 < 21 and < ^ ==

g < 1. Write g = gi • 92 with < g, < 1 and < g2 < 1. Then C3 = 21g =(3g,)(7g2) with < 3gi < 3 and < 7g2 < 7, i.e., 3gi G C(3) and 7g2 G C(7).Then C(21) C C(3) • C(7) and C(3) • C(7) = C(21) as required.

The laws A1-A4, Mi-M4, D1-D2 in -^^ follow immediately from the definitions of additionand multiplication in -^^ and the fact that these laws hold in Q+ and, hence, in J{.

Moreover, it is easily shown that C(l) is the multiplicative identity, so that Ms also holds.


MULTIPLICATIVE INVERSES

Consider now an arbitrary C = Q^ U {0} U C ^X'^ and define

C"* = {b: bGQ^, b < a^' for some a G C'}


If C = Q- U {0} U C then C = Q- U {0} U C"' is a positive cut.


For each C Gf^*, its multiplicative inverse is C~^ Gf^^.

At this point we may move in either of two ways to expand %^ into a system in which

each element has an additive inverse:

(1) Repeat Chapter 4 using X^ instead of N.

(2) Identify each element of %' as the additive inverse of a unique element ofK^.We shall proceed in this fashion.

ADDITIVE INVERSES

The definition of the sum of two positive cuts is equivalent to

C1 + C2 = {C1 + C2: CiGCi, C2GC2). CuC2GX^

We now extend the definition to embrace all cuts by

C1 + C2 = {c, + C2 : ci G Ci, C2 G C2), Ci, C2GX (1)

Example 3: Verify C(3) + C(-7) = C(-4).

Let Cj + C2 e C(3) + C(-7), where Cj G C(3) and 02 e C(-7). Since Ci < 3

and C2 < -7, it follows that Ci + e2< -4 so that Cj + C2 G C(-4). Thus,

C(3) + C(-7) C C(-4).

Conversely, let Cg G C(-4). Then Cg < -4 and -4-C3=:dGQ + . NowC3 = -4 - d = (3 - ^d) + (-7 - \d)\ then since 3 - ^d G C(3) and -7 - ^d G C(-7),

it follows that C3 G C(3) + C(-7) and C(-4) c C(3) + C(-7). Thus, C(3) + C(-7) =C(—4) as required.

Now, for each C G%, define

—C — {x : x G Q, a; < —a for some a G C'}

For C = C(-3), we have -C = {x: xGQ, a; < 3} since -3 is the least element of CBut this is precisely C(3); hence, in general,

-C(r) = C{-r) when rGQ

In Problem 8 we show that -C is truly a cut, and in Problem 9 we show that -C is

the additive inverse of C. Now the laws A1-A4 hold in %.

In Problem 10 we prove

The Trichotomy Law. For any C GX, one and only one of

C = C(0) CGX^ -CGX*holds.

MULTIPLICATION ON XFor all C GX, we define

C > C(0) if and only if CGX^C<C(0) if and only if -C GX*


and \C\ = C when C ^ C(0)

\C\ = -C when C < C(0)

Thus, |C| ^ C(0), that is, \C\ = C(0) or \C\ GX^.

For all Ci,d e '^, we define

C, • C2 = C(0) when Ci = C(0) or d = C(0) (^)

Ci • C2 = |C,| • IC2I when Cj > C(0) and C2 > C(0) or when Ci < C(0) and C2 < C(0)

C, • C2 = -(|C,| • IC2I) when Ci > C(0) and C2 < C(0) or when Ci < C(0) and C2 > C(0)

Finally, for all C ^ C(0), we define

C-i = |C|-' whenC>C(0) and C"* = -(|q-i) when C < C(0)

It now follows easily that the laws Ai-Ae, Mi-Me, D1-D2 (see Page 71) hold in %.

SUBTRACTION AND DIVISIONParalleling the corresponding section in Chapter 6, we define for all Ci, CiG%

C1-C2 = Ci + (-C2) (3)and, when C2 ^ C(0),

Note. We now find ourselves in the uncomfortable position of having two completelydifferent meanings for Ci - C2. Throughout this chapter, then, we shall agree to considerCi — C2 and Ci n C2 as having different meanings.

ORDER RELATIONSFor any two distinct cuts Ci, C2 e %, we define

Ci < C2, also C2 > Ci, to mean C1-C2 < C(0)

In Problem 11, we show

Ci < C2, also C2 > Ci, if and only if Ci c C2

There follows easily

The Trichotomy Law. For any Ci, C2 G "K, one and only one of the following holds:

(a) C, = C2 (b) Ci < C2 (c) Ci > C2

PROPERTIES OF THE REAL NUMBERSDefine X* = {C{r) : C{r) GX, rGQ}. We leave for the reader to prove

Theorem V. The mapping C{r) G%* ^ r GQ is an isomorphism of %* onto Q.

The elements of % are called real numbers and, whenever more convenient, X will bereplaced by the familiar R while A,B, . . . will denote arbitrary elements of R. NowQ C R; the elements of the complement of Q in i2 are called irrational numbers.

In Problems 12 and 13, we prove

The Density Property. If A,B G R with A < B, there exists a rational number C(r)

such that A < C(r) < B.and

The Archimedean Property. If A,BgR +, there exists a positive integer C{n) such

that C{n) 'A>B.


In order to state below an important property of the real numbers which does not hold

for the rational numbers, we make the following definition:

Let S 7^ be a set on which an order relation < is well-defined, and let T be any

proper subset of S. An element s &S, if one exists, such that s — i for every t G T(s~t for every i G T) is called an upper hound {lower hound) of T.

Example 4: (a) U S - Q and T = {-5, -1, 0, 1,3/2}, then 3/2,2,102,... are upper bounds

of T -while -5, -17/3, -100, ... are lower bounds of T.

(b) When S = Q and T = C&X, then T has no lower bound while any «' G T' = Cis an upper bound. On the other hand, 2" has no upper bound while any

t€.T is a lower bound.

If the set of all upper bounds (lower bounds) of a subset T of a set S contains a least

element (greatest element) e, then e is called the least upper bound {greatest lower hound)

of T.

Let the universal set be Q and consider the rational cut C{r) G %. Since r is the

least element of C'{r), every s ^ r in Q is an upper bound of C{r) and every t^ r in Q is

a lower bound of C'{r). Thus, r is both the least upper bound (l.u.b.) of C{r) and the

greatest lower bound (g.l.b.) of C'{r).

Example 5: (a) The set T of Example 4(a) has 3/2 as l.u.b. and -5 as g.l.b.

(6) Let the universal set be Q. The cut C of Problem 1 has no lower bounds and,

hence, no g.l.b. Also, it has upper bounds but no l.u.b. since C has no greatest

element and C has no least element.

(c) Let the universal set be R. Any cut C G •^, being a subset of Q, is a subset

of R. The cut C{r) then has upper bounds in R and r e B as l.u.b. Also, the

cut C of Problem 1 has upper bounds in R and yS G fi as l.u.b.

Example 5(c) illustrates

Theorem VI. If ^S is a non-empty subset of % and if S has an upper bound in %, it has

a l.u.b. in X- ^^j. ^ p^^^f^ ggg Problem 14.

Similarly, we have

Theorem VI'. If ^ is a non-empty subset of "K and if S has a lower bound in %, it has a

g.l.b. in X.

Thus, the set R of real numbers has the

Completeness Property. Every non-empty subset of R having a lower bound (upper

bound) has a greatest lower bound (least upper bound).

Suppose e" = a, where a,9GR* and n G /+. We call 6 the principal nth root of «

and write e = a'^". Then for r = m/n G Q, there follows a" - 0"".

Other properties of R are:

(1) For each a G R+ and each n G 7+, there exists a unique G 72+ such that i9" = a.

(2) For real numbers « > 1 and p, define «» as the l.u.b. of {a*-: rGQ, r<p}. Then «»

is defined for all « > 0, /? G /J by setting «» = (l/a)~^ when < « < 1.


SUMMARYAs a partial summary to date, there follows a listing of the basic properties of the

system R of all real numbers. At the right, in parentheses, is indicated other systemsN,I,Q for which each property holds.

Addition

r + s& R, for all r,s&R (iV, /, Q)

r-\-s = s + r, forallr.sGft (N,I,Q)

r + (8 + t) = {r + s) + t, ior all r,s,t e R (N,I,Q)

If r + t-s + t, then r = s for all r.s.teK. {N,I,Q)

There exists a unique additive identity element e i2

such that r + = + r = r, for every r S R. (/, Q)

For each rSR, there exists a unique additive inverse-rGR such that r + (-r) = (-r) + r = 0. (/, Q)

A,.

A2.

A3.

A4.

A5.

Closure Law

Commutative Law

Associative Law

Cancellation Law

Additive Identity

Ag. Additive Inverses

Multiplication

Mj. Closure Law

M2. Commutative Law

M3. Associative Law

M4. Cancellation Law

M.v

M„

Multiplicative Identity

Multiplicative Inverses

Distributive Laws

D2.

Density Property

Archimedean Property

Completeness Property

r's&R, for all r.s e Rr's-s'r, forallr.sGfi

r-{s't) = {r-s)-t, for all r,s,t G RIf m-p = n-p, then m = n, for all m,n G R andp¥'OGR.

There exists a unique multiplicative identity element1 G iZ such that 1 • r = r • 1 = r for every r G R.

For each r ^ G fi, there exists a unique multipli-cative inverse r~^GR such that r'r~'=r~*'r = l.

For every r, s, t G fi,

r«(s + t) = r'8 + V t

{s + t)-r = s"r + t»r

For each r,s G R, with r < s, there exists t G Qsuch that r < t < s.

For each r,s G ft + , with r < s, there exists n G /+such that M • r > 8.

Every non-empty subset of R having a lower bound(upper bound) has a greatest lower bound (least

upper bound).

{N,I,Q)

(N.I.Q)

(N.I.Q)

(N.I,Q)

(N,I,Q)

(Q)

(N,I,Q)

(Q)

(Q)

(N,I)

Solved Problems

Show that the set S consisting of Q-, zero, and all sGQ+ such that s^ < 3 is a cut.

First, S is a proper subset of Q since 1 G S and 2 « S. Next, let c G S and a G Q with a < c.Clearly, aS S when a =s and also when e ^ 0.

For the remaining case (0 < a < c), a^ < ac < c^ < 3; then a^ < 3 and a G S as required in (i)Page 66. Property (ii). Page 66, is satisfied by 6 = 1 when c ^ 0; then S will be a cut providedthat for each c> with c2 < 3 an w G Q+ can always be found such that (c + m)2 < 3. We cansimplify matters somewhat by noting that if p/g, where p,gGI+, should be such an m so also is

1/q. Now q

2c +11; hence ( c + = C2 + 2c 2c +1

1J 9 92 ^ q< 3-c2, that is, provided (3 - c2)g > 2c +1. Since (3-c2)G

thus (c + -j < 3 provided

and (2c + l)GQ +, the

existence of 9 G /+ satisfying the latter inequality is assured by the Archimedean Property of Q +Thus, S is a cut.


2. Show that the complement S' of the set S of Problem 1 has no smallest element.

For any rGS' = {a: aGQ +, a^ ^ 3}^ we are to show that a positive rational number m

can always be found such that (r - m)2 > 3, that is, the choice r will never be the smallest element

in S'. As in Problem 1, matters will be simplified by seeking an m of the form 1/q where qS I + .

Now (r--Y = r^-— + \ > r2-— ; hence, ( r --) > 3 whenever -f < r^-S, that

\ 9/ Q Q^ 9 \ 1/ '

is, provided {r^-S)q > 2r. As in Problem 1, the Archimedean Property of Q+ assures the existence

of qe 1+ satisfying the latter inequality. Thus S' has no smallest element.

3. Prove: If C is a cut and rGQ, then (a) D = {r + a: a GC) is a cut and (b) D' =

{r + a': a' e C'}.

(a) D # since C # 0; moreover, for any c' G C, r + c'€D and D ¥= Q. Thus I> is a proper

subset of Q.

Let b&C. For any s G Q such that s < r+b, we have s - r < 6 so that s - r G C and

then s = r + (s - r) G D as required in (i). Page 66. Also, for 6 G C there exists an element

c G C such that Ob; then r + 6, r + cGD and r + c > r+b as required in (ii). Page 66.

Thus Z? is a cut.

(6) Let 5' G C. Then r+b' € D since 6' € C; hence, r + 6' G D'. On the other hand, if

q> = r + p' S D' then p' € C since if it did we would have D n D' -^ 0. Thus D' is as

defined.

4. Prove: If C is a cut and r G Q+, there exists b gC such that r + b GC.

Prom Problem 3, D ^ {r + a: aSC} is a cut. Since r > 0, it follows that C Q D. Let q G Q

such that p = r + qeD but not in C. Then qr G C but r + q&C. Thus 9 satisfies the requirement

of 6 in the theorem.

5. Prove: If C e 'JC+, then C contains an infinitude of elements of Q + .

Since Cex+ there exists at least one r G Q+ such that rSC. Then for all qeN we have

r/q G C. Thus, no C G %+ contains only a finite number of positive rational numbers.

Prove: If C = Q" U {0} U C e ^^ then C'' = Q" U {0} U C~' is a positive cut.

Since C^Q*. it follows that C ^ 0; and since C ^ 0. it follows that C 1^ Q^.

Let d G C.

Then (d+l)-iGQ+ and (d+l)-»<d-i go that (d + D-iGC"' and C"'J^ 0- Also, if e^Cthen for every a G C we have c < a and c"' > a"'; hence, c"' « C"' and C ' ^ Q + - Thus C » is

a proper subset of Q.

Let c G C-i and r G Q+ such that r < c. Then r < c < d-' for some d G C and r G C"' as

required in (i), Page 66. Also, since c ^ d-i there exists sGQ+ such that c<s<d 1 and

8 G C"» as required in (ii). Thus C-Ms a positive cut.

7. Prove: For each C GTC , its multiplicative inverse is C ^ GTC

.

Let C = Q- U {0} U C so that C"! = Q- U {0} U C"'- Then Q-Q--^ - {c • 6 :c G C, be C->}.

Now 6 < d-i for some d G C, and so bd < 1; also, c < d so that be < 1. Thus, C'C ^ Q CW-

Let n G C(l) so that n"! > 1. By Theorem IV there exists c G C such that c-n"! G C. For

each aec such that a > c, we have n • a-> < n • c-i = (c • n-i)-«; thus, n • a^^ = e G C >. Then

nâeeC-C-^ and CWCC-C"'- Hence, C-C-' = C(1) and C.C-i = C(l). By Problem 6,

C-iG'K+.

CHAP. 7] THE REAL NUMBERS

8. When C GTî, show that -C is a cut.

73

Note first that -C ^ since C ¥- 0. Now let c S C; then -c € -C, for if it were we would have-c < -c' (for some c' G C) so that c' < e, a contradiction. Thus, —C is a proper subset of Q.Property (i), Page 66, is immediate. To prove property (ii), let x e -C, i.e., x < -c' for someC G C. Now «<!(«- c') < -c'. Thus, ^(x - c') > x and ^(a; - c') G -C.

9. Show that -C of Problem 8 is the additive inverse of C, i.e., C + (-C) = -C + C - C(0).

Let c + x e C+ (-C), where cGC and « G -C. Now if « < -c' for c' G C, we havec + a!<c-c'<0 since c<c'. Then C+(-C)cC(0). Conversely, let j/,zGC(0) with z > {/.Then, by Theorem III, there exist c G C and c' G C such that c + (z - j/) = c'. Since z - c' < -c'it follows that z-c' & -C. Thus, j/ = c + (z-c') G C + (-C) and C(0)cC+(-C). Hence,C + (-C) = -C + C = C(0).

10. Prove the Trichotomy Law: For C G.%, one and only one of

is true.^^^(') ^^^^ -^^^^

Clearly, neither C(0) nor -C(0) G •^+. Suppose now that C -^ C(0) and C g •J^+. Since everycGC is a negative rational number but C^Q", there exists c' G Q- such that c' G C. Sincec'<|c'<0, it follows that 0<-^c'<-c'. Then -^' G -C and so -Ce'!^+. On the contrary,if CG'^+, every c' G C is also G Q + . Then every element of -C is negative and -Cg7S:+.

11. Prove: For any two cuts Ci, Ca G •^, we have C, < C2 if and only if Ci c C2.

Suppose Cj c C2. Choose o' G CgnC,' and then choose 6 G C2 such that 6 > a'. Since -6 < -o',it follows that -6 G -Cj. Now -C, is a cut; hence there exists an element c G -C, such thatc > -6. Then 6 + c>0and6 + cGC2+ (-C,) = Cj - Ci so that C2 - C, G •3*:+. Thus C, - C, > C(0)and Cj < C2.

For the converse, suppose Ci < C2. Then C2-Ci>C(0) and C2-Ci67C+. Choose d&Q*such that d G C2 - C, and write d = h + a where 6 G Cg and a G -Cj. Then -h < a since d > 0;also, since o G -Cj, we may choose an a' G Cj such that a < -a'. Now -6 < -o'; then o' < 6so that 6 e Ci and, thus, C2 i C,. Next, consider any x G Cj. Then x < 5 so that x G C2 and, hence,Ci c C2.

12. Prove: If A.B&R with A < B, there exists a rational number C(r) such thatA < C{r) <B.

Since A < B, there exist rational numbers r and a with s <r such that r.sGB but not in AThen A ^ C(8) < C(r) < B, as required.

13. Prove: If A,BgR+, there exists a positive integer C{n) such that C(n)'A>B.Since this is trivial for A ^ B, suppose A < B. Let r,s be positive rational numbers such that

r G A and 8 G B'; then C{r) < A and C(s) > B. By the Archimedean Property of Q, there exists apositive integer n such that rtr > s, i.e., C(it) • C{r) > C(s). Then

. , C(w)'A ^ C(n)'C(r) > C{8) > Bas required.

14. Prove: If cS is a non-empty subset of % and if ^ has an upper bound (in %), it has al.u.b. in "K.

Let S = {<î. C^2. C3, . . . } be the subset and C be an upper bound. The union 1/ = Ci U C, U Co Uof the cuts of^ is itself a cut G «:; also, since C^ Q U, C^ c U, C3 Q U E7 is an upper boundof S- But CiCC, CzCC, C3CC, ...; hence VcC and i7 is the l.u.b. of S-



15. (o) Define C(3) and C(-7). Prove that each is a cut.

(6) Define C'(3) and C'(-7).

(c) Locate -10,-5,0,1.4' as G or € each of C(3), C(-7), C'(3), C'(-7).

(d) Find 5 rational numbers in C(3) but not in C(—7).

16. Prove: C(r) C C{8) if and only if r < s.

17. Prove: If A and B are cuts, then A c B implies A ¥= B.

18. Prove: Theorem II, Page 66.

19. Prove: If C is a cut and r € Q + , then C ^ D = {a + r: a&C}.

20. Prove: Theorem IV, Page 67.

21. Let r e Q but not in Ce-K. Prove: C ^ C(r).

22. Prove: (a) C(2) + C(5) = C(7) (c) C(r) + C(0) = C(r)

(b) C(2)-C(5) = C(10) (d) C(r)-C(l) = C(r)

23. Prove: (a) If C e %+ , then -C €•*:-.

(6) If Ce-K', then -CG'^+.

24. Prove: -(-€) = C

25. Prove: (a) If C„ Cg S •«:, then C, + C2 and |Ci| • IC2I are cuts.

(6) If Ci^C(0), then |Ci|-i is a cut.

(c) (C-»)-» = C for all C # C(0).

26. Prove: (o) If €&%+, then C-ie«:+.

(6) If CG-K-, then C-» G ]<:-.

27. Prove: If r,s G Q with r < s, there exists an irrational number a such that r < a < s.

Hint. Consider a = r -\ 7=-.

\/2

28. Prove: If A and B are real numbers with A < S, there exists an irrational number a such that

A <a<B.Hint. Use Problem 12 to prove: If A and B are real numbers with A < B, there exist rational

numbers t and r such that A < C(t) < C(r) < B.

29. Prove: Theorem V, Page 69.

chapter 8

The Complex Numbers

THE SYSTEM C OF COMPLEX NUMBERSThe system C of complex numbers is the number system of ordinary algebra. It is

the smallest set in which, for example, the equation x^ = a can be solved when a is anyelement of R. In our development of the set C, we begin with the product set R x R. Thebinary relation " = " requires

(a, h) - (c, d) if and only if o = c and b = d

Now each of the resulting equivalence classes contains but a single element. Hence, weshall denote a class as (a, b) rather than as [a, b] and so, hereafter, denote RxR hy C.

ADDITION AND MULTIPLICATION ON CAddition and multiplication on C are defined respectively by

(i) {a,b) + (c,d) = (a + c, 6 + d)

(ii) (a, b) • (c, d) - {ac - bd, ad + be)

for all (a, b), (c, d) G C.

The calculations necessary to show that these operations obey A1-A4, M1-M4, D1-D2 ofChapter 7, when restated in terms of C, are routine and will be left for the reader. It iseasy to verify that (0,0) is the identity element for addition and (1,0) is the identityelement for multiplication; also, that the additive inverse of (a, 6) is -(a, b) = {-a, -b)

and the multiplicative inverse of (0,6) ^ (0,0) is (a,b)-' = (-±-^,-^±^^, Hence,

the set of complex numbers have the properties As-As and Ms-Me of Chapter 7, restatedin terms of C.

We shall show in the next section that RcC, and one might expect then that C hasall of the basic properties of R. But this is false since it is not possible to extend(redefine) the order relation

" <" of R to include all elements of C.See Problem 1.

PROPERTIES OF COMPLEX NUMBERSThe real numbers are a proper subset of the complex numbers C. For, if in (i) and

(ii) we take 6 = d = 0, we see that the first components combine exactly as do the realnumbers a and c. Thus, the mapping a <^ (a, 0) is an isomorphism of R onto a certainsubset {(a, 6): a&R, 6 = 0} of C,

The elements (a, b)&C in which b¥'Q, are called imaginary numbers and thoseimaginary numbers (a, b) in which a-Q are called pure imaginary numbers.

For each complex number 2 = (a, 6), we define the complex number 2 = (a/b) = (a, -b)to be the conjugate of z. Clearly, every real number is its own conjugate while theconjugate of each pure imaginary is its negative.

75

76 THE COMPLEX NUMBERS [CHAP. 8

There follow easily

Theorem I. The sum (product) of any complex number and its conjugate is a real number.

Theorem II. The square of every pure imaginary number is a negative real number.

See also Problem 2.

The special role of the complex number (1,0) suggests an investigation of another,

(0,1). We find . , . ^{x, y) • (0, 1) = {-y, x) for every {x, y) e C

and, in particular,, , , , .

(i/,0)-(0,l) = (0,l)-(2/,0) = (0,2/)

Moreover, (0,1)^ = (0,1) -(0,1) = (-1, 0)<^-l in the mapping above so that (0,1) is a

solution of z^ — —1.

Defining (0, 1) as the pure imaginary unit and denoting it by i, vfe have

P = -1and, for every {x, y) G C,

{x.y) = {x,0) + iO,y) = (x, 0) + (i/, 0) • (0, 1) = x + yi

In this familiar notation, x is called the real part and y is called the imaginary part of the

complex number. We summarize:

the negative of z = x + yi is -z - -{x + yi) = -x-yi

the conjugate of z = x + yi is z = x + yi-x-yi

for each z — x-it-yi, Z'Z = x^ + y^ &R

foreach 2 7^ + 0-i = 0, z"' = - = ^ -z z-z x^ + y^ x^ + y''

SUBTRACTION AND DIVISION ON C

Subtraction and division on C are defined by

(iii) z-w = z + {-w), for all z,wGC

(iv) z -i- w — z'W^, for all w ¥- 0, z&C

TRIGONOMETRIC REPRESENTATIONThe representation of a complex number z by {x, y) and hy x + yi suggests the mapping

(isomorphism)X + yi *^ {x,y)

of the set C of all complex numbers onto the points {x,y) of the real plane. We may

therefore speak of the point P(x, y) or of P{x + yi) as best suits our purpose at the moment.

The use of a single coordinate, surprisingly, often simplifies many otherwise tedious

computations. One example will be discussed below; another will be outlined briefly in

Problem 20.

Consider in Fig. 8-1 the point P{x + yi) -¥- whose

distance r from O is given by r = ^/^T^. If is the, ^^^ ^ ^.^

positive angle which OP makes with the positive'"

X-axis, we have

X = r cos 0, y = r sin e

whence z = x + yi - r(cos 9 + i sin B) ^ \ i »- x

The right member of this equality is called the trigo-

nometric form (polar form) of z. The non-negative

real number Fig. 8-1.

CHAP. 8] THE COMPLEX NUMBERS 77

r — \z\ = yJz'Z = -y/xM^

is called the modtdus {absolute value) of z, and 6 is called the angle {amplitude or argument)of «. Now ^ satisfies x = r cos 6, y — r sin 0, tan = i/Za; and any two of these

determine up to an additive multiple of 2tt. Usually we shall choose as 6 the smallest

positive angle. {Note. When P is at O, we have r = and 9 arbitrary.)

Example 1 : Express (a) 1 + i, (6) —Vs + i in trigonometric form.

(a) We have r = Vl + 1 = ^2. Since tan e = 1 and cos e = 1/V2, we take 9 to be

the first quadrant angle 45° = v/i. Thus, 1 + t = \/2(coS7r/4 + Jsin7r/4).

(b) Here r = \/3 + 1 = 2, tan » = -l/\/3 and cos » = -|a/3. Taking 9 to be the

second quadrant angle 5!r/6, we have

-VS + i = 2(cos 5j!-/6 + 1 sin 5b-/6)

It follows that two complex numbers are equal if and only if their absolute values are

equal and their angles differ by an integral multiple of 27r, i.e., are congruent modulo 2ir.


Theorem III. The absolute value of the product of two complex numbers is the productof their absolute values, and the angle of the product is the sum of their

angles.

and

Theorem IV. The absolute value of the quotient of two complex numbers is the quotient

of their absolute values, and the angle of the quotient is the angle of the

numerator minus the angle of the denominator.

Example 2: (o) When Zj = 2(cos |^jr + i sin ^jr) and Z2 = 4(cos f ir + z sin fir), we have

Zi • Z2 = 2(cos îr + i sin Jtt) • 4(cos fir + i sinf??)

=: 8(cos TT + i sin v) = —8

Z2/Z1 = 4(cos Jtt + i sin |jr) -r- 2(cos ^t + i sin Jjt)

= 2(cos ^v + i sin ^tt) = 2i

Zi/22 = ^(cos (-^B-) + i sin (-^r)) = i(cos 3s-/2 + i sin Sir/2) - -î

(h) When z = 2(cosjr/6 + isin7r/6),

«2 = z-z = 4(cos jr/3 + i sin jr/3) = 2(1 + iyjz

)

and z^ = z^'z = 8(cos îr + i sin ^tt) = 8j

As a consequence of Theorem IV, we have

Theorem V. If n is a positive integer,

[r-(cos 9 + i sin e)Y = r"(cos nO + i sin n0)

ROOTS

The equation z" = A = p(cos<l>+ i sin <j>)

where w is a positive integer and A is any complex number, will now be shown to haveexactly n roots. If z = r(cos ^ + i sin 0) is one of these, we have by Theorem V,

r-"(cos nO + i sin nO) = p(cos<l>+ i sin <^)

Then r" = p and n0 = 4, -\- Zk-n- {k, an integer)

so that r - p^'" and = <t>/n + 2kTr/n


The number of distinct roots are the number of non-coterminal angles of the set

{4>/n + 2kTr/n} . For any positive integer k = nq + m, 0^m<n, it is clear that

4,/n + 2kiT/n and ^/n + 2mw/n are coterminal. Thus, there are exactly n distinct roots,

given byp'^" [cos i^/n + 2kTzln) + i sin (</./n + 2fe7r/n)], fc = 0, 1, 2, 3, . . . , n - 1

These n roots are coordinates of n equispaced points on the circle, centered at the origin,

of radius y/\A\. If then z = \/\A\{co&B + isin^) is any one of the nth roots of A, the

remaining roots may be obtained by successively increasing the angle 6 by 27r/n and

reducing modulo 27r whenever the angle is greater than 2-!^.

Example 3: (a) One root of z* = 1 is rj = 1 = cosO + isinO. Increasing the angle successively

by 2s-/4 = jr/2, we find r^ = cos îr + i sin îr, ra = cos v + i sin ir, and

r^ = cos 3n-/2 4- i sin Zvl2. Note that had we begun with another root, say

— 1 = cosjT + tsinjT, we would obtain cos 3?r/2 + i sin 3^/2, cos27r + i sin 2:7 =cos + t sin 0, and cos ^t + i sin îr. These are, of course, the roots obtained

above in a different order.

(b) One of the roots of z* = -4\/3 - 4i = 8(cos 7jr/6 + i sin Tir/B) is

rj = V2(cos77r/36 + tsin77r/36)

Increasing the angle successively by 2ir/6, we have

r2 - V2 (cos 19W36 + i sin 197r/36)

rs = Va (cos 31)r/36 + i sin 3l7r/36)

r4 = ^2 (cos 43jr/36 + i sin 43)^/36)

rs = V2 (cos 55B-/36 + i sin 5577-/36)

U = y/2 (cos 67S-/36 + t sin 677r/36)

As a consequence of Theorem V, we have

Theorem VI. The n nth roots of unity are

p = cos 27r/n + i sin 27r/n, p^, p^ p*, . . .,p"~', p" = 1

PRIMITIVE ROOTS OF UNITY

An nth root z of 1 is called primitive if and only if z"* ¥^1 when < m < n. Using

the results of Problem 5, it is easy to show that p and p^ are primitive sixth roots of 1

while p^,p^,p*,p^ are not. This illustrates

Theorem VII. Let p = cos 27r/n + i sin 27r/n. If {m,n) = d > \, then p™ is an n/dth

^^°* °^ ^-For a proof, see Problem 6.

There follows

Corollary. The primitive nth roots of 1 are those and only those nth roots p, p^, p^, . .. ,

p"

of 1 whose exponents are relatively prime to n.

Example 4: The primitive 12th roots of 1 are

p = cos 2,7-/12 + i sin 2n-/12 = ^^3 + î

p5 = cos5ir/6 + isin5ff/6 = -^V^ + |i

p7 = cos77r/6 + tsin77r/6 - -^y/l - ppii = coslWe + isinlWe = ^y/s - î


Solved Problems

1. Express in the form x + yi:

(a)3-2v/=l, {6)3 + a/=4, (c) 5. (d) 3^, («)^' (/) *^-

(a) 3 - 2V=i = 3 - 2i (e)^-^" = (5 - »)(2 + 3i) _ 13 + 13i _ , , .

^^ 2 - 3t (2 - 3i)(2 + 3i) - 13 - ^ + ^

(6) 3 + V^ = 3 + 2i (/) is = i2 . i = _j = _ J

(c) 5 = 5 + • i

(d\1 = 3-4i _ 3-4j _ 3 4 .

^ ' 3 + 4i (3 + 4i)(3-4i) ~ "25"" ~ 25 25*

2. Prove: The mapping z<^z, z&C is an isomorphism of C onto C.

We are to show that addition and multiplication are preserved under the mapping. This followssince for Zj = a;j + yî, 22 = ajg + j/ji e C,

zi + zj = (a;, + j/,i) + (a;2 + yî) = (x^ + aia) + (y^ + ^2)1

= («i + X2) ~ (Vi + y^H = (xi - j/ji) + (X2 - 3/2*) = z, + Z2

^^^ ^1 • ^2 - («i»2 - ViVi) + («i3/2 + a;23/i)t = (a;ia;2 - J/iJ/2) - (a;iJ/2 + X2J/i)»

= (a;, - j/ii) • (a;2 - j/2t) = Zj • Zg

3. Prove: The absolute value of the product of tviô complex numbers is the product oftheir absolute values, and the angle of the product is the sum of their angles.

Let z, = r,(cos», + I sintf,) and Zj = rg (cos fl2 + i sin flg)- Then

z, • Z2 = r,r2[(cos #1 • cos »2 - sin ff, • sin #2) + i(sin ff, • cos »2 + sin $2 • cos »,)]

= r,r2[cos {01 + »2) + i sin (Sj + tfg)]

4. Prove: The absolute value of the quotient of two complex numbers is the quotient oftheir absolute values, and the angle of the quotient is the angle of the numerator minusthe angle of the denominator.

For the complex numbers z, and «2 of Problem 3,

^ _ ri(cos g, + i sin tf,) _ ri(cos », + i sin #j)(cos 62-1 sin $2)

Z2 r2(cos »2 + i sin #2) ~ r2(cos »2 + i sin »2)(cos »2 - i sin ffg)

r,= — [(cos», cos»2 + sin »i sin #2) + i(sin», cos#2 - sinsacosfi,)]

= — [cos (»i— 82) + % sin (»i

-»2)]

''2

5. Find the 6 sixth roots of 1 and show that they include both the square roots and thecube roots of 1.

The sixth roots of 1 are

p = cos 7r/3 + i sin 7r/3 = ^ + ^Vs t p* = cos 4^/3 + i sin 4b-/3 = -^ - ^y/3i

p2 = cos 27r/3 + t sin 27r/3 = -| + ^Vs i p5 = cos 5i-/3 + i sin 5j7-/3 = | - ^-/Si

p3 = cos JT + i sin jr = -1 p6 -^.^s 2v + t sin 2i7- = 1

Of these, p^ = —1 and p* = 1 are square roots of 1 while p2 = -^ + ^V3i, p* = —^ — A\/3i, andp6 — 1 are cube roots of 1.


6. Prove: Let p = cos27rM + i sin27rM. If {m,n) = d> 1, then p"" is an n/dth root of 1.

Let m = mid and n = n^d. Since p"" = cos 2mjr/w + i sin 2rmr/n = cos 2miir/mi + t sin EwiTr/ni

and (p™)"! = cos2m,7r + i sin 2mi7r = 1, it follows that p"" is an rij = n/d th root of 1.


7. Express each of the following in the form x + yi: (a) 2 + ^^-^, (b) (4 + \^ ) + (3 - 2V^ ),

(c) (4+/=5) - (3-2/=5), (d) (3 + 4t) • (4 - 5t), (e) g^s^. (/) yz^. <«') 2+^' ^''^ **'

(t) js,(i) i6, (fe) iS.

Aws. (a) 2 + V5t, (6) 7 - \/5 t, (c) l + Sy/Si, (d) 32 + i, (e) 2/13 + 3i/13, (/) 4/29 + 19i/29,

(g) 4/13 - 19i/13, (h) 1 + • t, (i) + t, {}) -1 + • i, (fe) 1 +'

' I

8. Write the conjugate of each of the following: (a) 2 + Si, (6) 2 - 3i, (c) 5, (d) 2t.

Aws. (a) 2-3i, (6) 2 + Si, (c) 5, (d) -2i

9. Prove: The conjugate of the conjugate of z is z itself.

10. Prove: For every « # e C, (z-i) = (z)-*.

11. Locate all points whose coordinates have the form (a) (a + • i), (b) (0 + bi), where a, 6 G fi.

Show that any point z and its conjugate are located symmetrically with respect to the a;-axis.

12. Express each of the following in trigonometric form:

(a) 5 (c) -1 - VSi (e) -6 (g) -3 + \/3i (i) 1/i

(6) 4 - 4i (d) -3i (/) ^/2 + ^/2i (h) -1/(1 + i)

Ans. (a) 5 cis (d) 3 cis 3ir/2 (flf) 2\/3 cis 5jr/6

(6) 4V2 cis 7j7/4 (e) 6 cis ir {'^) V^/2 cis 3)7/4

(c) 2 cis 4ir/3 (/) 2 cis jr/4 (i) cis 37r/2

where cis 9 = cos e + i sin 9.

13. Express each of the following in the form a+ bi:

(a) 5 cis 60° (e) (2 cis 25°) • (3 cis 335°)

(6) 2 cis 90° (/) (10 cis 100°) • (cis 140°)

(c) cis 150° {g) (6 cis 170°) H- (3 cis 50°)

(d) 2 cis 210° (h) (4 cis 20°) - (8 cis 80°)

Ans. (a) 5/2 + 5\/3i72 (c) -^Vs + î (e) 6 (ff) -1 + \/3i

(6) 2i (d) -V3-t (/) -5-5V3i (h) i- iVSi

14. Find the cube roots of: (a) 1, (6) 8, (c) 27t, (d) -Si, (e) -4\/3 - 4i.

Aws. (a) -i ± iVSi, 1; (6) -1 ± Vsi, 2; (c) ±^ + |j, -3i; (d) 2i, ±a/3 - i; (e)2cis7W18,

2 cis 1977/18, 2 cis3l77/18

15. Find (a) the primitive fifth roots of 1, (6) the primitive eighth roots of 1.

16. Prove: The sum of the n distinct nth roots of 1 is zero.


17. Use Fig. 8-2 to prove:

(a) |zi + Zi\ ^ |2,| + \z^

(6) |2, - ^21 ^ |z,| - \z^\

18. If r is any cube root of a&C, then r, or, ui^r, where" — ~i + 2^3 1 and w^ are the imaginary cube roots î

~^2

of 1, are the three cube roots of o.

19. Describe geometrically the mappings

(o) z-*z (6) z-^zi (c) z-*zi

Z,+Zo

Fig. 8-2.

20. In the real plane let K be the circle with centerat and radius 1 and let AiA2A3, where Aj(xj,yj) =Aj(Zj) = AjiXj + yji), j = 1, 2, 3, be an arbitrary in-scribed triangle. Denote by P{z) = P{x + yi) anarbitrary (variable) point in the plane.

(a) Show that the equation of A^ is Z'z = 1.

i -I- bxv.(b) Show that P^{Xr,y^), where x^

o-Vj + 63/fc

„+ j,— . divides the line segment

a + h

and y^ =

AjA^ m the ratio 6: a. Then, since Aj.A^ and/aZj-f6zfcN

( a+h )'^® °" ^^'^ ''"^ "^J^fc- verify that

its equation is z -I- ZjZ^,z = Zj 4- z^.

,^1(^1)

Fig. 8-3.

(c) Verify: The equation of any line parallel to A,A, has the form z + z,z,z = , by showing thatthe midpomts 5, and B, of A,A, and A,4, lie on the line z + z^,z = |(z< + zj+ zl+ z,z,z,)

id) Verify: The equation of any line perpendicular to AÂ^ has the form z - zz.zthat and the midpoint of AjA^ lie on the line z - ZjZ^z = 0.

fi by showing

^11 thrnL% 't^"

V . *v***' '''"^"°" '-''"'^^ = ^'-W. Of the altitude of

^1,^2^3 through Aj. Then elimmate z between the equations of any two altitudes to obtain theircommon point ^(z, -)- z^ -)- 23). Show that H also lies on the third altitude

chapter 9

Groups

GROUPSA non-empty set q on which a binary operation <> is defined is said to form a group

with respect to this operation provided, for aiTteiteary a, b, c € Q, the following properties

hold:

Pi: (a o 6) o c = a o (6 o c) (associative law)

P2: There exists uG Q such that aou = uoa = a(existence of identity element)

Ps: For each aGQ there exists a"' e Q such that aoa~' = a^'oa = u(existence of inverses)

Note 1. The reader must not be confused by the use in P3 of a~' to denote the inverse

of a under the operation o . The notation is merely borrowed from that previously used

in connection with multiplication. Whenever the group operation is addition, o~' is to be

interpreted as the additive inverse —a.

Note 2. The preceding chapters contain many examples of groups for most of which

the group operation is commutative. We therefore call attention here to the fact that

the commutative law is not one of the requisite properties listed above. A group is called

abelian or non-abelian according as the group operation is or is not commutative. For

the present, however, we shall not make this distinction.

Example 1: (a) The set / of all integers forms a group with respect to addition; the identity

element is and the inverse of o e / is -a. Thus, we may hereafter speak of

the additive group /. On the other hand, / is not a multiplicative group since,

for example, neither nor 2 has a multiplicative inverse.

(6) The set A of Example 12(c), Chapter 2, forms a group with respect to °. The

identity element is a. Find the inverse of each element.

(c) The set A - {-3, -2, -1, 0, 1, 2, 3} is not a group with respect to addition on

/ although is the identity element, each element of A has an inverse, and addi-

tion is associative. The reason is, of course, that addition is not a binary opera-

tion on A , that is, the set A is not closed with respect to addition.

(d) The set A = {u, - - ^ + i\/3 i, «2 = - ^ - \yfî, "3 =^ 1} of the cube roots

of 1 forms a group with respect to multiplication on

the set of complex numbers C since (i) the product of

any two elements of the set Is an element of the set,

(ii) the associative law holds in C and, hence, in A,

(iii) <03 is the identity element, and (iv) the inverses

of uj, «2, "3 are uj, «i> "3 respectively.

This is also evident from (ii) and the adjacent Table 9-1

operation table.

See also Problems 1-2.

82

•(0, (02 "3

OJl <02 W3 "1

**>'! 0)3 "1 "2

U3 «1 0)2 "3

CHAP. 9] GROUPS 83

SIMPLE PROPERTIES OF GROUPS

The uniqueness of the identity element and of the inverses of the elements of a group

were established in Theorems III and IV, Chapter 2, Page 20. There follow readily

Theorem I (Cancellation Law). If a,b,c G g, then aob = aoc, (also, boa = coa), implies

b = cFor a proof, see Problem 3.

Theorem 11. For a,b G g, each of the equations aox = b and yoa = b has a unique


Theorem III. For every a E g, the inverse of the inverse of a is a, i.e., (a"*)~' = a.

Theorem IV. For every a,b G g, {aob)~^ = b~ôa~^.

Theorem V. For every a,b, . . .,p,q G g, {aobo • opoq)~^ = q-ôp-ô • • ob'ôa~^.

For any aG g and to e /+, we define

a^ = aoaoao • • • oa to m factors

a" = u, the identity element of ga""" = (a"')"" = a"'oa"*oa~*o • • • oa~* to m factors

Once again the notation has been borrowed from that used when multiplication is the

operation. Whenever the operation is addition, a" when n > is to be interpreted as

na = a + a + a+---+a ton terms, a" as u, and a~" as n(—a) = —a + (—a) + (—a) + • • • + (—a)

to n terms. Note that na is also shorthand and is not to be considered as the product of

nG I and aG g.

In Problem 5 we prove the first part of

Theorem VI. For any a G c;, (i) aôa" = 0""+" and (ii) (a"*)" = a""", where m,nGl.

By the order of a group g is meant the number of elements in the set g. The additive

group / of Example 1(a) is of infinite order; the groups of Example 1(6) and 1(d) arefinite groups of orders 5 and 3 respectively.

By the order of an element a G ^ is meant the least positive integer n, if one exists,

for which a" = u, the identity element of ^. If a # is an element of the additive group /

then, since na¥= for every positive integer n, the order of a is infinite. The element (a^

of Example 1(d) is of order 3 since <0j and <j>\ are different from 1 while w^ = 1, the identity

element.

SUBGROUPSLet g = {a,b,c, . . .} be a group with respect to o . Any non-empty subset g' of g is

called a subgroup of g if g' is itself a group with respect to o . Clearly g' = {u}, whereu is the identity element of g, and g itself are subgroups of any group g. They will becalled improper subgroups; other subgroups of g, if any, will be called proper. We notein passing that every subgroup of a group g contains u as its identity element.

Example 2: (a) A proper subgroup of the multiplicative group g = {1,-1, i,—i} is Q' = {1,-1}.(Are there others?)

(6) Consider the multiplicative group g = {p, p2, p3, p*, pS, pS = 1} of the sixth roots

of unity (see Problem 5, Chapter 8, Page 79). It has g' = {p^,p^} andg" = {p^, p*, p*} as proper subgroups.

The next two theorems are useful in determining whether a subset of a group g withgroup operation o is a subgroup.

84 GROUPS [CHAP. 9

Theorem VII. A non-empty subset Q' of a group g is a subgroup of Q if and only if

(i) Q' is closed with respect to o, (ii) q' contains the inverse of each of its

elements.

Theorem VIII. A non-empty subset Q' of a group g is a subgroup of g if and only if

for all a,b &G', a-^ oh &G'. ^ , „ ,,^ ^ For a proof, see Problem 6.

There follow

Theorem IX. Let a be an element of a group g. The set g' = {a" : n e /} of all integral

powers of a is a subgroup of g.

Theorem X, If S is any set of subgroups of a group g, the intersection of these sub-

groups is also a subgroup of Q. _, j, t^ , , „^ For a proof, see Problem 7.

CYCLIC GROUPSA group g is called cyclic if, for some a e g, every a; £ g is of the form a"*, where

m G I. The element a is then called a generator of g. Clearly, every cyclic group is

abelian.

Example 3: (a) The additive group / is cyclic with generator a = 1 since, for every m G /,

a™ = ma = m.

(b) The multiplicative group of fifth roots of 1 is cyclic. Unlike the group of (a)

vyhich has only 1 and —1 as generators, this group may be generated by anyof its elements except 1.

(c) The group g of Example 2{b) is cyclic. Its generators are p and p^.

Examples 3(6) and 3(c) illustrate

Theorem XI. An element a' of a finite cyclic group g of order w is a generator of g if

and only if (%,<) = !.


Theorem XII. Every subgroup of a cyclic group is itself a cyclic group.

PERMUTATION GROUPSThe set /S„ of the n ! permutations of n symbols was considered in Chapter 2. A review

of this material shows that S„ is a group with respect to the permutation operation o.

Since o is not commutative, this is our first example of a non-abelian group.

It is customary to call the group Sn the symmetric group on n symbols and to call anysubgroup of Sn a permutation group on n symbols.

Example 4: (a) S4 = {(1), (12), (13), (14), (23), (24), (34), a = (123), a^ = (132), /3 = (124),

;82 = (142), y = (134), y^ = (143), 8 = (234), S^ = (243), p = (1234),

p2 = (13)(24), p3 = (1432), a = (1243), o^ = (14)(23), a^ ^ (1342),

T = (1324), t2 = (12)(34), t3 = (1423)}.

(6) The subgroups of S4 : (i) {(1), (12)}, (ii) {(1), a, a^}, (iii) {(1), (12), (34), (12)(34)},

and (iv) A^ = {(1), a, a^ p, /S^, y, Y^. 8. S^, p^, a^, t^} are examples of permuta-tion groups on 4 symbols. (A4 consists of all even permutations in S4 and is

known as the alternating group on 4 symbols.) Which of the above subgroupsare cyclic? which abelian? List other subgroups of S4.

See Problems 9-10.

HOMOMORPHISMSLet g, with operation o

, and g' with operation a , be two groups. By a homomorphismof g into g' is meant a mapping

CHAP. 9] GROUPS 85

such that

(i) every g & Q has a unique image g' G Q'

(ii) if a^ a' and b -* h', then a o 6 -» o' a 6'

If, in addition, the mapping satisfies

(iii) every g' G Q' is an image

we have a homomorphism of g onto q' and we then call Q' a homomorphic image of g.

Example 5: (a) Consider the mapping n-*i" of the additive group / onto the multiplicative

group of the fourth roots of 1. This is a homomorphism since

and the group operations are preserved.

(6) Consider the cyclic group g = {a, a?, a?, . . ., a'2 — u) and its subgroupQ' = {o2, a^, a*, . .

.,ai2}. It follows readily that the mapping

a" -» a^"

is a homomorphism of Q onto Q' while the mapping

a" -> a"is a homomorphism of Q' into ^.

See Problem 11.


Theorem XIII. In any homomorphism between two groups g and g', their identity

elements correspond; and if a; G (^ and x' G ^' correspond, so also do

their inverses.

There follows

Theorem XIV. The homomorphic image of any cyclic group is cyclic.

ISOMORPHISMSIf the mapping of the section above is one-to-one, i.e.,

g <^ g'

such that (iii) is satisfied, we say that g and g' are isomorphic and call the mapping anisomorphism.

Example 6: Show that g, the additive group //(4), is isomorphic to g', the multiplicative groupof the non-zero elements of 7/(5).

Consider the operation tables

g g'

+ 1 2 3

1 2 3

1 1 2 3

2 2 3 1

3 3 1 2

13 4 2

113 4 2

3 3 4 2 1

4 4 2 132 2 13 4

Table 9-2 Table 9-3

in which, for convenience, [0], [1], ... have been replaced by 0, 1, . .

.

readily that the mapping

g^ff' : 0^1, 1^3, 2-»4, 3-»2

is an isomorphism. For example, l=2 + 3-»4'2 = 3, etc.

Rewrite the operation table for g' to show that

g-*g' : 0-»l, 1^2, 2^4, 3^3is another isomorphism of g onto g'. Can you find still another?

It follows

o 9i 92 9s 9i 95 96

9i 9\ 92 93 94 95 96

9i 9i 9i 95 96 93 94

93 93 96 9x 95 94 92

9-4 9i 95 96 9\ 92 93

95 95 94 92 93 96 9i

96 9s 93 94 92 9i 95

86 GROUPS [CHAP. 9


Theorem XV. (a) Every cyclic group of infinite order is isomorphic to the additive

group /.

(6) Every cyclic group of finite order n is isomorphic to the additive

group I/(n).

The most remarkable result of this section is

Theorem XVI (Cayley). Every finite group of order n is isomorphic to a permutationgroup on n symbols.

Since the proof, given in Problem 14, consists in showing how to construct an effective

permutation group, the reader may wish to examine first

Example 7: Consider the adjacent operation table for

the group g = {gi, g^, g^, ^4, g^, g^ withgroup operation ^ .

The elements of any column of the ta-

ble, say the fifth: gy^g^- g^, g2'=' 9s = 93,

93^95- 94, 94° 95 = 92, 95° 95 = 96,

96° 95 = 91 are the elements of the rowlabels (i.e., the elements of g) rearranged.

This permutation will be indicated by

9i 92 93 94 95 96\ ^ (i56)(234)Table 9-4

95 93 94 92 96 9\j= V5

It follows readily that g is isomorphic to P = {Pi,P2>P3«P4' Ps. Pe) where pj — (1),

V2 = (12)(36){45), P3 = (13)(25)(46), V4 = (14)(26)(35), Ps = (1561(234), pg = (165)(243)

under the mappingffi-^Pi (i = 1,2,3, ...,6)

COSETSLet g be a finite group with group operation o, H be a subgroup of Q, and a be an

arbitrary element of g. We define as the right coset Ha of H in g, generated by a, the

and as the left coset aH of H in g, generated by a, the subset of gaH = {aoh: hGH)

Example 8: The subgroup H = {(1), (12), (34), (12)(34)} and the element a = (1432) of S4 gen-erate the right coset

Ha = {(l)o(1432), (12)0(1432), (34) o (1432), (12)(34) <= (1432)}

= {(1432), (243), (142), (24)}

and the left coset

aH = {(1432) o(l), (1432)0(12), (1432) o (34), (1432) o (12)(34)}

= {(1432), (143), (132), (13)}

In investigating the properties of cosets, we shall usually limit our attention to right

cosets and leave for the reader the formulation of the corresponding properties of left

cosets. First, note that a G Ha since u, the identity element of g, is also the identity

element of H. If H contains m elements so also does Ha, for Ha contains at most melements and hi°a - h2°a for any huh2GH implies hi = h2. Finally, if Cr denotesthe set of all distinct right cosets of H in g, then H G Cr since Ha = H when a G H.

Consider now two right cosets Ha and Hb, a¥'b, of H in g. Suppose that c is acommon element of these cosets so that for some hi,h2 G H we have c = hioa — h2ob.Then a = hrô{h2ob) = {hr^°h2)ob and, since hr^°h2GH (Theorem VIII), it follows

that aGHb and Ha = Hb. Thus Cr consists of mutually disjoint right cosets of g and so is

a partition of g. We shall call Cr a decomposition of g into right cosets with respect to H.

CHAP. 9] GROUPS 87

Example 9: (a) Take Q as the additive group of integers and H as the subgroup of all

integers divisible by 5. The decomposition of g into right cosets with re-

spect to H consists of the five residue classes modulo 5, i.e., H - {x: 5|a;},

H\ = {a; : 5I

(X - 1)}, H2 = {x : 5\ {x -2)}, H3 = {x : 5\ {x - 3)}, andHi = {x: 5

I

(a: — 4)}. There is no distinction here between right and left cosetssince (j is an abelian group.

(6) Let ff = S4 and H = A4, the subgroup of all even permutations of S4. Thenthere are only two right (left) cosets of g generated by H, namely, A^ and thesubset of odd permutations of S4. Here again there is no distinction betweenright and left cosets but note that S4 is not abelian.

(c) Let g = S4 and H be the octic group of Problem 9. The distinct right cosetsgenerated by H are

H = {(1), (1234), (13)(24), (1432). (12)(34), (14)(23), (13), (24)}

H(12) = {(12), (234), (1324), (143), (34), (1423), (132), (124)}

H(23) = {(23), (134), (1243), (142), (1342), (14), (123), (243)}

and the distinct left cosets generated by H are

H = {(1), (1234), (13)(24), (1432), (12)(34), (14)(23), (13), (24)}

{12)H = {(12), (134), (1423), (243), (34), (1324), (123), (142)}

{2S)H = {(23), (124), (1342), (143), (1243), (14), (132), (234)}

Thus, G = Hu H(12) U H(23) = H u (12)H u (23)H. Here, the decompositionof g obtained by using the right and left cosets of H are distinct.

Let g be a finite group of order n and H be a subgroup of order m of g. The numberof distinct right cosets of H in g (called the index of H in g) is r where n = mr; hence,

Theorem XVII (Lagrange). The order of each subgroup of a finite group g' is a divisor

of the order of g.

As consequences, we have

Theorem XVIII. If ^ is a finite group of order n, then the order of any element a G ^(i.e., the order of the cyclic subgroup generated by a) is a divisor of n.

and

Theorem XIX. Every group of prime order is cyclic.

INVARIANT SUBGROUPSA subgroup H of a group g is called an invariant subgroup (normal subgroup or

normal divisor) of g if

(i) gH = Hg for every g Gg

Since g~^ E g whenever g G g, (i) may be replaced by

(i') 9~^Hg = H for every geg

Now (i') requires

(ii) for any g eg and any hG H, then g-ôhog GHand

(iz) for any gGg and each hGH, there exists some kGH such that g'ôkog -hor kog — goh.

We shall show that (ii) implies (iz). Consider any hGH. By (ij'), (g-^)-ôhog-^ =gohog-^ = kGH since g'^ G g; then g'ôkog ^ h as required. We have proved

88 GROUPS [CHAP. 9

Theorem XX. If H is a subgroup of a group ^ and if g-ôhog&H for all gf G g and

all h G H, then H is an invariant subgroup of Q.

Example 10: (a) Every subgroup H of an abelian group g is an invariant subgroup of g since

goh = h°g, for any g & g and every h& H.

(b) Every group g has at least two invariant subgroups: {u}, since uog = gou for

every g & g, and .g itself, since for any g,h& g we have

goh = g°h°{g^^°g) = {g °h° g~^)o g — k°g and k = g °h° g"^ Gg

(c) It H is a. subgroup of index 2 of (? [see Example 9(6)] the cosets generated by Hconsist of H and g — H. Hence, H is an invariant subgroup of g.

(rf) For g = {a, a2, a^, ..., a»2 = u), its subgroups {«, a^, ai,. . . ,

ai«}, {m, a^, a^, a^},

{u,a*,a^), and (m, «•*} are invariant.

(e) For the octic group (Problem 9), {u,p^,a^,T^, {w.p^, 6, e} and {u,p,p^,p^

are invariant subgroups of order 4 while {m, p^} is an invariant subgroup of

order 2. (Use Table 9-7 to check this.)

(/) The octic group P is not an invariant subgroup of S4 since for p = (1234) S Pand (12) e S4, we have (12)" 1 p (12) = (1342) € P.


Theorem XXI. Under any homomorphism of a group g with group operation o and

identity element u into a group Q' with group operation a and identity

element u', the subset S of all elements of g which are mapped onto u'

is an invariant subgroup of g.

The invariant subgroup of g defined in Theorem XXI is called the kernel of the

homomorphism.

In Example 10(6) it was shown that any group g has {u) and g itself as invariant

subgroups. They are called improper while other invariant subgroups, if any, of g are

called proper. A group g having no proper invariant subgroups is called simple.

QUOTIENT GROUPSLet H be an invariant subgroup of a group g with group operation o and denote by

glH the set of (distinct) cosets of H in g, i.e.,

g/H = {Ha, Hb, He, . . .

}

We define the "product" of pairs of these cosets by

iHa)(Hb) = {(hioa)o {h^ o b) : hi, hi G H} for all Ha, Hb G g/H

and, in Problem 16, prove that this operation is well-defined.

Now g/H is a group with respect to the operation just defined. To prove this, wenote first that

(hi o a) o (/i2 06) — hio{aoh2)ob = hio (ha » a) <> 6

= (fei o fes) o (o o &) = h4°{aob), h3,h4GH

Then (Ha){Hb) = H{aob)Gg/H

and [{Ha)iHb)]{Hc) = H[(ao6)oc] = H[ao{boc)] = (Ha)[(Hb){Hc)]

Next, for u the identity element of g, (Hu){Ha) = {Ha){Hu) = Ha so that Hu = H is the

identity element of g/H. Finally, since {Ha)(Ha-') = (Ha-^){Ha) = Hu = H, it follows

that g/H contains the inverse Ha'^ of each Ha G g/H.

The group g/H is called the quotient group (factor group) of g by H.

CHAP. 9] GROUPS 89

Example 11: (a) When g is the octic group of Problem 9 and H = {u,pi,b,e}, then QlH -{H,Hp}. This representation of QlH is, of course, not unique. The readerwill show that glH = {H,Hp3} - {H,Ha^ = {H,Ht^ = {H,Hp}.

(6) For the same g and H = {u, p^}, we have

g/H = {H,Hp,Hu^Hb} = {H,Hp3,HT\He}

The examples above illustrate

Theorem XXII. If H, of order m, is an invariant subgroup of q, of order n, then thequotient group Q/H is of order n/m.

From (Ha){Hb) = H{a o 6) e Q/H, obtained above, there follows

Theorem XXIII. If H is an invariant subgroup of a group g, the mapping

g^g/H: g^Hgis a homomorphism of g onto glH.


Theorem XXIV. Any quotient group of a cyclic group is cyclic.

We leave as an exercise the proof of

Theorem XXV. If H is an invariant subgroup of a group g and if H is also a subgroupof a subgroup K of g, then H is an invariant subgroup of K.

PRODUCT OF SUBGROUPSLet H = {huh2, . . .,hr} and K = {6i, 62, . . .,6p} be subgroups of a group g and

define the "product"HK = {hiobji hiGH, bjGK)

In Problems 65-67, the reader is asked to examine such products and, in particular,

to prove

Theorem XXVI. If H and K are invariant subgroups of a group g, so also is HK.

COMPOSITION SERIES

An invariant subgroup H of a group g is called maximal provided there exists noproper invariant subgroup K of g having H as a proper subgroup.

Example 12: {a) A^ of Example 4(6) is a maximal invariant subgroup of S4 since it is a sub-group of index 2 in S4. Also, {u,p^,a^,T^} is a maximal invariant subgroupof A4. (Show this.)

(6) The cyclic group g - {w.a.a^, . . .,a»} has H = {u,a^,a* ai"} andK — {u,a^,a^,a^} as maximal invariant subgroups. Also, J = {m, a*, a^} jg ^maximal invariant subgroup of H while L = {u, a^} is a maximal invariantsubgroup of both H and K.

For any group g a sequence of its subgroups

g.H,J,K, ...,U = {u}

will be called a composition series for g if each element except the first is a maximalinvariant subgroup of its predecessor. The groups g/H, H/J, JIK, ... are then called

the quotient groups of the composition series.


Theorem XXVII. Every finite group has at least one composition series.

90 GROUPS [CHAP. 9

Example 13: (a) The cyclic group Q = {u, a, a^, a^, a*) has only one composition series: g,V = {«}.

(6) A composition series for g — S4 is

S4, A,. {(1), p2, <r2, r2}, {(1), p2}. U = {(1)}

Is every element of the composition series an invariant subgroup of ^ ?

(c) For the cyclic group of Example 12(6), there are three composition series:

(i) g,H,J, U, (ii) g,K,L, C/, -(iii) g,H,L,U. Is every element of each composi-

tion series an invariant subgroup of ff ?

In Problem 19, we illustrate

Theorem XXVIII (The Jordan-Holder Theorem). For any finite group with distinct

composition series, all series are of the same length, i.e., have the same

number of elements. Moreover, the quotient groups for any pair of

composition series may be put into one-to-one correspondence so that

corresponding quotient groups are isomorphic.

Before attempting a proof of Theorem XXVIII (see Problem 23) it will be necessary

to examine certain relations which exist between the subgroups of a group g and the

subgroups of its quotient groups. Let then H, of order r, be an invariant subgroup of a

group Q of order n and write

S = g/H = {HauHa^.Haz, ...,Has), cuGg (1)

where, for convenience, ai = u. Further, let

P = {HbuHb2,Hb3, ,Hb^} (2)

be any subset of S and denote by

K = HbiUHbiU HbaU U Hbp (3)

the subset of g whose elements are the pr distinct elements (of g) which belong to the

cosets in P.

Suppose now that P is a subgroup of index t of S. Then n - prt and some one of the

b's, say 6i, is the identity element u of g. It follows that X is a subgroup of index t of gand P = K/H since

(i) P is closed with respect to coset multiplication; hence, K is closed with respect to

the group operation on g.

(ii) The associative law holds for g and thus for K.

(iii) H GP; hence, u G K.

(iv) P contains the inverse Hbf^ of each coset Hbi G P; hence, K contains the inverse of

each of its elements.

(v) K is of order pr; hence K is of index t in g.

Conversely, suppose K is a subgroup of index t of g which contains H, an invariant

subgroup of g. Then, by Theorem XXV, H is an invariant subgroup of K and so P = K/His of index t in S = g/H.

We have proved

Theorem XXIX. Let H be an invariant subgroup of a finite group g. A set P of the

cosets of S = g/H is a subgroup of index t of S if and only if K, the

set of group elements which belong to the cosets in P, is a subgroup

of index t of g.

We now assume bi = u in (2) and (3) above and state

Theorem XXX, Let g' be a group of order n - rpt, iiT be a subgroup of order rp of g,

and H be an invariant subgroup of order r of both K and g. ThenK is an invariant subgroup of g if and only if P = K/H is an invariant

subgroup of S = g/H. p^j. ^ p^^^f ^ ggg Problem 20.

CHAP. 9] GROUPS 91

Theorem XXXI.

Theorem XXXII.

Let H and K be invariant subgroups of Q with H an invariant sub-group of K, and let P = K/H and S = Q/H. Then the quotient groupsSIP and QlK are isomorphic.


If ^ is a maximal invariant subgroup of a group g then QlH is simple,and conversely.

Theorem XXXIII. Let H and K be distinct maximal invariant subgroups of a group Q.Then

(a) D = H r\K is &n invariant subgroup of Q, and(6) H/D is isomorphic to g/K and K/D is isomorphic to glH.


Solved Problems

Does 7/(3), the set of residue classes modulo three, form a group with respect to addi-tion? with respect to multiplication?

From the addition and multiplication tables for 7/(3) in which [0], [1], [2] have been replaced byVf -I, £if

+ 1 2

1 2

1 1 2

2 2 1

• 1 2

1 1 2

2 2 1

Table 9-5 Table 9-6

It IS clear that //(3) forms a group with respect to addition. The identity element is and the inversesof 0,1,2 are respectively 0,2,1. It is equally clear that while these residue classes do not form agroup with respect to multiplication, the non-zero residue classes do. Here the identity element is 1and each of the elements 1,2 is its own inverse.

2. Do the non-zero residue classes modulo four form a group with respect to multiplication ?From Table 5-2 of Example 12, Chapter 5, Page 53, it is clear that these residue classes do notform a group with respect to multiplication.

3. Prove: If a,b,cGg, then aob = aoc {also, boa ^ coa) implies b = c.

Consider aobâoc. Operating on the left with o-> e g, we have a-i o(ao6) rz: ^-lo (aoc)Usmg the associative law, {a~i°a)ob = (a-^°a)°c; hence, u°b = uoc and so b = c. Similarly(6oa)oa-i = (coa)oa-> reduces to b = c.

4. Prove: When a.bGg, each of the equations aox = b and yoa ^ b has a uniquesolution.

We obtain readily x = a'ôb and j/ = 6oa-i as solutions. To prove uniqueness, assumeX and y to be a second set of solutions. Then a<>x = aox' and yoa-y'oa whence by Theorem IX — x' and 1/ — y'. > j ,

92 GROUPS [CHAP. 9

5. Prove: For any a&Q, a'"oa" = a'"+" when m.nEl.

We consider in turn all cases resulting when each of m and n are positive, zero, or negative.

When m and n are positive,

m factors n factors m + n factorsA A -^ ^

A^

(jmoa." = {aoa°---°a)o(a°a°---°a) = a° a') • • • °a = «" + "

When m = —r and n = s, where r and s are positive integers,

amoo" = a-''oos = (a"i)''°a'' = (a-^ °a-^ ° • °a~^)°(a°a° ° a)

[a^-r = a'»+" when s — r

(a"!)"""* = a^"*^ = a'™+" when « < r

The remaining cases are left for the reader.

6. Prove: A non-empty subset g' of a group ^ is a subgroup of g if and only if, for all

a,b G g', a-'ob G Q'.

Suppose g' is a subgroup of Q. If a, 6 e Q', then a'^ & g' and, by the Closure Law, so also

does a~^°h.

Conversely, suppose g' is a non-empty subset of g for which a-^°b&g' whenever a,he.g'.

Now a-^°a = u&g'. Then Moa-i = a-ie&' and every element of g' has an inverse in g'.

Finally, for every a,be g', {a-^)-^°b = a°b S g', and the Closure Law holds. Thus, g' is a group

and, hence, a subgroup of g.

7. Prove: If S is any set of subgroups of a group g, the intersection of these subgroups

is also a subgroup of g.

Let a and b be elements of the intersection and, hence, elements of each of the subgroups which

make up S. By Theorem VIII, a-^°b belongs to each subgroup and, hence, to the intersection.

Thus, the intersection is a subgroup of g.

8. Prove: Every subgroup of a cyclic group is itself a cyclic group.

Let g' be a subgroup of a cyclic group g whose generator is a. Suppose that m is the least

positive integer such that a™ S g'. Now every element of g', being an element of g, is of the form a'',

k G I. Writing „ ^k = mq + r, ~ r < m

we have a" = «""' + '• = (a™)«oa'-

and, hence, a'' = (a'")-'>oafc

Since both a"' and a^ G g', it follows that a^ G g'. But since r < m, r = 0. Thus k = mq, every

element of g' is of the form (a"')", and g is the cyclic group generated by a™.

9. The subset {u = (1), p, p\ p\ <t\ r^, b = (13), e = (24)} of Si is a group (see the operation

table below), called the octic group of a square or the dihedral group. We shall now

show how this permutation group may be obtained using properties of symmetry of a

square.

Consider the square (Fig. 9-1) with vertices denoted by 1,2,3,4;

locate its center O, the bisectors AOB and COD of its parallel sides,

and the diagonals 103 and 204. We shall be concerned with all rigid

motions (rotations in the plane about O and in space about the bisectors

and diagonals) such that the square will look the same after the motion

as before.

Denote by p the counterclockwise rotation of the square about Othrough 90°. Its effect is to carry 1 into 2, 2 into 3, 3 into 4, and

4 into 1; thus, p = (1234). Now p2 = p°p = (13)(24) is a rotation

CHAP. 9] GEOUPS 93

^^"no ^5 ^^^°' "' " ^"^^^ '" ^ ''°*^*'°" °' 270°, and p* = (1) = M is a rotation about O of 360°

''^-/1.^ô!^°*^*'"'"^ through 180° about the bisectors AOB and COi? give rise respectively to

a - (14)(23) and r2 = {12)(34) while the rotations through 180° about the diagonals 103 and 204 giverise to e = (24) and b = (13).

^

The operation table for this group isUp ©2 „3 „2 9 r

u

p

p"

r2

6

e

Up p2 p3 ff2 ^2 6g

P p2 p3 M 6 e t2 ,,2

P^ P^ u p T^ a^ e b

P^ u p p2 e 6 (,2 ^2

"^ e t2 6 M p2 p3 p

T^ 6 or2 e p2 It p p3

^ "^ e t2 p p3 u p2

e t2 6 ^2 p3 p p2 M

Table 9-7

In forming the table

(1) fill in the first row and first column and complete the upper left 4x4 block,

(2.) complete the second row, (p o ,,2 = (1234) o (14)(23) = (13) = 6, . . .)

and then the third and fourth rows,

(p2 o „2 = p o (p o ,,2) = p o 6 = ^2^-J

(3) complete the second column and then the third and fourth columns,

((r2 o p2 = („2 o p) o p = e o p = t2, . .

.

)

(4) complete the table, (a^ o ^2 = ,,2 o („2 o p2) = p2_ \

10. A permutation group on n symbols is called regular if each of its elements except theIdentity moves all n symbols. Find the regular permutation groups on four symbols.

Using Example 4, the required groups are:

{p. p2, p^ p* = (1)}, {<r, <72, <t3, o* =: (1)}, and {r, t2, r3, t* = (1)}

11. Prove: The mapping / ^ I/(n) -.m-^lm] is a homomorphism of the additive group /onto the additive group I/{n) of integers modulo n.

HowS? J'"'" ^'^

Z^""^"'"^ = m + r,Q^r<n.itis evident that the mapping is not one-to-one.

moIuTo ; ir/'"^^^'"^

f rr T^" '" *^' ''' m,nm [«-l]} of residue classesmodulo n and every element of this latter set is an image. Also, if a - \r] and b -* M then

r M + [s] = Itl the residue class modulo n of c = „+ 6. Thus the groVp operations aê pre"served and the mappmg is a homomorphism of / onto I/{n).

12. Prove: In a homomorphism between two groups g and Q', their identity elementscorrespond, and if a; G ^ and x' G g' correspond so also do their inverses.

Denote the identity elements of ff and g' by M and M' respectively. Suppose now that u^v' and

Z It^: ^^: «•. . I 1 "? ^v'^x' = cc' = u'^^' whence, by the Cancellation Law, v' = u'and we have the first part of the theorem.

For the second part, suppose x -- x' and a;-' -^ y'. Then u = xox'i -* x' '^ y' = u' = x' c. (a;')-i

13. Prove: Every cyclic group of infinite order is isomorphic to the additive group /.Consider the infinite cyclic group g generated by a and the mapping

n ^ o", n& Ioil into g Now this mapping is clearly onto; moreover, it is one-to-one since, if for s>t we hads ^ as and t «• a* with a» = a', then a^-t = u and g would be finite. Finally, s + t *^ a» + t = o» • a'and the mappmg is an isomorphism.

94 GROUPS [CHAP. 9

14. Prove: Every finite group of order n is isomorphic to a permutation group on n

symbols.

Let g = {91,92,93, ,9n} with group operation = and define

r, = f "' \ ^ { ^' ^^ ^' ^" V (i= 1.2.3 n)'

\9i'^9jJ \9i'='ffj 92° 9j 9i°gi 9n°9il

The elements in the second row of Vj are those in the column of the operation table of G labeled g^

and, hence, are a permutation of the elements of the row labels. Thus P = {pi.Pz.Pa. • • -.Pn) is a

subset of the elements of the symmetric group S„ on n symbols. It is left for the reader to show

that P satisfies the conditions of Theorem VII for a group. Now consider the one-to-one cor-

respondence^^j ^.^p, iî,2,3,...,n

If 9t = 9r° 9», then fft *^ Pr ° Ps 8° that

^' ^ \9i^9r)° \9i°9s)^ \9i°9r)\gi°9r°9s) " \9i°9t)

and (a) is an isomorphism of g onto P. Note that P is regular.

15. Prove: Under any homomorphism of a group Q with group operation o and identity

element u into a group g' with group operation ^ and identity element W, the subset

S of all elements of g which are mapped onto u' is an invariant subgroup of g.

As consequences of Theorem XIII, we have

(a) U-* u'; hence, S is non-empty.

(6) if aeS, then a-» ^ (m')~' = «'; hence, a-^ G S.

(c) if a.beS, then a'ôb ^ u' <=• u' = u'; hence, a-io6GS.

Thus S is a subgroup of g.

For arbitrary aGS and g ^ g,

g-ioaog -> (g')-^°u' og' = u'

so that g-ôa°gSS. Then by Theorem XX, S is an invariant subgroup of g as required.

16. Prove: The product of cosets

{Ha){Hb) = {(hi o a) o {hi o b) : hi, hi G H} for all Ha, Hb G g/H

where H is an invariant subgroup of g, is well defined.

First, we show: For any x,x' e g, Hx' = Hx if and only if x' = v ° x for some vê H.

Suppose Hx' = Hx. Then x' S Hx requires x' = vox for some vGH. Conversely, if x' -vox

with vGH, then Hx' = H(v °x) = (Hv)x = Hx.

Now let Ha' and Hb' be other representations of Ha and Hb respectively with a' = a°r,

5' = 5og, and r.sGH. In (Ha')(Hb') = {[hi°(a°r)]o[h2o{bo s)]: hi,h.,e H} we have, using

(ia). Page 87,

[hi o (a o r)] ° [h^ °(b° s)] = {hi oa)°{ro h^) o (6 o g)

= (ftj o a) o /13 o (t o 6) = (fcj o a) o (Tig o t) o 6

= (fci o a) o (^4 ° 6) where /13, t, h^ G ii/

Then (Ha'KHb') = (Ha)(Hb)

and the product {Ha)(Hb) is well-defined.

17. Prove: Any quotient group of a cyclic group g is cyclic.

Let H be any (invariant) subgroup of the cyclic group g = {u,a,a\ . ..,ar} and consider the

homomorphism g -^ g/H : a* ^ Ha^

Since every element of g/H has the form Ha' for some o' e ff and Ha' = (Ha)* (prove this) it fol-

lows that every element of g/H is a power of 6 = Ha. Hence, g/H is cyclic.

CHAP. 9] GROUPS 95

18. Prove: Every finite group g has at least one composition series.

(i) Suppose g is simple; then g, 17 is a composition series.

(ii) Suppose g is not simple; then there exists an invariant subgroup H ¥' g,U ot g. U H is maxi-mal in g and U is maximal in H, then g, H, 17 is a composition series. Suppose H is notmaximal in g but U is maximal in H; then there exists an invariant subgroup K of g such thatH is an invariant subgroup of K. If K is maximal in g and H is maximal in K, there g, K, H, Uis a composition series. Now suppose H is maximal in g but U is not maximal in H; thenthere exists an invariant subgroup J of H. If J is maximal in H and U is maximal in J, theng, H, J, C7 is a composition series. Next, suppose that H is not maximal in g and U is notmaximal in H; then .... Since g is finite, there are only a finite number of subgroups andultimately we must reach a composition series.

19. Consider two composition series of the cyclic group of order 60: g = {u,a,a?, . . .,a^^}:

g,H ^ [u, a\ a\..., a^«}, / = {u, a\ a», . . ., a^}, K = {u, a^^ a^\ a^\ a«}, U = {u}

and

g.M = {u,a\a\ . . .,a"}, N = [u,a'\a''o,a*^}, P = {u.a^"}, U

The quotient groups are:

g/H = {H, Ha}, H/J = {J, Ja2}, J/K = {K, Ka*, Ka»}, K/U = {U, Ua>^, Ua^, Uaô, Ua*»}and

g/M = {M, Ma, Ma^}, M/N = {N, Na^, Na^, Na», Na^^, NIP = {P, Pa^^}, P/U = {U, UaÔ}

Then in the one-to-one correspondence: g/H *^ NIP, HIJ <^ P/U, JIK <^ gIM, KlU *^ MIN, cor-responding quotient groups are isomorphic under the mappings:

H<r^P JÛ K^M U<r^N

Ha <^ Poi5 Jat ^ Ua»« Ka* ^ Ma Ua^^ <-» Na^

Kali <^ Afo2 (/a2* <^ Na^

t7a3« <^ Na9

Ua*» <^ iVa>2

20. Prove: Let g be a group of order n = rpt, K he a. subgroup of order rp of g, and H bean invariant subgroup of order r of both K and g. Then K is an invariant subgroupof g if and only if P = K/H is an invariant subgroup of S = g/H.

Let g be an arbitrary element of g and let K = {ft,, b^, . ..,b^}.

Suppose P is an invariant subgroup of S. For Hg G S, we have

(i) {Hg)P = P{Hg)

Thus for any H6j e P, there exists Hbj G P such that

(ii) {Hg)(Hbt) = (Hbj)(Hg)

Moreover, (Hg)(Hb) = (Hbj)(Hg) = (Hg)(Hbk) implies Hftj = Hb^. Then

(«i) Hbi = {Hg-i){Hbj){Hg) =. g-HHbi)g

(iv) K = HbiUHbÛ U Hbp = g-^ Kg

and

(v) fl-X = Kg

Thus if is an invariant subgroup of g.

Conversely, suppose K is an invariant subgroup of g. Then, by simply reversing the steps above,we conclude that P is an invariant subgroup of S.

96 GROUPS [CHAP. 9

21. Prove: Let H and K be invariant subgroups of Q with H an invariant subgroup of K,

and let P = KIH and S = QlH. Then the quotient groups SIP and qlK are isomorphic.

Let g, K,H have the respective orders n = rpt, rp, r. Then K is an invariant subgroup of

index t in ^ and we define^ , „ ^ __

g/K = {Kci, Kc^ Kct)

,

CiGg

By Theorem XXX, P is an invariant subgroup of S; then P partitions S into t cosets so that we

may write^^^ ^ {P{Ha^^), PiHUi^) P(//ai,)} ,

Ha;. G S

Now the elements of g which make up the subgroup K, when partitioned into cosets with respect

to H, constitute P. Thus each c^ is found in one and only one of the Hat.. Then, after rearranging

the cosets of S/P when necessary, we may write

S/P - {P(Hci), P(Hc2), ..., P{Hci)}

The required mapping is g/K <^ S/P : Kci *^ P{Hci)

22. Prove: Let H and K be distinct maximal invariant subgroups of a group g. Then

(a) D - H n K is an invariant subgroup of g and (5) i?/Z? is isomorphic to g/K and

K/D is isomorphic to ^/H.

(a) By Theorem X, D is a subgroup of g. Since /f and K are invariant subgroups of g, we have for

each dG D and every g S g,

g-iodog e. H, g-ôdog G K and so g-^°d°g G D

Then, for every g G g, g-^Dg = D and D is an invariant subgroup of g.

(6) By Theorem XXV, D is an invariant subgroup of both H and K. Suppose

(i) H = Dhi U Dh2 u . , . u Dhn, h&Hthen, since K(Dhi) = (KD)ht = Khi (why?),

(ii) KH = Kh^ U Kh^ U . . . U Khn

By Theorem XXVI, HK = KH is a subgroup of g. Then, since H is a. proper subgroup of HKand, by hypothesis, is a maximal invariant subgroup of g, it follows that HK = g.

From (i) and (ii), we have

H/D = {Dhi,Dh2, ...,Dh„} and g/K = {Khi,Kh2, . . .,KhJ

Under the one-to-one mapping

Dhi «• Khi, (i = 1,2,3, ...,w)

{Dhi){Dhj) := D(hiOhj) «• K(hiohj) = (Khi)(Khj)

and H/P is isomorphic to g/K. It will be left for the reader to show that K/D and g/H are

isomorphic.

23. Prove: For a finite group with distinct composition series, all series are of the same

length, i.e., have the same number of elements. Moreover the quotient groups for any

pair of composition series may be put into one-to-one correspondence so that correspond-

ing quotient groups are isomorphic.

Let (a) g, Hi, Hi, Ha, ..., H^ = U

(6) g, K„ K2, Kg, ..., K, = U

be twx) distinct composition series of g. Now the theorem is true for any group of order one. Let us

assume it true for all groups of order less than that of g. We consider two cases:

(i) Hi = Ki- After removing g from (o) and (6), we have remaining two composition series of H^

for which, by assumption, the theorem holds. Clearly, it will also hold when g is replaced in each.

CHAP. 9] GROUPS 97

(ii) Hi ¥= Ki. Write D = H^n K,. Since g/Hi (also g/K{) is simple and, by Theorem XXXIII,IS isomorphic to KÎD (also glK^ is isomorphic to HÎD), then KÎD (also HJD) is simple. ThenD is the maximal invariant subgroup of both H^ and K^ and so g has the composition series

(a') g, Hi, D, Dy, D^, D^, ..., D, = Uand

(6') g, if,, D, Z)„ Z>2, D3, ..., D, = UWhen the quotient groups are written in the order

g/Hi, HJD, DIDi, DJD^, DÎD^, ..., D,_JDtand

KJD, GIKi, D/Di, Di/D^, DÎD^ D,_JD,,corresponding quotient groups are isomorphic, that is, g/H, and KJD, Hi/D and glK^, DID, andD/Dy, . . . are isomorphic.

Now by (i) the quotient groups defined by (a) and (a') [also by (6) and (6')] may be put intoone-to-one correspondence so that corresponding quotient groups are isomorphic. Thus the quotientgroups defined by (a) and (6) are isomorphic in some order, as required.


24. Which of the following sets form a group with respect to the indicated operation:

(o) S = {x: a; G /, a; < 0} ; addition

(6) S = {hx: a; e 7} ; addition

(c) S = {x : « G /, a; is odd} ; multiplication

(d) The n nth roots of 1; multiplication

(e) S = {-2, -1, 1, 2} ; multiplication

(/) S = {1, —1, i, —i) ; multiplication

(g) The set of residue classes modulo m; addition

(h) S - {[a] : [a] G //(m), (a,m) = 1} ; multiplication

(i) S = {z: z G C, (zl = 1} ; multiplication

Ans. (a), (c), (e), do not.

25. Show that the non-zero residue classes modulo p form a group with respect to multiplication if andonly if p is a prime.

26. Which of the following subsets of 7/(13) is a group with respect to multiplication: (a) {fll, fl21}-(6) {[1], [2], [4], [6], [8], [10], [12]}; (c) {[1], [5], [8], [12]} ? Ans. {a), {c).

^ ' U J.l \i,

27. Consider the rectangular coordinate system in space. Denote by a,b,crespectively clockwise rotations through 180° about the X^,K,Z-ax'isand by u its original position. Complete the adjacent table to showthat {u,a,b,e} is a group, the Klein 4-group.

28. Prove Theorem III, Page 83.

Hint. a-ôx = u has x = a and a; = (o-i)-i as solutions.

29. Prove Theorem IV, Page 83.'^"'''* ^'^

Hint. Consider (ao 6) o (6-1 oo-») = 00 (ft o ft-i) oa-i

30. Prove: Theorem V, Page 83.

31. Prove: «-•" = (o"")-!, mBI.

32. Complete the proof of Theorem VI, Page 83.

u a b c

u u a b c

a a

b b c

c e b a

98 GROUPS [CHAP. 9

33. Prove: Theorems IX, XI on Page 84 and Theorem XIV on Page 85.

34. Prove: Every subgroup &' of a group g has u, the identity element of g, as identity element.

35. List all of the proper subgroups of the additive group //(18).

36. Let S^ be a group with respect to ° and o be an arbitrary element of g. Show that

H = {x: x& g, xoa — a°x}is a subgroup of g.

37. Prove: Every proper subgroup of an abelian group is abelian. State the converse and show by an

example that it is false.

38. Prove: The order of a e ff is the order of the cyclic subgroup generated by a.

39. Find the order of each of the elements (a) (123), (b) (1432), (c) (12)(34) of S4.

Ana. (a) 3, (6) 4, (c) 2

40. Verify that the subset i4„ of all even permutations in S„ forms a subgroup of S„. Show that each

element of A„ leaves the polynomial of Problem 12, Chapter 2, Page 27, unchanged.

41. Show that the set {a; : a; £ /, 5|x) is a subgroup of the additive group /.

42. Form an operation table to discover whether {(1), (12)(34), (13)(24), (14)(23)} is a regular permutation

group on four symbols.

43. Determine the subset of S4 which leaves (a) the element 2 invariant, (6) the elements 2 and 4 invari-

ant, (c) x,a!2 + xgpci invariant, (rf) XxX2 + ^s + ^4 invariant.

Ans. (a) {(1), (13), (14), (34), (134), (143)} (c) {(1), (12), (34), (12)(34), (13)(24), (14)(23), (1423), (1324)}

(6) {(1),(13)} (d) {(1),(12),(34),(12)(34)}

44. Prove the second part of Theorem XV, Page 86. Hint. Use [m] <-> a-".

45. Show that the Klein 4-group is isomorphic to the subgroup P = {(1), (12)(34), (13)(24), (14)(23)} of S4.

46. Show that the group of Example 7 is isomorphic to the permutation group

P = {(1), (12)(35)(46), (14)(25)(36), (13)(26)(45), (156)(243), (165)(234)}

on six symbols.

47. Show that the non-zero elements 7/(13) under multiplication form a cyclic group isomorphic to

the additive group //(12). Find all isomorphisms between the two groups.

48. Prove: The only groups of order 4 are the cyclic group of order 4 and the Klein 4-group.

Hint, g - {u, a, h, c) either has an element of order 4 or all of its elements except u have order 2.

In the latter case, a°h ¥= a,b,u by the Cancellation Laws.

49. Let S be a subgroup of a group g and define T = {x: xGg, Sx = xS}. Prove that T is a sub-

group of g.

50. Prove: Two right cosets Ha and Hb of a subgroup H of a. group g are identical if and only if

51. Prove: a&Hb implies Ha = Hb where H is a subgroup of g and a,6 e g.

52. List all cosets of the subgroup {(1), (12)(34)} in the octic group.

53. Form the operation table for the symmetric group S3 on three symbols. List its proper subgroups

and obtain right and left cosets for each. Is S3 simple?

54. Obtain the symmetric group of Problem 53 using the properties of symmetry of an equilateral triangle.

55. Obtain the subgroup {u,p^,a^,T^ of S4 using the symmetry properties of a non-square rectangle.

56. Obtain the alternating group A4 of S4 using the symmetry properties of a regular tetrahedron.

57. Prove: Theorem XXV, Page 89.

CHAP. 9] GROUPS99

58. Show that X = {M,p2 2^} is an invariant subgroup of S,. Obtain S^/K and write out in full thehomomorphism S4 - SJK : x - Kx.

Partial answer. U -^ K, (12) ^ K{12), (13) ^ Jf(13), .... (24) -» «:(13), (34) -^ «-(12),

59. Use K = {upâ\r^, an invariant subgroup of S4 and H = {u,a^, an invariant subgroup of K,to show that a proper invariant subgroup of a proper invariant subgroup of a group g is not neces-sarily an invariant subgroup of g.

60. Prove: The additive group //(m) is a quotient group of the additive group /.

61. Prove: If H is an invariant subgroup of a group G, the quotient group g/H is cyclic if the index ofW m ff IS a prime.

62. Show that the mapping \a,p\y,S^ â defines a homomorphism of A4 onto g = {u,a,a^}. Note

that the subset of A4 which maps onto the identity element of g is an invariant subgroup of A4.

63. Prove: In a homomorphism of a group g onto a group g', let H be the set of all elements of gwhich map into u' G g'. Then the quotient group of g/H is isomorphic to g'.

64. Set up a homomorphism of the octic group onto {u,a).

65. When H = {u, a, a^ and K = {u, p, ^2} are subgroups of S4, show that HK ^ KH. Use HK and KHto verify: In general, the product of two subgroups of a group g is not a subgroup of g.

66. Prove: " « =; {'^1.^2. • -Â} and K = {h„b^ h^} are subgroups of a group g and one isinvariant, then (a) HK = KH, (b) HK is a subgroup of ^.

67. Prove: If H and K are invariant subgroups of g, so also is HK.

68. Let g, with group operation o and identity element u, and g', with group operation o and unity ele-ment u , be given groups and form

J = g X g' = {(g.g') : g&g, g'& g'}

Define the "product" of pairs of elements (g,g'), (h, h') & J by

(') (g,g'){h,h') ^ (g°h,g' ^ h')

(a) Show that J is a group under the operation defined in (i).

(6) Show that S = {(g.u') : g & g} and T = {{u.g') : g' e g'} are subgroups of J.

(c) Show that the mappings

S-^ g: (g.u')^ g and T ^ g' : (u,g')->g'are isomorphisms. v > w / v

69. For g and (?' of Problem 68, define U = {«} and U' = {«'}; also g = g x U' and c ' = t; v (?'Prove: 'n w .

(a) 6^ and g' are invariant subgroups of J.

(6) .//^ is isomorphic to Uxg', and J/^' is isomorphic to &' X [/'.

(c) ^ and g' have only (m, u') in common.

(d) Every element of g commutes with every element of g'.

(e) Every element of J can be expressed uniquely as the product of an element of g by an elementof g .

70. Show that S4, ^4, {«,Pâ2,r2}, {M,a2}, 17 is a composition series of S4. Find another in addition tothat of Example 13(6), Page 90.

^00GROUPS [CHAP. 9

71. For the cyclic group G of order 36 generated by a:

(i) Show that a^, a?, a*, o*, o^, a>'^, oi* generate invariant subgroups Qi^, Qx%, <S9< Cs. Qf Gs, Q% respec-

tively of Q.

(ii) g,gis,G9,G3.U is a composition series of g. There are six composition series of g in all;

list them.

72. Prove: Theorem XXXII, Page 91.

73. Write the operation table to show that Q = {l,-l,i.-i,},-},k,-k} satisfying i^ - p = fc2 = -1,

ij ^ fc = -ji, }k = i = -kj, ki - j - -ik forms a group.

74. Prove: A non-commutative group g, with group operation ° has at least six elements.

Hint. (1) g has at least 3 elements: u, the identity, and two non-commuting elements a and b.

(2) g has at least 5 elements: u,a,b,aob,boa. Suppose it had only 4. Then ao6#6oa

implies a°b or 6°o must equal some one of u,a,b.

(3) g has at least 6 elements: M,a,6,a ° 6,6 ° a, and either a^ or a°b° a.

75. Construct the operation tables for each of the non-commutative groups with 6 elements.

76. Consider S =^ {u,a,a^,a»,b,ab,a^b,a?b} with a* = m. Verify:

(a) If 52 = u, then either 60 = ab or 60 = a»6. Write the operation tables Ag, when 60 = ab,

and Dg, when 60 = a^b, of the resulting groups.

(6) If 62 = a or 6^ = a^, the resulting groups are isomorphic to Cg, the cyclic group of order 8.

(c) If 62 = a2, then either 60 = ab or 60 = 0^6. Write the operation tables K, when 6a = 06, and

Qs, when 60 = 0^6.

(d) A 8 and Ag are isomorphic.

(e) Dg is isomorphic to the octic group.

(/) Qg is isomorphic to the (quaternion) group Q of Problem 73.

(g) Qs has only one composition series.

77 Obtain another pair of composition series of the group of Problem 19; set up a one-to-one correspond-

ence between the quotient groups and write the mappings under which correspondmg quotient groups

are isomorphic.

chapter 10

Rings

(distributive laws)

RINGS

A non-empty set % is said to form a ring with respect to the binary operations addi-tion (+) and multiplication (•) provided, for arbitrary a,b,c G %, the following propertieshold:

Pi: {a + b) + c = a + {b + c) (associative law of addition)

P2: a + b = b + a (commutative law of addition)

P3: There exists zG% such that a + z - a.

(existence of an additive identity (zero))

P4: For each ae% there exists -a € "5^ such that a + (-a) = z.

(existence of additive inverses)

Ps: {a'b)'c = a-{b-c) (associative law of multiplication)

Per a{b + c) = a-b + tt'c

Pt: {b + c)a = b-a + caExample 1: Since the properties enumerated above are only a partial list of the properties com-

mon to I,Q,R, and C under ordinary addition and multiplication, it follows thatthese systems are examples of rings.

Example 2: The set S = {x + yy/J + z^: x,y,z & Q} is a ring with respect to addition andmultiplication on R. To prove this, we first show that S is closed with respect tothese operations. We have, for a + by/3 + cy/9, d+ey/3 + fy/d G S,

ia+by/B + ey/9) + (d + ey/3 + f^) = (a + d) + {b + e)^ + (c + f)y/9 G Sand

(a + 6^ + e^){d + ev/3 + fy/d) = {ad + Zbf + Zee) + (ae + hd + Zef)^

+ (af + 6e + ed)^/^ G SNext, we note that Pi,P2,P5-Pi hold since S is a subset of the ring B. Finally,

= + 0\/3 + 0\/9 satisfies P3, and for each x + y^+z^&S there exists-X - yy/3 - 2V9 G S which satisfies P4. Thus S has all of the required propertiesof a ring.

Example 3: (o) The set S = {a, 6} with addition and multiplication defined by the tables

+ a b

a a b

b b a

and a

a

a

b

is a ring.

(6) The set T = {a, b, e, d) with addition and multiplication defined by

and

+ a b c d

a a b e d

b b a d e

c e d a b

d d c b a

• a b c d

a a a a a

b a b a b

e a e a e

d a d a dis a ring.

101

102RINGS [CHAP. 10

For these rings the zero element is a and each element is its own additive inverse.


In Examples 1 and 2 the binary operations on the rings (the ring operations) coincide

with ordinary addition and multiplication on the various number systems mvolved; mExample 3 the ring operations have no meaning beyond the given tables. In this example

there can be no confusion in using familiar symbols to denote the ring operations. How-

ever, when there is a possibility of confusion, we shall use and O to indicate the ring

operations.

Example 4: Consider the set of rational numbers Q. Clearly addition (0) and multiplication (O)

defined by _, ,. xii»i,<=na 6 = a • b and o O b = a + 6 for all a,6 G «

where + and • are ordinary addition and multiplication on rational numbers, are

binary operations on Q. Now the fact that Pi, Pg, and P5 hold is immediate; also,

P3 holds with 2 = 1. We leave it for the reader to show that P4, Pg, and P7 do not

hold and so Q is not a ring with respect to and O .

PROPERTIES OF RINGS

The elementary properties of rings are analogous to those properties of / which do

not depend upon either the commutative law of multiplication or the existence of a

multiplicative identity element. We call attention here to some of these properties:

(i) Every ring is an abelian additive group.

(ii) There exists a unique additive identity element z, (the zero of the ring).

See Theorem III, Chapter 2, Page 20.

(iii) Each element has a unique additive inverse, (the negative of that element).

See Theorem IV, Chapter 2, Page 20.

(iv) The Cancellation Law for addition holds.

(v) -(-a) = a, -{a + b) = (-a) + (-b) for all a,b of the ring.

(vi) a-z = z-a = z (For a proof, see Problem 4.)

(vii) a{-b) = -(ab) = (-a)&

SUBRINGSLet 'R be a ring. A non-empty subset S of the set %, which is itself a ring with

respect to the binary operations on % is called a subnng of %. When S is a subring of

a ring %, it is evident that S is a subgroup of the additive group %.

Example 5: (a) From Example 1 it follows that / is a subring of the rings Q.R.C; that Q is a

subring of R, C; and fi is a subring of C.

(b) In Example 2, S is a subring of R.

(c) In Example 3(6), T, = {a}, T^ = {a, b} are subrings of T. Why is T^ = {a, b, c}

not a subring of T?

The subrings {z} and % itself of a ring Ii are called improper; other subrings, if any,

of % are called proper.

We leave for the reader the proof of

Theorem I. Let 9^ be a ring and S be a proper subset of the set %. Then S is a subring

of % if and only if

(a) S is closed with respect to the ring operations.

(b) for each a e S, we have -a G S.

CHAP. 10] RINGS 103

TYPES OF RINGS

A ring for which multiplication is commutative is called a commutative ring.

Example 6: The rings of Examples 1,2, 3(a) are commutative; the ring of Example 3(5) is non-commutative, i.e., b'c = a but ch = c.

A ring having a multiplicative identity element {unit element or unity) is called aring with identity element or ring with unity.

Example 7: For each of the rings of Examples 1 and 2, the unity is 1. The unity of the ringof Example 3(a) is 6; the ring of Example 3(6) has no unity.

Let <7i be a ring with unity u. Then u is its own multiplicative inverse (m-i = u), butother non-zero elements of % may or may not have multiplicative inverses. Multiplicativeinverses, when they exist, are always unique.

Example 8: (a) The ring of Problem 1 is a non-commutative ring without unity.

(6) The ring of Problem 2 is a commutative ring with unity u = h. Here the non-zero elements h,e,f have no multiplicative inverses; the inverses of e,d,g,h areg,d,e,h respectively.

(c) The ring of Problem 3 has as unity u = (1, 0, 0, 1). (Show this.) Since(1, 0, 1, 0)(0, 0, 0, 1) = (0, 0, 0, 0) while (0, 0, 0, 1)(1, 0, 1, 0) = (0, 0, 1, 0), the ring isnon-commutative. The existence of multiplicative inverses is discussed inProblem 5.

CHARACTERISTIC

Let ?( be a ring with zero element z and suppose that there exists a positive integer nsuch that mt = a + a + a+---+a ^ z for every ae%. The smallest such positiveinteger n is called the characteristic of -5^. If no such integer exists, "5^ is said to havecharacteristic zero.

Example 9: (a) The rings /, Q, R, C have characteristic zero since for these rings na = n'a.

(6) In Problem 1 we have a + a = b + b = = h + h = a, the zero of the ring,and the characteristic of the ring is two.

(c) The ring of Problem 2 has characteristic four.

DIVISORS OF ZEROLet g^ be a ring with zero element z. An element a 9^ z of -^ is called a divisor of zero

if there exists an element b ^ z of % such that a-b = z or b-a = z.

Example 10: (a) The rings I,Q,R,C have no divisors of zero, that is, in each system ab ="' implies a = or 6 = 0.

(6) For the ring of Problem 3, we have seen in Example 8(c) that (1, 0, 1, 0) and(0, 0, 0, 1) are divisors of zero.

(c) The ring of Problem 2 has divisors of zero since b • e = a. Find all divisorsof zero for this ring.

HOMOMORPHISMS AND ISOMORPHISMSA homomorphism (isomorphism) of the additive group of a ring % into (onto) the

additive group of a ring %' which also preserves the second operation, multiplication, iscalled a homomorphism (isomorphism) of % into (onto) %'.

Example 11: Consider the ring "5^ = {a,b,c,d} with addition and multiplication tables

104 RINGS [CHAP. 10

4- a b c d

o a b c d

b b a d c

c c d a b

d d c b a

• a h c d

a a a a a

b a b c d

c a c d b

d a d b c

and the ring %' = {p, q, r, s} with addition and multiplication tables

+ P 9 r s

p r 8 P 9

« s r 1 P

r P 9 r s

s 1 P s r

• P 1 r s

P s P r Q

1 P 1 r s

r r r r r

s 1 s r P

The one-to-one mapping

o<-*r, b <r^ q, c *^ s, d *^ p

carries % onto %' (also, %' onto %) and at the same time preserves all binary opera-

tions; for example,d = b + c*^g + s = p

b = c'd*-*s'p = q, etc.

Thus, % and %' are isomorphic rings.

Using the isomorphic rings '5^ and %' of Example 11, it is easy to verify

Theorem II. In any isomorphism of a ring % onto a ring %':

(a) if z is the zero of % and z' is the zero of %', we have 2 «-» «'.

(b) if %<^%': a<^ a', then -a <-* -a'.

(c) if u is the unity of % and W is the unity of %', we have u «-» u'.

(d) if "^ is a commutative ring, so also is %'.

IDEALSLet 'J^ be a ring with zero element z. A subgroup S of %, having the property

r-xGS (x-rGS) for all a; G S and r G %, is called a left (right) ideal in %. Clearly,

{z} and % itself are both left and right ideals in %; they are called improper left (right)

ideals in %. All other left (right) ideals in %, if any, are called proper.

A subgroup ^ of "5^, which is both a left and right ideal in %, that is, for &\l x G Sand rG% both r-xGj and x-rGj, is called an ideal (invariant subring) in "5^.

Clearly, every left (right) ideal in a commutative ring '5^ is an ideal in %.

For every ring %, the ideals {z} and % itself are called improper ideals in %; any other

ideals in % are called proper. A ring having no proper ideals is called a simple ring.

Example 12: (a) For the ring S of Problem 1, {a, b, e, d} is a proper right ideal in S (examine the

first four rows of the multiplication table) but not a left ideal (examine the first

four columns of the same table). The proper ideals in S are {a,e}, {a,e}, {a,g},

and {a,e,e,g}.

(6) In the commutative ring /, the subgroup P of all integral multiples of any

integer p is an ideal in /.

(c) For every fixed a,6GQ, the subgroup J = {{ar,br,as,bs) : r,8 G Q} is a

left ideal in the ring M of Problem 3 and K - {{ar,as, br, bs) : r.s e Q} is a

right ideal in M since, for every («t, n, p, q) G M,

(m,n,p,q)'(ar,br,as,bs) - (o(wir + ns), 6(«tr -I- ns), o(pr+ qs), 6(pr -I- qs)) G J

and

{ar, as, br, bs) • (m, n, p, q) - (a(mr + ps), a{nr + qs), b{mr + ps), b(nr + qs)) G K

CHAP. 10] RINGS 105

Example 12(6) illustrates

Theorem III. If p is an arbitrary element of a commutative ring %, then P =[p-r: re:%} is an ideal in %.


In Example 12(a), each element x of the left ideal {a,c,e,g} has the property that itis an element of S for which r-x = a, the zero element of S, for every r&S. Thisillustrates

Theorem IV. Let "T^ be a ring with zero element z; then

T = {x: x&%, r-x^z {x-r^z) tor&W r&%)is a left (right) ideal in %.

Let P, Q,S,T, . .. be any collection of ideals in a ring % and define S =

PnQnSnrn -. since each ideal of the collection is an abelian additive group, soalso, by Theorem X, Chapter 9, Page 84, is S. Moreover, for any a; e ^ and rG%, theproducts x-r and r-a; belong to each ideal of the collection and, hence, to S. We haveproved

Theorem V. The intersection of any collection of ideals in a ring is an ideal in the ring.


Theorem VL In any homomorphism of a ring % onto another ring %' the set S ofelements of % which are mapped on z', the zero element of %', is an idealin %.

Example 13: Consider the ring G = {a+hi: a,b G 1} of Problem 8.

(o) The set of residue classes modulo 2 of G is H = {[0], [1], [t], [i + a]}. (Note that1 - i = 1 + i (mod 2).) From the operation tables for addition and multiplica-tion modulo 2, it will be found that H is a. commutative ring with unity; also, Hhas divisors of zero although G does not.

The mapping G -* H : g ^ [g] is a homomorphism in which S = {2g: g& G},an ideal in G, is mapped on [0], the zero element of H.

(6) The set of residue classes modulo 3 of G is

K = {[0], [1], [x\, [2], [2t], [1 + 1], [2 + i], [1 + 2i], [2 + 2i]}

It can be shown as in (a) that K is a commutative ring with unity but is withoutdivisors of zero.

PRINCIPAL IDEALSLet %h&& ring and /f be a right ideal in % with the further property

K = {a-r: r&%,ais some fixed element of K)

We shall then call K a principal right ideal in % and say that it is generated by theelement a of K. Principal left ideals and principal ideals are defined analogously.

Example 14: (a) In the ring S of Problem 1, the subring {a,g} is a principal right ideal in Sgenerated by the element g (see the row of the multiplication table opposite g)Smce r-gâ for every rGS (see the column of the multiplication tableheaded g), {a,g} is not a principal left ideal and, hence, not a principal ideal in S.

(b) In the commutative ring S of Problem 2, the ideal {a,b,e,f} in S is a principalideal and may be thought of as generated by either 6 or /.

(c) In the ring S of Problem 1, the right ideal {a.b.c.d) in S is not a principalright ideal since it cannot be generated by any one of its elements.

(d) For any w 6 /, J - {mx : a; G /} is a principal ideal in /.

106 RINGS [CHAP. 10

In the ring /, consider the principal ideal K generated by the element 12. It is clear

that K is generated also by the element -12. Since K can be generated by no other of its

elements, let it be defined as the principal ideal generated by 12. The generator 12 of K,

besides being an element of K, is also an element of each of the principal ideals: A generated

by 6, B generated by 4, C generated by 3, D generated by 2, and / itself. Now K C A,

K C B, K C C, K C D, K C I; moreover, 12 is not contained in any other principal ideal

of /. Thus K is the intersection of all principal ideals in / in which 12 is an element.

It follows readily that any principal ideal in / generated by the integer m is contained

in every principal ideal in / generated by a factor of m. In particular, if to is a prime

the only principal ideal in / which properly contains the principal ideal generated by m is /.

Every ring % has at least one principal ideal, namely, the null ideal {z} where z is

the zero element of %. Every ring with unity has at least two principal ideals, namely, (z)

and the ideal % generated by the unity.

Let 'R be a commutative ring. If every ideal in 'R is a principal ideal, we shall call %a principal ideal ring. For example, consider any ideal J) ^ {0} in the ring of integers /.

If ajÔG S so also is -a. Then J) contains positive integers and, since /+ is well-

ordered, contains a least positive integer, say e. For any b G S, we have by the Division

Algorithm of Chapter 5, Page 50,

b = e-q + r, q.r&I, 0^r<e

Now e-q&S; hence, r = and b = e-q. Thus, ^ is a principal ideal in / and we

have provedThe ring / is a principal ideal ring.

PRIME AND MAXIMAL IDEALS

An ideal ^ in a commutative ring % is said to be a prime ideal if, for arbitrary elements

r, 8 of %, the fact that r-sGS implies either r&S or sGS.

Example 15: In the ring /,

(o) the ideal 7 = (7r : r £ /}, also written as J = (7), is a prime ideal since if

a'b&J either 7|a or 7 | 6; hence, either a G J or b e J.

(b) the ideal K = {14r : r e /} or K = (14) is not a prime ideal since, for example,

28 = 4 • 7 G X but neither 4 nor 7 is in K.

Example 15 illustrates

Theorem VII. In the ring / a proper ideal J = {mr : r e /, m ^ 0} is a prime ideal if

and only if m is a prime integer.

A proper ideal ciJ in a commutative ring % is called maximal if there exists no proper

ideal in % which properly contains J.

Example 16: (o) The ideal J of Example 15 is a maximal ideal in / since the only ideal in / which

properly contains / is / itself.

(6) The ideal K of Example 15 is not a maximal ideal since K is properly contained

in J which, in turn, is properly contained in /.

QUOTIENT RINGS

Since the additive group of a ring % is abelian, all of its subgroups are invariant

subgroups. Thus, any ideal J in the ring is an invariant subgroup of the additive group %and the quotient group %/J = {r + ^ : rG%} is the set of all distinct cosets of S in %.

CHAP. 10] RINGS 107

(Note. The use ot r + J instead of the familiar rj for a coset is in a sense unnecessarysince, by definition, rj = [roa: aGj} and the operation here is addition. Nevertheless,we shall use it.) In the section titled Quotient Groups of Chapter 9, Page 88, addition (+)on the cosets (of an additive group) was well-defined by

ix + J) + (y + ^) = {x + y) + ^

We now define multiplication (•) on the cosets by

ix + S)-iy + Sj = {x-y) + S

and establish that it too is well-defined. For this purpose suppose x' = x-\-s and y' = y + t

are elements of the additive group % such that x' + S and y' + S are other representationsof x + S and y + S respectively. From

X' + S = (x + s) + ^ = (x + S) + (s + J) =. x + Sit follows that s (and similarly t) G S. Then

(X' + S) '{y' + S) = (x''y') + S = [{x-y) + (x-t) + (s-y) + {s-t)] + J = (x-y) + S

since x-t, s-y, s-t G J, and multiplication is well-defined. (We have continued to calla; + J a coset; in ring theory, it is called a residue class of J in the ring %.)

Example 17: Consider the ideal J = {3r : r e /} of tlie ring / and the quotient group I/J =(o/> 1 + J,2 + J}. It is clear that the elements of I/^ are simply the residue classesof 1/(3) and, thus, constitute a ring with respect to addition and multiplicationmodulo 3.


Theorem VIII. If J is an ideal in a ring %, the quotient group 'J(^/S is a ring with respectto addition and multiplication of cosets (residue classes) as defined above.

It is customary to designate this ring by %/^ and to call it the quotient or factor ringof % relative to J.

From the definition of addition and multiplication of residue classes, it follows that

(c) The mapping % ^ %/cS : a->-a + ^ is a homomorphism of % onto %/J.

(b) S is the zero element of the ring %IS.

(c) If <?( is a commutative ring, so also is %IS.

(d) If % has a unity element u so also has %/3, namely, u-\- S-

{e) If % is without divisors of zero, %IS may or may not have divisors of zero. For, while

{a -\- S) • {h + S) = a-h + S = Sindicates a-h&S, it does not necessarily imply either a G ^ or 6 G ^.

EUCLIDEAN RINGSIn the next chapter we shall be concerned with various types of rings, for example,

commutative rings, rings with unity, rings without divisors of zero, commutative ringswith unity, . . . obtained by adding to the basic properties of a ring one or more otherproperties (see Page 71) of R. There are other types of rings and we end this chapterwith a brief study of one of them. By a Euclidean ring is meant:

Any commutative ring % having the property that to each xG% a, non-negativeinteger e{x) can be assigned such that

108 RINGS [CHAP. 10

(i) 0(x) = if and only if x = z, the zero element of %.

(ii) 0{x • y) ^ e{x) when x"y¥'Z.

(iii) For every x G% and y ¥= z e%,

X z:^ yq + r q,re%, 0ê{r)< 9{y)

Example 18: / is a Euclidean ring. This follows easily by using »(x) = \x\ for every « € /.

See also Problem 12.

There follow

Theorem IX. Every Euclidean ring <2?, is a principal ideal ring.

Theorem X. Every Euclidean ring has a unity.

Solved Problems

1. The set S = {a,b,c,d,e,f,g,h} with addition and multiplication defined by

+ a b c d e f 9 h

a a b e d e f g h

b b a d e f e h 9

c e d a b 9 h e f

d d c b a h 9 f e

e e f 9 h a b e d

f f e h 9 b a d e

9 a h e f e d a b

h h 9 f e d e b a

• a b e d e f 9 h

a a a a a a a a a

b a b a b a b a b

c a e a c a e a e

d a d a d a d a d

e a e a e a e a e

f a f a f a f a f

9 a 9 a 9 a 9 a 9

h a h a h a h a k

is a ring. The complete verification that Pi and Ps-Pt, Page 101, are satisfied is a

considerable chore but the reader is urged to do a bit of "spot checking". The zero

element is a and each element is its own additive inverse.

2. The set S of Problem 1 with addition and multiplication defined by

+ a b c d e f 9 h

a a b c d e f 9 h

b b a d e f e h 9

c e d e f 9 h a b

d d c f e h 9 b a

e e f 9 h a b e d

f f e h 9 b a d e

9 9 h a b e d e f

h h 9 b a d e f e

• a b c d e f g h

a a a a a a a a a

b a e f b a e f b

c a f d 9 e b h e

d a b 9 h e f c d

e a a e e a a e e

f a e b f a e b f

9 a f h e e b d g

h a b e d e f g h

is a ring. What is the zero element? Find the additive inverse of each element.

3. Prove: The set M - {(a,b, c,d): a,h,e,& G Q) with addition and multiplication

defined by(a,h,c,d) -y {e,f,g,}i) = (a + e, b + /, c + flf, d + î)

(a, h, c, d)(e, f, g, h) = {ae + bg, af + bh, ce + dg, cf + dh)

for all (a, b, c, d), (e, f, g,h)eM is a ring.

CHAP. 10] RINGS 109

The associative and commutative laws for ring addition are immediate consequences of the asso-ciative and commutative laws of addition on Q. The zero element of M is (0, 0, 0, 0) and the additiveinverse of (a, b, c, d) is (-o, -6, -c, -d) e M. The associative law for ring multiplication is verifiedas follows:

[(a,b,e,d){e,f,g,h)]{i,j,k,l)

= ({ae + bg)i + (af + bh)k, (ae + bg)j + (a/ + bh)l, (ce + dg)i + (cf + dh)k, (ce + dg)j + (c/ + dh)l)

= {a(ei + fk) + b{gi + hk), a(ej + fl) + b{gj + hi), c{ei + fk) + d{gi + hk), c(ej + fl) + d(gj + hi))

= (a, 6, c, d)(ei + fk, ej + fl, gi + hk, gj + hi)

= (a,b,c,d)[(e,f,g,h)(i,i,k,l)]

for all (a, b, c, d), (e, /, g, h), (i, j, k, I) G M.

The computations required to verify the distributive laws will be left for the reader.

4. Prove: If "5^ is a ring with zero element z, then for all aeL%, a-z = z-a = z.

Since a + z = a, it follows that

a' a = (a + z)a — (a- a) + z' a

Now a' a = (a-a) + z; hence, (a'a) + z-a = (a-a) + z. Then, using the Cancellation Law, wehave z'a = z. Similarly, o • a = a{a + z) = a- a + a- z and a' z = z.

5. Investigate the possibility of multiplicative inverses of elements of the ring M ofProblem 3.

For any element (a, b, c, d) ¥- (0, 0, 0, 0) of M, set

{a,b,c,d)(,'p,q,r,s) - {ap + br, aq + bs, cp + dr, cq + ds) = (1,0,0,1)

the unity of M, and examine the equations

(i)hP +br = 1 Uq + bs =[ep + dr = [cq + ds = I

for solutions p, q, r, s.

dFrom (i), we have {ad — bc)p = d; thus, provided ad~bc ¥= 0, p = -—: r- and r = —

c

_5 jj

' ad— be ad — be

'

J—;— and s = —^—;— . We i

t — be ad— be(a, b, c,d) e M for which ad - 6c t^ have multiplicative inverses.

Similarly, from (ii), we find q =^^_^^ and s =

^^ _ ^. We conclude that only those elements

Shovi^ that P = {{a,b,-b,a): a,bGl} with addition and multiplication defined by

(a, b, -b, a) + (c, d, -d, c) = {a + c,b + d, -b-d,a + c)

and {a,b,-b,a){c,d,-d,c) = {ac-bd, ad + bc, -ad-bc,ac-bd)is a commutative subring of the non-commutative ring M of Problem 3.

First, we note that P is a subset of M and that the operations defined on P are precisely thosedefined on M. Now P is closed with respect to these operations; moreover, {—a, —b, b, —a) G Pwhenever {a, b, -b, a) G P. Thus, by Theorem I, P is a subring of M. Finally, for arbitrary(a, b, —b, a), (c, d, —d, c) G P we have

{a,b,-b,a){e,d,-d,e) = (e,d,-d,e)(a,b,—b,a)

and P is a commutative ring.

Consider the mapping (a, b, -b, a)-â of the ring P of Problem 6 into the ring / ofintegers.

The reader will show that the mapping carries

(a, b, —b, a) + {e, d, —d, c) -» a + e

and (a, 6, —6, a) • (c, d, —d, e) -* ac-bd

Now the additive groups of P and / are homomorphic. (Why not isomorphic?) However, sinceae — bd ¥- ae generally, the rings P and / are not homomorphic under this mapping.

110 RINGS [CHAP. 10

8. A complex number a + hi, where a,b G I, is called a Gaussian integer. (In Problem 26,

the reader is asked to show that the set G = {a + bi: a,b Gl} of all Gaussian integers

is a ring with respect to ordinary addition and multiplication on C) Show that the

ring P of Problem 6 and G are isomorphic.

Consider the mapping (o, 6, —6, a) -* a + bi of P into G. The mapping is clearly one-to-one;

moreover since

{a,b,—b,a) + (c,d,—d,c) = {a + c, b + d, —b — d, a+ e)->

(a + c) + (b + d)i = (a+bi) + {c + di)

and (a, b, —b, o)(c, d, —d, c) = (ae — bd, ad + be, —ad — be, ae — bd) -^

(ac — bd) + {ad + bc)i = (a+bi){c + di}

all binary operations are preserved. Thus P and G are isomorphic.

9. Prove: If p is an arbitrary element of a commutative ring %, then P = {p-r: r E%}is an ideal in %.

We are to prove that P is a subgroup of the additive group % such that (p ' r)s & P for all s & %.

For all r,s & %, we have

(i) p'r + p' 8 = p(r + s) G P, since r + sG%; thus P is closed with respect to addition.

(ii) —(p • r) = p(—r) € P whenever p'r & P, since —r&% whenever r G '^; by Theorem VII, Chap-

ter 9, Page 84, P is a subgroup of the additive group.

(iii) (p • r)8 = p(r • s) G P since {r's)G %.

The proof is complete.

10. Prove: In any homomorphism of a ring % with multiplication denoted by • , into

another ring '5^' with multiplication denoted by ^ , the set S of elements of % which

are mapped on z', the zero element of %', is an ideal in %.

By Theorem XXI, Chapter 9, Page 88, S is a subgroup of %'; hence, for arbitrary a,b,e B S,

Properties P1-P4, Page 101, hold and ring addition is a binary operation on S.

Since all elements of S are elements of %, Properties P5-P7 hold. Now for all 0,6 e S, o« 6 -> z';

hence, a' b & S and ring multiplication is a binary operation on S.

Finally, for every aG S and g G%, we have

a •fir

-* 2' Dfir' = z' and g ' a -* g' <^ z' — z'

Thus S is an ideal in %.

11. Prove: The set %/S - {r + JI: rG%} of the cosets of an ideal c5? in a ring % is

itself a ring with respect to addition and multiplication defined by

{x + J) + (y + J) = (x + y) + JI ^,, ^ ^ ^,^

and forsAl x + ^,y + S & RI3{x + S)'{y + S) = {x-y) + S

Since S J* »" invariant subgroup of the group %, it follows that %IS is a group with respect to

addition. It is clear from the definition of multiplication that closure is assured. There remains

then to show that the Associative Law and the Distributive Laws hold. We find for all w-'r ^,

[(w + S)' (•« + 3)]' (V + S) = (<»'<« + S)'(y + S) = {.wx)'V+ S - w{x-y) + J= {w + J)-(x'y + J) = (w + J)'[{x + S)-(y + S)].

(w + S)- [{x + 3 + (y + J)] = iw + J)- i(x + y) + J] = [w'{x + v)]+ S= {wx + wy) + ^ — (wx + ^) + (wy + ^)

= (w + s)' ('^ + S) + iv> + S)- (y + S)and, in a similar manner,

[(« + J) + (3/ + 3] • (w + ^) = (X + S)'^^ + S) + (v + ^-iw + S)

CHAP. 10] RINGS 111

12. Prove: The ring G — {a + bi: a,b G 1} is a Euclidean ring.

Define 0{a + /3t) = tt^ + p^ for every a + piG G. It is easily verified that the properties (i) and(ii), Page 108, for a Euclidean ring hold. (Note also that e(a + bi) is simply the square of the ampli-

tude of a + bi and, hence, is defined for all elements of C.)

For every x&G and y¥'z<^G, compute x • y^'^ = s + ti. Now if every « + tt G G, the theorem

would follow readily; however, this is not the case as the reader will show by taking x = 1 + 1 and

J/= 2 + 3t.

Suppose then for a given x and y that s-'rti^G. Let e-Vdi^G be such that \c — b\ —-J

and\d—i\ — ^, and write x = y(e + di) + r. Then

#(r) = «[x — y(c + di)] = e\x — y{s + ti) + y{s + ti) — y(e + di)]

= 0[y{{s-e) + (t-d)i}] ^ ê{y) < e(y)

Thus (iii) holds and G is a Euclidean ring.


13. Show that S = {2x : x B 1} with addition and multiplication as defined on / is a ring while

r = {2« + 1 : X G /} is not.

14. Verify that S of Problem 2 is a commutative ring with unity = h.

15. When a,b G I define a® b = a+ b + 1 and o06 = o+6 + a6. Show that / is a commutative ring

with respect to © and O . What is the zero of the ring? Does it have a unit element?

16. Verify that S = {o, 6, c, d, e, f, g) with addition and multiplication defined by

+ a b c d e f 9

a a b c d e f 9

b b c d e f 9 a

c c d e f 9 a b

d d e f 9 a b e

e e f 9 a b c d

f f 9 a b e d e

g a b e d e f

• a b c d e f 9

a a a a a a a a

b a b c d e f 9

c a e e 9 b d f

d a d a e f b e

e a e b f e 9 d

f a f d b 9 e c

9 a 9 f e d e b

is a ring. What is its unity? its characteristic? Does it have divisors of zero? Is it a simple ring?

Show that it is isomorphic to the ring 7/(7).

17. Show that Q = {(zi, Zj, —z^, Zi) : z^, z^ G C} with addition and multiplication defined as in Problem 3

is a non-commutative ring with unity (1, 0, 0, 1). Verify that every element of Q with the exception

of the zero element (zj = Z2 = + Oi) has an inverse in the form {zjM, —Z2/^, z^^, ZiM}, where

A — |zip + \z^, and thus the non-zero elements of Q form a multiplicative group.

18. Prove: In any ring %,

(o) —(—a) = a for every o G "^

(6) a(-6) = -(a6) = (-0)6 for all 0,6 G <!^.

Hint, (a) a + [{—a) — (—a)] = a + z = a

19. Consider "3^, the set of all subsets of a gfiven set S and define, for all A,B G %,

A @ B = Av B - AnB AQB = AnBShow that <^ is a commutative ring with unity.

112 RINGS [CHAP. 10

20. Show that S = {{a,b,—b,a): a,b G Q} with addition and multiplication defined as in Problem 6 is

a ring. What is its zero? its unity? Is it a commutative ring? Follow through as in Problem 6 to

show that every element except (0, 0, 0, 0) has a multiplicative inverse.

21. Complete the operation tables for the ring % = {a,b,c,d}:

+ a b e d

a a b e d

b b a d e

c c d a b

d d e b a

• a b e d

a a a a a

b a b

e a a

d a b e

Is <^ a commutative ring? Does it have a unity? What is its characteristic?

Hint. c«6 = (6 + d)«J>; cc = c-(b + d); etc.

22. Complete the operation tables for the ring 'B = {a,b,c,d}:

+ a b e d

a a b c d

b b a d e

c c d a b

d d e b a

• a b c d

a a a a a

b a b

c a c

d a b c

Is 'B a commutative ring? Does it have a unity? What is its characteristic? Verify that x^ = xfor every « € 'B. A ring having this property is called a Boolean ring.

23. Prove: If 'B is a Boolean ring, then (a) it has characteristic two, (b) it is a commutative ring.

Hint. Consider (x + y)^ — x + y when y — x and when y ¥= x.

24. Let ^ be a ring with unity and let a and 5 be elements of %, with multiplicative inverses o~* and6-1 respectively. Show that (a'5)-i = 6-io-i.

25. Show that {o}, {o, 6}, {o, 6, c, d} are subrings of the ring S of Problem 1.

26. Show that G = {a+ bi: a,b e 1} with respect to addition and multiplication defined on C is asubring of the ring C.

27. Prove Theorem I, Page 102.

28. (o) Verify that % - {(zj, z^, Z3, z^) : z„ z^, Zg, z^ G C} with addition and multiplication defined as in

Problem 3 is a ring with unity (1, 0, 0, 1). Is it a commutative ring?

(6) Show that the subset S = {(zi, Zg, —Z2, Zj) : z^, z^ G C} of "^ with addition and multiplication

defined as on ^ is a subring of %^.

29. List all 15 subrings of S of Problem 1.

30. Prove: Every subring of a ring '^ is a subgroup of the additive group %.

31. Prove: A subset S of a ring "3^ is a subring of % provided 0—6 and a • 6 G S whenever a,6 G S.

32. Verify that the set I/(n) of integers modulo w is a commutative ring with unity. When is the ringwithout divisors of zero? What is the characteristic of the ring //(5)? of the ring 7/(6)?

CHAP. 10] RINGS 113

33. Show that the ring 11(2) is isomorphic to the ring of Example 3(a).

34. Prove Theorem II, Page 104.

35. (a) Show that Af j = {(a, 0, c, d): a,c,dS Q} and Mj = {(a, 0, 0, d) : a,de.Q} with addition andmultiplication defined as in Problem 3 are subrings of M of Problem 3.

(6) Show that the mapping Af j- Mj : (x, 0, y, w) -» (x, 0, 0, w)

is a homomorphism.

(c) Show that the subset {(0, 0,y,0): y & Q) of elements of Af j which in (6) are mapped into

(0,0,0,0) G Afj is a proper ideal in Af,.

(d) Find a homomorphism of Af, into another of its subrings and, as in (c), obtain another properideal in Afj.

36. Prove: In any homomorphism of a ring % onto a ring %', having z' as identity element, let

JJ = {x: xe%, x-*z'}

Then the ring %IS is isomorphic to %'.

Hint. Consider the mapping a + î?-^ a' where o' is the image of a&% in the homomorphism.

37. Let a, b be commutative elements of a ring % of characteristic two. Show that (a + 6)* — o* + 6* =(o-6)2.

38. Let <^ be a ring with ring operations + and • ; let (a, r), (6, a) e %xl. Show that

(i) ^ X / is closed with respect to addition (®) and multiplication (O) defined by

(o,r)©(6,«) = (a + b.r + a)

(a, r) Q (b,s) — {a-b + rb + sa, rs)

(ii) ^ X / has (z, 0) as zero element and (z, 1) as unity,

(iii) "^ X / is a ring with respect to © and O

.

(iv) "3^ X {0} is an ideal in %XI.

(v) The mapping %*r*%x{0}: » <-» (ac, 0) is an isomorphism.

39. Prove Theorem IX, Page 108. Hint. For any ideal J ¥- {z} in %, select the least e{y), say e(b),

for all non-zero elements y ^ S- For every x & S write x — 6 • g + r with q,r & ^ and either r — z

or #(r) < e(b).

40. Prove Theorem X, Page 108. Hint. Suppose % is generated by o; then o = o • s = g • o for some8^%. For any b &%, b = q'a = q(a ' s) = b • s, etc.

chapter 11

Integral Domains, Division Rings, Fields

INTEGRAL DOMAINSA commutative ring 2D, with unity and having no divisors of zero, is called an integral

domain.Example 1 : (a) The rings /, Q, R, and C are integrral domains.

(6) The rings of Problems 1 and 2, Chapter 10, Page 108, are not integral domains;

in each, for example, f • e = a, the zero element of the ring.

(c) The set S = {r + sy/Vl: r,8 S /} with addition and multiplication defined as

on R is an integral domain. That S is closed with respect to addition and multi-

plication is shown by

{a + bVrJ) + (c + dy/ri) = {a + c) + {b + d)y/rf £ S

(a + by/rf)(c + dy/l7) = (ae + llbd) + (ad + hc)y/r! e S

for all {a + bVvi),{e + dy/Tf)eS. Since S is a subset of R, S is without

divisors of zero; also, the associative laws, commutative laws and distributive

laws hold. The zero element of S is G jB and every a + by/Vj G S has an

additive inverse —a — by/Tl € S. Thus, S is an integral domain.

(d) The ring S = {a, b, e, d, e, f, g, h) with addition and multiplication defined by

+ a b c d e f g h

a a b c d e f g h

b b a d c f e h

e c d a b g h e f

d d c b a h g / e

e e f h a b c d

f f e h 9 b a d e

g a h e f e d a b

h h a f e d e b a

• a b c d e f g h

a a a a a a a a a

b a b c d e f g h

e a e h f g e b d

d a d f 9 e b h e

e a e a e d h f b

f a f e b h c d g

g a g b h f d e e

h a h d e b g c f

Table 11-1

is an integral domain. Note that the non-zero elements of S form an abelian

multiplicative g^roup. We shall see later that this is a common property of all

finite integfral domains.

A word of caution is necessary here. The term integral domain is used by some to

denote any ring without divisors of zero and by others to denote any commutative ring with-

out divisors of zero. ^^ Problem 1.

The Cancellation Law for Addition holds in every integral domain ^ since every element

of 2) has an additive inverse. In Problem 2 we show that the Cancellation Law for Multi-

plication also holds in ^ in spite of the fact that the non-zero elements of ^ do not neces-

sarily have multiplicative inverses. As a result "having no divisors of zero" in the

definition of an integral domain may be replaced by "for which the Cancellation Law for

Multiplication holds".


Theorem I. Let .2> be an integral domain and ci? be an ideal in ^. Then :D/Jj is an integral

domain if and only if ^ is a prime ideal in J).

114

CHAP. 11] INTEGRAL DOMAINS, DIVISION RINGS, FIELDS 115

UNIT, ASSOCIATE, DIVISOR

Let ^ be an integral domain. An element v of .2) having a multiplicative inverse in 3) is

called a unit {regular element) of .2). An element 6 of ^ is called an associate oi a G J)

if b — va where v is some unit of JD.

Example 2: (o) The only units of / are ±1; the only associates of o G / are ±a.

(6) Consider the integral domain JD - {r + 8\/l7 : r,« G /}. Now a = a + 6\/l7 S îs a unit if and only if there exists x + yy/vf G ^ such that

{a + b^/vj){x + y^/l^) = (ax + lUy) + {bx + ay)y/ri = 1 = 1 + Oy/vf

{ax + :

\bx + t

From < we obtain x ^ and v ^ Nowi»--ôj/ =0 *"'""""" •* o2_i752 »"° y a2_i752- "°^

X + yy/vi G JZ), i.e., a!,y G /, if and only if o* - 1762 = ±i; hence, a is a unit

if and only if a^ - 1762 = ±1. Thus, ±1, 4 ± y/VJ, -4 ± y/vf are units in ^while 2 - Vl7 and -9 - 2^17 = (2 - Vl7 )(4 + ^/l7 ) are associates in JZ).

(c) Every non-zero element of 7/(7) - {0, 1, 2, 3, 4, 6, 6} is a unit of 7/(7) since1*1 = l(mod 7), 2 • 4 = l(mod 7), etc.

See Problem 4.

An element a of .0 is a divisor at b G 3) provided there exists an element c ot 3)

such that b = a'C. Every non-zero element b of 3) has as divisors its associates in J) andthe units of 3). These divisors are called trivial (improper); all other divisors, if any, arecalled non-trivial (proper). A non-zero, non-unit element b of 3), having only trivial divisors,is called a prime (irreducible element) of 3). An element 6 of 3), having non-trivial divisors,is called a reducible element of 3). For example, 15 has non-trivial divisors over / but notover Q; 7 is a prime in / but not in Q.

See Problem 5.

There follows

Theorem II. If 3) is an integral domain which is also a Euclidean ring then, for a ^ «,

b^-z ot 3),

e(a ' b) = 0(a) if and only if 6 is a unit of 3)

SUBDOMAINSA subset 3>' of an integral domain 3), which is itself an integral domain with respect

to the ring operations of 3), is called a subdomain of 3). It will be left for the reader toshow that z and u, the zero and unity elements of 3), are also the zero and unity elementsof any subdomain of 3).

One of the more interesting subdomains of an integral domain 3) (see Problem 6) is

3)' = {nu: nGl}

where nu has the same meaning as in Chapter 10. For, if 3)" be any other subdomain of 3),

then 3)' is a subdomain of <Z)" and hence, in the sense of inclusion, 3)' is the least subdomainin 3). Thus,

Theorem III. If 3) is an integral domain, the subset 3)' = {nu : nGl} is its least sub-domain.

By the characteristic of an integral domain 3) we shall mean the characteristic, asdefined in Chapter 10, of the ring 3). The integral domains of Example 1(a) are then ofcharacteristic zero while that of Example 1(d) has characteristic two. In Problem 7,we prove

Theorem IV. The characteristic of an integral domain is either zero or a prime.

116 INTEGRAL DOMAINS, DIVISION RINGS, FIELDS [CHAP. 11

Let X> be an integral domain having £D' as its least subdomain and consider the mapping

/ -» <Z)' : n -» nM

If 3) is of characteristic zero, the mapping is an isomorphism of / onto X)'\ hence, in 3) we

may always replace 3)' by /. If 3) is of characteristic p (a prime), the mapping

//(p) ^ 3)': [n] -^ nu

is an isomorphism of I/{p) onto 3)'.

ORDERED INTEGRAL DOMAINSAn integral domain 3) which contains a subset 3)-^ having the properties:

(i) ^+ is closed with respect to addition and multiplication as defined on 3),

(ii) for every aG 3), one and only one of

a = z aG 3)+ -aG 3)+

holds,

is called an ordered integral domain. The elements of ^+ are called the positive elements

of D; all other non-zero elements of 3) are called negative elements of =».

Example 3: The integral domains of Example 1(a) are ordered integral domains. In each, the

set 3)+ consists of the positive elements as defined in the chapter in which the

domain was first considered.

Let 3) be an ordered integral domain and, for all a,b G 3), define

a>b when a-bG3)+

and a a

Since a>z means a G 3)+ and a<z means -a e iZ) + , it follows that, if a¥=z,then

a^G3)+. In particular, uG3)^.

Suppose now that 3) is an ordered integral domain with 3)+ well-ordered; then u is the

least element of ^+. For, should there exist a G 3)+ with z<a<u, then z<a^<au = a.

Now a^ G 3)-^ so that 3)+ has no least element, a contradiction.


Theorem V. If 3) is an ordered integral domain with =Z)+ well-ordered, then

(i) 3)* = {pu: pGl+}

(ii) 3) = [mu : m G 1}

Moreover, the representation of any aG 3) as a = mu is unique.

There follow

Theorem VI. Two ordered integral domains î and 3)^, whose respective sets of positive

elements X)^ and 3)^ are well-ordered, are isomorphic.

and

Theorem VII. Apart from notation, the ring of integers / is the only ordered integral

domain whose set of positive elements is well-ordered.

DIVISION ALGORITHM

Let ^ be an integral domain and suppose dG 3) is a common divisor of the non-zero

elements a,b G 3). We call d a greatest common divisor of a and b provided for any


other common divisor d' G J), we have d' \ d. When JD is also a Euclidean ring, d'\d is

equivalent to 0{d) > 6(d').

(To show that this definition conforms with that of the greatest common divisor of twointegers as given in Chapter 5, suppose ±d are the greatest common divisors of a,b G I

and let d' be any other common divisor. Since for n G /, e(n) = \n\, it follows thate{d) = e(-d) while e{d) > d(d').)

We state for an integral domain which is also a Euclidean ring

The Division Algorithm. Let a^ z and b be in ^, an integral domain which is alsoa Euclidean ring. There exist unique q,r G H) such that

h = q-a + r, 0^ d{r) < e{q)

See Problem 5, Chapter 5.

UNIQUE FACTORIZATIONIn Chapter 5 it was shown that every integer a > 1 can be expressed uniquely (except

for order of the factors) as a product of positive primes. Suppose a = pi • p2 • Ps is sucha factorization. Then

-a = -P1-P2-P3 - Pl(-P2)P3 = Pl-P2(-P3) = (-1)P1-P2-P3 = (-l)Pl * (-1)P2 ' (-1)P3

and this factorization in primes can be considered unique up to the use of unit elementsas factors. We may then restate the unique factorization theorem for integers as follows:

Any non-zero, non-unit element of / can be expressed uniquely (up to the order offactors and the use of unit elements as factors) as the product of prime elements of /.

In this form we shall show later that the unique factorization theorem holds in anyintegral domain which is also a Euclidean ring.


Theorem VIII. Let J and K, each distinct from {«}, be principal ideals in an integral do-main 2). Then J = K if and only if their generators are associate ele-

ments in 3).


Theorem IX. Let a,b,p G X), an integral domain which is also a principal ideal ring,

such that pI

tt'b. Then if p is a prime element in >2), p |a or p |

6.

A proof that the unique factorization theorem holds in an integral domain which is also

a Euclidean ring (sometimes called a Euclidean domain) is given in Problem 11.

As a consequence of Theorem IX, we have

Theorem X. In an integral domain 3) in which the unique factorization theorem holds,

every prime element in X) generates a prime ideal.

DIVISION RINGS

A ring d", whose non-zero elements form a multiplicative group, is called a division ring(skew field or sfield). Every division ring has a unity and each of its non-zero elements hasa multiplicative inverse. Multiplication, however, is not necessarily commutative.

Example 4: (a) The rings Q, R, and C are division rings. Since multiplication is commutative,they are examples of commutative division rings.

(6) The ring Q of Problem 17, Chapter 10, is a non-commutative division ring.

(c) The ring / is not a division ring. (Why?)

llg INTEGRAL DOMAINS, DIVISION RINGS, FIELDS [CHAP. 11

Let ^ be an integral domain having a finite number of elements. For any b ¥- z G ^,

since otherwise b would be a divisor of zero. Thus, b-xû for some x G ^D and b has

a multiplicative inverse in £>. We have proved

Theorem XI. Every integral domain, having a finite number of elements, is a commuta-

tive division ring.

We now prove

Theorem XII. Every division ring is a simple ring.

For, suppose ^ ¥= {z} is an ideal of a division ring el". If a # z G J", we have a"* G of

and wa-^-uGj. Then for every bGef, b-u-bGS; hence, J = ef.

FIELDS

A ring f whose non-zero elements form an abelian multiplicative group is called a field.

Example 5: (a) The rings, Q, R, and C are fields.

(6) The ring S of Example 1(d) is a field.

(c) The ring M of Problem 3, Chapter 10, Page 108, is not a field.

Every field is an integral domain; hence, from Theorem IV, Page 115, there follows

Theorem XIII. The characteristic of a field is either zero or is a prime.

Since every commutative division ring is a field, we have (see Theorem XI)

Theorem XIV. Every integral domain having a finite number of elements is a field.

Any subset f of a field f, which is itself a field, is called a subfield of f.

Example 6: Q is a subfield of the fields R and C; also, B is a subfield of C.


Let 7^ be a field of characteristic zero. Its least subdomain, /, is not a subfield. How-

ever, for each 6^0, be/, we have b"' e f ; hence, for all a,b G I with b ^ 0, it follows

that a'b-^ = a/bGf. Thus, Q is the least subfield of f. Let f be a field of character-

istic p, a prime. Then I/(p), the least subdomain of f is the least subfield of f.

A field f which has no proper subfield f is called a prime field. Thus Q is the prime

field of characteristic zero and //(p) is the prime field of characteristic p, where p is a

prime.

We state without proof

Theorem XV. Let 7^ be a prime field. If f has characteristic zero, it is isomorphic to Q;

If f has characteristic p, a prime, it is isomorphic to I/(p).


Theorem XVI. Let J) be an integral domain and J an ideal in ^. Then :D/S is a field

if and only if J7 is a maximal ideal in J).


Solved Problems

1. Prove: The ring Il(m) is an integral domain if and only if m is a prime.

Suppose TO is a prime p. If [r] and [s] are elements of //(p) such that [r] • [s] = [0] , thenr-8 = 0(modp) and r = 0(modp) or 8 = 0(modp). Hence, [r] = [0] or [s] = [0]; and //(p), havingno divisors of zero, is an integral domain.

Suppose m is not a prime, that is, suppose m = m^- m^ with 1 < m,,?n2 < m. Since [m] —[mj] • [mg] = [0] while neither [m,] = nor [wg] = [0], it is evident that I/(m) has divisors of zero

and, hence, is not an integral domain.

2. Prove: For every integral domain the Cancellation Law of Multiplication

If a'c = b' c and c ¥= z, then a — hholds.

From o • c = 5 • c we have a-c— 6-c = {a—h)'c = z. Now D has no divisors of zero; hence,

a — b = z and a = 6 as required.

3. Prove: Let .2) be an integral domain and J be an ideal in ^. Then ^/J is an integral

domain if and only if c5 is a prime ideal in .3).

The case ^^ = JD is trivial; we consider ^ G J).

Suppose<i? is a prime ideal in ^. Since JZ) is a commutative ring with unity, so also is JD/^.

To show that ^/J is without divisors of zero, assume a+ J,b + ^ G ^/J such that

(a + S,{b + J) = a-b + ^ = SNow a'b & ^ and, by definition of a prime ideal, either o G ^ or b G ^. Thus either a + ^ or

6 + <5 is the zero element in ID/J; and =Z)/<5. being without divisors of zero, is an integral domain.

Conversely, suppose JD/J is an integral domain. Let a ¥= z and b ¥^ z of ^ be such that

a'b e. Jj. FromJ z= a-b + J = ia + Mb + S)

it follows that a + J = J or b + S = S- Thus a-b& S implies either a e ^^ or b & S> and Sis a prime ideal in J).

Note. Although 2) is free of divisors of zero, this property has not been used in the above proof.

Thus, in the theorem, "Let 3) be an integral domain" may be replaced by "Let "^ be a commutativering with unity".

4. Let t be some positive integer which is not a perfect square and consider the integral

domain 3) = {r-\-s^: r,s G/}. For each p - r + s\/t G 3), define p = r-sy/t andthe norm of p as N(p) = p-p. From Example 2(6), Page 115, we infer that p = a-{^ hy/t

is a unit of 3) if and only if iV(p) = ±1. Show that- for a = a + h\/t G 3) and

i8= c + dsft e O), N{a'l3) = N(a)-N(fi).

We have a'P = {ac + bdt) + (ad + bc}-/i and a^8 = (ae + bdt) - (ad + be)-/t. Then N(a'P) ={a • j8)(a • /8) = {ae+ bdt)^ - (ad+ bc)H = (a^ - bH){c^ - d?t) = N{a) • N{p), as required.

5. In the integral domain 3) = {r + sy/Tf: r.sGl}, verify: (o) 9 - 2vT7 is a prime,

(6) y = 15 + 7\/i7 is reducible.

(a) Suppose a.p £ 3) such that a • /S = 9 — 2^17- By Problem 4,

N{a-p) = N(a)-N(p) = iV(9-2Vl7) = 13

Since 13 is a prime integrer, it divides either N(a) or N{p); hence either j3 or a is a unit of 3), and

9 — 2V17 is a prime.

(6) Suppose a-a+ by/vi, p = e + dyfvj G 3) such that a • ;8 = y = 15 + 7\/l7; then N{,a) • N{p) =-608. From N(a) = a^- 1762 = 19 and iV(/3) = o^ - 17(P = -32, we obtain a = 6 - -y/l7 and

)3 = 11 + 3\/l7. Since a and p are neither units of 3) nor associates of 7, 15 + lyfvj is reducible.


6. Show that CD' — {nu: n G /}, where u is the unity of an integral domain .2), is a sub-

domain of <2).

For every ru, su € jZ)', we have

ru + su = (r + 8)m e ^' and {ru){su) = rsu G ^'

Hence ^' is closed with respect to the ring operations on J2). Also,

Om = « e ^' and 1m = u e ID'

and for each rw G ^' there exists an additive inverse —ru G J)'. Finally, {ru)(su) = z implies

ru = z or 8U — z. Thus JD' is an integral domain, a subdomain of ^.

7. Prove: The characteristic of an integral domain .2) is either zero or a prime.

From Examples l(o) and 1(d) it is evident that there exist integral domains of characteristic zero

and integral domains of characteristic m > 0.

Suppose j2) has characteristic m = mj • wi2 with I < mi,m2 < m. Then mu = {miu)(,m2u) = z and

either rriiu = « or mû = z, a, contradiction. Thus m is a prime.

8. Prove: If -2) is an ordered integral domain such that .2)+ is well-ordered, then

(i) :d+ = (pm: pG/+} (ii) O) = {mu: m G /)

Moreover, the representation of any aG. 3) as a = mu is unique.

Since m G ^+ it follows by the closure property that 2u — u + uS ^+ and, by induction, that

puG^+ for all pG/+. Denote by E the set of all elements oi ^+ not included in the set

{pu : p G /+} and by e the least element of E. Now u ^ E so that e> u and, hence, e — uG ^ +

but e-uÊ. (Why?) Then e

-

m = pjU, for some PiGl*, and e - u + piU = (1 + Pi)u = paM,

where pj G /+. But this is a contradiction; hence E = 0, and (i) is established.

Suppose aGJ) but oS,2) +; then either a = z or —aG^ + . If a = «, then o = Om.

If -aG^+ then, by (i), —a = mu for some mGI+ so that a = (-m)u, and (ii) is established.

Clearly, if for any a G JZ) we have both a = ru and a = su, where r,s S I, then z = a — a =ru — 8U — (r — s)u and r = s. Thus the representation of each a C JD as a = mw is unique.

9. Prove: Let / and K, each distinct from {z}, be principal ideals in an integral domain .2).

Then J = K if and only if their generators are associate elements in ID.

Let the generators of J and Khe a and 5 respectively.

First, suppose a and 6 are associates and 6 = a-v, where v is a unit in ID. For any cG Kthere exists some a €. ID such that

c = b's — (a'v)s = a{v • s) — a's', where a' e ID

Then e&J and KqJ. Now b — a-v implies a = b'v-^; thus, by repeating the argument with

any dS J, we have J Q K. Hence, J = K a,a required.

Conversely, suppose J = K. Then for some s,tG ID we have a~b-s and b = a't. Now

a = 6 • 8 = (a' t)s = a{t • s)

so that a — a(t- s) = a(u — t's) = z

where u is the unity and z is the zero element in ^. Since a ¥= z, by hypothesis, we have w — t • s = z

so that t • 8 = M and 8 is a unit in ID. Thus a and 6 are associate elements in ID, as required.

10. Prove: Let a,b,p G ID, an integral domain which is also a principal ideal ring, and

suppose pI

a • 6. Then if p is a prime element in ^, p |

a or p |6.

If either o or 6 is a unit or if a or 6 (or both) is an associate of p, the theorem is trivial.

Suppose the contrary and, moreover, suppose p/o. Denote by J the ideal in ID which is the inter-


section of all ideals in 3) which contain both p and a. Since ^ is a principal ideal, suppose it is

generated by c G ^^ so that v = C'x for some x G ^. Then either (i) a; is a unit in X> or

(ii) c is a unit in 2).

(i) Suppose X is a unit in 3); then, by Theorem VIII, p and its associate c generate the sameprincipal ideal S- Since a€ JJ, we must have

a — eg = p'k for some g,h& 2)

But then p |a, a contradiction; hence, x is not a unit.

(ii) Suppose c is a unit; then c • c~i = m G JJ and S — 3). Now there exist 8,t G .2) such thatM = p • 8 + t • a, where u is the unity of 3). Then

6 = M . 6 = (p • «)6 + (t • o)6 = p(« • 6) + t(a, • V)

and, since p |a • 6, we have p |

6 as required.

11. Prove: The unique factorization theorem holds in any integral domain 3) which is also

a Euclidean ring.

We are to prove that every non-zero, non-unit element of 3) can be expressed uniquely (up to

the order of the factors and the appearance of unit elements as factors) as the product of primeelements of 3).

Suppose a ?^ G ^ for which «(o) = 1. Write a = 6 • c with 6 not a unit; then c is a unit anda is a prime element in 3), since otherwise

»(o) = 9(6 • c) > »(6) by Theorem II, Page 115

Next, let us assume the theorem holds for all b G 3) for which e{b) < m and consider c& 3)for which 9(c) = m. Now if c is a prime element in 3), the theorem holds for c. Suppose, on thecontrary, that c is not a prime element and write c = d' e where both d and e are proper divisors of c.

By Theorem II, we have e{d) < m and 8(e) < m. By hypothesis, the unique factorization theoremholds for both d and e so that we have, say,

c = d • e = Pi • P2 • P3 • • • Ps

Since this factorization of c arises from the choice d,e of proper divisors, it may not be unique.

Suppose that for another choice of proper divisors we obtained c = gj • gj • 93 • • • g^. Considerthe prime factor pi of c. By Theorem IX, Page 117, Pi | q^ or p, | (gg • gg • • •

gt); if Pi /gj, then Pi | qôr Pi I (93 • 9t); if Suppose Pi | gj. Then q, = f • Pi where / is a unit in 3) since, other-wise, q, would not be a prime element in 3). Repeating the argument on

P2*P3'"Ps = f~^'<Ii'<l2--- <lj-i'Qj+i--- Qt

we find, say, pj | g^ so that qk = O'Pz with g a unit in 3). Continuing in this fashion, we ultimatelyfind that, apart from the order of the factors and the appearance of unit elements, the factorization ofc is unique. This completes the proof of the theorem by induction on m (see Problem 27, Chapter 3,

Page 37).

12. Prove: S = {x + yy/3 + zy/9 : x,y,z E Q} is a subfield of R.

Prom Example 2, Chapter 10, Page 101, S is a subring of the ring R. Since the commutative

law holds in R and 1 = 1 + OyS + 0\/9 is the multiplicative identity, it is necessary only to verify

that for X + yy^+z^ ^ G S, the multiplicative inverse'^^ ~ ^^'^ + ^^^^-^^ +

^^~'^^^,

where D = x» + 3i/9 + 9z3 - 9xyz, is in S.

13. Prove: Let 3) be an integral domain and J an ideal in 3). Then 3)/S is a field if andonly if ,5 is a maximal ideal in 3).

First, suppose J? is a maximal ideal in 3); then J c 3) and (see Problem 3) 3)/JI is a commuta-tive ring with unity. To prove 3)/^^ is a field, we must show that every non-zero element has amultiplicative inverse.

For any g G ^ —^, consider the subset

S = {a + q-x: ae J, xe 3)}

of 3). For any y G 3) and a + g • x G S, we have (a + g • x)y = a-y + q(x'y) G S since

a'V ^ J; similarly, j/(o + g • x) G S. Then S is an ideal in 3) and, since ^ C S, we have S — 3).Thus any r & 3) may be written as r - a + q'e, where e G ^. Suppose for u, the unity of 3),we find

u = a + q'f, f& 3)


From u+S = (a+S) + {q+J)-{f +JJ) = (q+3-(f+S)

it follows that / +S is the multiplicative inverse of q +J. Since q is an arbitrary element of ^ -^,the ring of cosets ,5)/^? is a field.

Conversely, suppose JD/S >s a field. We shall assume ^fj not maximal in ^ and obtain a con-

tradiction. Let then J be an ideal in JZ) such that ^5 C J C JZ).

For any o e ,2) and any pBJ-J, define (p+^)~^' (a+^) = s + Jj; then

a+J = (p+JI)-is+S)

Now o — p • 8 G JJ and, since ^^cj, a — p-sGJ. But p G J; hence ae. J, and J = 3), & con-

tradiction of J Q 3). Thus ^<? is maximal in ^.

The note in Problem 3, Page 119, also applies here.


14. Enumerate the properties of a set necessary to define an integral domain.

15. Which of the following sets are integral domains, assuming addition and multiplication defined as on R:

(a) {2a-f-l: a&I) (e) {a+hy/Z: a,b S 1}

(b) {2o: oe/} (/) {r + sVS: r.s G Q}

(c) {ay/S : o e /} (jr) {a -I- by/2 + cy/5 + dy/W : a,b,e,d e /}

(d) {rA/3: r & Q}

16. For the set G of Gaussian integers (see Problem 8, Chapter 10, Page 110), verify:

(a) G is an integ:ral domain.

(6) a = a+bi is a unit if and only if N(a) = o^ + 6* = 1.

(c) The only units are ±1, —i.

17. Define S = {(«!, 02, 03, 04) : Oj S ft} with addition and multiplication defined respectively by

(ai, 02, as, a^) + (fej, 62, 63, 64) = (oj -I- fej, Oj -I- 62. <»3 + *3> «4 + ''4)

and (oi, 02. "S" «4)(6i, b^, 63. 64) = («i * î' *2' ^2. ^s * *3. »4 ' ''4)

Show that 5 is not an integral domain.

18. In the integral domain JZ) of Example 2(6), Page 115, verify:

(o) 33 ± 8y/vf and -33 ± Sv'l? are units.

(6) 48 - ll-\/l7 and 379 - 92-\/l7 are associates of 5 -I- 4\/l7.

(c) 8 = 2'2«2 = 2(8 + 2y/vf){—S + 2^17) = (5 -I- \/l7)(5 - \/i7), in which each factor is a prime;

hence, unique factorization in primes is not a property of JD.

19. Prove: The relation of association is an equivalence relation.

20. Prove: If, for a G =2), N(a) is a prime integer then a is a prime element of ^.

21. Prove: A ring % having the property that for each a¥' z,be.% there exists rG% such that

a-r — b is a division ring.

22. Let :D' = {[0], [5]} and JD" = {[0], [2], [4], [6], [8]} be subsets of .2) = 1/(10). Show:

(o) ,2)' and =Z)" are subdomains of .2).

(6) .2)' and //(2) are isomorphic; also, J)" and 7/(5) are isomorphic.

(c) Every aG S) can be written uniquely as a = a' + a" where a' € .2)' and a" € .2)".

(d) For 0,6 e^, with a = »'-!- o" and 6 = 6' + 6", o -h 6 = (a' + 6') + (»" + 6") and a-b = a''b' +a" • 6".


23. Prove: Theorem II.

Hint. If 6 is a unit then e{a) = »[6-i (a • b)] ^ e(a • 6). If 6 is not a unit, consider a = q{a-b) + r,

where either r = « or e(r) < e{a • b), for a ^ z G ^.

24. Prove: The set S of all units of an integral domain is a multiplicative group.

25. Let JZ) be an integral domain of characteristic p and ^' = {x" : a; G ^}. Prove:

(a) (a ± b)" = a" ± bP (6) The mapping £0^3)': x -* x^ is an isomorphism.

26. Show that for all a¥' z,b of any division ring, the equation ax — b has a solution.

27. The set Si = {(gj + qii + qaJ + <l^k) : q^, gj, «3. «4 ^ *} of quaternions with addition and multiplica-tion defined by

(oi + 021 + aî + a^k) + (5i + bî + b^j + b^k) = (a^ + 6,) + (02 + 62)1 + (03 + 63)^ + (04 + 64)*;and

(ffli + 02i + asj + 04A;) • (6, + bii + b^j + b^k) = (0,61 - 0362 - 0363 - a464) + (0162 + a^bi + O364 - 0463)1

+ (0163 — 0264 + O361 + 0462); + (0164 + ©263 — O362 + 0461)*;

is to be proved a non-commutative division ring. Verify:

(o) The subsets Q2 = {(9i + 92* + Oj + Ok)}, Q3 = {(g, + Oi + q^j + Ok)}, and Q4 = {(g, + Oi + Oi + 94*;)}of Si combine as does the set C of complex numbers; thus, i^ = p = k^ = —1.

(b) 91, g2> 93> 94 commute with i, i, fc.

(c) iy = k, jk — i, ki = j

(d) ji = —k, kj — —i, ik = —j(e) With Q as defined in Problem 17, Chapter 10, Page 111, the mapping

Q -» ^ : (9i + 92». 93 + 94^. -93 + 94«. 9i - Qui) "* (9i + 92* + 93^ + 94*^)

is an isomorphism.

(/) ^ is a non-commutative division ring (see Example 4, Page 117).

28. Prove: A field is a commutative ring whose non-zero elements have multiplicative inverses.

29. Show that P = {(o, 6, -5, o) : a,b e R} with addition and multiplication defined by

(o, 6, —6, o) + (c, d, —d, c) = (o + c, 6 + d, —6 — d, o + c)

and (o, 5, -5, o)(c, d, -d, c) = (ac — 6d, od + 6c, -od - 6c, oc — 5d)

is a field. Show that P is isomorphic to C, the field of complex numbers.

30. (o) Show that {0 + 6\/3 : 0,6 G Q} and {a + 6\/2 + c\/5 + d\/l0 : o,6,c,dGQ} are subfields of B.

(6) Show that {o + byfi : 0,6 G Q} is not a subfield of R.

31. Prove: S = {0 + 6r : a,6 G fi, r = -^(l + Vsi)} is a subfield of C.

Hint. The multiplicative inverse of o + 6r ^ G S is „"~^

,„ 5 , ,.. y G S.

O'' — 06 + 6' O'^ — 00 + 0^

32. (a) Show that the subsets S = {[0], [5], [10]} and T = {[0], [8], [6], [9], [12]} of the ring 7/(15) are

integral domains with respect to the binary operations on 7/(15).

(6) Show that S is isomorphic to 7/(3) and, hence, is a field of characteristic 3.

(c) Show that T is a field of characteristic 5.

33. Consider the ideals A = {2g. g^G}, B = {5g: gGG}, E = {lg:gGG}, and F = {(X + i)g : g&G}of G, the ring of Gaussian integers, (o) Show that G/A = G/(2) and GIB - G/{5) are not integraldomains. (6) Show that G/E is a field of characteristic 7 and G/F is a field of characteristic 2.

34. Prove: A field contains no proper ideals.

35. Show that Problems 3 and 13 imply: If ^ is a maximal ideal in a commutative ring % with unity,then t<? is a prime ideal in %.

chapter 12

Polynomials

INTRODUCTION

A considerable part of elementary algebra is concerned with certain types of functions,

for example, , ^ . » „ ,„1 + 2a; + 3a;2 x + x^ 5 - 4x^ + 3a;'"

called polynomials in x. The coefficients in these examples are integers although it is not

necessary that they always be. In elementary calculus, the range of values of x (domain of

definition of the function) is R. In algebra, the range is C; for instance, the values of x

1 V2for which 1 + 2a; + 3a;Ms are ~

o"- 3 ^

In the light of Chapter 2, any polynomial in x can be thought of as a mapping of a

set S (range of x) onto a set T (range of values of the polynomial). Consider, for example,

the polynomial 1 + \/2 a; - 3a;^ If S = I, then T cR and the same is true if S = Q or

S = R; if S = C, then TcC.

As in previous chapters, equality implies "identical with"; thus two polynomials in x

are equal if they have identical form. For example, a + bx = c + dx if and only if a = c

and b-d. (Note that a + bx = c + dx is never to be considered here as an equation in x.)

It has been our experience that the images of each value of x & S are the same ele-

ments of T when a(x) = p{x) and, in general, are distinct elements of T when a{x) ¥= p{x).

However, as will be seen from Example 1 below, this familiar state of affairs is somehow

dependent upon the range of x.

Example 1: Consider the polynomials a(x) = [l]x and p(x)-[l]x^, where [1] G 7/(5), and

suppose the range of a; to be the field //(5) = {[0], [1], [2], [3], [4]}. Clearly, a(x) and

P(x) differ in form (are not equal polynomials); yet, as is easily verified, their images

for each x e 7/(5) are identical.

Example 1 suggests that in our study of polynomials we begin by considering them

as forms.

POLYNOMIAL FORMSLet <2i be a ring and let x, called an indeterminate, be any symbol not found in %.

By a polynomial in x over % will be meant any expression of the form

a{x) = aoa;" + aa^ + a%x^ + = 2 f"-^^" > «* ^ "^

in which only a finite number of the a's are different from z, the zero element of %.

Two polynomials in x over %, a{x) defined above, and

p(x) = box" + 61a;' + 62a;' + • • • = 2 b"^" ' biG%

will be called equal, aix) = p{x), provided a^ = bk for all values of A;.

124

CHAP. 12] POLYNOMIALS 125

In any polynomial, as a(x), each of the components aoic", aix^, a2X^, . . . will be called aterm; in any term, as aiX\ a, will be called the coefficient of the term. The terms of a{x)

and p{x) have been written in a prescribed (but natural) order and we shall continue this

practice. Then i, the superscript of x, is merely an indicator of the position of the term UiX'

in the polynomial. Likewise, juxtaposition of ai and x" in the term aiX* is not to be con-strued as indicating multiplication and the plus signs between terms are to be thoughtof as helpful connectives rather than operators. In fact, we might very well have writtenthe polynomial a{x) above as « = (ao, ai, a2, . . .).

If in a polynomial, as a(x), the coefficient an^ z while all coefficients of terms whichfollow are z, we say that a(x) is of degree n and call a„ its leading coefficient. In particular,

the polynomial aox" + zx^ + zx^ + • is of degree zero with leading coefficient ao whenao¥= z and has no degree (and no leading coefficient) when a© = z.

Denote by %[x] the set of all polynomials in x over % and, for arbitrary a{x), p(x) &%[x],define addition ( + ) and multiplication (•) on '5?^[a;] by

a{x) + p{x) = (ao + 6o)a;» + (ai + 6i)a;> + (02 + h2)x^ +

and a{x) • /3(a;) = ao&ox" + (aobi + aibo)a;» + (ao62 + aihi + azbo)*^ + • • •

k

— ^CkX^, where Ck = 'âihk-i

(Note that multiplication of elements of % is indicated here by juxtaposition.)

The reader may find it helpful to see these definitions written out in full as

a{x) [±] p{x) = (ao e bo)x'' + (a, 6,)a;> + [a^ ® b2)x^ + • • •

and a{x) UD p{x) - (ao O bo)*" + (ao © &i © ai O bo)x'

+ (ao O 62 © ai O 61 © a2 O b<,)x^ +

in which [+] and CZl are the newly defined operations on %[x], © and O are the binaryoperations on % and, again, + is a connective.

It is clear that both the sum and product of elements of %[x] are elements of %[x],

i.e., have only a finite number of terms with non-zero coefficients G %. It is easy to verify

that addition on %[x] is both associative and commutative and that multiplication is asso-

ciative and distributive with respect to addition. Moreover, the zero polynomial

zx" + zx^ + zx^ + • - ^zx" G. %[x]

is the additive identity or zero element of %{x] while

-a{x) = -aox" + (-ai)a;> + (-az)^^ + • • • = 2 i-ak)x^ e %[x]

is the additive inverse of a{x). Thus,

Theorem I. The set of all polynomials %[x\ in x over "T^ is a ring with respect to addition

and multiplication as defined above.

Let a{x) and p{x) have respective degrees m, and n. If m ^ n, the degree of a{x) + p(x)

is the larger of m,n; if m = n, the degree of a(x) + p{x) is at most m (why?). The degree

of a{x) • /3(x) is at most m + n since amb„ may be z. However, if % is free of divisors of zero,

the degree of the product is m + n. (Whenever convenient we shall follow practice andwrite a polynomial of degree m as consisting of no more than m + 1 terms.)

126 POLYNOMIALS [CHAP. 12

Consider now the subset S - {rx": rG%} of %[x] consisting of the zero polynomial

and all polynomials of degree zero. It is easily verified that the mapping

%^S: r^rx"

is an isomorphism. As a consequence, we may hereafter write Co for Uox" in any poly-

nomial a{x) G%[x].

MONIC POLYNOMIALS

Let <!( be a ring with unity u. Then u = ua;* is the unity of %[x] since ux" • a{x) - a{x)

for every a{x)G'=S\x\. Also, writing x = ux^ - zx'^ \- ux^ , we have xG%\x\. Nowa\,{x -x'X • • \jak factors) = o-yo^ e %{x\ so that in a.{x) = ao + a.xx + a^x^ + • • • we may con-

sider the superscript i in aix* as truly an exponent, juxtaposition in any term ttio;* as

(polynomial) ring multiplication, and the connective + as (polynomial) ring addition.

Any polynomial a{x) of degree m over % with leading coefficient m, the unity of %., will

be called monic.

Example 2: (a) The polynomials 1, x + S, and 3:2-535 + 4 are monic while 2a;2 - a; + 5 is not a

monic polynomial over / (or any ring having / as a subring).

(5) The polynomials 6, bx + f, and bx^ + dx + e are monic polynomials in S[x] over

the ring S of Example 1(d), Chapter 11, Page 114.

DIVISION


Theorem II. Let <?( be a ring with unity u, a(x) = ao + aiX + a2X^+ +amX'" G %[x] be

either the zero polynomial or a polynomial of degree m, and p{x) -

bo + bix + biX^ + • • • + ux" e %[x] be a monic polynomial of degree n. Then

there exist unique polynomials q„{x),r^(x);q^{x),r^^{x) G%[x] with r^{x),r^{x)

either the zero polynomial or of degree < n such that

(i) «(«) = Qr{x) ' P{x) + r^(x)

and (ii) cc{x) = p{x) • q^îx) + r^îx)

In (i) of Theorem II we say that a{x) has been divided on the right by I3{x) to obtain the

right quotient qJ^{x) and right remainder r^îx). Similarly, in (ii) we say that a(x) has been

divided on the left by /8(x) to obtain the left quotient q^^^x) and left remainder rj^{x).

When r^îx) = z (r^îx) = z), we call I3{x) a right (left) divisor of a{x).

For the special case j3(x) = ux-b - x-b, Theorem II yields (see Problem 2),

Theorem III. The right and left remainders when a{x) is divided hy x-b, b G%, are

respectivelyrR

= ao + aib + ttzb^ + • • • + anb"

and r^ = ao + bai + bHi + • • • + &"««

There follows

Theorem IV. A polynomial a(x) has x - b as right (left) divisor if and only if r^ = z

Examples illustrating Theorems II-IV when % is non-commutative will be deferred

until Chapter 15. The remainder of this chapter will be devoted to the study of certain

polynomial rings %[x] obtained by further specializing the coefficient ring %.


COMMUTATIVE POLYNOMIAL RINGS WITH UNITYLet g^ be a commutative ring with unity. Then %[x] is a commutative ring with unity

(what is its unity?) and Theorems II-IV may be restated without distinction between rightand left quotients (we replace q^(x) = q^{x) by q{x)), remainders (we replace r^{x) = r^{x) byr{x)), and divisors. Thus (i) and (ii) of Theorem II may be replaced by

(iii) a{x) = q{x)'P{x) + r{x)

and, in particular, we have

Theorem IV'. In a commutative polynomial ring with unity, a polynomial a{x) of degreem has x - 6 as divisor if and only if the remainder

(a) r = ao + a,b + azh^ + • • • + Omb'" = z

When, as in Theorem IV', r = z then h is called a zero (root) of the polynomial a(x).

Example 3: (a) The polynomial a;2-4 over / has 2 and -2 as zeros since (2)2-4 - and(-2)2 -4 = 0.

(6) The polynomial [3]a;2 - [4] over the ring 7/(8) has [2] and [6] as zeros while thepolynomial [l]a;2-[l] over 7/(8) has [1], [3], [5], [7] as zeros.

When % is without divisors of zero so also is %[x]. For, suppose a(x) and |8(x) areelements of %[x], of respective degrees m and n, and that

a{x)-p{x) = aobo + (ao6i + ai6o)a; + •• + am6„x'"+» = z

Then each coefficient in the product and, in particular ajjn, is z. But % is without divisorsof zero; hence ambn = z if and only if a,n = z or hn = z. Since this contradicts the assump-tion that a{x) and ^(a;) have degrees m and n, %\x] is without divisors of zero.

There follows

Theorem V. A polynomial ring ^^[a;] is an integral domain if and only if the coefficientring % is an integral domain.

SUBSTITUTION PROCESSAn examination of the remainder

(a) r - ao + ttib + azft^^ + • • • + aj)"^

in Theorem IV' shows that it may be obtained mechanically by replacing x by 6 throughouta(x) and, of course, interpreting juxtaposition of elements as indicating multiplication in %.Thus, by defining f{b) to mean the expression obtained by substituting 6 for x throughoutfix), we may (and will hereafter) replace r in (a) by «(6). This is, to be sure, the familiarsubstitution process in elementary algebra where (let it be noted) x is considered a variablerather than an indeterminate.

It will be left for the reader to show that the substitution process will not lead tofuture difficulties, that is, to show that for a given h G%, the mapping

f{x) -* f(b) for all fix) e %[x]

is a homomorphism of %[x] onto %.

THE POLYNOMIAL DOMAIN f'[x]

The most important polynomial domains arise when the coefficient ring is a field f.We recall that every non-zero element of a field ^ is a unit of f and restate for the integraldomain fi^] the principal results of the sections above as follows:


The Division Algorithm. If a{x),l3{x) G f[x] where p{x) ^ z, there exist unique poly-

nomials q{x), r{x) with r{x) either the zero polynomial or of degree less than that of p{x),

such that , ,/ \ />/ \ , „/„\

a{x) = q{x)-fi{x) + r{x)


When r{x) is the zero polynomial, p{x) is called a divisor of a(x) and we write /3{a;)|

a(x).

The Remainder Theorem. If «(a;), x-bG f[x], the remainder when a{x) is divided by

X — b is a{b).

The Factor Theorem. If a{x) G f [a;] and bGf, then a; - 6 is a /actor of a(x) if and

only if a(b) = z, that is, a; - 6 is a factor of a(x) if and only if b is

a zero of a{x).

There follow

Theorem VI. Let a{x) G f[x] have degree m > and leading coefficient a. If the dis-

tinct elements 6i, ba, . . ., bm of f are zeros of a(x), then

a(a;) = a(x - bi)(a; - b2) • • • (x - bm)


Theorem VII. Every polynomial a{x) G f[x] of degree m > has at most m distinct

zeros in f.

Example 4: (a) The polynomial 2x' + 'Ix-15 e Q[x] has the zeros 3/2, -5, G Q.

(6) The polynomial x^ + 2a; + 3 e C[x] has the zeros -l + Vî and -l-VZt

over C. However, a;^ + 2x + 3 6 Q[a;] has no zeros in Q.

Theorem VIII. Let a(a;),i8(a;) G f[x] be such that a(s) = ^(s) for every sGf. Then, if

the number of elements in f exceeds the degrees of both a{x) and p{x), we

have necessarily a(a;) = p{x). p^^ ^ pj.QQf^ ggg Problem 6.

Example 5: It is now clear that the polynomials of Example 1 are distinct, whether considered

as functions or as forms, since the number of elements of y = 1/(5) does not exceed

the degree of both polynomials. What then appeared in Example 1 to be a contra-

diction of the reader's past experience was due, of course, to the fact that his past

experience had been limited solely to infinite fields.

PRIME POLYNOMIALS

It is not difficult to show that the only units of a polynomial domain f[x] are the

non-zero elements (i.e., the units) of the coefficient ring T- Thus the only associates of

a(x) G f[x] are the elements v • a{x) of f[x] in which v is any unit of f.

Since for any v¥'zGf and any a(x) Gf[x],

a(x) - V-^-a{x)-V

while, whenever a{x) = q{x) • p{x),

a{x) = [V-' q{x)] - [v ^{x)]

it follows that (a) every unit of f and every associate of a(x) is a divisor of a{x) and

(b) if p{x)I

a(x) so also does every associate of p(x). The units of T and the associates of

«(x) are called trivial divisors of a(x). Other divisors of a{x), if any, are called non-trivial

divisors.

A polynomial a(x) G f[x] of degree m ^ 1 is called a prime (irreducible) polynomial

over f if its only divisors are trivial.


Examples: (a) The polynomial Sx^ + 2x + 1 e R[x] is a prime polynomial over ft.

(b) Every polynomial ax + b&f[x], with a ¥= z, is a prime polynomial over f.

THE POLYNOMIAL DOMAIN C[x]

Consider an arbitrary polynomial

p{x) = bo + bix + bzx^ + • • + bmX"" G C[x]

of degree m^l. We shall be concerned in this section with a number of elementarytheorems having to do with the zeros of such polynomials and, in particular, with thesubset of all polynomials of C[x] whose coefficients are rational numbers. Most of thetheorems will be found in any College Algebra text stated, however, in terms of roots ofequations rather than in terms of zeros of polynomials.

Suppose r e C is a zero of p{x). Then pir) = and, since 6~' e C, also b~' • p{r) = 0.Thus the zeros of p{x) are precisely those of its monic associate

»i^) - b~^'P{x) = Co + aix + azx^ + • • • + Um-ix""-^ + a;"*

Whenever more convenient, we shall deal with monic polynomials.

It is well-known that when m^l, a{x) -ao + x has -a« as zero and when m = 2,a(x) = ao + aix + x^ has i(-ai - Va?-4ao) and i{-a, + v'ai-4ao) as zeros. In Chapter 8,it was shown how to find the n roots of any a G C; thus every polynomial x^-aG C[x]has at least n zeros over C. There exist formulas (see Problems 16-19) which yield thezeros of all polynomials of degrees 3 and 4. It is also known that no formulas can bedevised for arbitrary polynomials of degree w ^ 5.

By Theorem VII any polynomial «(«) of degree m^l can have no more than m distinctzeros. In the paragraph above, a(x) = ao + aix + x^ will have two distinct zeros if andonly if its discriminant a?-4a«^0. We shall then call each a simple zero of a{x).However, if a? - 4ao = 0, each formula yields -âi as a zero. We shall then call -âia zero of multiplicity two of a{x) and exhibit the zeros as -â,, -âi.

Example 7: (o) The polynomial x^ + x^-Sx + S = (x-l)Hx + 3) has -3 as simple zero and1 as zero of multiplicity two.

(b) The polynomial x*-x3-3x^ + 5x-2 = (x-l)Hx + 2) has -2 as simple zeroand 1 as zero of multiplicity three.

The so-called

Fundamental Theorem of Algebra. Every polynomial a{x) G C[x] of degree m ^ 1 hasat least one zero in C.

will be assumed here as a postulate. There follows, by induction

Theorem IX. Every polynomial a{x) G C[x] of degree m ^ 1 has precisely m zeros over C,with the understanding that any zero of multiplicity n is to be counted asn of the m zeros.

and, hence.

Theorem X. Any a{x) G C[x] of degree w ^ 1 is either of the first degree or may bewritten a? the product of polynomials G C[x] each of the first degree.

Except for the special cases noted above, the problem of finding the zeros of a givenpolynomial is a difficult one and will not be considered here. In the remainder of thissection we shall limit our attention to certain subsets of C[x] obtained by restricting thering of coefficients.


First, let us suppose that

a(x) = ao + aix + aiX^ + • • • + amX"^ G R[x]

of degree m ^ 1 has r = a + hi as zero, i.e.,

a(r) = ao + air + air^ + • • • + amr'^ - s + ti =

By Problem 2, Chapter 8, Page 79, we have

a{f) = tto + ttif + a2'F + • • • + amf'" = s + ti -

so that

Theorem XI. If rGC is a zero of any polynomial a{x) with real coefficients, then f is

also a zero of a{x).

Let r=â + bi, with b # 0, be a zero of a{x). By Theorem XI f = a-bi is also a zero

and we may writea(x) = [a; - (a + 6i)] [x-{a- bi)] • aj(a;)

= [x^ - 2ax + a^ + b^] • a^{x)

where «,(a;) is a polynomial of degree two less than that of a(x) and has real coefficients.

Since a quadratic polynomial with real coefficients will have imaginary zeros if and only if

its discriminant is negative, we have

Theorem XII. The polynomials of the first degree and the quadratic polynomials with

negative discriminant are the only polynomials G R[x] which are primes

over R.

and

Theorem XIII. A polynomial of odd degree G R[x] necessarily has a real zero.

Suppose next that

p(x) = 6o + bix + b2X^ + • • • + bmx"' e Q[x]

Let c be the greatest common divisor of the numerators of the b's and d be the least

common multiple of the denominators of the b's; then

a{x) = - • ;8(a;) = oo + aix + a^x^ + • • • + amX'" G Q[x]

has integral coefficients whose only common divisors are ±1, the units of /. Moreover,

j8(x) and a(x) have precisely the same zeros.

If r G Q is a zero of a{x), i.e. if

«(r) = a« + air + ttar" + • • + an.r'^ =

there follow readily

(i) if r G /, then r]ao

(ii) if r = sit, a common fraction in lowest terms, then

f^-ais/t) = aoi™ + aisf"-! + azS^t""-" + . • + a^-is'^-H + a„,s'" =

so that sI

ao and t\am. We have proved

Theorem XIV. Let »(x) = ao + atx + a^x^ + + amX^

be a polynomial of degree m ^ 1 having integral coefficients. If s/t G Q,

with (s, f) = 1, is a zero of aix), then s]Oo and t

\

Om-


Example 8: (a) The passible rational zeros of

a(x) = Sa:* + 2«2 - 7x + 2

are ±1, ±2, ±J, ±|. Now «(!) = 0, a(-l) # 0, <«(2) ^ 0, «(-2) = 0, a(^) = 0,

«(—i) 'i* 0, a(|) # 0, o(—§) #0 SO that the rational zeros are 1, -2, J and

o(*) = 3(« - !)(« + 2){x - I).

Note. By Theorem VII, a{x) can have no more than three distinct zeros.

Thus once these have been found, all other possibilities untested may be dis-

carded. It was not necessary here to test the possibilities —^, §, —§.

(b) The possible rational zeros of

a{x) = 4x^ - 4x* — Bafl + 5x^ + X - 1

are ±1, ±^, ±^. Now a(l) = 0, «(-!) = 0, «(^) = 0, ai-^) = so that

a{x) = Mx-l)(x + l){x-:^)(x + ^)-{x-l)and the rational zeros are 1, 1, —1, ^, —^.

(e) The possible rational zeros of

a{x) = x*-2a^-5x^ + 4x + 6

are ±1, ±2, ±3, ±6. For these, only a(-l) = and a(3) = so that

a{x) - {x+l){x-Z)(x!t-2)

Since none of the possible zeros ±1, ±2 of x* — 2 are zeros, it follows thatx* — 2 is a prime polynomial over Q and the only rational zeros of a(x) are -1, 3.

(d) Of the possible rational zeros: ±1, ±^, ±^, ±^, of a(x) = 6x* - 5x» + Tw^ - 5x + 1

only ^ and | are zeros. Then a{x) = 6(x- ^)(x - ^)(x2 + 1) so that x" + 1

is a prime polynomial over Q, and the rational zeros of a{x) are ^, ^.

(e) The possible rational zeros of

a{x) = 3x* - 6x3 + 4a,2 _ iqx + 2

are ±1, ±2, ±J, ±|. Since none, in fact, is a zero, a(x) is a prime polynomialover Q.

GREATEST COMMON DIVISORLet a(x) and p{x) be non-zero polynomials in f[z]. The polynomial d{x) G f[x] having

the properties

(1) d{x) is monic,

(2) dix)\aix) and d{x)\p{x),

(3) for every c{x)Gf'[x] such that c{x)\a{x) and c(a;)|^(a;), we have c{x)\d{x),

is called the greatest common divisor of a(x) and p{x).

It is evident (see Problem 7) that the greatest common divisor of two polynomials infix] can be found in the same manner as the greatest common divisor of two integers inChapter 5. For the sake of variety, we prove in Problem 8

Theorem XV. Let the non-zero polynomials a{x) and p(x) be in f[x]. The monic poly-nomial

(6) d{x) = six) ' a{x) + t{x) -p{x), s{x), t{x) e r[x]

of least degree is the greatest common divisor of a{x) and p{x).

There follow

Theorem XVI. Let a{x) of degree m ^ 2 and /8(a;) of degree n ^ 2 be in 7^ [a;]. Then non-zero polynomials fi{x) of degree at most n - 1 and v{x) of degree at mostm - 1 exist in f[x] such that

(c) ia(«)'«(a;) + v(a;)-j8(a;) = z

if and only if a{x) and p{x) are not relatively prime.



and

Theorem XVII. If a{x), ^(x), p(x) G f[x] with a{x) and p{x) relatively prime, then

p(x)I

a(x) • p(x) implies p{x)\

p{x).


The Unique Factorization Theorem. Any polynomial a(x), of degree m ^ 1 and with

leading coefficient a, in f[x] can be written as

a{x) = a '[Pi(a;)]'».

• [p^{x)]'"' • • • [pîx)]^'

where the pJl^x) are monic prime polynomials over fand the 'm^ are positive integers. Moreover, except for

the order of the factors, the factorization is unique.

Example 9: Decompose a{x) = 4x* + 3a;» + ix" + 4x + 6 over 1/(7) into a product of prime

polynomials.

We have, with the understanding that all coefficients are residue classes

modulo 7,

a{x) = Ax* + 3*3 + 4«2 + 4a: + 6 = 4x* + 24«» + 4a;2 + 4as + 20

= 4(x4 + 6*8 + «2 + « + 5) = 4(x + l)(x» + 5x2 + 3a! + 5)

= 4(x + l)(x + 3)(x2 + 2x + 4) = 4(x + l)(x + 3)(x + 3)(* + 6)

= 4(a + l)(x + 3)2(x + 6)

PROPERTIES OF THE POLYNOMIAL DOMAIN f[x]

The ring of polynomials T[^] over a field T has a number of properties which parallel

those of the ring / of integers. For example, each has prime elements, each is a Euclidean

ring, (see Problem 11), and each is a principal ideal ring (see Theorem IX, Chapter 10,

Page 108). Moreover, and this will be our primary concern here, TM may be partitioned

by any polynomial X{x) G f[x] of degree « — 1 into a ring

n^]/{\ix)) = {[a{x)].[m].---}

of equivalence classes just as / was partitioned into the ring //(m). For any a{x),p{x) G f[x]

we define

(i) [«(«)] = {«(«) + /x(a;) • \(x) : ti{x) G f [a;]}

Then a{x) G [a(x)], since the zero element of f is also an element of TM, and [a{x)] = [fi{x)]

if and only if a{x) = p{x) (mod X{x)), i.e. if and only if X{x)|

{a{x) - pix)).

We now define addition and multiplication on these equivalence classes by

[a{x)] + [Pix)] = HX) + P{X)]

and [a(x)]'[p{x)] = Hx)'P(x)]

respectively and leave for the reader to prove

(a) Addition and multiplication are well-defined operations on T[^]/{Hx)).

(6) f[x]/(k{x)) has [z] as zero element and [u] as unity, where z and u respectively are

the zero and unity of f.

(c) T[x]/{x.(x)) is a commutative ring with unity.


Theorem XVIII. The ring T[^]/{x{x)) contains a subring which is isomorphic to the

field f.


If \{x) is of degree 1, it is clear that T[x]/{\{x)) is the field f; if \{x) is of degree 2,

T[^]/{H^)) consists of 7^ together with all equivalence classes {[ao + aix]: ao.aiE^f.au^z};

in general, if X{x) is of degree n, we have

f [x]/(a.(x)) = {[oo + aix + aiX^ + • • • + a„-ia;"-'] : Oi^f)Now the definitions of addition and multiplication on equivalence classes and the isomor-phism: tti <^ [Oi] imply

[tto + aix + a^x^ + • • • + a„-ia;"-i] = [oo] + [ai][x] + [cz] • [x]^ + • • • + [a„-i] • [x]""'

= Co + ai[x] + a2[x]^ + • • • + a„-i [x]"~>

As a final simplification, let [x] be replaced by | so that we have

T[x]l{\{x)) = {tto + ail + a2|^ + • • • + a„-ir-i : aj G ^}


Theorem XIX. The ring T[x\l{\(x)) is a field if and only if A,(x) is a prime polynomial

over f.

Example 10: Consider \{x) - x^~Z€.Q[x], a prime polynomial over Q. Now

Q[x]/(aj2-3) = {ao + ai«: aa,ai&Q}

is a field with respect to addition and multiplication defined as usual except that in

multiplication {^ jg ^ be replaced by 3. It is easy to show that the mapping

Oo + "li *^ fflo + "iVs

is an isomorphism of Q[x\l(x^ — 3) onto

Q[V3] = {ao + OiA/3: a^.a^^Q},

the set of all polynomials in Vs over Q. Clearly Q[V3 ] C fi so that Q[\/3 ] is thesmallest field in which x^ — Z factors completely.

The polynomial x*-3 of Example 10 is the monic polynomial over Q of least degree

having \/3 as a root. Being unique, it is called the minimum polynomial of )/3 over Q.

Note that the minimum polynomial of \/3 over i2 is x - y/S.

Example 11: Let ^ = 7/(3) = (0, 1, 2} and take X(x) = x^ + l, a prime polynomial over f- Con-struct the addition and multiplication table for the field TM/iHx)).

Here f[x]/{\{x)) = {oq + OiJ : ao.UiGf}

= {0,l,2,{,2j,l + |,l + 2c.2 + i,2 + 2j}

Since X(f) = j2 + l = [o], we have |2 = [-1] = [2], or 2. The required tables are:

1 + i 1 + 2J 2 + £ 2 + 2J+ 1 2 { 21

1 2 i 2£

1 1 2 1 + i 1 + 2J

2 2 1 2 + i 2 + 2J

£ 1 l + { 2 + f 2{

2£ 2{ l + 2{ 2 + 24 «

1 + i 1 + 1 2 + C i i + 2i 1

l + 2{ 1 + 2| 2 + 2| 2{ 1 1 + 1

2 + 4 2 + { { l + « 2 + 2{ 2

2 + 2J 2 + 2£ 2{ 1 + 2J 2 2 + £

1 + 1

2 + J

1 + 2£

1

2 + 2J

2

2{

1 + 2J

2 + 2|

2i

1

l+{

2

2 + {

i

2 + i

2 + 24

2

2£

l + 2{

1

2 + 24

24

1 + 24

2

2 + 4

1

l + £

Table 12-1


« 1 2 £ 2£ l + £ 1 + 24 2 + 4 2 + 24

1 1 2 { 21 1 + i 1 + 24 2 + 4 2 + 24

2 2 1 2{ £ 2 + 24 2 + 4 1 + 24 l + {

i £ 2« 2 1 2 + 4 l + £ 2 + 24 1 + 24

2« 2i « 1 2 1 + 24 2 + 24 l + £ 2 + 4

l + £ 1 + C 2 + 24 2 + £ 1 + 24 24 2 1 i

l + 2£ l + 2£ 2 + J l + « 2 + 24 2 « 24 1

2 + i 2 + £ 1 + 2J 2 + 24 l + £ 1 24 i 2

2 + 2£ 2 + 2? l + £ 1 + 2C 2 + 4 £ 1 2 24

Table 12-2

Example 12: Let f-Q and take \(x) = x^ + x + 1, a prime polynomial over f. Find the mul-

tiplicative inverse of 42 + 4 + 1 G f[x]/(\(x)).

Here ?"[a]/(x(«)) = {ao + «il + «2£^: 09,01,02 eQ} and, since X(4) = 4* +4 + 1 = 0, we have 4* = —1 — 4 and 4* = —4 — 4*. One procedure for finding the

required inverse is:

set (ao + aii + a2i^)(l + i + e) = l,

multiply out and substitute for 4^ and 4*,

equate the corresponding coefficients of 4", 4,4^

and solve for Oq, Oi, a^.

This usually proves more tedious than to follow the proof of the existence of the

inverse in Problem 13. Thus, using the division algorithm, we find

1 = ^(43 + 4+l)(l-4) + i(42 + 4 + l)(£^-24 + 2)

Then (43 + 4 + 1)I[1 - ^(1" + £ + 1)(£' - 2| + 2)]

so that i(42 + 4 + 1)(£* - 24 + 2) = 1

and ^(4^ — 24 + 2) is the required inverse.

Example 13: Show that the field R[x]/{x^ + l) is isomorphic to C.

We have R[x]/(x^ + D = {ô + Oil = «o.ai ^ «}• Since 4^ = -1, the mapping

is an isomorphism of R[x]/(x^ + 1) onto C. We have then a second method of con-

structing the field of complex numbers from the field of real numbers. It is, how-

ever, not possible to use such a procedure to construct the field of real numbers

from the rationals.


Theorem XX. If a{x) of degree m^2 is an element in f[x], then there exists a field f,

For a proof, see Problem 14.where f C f, in which a(x) has a zero.

It is to be noted that over the field f of Theorem XX, a{x) may or may not be written

as the product of m factors each of degree one. However, if a(x) does not factor completely

over f, it has a prime factor of degree n — 2 which may be used to obtain a field f",

with f Cf C f", in which a(x) has another zero. Since a(x) has only a finite number of

zeros, the procedure can always be repeated a sufficient number of times to ultimately

produce a field J"'" in which a(x) factors completely.g^^ Problem 15.


Solved Problems

1. Prove: Let <3^ be a ring with unity u; let

a{x) = ao + aix + a^x^ + • • • + Oma;"' e %[x]

be either the zero polynomial or of degree m; and let

P{x) = 6o + 6ia; + hiX^ +•••+««» G %[x]

be a monic polynomial of degree n. Then there exist unique polynomials qj^x),r^{x) £ %[x]with r^(a;) either the zero polynomial or of degree < n such that

(i) «{x) = q^(x)-p{x) + r^{x)

If a{x) is the zero polynomial or if n> m, then (i) holds with qîx) = z and rîx) = a(x).

Let n ^ m. The theorem is again trivial if m = or if to = 1 and m = 0. For the casein = n = l, take a(a;) = Oq + aix and ;8(a;) = bo + ««• Then

a(x) = oo + oja; = ai(6o + ux) + (oq - a^bo)

and the theorem is true with ^^(a;) = Oj and r„(x) = Oq - ajfeo-

We shall now use the induction principle of Problem 27, Chapter 3, Page 37. For this purposewe assume the theorem true for all a{x) of degree ^ m-1 and consider a{x) of degree m. Nowy(x) - a(a;) - o„a;» >» • ^(x) € '^[a;] and has degree < to. By assumption,

y(x) = S(x) • p(x) + r{x)

with r(x) of degree at most n — 1. Then

a{x) = y(x) + a^x"-"' p(x) = (s(x) + o,„a;»-'») • /8(x) + r{x)

= ««(«) • Pix) + rR(«)

where 9rR(a;) = 8(x) + a„xn-m and rn(x) = r{x), as required.

To prove uniqueness, suppose

«(*) = <1r(x) ' p{x) + rîx) = q'nix) ' p{x) + r'nix)

'^•'«'»(9R{x)~g'J^{x))'P{x) = r'jîx) ~ rîx)

Now r;j(x)-rR(x) has degree at most n-1 while, unless qnix) - q'j,(x) = z, (^(a:) -««(«))• j8(«)has degree at least n. Thus qîx) ~ g'j^(x) ^ z and then r^ (x) - r^Cx) = z, which establishesuniqueness.

2. Prove: The right remainder when a{x) = ao + aix + a^x^ + • • • + Oma;"* e <5^[«] is dividedby a; - 6, 6 G g^, is r^ = ao + ai6 + 026* + • • • + amft".

Consider

«(») - '•r == Oi(« - 6) + a2(«'' - 6'') + • • • + a„(x"' - 6"»)

= {oi + 02(« + 6) + • • • + a„(x">-i + 6x">-2 + • . . + ftm-i)} . (x- 6)

= 9r(«) • (x - b)

"^•i^n «(«) = gR(«)'(«-6) + th

By Problem 1 the right remainder is unique; hence r^ is the required right remainder.

3. In the polynomial ring S[x] over S, the ring of Example 1(d), Chapter 11, Page 114,

(a) Find the remainder when «(«) = ex* + dx^ + cx^ + hx + g is divided by Bix") =hx^ + fx + d.

' f'y I

(b) Verify that / is a zero of y{x) = ex* + dx^ + cx"^ + hx +^d.

+ d

cx^ + Ax + 6

bx^ + fx )ca;4 + dx3 + ca;2 + fcac + 9

c«* + ex3 + /x2

fcx3 + ;ia;2 + hx

hx^ + gx^ + ex


(a) We proceed as in ordinary division, with one

variation. Since S is of characteristic two,

s + {—t) = 8-Vt for all s,t G S; hence, in the

step "change the signs and add" we shall

simply add.

The remainder is r{x) = gx + f.

6x2 + dx + g(b) Here P = c, p = of = e, and /< = h. Then

j^^2 + fx + d

y(f) - c/i + de + c2 + 6/ + d jrx + /

z= d+c+fc + / + d = o

as required.

4. Prove the Division Algorithm as stated for the polynomial domain f\x\.

Note. The requirement that p(x) be monic in Theorem II, Page 126, and its restatement for

commutative rings with unity was necessary.

Suppose now that a(x),y3(x) G f[x] with 6„^m the leading coeflScient of /3(x). Then, since 6"'

always exists in f and p'(x) = 6„ ' • /3(x) is monic, we may write

a(x) = q'(x)«/8'(x) + r(x)

= [b„''9'(*)] • [^ • i8'(*)] + »-(x)

= q(x)'P{x) + r(x)

with r(x) either the zero polynomial or of degree less than that of /3(x).

5 Prove- Let a{x) G f[x\ have degree m > and leading coefficient a. If the distinct

elements bi, h^ 6m of f are zeros of a{x), then a{x) = a{x - h,){x - b^) • • • (a; - M-

Suppose m = 1 so that a(x) = ax + a, has, say, b^ as zero. Then a(6i) = a6, + ai = «, "i = -««>i.

a(x) = ax + ai = ax - o6i = a(x - 6i)

The theorem is true for m = 1.

Now assume the theorem to be true for m = fe and consider a(x) of degree fc + 1 with zeros

h^, &2 6„ + i.Since 6i is a zero of a(x), we have by the Factor Theorem

a(x) = g(x) • (x - &i)

where g(x) is of degree fe with leading coefficient a. Since a{b,) = q(h,)' (hj-y = z for

j = 2,3 fe+ 1 and since b^-b^^z for all j ^ 1, it follows that 63,63 6^ + 1are ft

distinct zeros of g(x). By assumption,

g(x) = a(x - 62)(x - 63) • • • (x - 6^ + 1)

Then «("^) = »(* ~ î)(* - 62) • • • (x - 6fc + 1)

and the proof by induction is complete.

6 Prove: Let a{x),p{x) &f[x] be such that a{s) =m for every s&f. Then, if the

number of elements of f exceeds the degree of both a{x) and p{x), we have necessarily

a{x) = ^(x).

Set y(x) = a(x) - B(x). Now y(x) is either the zero polynomial or is of degree p which certainly

does nit exceed h greaier of th'e degrees of .(x) and ^(x). By hypothesis, y(«) = .(s) -m for

every gey Then y(x) = z (otherwise y(x) would have more zeros than its degree, contrary to

Theorem VII, Page 128) %nd a(x) = /3(x) as required.


7. Find the greatest common divisor of a{x) = 6x^ + Ix* -bx^ -2x^ ~ x + 1 and p{x) =6x* - 5a;3 - \Qx^ - 13a; - 5 over Q and express it in the form

d{x) = s{x)-a{x) + t{x)-p{x)

Proceeding as in the corresponding problem with integers, we find

a(x) = (x + 2)' p(x) + (24a:3 + 49a;2 + ZQx + 11) = q^(x) • ^(x) + ri(x)

^^*^ " 32 ^^'^ ~ ^^^ * ''^*^ + H <^*^ + 2a; + 1) = q^(x) • rîx) + rîx)

ri(x) = ^(256x + 352)-r2(a;)

Since r2ix) is not monic, it is an associate of the required greatest common divisor d(x) = x^ + %x + 1.

Now Viix) = fi(x) - q2(x)'ri(x)

= p{x) - q2{x)'a(x) + qi(x) • q2(x) ' p{x)

= - g2(«) • cc(x) + (1 + qi(x) • q2(x)) • j8(«)

^(8» - 23) • a(x) + ^ (8x^ -7x- 14) • p(x)32^- ""' """

' 32

279 ' ^"" = ~ 279 ^^'^ ~ ^^> • "^*^' "^279

and d(a;) = —- r2(x) = -5™ (8x - 23) • a{x) + ~ (8x2 _ 7^ _ 14) . ^(^)

8. Prove: Let the non-zero polynomials a(x) and /8(a;) be in f[x]. The monic polynomial

d{x) = So(x)-a(a;) + to{x)'l3{x), So{x),U{x) G f[x]

of least degree is the greatest common divisor of a(x) and /?(«).

Consider the set

S = {«(») • a{x) + t(x) • p(x) : s(a;), t{x) G y[x]}

Clearly this is a non-empty subset of f[x] and, hence, contains a non-zero polynomial S(x) of leastdegree. For any b(x) e S, we have, by the Division Algorithm, b(x) - q(x) • S(x) + r{x) wherer(x) G S (prove this) is either the zero polynomial or has degree less than that of S{x). Then r(x) = zand b{x) = q{x) • S(x) so that every element of S is a multiple of S{x). Hence, 8{x)

\a(x) and

S{x)\p(x).^ Moreover, since 8(x) = Ux) ' cc{x) + iîx) • ^{x), any common divisor e(x) of a(x) and^(x) is a divisor of S(x). Now if 8(x) is monic, it is the greatest common divisor d(x) of o(x) and /8(x);otherwise, there exists a unit v such that v - S(x) is monic and d(x) = v • S(x) is the required greatestcommon divisor.

9. Prove: Let a{x) of degree m^2 and p{x) of degree n^2 be in f[x]. Then non-zeropolynomials ^(x) of degree at most n-1 and v{x) of degree at most m- 1 exist in f[x]such that

(c) ;Li(a;) • «(a;) + v{x) • ;8(a;) = z

if and only if a{x) and /8(a;) are not relatively prime.

Suppose S(x) of degree p - 1 is the greatest common divisor of a(x) and y8(x), and write

o(x) - ao(x)'S(x), p(x) = i8o(x)'8(x)

Clearly ao(x) has degree - wi — 1 and /8o(x) has degree -n-1. Moreover,

ô(») • a(x) = ô(«) • ao(x) • 8(x) r= ao(«) ' [;3o(x) • 8(x)] = oo(x) • /3(x)

so that p^(x) • a(x) + [-ao(«) • j8(x)] = z

and we have (c) with /((x) = /3o(x) and i'(x) = —a(,(x).


Conversely, suppose (c) holds with a{x) and I3{x) relatively prime. By Theorem XV, Page 131,

we haveu - s{x) • a(x) + t{x) ' fi(x) for some s(x), t(x) G f[x]

Then ii{x) = ft(x) - s(x) • a(x) + ^l(x) - t(x) • p(x)

= 8(X)[-V(X)-P(X)] + ,,(x)'t{x)'P(x)

= P(x){^(x)'t{x) - v{x)-s(x)]

and P(x)I/t(x). But this is impossible; hence (c) does not hold if a(x) and p{x) are relatively prime.

10. Prove: The unique factorization theorem holds in T[A-

Consider an arbitrary a(x) € f[x\. If o(x) is a prime polynomial, the theorem is trivial. If a(x)

is reducible, write a(x) = a • p(x) • y(x) where p(x) and y{x) are monic polynomials of positive degree

less than that of a(x). Now either p(x) and y(a;) are prime polynomials, as required in the theorem,

or one or both are reducible and may be written as the product of two monic polynomials. If all factors

are primes, we have the theorem; otherwise This process cannot be continued indefinitely

(for example, in the extreme case we would obtain a(x) as the product of m polynomials each of

degree one). The proof of uniqueness is left for the reader, who also may wish to use the induction

procedure of Problem 27, Chapter 3, Page 37, in the first part of the proof.

11. Prove: The polynomial ring f\x] over the field f is a Euclidean ring.

For each non-zero polynomial a(x) e f[x\, define e(a) - m where m is the degree of «(»).

If o(x), i8(x) e T[x] have respective degrees m and n, it follows that »(a) = m, e(l3) = n, e(a • p) -

m^n and, hence, »(« • j8) ^ 9(a). Now we have already established the division algorithm:

a(a;) = q(x) • P(x) + r(x)

where r(x) is either z or of degree less than that of fi{x). Thus, either r{x) = z or »(r) < e(P) as

required.

12. Prove: The ring T[x]/{\{x)) contains a subring which is isomorphic to the field f.

Let a,b be distinct elements of f; then [a], [b] are distinct elements of f[x]/(\{x}) since [a] = [b]

if and only if \{x)|(o - 6).

Now the mapping a -» [o] is an isomorphism of T onto a subset of f[x]/(\(x)) since it is

one-to-one and the operations of addition and multiplication are preserved. It will be left for the

reader to show that this subset is a subring of f[x]/(\(x)).

13. Prove: The ring f[x]/(kix)) is a field if and only if X{x) is a prime polynomial over f.

Suppose X(a;) is a prime polynomial over f. Then for any [a(x)] ^ [z] of f[x]/(x(x)), we have

by Theorem XV, Page 131,

u - a(a;)')8(«) + \(x)'-)'(a;) for some ^{x),y(x) & f\x\

Now X(x) \u-ci^x)-p(x) so that [a(a;)] • \fi(x)\ = \u\. Hence, every non-zero element [«(*)] e y[x]/(x(x))

has a multiplicative inverse and y[x]/(x(x)) is a field.

Suppose X(x) of degree m s= 2 is not a prime polynomial over f, i.e. suppose X(x) = /i(x) • i'(x)

where m(«). »(«) ^ f\A have positive degrees s and t such that s 4-1 = m. Then s < m so that

X(x)/M«) and [;»(x)]^[2]; similarly, \v{,x)\ ¥ \z\. But [/.(x)] • [Kx)] = W^^) • K*)] = [X(x)] = [z].

Thus, since [/.(x)], [»(»)] e y[*]/(?^(a')). i* '"Hows that f[x]/(x(x)) has divisors of zero and is not

a field.


14. Prove: If a(x) of degree m^2 is an element of f[x], there exists a field J^', where

T c f, in which a{x) has a zero.

The theorem is trivial if a(x) has a zero in f; suppose that it does not. Then there exists a monicprime polynomial \(x) G f[x] of degree w - 2 such that \(x)

|a(x). Since Mx) is prime over f,

define T' = fix] / (\{x)).

Now by Theorem XVIII, Page 132, f G T' so that a(x) e. f'[x]. Also, there exists | e f' suchthat Mi) — [«]• Thus J is a zero of a{x) and y" is a field which meets the requirement of the theorem.

15. Find a field in which x^-SgQ[x] (a) has a factor, (6) factors completely.

Consider the field Q[x]/{x^ - 3) = {oq + ajl + Oal^ : a^,o,,a2 ^ Q}

(o) The field just defined is isomorphic to

T = {oo + OiV'3 + O2V9 : Oo,0i,02 e Q}

in which a;* — 3 has a zero.

(b) Since the zeros of x^-3 are v'S, s/Sa.y/s u*, it is clear that «» - 3 factors completely in

r" = rR.

16. Derive formulas for the zeros of the cubic polynomial a{x) = Oo + aix + azx^ + x^ overC when Oo ^ 0.

The derivation rests basically on two changes to new variables:

(i) If 02 = 0, use X = y and proceed as in (ii) below; if 02 ¥= 0, use x = y + v and choose v sothat the resulting cubic lacks the term in y^. Since the coefficient of this term is aj + 3v, theproper relation is x = y — a^/i. Let the resulting polynomial be

P(y) = a(y - a^Z) = q + py + y»

If 9 = 0, the zeros of p(y) are 0, V—P. — V—p and the zeros of a{x) are obtained by decreasingeach zero of p(y) by 02/8. If g ^ but p — Q, the zeros of p(y) are the three cube roots p,wp, u^p(see Chapter 8) of -q from which the zeros of a(x) are obtained as before. For the case pq ¥= 0,

(ii) Use 1/ = z — p/3z to obtain

r(2) = p(z-plZz) = z3 + ,_p3/2723 = z^ + q^- p^m

Now any zero, say s, of the polynomial S(z) = z* + g«3 - p3/27 yields the zero s - p/3s - aj/Sof a(x); the six zeros of S(z) yield, as can be shown, each zero of a(x) twice. Write

&{z) = [z3 + ^{q - y/q2 + 4p3/27 )]. [«3 + ^(g + Vg^' + 4p»/27

)]

and denote the zeros of «* + ^{q - y/q^ + 4p*/27 ) by A,«A,u2A. The zeros of a(a;) are then:

A - p/BA — 02/3, <oA - a^p/SA - 02/3, and uÂ - wp/3A - a^/S.

17. Find the zeros of a{x) - -11 - 3x + Sx^ + x^.

The substitution x = y — 1 yields

P{y) = a(y-l) = -6 -63/ + 1/3

In turn, the substitution y = z+ 21z yields

y(z) = P(z + 2lz) = z^ + S/z»-6 = '" " ^'^ + « = (z^-2)i^-4)

8 ^ '^

Take A = V2; then the zeros of a(x) are:

^+\/4-l, «v^ + o,2\/4-l, <o2v^+a>\/4-l

The reader will now show that, apart from order, these zeros are obtained by taking A = \/4.

140 POLYNOMIALS (CHAP. 12

18. Derive a procedure for finding the zeros of the quartic polynomial

aix) = ao + aix + aix^ + as** + a;* € C{x\, when ao ¥- 0.

If 03 # 0, use X = 3/— 03/4 to obtain

/8(j/) = a{y - 03/4) = 60 + 6iJ/ + *2J/^ + a/*

Now /8(j/) = (p + 293/ + y^W - 2gy + j/2)

= pr + 2g(r-p)y + (r + p- 4q2)5^2 ^. ^4

provided there exist p,q,rec satisfying

pr = 60. 2g(r - p) = fci, r + p - 4g2 = 6^

If 61 = 0, take q = 0; otherwise, with g j^ 0, we And

2p = 62 + 492 - 6i/2g and 2r - 62 + ^g^ + bi/2q

Since 2p • 2r = 46o, we have

(i) 64g6 + 32529* + 4(6|-46o)g2-b? =

Thus, considering the left member of (i) as a cubic polynomial in q\ any of its zeros different from

will yield the required factorization. Then, having the four zeros of p{y), the zeros of a(x) are obtained

by decreasing each by O3/4.

19. Find the zeros of a(x) = 35 - 16a; - 4a;» + x\

Using X = y + 1, we obtain

P(V) = «(3/ + l) = 16 - 243/ - 63/2 + J/*

Here, (i) of Problem 18 becomes

6496 - 1929^ - 11292 - 576 = 16(92 - 4)(49* + 492 + 9) =

Take 9 = 2; then p = 8 and r = 2 so that

16 - 243/ - 63/2 + 3/* = (8 + 43/ + 3/2)(2 - 4j/ + 3/2)

with zeros -2 ± 2i and 2 ± V2. The zeros of a(x) are: -1 ± 2j and 3 ± ^2.


20. Give an example of two polynomials in x of degree 3 with integral coefficients whose sum is of degree 2.

21. Find the sum and product of each pair of polynomials over the indicated coefficient ring. (For con-

venience, [a], [b], .. . have been replaced by a,b, . .. .)

(o) 4 + X + 2a2, 1 + 2« + 3a;2; 1/(5)

(6) 1 + 5a; + 2a;2, 7 + 2a; + 3a;2 + 4a;3; 7/(8)

(c) 2 + 2x + aS, 1 + a; + x2 + x*; 1/(3)

Ans. (a) 3x; 4 + 4a; + x2 + 2x3 + x*

(c) x2 + x3 + a;*; 2 + x + x2 + x'


22. In the polynomial ring S[x] over S, the ring of Problem 2, Chapter 10, Page 108, verify:

(a) (b + gx + /a;2) -^ (d + gx) = c + ex + fx^

(b) (b + gx + fx^){d + ex) = b + hx + cx^ + bx^

(c) (b + gx + fx^)(d + ex) = b + ex + bx^

(d) f + bx and e + ex are divisors of zero.

(e) c is a zero of f + ex + fx^ + ex"^ + dx^.

23. Given a(,x),^(x),y{x) &f\x\ with respective leading coefficients a,b,e and suppose a(x) = p{x)'y{x).Show that a{x) = a • fi'(x) • y'{x) where p'{x) and y'{x) are monic polynomials.

24. Show that ^[x] is not a field for any integral domain ^.Hint. Let a(x) e ^[a;] have degree > and assume p(x) G ^[x] a multiplicative inverse of a(x).

Then a(x) • pix) has degree > 0, a contradiction.

25. Factor each of the following into products of prime polynomials over (i) Q, (ii) R, (iii) C.

(a) x*-l (c) 6a;* + 5a;3 + Ax^ - 2x - 1

(6) a;4 - 4a;2 - a; + 2 (d) 4a;5 + 4x* - 13a;3 - lla;2 + lOx + 6

Ans. (a) (x - l)(x + l){x^ + 1) over Q, R; (x - l)(x + l)(x - i)(x + i) over C(d) (x - l)(2a; + l)(2a; + 3)(x' - 2) over Q; (x ~ l)(2a; + l)(2a; + 3){x - y/2)(x + VI ) over R,

C

26. Factor each of the following into products of prime polynomials over the Indicated field. (See notein Problem 21.)

(o) a;2 + 1; //(5) (c) 2a;2 + 2x + 1; //(5)

(6) «2 + X + 1; 7/(3) (d) 3x3 + 4^.2 + 3. 7/(5)

Ans. (a) (a! + 2)(« + 3), (d) (x + 2)2 (3* + 2)

27. Factor x*-l over (a) 7/(11), (6) 7/(13).

28. In (d) of Problem 26 obtain also 3x3 + 4^.2 + 3 = (j. + 2)(x + 4)(3x + 1). Explain why this does notcontradict the Unique Factorization Theorem.

29. In the polynomial ring S[x] over S, the ring of Example 1(d), Chapter 11, Page 114,

(a) Show that 6x2 -|- ex + jr and Sfx2 + dx + b are prime polynomials.

(b) Factor hx* + eafl + cx^ + b.

Ans. (b) {bx + b)(ex + g)(gx + d)(hx + e)

30. Find all zeros over C of the polynomials of Problem 25.

31. Find all zeros of the polynomials of Problem 26. Ans. (o) 2, 3; (d) 1, 3,

3

32. Find all zeros of the polynomial of Problem 29(6). Ans. 6, e, /, d

33. List all polynomials of the form 3x2 + ex + 4 which are prime over 7/(5).

Ans. 3x2 + 4^ 3a;2 + a; + 4^ 34.2 + 4j. + 4

34. List all polynomials of degree 4 which are prime over 7/(2).

35. Prove: Theorems VII, IX, and XIII.

36. Prove: If o + 6\/c, with a,6,c € Q and c not a perfect square, is a zero of a(x) e 7[x], so also isa — byfe.


37. Let "^ be a commutative ring with unity. Show that ^[ar] is a principal ideal ring. What are its

prime ideals?

38. Form polynomials a{x) G I[x] of least degree having among its zeros:

(a) V3 and 2 (c) 1 and 2 + 3-\/5 (e) 1 + i of multiplicity 2

(6) i and 3 (d) -1 + i and 2 - 3i

Ans. (a) a;3 - 2x2 _ 33; + e, (d) x* - 2x3 + 7-152 + iga- + 26

39. Verify that the minimum polynomial of Vs + 2t over B is of degree 2 and over Q is of degree 4.

40. Find the greatest common divisor of each pair a(x), j8(x) over the indicated coefficient ring and express

it in the form s(x) • a(x) + t(x) • p(x).

(a) x« + X* - x3 - 3x + 2, x3 + 2x2 - X - 2 ; q

(6) 3x4 _ 6x* + 12x2 + Sx-6, x^ - 3x2 + 6x - 3 ; q

(c) xS - 3tx3 - 2ix2 - 6, x2 - 2t ; C

(d) x5 + 3x3 + x2 + 2x + 2, X* + 3x3 + 3x2 + X + 2 ; 7/(5)

(e) x5 + x3 + X, x* + 2x3 + 2x ; 7/(3)

(/) ex* + fcx3 + ax2 + jrx + e, ^x3 + Ax* + dx + fif ; S of Example 1(d), Chapter 11, Page 114.

Ans. (b) ^ (37x2 -108X + 177) -aCx) + ^(-111x3 + 213x2- 318x - 503) • j8(x)

(d) (x + 2) • «(x) + (4x2 + X + 2)' p(x)

(/) (flrx + h) • a(x) + (cx2 + hx + e)' p(x)

41. Prove Theorem XVII, Page 132.

42. Show that every polynomial of degree 2 in T[x\ of Example 11, Page 133, factors completely in

y[x]/(x2 + 1) by exhibiting the factors.

Partial Ana. x2 + 1 = (x + £)(x + 2{); x2 + x + 2 = (x + { + 2)(x + 2{ + 2);

2x2 + X + 1 = (a, + J + l)(2x + 1 + 2)

43. Discuss the field Q[x]/(x3 - 3). (See Example 10, Page 133.)

44. Find the multiplicative inverse of {2 + 2 in (a) Q[x]/(x3 + x + 2), (6) f[x]/(x2 + x + 1) when f = 7/(5).

Ans. (o) i(l-2j-{2), (6) id + 2)

chapter 13

Vector Spaces

(a+c,b + d)

Fig. 13-1

INTRODUCTIONIn this chapter we shall define and study a type

of algebraic system called a vector space. Beforemaking a formal definition, we recall that in elemen-tary physics one deals with two types of quantities:

(a) scalars (time, temperature, speed) which havemagnitude only and (6) vectors (force, velocity,

acceleration) which have both magnitude and direc-

tion. Such vectors are frequently represented byarrows. For example, consider in Fig. 13-1 a givenplane in which a rectangular coordinate system hasbeen established and a vector i^ = OP= (a, 6) join-ing the origin to the point P(a, b). The magnitudeof I, (length of OP) is given by r = \/aFT¥ andthe direction (the angle 6, always measured from thepositive a;-axis) is determined by any two of therelations sin 6 = b/r, cos 6 = a/r, tan 9 = b/a.

Two operations are defined on these vectors:

Scalar Multiplication. Let the vector |j = (o, b) represent a force at 0. The productof the scalar 3 and the vector i^ defined by 3f, = (3a, 36) represents a force at O having thedirection of |, and three times its magnitude. Similarly, -2|, represents a force at O havingtwice the magnitude of |, but with direction opposite that of |

.

Vector Addition. If ^, = (a, 6) and 1^ = (c, d) represent two forces at O, their result-ant I (the single force at having the same eflfect as the two forces |, and IJ is given by^ = li + ^2 = ("^ + c, 6 + d) obtained by means of the parallelogram law.

In the example above it is evident that every scalar sER and every vector êRxR.There can be no confusion then in using (+) to denote addition of vectors as well as additionof scalars.

Denote by V. the set of all vectors in the plane, (i.e., V = RxR). Now V has a zeroelement C = (0, 0) and every | = (a, 6) e F has an additive inverse -^ = (-a, -b) G Vsuch that ^ + (-1) = C; in fact, V is an abelian additive group. Moreover, for all s,t G Rand $,rj gV, the following properties hold:

Sd + V) = Si + Sr,

sm = (St)^

{S + «)| = Si + ti

u = $

Example 1: Consider the vectors { = (1,2), 17 = (|,0), o = (0,-3/2). Then

(o) 3{ = 3(1,2) = (3,6), 2, = (1,0), and 3j + 2, = (3,6) + (1,0) = (4,6).

(6) i + 2v = (2,2), „ + a = (^, -3/2), and 5(J + 2,) - 4(, + a) = (8, 16).

143

144 VECTOR SPACES [CHAP. 13

VECTOR SPACES

Let 7^ be a field and V be an abelian additive group such that there is a scalar multipli-

cation of V by f which associates with each s ef and | G 7 the element s| € V. Then

V is called a vector space over f provided, with u the unity of f,

(i) s{i + ri) = si + srj (iii) s(U) = (st)t

(ii) (s + *)| = s| + f| (iv) Ml = I

hold for all s,tGf and I,,, G F.

It is evident that, in fashioning the definition of a vector space, the set of all plane

vectors of the section above was used as a guide. However, as will be seen from the

examples below, the elements of a vector space, i.e. the vectors, are not necessarily quantities

which can be represented by arrows.

Example 2: (a) Let f = fi and y = Vg (fl) = {(a„ 02) : Oi. <»2 ^ ^} with addition and scalar

multiplication defined as in the first section. Then, of course, V is a vector

space over f\ in fact, we list the example in order to point out a simple generali-

zation: Let f = R and V = V„(R) = {(oj, O2, . .., o„) : Oj G B} with addition

and scalar multiplication defined by

£ + 1, = (Oi, 02 «„) + (61, 62 ''n)

= (Oi + 6|, 02 + ''2. • • • . On + *n)t ?. 17 S ^n (*)

and8{ = 8(Oi, 02. • • • . »n) == (»*i> ^'Ht ••> «««)

'

sBR, i&V„(B)

Then y„ (R) is a vector space over R.

(b) Let r = R and y„(0 = {(ai.Oz «n) : »i e C} with addition and scalar

multiplication as in (a). Then V„ (C) is a vector space over R.

(c) Let f be any field, 7 = f[x] be the polynomial domain in x over y, and define

addition and scalar multiplication as ordinary addition and multiplication in

f[x]. Then V is a vector space over f.

Let 7^ be a field with zero element « and V be a vector space over f. Since V is an

abelian additive group, it has a unique zero element C and, for each element I G V, there

exists a unique additive inverse -| such that | + (-1) = C- By means of the distributive

laws (i) and (ii), we find for all sGf and IGF,

s^ + z^ = (s + z)$ = s^ ^ s^ + C

and si + sC = s(| + = si = s$ + C

Hence, zi = C and sC = C-

We state these properties, together with others which will be left for the reader to estab-

lish, as

Theorem I. In a vector space V over f with z the zero element of f and g the zero element

of V, we have

(1) sC = C for all sGT(2) z| = C for all | G V

(3) (-s)l = s(-|) = -(s|) for all s G f and I G 7

(4) If si = C. then s = z or i = C

SUBSPACE OF A VECTOR SPACE

A non-empty subset C7 of a vector space V over ^ is a subspace of V provided U is itself

a vector space over f with respect to the operations defined on V. There follows

CHAP. 13] VECTOR SPACES 145

Theorem II. A non-empty subset C7 of a vector space V over 7 is a subspace of V if andonly if U is closed with respect to scalar multiplication and vector additionas defined on V.


Example 3: Consider the vector space V = V3 (R) = {(a, b, c) : a,b,eG R} over R. By TheoremII, the subset U = {{a, 6,0): a,bS R} is a subspace of V since for all 8 G B and(a, b, 0), (c, d, 0) G U, we have

(a, 6, 0) + (c, d, 0) = (o + c, 6 + d, 0) € U

and 8{a, b, 0) = {sa, sb, 0) G 17

In Example 3, V is the set of all vectors in ordinary space while U is the set of all suchvectors in the ÔY-plane. Similarly, W ^ {(a,0,0): a G R} is the set of all vectors alongthe Z-axis. Clearly, W is a subspace of both U and V.

Let Ij, i^, .. .,i^GV, a vector space over f. By a linear combination of these m vec-tors is meant the vector iGV given by

i = Xcfi, = cj, + c,^^+ ... + cj^, c^^r

Consider now two such linear combinations 2 ('A and '^d^^.. Since

and, for any s e f

,

s 2 c.^^ = 2 (sc.)|.

we have, by Theorem II,

Theorem III. The set U of all linear combinations of an arbitrary set S of vectors of a(vector) space F is a subspace of V.

The subspace U of V defined in Theorem III is said to be spanned by S. In turn, thevectors of S are called generators of the space U.

Example 4: Consider the space VîR) of Example 3 and the subspaces

U = {«(1, 2, 1) + «(3, 1, 5) : 8, t G R}

spanned by {1 = (1,2,1) and |2 = (3,1,5)

and W = {0(1, 2, 1) + 6(3, 1, 5) + c(3, -4, 7) : o, 6. c G B}

spanned by ii, {2. and £3 = (3, -4, 7).

We now assert that U and W are identical subspaces of V. For since(3, -4, 7) = -3(1, 2, 1) + 2(3, 1, 5), we may write

W = {{a- 3e)(l, 2, 1) + (6 + 2c)(3, 1, 5) : a.b.e G R]

= {8'(1. 2, 1) + t'(3, 1, 5) : 8'. t' G R}

- U

Let u = {K^, + K^.+ ---+kjm- K^T}be the space spanned by S = {^j, Ig, . . . , ^„}, a subset of vectors of V over 5^. Now U con-tains the zero vector êV (why?); hence, if C G S, it may be excluded from S leavinga proper subset which also spans U. Moreover, as Example 4 indicates, if some one of thevectors, say $., of iS can be written as a linear combination of other vectors of S, then | mayalso be excluded from S and the remaining vectors will again span U. This raises questionsconcerning the minimum number of vectors necessary to span a given space U and thecharacteristic property of such a set.

See also Problem 2.


LINEAR DEPENDENCEA non-empty set S = {l,,^^ ^J of vectors of a vector space V over f is called

linearly dependent over f if and only if there exist elements k^, k^, . ..,k^ of f, not all

equal to z, such that

SÂ = M1 + M2+ ••• +M™ = ^

A non-empty set S ={|,, 1^,..., ij of vectors of V over f is called linearly independent

over 7^ if and only if

S fciî = ^1^1 + M2 + • • • + fcJ™ = ^

implies every k^ = z.

Note. Once the field f is fixed, we shall omit thereafter the phrase "over f"; moreover,

by "the vector space F„(Q)" we shall mean the vector space Vn{Q) over Q, and similarly

for Vn (R). Also, when the field is Q or R, we shall denote the zero vector by 0. Although

this overworks 0, it will always be clear from the context whether an element of the field

or a vector of the space is intended.

Example 5: (a) The vectors ii= (1,2,1) and i2 = (3,l,5) of Example 4 are linearly inde-

pendent, since if

fcih + fc2«2 = (fci + 3*=2. 2A:, + fca. fei + Bfcj) = = (0, 0, 0)

then fc, + 3fe2 = 0, 2fci + fcz = 0, fei + 5^2 = 0. Solving the first relation for

fej = -Zki and substituting in the second, we find -5^2 = 0; then k^ - and

fcj = -3fc2 = 0.

(6) The vectors £1 = (1. 2, 1), £2 - (3. L 5) and £3 = (3, -4. 7) are linearly dependent,

since 3{i- 2^2 + £3 = 0.

There follow

Theorem IV. If some one of the set S = (|,, I2 U ©^ vectors in V over f is the zero

vector ^, then necessarily S is a linearly dependent set.

Theorem V. The set of non-zero vectors S = {i„i^ U °^ ^ oêr f is linearly

dependent if and only if some one of them, say 1^, can be expressed as a

linear combination of the vectors ^^,^^, . . ..tj-^ which precede it.


Theorem VI. Any non-empty subset of a linearly independent set of vectors is linearly

independent.

Theorem VII. Any finite set S of vectors, not all the zero vector, contains a linearly inde-

pendent subset U which spans the same vector space as S.


Example 6: In the set S = {Ji.h.fs) »{ Example 5(6), £, and £2 are linearly independent while

I3= 2^2 - 34i- Thus, r, = {£1, £2} is a maximum linearly mdependent subset of S.

But, since Ji and I3 are linearly independent (prove this) while {2 = ^(£3 + 3£i),

T2 = {£1, ia) is also a maximum linearly independent subset of S. Similarly,

^3 = {J2, fs) is another. By Theorem VII each of the subsets Ti, T2, T3 spans the

same space as does S.

The problem of determining whether a given set of vectors is linearly dependent or

linearly independent (and, if linearly dependent, of selecting a maximum subset of linearly

independent vectors) involves at most the study of certain systems of linear equations.

While such investigations are not difficult, they can be extremely tedious. We shall post-

pone most of these problems until Chapter 14 where a neater procedure will be devised.

See Problem 5.


BASES OF A VECTOR SPACEA set S = {Ij.lj, •..,!„} of vectors of a vector space V over f is called a basis of V

provided:

(i) S is a linearly independent set,

(ii) the vectors of S span V.

Define the unit vectors of Vn {f) as follows:

£, = {u, 0, 0, 0, . ..,0,0)

e^ = (0, U, 0, 0. . ..,0.0)

.3 = (0, 0, u, 0. . ..,0.0)

£„ = (0.0,0,0. ...,0,u)

and consider the linear combination

I = a,c, + a3t,+ ••• +a„e„ = (a,.a„...,oJ. a.&f WIf I = ^, then ci = 02 = • • • = a„ = z; hence, £? = {«,, t^ «„} is a linearly independent set.

Also, if I is an arbitrary vector of Vn{f), then (1) exhibits it as a linear combination of theunit vectors. Thus the set E spans Vn (f) and is a basis.

Example 7 : One basis of V4 (R) is the unit basis

E = {(1,0,0,0), (0,1,0,0), (0,0,1.0), (0,0,0,1)}

Another basis is

F = {(1,1,1,0), (0,1,1,1), (1.0, 1.1). (1,1, 0,1)}

To prove this, consider the linear combination

J = 01(1, 1.1,0) + 02(0. 1,1,1) + 03(1,0.1,1) + 04(1,1,0.1)

= (oj + 0.3 + 04, a, + 02 + 04, o, + ttj + Og, 02 + 03 + 04) ; Oj G fi

If { is an arbitrary vector (p, g, r, s) G V^ (R), we find

01 = (p + g + r - 2s)/3 03 = (p + r + 8- 29)/3

02 = (g + r + 8-2p)/3 04 = (p + q + s-2r)/3

Then F is a linearly independent set (prove this) and spans VîR).


Theorem VIII. If S = {^,, ^^, .... |^} is a basis of the vector space V over f and^= {71,72' • • 'Vn) is any linearly independent set of vectors of F, thenn^m.

As consequences, we have

Theorem IX. US- {|,. I^, ...,!„} is a basis of V over 7, then any to + 1 vectors of Vnecessarily form a linearly dependent set.

and

Theorem X. Every basis of a vector space V over f has the same number of elements.

The number defined in Theorem X is called the dimension of V. It is evident thatdimension, as defined here, implies finite dimension. Not every vector space has finite

dimension, as shown in

Example 8: (a) From Example 7, it follows that VîR) has dimension 4.

(6) Consider V = {oo + a^x + 02*2 + a^x^ + 04a;* : Oj G fi}

Clearly, B = {1, x, x^, gfi, x*} is a basis and V has dimension 5.

(c) The vector space V of all polynomials in x over R has no finite basis and, hence,is without dimension. For. assume B, consisting of p linearly independentpolynomials of V with degrees - g, to be a basis. Since no polynomial of V ofdegree > q can be generated by B, it is not a basis. o n ui r.See Problem 7.


SUBSPACES OF A VECTOR SPACE

Let V, of dimension n, be a vector space over f and U, of dimension m <n having

S = (Ij, Ij, . .. , !„} as a basis, be a subspace of F. By Theorem VIII, only m of the unit

vectors of V can be written as linear combinations of the elements of B; hence there exist

vectors of V which are not in U. Let rj^ be such a vector and consider

k^$, + kî^+ + kj^ + krj, = i, kuk e f {2)

Now fe = «, since otherwise k~^ G f,

Vi = k~^ (~îî"" K^2 — ... — kîj)

and T/j e U, contrary to the definition of t/j. With k = z, {2) requires each ki = z since

S is a basis, and we have proved

Theorem XI. If S = {|,, i^,...,ij is a basis of U c V and if ri^GV but v, « U, then

S U {t^j} is a linearly independent set.

When, in Theorem XI, m + l - n, the dimension of V, J5i = J5 U {i/,) is a basis of V;

when m + Kn, Bi is a basis of some subspace C/i of V. In the latter case there exists

a vector r)^ in V but not in Ui such that the space Ui, having B U {r/j, t^^} as basis, is either Vor is properly contained in F, . . . . Thus we eventually obtain

Theorem XII. If B = {|,, l^, . .. , |„} is a basis of U CV having dimension n, there exist

vectors r,^, -q^, ..., Vn-m J" ^ such that B U {ijj, t,^, . . . , ^„_,„} is a basis of F.

See Problem 8.

Let U and W be subspaces of F. We define

J7 n IF = {I : I € [/, I € IF}

f/ + T7 = {| + r,:|GC/, r,GlF}

and leave for the reader to prove that each is a subspace of F.

Example 9: Consider V = VîR) over R with unit vectors cj, eg, £3, £4 as in Example 7. Let

and W - {61C2 + ''2«3 + ^3*4 : &i ^ B}

be subspaces of dimensions 3 of V. Clearly,

U nW = {ci«2 + 02^3 : CiSR} , of dimension 2

and C7 + W = {fflie, + a2«2 + «3«3 + î«2 + *2f3 + ''3^4 : «t. &t ^ *>


Theorem XIII. If U and IF, of dimension r^n and s^n respectively, are subspaces of a

vector space F of dimension n and if f/ n PF and f/ + PF are of dimensions

p and t respectively, then t = r + s-p.^^^ ^ ^^^^^^ ^^ p^^^j^^ 9

VECTOR SPACES OVER R

In this section, we shall restrict our attention to vector spaces V = Vn (R) over 22.

This is done for two reasons: (1) our study will have applications in geometry and (2) of all

possible fields, R will present a minimum in difficulties.


Consider in V = V2(R) the vectors |= (01,02)

and r, = (61, 62) of Fig . 13-2. The length ||| of $ is

given by||| = y/gf + a\ and the length of r, is given

by \r}\ = y/bf+bl. By the law of cosines we have

|^-,p = IIP + |,|2 _ 2 III- H cose

{al + aD + jbl + bl) - [{at - h.f + (02 - 62)"]

so that

cos

aibi + 02^2

lll-M

2 III'

Fig. 13-2

The expression for cos e suggests the definition ofthe scalar product (also called dot product and innerproduct) of I and t) by

^•q = O161 4- O262

Then ||| - vT^, cos e = ^^'' and the vectors | and i? are orthogonal (i.e., mutually per-

pendicular, so that cos e = 0) if and only if | • i; = 0.

In the vector space V = Vn (R) we define for all | = (a,, 02, .... a„) and v = (6i, 62, ... , 6«),

i'V — 2 "^î — Ol^l + C'^^2 + • • • + Onbn

III = VFl = Val + al+---+al

There follow

(1) |s|| = \s\ •HI for alU e F and s e /J

(2) HI ^ 0, the equality holding only when $ =

(3) \i'r)\ ^ \i\'\if\ (Schwarz inequality)

(4) \i + rj\^ HI + |,,| (Triangle inequality)

(5) I and t) are orthogonal if and only if | •1; = 0.

For a proof of (3), see Problem 10.


Suppose in F„(/e) the vector 7; is orthogonal to each of I,,l2'- -'Im- Then, since

Vli = '?-l2= •• ='/*l„ = 0, wehave v*(c,l, + C2l2+ •• • +cJJ = forany CtGR and

have proved

Theorem XIV. If, in Vn (R), a vector i? is orthogonal to each vector of the set (I,, i^ |„},then ri is orthogonal to every vector of the space spanned by this set.

See Problem 14.

LINEAR TRANSFORMATIONSA linear transformation of a vector space V{f) into a vector space W{f) over the same

field f is a mapping T of V{f) into W(f) for which

(i) (|. + |.)r = i,r + i.r for all i,. i, g Vif)

(ii) (sljr = sii.T) for all |, G V(f) and s G ^We shall restrict our attention here to the case W(f) = V{f), i.e. to the case when T

is a mapping of V{f) into itself. Since the mapping preserves the operations of vector


addition and scalar multiplication, a linear transformation of V{f) into itself is either an

isomorphism of V{f) onto Vif) or a homomorphism of V{f) into V{f).

Example 10: In plane analsrtic geometry the familiar rotation of axes through an angle a is a

linear transformation

T : (x, y) -» (x eoaa — y sin a, x sin a + y cos a)

of ViiR) into itself. Since distinct elements of V-i(R) have distinct images and

every element is an image (prove this), T is an example of an isomorphism of V(R)

onto itself.

Example 11 : In Vg (Q), consider the mapping

T: (o,5,c) - (o+6 + 5c, a + 2cj 2b + 6c), (a.b.c) G VgiQ)

For (a, 6,c),(d,e,/) 6 VglQ) and s G Q, we have readily

(i) (a,h,c) + (d,e,f) = (a + d,b + e,e + f) -*

(a + d + b + e + 5e + 5f, o + d + 2c + 2/, 26 + 2e + 6c + 6/)

= {a+b + 5e, a + 2e,2b + 6e) + (d + e + 5/, d + 2/, 2e + 6/)

i.e.. [{a, b, c) + {d, e,f)]T = (o, b,e)T + (d, e, f)T

and

(ii) «(a, b, c) = («o, ab, se) -» (sa + sb + bse, sa + 2«c, 286 + 6«c)

= s(o+6 + 6c, a + 2c, 26 + 6c)

i.e., [8{a,b,e)]T = «[(a,6,c)r]

Thus r is a linear transformation on V3 (Q).

Since (0,0,1) and (2,3,0) have the same image (5,2,6), this linear transforma-

tion is an example of a homomorphism of VgiQ) into itself.

The linear transformation T of Example 10 may be written as

x(l, 0) + 2/(0, 1) -* a;(cos a, sin a) + y{- sin a, cos a)

suggesting that T may be given as

T : (1, 0) -* (cos a. sin «), (0, 1) -> (- sin a, cos a)

Likewise, T of Example 11 may be written as

a(l,0,0) + 6(0,1,0) + c(0,0,l) -> c(l,l,0) + 6(1,0,2) + c(5,2,6)

suggesting that T may be given as

T : (1, 0, 0) -* (1, 1. 0), (0, 1, 0) -* (1, 0, 2), (0, 0, 1) -» (5, 2, 6)

Thus we infer:

Any linear transformation of a vector space into itself can be described completely

by exhibiting its effect on the unit basis vectors of the space.^^ Problem 15

In Problem 16, we prove the more general

Theorem XV. If U,,^^, ...,!„} is any basis of V = Vif) and if {v,,V2' •••''»»} ^^ any

set of n elements of V, the mapping

T: l^^ Vi. (i = 1,2, ...,«)

defines a linear transformation of V into itself.



Theorem XVI. If T is a linear transformation of V(f) into itself and if W^ is a subspaceof Vif), then W^ = {^T : | G W], the image of W under T, is also a sub-space of V{f).

Returning now to Example 11, we note that the images of the unit basis vectors of Fa (Q)are linearly dependent, i.e., 2^J + 3€^T- ,^T = (0,0,0). Thus, V^cV; in fact, since(1,1,0) and (1,0,2) are linearly independent, V^ has dimension 2. Defining the rank of alinear transformation T of a vector space V to be the dimension of the image space V^ of Vunder T, we have by Theorem XVI,

Vj. = rank of T ^ dimension of VWhen the equality holds, we shall call the linear transformation T non-singular; otherwisewe shall call T singular. Thus T of Example 11 is singular of rank 2.

Consider next the linear transformation T of Theorem XV and suppose T is singular.Since the image vectors r,. are then linearly dependent, there exist elements s^ € f, notall 2, such that '^s^r,. = C. Then, for I = 2)«iî, we have ^r = C. Conversely, suppose>? = 2 tA ^ C and

vT ^ XmT = tîi^T) + m^T) +..+ tîij) = c

Then the image vectors ^.T, (i = 1,2, . . .,w) must be linearly dependent. We have proved

Theorem XVII. A linear transformation T of a vector space V{f) is singular if and only if

there exists a non-zero vector $ G V(f) such that IT = ^.

Example 12: Determine whether each of the following linear transformations of V4 (Q) into itselfis singular or non-singular:

€, ^ (1,1,0,0) fe, -» (1,1,0,0)

(a) A: r^" (0.1.1,0) L, -* (0,1,1,0)

1.3 ^ (0.0,1,1)' ^"^ ^- 1.3 ^ (0,0,1,1)

U -* (0.1,0,1) [^4-* (1,1,1,1)

Let i = (a, 5, c, d) be an arbitrary vector of V4 (Q).

(a) Set |A = (aei + 6^2 + ceg + dti)A = {a, a + b + d, b + c, c + d) ~ 0. Since this re-quires a = 6 = c = d = 0, that is, £ = 0, A is non-singular.

(6) Set iB = (a + d,a+b + d,b + c + d,c + d) = 0. Since this is satisfied wheno = c = 1, 6 = 0, d = -1, we have (1, 0, 1, -1)B = and B is singular. Thisis evident by inspection, i.e., cjB + e^B = C4B. Then, since e,B, €28, e^B areclearly linearly independent, B has rank 3 and is singular.

We shall again (see the paragraph following Example 6) postpone additional examplesand problems until Chapter 14.

THE ALGEBRA OF LINEAR TRANSFORMATIONSDenote by c4 the set of all linear transformations of a given vector space V(f^ over

f into itself and by ffH the set of all non-singular linear transformations in cA. Let addition(-I-) and multiplication (•) on o// be defined by

A+B: HA + B) =^ $A + $B, iG F(f

)

and A'B: Â • B) = {U)B, ^ G ViT)

for all A,B G c/l. Let scalar multiplication be defined on cA by

kA : l(fcA) = (fc|)A, i G Vif)for all A G o/? and kGf.


Example 13: Let ^ r , «

A: ^^!"'t^ and B :

^'^^^''^^

[e^ -* ic,d) 1^2 -* (9,h)

be linear transformations of Vg (*) into itself. For any vector { = (s, t) e V2 (B),

we find

iA = (s, t)A = (8e, + te2)A = 8(a, b) + t(c, d)

= (sa + tc, sh + td)

IB = (se + tflT, 8/ + th)

and

{(A + B) = (8, t)A + (8, t)B = (s(a + e) + t(c + pr), 8(6 + /) + t(d + fc))

Thus we have ^ , . ,.

[eg -» (e-¥g,d+h)

Also,

5(A • B) = ((8, t)A)B = (80 + te, sb + td)B

= (sa + te) • (e, f) + (86 + td) • (g, h)

= (siae + bg) + t(ee + dg), s{af + bh) + t(cf + dh))

Jej - (ae + bg, af + bh)

\^2 * (ee + dg, ef + dh)

Finally, for any k G R, we find

{k^)A = {ks,kt)A = {k{sa + tc),k(sb + td))

kA: h" ^^'^^^

\t^ -* (fee, fed)


Theorem XVIII. The set qA of all linear transformations of a vector space into itself

forms a ring with respect to addition and multiplication as defined

above.


Theorem XIX. The set ffH of all non-singular linear transformations of a vector space

into itself forms a group under multiplication.

We leave for the reader to prove

Theorem XX. If o/i is the set of all linear transformations of a vector space V(f) over

f into itself, then c4 itself is also a vector space over f.

Let A,B GeA. Since, for every I G V,

${A + B) = U + ^B

it is evident that 'îA+ay ^ â + ^b

Then dimension of T^f^+Bj - dimension of V^ + dimension of V^

and ^(A+B) ~ Â + *"»


Since for any linear transformation T GcA,

dimension of V^. ^ dimension of V

we have dimension of Fj^.3 J =^ dimension of F^

Also, since V^ C V, dimension of F,^.^, ^ dimension of F^

Thus'''(A-B^ ~ Â' ^(A-B) "~ **»

Solved Problems

1. Prove: A non-empty subset f/ of a vector space F over 7=" is a subspace of F if andonly if U is closed with respect to scalar multiplication and vector addition as definedon F.

Suppose 17 is a subspace of V; then U is closed with respect to scalar multiplication and vectoraddition. Conversely, suppose U is a non-empty subset of V which is closed with respect to scalarmultiplication and vector addition. Let J 6 J7; then (-m)j = -(m{) = -ê U and J + (-J) = f G {/.

Thus U is an abelian additive group. Since the properties (i)-(iv). Page 144, hold in V, they also holdin U. Thus U is a. vector space over y and, hence, is a subspace of V.

2. In the vector space VsiR) over R (Example 3, Page 145), let U be spanned byI, = (1, 2, -1) and ^ = (2, -3, 2) and W be spanned by $^ = (4, 1, 3) and |^ = (-3, 1, 2).Are U and W identical subspaces of F?

First, consider the vector j = jg - j^ = (7, 0,1) ew and the vector

iJ = «li + J/?2 = (« + 2y, 2x - 3j/, -X + 2y) e UNow I and ij are the same provided x,y e R exist such that

(« + 2y, 2x - Sy, -x + 2y) = (7, 0, 1)

We find x = 3, y = 2. To be sure, this does not prove U and W identical; for that we need to be ableto produce x,y G R such that

«li + yi2 = «l3 + Hifor arbitrary a,b G R.

{ x + 2y = 4a-SbFrom i ^^2^ ^ 3a + 26 '^^ *'"*^ '^ " i(o-56), 1/ = i(7a-6). Now 2x-3y # o + 5;

hence U and W are not identical.

Geometrically, U and PF are distinct planes through O, the origin of coordinates, in ordinary space.They have, of course, a line of vectors in common; one of these common vectors is (7, 0, 1).

3. Prove: The set of non-zero vectors S = {|j, $^ ^J of F over f is linearly dependentif and only if some one of them, say ^., can be expressed as a linear combination of thevectors ^,,^2' • • '^j-i which precede it.

Suppose the m vectors are linearly dependent so that there exist scalars ki, k^, ..., k^, not all

equal to z, such that 2 *;& = ?• Suppose further that the coefficient kj is not z while the coefficients

*^j+i''fj+2 fcm are z (but not excluding, of course, the extreme case j — m). Then in effect

*;i?i + fe2i2 + • + kjij = f (1)


and, since kfij # J, we have

Hi = ~^lii ~ *'2C2 - • • • - fcj-llj-1

or £j = gi£i + qi(,i + ••• + qj-iii-i. (2)

with some of the gj ¥- a. Thus, £j is a linear combination of the vectors which precede it.

Conversely, suppose (2) holds. Then

fci«i + fc2& + • • • + kfij + a|i+i + «£j+2 + • • • + ««m = ?

with kj¥= z and the vectors ii,£2. • • -.In. are linearly dependent.

4. Prove: Any finite set S of vectors, not all the zero vector, contains a linearly independ-

ent subset U which spans the same space as S.

Let S = {{i, i2. • • •, £m}- The discussion thus far indicates that U exists while Theorem V suggests

the following procedure for extracting it from S. Considering each vector in turn from left to right,

let us agree to exclude the vector in question if (1) it is the zero vector or (2) it can be written as

a linear combination of all the vectors which precede it. Suppose there are n =s w vectors remammg

which, after relabeling, we denote by C7 = {ii, 12. • • • . >/«}• By its construction, C/ is a linearly

independent subset of S which spans the same space as S.

Example 6, Page 146, shows, as might have been anticipated, that there will usually be linearly

independent subsets of S, other than V, which will span the same space as S. An advantage of the

procedure used above lies in the fact that, once the elements of S have been set down in some order,

only one linearly independent subset, namely U, can be found.

5. Find a linearly independent subset U of the set S = {î.^^ykÂ^}, where

I, = (1, 2, -1), I2 = (-3. -6. 3), I3 = (2. 1. 3), I4 = (8. 7, 7) £ R,

which spans the same space as S.

First, we note that £, # ? and move to fa- Since I2 = -3Ji, (i.e. I2 is a linear combination of y,

we exclude {2 and move to {3- Now U ^ «£i. for any s e fi; we move to U- Since ^4 is neither a scalar

multiple of |i nor of fg (and, hence, is not automatically excluded), we write

8f, + t|3 = 8(1,2,-1) + t(2, 1,3) = (8,7,7) = U

and seek a solution, if any, in R of the system

8 + 2t = 8, 2s+t = 7, -s + 3t = 7

The reader will verify that U = 2f 1 + Sfg; hence, U = {h.lg} is the required subset.

6. Prove: If S = {i,A^ U is » basis of a vector space V over f and if

T =(vi. nv ••' >?„) is a linearly independent set of vectors of V, then n^m.

Since each element of T can be written as a linear combination of the basis elements, the set

S' = {lilt fli ii> • > Im)

spans V and is linearly dependent. Now ,1 # T, hence some one of the f's must be a linear combina-

tion of the elements which precede it in S'. Examining the «'s in turn, suppose we find {jsatisfies

this condition. Excluding ii from S', we have the set

Si = {ui, £l> l2> • • •> 4i-l' î+l' • • •' ^""^

>e a basis of V. It is clear that Si spans

o show that Si is a linearly independent »

11 = o,\h + a2«2 + • • • + «i«i

'

Oj e f,oj # z

which we shall now prove to be a basis of V. It is clear that Si spans the same space as S', i.e. Si

spans V. Thus we need only to show that Si is a linearly independent set. Write


If Sj were a linearly dependent set there would be some £j, ; > i, which could be expressed as

ij = Ml + âii + 63J2 + ••• + 6{{i-i + bi+ifi+i + ••• + 6j_,fj_i, bi¥'z

whence, upon substituting for j/j,

ii = «ifi + Czfz + ••• + «,-i4i-i

contrary to the assumption that S is a linearly independent set.

Similarly, S[ = {,2, 1,1, ii, £2. • • •. fi-i, £i + i, • • •, U)is a linearly dependent set which spans V. Since vi and V2 are linearly independent, some one of theJ's in Sj, say ij, is a linear combination of all the vectors which precede it. Repeating the argumentabove, we obtain (assuming j > i) the set

Si = {'»2. 71, Ji, £2. •••. Ji-i. £t+i. ••, £i-i. i, + i, ..., U)as a basis of V.

Now this procedure may be repeated until T is exhausted provided n ^ m. Suppose n > m andwe have obtained

Sm — {Vmt Vm-V • • • , '»2, "^l)

as a basis of V. Consider S^ = S U {i,„+i>. Since S„ is a basis of V and i,„ + ,Gy, thenVm+i is & linear combination of the vectors of 5„. But this contradicts the assumption on T. Hencen — m, as required.

(a) Select a basis of Va {R) from the set

S = {11.12.^3.^4} = {(1,-3, 2), (2, 4,1), (3, 1,3), (1,1,1)}

(b) Express each of the unit vectors of Va {R) as linear combinations of the basis vectorsfound in (a).

(a) If the problem has a solution, some one of the {'s must be a linear combination of those whichprecede it. By inspection, we find £3 = «i + £2- To prove {£i,£2,£4} a linearly independent setand, hence, the required basis, we need to show that

«£i + H2 + «£4 - (a + 2b + e, -3a + 4b + c, 2a + b + c) - (0, 0, 0)

requires a = b - c = 0. We leave this for the reader.

The same result is obtained by showing that

»£i + '£2 = £4. 8,teRr 8 + 2t - 1

i.e., < —3s + 4t = 1 is impossible. Finally any reader acquainted with determinants will recall

2s + t = 1

12 1

that these equations have a solution if and only if —3 4 1

2 11= 0.

(5) Set a£i + 5£2 + ci^ equal to the unit vectors ej, ea, «3 in turn and obtain

a + 26 + c = l r o + 26 + c = f a + 2b + c =-3a + 46 + c = I -3a + 46 + c = 1 J -3a + 46 + c =2a+6 + c = [20+6 + = [20+6 + = 1

having solutions:

a = 3/2, 6 = 5/2, c = -11/2 a = 6 = -1/2, c = 3/2 o = -1, 6 = -2, c = 5

Thus ei = |(3£i + 5£2 - ll£4), «2 - i(-£i - £2 + 3£4), and eg = -£i - 2£2 + 5£4.

8. For the vector space VtiQ) determine, if possible, a basis which includes the vectorsi, = (3, -2, 0, 0) and |, = (0, 1, 0, 1).

Since £1 ?* f, £2 ^ f, and £2 9^ «£i for any e e Q, we know that £i and £2 can be elements of abasis of V4 (Q). Since the unit vectors e„ ej, 63, 84 (see Example 7, Page 147) are a basis of V^ (Q),


the set S = {Ji, {2. 'i. ^2> '3. '^ spans V^ (Q) and surely contains a basis of the type we seek. Now e^

will be an element of this basis if and ohly if

ail +. H2 + cei = (3o + c, -2o+ 6, 0, 6) = (0, 0, 0, 0)

requires a = b = c = 0. Clearly, ej can serve as an element of the basis. Again, t2 will be an element

if and only if

fflî + Hi + C'l + rf'2 = (3a + e, -2a + b + d, 0, b) = (0, 0, 0, 0) (J)

requires a = 5 = c = d = 0. We have 6 = = 3a + c = -2a + d; then (1) is satisfied by a = 1,

6 = 0, c = —3, d = 2 and so {iuii'^v^} is not a basis. We leave for the reader to verify that

{£1. I2. H> ^3} is a basis.

9. Prove: If U and W, of dimensions r ^ n and s^n respectively, are subspaces of a

vector space V of dimension n and if C/ n W' and U + W are of dimensions p and t

respectively, then t — T-\-s — 'p.

Take A = {j„ J2 £p} as a basis of 17 n W and, in agreement with Theorem XII, Page 148,

take B = A U {Xi, X2, . . .,Xr-p} as a basis of 17 and C = A U {/ii,/t2. • • •./'s-p} as a basis of W.

Then, by definition, any vector of 17 + W can be expressed as a linear combination of the vectors of

D = {Ji, 42> • • •> ^p> ^l> ^2> • •! V-P> V-\> ''2> • • •> /"s-p/

To show that D is a linearly independent set and, hence, is a basis of V -^-W, consider

Ol£l + «2?2 + • • • + Op^p + frl?'! + M2 + • • • + t»T-pV-p + Cl/*! + <'2/'2 + • • • + C^-p/ls-p = f (i)

where aj, 6j, c^ G y.

Set ir = Cj/ii + C2/12 + • • • + Cs-p/'s-p- Now >r e W and by (1), irGU; thus, tt S ?7 n W and

is a linear combination of the vectors of A, say, ?r = diî + ^212 + • • • + d^^v- Then

Ci/li + 02/12 + • • + Cs_p/*s_p — dlJl — ^2^2 — • • • — dp^p = J

and, since C is a basis of W, each Cj = 2 and each dj = «. With each Cj = z, (Jf) becom«s

aiii + 0.2^2 + • • • + ap?p + 61X1 + 62^2 + • • • + 6r-p^r-p = f i^')

Since B is a basis of U, each aj = « and each bi = z in (i'). Then O is a linearly independent set

and, hence, is a basis of U + W of dimension t = r + s-p.

10. Prove: \i-v\- HI*hi for all 1, 7/

G 7„ (i2).

For J= or 1) = 0, we have 0-0. Suppose 1^0 and 1? # 0; then |ij| = k ||| for some k G i2 + ,

and we have^^^ ^ ^^^^

^ ^^^^^^^^ ^ ^^^^^^l^l ^ ^^,|^l^^ ^^^^ .

^^

and ^ (fcj ± i;) • (fce ± -)) = feÛ'i) * 2fcU • ,) + ,-7,

= 2fc-|«|-|ll ± 2fea-,)

Hence ± 2&U • ,) ^ 2fc |j|•

l-zl

ia-ij) - lll-hl

and \i'v\ - III -It/I

11. Given | = (1,2,3,4) and , = (2,0,-3,1), find

(a) ^'v, (P) 1^1 and hi, (c) |5^1 and |-3r,|, (d) \i + r,\

(a) i'n = l'2 + 2«0 + 3(-3) + 4*1 = -3

(6) IjI= Vl'1+2-2 + 3'3 + 4'4 = y/SO, \v\ - V4 + 9 + 1 = Vl4

(c) |6{| = V25 + 25 • 4 + 25 • 9 + 25 • 16 = 5^30, |-3»,| = V9 • 4 + 9 • 9 + 9 • 1 = S-y/li

(d) C + i = (3,2,0,5) and U + nl = V9 + 4 + 25 = V^


12. Given | = (1, 1, 1) and ^ = (3,4,5) in Vs{R),find the shortest vector of the form X = i + srj.

Here, \ = (1 + 3s, 1 + 4s, 1 + 5s)

and |\|2 = 3 + 24s + 508^

Now |\| is minimum when 24 + 100s = 0. Hence,\ = (7/25, 1/25, -1/5). We find easily that \ and vare orthogonal. We have thus solved the problem:Find the shortest distance from the point P(l, 1,1)in ordinary space to the line joining the origin andQ(3,4,5). Fig. 13-3

13. For $ = iai,a2,a3), -qîhiMM) e Vz{R), define

I X 77 = (0263 — 0362, 0361 — aiba, 0162 — aibi)

(a) Show that | x ,, is orthogonal to both | and r,.

(h) Find a vector A orthogonal to each of | = (1, 1, 1) and ^ = (1. 2, -3).

(a) {• (I X ,) - aiia^h^ - a^h^) + 02(0361 - 0,63) + 03(0162 - o-iby) = and similarly for 7 •

({ X v).

(b) \ = jx, = (l(-3)-2-l, l.l-l(-3), 1-2-1-1) = (-5,4,1)

14. Let I = (1,1, 1,1) and ,? = (1, 2, -3, 0) be given vectors in V4{R).

(a) Show that they are orthogonal.

(6) Find two linearly independent vectors A and ju, which are orthogonal to both ^ and ,.

(c) Find a non-zero vector v orthogonal to each of ^, r,, A and show that it is a linear com-bination of A and II.

(a) i-r, = 1.1 + 1'2 + l(-3) = ; thus { and v are orthogonal.

(6) Assume {a,b,c,d) e VîR) orthogonal to both J and ti; then

(i) o + 6 + c + d = and o + 26 - 3c =

First, take c = 0. Then o + 26 = is satisfied by o = 2, 5 = -1; and a+6 + c + d = nowgives d = -1. We have X = (2,-1,0,-1).

Next, take 6 = 0. Then o- 3c = is satisfied by o = 3, c = 1; and a+ 6 + c + d = nowgives d - -4. We have /» = (3, 0, 1, -4). Clearly, x and ^ are linearly independent.

Since an obvious solution oi the equations (i) is a = 6 = c = d = 0, why not take X = (0, 0, 0, 0)?

(c) If K = (o, 6, c, d) is orthogonal to J, i/ and X, then

a + 6 + c + d = 0, o + 26-3c = and 2o - 6 - d =

Adding the first and last equation, we have 3o + c = which is satisfied by a = 1 c = -3Now 6 = -5, d = 7, and , = (1, -5, -3, 7). Finally, „ = 5\-3^^.

Note. It should be clear that the solutions obtained here are not unique. First of all anynon-zero scalar multiple of any or all of X,^,,- is also a solution. Moreover, in finding X (also, m) wearbitrarily chose c = (also, 6 = 0). However, examine the solution in (c) and verify that v is uniqueup to a scalar multiplier.


15. Find the image of | = (1, 2, 3, 4) under the linear transformation

\, -* (1,-2,0,4)

A:e, - (2,4,1,-2)

^3- (0,-1,5,-1)

e, -* (1,3,2,0)of V4 (Q) into itself.

We have

(1,-2,0,4) + 2(2, 4, 1,-2) + 3(0, -1,5,-1) + 4(1,3,2,0) = (9,15,25,-3)

16. Prove: If {l,,!^, . . .,|J is any basis of V = V{f) and if (vi,'?^, •••,'?„} is any set

of n elements of V, the mapping

T: I, -»>?;. (i=l,2, ...,n)

defines a linear transformation of V into itself.

Let £ = 2 Sili and 7 = 2 *i£i be any two vectors in V. Then

I + >; = 2 (Si + tiHi - 2 («( + ti)vi - 2 Sj'/i + 2 t{<n

so that (i) (I + 'j)r = ir + ^r

Also, for any s&f and any | G V,

so that (ii) (s?)r = 8((,T)

as required.

17. Prove: If T is a linear transformation of V{f) into itself and if W' is a subspace of

V{f), then W^= {^T: i& W}, the image of W under T, is also a subspace of V{f).

For any jf, i,r G Wr, we have jT + ^r = (J + 7)7. Since i,r,eW implies J + d G W, then

(j + .,)2'G T^r- Thus Wj. is closed under addition. Similarly, for any iT^W-r, sGf, we have

s(iT) - (s|)r G Wr since J G W implies s? G W. Thus W-j. is closed under scalar multiplication.

This completes the proof.

18. Prove: The set cA of all linear transformations of a vector space V{f) into itself forms

a ring with respect to addition and multiplication defined by

A+B: ^(A+B) = |A+|S, I G F(f

)

A-B: l(A-B) = iU)B, ^&V(r)for all A.BecA.

Let J,.; G y(f), fc G f, and A.B.C G c^. Then

({ + ,)(A + B) = (I + .;)A + (I + .))B = |A + ,A + |B + ,B

= i{A +B) + v{A + B)

and {ki){A + B) = (fc«)A + (fc£)B = fc(cA) + k^B)

= kÂ + £B) = fc£(A + B)

Also, (£ + ,)(A-B) = [(J + ,)A]B = (CA + ,A)B

= (|A)B + (i,A)B = f(A'B) + ,(A-B)

CHAP. 13] VECTOR SPACES I59

Thus A+ B,A-B e cA and a4 is closed with respect to addition and multiplication.

Addition is both commutative and associative since

i{A+B) = iA+ iB = iB + iA = i{B + A)

and i[(A + B) + C] = i(A + B) + iC = iA + iB + iC

= Â + i{B + C) = i[A + (B + C)]

Let the mapping which carries each element of V(f) into f be denoted by 0; i.e.,

0: |0 = f, ieVif)

Then e o/^ (show this), i{A + 0) = ?A + |0 = {A + f = |A

and is the additive identity element of c4.

For each A £ c/f , let -A be defined by

-A: |(-A) = -(|A), J6y(y)It follows readily that -A G o/f and is the additive inverse of A since

f = fA = (J - {)A = JA + [-(|A)] = j[A + (-A)] = £ .

We have proved that c/i is an abelian additive group.

Multiplication is clearly associative but, in general, is not conmiutative (see Problem 55). To com-plete the proof that o/f is a ring, we prove one of the distributive laws

A- (B + O = A'B + A-C

and leave the other for the reader. We have

i[A-(B + C)] = {iA)(B + C) = (iA)B + (iA)C

= i{A-B) + i(A'C) = i{A-B + A'C)

19. Prove: The set 31 of all non-singular linear transformations of a vector space V{f)into itself forms a group under multiplication.

êt A,B e <^. Since A and B are non-singular they map V{T) onto V(f), i.e., Vj^ = V andVb - V. Then VfA-B) = (Va)b = Vb = V and A • B is non-singular. Thus A-BG^ and <^5i/ isclosed under multiplication. The associative law holds in zM since it holds in c4.

Let the mapping which carries each element of V{f) into itself be denoted by /, i.e.,

/: il = i, {GV(y)

Evidently, / is non-singular, belongs to <vl/, and since

f(/-A) = (|/)A = iA = (jA)/ = j(A-/)

is the identity element in multiplication.

Now A is a one-to-one mapping of Vif) onto itself; hence it has an inverse A"! defined by

A-i; (?A)A-i = J, iGVif)

For any |, , G V(f), A G ^/, and k&f, we have JA,,A G V(r). Then, since

(iA-l-,A)A-» = a + ,)A-A-i = 1 + , = (jA)A-» -f- (,A)A-i

and [fc(|A)]A-i = [(fe|)A]A-i = ki = k[{iA)A-i]

it follows that A-»Gci^. But, by definition. A"! is non-singular; hence A-ie.,M. Thus eachelement of cM has a multiplicative inverse and ^M is a multiplicative group.

IQQ VECTOR SPACES [CHAP. 13


20. Using Fig. 13-1, Page 143:

(a) Identify the vectors (1, 0) and (0, 1); also (a, 0) and (0, b).

(6) Verify (a,b) = (a,0) + (0,6) = a(l, 0) + 6(0, 1).

21. Using natural definitions for scalar multiplication and vector addition, show that each of the following

are vector spaces over the indicated field:

(a) V = R: T = Q ib) V = C; f = R

(c) V - {a + b^ + cy/3 : a.b.e e Q} ; f^Q(d) V - all polynomials of degree - 4 over B, including the zero polynomial; f = Q

(e) V = {cie^ + CjeS^ : Ci, C2 € B} ; J = fi

(/) V = {(oi, 02, 03) : Oi e Q, oi + 2a2 = Sas) ; 7" = Q

(p) y = {a + 6a; : 0,6 e 7/(3)} ; f = 7/(3)

22. (a) Why is the set of all polynomials in x of degree > 4 over R not a vector space over Rl

(6) Is the set of all polynomials R[x] a vector space over Q? over C?

23. Let £,D G V over f and s,t G f . Prove:

(o) When J # L then «£ = t| implies s = t.

(6) When s ¥= z, then si = sn implies 4 = 1-

24. Let i,Tiî^V over f. Show that J and v are linearly dependent if and only if i = sv for some s e f

.

25. (a) Let |,i) € V(R). If £ and 5 are linearly dependent over R, are they necessarily linearly dependent

over Q? over C?

(6) Consider (a) with linearly dependent replaced by linearly independent.

26. Prove Theorem IV and Theorem VI, Page 146.

27. For the vector space V = V, (R) = {{a, b, e. d) : a,b,c,d& R} over R, which of the following subsets

are subspaces of V?

(a) 17 = {(a, a, a, a): a £ R}

(b) U = {(o,b,o,b): a,bG7}

(c) U - {(0,20,6,0+6): 0,6 eii}

id) U = {(01,02,03,04): 0,1^ R, 202 + 303 = 0}

(e) U = {(oi, 02, 03, 04) : Oj G 72, 2o2 + 303 = 5}

28. Determine whether the following sets of vectors in Vg (Q) are linearly dependent or independent over Q:

(o) {(0,0,0), (1,1,1)} (d) {(0,1,-2), (1,-1,1), (1,2,1)}

(6) {(1,-2.3), (3,-6,9)} (e) {(0,2,-4), (1,-2,-1), (1,-4,3)}

(e) {(1, -2, -3), (3, 2, 1)} (/) {(1, -1. -1), (2, 3, 1), (-1, 4, -2), (3, 10, 8)}

Ans. (c), (d) are linearly independent.

29. Determine whether the following sets of vectors in ^3(^/(5)) are linearly dependent or independent

over 7/(5).

(o) {(1,2,4), (2,4,1)} (c) {(0,1,1), (1,0,1), (3,3,2)}

(6) {(2, 3, 4), (3, 2,1)} (d) {(4, 1,3), (2,3, 1), (4,1, 0)}

Ans. (a), (c) are linearly independent.

30. For the set S = {(1, 2, 1), (2, 3, 2), (3, 2, 3), (1, 1, 1)} of vectors in V(I/(5)) find a maximum linearly inde-

pendent subset T and express each of the remaining vectors as linear combinations of the elements of T.


31. Find the dimension of the subset of V3 (Q) spanned by each set of vectors in Problem 28.

Ana. {a),(b) 1; (c),(e) 2; (d), (/) 3

32. For the vector space C over R, show

(a) {l,t} is a basis.

(6) {a +bi,c + di} is a basis if and only if ad—he¥' 0.

33. In each of the following, find a basis of the vector space which includes the griven set of vectors:

(a) {(1,1,0). (0,1,1)} inVsiQ).

(6) {(2, 1, -1, -2), (2, 3, -2, 1), (4, 2, -1. 3)} in ^4 (Q).

(c) {(2,1.1,0), (1,2,0,1)} in F4 (7/(3)).

(d) {(t,0,l,0), (O.i.1.0)} mViiC).

34. Show that S = {{1, ij, £3} = {(i, 1 + 1, 2), (2 + i, i, 1), (3, 3 + 2t, -1)} is a basis of Va{C) and expresseach of the unit vectors of V3 (C) as a linear combination of the elements of S.

Ana. ei = [(-39 - 6i)Ji + (80 - t)£2 + (2 - 13i){3]/173

ea = [(86 - 40t){, + (-101 + 51i){2 + (71 - 29t)J3]/346

es = [(104 + 16i)£i + (75 - 55i)|2 + (-63 - 23t)j3]/346

35. Prove: If fejjj + fcjjg + fc3{3 = f, where fejfca # 2, then {£„ £3} and {I2, J3} generate the same space.

36. Prove Theorem IX, Page 147.

37. Prove: If {£1, i^, is) is a basis of V3 (Q), so also is {£, + £2, {2 + is. {3 + ij- Is this true in V3 (7/(2)")?

In V3(//(3))?

38. Prove Theorem X. Page 147. Hint. Assume A and B, containing respectively m and n elements, arebases of V. First, associate S with A and T with B, then S with B and T with A, and apply TheoremVIII. Page 147.

39. Prove: If V is a vector space of dimension n ^ 0. any linearly independent set of n vectors of V is

a basis.

40. Let ii,i2' ••im^V and S = {fi,£2 {«} span a subspace UcV. Show that the minimumnumber of vectors of V necessary to span U is the maximum number of linearly independent vectorsin S.

41. Let {{i,42. ,in} be a basis of V. Show that every vector | e V has a unique representation as alinear combination of the basis vectors.

Hint. Suppose £ = Scjij = 2^^; then 2cj|j-2dA =f.

42. Prove: If U and W are subspaces of a vector space V, so also are U n W and U + W.

43. Let the subspaces U and W of V4 (Q) be spanned by

A = {(2,-1,1,0), (1,0,2,1)} and B = {(0,0, 1.0). (0,1.0,1). (4,-1,5,2)}

respectively. Verify Theorem XIII, Page 148. Find a basis of 17 + W which includes the vectorsof A ; also a basis which includes the vectors of B.

44. Show that P = {(a,b,-b,a) : a,b G R} with addition defined by

(a, 6, -6, a) + (c, d, -d, c) = (a, + e,b + d, -(6 + d),a + e)

and scalar multiplication defined by k(a, b, -b, a) - {ka, kb, -kb, ka), for all (a, 6, -6, o), (c, d, -d, c) e Pand k G R, is a vector space of dimension two.

45. Let the prime p be a prime element of G of Problem 8, Chapter 10. Denote by f the field G/{p) andby f the prime field I/(p) of f. The field f, considered as a vector space over f, has as basis {l.»};hence, f = {o, 'l + aa't: aj.Oa e J'}, (a) Show that f has at most p* elements. (6) Show thatf has at least p2 elements, (that is, o, • 1 + ag • i == ft, • 1 + 62 • t if and only if a, - 61 = Og - 62 = 0)and hence exactly p^ elements.

IQ2 VECTOR SPACES [CHAP. 13

46. Generalize Problem 45 to a finite field f of characteristic p, a prime, over its prime field having n

elements as basis.

47. For £, 7), /I G V„ (R) and k G R, prove:

(a) |., = ,.t, (6) (4 + r,) •/» = £•<. + .!•/«, («)(*;«•'() = Hi-V)-

48. Obtain from the Schwarz inequality -1 ^ (« • v)/(\i\ • \v\) - 1 to show that cos e = ^-^^deter-

mines one and only one angle between 0° and 180°.

49. Let length be defined as in V„(ft). Show that (1,1) e VîQ) is without length while (hi)^J^((^

is of length 0. Can you suggest a definition of | • i; so that each non-zero vector of V2 (C) will have

length different from 0?

50. Let i,, eV„(fl) such that ||1= |i,|. Show that |-, and i + v are orthogonal. What is the geo-

metric interpretation?

51. For the vector space V of all polynomials in x over a field f , verify that each of the following mappings

of V into itself is a linear transformation of V.

(a) a(x) ^ a(x) (e) a(x) -» a(-x)

(b) a(x) -» -o(a;) (d) «(«) ^ «(0)

52. Show that the mapping T : {a,b) ^ (a + l,b + 1) of V^ (R) into itself is not a linear transformation.

Hint. Compare (ei-t-ej)? and {eiT + CiT).

53. For each of the linear transformations A, examine the image of an arbitrary vector { to determine

the rank of A and, if A is singular, to determine a non-zero vector whose image is 0.

(a) A: H^ (2.1), -2"^ (1,2)

(b) A: 6, ^ (3,-4), e2^ (-3,4)

(e) A: e, ^(1,1,2), e^^ (2,1,3), .3-* (1,0,-2)

(d) A: M ^ (1,-1,1), ^2-* (-3,3,-3), eg ^ (2,3,4)

(e) A : ei-* (0, 1, -1), ea ^ (-1, 1, 1), ^3

^ (1, », "2)

(/) A: e, ^ (1,0,3), £2-* (0,1,1). C3^ (2,2,8)

Ans. (a) non-singular; (6) singular, (1, 1); (c) non-singular; (d) singular, (3, 1,0); (e) singular, (-1, 1,1);

(/) singular, (2, 2, -1)

54. For all A.B&cA and k.ief, prove

(o) fcA g o/f

(5) fc(A -I- B) = &A + kB; {k + l)A = kA + lA

(c) k(A'B) = {kA)B ^ A(kB); {k-l)A = k{lA)

(d) O'A = fcO = 0; uA = A

with defined on Page 159. Together with Problem 18, this completes the proof of Theorem XX,

Page 152.

55. Compute B-A for the linear transformations of Example 13, Page 152, to show that, in general,

A'B^-B-A.

56. For the linear transformations on V3 (R)

(ti^{a,b,c) {nÛ.k.l)

A: \€2^(d,e,f) and B : U^îm.n.p)

fej -* (a + },b + k,c + l)

obtain A + B: L^ -* (d + m, e + n, f + p)

I £3 - (g + Q.h + r.i + s)


«i-* (aJ +bm + eq,^ + bn + cr, al + bp + cs)

A'B : -j «£ "*(<î + em + fq, dk + en + fr, dl+ ep + fa)

J3 -* {gj + hm + iq, gk + hn-\- ir, gl + hp + is)

and, for any kG R,

iei-* (ka,kb,kc)

«2 -» (kd,ke,kf)

«3 ~* (kg,kh,ki)

57. Compute the inverse A-i of A: i *i "* (1' ^) of V,(R).

Hint. Take A-> : V^^ J*'^] and consider ({A)A-i = £ where f = (o, 6).1^C2 -» (r, 8)

Ana.^''-^<^'-l)

62 ^(-2,1)

58. For the mappingfe, = (1,0) - (1,1,1)

"^^ -

il = (0,1) -* (0,1,2)"' ^ = ^^(«) '"*» ^ = ^3(ft)

verify:

(a) Ti is a linear transformation of V into W.(b) The image of f = (2, 1) € V is (2, 3, 4) e W.

(c) The vector (1, 2, 2) G W' is not an image.

(d) Vf has dimension 2.

59. For the mappingfei = (1,0,0) -* (1,0,1,1)

^2= -^£2= (0,1,0) -» (0,1,1,1) of V=V3(ft) into W::^Vi(R)

[63 = (0,0,1) ^ (1,-1,0,0)verify:

(a) T2 is a linear transformation of V into W.

(6) The image of { = (1,-1,-1) G V is (0, 0, 0, 0) G W'.

(c) ^Tj has dimension 2 and r^. = 2.

60. For Ti of Problem 58 and Tg of Problem 59, verify

T .T /*» = <1'0) ^ (2,0,2,2)

„,, .

' ' 1*2 = (0.1) ^ (2,-1,1,1)What IS the rank of Ti • Tj?

chapter 14

Matrices

INTRODUCTION

Consider again the linear transformations on Vs (R)

,^^(a,b,c) U^{j,k,l)

A: \e^-*(d,e,f) and B : U^-*{m,n,p)

for which (see Problem 56, Chapter 13, Page 162)

^ {a + j,b + k,c + l)

A+B : U«-* (d + m,e + n,f + p)

,^ {g + q,h + r,i + s)

W

A-B(aj + 6w + cq, ak-¥bn + cr, al-¥hp + cs)

(dj + em + fq, dk + en-\- fr, dl + ep + fs)

(gj + hm + iq, gk + hn + ir, gl + hp + is)

kA

{ka, kb, kc)

{kd, ke, kf)

(kg, kh, ki)

kGR

As a step toward simplifying matters, let the present notation for linear transforma-

tions A and B be replaced by the arrays

A =a b c

d e f

9 h i

and B =i k I

m n V

q r s

effected by enclosing the image vectors in brackets and then removing the excess paren-

theses and commas. Our problem is to translate the operations on linear transformations

into corresponding operations on their arrays. We have:

The sum A + B ot the arrays A and B is the array whose elements are the sums of

the corresponding elements of A and B.

The scalar product kA of k, any scalar, and A is the array whose elements are k

times the corresponding elements of A.

In forming the product A • B think of A as consisting of the vectors py p^, p^ (the row

vectors of A) whose components are the elements of the rows of A and think of B as con-

sisting of the vectors y^, y^, y^ (the column vectors of B) whose components are the elements

of the columns of B. Then

164

CHAP. 14] MATRICES 165

A'B =Pi

Pa

Pa

bi 72 Ys]

Pi' ri Pi- ya Pi- Ys

P2' Yi Pa- Va Pa- Ys

Pa- Vi pa- Ya Ps- Ys

where p^.y^ is the inner product of p^ and y^. Note carefully that in A • B the inner productof every row vector of A with every column vector of B appears; also, the elements of anyrow of A'B are those inner products whose first factor is the corresponding row vector of A.

Example 1:

(o) When A -

A+B

A-B

1 2and B - 5 6

3 4 7 8

1 + 5 2 + 6 6 8

3 + 7 4 + 8 10 12

1 2 5 1 6t

"

_3 4 _7l8

over Q, we have

lOA

1-5 + 2-7

3-5 + 4«7

10-1

10 '3

10 '2

10 '4

10 20

30 40

1-6 + 2'8

3'6 + 4«8

19 22

43 50

and B'A = 5 6

7 8

5 + 18 10 + 24

7 + 24 14 + 32

23 34

31 46

(6) When A =1 -2-3 1 and B =

3-1-2 3 over Q, we have

2 1 2-1

3 -1-4 2 3-5 2 3 3 -7A-B = 6 2 -9-1 = 6 2 -10 and B'A = 4 3 4

6- 2 -2 3 4-2 3 -2 -7 2

SQUARE MATRICESThe arrays of the preceding section, on which addition, multiplication, and scalar mul-

tiplication have been defined, are called sqûtre matrices; more precisely, they are squarematrices of order 3 since they have 3^ elements. (In Example 1(a), the square matrices areof order 2).

In order to permit the writing in full of square matrices of higher orders, we now intro-duce a more uniform notation. Hereafter, the elements of an arbitrary square matrix willbe denoted by a single letter with varying subscripts; for example^

A =Oil fflia ttis

Oai Oaa (I23

Osi Asa ttss

and B =

bii 612 613 614

621 62a &a3 &a4

bai &32 &33 &34

&41 &42 &43 b44

Any element, say, 624 is to be thought of as 6a,4 although unless necessary (e.g., 6121 whichcould be 612.1 or 61.21) we shall not print the comma. One advantage of this notation is thateach element discloses its exact position in the matrix. For example, the element 624 standsin the second row and fourth column, the element 632 stands in the third row and secondcolumn, and so on. Another advantage is that we may indicate the matrices A and B aboveby writing

A = [an], {i= 1,2.3; ; = 1,2,3)

and B = [bii], (i = 1. 2, 3, 4; j = 1, 2, 3, 4)

166 MATRICES [CHAP. 14

Then with A defined above and C = [dj], (i, j = 1,2,2), the product

A-C = "âzjCn ânCn 2<î<'j3

where in each case the summation extends over all values of j; for example,

2 ffl2jCj3 = a2iCi3 + a22C23 + flzsCas , etc.

Two square matrices L and M will be called equal, L = M, if and only if one is the

duplicate of the other; i.e., if and only if L and M are the same linear transformation. Thus

two equal matrices necessarily have the same order.

In any square matrix the diagonal drawn from the upper left corner to the lower right

comer is called the principal diagonal of the matrix. The elements standing in a principal

diagonal are those and only those (an, an, a^s of A, for example) whose row and column

indices are equal.

By definition, there is a one-to-one correspondence between the set of all linear transfor-

mations of a vector space over f of dimension n into itself and the set of all square matrices

over f of order n (set of all n-square matrices over f). Moreover, we have defined addition

and multiplication on these matrices so that this correspondence is an isomorphism. Hence,

by Theorems XVIII and XIX of Chapter 13, Page 152, we have

Theorem I. With respect to addition and multiplication, the set of all n-square matrices

over 7=" is a ring % with unity.

As a consequence:

Addition is both associative and commutative on %.

Multiplication is associative but, in general, not commutative on %.

Multiplication is both left and right distributive with respect to addition.

There exists a matrix 0„ or 0, the zero element of %, each of whose elements is the zero

element of f. For example, 02="" and O3 = are zero matrices over

R of orders 2 and 3 respectively.

There exists a matrix /„ or /, the unity of %, having the unity of f as all elements along

, ,1

0^

the principal diagonal and the zero element of f elsewhere. For example, h

and I3 =1

1

1

1

are identity matrices over R of orders 2 and 3 respectively.

For each A = [an] e %, there exists an additive inverse -A = (-l)[cuj] = [-On] such

that A + (-A) = 0.

Throughout the remainder of this book we shall use and 1 respectively to denote the

zero element and unity of any field (see Page 118). Whenever z and u, originally

reserved to denote these elements, appear they will have quite different connotations.

Also, and / as defined above over R will be used as the zero and identity matrices over

any field f.

By Theorem XX, Chapter 13, Page 152, we have

Theorem II. The set of all n-square matrices over f is itself a vector space.


A set of basis elements for this vector space consists of the n^ matrices

Eii, (i,} = l,2,S,...,n)

of order n having 1 as element in the (i.j) position and O's elsewhere. For example,

{^11, Ei2, Ezi, B22} = 1 1 "0 0"

t

1»

1_

is a basis of the vector space of all 2-square matrices over T- and for any A =we have A = aEn + bE,i + cE^ + dE22.

a b

c d

TOTAL MATRIX ALGEBRAThe set of all n-square matrices over f with the operations of addition, multiplication

and scalar multiplication by elements of f is called the total matrix algebra 31 (T) Nowjust as there are subgroups of groups, subrings of rings , so there are subalgebras of

^„(f). For example, the set of all matrices 31 of the form

a b

2c a

26 2c

where

a.b,c e Q IS a subalgebra of 3I,{Q). All that is needed to prove this is to show that addi-tion multiplication and scalar multiplication by elements of Q on elements of ^ invariablyyield elements of 3f. Addition and scalar multiplication give no trouble, and so ^ is Isubspace of the vector space 31,(Q). For multiplication, note that a basis ofTLê set

I, X =10

1

2

Y =1

2

2We leave for the reader to show that for A, B e 31,

A'B = (aI + bX + cY)(xI + yX + zY)

= iax + 2bz + 2cy)I + (ay + bx + 2cz)X-h(az + by + cx)Y Galso, that multiplication is commutative on 31.

(M;

A MATRIX OF ORDER w x «By a matrix over f we shall mean any rectangular array of elements of f; for example,

A =On a\2 ffll3 bu 612 613 6i4l Cii C12

0.2\ aw, O23 B = &21 622 623 624 C = C21 C22

O31 ttsz das _&3i &32 633 634Csi

C41

C32

C42

°'' ^ = N. (i. J = 1. 2, 3) B= [6„], (i = 1, 2, 3; j = 1, 2, 3. 4)

C = N, (i= 1,2.3,4; ; = 1,2)

Any such matrix of m rows and n columns will be called a matrix of order mxn.For fixed w and n, consider the set of all matrices over f of order mxn. With additionand scalar multiplication defined exactly as for square matrices, we have

Theorem IF. The set of all matrices over f of order m x » is a vector space over f.Multiplication cannot be defined on this set unless m^n. However, we may. as Problem 60Chapter 13. Page 163. suggests, define the product of certain rectangular arrays Forexample using the matrices A,B,C above, we can form A -5 but not B-A; B-Chut notC-B; and neither A -C nor C- A. The reason is clear; in order to form L-W the num^rof columns of L must equal the number of rows of M. For B and C above we have

168

B-C =Pi

Pi

Pz

[vi yî -

MATRICES

Pi • Vi Pi • Va 2 ^«Cji 2 &ljCj2

Pi -71 Pi- 72= 2 îif'n 2 ^ZJ^'iz

Pa • Vi P3 • Yz 2 &3jCjl 2 f'33Cj2

[CHAP. 14

Thus the product of a matrix of order to x n and a matrix of order n x p, both over the

same field f, is a matrix of order to x p. g^g problems 2-3.

SOLUTIONS OF A SYSTEM OF LINEAR EQUATIONS

The study of matrices, thus far, has been dominated by our previous study of linear

transformations of vector spaces. We might, however, have begun our study of matrices

by noting the one-to-one correspondence between all systems of homogeneous linear equa-

tions over R and the set of arrays of their coefficients; for example

(i) 2x+ Zy + z = Q

(ii) X- 1/ + 42 =

(iii) 4a; + llj/ - 5z =

and

2 3 1

1-144 11 -5

{3)

What we plan to do here is to show that a matrix, considered as the coefficient matrix of

a system of homogeneous equations, can be used (in place of the actual equations) to obtain

solutions of the system. In each of the steps below, we state our "moves", the result of

these "moves" in terms of the equations, and finally in terms of the matrix.

The given system {3) has the trivial solution a; = ?/ = z = 0; it will have non-trivial

solutions if and only if one of the equations is a linear combination of the other two, i.e.

if and only if the row vectors of the coefficient matrix are linearly dependent. The procedure

for finding the non-trivial solutions, if any, is well known to the reader. The set of "moves"

is never unique; we shall proceed as follows: multiply the second equation by 2 and subtract

from the first equation, then multiply the second equation by 4 and subtract from the third

equation to obtain

(i)-2(ii)

(ii)

(iii)-4(ii)

Oa; + by- 7z =

X - y + 4z =

0x + 15y-21z =

5 -7

1 -1 4

15 -21

(-4)

In (4), multiply the first equation by 3 and subtract from the third equation to obtain

(5)

(i)-2(ii)

(ii)

(iii) + 2(ii)-3(i)

0x + 5y-7z =

x — 3/ + 42 =

Ox + 0y + 0z =

5

1 -1

-7

4

Finally, in (5), multiply the first equation by 1/5 and, after entering it as the first

equation in (6), add it to the second equation. We have

i[(i)-2(ii)]

i[3(ii) + (i)l

(iii) + 2(ii)-3(i)

Oa; + y - 72/5 =

x + Oy + 132/5 =

Ox + Oy + Oz -

1 -7/5"

13/5 («)

Now if we take for z any arbitrary r G R, we have as solutions of the system:

a; = -13r/5, y = lr/5, z = r.

CHAP. 14] MATRICES169

We summarize: from the given system of equations (3), we extracted the matrix

A =2 3

1 -1

4 11

1

4

-5

;by operating on the rows of A we obtained the matrix B =

1 -7/5

1 13/5

Example 3:

considering B as a coefficient matrix in the same unknowns, we read out the solutions ofthe origmal system. We give now three problems from vector spaces whose solutions followeasily.

Example 2: Is |, = (2,1,4), i^ = (3,-1,11), I3 = (1,4,-5) a basis of V,{R)-!

We set x|i + j/|2 + 2|3 = (2x + 3y + z, x-y + iz, Ax + lly-5z) = =(0,^, 0) and obtain the system of equations (3). Using the solution x = -13/5y- 7/5 z = l, we find I3 = (13/5)4i - (7/5)62- Thus the given set is not a basis.'Ihis, of course, is implied by the matrix (5) having a row of zeros.

Is the set pi = (2,3,1), p^ = (1,-1,4), p, = (4,11,-5) a basis of V,{R)1The given vectors are the row vectors of (3). From the record of moves in (5)we extract (iii) + 2(ii) - 3(i) = or P3 + 2p2 - 3p, = 0. Thus the set is not a basis.'

Note. The problems solved in Examples 2 and 3 are of the same type and the com-putations are identical; the initial procedures, however, are quite different. In Example 2the given vectors constitute the columns of the matrix and the operations on the matrixinvolve linear combinations of the corresponding components of these vectors. In Example 3the given vectors constitute the rows of the matrix and the operations on this matrixinvolve linear combinations of the vectors themselves. We shall continue to use thenotation of Chapter 13 in which a vector of F„(f) is written as a row of elements andthus hereafter use the procedure of Example 3.

Example 4: Show that the linear transformation

fei ^ (2,3,1) = pi

T: L^ ^ (1.-1,4) =p2[eg ^ (4,ll,-5)=p3

of V - V3 (R) is singular and find a vector of V whose image is 0.

We write T =2 3 1

1 -1 4

4 11 -5

Pi

P2

Ps

By Example 3, pg + 2p2 - 3pi = 0;

thus the image of any vector of V is a linear combination of the vectors pi and pj.Hence V^ has dimension 2 and T is singular. This, of course, is implied by thematrix (5) having a single row of zeros.

Since Spj - 2p2 - pg = 0, the image of ij = (3, -2, -1) is 0, i.e.

(3,-2,-1)

2 3 1

1 -1 4

4 11 -5

=

Note. The vector (3, -2, -1) may be considered as a 1 x 3-matrix; hence the indicatedproduct above is a valid one.

See Problem 4.

ELEMENTARY TRANSFORMATIONS ON A MATRIXIn solving a system of homogeneous linear equations with coefficients in f certain

operations may be performed on the elements (equations) of the system without changingits solution or solutions:

Any two equations may be interchanged.

Any equation may be multiplied by any scalar fc# of f.


Any equation may be multiplied by any scalar and added to any other equation.

The operations, called elementary row transformations, thereby induced on the coeffi-

cient matrix of the system are:

The interchange of the ith and yth rows, denoted by Ha.

The multiplication of every element of the ith row by a non-zero scalar k,

denoted by Hi(k).

The addition to the elements of the ith row of k (a scalar) times the correspond-

ing elements of the ^th row, denoted by Hij{k).

Later we shall find useful the elementary column transformations on a matrix which

we now list:

The interchange of the ith and yth columns, denoted by Ka.

The multiplication of every element of the ith column by a non-zero scalar k,

denoted by Ki{k).

The addition to the elements of the ith column of k (a scalar) times the cor-

responding elements of the yth column, denoted by Kij{k).

Two matrices A and B will be called row (column) equivalent if B can be obtained

from A by a sequence of elementary row (column) transformations. Similarly, two

matrices A and B will be called equivalent if B can be obtained from A by a sequence of

row and column transformations. When B is row equivalent, column equivalent or equiva-

lent to A, we shall write B ~ A. We leave for the reader to show that - is an equivalence

relation.

(a) Show that the set {(1, 2, 1, 2). (2, 4, 3, 4), (1, 3, 2, 3), (0, 3, 1, 3)} is not a basis of

V4 (Q). (b) If T is the linear transformation having the vectors of (a) in order as

images of ej, eg. *3, U, what is the rank of T1 (e) Find a basis of V^ (Q) containing

a maximum linearly independent subset of the vectors of (a).

(a) Using in turn Hîi-Z), H^î-l); Hi3(-2), H^si-S); H^^m,

Example 5:

we have

A =

12 122 4 3 4 _13 2 3

3 13

12 1210

111~

3 13

-1

1

1 1

The set is not a basis.

(h) Using Hi2(l), H^^^-l) on the final matrix obtained in (a), we have

1 -1

1

1 1 1

1010

10 1B

(c)

Now B has the maximum number of zeros possible in any matrix row equivalent

to A (verify this). Since B has 3 non-zero row vectors, V-j- is of dimension 3 and

Tr = 3.

A check of the moves will show that a multiple of the fourth row was never

added to any of the other three. Thus the first three vectors of the given set

are linear combinations of the non-zero row vectors of B. The first three

vectors of the given set together with any vector not a linear combination of

the non-zero row vectors of B, for example e^ or £4, is a basis of V4 (Q).

Consider the rows of a given matrix A as a set S of row vectors of Vn if) and interpret

the elementary row transformations on A in terms of the vectors of S as:


The interchange of any two vectors in S.

The replacement of any vector i G S by a non-zero scalar multiple a|.

The replacement of any vector | G S by a linear combination 1 + 6?; of | andany other vector rj E S.

The foregoing examples illustrate

Theorem III. The above operations on a set S of vectors of V„{f) neither increase nordecrease the number of linearly independent vectors in S.

See Problems 5-7.

UPPER TRIANGULAR, LOWER TRIANGULAR AND DIAGONAL MATRICESA square matrix A = [ay] is called upper triangular if Of,- = whenever i > j, and

is called lower triangular if an = whenever i<j. A square matrix which is both

upper and lower triangular is called a diagonal matrix. For example, 4 is upper

1 2 3

4

2

triangular,

1

2 3

3 4 5

is lower triangular, while

10and

2

-2

3

1

are diagonal.

By means of elementary transformations, any square matrix can be reduced to uppertriangular, lower triangular, and diagonal form.

Example 6:

1 2 3

4 5 6 over Q to upper triangular, lower triangular, and diagonal form.

5 7 8

Reduce A =

(o) Using H2i(-4), H3i(—5); HjaC— 1), we obtain

12 3

4 5 6 ~

5 7 8

12 3

-3 -6 ~

-3 -7

12 3

-3 -6

0-1which is upper triangular.

(6) Using fli2(-2/5), H^{-m); Hi2{-21/10). H23(-l/28)

A =12 3

4 5 6 ~

5 7 8

-3/5 3/5

3/7 2/7 ~

5 7 8

-3/2

1/4 -1/4

5 7

which is lower triangular.

(c) Using H2i(-4), H3i(-5), Hîi-l); H,2(2/3); H,3{-1), H^(-6)

1 2 3

-3 -6 ~

-1

1 -1

-3 -6 ~

-1

100-3

0-1which is diagonal.

See also Problem 8.

A CANONICAL FORMIn Problem 9, we prove

Theorem IV. Any non-zero matrix A over f can be reduced by a sequence of elementaryrow transformations to a row canonical matrix {echelon matrix) C havingthe properties:

(i) Each of the first r rows of C has at least one non-zero element; the

remaining rows, if any, consist entirely of zero elements.


(ii) In the ith row (i = 1,2, . . .,r) of C, its first non-zero element is 1.

Let the column in which this element stands be numbered ju

(iii) The only non-zero element in the column numbered ju (^ = 1,2, . . .,r)

is the element 1 of the ith row.

(iv) jl<}2< < jr.

Example 7: (a) The matrix B of Problem 6, Page 187, is a row canonical matrix. The first

non-zero element of the first row is 1 and stands in the first column, the first

non-zero element of the second row is 1 and stands in the second column, the

first non-zero element of the third row is 1 and stands in the fifth column.

Thus, ii = 1, J2 — 2, Ja = 5 and Ji < j^ < j^ is satisfied.

(6) The matrix B of Problem 7, Page 187, fails to meet condition (iv) and is not a

1 l"

row canonical matrix. It may, however, be reduced to

1 -1

1 1 = c.

a row canonical matrix, by the elementary row transformations H^^, Hj^.

In Problem 5, Page 186, the matrix B is a row canonical matrix; it is also the identity

matrix of order 3. The linear transformation A is non-singular; we shall also call the

matrix A non-singular. Thus,

An n-square matrix is non-singular if and only if it is row equivalent to the

identity matrix /„.

Any n-square matrix which is not non-singular is called singular. The terms singular and

non-singular are never used when the matrix is of order mxn with m¥'n.

The rank of a linear transformation A is the number of linearly independent vectors

in the set of image vectors. We shall call the rank of the linear transformation A the

row rank of the matrix A. Thus,

The row rank of an w X n matrix is the number of non-zero rows in its rowequivalent canonical matrix.

It is not necessary, of course, to reduce a matrix to row canonical form to determine its

rank. For example, the rank of the matrix A in Problem 7 can be obtained as readily

from B as from the row canonical matrix C of Example 7(6).

ELEMENTARY COLUMN TRANSFORMATIONSBeginning with a matrix A and using only elementary column transformations, we

may obtain matrices called column equivalent to A. Among these is a column canonical

matrix D whose properties are precisely those obtained by interchanging "row" and"column" in the list of properties of the row canonical matrix C. We define the columnrank of A to be the number of columns of D having at least one non-zero element. Ouronly interest in all of this is

Theorem V. The row rank and the column rank of any matrix A are equal.


As a consequence, we define

The rank of a matrix is its row (column) rank.

Let a matrix A over f of order mxn and rank r be reduced to its row canonical form C.

Then using the element 1 which appears in each of the first r rows of C and appropriate

transformations of the type Kaik), C may be reduced to a matrix whose only non-zero


elements are these I's. Finally, using transformations of the type Ka, these I's can bemade to occupy the diagonal positions in the first r rows and first r columns. The resultingmatrix, denoted by N, is called the normal form of A.

Example 8:

(a) In Problem 4 we have

A = 1

-2

1= C

Using Ksii-4), if4i(2); XgaCl), K^î-l) on C, we obtain

1 4 -2

1 -1 1

oj

1 -1

1010 h

the normal form.

(6) The matrix B is the normal form of A in Problem 5.

(c) For the matrix of Problem 6, we obtain using on B the elementary column transformations

A ~10 4

1-121

101010

= [/3 0]

Note. From these examples it might be thought that, in reducing A to its normal

form, one first works with row transformations and then exclusively with column trans-

formations. This order is not necessary. ggg Problem 11

ELEMENTARY MATRICESThe matrix which results when an elementary row (column) transformation is applied

to the identity matrix !„ is called an elementary row (column) matrix. Any elementary

row (column) matrix will be denoted by the same symbol used to denote the elementary

transformation which produces the matrix.

Example 9:

When / =

Htt, —

1

1

1

1

1

1

K,

we have

Hîk)

1

k

1

K^W, H^sik) =1

1 k

1

= Ks2(k)

By Theorem III, we have

Theorem VI. Every elementary matrix is non-singular,

and

Theorem VII. The product of two or more elementary matrices is non-singular.


Theorem VIII. To perform an elementary row (column) transformation H (K) on a

matrix A of order mx n, form the product H -A (A-K) where H (K)

is the matrix obtained by performing the transformation H (K) on I.


The matrices H and K of Theorem VIII will carry no indicator of their orders. If

A is of order to x w, then a product such as His • A • K2s(k) must imply that H^ is of order mwhile K23(k) is of order n since otherwise the indicated product is meaningless.

Example 10: Given A =

(o) Hi3-A

12 3 4

5 6 7 8

2 4 6 8

over Q, calculate

1

1 •

10

12 3 4

5 6 7 8 =

2 4 6 8

2 4 6 8

5 6 7 8

12 3 4

(6) Hi(-3)-A

(c) A • X4,(-4) =

3

10 •

1

12 3 4

5 6 7 8 —

2 4 6 8

12 3 4

5 6 7 8 •

2 4 6 8

1 -4

1

1

1

-3 -6 -9 -12

5 6 7 8

2 4 6 8

12 3

5 6 7-122 4 6

Suppose now that Hi, H2, . . .,Hs and Ki, K2, . . .,Kt are sequences of elementary trans-

formations which when performed, in the order of their subscripts, on a matrix A reduce

it to B, i.e.,

Hs' ... -Hi-Hi-A'Ki-Kz' ... 'Kt = B

Then, defining S = Hs- . .. -Hz- Hi and T = Ki-K^- ...-Kt, we have

S'A'T = B

Now A and B are equivalent matrices. The proof of the converse

Theorem IX. If A and B are equivalent matrices there exist non-singular matrices Sand T such that S-A-T = B.

will be given in the next section.

As a consequence of Theorem IX, we have

Theorem IX'. For any matrix A there exist non-singular matrices S and T such that

S'A'T = N, the normal form of A.

Example 11: Find non-singular matrices S and T over Q such that

S'A'T = S1 2 -1

3 8 2

4 9 -1T = N, the normal form of A.

Using Hii(-3), H3i(-4). K2i(-2), Xjid), we find A =

using H^si—l), we obtain A

Thus,

1

1

1

1

1 2

1 3

2 -1 108 2 ~ 2 5

9 -1 13Then,

and finally i/32(— 1), A'32(—2) yield the normal form

Hs^{-1) ' Hisi-1) • H3i(-4) • Hîi-S) ' A ' Kî(-2) • K^.il) • Ksi{-2)

1 1 10 10 1 -2 1 1 1

1 • 1 -1 • 10 • -3 10 •A- 1 • 10 • 1 -2-1 1 1 -4 1 1 1 1 1


1011-1-5 -1 2

1 2 -1

3 8 2 •

4 9 -1

1-2 5

1-21

10= 10

1

An alternate procedure is as follows: We begin with the array

hA h =

1

1

1

1 2 -1 1

3 8 2 1

4 9 -1 1

and proceed to reduce ^ to /. In doing so, each row transformation is performed on the rows of

six elements and each column transformation is performed on the columns of six elements. UsingH2i(-3), H3j(-4); X2i(-2), Kgid); Hî(-\); Hgî-l); Ks2(-2) we obtain

1 1 1 -2 1

1 1 1

1 1 1

1 2 -1 1 1 2 -1 1 1 1

3 8 2 1 2 5 -3 1 2 5 -3 1

4 9 -1

1 --2

1

1

1

1 -» 1 3

1 -

-4

-2

1

1

1

1 -> 1 3 -4

1 -2

1 -

5

-2

1

1

1 1 1 1 1 1

1 2 1 1 --1 1 2 1 1 --1 1 1 1 -1 T- 1 3 --4 1 -» 1 --5 --1 2 -» 1 --5 -1 2 =-- I S

and S'A'T = I, as before.See also Problem 12.

INVERSES OF ELEMENTARY MATRICESFor each of the elementary transformations there is an inverse transformation, that is,

a transformation which undoes whatever the elementary transformation does. In termsof elementary transformations or of elementary matrices, we find readily

Hij = Hij Kij = Kij

H^'ik) = Hi{l/k) Kr\k) = Ki{l/k)

H;;'{k) = Hui-k) K^'{k) = Kui-k)Thus,

Theorem X. The inverse of an elementary row (column) transformation is an elemen-

tary row (column) transformation of the same order,

and

Theorem XI. The inverse of an elementary row (column) matrix is non-singular.


Theorem XII. The inverse of the product of two matrices A and B, each of which has aninverse, is the product of the inverses in reverse order, that is,

(A-B)-' = B-i-A-i


Theorem XII may be extended readily to the inverse of the product of any number of

matrices. In particular, we have:

if S ^ Hs- ...•H2-Hi then S"' = H^' H^' . . .- H7'

,

if T = Ki-K2' ... -Kt then T-' = K;' - . . . K'' • KT'

.

Suppose A of order mxn. By Theorem IX', there exist non-singular matrices S of

order m and T of order n such that S' A'T = N, the normal form of A. Then

A = S-^{S-A'T)T-' = S-'-iV-r-»In particular, we have

Theorem XIII. If A is non-singular and if S • A • T = /, then

that is, every non-singular matrix of order n can be expressed as a

product of elementary matrices of the same order.

Example 12: In Example 11, we have

S = H32(-l) • •ff23(-l) • «3i(-4) • H2i(-3) and T = Kîi-Z) K^^W - Kîi-i)

Then

S-i = H-M-i)-H-M-^)-H^H-^)'H-'(-\) = H2,(3) • H3i(4) . H23(l) . H32(l)

10 10 1 10 103 10 • 10 • 1 1 • 10 = 3 2 1

1 4 1 1 oil 4 11

r-l = ^32(2) • XgiC-l) • X2i(2) =1012 •

1

1 -1 12 0^

1 • 10 =

1 1

12-112

1

and A = S-i • r-i =103 2 1

4 11

1 2 -1

1 2 =1

1 2 -1

3 8 2

4 9 -1

Suppose A and B over f of order mxn have the same rank. Then they have the same

normal form N and there exist non-singular matrices Si, Ti; S2, T2 such that SiATi -

N = S2BT2. Using the inverse Si^ and Tr^ of Si and Ti, we obtain

A = 5r' • SiATi • rr' = sr''S2BT2'Tr' = {s:'-S2)b{T2-t:') = s-b-t

Thus A and B are equivalent. We leave the converse for the reader and state

Theorem XIV. Two mxn matrices A and B over f are equivalent if and only if they

have the same rank.

THE INVERSE OF A NON-SINGULAR MATRIXThe inverse A~\ if it exists, of a square matrix A has the property

A- A-' = A-'- A = /

Since the rank of a product of two matrices cannot exceed the rank of either factor

(see Chapter 13), we have

Theorem XV. The inverse of a matrix A exists if and only if A is non-singular.

Let A be non-singular. By Theorem IX' there exist non-singular matrices S and T

such that S-A-T = I. Then A = S'-'-T-'' and, by Theorem XII,

A-i = (s-'-r->)-' = T-S


Example 13:

Using the results of Example 11, we find

A-i = T-S1 -2 6

1 -2 •

1

1

1 1 -1 =-5 -1 2

-26

11

-5

-7 12

3 -5

-1 2

In computing the inverse of a non-singular matrix it will be found simpler to useelementary row transformations alone.

of Example 11 using only elementary row transfor-

Example 14: _1 2 -1

Find the inverse of A = 3 8 2mations.

4 9 -1

We have

12-11 —

1

[A I] = 3 8 2 1 ~

4 9-10 1

12-112 5-313-4

1

1 2 -1 1

1 3 -4 1

2 5 -3 1

1 0-7 9 0-213-4010-1 5 1-2

10 -26 -7 12

1 11 3 -5

1-5-1 2

[/ A->].


MINIMUM POLYNOMIAL OF A SQUARE MATRIXLet A vÔ be an n-square matrix over f. Since A G S^JJ), the set {I,A,A^, . . ., A"^}

is linearly dependent and there exist scalars ao, ai, ch, . . ., a^2 not all such that

(A) = oo/ + aiA + 02^2 + • • • + a„2A"^ =

In this section we shall be concerned with that monic polynomial m(A.) G f[\] of

minimum degree such that m(A} - 0. Clearly, either to(A) = ^(A) or m{X) is a properdivisor of ^(\). In either case, m(X) will be called the minimum polynomial of A.

The most elementary procedure for obtaining the minimum polynomial of A ^^ involves

the following routine:

(1) If A = ool, OoGf, then m(\) = A - oo.

(2) If A¥=al for all aef but A^ = aiA + ao/ with ao, a, e f, then m(A) = A^ _aiA — ao.

(3) If A''âA + bl for all a,bGf but A^ = a2A^ + aiA + aj with ao, ai, aa e f,then w(A) = A^ — a2A^ — aiA — ao

and so on.

Example 15:

Find the minimum polynomial of A =

Since A ¥= aÎ for all a^G Q, set

A2

12 2

2 122 2 1

over Q.

9 8 8 12 2 108 9 8 = a, 2 12 + 0-0 10 =

8 8 9 2 2 1 1

aj + Oo 2a, 2oi

2oi Oi 4- Oo 2a,

2oi 2oi Oi + Oo

After checking every entry, we conclude that A^ = 4A + 51 and the minimum polynomial is



Example 15 and Problem 15 suggest that the constant term of the minimum polynomial

of A =7^ is different from zero if and only if A is non-singular. A second procedure for

computing the inverse of a non-singular matrix follows.

12 2

Example 16: Find the inverse A~^, given that the minimum polynomial of A(see Example 15) is \^ — A\ — 5.

2 122 2 1

Since A^ - 4A - 5/ = we have, after multiplying by A ', A - 4/ - 5A > = 0;

-3/5 2/5 2/5

hence, A-» = i(A-4/) = 2/5 -3/5 2/5

2/5 2/5 -3/5

SYSTEMS OF LINEAR EQUATIONS

Let f be a given field and xi, Xi., .. .,Xnhe indeterminates. By a linear form over fin the n indeterminates we shall mean a polynomial of the type

axi + hxi + • • • + pxn

in which a,b, ...,pE.f. Consider now a system of m linear equations

OiiXi + ai2X2 + • • + aina:n = hi

0,21X1 + 022X2 + • • + a2nXn = h2(7)

= h„dmlXi + a.m2X2 + • • ' + dmnXn

in which both the coefficients a,j and the constant terms hi are elements of f. It is to be

noted that the equality sign in (7) cannot be interpreted as in previous chapters since in

each equation the right member is in f but the left member is not. Following common

practice, we write (7) to indicate that elements n.ra, . . .,r„ G y" are sought such that

when Xi is replaced by r^ (i = 1,2, . . .,w), the system will consist of true equalities of

elements of f. Any such set of elements n is called a solniiim of (7).

Denote by A = [ay], (i = 1,2, . . .,m; y = 1,2, . . .,n)

the coefficient matrix of (7) and by S = {Ij,^^, ...AJ the set of row vectors of A. Since

the Ij are vectors of Yn{f), the number of linearly independent vectors in S is r^n. With-

out loss of generality, we can (and will) assume that these r linearly independent vectors

constitute the first r rows of A since this at most requires the writing of the equations

of (7) in some different order.

Suppose now that we have found a vector p = (ri, r2, . . . , r„) G Vn {f) such that

Ij . p — fej, I2 . p - h^. ^r-P = KSince each t, {i - r + l, r + 2, . . ., m) is a linear combination with coefficients in f of the

r linearly independent vectors of A, it follows that

^r+l-P='^r+l' ^r+2-P- Kr + 2» L = fe„ («)

and Xi = n, X2 = r2, . . ., x„ = r„ is a solution of (7) if and only if in (8) each hi is the same

linear combination of hi, fea, . . . , fer as $. is of the set ?,, I^. • • • . ^r'that is, if and only if

hi

the row rank of the augmented matrix [A H]

an

a2i

ai2

a22

ain

din hi

fflml fflm2 hm

is also r.


We have proved

Theorem XVI. A system (7) of m linear equations in n unknowns will have a solution if

and only if the row rank of the coefficient matrix A and of the augmentedmatrix [A H] of the system are equal.

Suppose A and [A H] have common row rank r<n and that [A H] has been reducedto its row canonical form

10 0..

10... Ci,r+1 Ci.r + 2 ... Cm fci

. C2.,-+l C2.r+2 ... Cin ICi

0..

0..

. 1 Cr,r + 1 Cr,r+2 ... Cm kr

.00 ...

0...00 ...

Let arbitrary values Sr+i,Sr+2, . . .,s„Gf be assigned to Xr+i,a;r+2, . . .,Xn; then

Xi = fci — Ci.r+l"Sr+l — Cl,r + 2*Sr+2 — ••• — Cm'Sn

X2 = k2 — C2,r+l'Sr+l — C2.r+2'Sr + 2 — •• — C2„'Sn

— tCr Cr,r+l'Sr+l — Cr,r + 2'Sr + 2 ~ - Cr ' Sn

are uniquely determined. We have

Theorem XVI'. In a system (7) in which the common row rank of A and [A H] is r < n,

certain n-r ot the unknowns may be assigned arbitrary values in fand then the remaining r unknowns are uniquely determined in terms of

these.

Systems of Non-homogeneous Linear Equations

We call (7) a system of non-homogeneous linear equations over f provided not everyAi = 0. To discover whether or not such a system has a solution as well as to find thesolution (solutions), if any, we proceed to reduce the augmented matrix [A H] of the systemto its row canonical form. The various possibilities are illustrated in the examples below.

Example 17:

Consider the system

aij + 2a52 — 3x3 + x^ = X

, 2xi — xa + 2«3 — X4 = 1 over Q.

y^x-i + Zxi — Ax^ + x^ = 2

We have

[A H] =12-312-1 2-14 3-41

J

12-31 1

0-5 8-3 -1

0-5 8-3 -2

1 2 -3 1 1

-5 8 -3 -1

-1

Although this is not the row canonical form, we see readily that

rA = 2 < S = Via Hj

and the system is incompatible, i.e., has no solution.

Example 18:

ixi+ 2*2 - Xa = -1

SiTj + 8a:2 + 2x3 - 28

4a;, + 9x2 — x^ — lA

over Q.


We have

[A H] =2 -1 -1

8 2 28 ~

9 -1 14

1 2-12 5

13

-1

31

18

12-1132 5

-1

18

31

10-7 -37

13 18

0-1 -5

1010

1

Here, r^ = r[A ^j = S — the number of unknowns. There is one and only one solution: Xj

«2 = 3, Xs = 5.

= -2,

Example 19:

Consider the system

Xi + X2+ X3+ X^+ X^ = S

2*1 + 3«2 + 3a;3 + X4 — x^ =

—Xi + 2X2 — 5X3 + 2X4 — X5 = 1

3xi — X2 + 2x3 — 3x4 — 2x5 = —

:

over Q.

We have

[A H] =

1 1

3 3

2 -5

-1 2

1 1

1 -1

2 -1

-3 -2

3

1

-1

11111 3

11-1-3 -6

3-430 4

-4 -1 -6 -5 -10

10 2 4

1 1-1-30-769

3 -10 -17

f~l 2 4

1 1-1-31 14 25

3 -10 -17

9

-6

22

-34

9

-6

46

-34

10 2 4

1 1-1-30-1 -14 -25

3-10 -17

9

-6

-46

-34

10 2 4 9

10 -15 -28 -52

1 14 25 46

-52 -92 -172

9

-52

46

43/13

Here both A and [A H] are of rank 4; the system is compatible, i.e., has one or more solutions.

Unlike the system of Example 18, the rank is less than the number of unknowns. Now the given

1 2 4

1 -15 -28

1 14 25

1 23/13

1 6/13 31/13

1 -19/13 -31/13

1 3/13 -4/13

1 23/13 43/13

system is equivalent to

«i + TSi^s = 31/13

«2-i§*5 = -31/13

a^3 + ra*5 = -4/13

»4+Tf«5 = 43/13

and it is clear that if we assign to X5 any value

r&Q then Xj = (31 - 6r)/13, Xj = (-31 + 19r)/13, X3 = (-4 - 3r)/13, X4 = (43 - 23r)/13, X5 = r

is a solution. For instance, Xj = 1, x^ — 2, X3 = —1, X4 = —2, X5 = 3 and xj = 31/13, Xj = —31/13,

X3 = —4/13, X4 = 43/13, X5 = are particular solutions of the system.


These examples and problems illustrate

Theorem XVII. A system of non-homogeneous linear equations over f in n unknownshas a solution in f if and only if the rank of its coefficient matrix is

equal to the rank of its augmented matrix. When the common rank

is n, the system has a unique solution. When the common rank is r<n,


certain n — r of the unknowns may be assigned arbitrary values in fand then the remaining r unknowns are uniquely determined in terms

of these.

When w = « in system (7), we may proceed as follows:

(i) Write the system in matrix form

an Oi2

fflZl ffl22

amdin

ftnl ttn2 . On

Xi fel

•X2 = fe2

Xn hn

or, more com-

pactly, as A'X = H where X is the « x 1 matrix of unknowns and H is the « x 1

matrix of constant terms.

(ii) Proceed with the matrix A as in computing A"'. If, along the way, a row or columnof zero elements is obtained, A is singular and we must begin anew with the matrix[A H] as in the first procedure. However, if A is non-singular with inverse A~', thenA-\A-X) ^ A-^-H and X = A'^-H.

Example 20:

For the system of Example 18, we have from Example 14, A~i =

«i

= af2 = A-i'H =

^3

bef are.

26 -7 12 r-iH -2

11 3 -5 • 28 = 3

-5 -1 2L^^j

5

-26 -7 12

11 3 -5

-5 -1 2

then

and we obtain the unique solution

Systems of Homogeneous Linear Equations

We call (7) a system of homogeneous linear equations provided each hi = 0. Since thenthe rank of the coefficient matrix and the augmented matrix are the same, the system alwayshas one or more solutions. If the rank is n, then the trivial solution Xi = X2= • • = x„ =is the unique solution; if the rank is r < w. Theorem XVI' assures the existence of non-trivial

solutions. We have

Theorem XVIII. A system of homogeneous linear equations over f in n unknowns alwayshas the trivial solution Xi = X2 - - Xn - 0. If the rank of thecoefficient matrix is n, the trivial solution is the only solution; if the rankis r < n, certain n — r of the unknowns may be assigned arbitrary valuesin f and then the remaining r unknowns are uniquely determined in

terms of these.

Example 21: Solve the system

(Xj + 2Xi — x^ —

3xi + 8x2 + 2x3 =4xi + 9X2 ~ *3 =

over Q.

By Example 18, A ~ /j. Thus, Xi = Xg = X3 = is the only solution.

Example 22: Solve the system

r xj + X2 + xj + X4 =

2xi + 3x2 + 2x3 + X4 =3x, + 4x2 + 3x3 + 2x4 =

over Q.

We have

1

-1

-1of rank 2

Setting X3 = 3, X4 = t where s.tGQ, we obtain the required solutions as:

X. = -« - 2t, X2 = t, X3 := S, X4 = t. g^ ^,g^ p^^^j^^ jg_


DETERMINANT OF A SQUARE MATRIX

To each square matrix A over f there may be associated a unique element aGf. This

element a, called the determinant of A is denoted either by det A or \A\. Consider the

«-square matrix

ail ttl2 <Il3 . . . din

0,21 Ct22 ffl23 . . . tt2n

tt31 ft32 ft33 . . . dan

fflnl CLn2 (ln3 . • • dm

and a product

flljl tt2J2 ttsjg . . . anj„

of n of its elements selected so that one and only one element comes from any row and one

and only one element comes from any column. Note that the factors in this product have

been arranged so that the row indices (first subscripts) are in natural order 1, 2, 3, . .. , n.

The sequence of column indices (second subscripts) is some permutation

P — Wl» J2' •'3' • • • ' J nl

of the digits 1,2,3 n. For this permutation, define cp = +1 or -1 according as p is

even or odd and form the signed product

(O) Cp ttljj ft2J2 0^3J3 . . . dn}„

The set Sn of all permutations of n symbols contains n! elements; hence, n\ distinct signed

products of the type (a) can be formed. The determinant of A is define(i to be the sum of

these n\ signed products (called terms of \A\), i.e.,

(b) \A\ = 2 «P <*lJi d2i2 "3J3 • • ""Jn

Example 23:

(i)

Oil <*12

"21 "âej2aiia22 + e21<*12<*21 — ''•ll'*22 °'12<''21

Thus, the determinant of a matrix of order 2 is the product of the diagonal elements of the

matrix minus the product of the off-diagonal elements.

(ii)

Oil «12 "13

O21 O22 O23

131 O32 O33

— «123<*ll'*22<*33 + ei32<*ll'''23<''32 + «213<''12''^2l''33

+ «23l'''l2<''23<'''31 + f312<*13'''2l"32 + «321°^13''22*31

— Olltt22*33 ~ <»11<»23<'''32 " <»12"21»33 + <»12<»23"31 + ''•13"2l"32 ~ <»13"22»31

= <»ll(022*33 ~ <*23<»32) ~" ''tl2('»21<*33 "" «23*3l) + ttl3(«21<'32 ""'»22"'3l)

O22 ''•23

132 "^33

O21 O23

O31 O33

-1)1O22 O23

'''32 *33

O21, O22

031 O32

0-21 O23

fâi <*33

+ (-1)1+3 oi

O21 O22

031 O32

called the expansion of the determinant along its first row. It will be left for the reader to

work out an expansion along each row and along each column.


PROPERTIES OF DETERMINANTSThroughout this section, A is the n-square matrix whose determinant |A| is given by

(6) of the preceding section.

From (6) there follow easily

Theorem XIX. If every element of a row (column) of a square matrix A is zero, then\A\ = 0.

Theorem XX. If A is upper (lower) triangular or is diagonal, then |A| = 011022033 • • • a„„,

the product of the diagonal elements.

Theorem XXI. If B is obtained from A by multiplying its ith row (ith column) by anon-zero scalar k, then |B| = k\A\.

Let us now take a closer look at (a). Since p is a mapping of S = (1,2,3, . . .,n} intoitself, it may be given (see Chapter 1) as

p: lp = 3\, 2p = j^, Sp = j^, ..., np = j^

With this notation, (a) takes the form

(a') €p Oi.ip 02, 2p 0,3.30 • On.np

and (b) takes the form

("')l-^l — ^ fp 0,1, ip a2,2t> cia.sp . a„,np

Since S„ is a group, it contains the inverse

p-': j,p-' = 1, ;>-> = 2, ;>-> = 3, . . ., )\p-^ = n

of p. Moreover, p and p-' are either both odd or both even. Thus (a) may be written as

and, after reordering the factors so that the column indices are in natural order, as

(a") tp-i fflip-i,ia2p-i.2a3p-i.3 . . . a„p-i.„

and (6) as (b") \A\ = D cp-. a,p-.., a2p-..2a3p-..3 . . . a„p-..„

For each square matrix A = [an] define transpose A, denoted by A^, to be the matrixobtained by interchanging the rows and columns of A. For example.

when A =fflll ffll2 ffll3

O2I ffl22 ffl23

O3I tt32 ffl33

then

Oil O21 O31

O12 022 032

Ol3 O23 O33

Let us also write A^ = [oj], where oj = an for all i and j. Then the term of \A'^\

£p ftl.j, ali2 "-^-k • "'în = *" %.^ %.2 %.s . . aj^_„

— *p *ip.i o,2p,i a^p,3 . . . a„p_„

is by (6") a term of |A|. Since this is true for every p G S„, we have proved

Theorem XXII. If A'" is the transpose of the square matrix A, then \A'^\ = |A|.

Now let A be any square matrix and B be the matrix obtained by multiplying the ethrow of A by a non-zero scalar k. In terms of elementary matrices B = Hdk)-A: and. bvTheorem XXI, ,„, ,,B = \Hi{k)-A = k\A\


But \Hik)\ = k; hence, |Hi(fc)-A| = \Hik)\ • \A\. By an independent proof or by Theorem

XXII. we have also\A'K,{k)\ = \A\-\Kik)\

Next, denote by B the matrix obtained from A by interchanging its iih and ^th columns

and denote by t the corresponding transposition (t, j). The effect of t on (a') is to produce

(ffl'") ipT0'l,lpTO'2,2prO'3,3pT • • • C^n.npT

hence, \B\ = ^ epT*''l.lpT'''2,2pT'^3,3pT • • • 0,n.npT

Now CT = pT e Sn is even when p is odd and is odd when p is even; hence, £„ == -tp. More-

over, with r fixed, let p range over Sn; then a ranges over S„ and so

l-^l ~ ^ V ^'l.lo- *'2,2(r *'3,3i7 • • • ^n.rUT ~ ~|-^l

We have proved

Theorem XXIII. If B is obtained from A by interchanging any two of its rows (columns),

then \B\ = -\A\.

Since in Theorem XXIII B = A-Ka and |X«| = -1, we have \A'Ki,\ = \A\'\Kii\

and, by symmetry, \Hij-A\ - \Hij\-\A\.

There follows readily, excluding all fields of characteristic two.

Theorem XXIV. If two rows (columns) of A are identical, then \A\ = 0.

Finally, let B be obtained from A by adding to its ith row the product of k (a scalar)

and its jth row. Assuming j < i,

\B\ = 2 *p'''l.lp • • • î,iP • • ^-lAi-lIp i'î.ip'^'^î.ip) *t+l,(i + l)p • • ^n.npS„

~ 2j ^p ^l.lP î.2p *^3,3p • • • *^n.nps„

+ 2 ^p'^l.lp • • • î.iP • • î-l,(i-l^py"'î.ip) *t+l,(i + l)p • • *n,nps„

- |A| + = \A\ (using {b') and Theorems XXI and XXIV)

We have proved (the case j>i being left for the reader)

Theorem XXV. If B is obtained from A by adding to its ith row the product of k

(a scalar) and its jth row, then |S| = \A\. The theorem also holds when

row is replaced by column throughout.

Since, in Theorem XXV, B = Hij{k)'A and |i?i;(fe)l = |/| = 1, we have

\Hii(k)'A\ ^ \Hii{k)\'\A\ and \A - Kii{k)\ ^ \A\ - \Kiik)\

We have now also proved

Theorem XXVI. If A is an n-square matrix and H{K) is any n-square elementary row

(column) matrix, then

\H-A\ = \H\-\A\ and \A-K\ = \A\-\K\

By Theorem IX', any square matrix A can be expressed as

(c) A = HT' . H^' ...H7''N- K;' . . . K^' • Kr'

Then, by repeated applications of Theorem XXVI, we obtain


\Hr'\ ' \H^' ... H7'-N'K;' k^'-kth

= \H7'\'\m'\-\H^' ...H:''N'K-,' Ki' ' KT'\

= \h;'\-\h..-ll2 . 1^; lî". .

\N\ . \KJ'\ . . . \K^'\

If A is non-singular, then iV = / and |Ar| = 1; if A is singular, then one or more of thediagonal elements of AT is and \N\ = 0. Thus,

Theorem XXVU. A square matrix A is non-singular if and only if \A\¥'0.

and

Theorem XXVIII. If A and B are n-square matrices, then \A-B\ = |A|- ISI.

EVALUATION OF DETERMINANTSUsing the result of Example 23 (ii), we have

2

5

7(1) (2) + (3)

4 5

5 7

- (40 - 42) - 2(32 - 30) -I- 3(28 - 25)

= -2-4 + 9 = 3

The most practical procedure for evaluating \A\ of order n^3, consists in reducing Ato triangular form using elementary transformations of the types Hij{k) and Kij{k) ex-clusively (they do not disturb the value of \A\) and then applying Theorem XX. If otherelementary transformations are used, careful records must be kept since the effect of Huor Kij is to change the sign of \A\ while that of Hi{k) or Ki{k) is to multiply \A\ by k.

Example 24: An examination of the triangvdar forms obtained in Example 6 and Problem 8 showsthat while the diagonal elements are not unique, the product of the diagonal ele-ments is. From Example 6(a), Page 171, we have

= (l)(-3)(-l) = 3


1 2 3

4 5 6 =

6 7 8

1 2 3 1 2 3

-3 -6 = -3 -6-3 -7 -1

Solved Problems

1. Find the image of | = (1, 2, 3, 4) under the linear transformation A =of F4 (Q) into itself.

iA = €[yi V2 73 74] = (I • Yi, « • 72, « • 73. f • 74)

= (9, 15, 25, -3) (See Problem 15, Chapter 13, Page 158.)

1 -2 4

2 4 1 -2-1 5 -1

1 3 2

186 MATRICES

2. Compute A • B and B-A, given A =

and

A • B [4 5 6]

B-A = [4 5 6]

and B = [4 5 6].

1-4

2-4

3-4

[CHAP. 14

1-5 1'6

2-5 2-6 =

3*5 3-6

4 5 6

8 10 12

12 15 18

= [4-1 + 5-2 + 6-3] = [32]

3. When A12-23 1

A- B =

and B =2 10-113-20

1-1-1, find A • B.

2+2 1+6-6 3 + 1

2 -4 + 2 -1 + 2

-1 -3-1= 4 5-2 1

6 4-1-4

Show that the linear transformation

whose image is 0.

12 2

2 5 3 1

3 8 4 2

2 7 13

in Vt (R) is singular and find a vector

Using in turn H2,(-2), H3,(-3), H4i(-2); Hi2(-2), H32(-2). H^î'^), we have

1220 104-21-11^01-11

02-2 2 000003-3 3 0000

The transformation is singular, of rank 2.

Designate the equivalent matrices as A,B,C respectively and denote by pi,P2,Ps'P4 the row

vectors of A, by P[, p; P'^, P; the row vectors of B, and by p[', p'^, p'^, p'; the row vectors of C.

Using the moves in order, we have

12 2

2 5 3 1 ^^

3 8 4 2

2 7 13

— Pi — 2pi, ^^3

J' —— P3 — 3pi, P4 ~ 2pi

Now

while

p;' = pi-2p;, p;' = p;-2p;, p;' = p;-3p;

= p; - 2p; = (P3-

3pi)- 2(p2 - 2pi) = P3 - 2p2 + Pi =

=3p' = (P4 - 2pi)

- 3(p2 - 2p,) = P4 - 3p2 + 4pi

Thus, the image of £ = (1,-2,1,0) is 0; also, the image of „ = (4,-3,0,1) is 0. Show that the

vectors whose images are fill a subspace of dimension 2 in Vt^^ («).

5. Show that the linear transformation A =

We find

A =12 3

2 133 2 1

1 2 3

-3 -3 ~

-4 -8

1 2 3

1 1 ~

-4 -8

12 3

2 133 2 1

is non-singular.

1—

1

— —

1 1 1 101 1 ~ 1 1 ~ 10-4 1 1

_ _ — —

'

= B

The row vectors of A are linearly independent; the linear transformation A is non-singular.


6. Find the rank of the linear transformation A =12 2 4 2

2 5 3 10 7

3 5 7 10 4

of F8(iJ)intoF5(/2).

We find

12 2 4 2

2 5 3 10 7

3 5 7 10 4

2

1

-1

2 4 2

12 3-1 -2 -2

1 4 -4 1 4

1 -1 2 3 ~ 1 -1 2

1 1

= B

The image vectors are linearly independent; r^ = 3.

7. From the set {(2,5,0,-3), (3,2,1,2), (1.2,1,0), (5,6,3,2), (1,-2,-1,2)} of vectorsin Vi (R), select a maximum linearly independent subset.

The given set is linearly dependent (why?). We find

A =

2 5 -3 I -2 -31 ~0 1 -2 -33 2 1 2 0-4-2 2 -10 -101 2 1 ~ 12 10 ~ 1 5 6

5 6 3 2 0-4-2 2 -10 -101 -2 --1 2 -4 -2 2 -10 -10_

~01 -2 -3~

1 -1~1 1 11

'*' 1 5 6

-10 -10

"^ 10 1 r: B

-10 -10[)

From an examination of the moves, it is clear that the first three vectors of A are linear combinationsof the three linearly independent vectors of B (check this). Thus, {(2,5,0,-3), (3,2,1,2), (1,2,1,0)}is a maximum linearly independent subset of A. Can you conclude that any three vectors of A arenecessarily linearly independent? Check your answer by considering the subset (1, 2, 1, 0), (5, 6, 3 2)(1,-2.-1,2).

8. By means of elementary column transformations, reduce A =triangular, lower triangular, and diagonal form.

Using K:,3(-2/3), «-23(-5/6); Ky^{\\ K^{r-\m), we obtain

12 3

4 5 6

5 7 8

to upper

12 3

4 5 6~

5 7 8

-1 -1/2 3

6 ~

-1/3 1/3 8

3/2 -5/8 3

-1/4 6

8

which is upper triangular.

Using ^2i(-2), ^si(-3); ^s2(-2) we obtain

A =12 3

4 5 6 ~

5 7 8

-3 -6

-3 -7which is lower triangular.

Using «'2i(-2), «:3i(-3), A'32(-2); K,3(5), «-23(-3); K^^^MZ) we obtain

which is diagonal.A ~1

4 -3 ~

5 -3 -1

1

4 -3 ~-1


Prove: Any non-zero matrix A over f can be reduced by a sequence of elementary

row transformations to a row canonical matrix (echelon matrix) C having the properties:

(i) Each of the first r rows of C has at least one non-zero element; the remaining

rows, if any, consist entirely of zero elements.

(ii) In the ith row (i = 1, 2, . . . , r) of C, its first non-zero element is 1, the unity of f.Let the column in which this element stands be numbered ji.

(iii) The only non-zero element in the column numbered ji (i

element 1 in the ith row.

1, 2, . ..

, r) is the

(iv) jl<J2< • < jr.

Consider the first non-zero column, numbered Ji, of A;

(a) If ttij ^ 0, use Hi(d^j^) to reduce it to 1, if necessary.

(6) If ttjj = but Cpj y^ 0, use Hip and proceed as in (o).

(c) Use transformations of the type Hîk) to obtain zeros elsewhere in the ii column when necessary.

If non-zero elements of the resulting matrix B occur only in the first row, then B = C; otherwise,

there is a non-zero element elsewhere in the column numbered J2 > ii- If b2j "^ 0, use H2(62j„) as

in (a) and proceed as in (c); if b2j — but b^j ¥= 0, use H^q and proceed as in (a) and (c).

If non-zero elements of the resulting matrix occur only in the first two rows, we have reached C;

otherwise, there is a column numbered is > j^ having non-zero elements elsewhere in the column.

If . . . and so on; ultimately, we must reach C.

10. Prove: The row rank and column rank of any matrix A over f are equal.

Consider any w X n matrix and suppose it has row rank r and column rank s. Now a maximumlinearly independent subset of the column vectors of this matrix consists of s vectors. By inter-

changing columns, if necessary, let it be arranged that the first s columns are linearly independent.

We leave for the reader to show that such interchanges of columns will neither increase nor decrease

the row rank of the given matrix. Without loss in generality, we may suppose in

A =

0'2i tt22 "'is ^2.8+1 'hm

O'sl "-si

"'ml <*m2

a. "•s.s+l

^ms '^m, « + 1 "w.n

the first s column vectors Yi, y2> • • • i Ts *'e linearly independent while each of the remaining n — 8

column vectors is some linear combination of these, say,

Ys + t= CtiYi + Ct2Y2 + • • • + CtsVs. (t = 1, 2, . . ., n - s)

with cj^ e f. Define the following vectors:

Pi — (.0,11, a^2 '"'Isl' 92, — ('''21> <''22> • • • > <''2s)> ' Pm — (O'mli f^m2' • > ""msl

, ff„ — (Oln. l2n> ••>'*« + l,n)and "l = (Oll. <l21> • • •>''^s + l,l)> "2 — (<*12><*22> • • -(^S + La);

Since the p's lie in a space V^ (T), any s + 1 of them forms a linearly dependent set. Thus there exist

scalars 6i, 62, . . ., &s + i in f not all such that

6iPj + 62P2 + ••• + fcs + iPs + i = (biiii + &2«2i+•••+ *s + !«» + 1,1. &iai2 + M22 + • •

•

+ hs^l"'s+l.2> •••> ?'l«ls + ''2"2s + • + ''s+l'^s + l.s)

= (£•"!. i'02, , i'Os) - ?

where J=:: (0, 0, . .

., 0) = is the zero vector of V, (f) and i = (ftj, 62. • • • . &s + 1)- Then


i' Ol = f • "2 - • • • = i- Ob =

Consider any one of the remaining <t's, say,

•^s + fc = («l.s + fc. «2.s+lc. , O-s+l.s + k)

= (Cfciflji + Cfcadja + • • + Cfcâij, c,,ia2i + 0^2122 + • • + 0^502,

CkA + i,! + Ck2as + i,2 + • • • + Ck5a, + ,_j)

Then «*<'s + fc = Cfci(i*<'i) + Cfc2U'<r2) + •• + Cfcâ-<T,) =

Thus any set of s + 1 rows of A is linearly dependent; hence » — r, that is,

the column rank of a matrix cannot exceed its row rank.

To complete the proof, we must show that r^s. This may be done in either of two ways:

(i) Repeat the above argument beginning with A, having its first r rows linearly independent, andconcluding that its first r + 1 columns are linearly dependent.

(ii) Consider the transpose of A

AT =

Ojj 021

fl'12 ^2

"In "211

whose rows are the corresponding columns of A. Now the row rank of A^ is s, the columnrank of A, and the column rank of A ^ is r, the row rank of A. By the argument above thecolumn rank of A''' cannot exceed its row rank; i.e. r — s.

In either case we have r = s, as was to be proved.

11. Reduce A3 2 3 4 5

2-14514 5 12-3

over R to normal form.

First we use Hi2(-1) to obtain the element 1 in the first row and first column; thus

3 2 3 4 5

2-14514 5 12-3

1 3-1-1 4

2-14514 5 12-3

Using H2i(-2), //3i(-4), X2,(-3), Ksid), K^â), Kî(-4), we have

A ~1

-7 6 7 -7-7 5 6 -19

Then using Hsî-l), Kî-l/l), Ksî-G), K^î-I), Kî(l), we have

A ~

52V

10100-1-1 -12

and finally, using H^^-X), K43(-l), A^53(-12),

A ~101010


12. Reduce A11122 1-3-63 3 12

over R to normal form N and find matrices S and T

such that S'A-T^N.We find

I4

A I3

1

1

2

3

1

2

-6

2

-1 -5 -10 -20-2-4 -3

1 -1 -110

1

10 1

-1 -5 -10 -20-2-4 -3

-2

1

1

-2

1

10

2

1

1

2

3/2 -1/2

-2

1 10-11/2 -1 5/2

2 3/2 -1/2

1 -1 -1

1

1 -21

1

10-11/2 -1 5/2

3/2 -1/2

TN S

Hence S =10

-11/2 -1 5/2

3/2 -1/2

and T =

1-1-110

1-21

13. Prove: The inverse of the product of two matrices A and B, each of which has an

inverse, is the product of the inverses in reverse order, that is,

(A'B)-' = 5-1 -A-i

By definition, (A' B)-^ • (A- B) = (A • B)(A • B) - 1 = /. Now(B-i«A-i) • (A'B) = B-»(A-i'A)J? = B-^'I'B = B'^ ' B = I

and (A-B)(B-»'A-i) = A(B'B-i)A-» = A • A-^ = /

Since {A'B)-^ is unique (see Problem 33), we have (A 'B)-^ = B"* • A"*.

14. Compute the inverse of A =12 4

3 102 2 1

over 1/(5).

We have1 2 4 10 1 2 4 10 1 2 4 10

[A h] = 3 10 102 2 10 1

3 2 103 3 3 1

3 3 3 1

3 2 10

12 4 10 10 2 4 1 10 111~ 1110 2 ~ 1110 2 ~ 10 2 3 2

1 4 2 1 4 2 1 4 2

and A-» =1112 3 2

4 2


15. Find the minimum polynomial of A =

Clearly, A ¥= UqI for all Oq G R. Set

1 1 1

2

111over R.

A^ =2 4 2 1 1 1 10

4 = 0,1 2 + Oo 10 =2 4 2 111 1

O, +Oo «i Oj

2ai + oo

Oj a, 0, + 0o

which is impossible. Next, set

A3 =4 12 4

8

4 12 4

2 4 2 1 1 1 10= 02 4 + ai 2 + ao 10

2 4 2 111 1

2o2 + Oi + Oq 4a2 + Oi 2a2 + Oj

4o2 + 2o, + Ofl

2a2 + Oi 4o2 + o-i 2o2 + Oj + Oq

i2o2

+ ai + Of, = 4

4a2 + a.1 = 12 , we obtain a^ = 0, Oj = —4, a.2 = 4. After checking for every element

2a2 + Oj =4of i4* and not before, we conclude w(X) — \^ — 4\^ + 4\.

16. Find all solutions, if any, of the system

2a;i + 2a;2 + 3x3 + Xi = 1

3xi - X2+ X3 + 3x4 = 2

-2xi + 3x2 - Xs — 2x4 = 4

Xi + 5x2 + 3x3 - 3x4 = 2

2xi + 7x2 + 3x3 - 2x4 = 8

over R.

We have

[AH] =

2 2 3 1

3-113-2 3 -1 -2

15 3-32 7 3-2

1 5

3 -1

-2 3

2 2

2 7

3 -3

1 3

-1 -2

3 1

3 -2

1 5

-16

13

-8

-3

-3

12

-8

7

4

2

-4

8

-3

4

15 3-31 1/2 -3/4

13 5-80-8-3 7

0-3-3 4

10 1/4 5/4"

10 -5/4 3/4

1 1 -1

13/4 13/4

2

1/4

8

-3

1 1/2 3/4 3/4 1 1/2 3/4 3/4

1 1/2 -3/4 1/4 1 1/2 -3/4 1/4

-3/2 7/4 19/4 ~ 1 1 -1

1 1 -1 -3/2 7/4 19/4

-3/2 7/4 19/4

10 1/4

10 -5/4

1 1

1

5/4

3/4

-1

1

10 1

10 2

10-211

Both A and [A H] have rank 4, the number of unknowns. There is one and only one solution:

«! = 1, X2- 2, «3 = -2, Xi = 1.

Note. The first move in the reduction was H^. Its purpose, to obtain the element 1 in the first

row and column, could also be realized by the use of Hi(^).


17. Reduce

3 2 1

6 5 4

4 2 5

over 7/(7) to normal form.

Using Hi(5); ^2,(1), i?3i(3); H^^W, HsîS); Hs(S); Hi3(l), ^23(5), we have

3 2 1 1 3 5 13 5 10 6 10 6 1

6 5 4 ~ 6 5 4 ~ 12 ~ 12 ~ 12 ~ 104 2 5 4 2 5 4 6 5 1 1

18. Find all solutions, if any, of the system

We have

[A H]

12 13 4

2 13 2 1

2 11 3

3 13 4 2

Xi + 2a;2 + X3 + 3x4 = 4

2xi + X2 + Sxa + 2x4 = 1

2*2 + Xa + X4 = 3

3«i + X2 + 3a;3 + 4ii;4 = 2

12 132 112 11

12 13 4

13 3

2 114

3~

over //(5).

10 2 1

13 3 4

Here, r^ = rj^ jj] = 2; the system is compatible. Setting a

solutions are given byXi = 1 + St, X2 = ^ + 2s + 2t, x^ — s, x^ = t

Since 7/(5) is a finite field, there is only a finite number (find it) of solutions.

s and Xi = t, with s,tG//(5), all

19. Solve the system

We have

2a;i + a;2 + a;3 =

xi +a;3 = over 7/(3).

2a;2 + 0:3 =

A =2 11 1 2 2 12 2 1 1

10 1 ~ 10 1 ~ 1 2 ~ 122 1 2 1 2 1

Then assigning aig = s e 7/(3), we obtain Xi = 2s, X2 = X3 = s as the required solution.

20. With each matrix over Q, evaluate:

(a)

1 -3

2 5 4 =-3 2 -2

-1 1 -3

-354 =-5 2 -2

1

3 2 13 =

5 -3 13

-102 5

-5 -3 13

-65

Ki2(—1) is used to replace a^ = by a non-zero element. The same result can be obtained by

using X,2; then

1 -3

5 4

2 -2

-3

2 4

-3 -22 19 = —

3 4

105 2

2 -3 65/2

= -65

An alternate evaluation is as follows:

102 6 19

-3 2 4

1 -3

2 5 4

-3 2 -2-d)

2 19

-3 4= -(8 + 57) = -65

CHAP. 14]

(b)

] MATRICES

2 3 -2 4 -1 3 -2 4 -1

3 -2 1 2 5 •-2 1 2 5 13 -9 22

3 2 3 4 1 2 3 4 1 5 1 8

-2 4 5 -6 4 5 -6 -14 12 -19

-1 -1

-1 -1 3 3 -1 -101 5 1 8 1 5 16 23

-6 -14 12 -19 -6 - 14 -30 -61

= (-1)(--1)(16) (-143/8) = -286

193

-10-1-101 5 16

-6 -14 -30 -143/8


21. Given A =10 2

*.

12 3

3 1 . B = 13 4 , c =4 2 14 3

''

-7 6 -1

10-11 -2 1

over Q, compute:

2 2 5

(o) A+B = 16 5

5 6 3

(b) SA3 6

9 3

12 6

(c) A'B

(d) B • C =

3 10 9

4 13 15

6 14 20

-2

0-20-2

(e) A • C =-5 2 1

4 -2 -2-26 24 -6

if) A2 A'A =9 4 2

4 11 3

4 6 10

22. For the arrays of Problem 21, verify: (a) (A + B)C = AC + BC, (6) (A • B)C = A{B • C).

23. For A = [ay], (i = 1,2,3; j = 1,2,3), compute Ig'A and A '/g (also Oj-A and A • O3) to verify:In the set % of all n-square matrices over f, the zero matrix and the identity matrix commute withall elements of %.

24. Show that the set of all matrices of the form^siQ).

25. Show that the set of all matrices of the form

a b

a+b0c

a b c

a + e

c b a

, where a, 6, c S Q, is a subalgebra of

where o, b, c G R, is a subalgebra of

26. Find the dimension of the vector space spanned by each set of vectors over Q. Select a basis for each,

(a) {(1,4,2,4), (1,3,1,2), (0,1,1,2), (3,8,2,4)}

(6) {(1, 2, 3, 4, 5), (5, 4, 3, 2, 1), (1, 0, 1, 0, 1), (3, 2, -1, -2, -5)}

(c) {(1,1,0,-1,1), (1,0,1,1,-1), (0,1,0,1,0), (1,0,0,1,1), (1,-1,0,1,1)}

Ans. (a) 2, (5) 3, (c) 4

27. Show that the linear transformation A =vector whose image is 0.

12 3

2 4 3 1

3 2 142 4 2

of V4 (R) into itself is singular and find a


28. Prove: The 3-square matrices /, H12, Hjs, Hâ, Hi2'Hj3, •Hi2*-H23 under multiplication form a

group isomorphic to the symmetric group on 3 letters.

29. Prove: Under multiplication the set of all non-singular n-square diagonal matrices over 7* is acommutative group.

30. Reduce each of the following matrices over R to its row equivalent canonical matrix:

(a)

(b)

1 2 -3

2 5 -4

1 1 1 2

2 1 -3 -6

3 3 1 2

(c)

id)

12 1213 2 2

2 4 3 4

3 7 4 6

12-22 5-4

-1 -3 2

2 4-1

(e)

3 4 5 6 7 8

4 5 6 7 8 9

6 6 7 8 9 8

10 11 12 13 14 15

Ana. (a)10-712 (6)

_1010 (c)

12—

'

10 2

1010

(d) h (e)

1 0-1-2-312 3 4

1

31. In Example 11, Page 176, use Hî^), H^(-\), H^sf.-^), Kî?) on

1 -2 1

1

1

1 1

2 5 -3 1

1 3 -4 1 and obtain S =1

11 3 -6

-5/2 -1/2 1

and T =1 -2

1

to show that the non-singular matrices S and T such that S'A-T = I are not unique.

32. Reduce A =

S'A'T = N.

over R to normal form N and compute matrices S and T such that

33. Prove that if A is non-singular, its inverse A"' is unique.

Hint. Assume A-B - C'A = I and consider (C • A)B = C(A • B).

34. Prove: If A is non-singular, then A' B - A-C implies B = C.

35. Show that if the non-singular matrices A and B commute, so also do (a) A-i and B, (b) A and B-i,

(c) A-i and B'KHint, (a) A-i(A'B)A-i = A-i(B«A)A-i.

36. Find the inverse of:

(a)

(6)

13 3

14 3

13 4

1 2 3

2 4 5

3 5 6

(c)

(d)

1 2 3

13 3

2 4 3

2 1-113 2

-1 2 1

ie)

if)

3 4 2 7

2 3 3 2

5 7 3 9

2 3 2 3

3 -4

1 1

3 -4

5 -5-5 8

over Q.


Ana. (a)

7 -3 -3

1 1 , (6)

1 1

1-3 2

-3 3 -1

2-10(0 i

3-6 3

-3 3

2 0-1

('^ ^13-53-16

-6 6 -5(e) i

-1 11 7 -26

-1 -7 -3 16

11-11-1-1 2

(/) 1^

2 16 -6 4

22 41 -30 -1

-10 -44 30 -2

4 -13 6 -1

37. Find the inverse of A10 1

1112 11

over //(3). Does A have an inverse over 7/(5)?

Ans. A-^ =2 1

2 10112

38. Find the minimum polsmomial of (a)

1 2 1 1 2 2 11

2

1

-1, (b) 10

1

, (e) 1 1 2

112, (d) 12 1

1 1 2

Ans. (a) \3 + X2 - 2X - 1, (6) X^ - 3X + 2, (c) X^ - 4X, (d) X^ - 5X -I- 4

39. Find the inverse of each of the matrices (a), (6), (d) of Problem 36, using its minimum polynomial.

40. Suppose X3 + aX2 + 6X is the minimum polynomial of a non-singular matrix A and obtain acontradiction.

41. Prove: Theorems XIX, XX and XXI, Page 183.

42. Prove Theorem XXIV (Hint. It the ith and jth rows of A are identical, |A| = \HJ • |AI) andTheorem XXVIII.

. Evaluate:

12 3 -7 6 -1

(a) 13 4 (c) 10-11 4 3 1-2 1

10 2 2 -1 1

(ft) 3 1 (d) 3 2 4

4 2 -10 3

(«)

{/)

11162 4 164 12 9

2 4 2 7

3 5 7 2

2 4 11-2000113 4

Ans. (a) -2, (6) -26, (c) 4, (d) 27, (e) 41, (/) 156

44. Evaluate: (a)

X - 2 -1 -4-1 X - 3 -5-4 -5 X - 6

X-1 2 3

1 X-3 4 , (6)

1 4 X-3Hint. Expand along the first row or first column.

Ans. (a) X» - 7X2 - 6X + 42, (6) X^ - llX^ - 6X -h 28

45. Denote the row vectors of A = [oj^], (t,i = 1,2,3), by p„p2,p3. Show that

(a) pi X p2 (see Problem 13, Chapter 13, Page 157) can be found as follows: Write the array

a/i Oi2 Oi3 ttjj o,2

C^l 022 '»23 <»21 O22

and strike out the first column. Then

Pi X P2 —

(6) \A\ = p, . (p2 X P3) = -p2 • (p, X Pa) = pj •(pi X P2).

«12 ttlS

O22 <ha

*1S Oil

"'23 021

«„ ai2

«21 <»22


46. Show that the set of linear forms

(a)

/l = ail'»:i + «12a^2 + • • • + a^Xn

/2 = a2iXi + 022«2 + • • + 02n«nover f

fm = «ml*l + Om2*2 + ' " ' + "mn^^n

is linearly dependent if and only if the coefficient matrix

A = [ay], (t = 1,2, ...,m; j = 1,2, . . .,n)

is of rank r < m. Thus (a) is necessarily linearly dependent if m > n.

47. Find all solutions of:

(a) xi — 2x2 + 3*3 — 5x4 = 1 (6)Xi + X2 + X3 = 4

2x, + 5x2 — 2x3 = 3

X, + X2 + X3 = 4

(c) \ 2x, + 5x2 - 2x3 = 3

[ X, + 7x2 - 7x3 = 5

(d)

(/)

2xi + X2 + 5x3 + X4 = 5

X, + X2 — 3x3 — 4x4 = —1

3xi + 6x2 - 2x3 + X4 = 8

2xi + 2*2 + 2x3 - 3x4 = 2

ixj+ X2 + 2x3 + X4 = 5

2x, + 3X2 - X3 - 2x4 = 2

4xi + 5x2 + 3x3 =7

Xj + X2 — 2x3 + X4 + 3x5 = 1

2xi - X2 + 2x3 + 2x4 + 6x5 = 2 (g) -

3xi + 2x2 - 4x3 - 3x4 - 9x5 = 3

Xi + 3x2 + X3 4- X4 + 2x5 =

2xi + 5x2 - 3x3 + 2x4 - X5 = 3

-Xi + X2 + 2x3 - X4 + X5 = 5

3xi + X2 + X3 — 2x4 + 3x5 =

over Q.

48.

Ans. (o) X, = 1 + 2r - 3» + U, x^ = r, X3 = s, X4 = t

(6) Xi = 17/3 - 7r/3, Xj = -5/3 + 4r/3, X3 = r

(d) Xj -2, X2 = 1/5, X3 ==0, X4 = 4/5

(/) Xj = 1, X2 = 2r, X3 = r, X4 = -36, x^ - b

(g) X, = -11/5 - 4r/5, X2 = 2, X3 = -1 - r, X4 = -14/5 - r/5, Xg = r

(a) Show that the set M^- {A,B, ...) of all matrices over Q of order 2 is isomorphic to the vector

space Vi (Q). Hint. Use A

(6) Show that /n =space.

1

. '12 —

"0,1

021

012

<h2

-^ (Oil, O12,

1' /21-

1

(Oil, O12, 021. <»22)- See Prob. 3, Chap. 10, Page 108.

I9.9. —1

a basis for the vector

(c) Prove: A commutes with B =

Hint. B

"u "12

621 ^22

*ll^ll + ^12-^12 + ''21-^21 + *22-^22-

if and only if A commutes with each /y of (6).

49. Define S, = X y

-y Xx,y &r\ . Show (o) S2 is a vector space over R, (h) S^ is a field.

Hint. In (b) show that the mapping S2 -» CX y

-y XX ¥ yi is an isomorphism.

50. Show that the set ^ = {(9i + 921 + 93^ + 94*^) = 9i, «2. 93. 94 ^ i2} with addition and multiplication

defined in Problem 27, Chapter 11, Page 123, is isomorphic to the set

S4 =

Is S4 a field?

9i 92 93 94

—92 9i —94 93

-93 94 9i —92

—94 ~93 92 9i

9i. 92. 93. 94 ^ ^


51. Prove: If ii, i2' • . im are m<n linearly independent vectors of V„ (7"), then the p vectors

ij = «;i?i + Sj2l2 + • • • + Sjmim , (; = 1, 2 p)

are linearly dependent if p > w or, when p - m, if [sy] is of rank r < p.

52. Prove: If {i.jg, ••,l„ are linearly independent vectors of V„ (7), then the n vectors

Vj = Oji£i + aj2i2 + • • • + Oj„|„, (;• = 1,2, . . .,n)

are linearly independent if and only if |oi,l # 0.

53. Verify: The ring Tj = a b

e

X: xGRi,

: a,b,cGRy has the subrings

X y: x,j/ G K I , and X

y: a;,3/Gft

as its proper ideals. Write the homomorphism which determines each as an ideal. (See Theorem VIChapter 10, Page 105.)

54. Prove: (A + B)t = A^ + fir and {A • B)^ = B^ • A^ when A and B are n-square matrices over f.

55. Consider the n-vectors X and F as 1 X n matrices and verify:

X •¥ = X -y^ = Y ' XT

56. (a) Show that the set of 4-square matrices

^ = {/, H,2, H,3, H,4, H23, H24, H34, H,2 • H,3, H,2 . H23, H12 • Hi4, Jtf,2• H24, «i3 • H,„

Hn-Hi3, His'Hii, Hî-Hii, Hi2-Hs^, H^^-H^, Hi^'Hâ, H^^'Hi^'Hi^,H,2-H,4-H,3, Hi3'Hi2-Hn, Hi3-Hii'Hy ^14 '^12 '^13. ^14 • ^13 * ^12}

is a multiplicative group. Hint. Show that the mapping:

Hij^(ij), Hij'Ht^-^(ijk), Htj'Hî^{i})(kl), Hij-Hik-Hii^(ijkl)

of •!M into iS„ is an isomorphism.

(6) Show that the subset {/, H^^, H^^, H^^-H.^^, H^^-H^, H.^-H^^, H^^'H^^-H^^, H^^-H^^-H^^}of «3/ IS a group isomorphic to the octic group of a square. (In Fig. 9-1, Page 92) replace thedesignations 1,2,3,4 of the vertices by (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1) respectively.)

57. Show that the set of 2-square matrices

1

-1 »

-1

-1»

-1

1'

1

-1t

-1

1»

1

1»

-1

-1

is a multiplicative group isomorphic to the octic group of a square.

Hint. Place the square of Fig. 9-1, Page 92, in a rectangular coordinate system so that the vertices1, 2, 3, 4 have coordinates (1, -1), (1, 1), (-1, 1), (-1, -1) respectively.

58. Let S spanned by (1,0,1,-1), (1,0,2,3), (3,0,2,-1), (1,0, -2, -7) and T spanned by (2,1,3,2),(0,4,-1,0), (2,3,-4,2), (2,4,-1,2) be subspaces of V4 (Q). Find bases for S, T, SnT, and S+T.

chapter 15

Matrix Polynomials

MATRICES WITH POLYNOMIAL ELEMENTS

Let f[\] be the polynomial domain consisting of all polynomials in X with coefficients

in 7=". An TOXn matrix over f[\], that is, one whose elements are polynomials of ^[A],

an(A) ai2(A) . . . Oin(A)

a2i{A) a22(A) . . . a2n(A)

A(A) = MA)] =

aml(A) am2(A) . . . amn{A)

is called a A-matrix (read, lambda matrix).

Since f G f[\], the set of all m x n matrices over f is a. subset of the set of all to x «

A-matrices over f[\]. It is to be expected then that much of Chapter 14 holds here with,

at most, minor changes. For example, with addition and multiplication defined on the

set of all n-square A-matrices over f[\] precisely as on the set of all w-square matrices

over f, we find readily that the former set is also a non-commutative ring with unity /«.

On the other hand, although A (A) = is non-singular, i.e. 1A(A)| — A(A + 1) ^ 0,

A + lJ

A(A) does not have an inverse over f[X.]. The reason, of course, is that generally a(A) does

not have a multiplicative inverse in f[\]. Thus it is impossible to extend the notion of

elementary transformations on A-matrices so that, for instance, A (A)A

A + 1

1

Ij

ELEMENTARY TRANSFORMATIONS

The elementary transformations on A-matrices are defined as follows:

The interchange of the tth and jth rows, denoted by Ha; the interchange of the

ith and yth columns, denoted by Ka.

The multiplication of the ith row by a non-zero element kGf, denoted by

Hi(k); the multiplication of the ith column by a non-zero element k&f, denoted

by Ki(k).

The addition to the tth row of the product of /(A) e f[\] and the yth row,

denoted by ffo(/(A)); the addition to the ith column of the product of /(A) G f [A]

and the yth column, denoted by Ki,{f(\)).

(Note that the first two transformations are identical with those of Chapter 14 while the

third permits all elements of f [A] as multipliers.)

An elementary transformation and the elementary matrix obtained by performing that

transformation on / will again be denoted by the same symbol. Also, a row transformation

on A(A) is effected by multiplying it on the left by the appropriate H, and a column trans-

formation is effected by multiplying it on the right by the appropriate K.

198

CHAP. 15] MATRIX POLYNOMIALS 199

Paralleling the results of Chapter 14, we state:

Every elementary matrix is non-singular.

The determinant of every elementary matrix is an element of f.

Every elementary matrix has an inverse which, in turn, is an elementary matrix.

Two mxn A-matrices A(A) and B(\) are called equivalent if one can be obtainedfrom the other by a sequence of elementary row and column transformations, i.e.

if there exist matrices S(A) = Hs ... Hi- Hi and T{X.) = Ki'Kz.. .Kt such that

Six.) • A{k) • T{\) = B{\)

The row (column) rank of a A-matrix is the number of linearly independentrows (columns) of the matrix. The rank of a A-matrix is its row (column) rank.

Equivalent A-matrices have the same rank. The converse is not true.

NORMAL FORM OF A A-MATRIX

Corresponding to Theorem IX', Chapter 14, Page 174, there is

Theorem I. Every mxn A-matrix A (A) over f[k] of rank r can be reduced by elementarytransformations to a canonical form {normal form)

N{X) =

A (A)

/,(A) .

. . /,(A) . .

. . .

.. . .

in which f^(k),f^{\),

/,+,(A) for i=l,2, .

. ., /,(A) are monic polynomials in f[\] and /.(A) divides.,r-l.

We shall not prove this theorem nor that the normal form of a given A(A) is unique(The proof of the theorem consists in showing how to reach N{\) for any given A(\yuniqueness requires further study of determinants.) A simple procedure for obtainingthe normal form is illustrated in the example and problems below.

Example 1:

Reduce A(\) =X + 3

2\2 + \ - 3

X3 + \2 + 6X + 3

\ + l

\2 + \ - 1

2X2 + 2X + 1

X + 2

2X2-2X3 + X2 + 5X + 2

over R(\) to normal form.

The greatest common divisor of the elements of A(\) is 1; take /,(X) = 1 Now use K (-1)to replace a„(X) by /,(X), and then by appropriate row and column transformation obtain anequivalent matrix whose first row and first column have zero elements except for the commonelement /i(X); thus,

A(X) ~1

X-1X + 1

X + 1

X2 + X - 1

2X2 + 2X + 1

1

X-1 X

x + 1 X2

x + 2

2X2-2X3 + X2+5X + 2

X2-\X3 + 2X

10X X2 - X

X2 X3 + 2X

= B(\)

200 MATRIX POLYNOMIALS [CHAP. 15

Consider now the submatrixX X2_x\2 \» + 2\

The greatest common divisor of its elements is \;

set fzM = X. Since /2(X) occupies the position of 622^^) in B(\), we proceed to clear the second

row and second column of non-zero elements, except of course for the common element /2(X),

and obtain

A(X) ~ X X2-X ~

X2 X3 + 2X

X2-XX2 + 2X X2 + 2X

= N{\)

since X^ + 2X is monic. See also Problems 1-3.

The non-zero elements of N{k), the normal form of A(X.), are called invariant factors

of A{\). Under the assumption that the normal form of a A-matrix is unique, we have

Theorem II. Two mxn A-matrices over ^^'[A] are equivalent if and only if they have the

same invariant factors.

POLYNOMIALS WITH MATRIX COEFFICIENTS

In the remainder of this chapter we shall restrict our attention to «-square A-matrices

over f[\]. Let A (A) be such a matrix and suppose the maximum degree of all polynomial

elements aij(A) of A(A) is p. By the addition, when necessary of terms with zero coefficients,

A(A) can be written so each of its elements has p + 1 terms. Then A (A) can be written

as a polynomial of degree p in A with w-square matrices Ai over f as coefficients, called

a matrix polynomial of degree p in A.

Example 2:

For the X-matrix A{\) of Example 1, we have

A(\) =X + 3

2X2 + X - 3

X» + X2 + 6X + 3

0X3 + o\2 + X + 3

0X3 + 2X2 + X - 3

X* + X2 + 6X + 3

X + 1

X2 + X - 1

2X2 + 2X + 1

X + 2

2X2-2

X3 + X2 + 5X + 2

0X3 + 0\2 + \ + 1

0X3 + X2 + X - 1

0X3 + 2X2 + 2X + 1

0X3 + 0X2 + X + 2

0X3 + 2X2 + OX - 2

X3 + X2 + 5X + 2

1 1 1

X3 + 2 12 X2 + 110 X +10 1 12 1 6 2 5

1 2

-1 -2

1 2

Consider now the n-square A-matrices or matrix polynomials

A(A) = ApA" + Ap-iA"-* + ••• + AiA + Ao (1)

and B(\) = B,A« + B,-iA'-» + ••• + BiA + Bo (2)

The two A-matrices (matrix polynomials) are said to be equal when p = q and Ai = Bi

for i = 0,1,2, ...,p.

The sum A (A) + B(A) is a A-matrix (matrix polynomial) obtained by adding correspond-

ing elements (terms) of the A-matrices (matrix polynomials). If p>q, its degree is p;

if p = q, its degree is at most p.

The product A(A)-J5(A) is a A-matrix (matrix polynomial) of degree at most p + q.

If either A(A) or B(A) is non-singular (i.e., either |A(A)| 7^ or 1B(A)| ^ 0) then both

A(A)'B(A) and B(k)-A{\) are of degree p + q. Since, in general, matrices do not com-

mute, we shall expect A(A)-B(A) ^ B(A)-A(A).


The equality in {1) is not disturbed if A is replaced throughout by any k Gf. Forexample,

A(k) = Apk" + Ap-ife"-' + ••• + Aik + Ao

When, however, A. is replaced by an w-square matrix C over f, we obtain two results

which are usually different

Aj^(C) = ApC" + Ap-iC"-' + ••• + AiC + Ao (3)

and A^^(C) = CMp + C'Âp-i + + CAi + Ao {3')

called, respectively, the right and left functional values of A (A.) when X = C.

Example 3:

When A{\) =

then An(C) =

and Ai,(C) =

See also Problem 4.

\2

X + 3

X-1X2 + 2

= 1

1X2 4

1

1X f

-

3

-1

2and C -

12"

2 3

1 o1

1

•1

2"

2 3

2

+"0 1'

1

•

"1 2^

2 34

"0 -1"

3 2= 7

12

10

17

1 2

2 3

2

•

1

14

1

2

2"

3•

1

1

4r

3

-1

2= 7

14

8

17

DIVISION ALGORITHMThe division algorithm for polynomials a{x),p(x) in x over a non-commutative ring %

with unity was given in Theorem II, Chapter 12, Page 126. It was assumed there that thedivisor ^{x) was monic. For a non-monic divisor, that is, a divisor ^{x) whose leadingcoefficient is 6„t^1, the theorem holds only if h'^ G%.

For the coefficient ring considered here, every non-singular matrix A has an inverseover f; thus the algorithm may be stated as

If A(X) and B{\) are matrix polynomials (1) and (2) and if B, is non-singular,then there exist unique matrix polynomials Qi(A),i?i(A); Q2(A), /22(A) G f[X.], whereRii\) and R2(x.) are either zero or of degree less than that of B{\), such that

A(A) = Q,(A) • B{x) + /2,(A) (.4)

and A(A) = B(A) • Q2(a) 4- i22{A) (^')

If in (4) Ri(\) = 0, B(\) is called a WflfAi divisor of A(A); if in (V) ^2(A) = 0, J5(A) is

called a left divisor of A (A).

Example 4:

Given

A(X) =

and

X3 4 3X2 + 3x 2X2 4 5x + 4

X2 4 X - 1 X2 + 1

1

B(X) = X41 1

X X42

X3 +

1

1 1

3 2

1 1

X 4-

X2 4

1 1

2

3 5

1X 4

4

-1 1

find Qi(X),Bi(X);Q2(X),B2(X) such that

(o) A(\) = Qi(X)'B(X) 4 Bi(X), (b) A(\) = BM'QîX) 4 RîX)

Here B, =1

1 1¥^ and J?r*

1

-1 1


(a) We compute2 1

1 1

Then

C(\) - C2B~^\B(\) =

D{\) - DiB-^B(\)

Qi(X) = (A3X2 + C2X + r»i)Br'

X2 +

2 2

1 -2

3 5

1

X +

X +

4

-1 1

L

4

-1 1C(x)

= D(\)

-4 2= i?l(X)

1 o'X2 + 1 1^

1X +

2

3 -2

X2 + X X + 2

3 X-2

(6) We compute

A(X) - BMB-ÂsiK^ =3 2

3 1X2 +

JS7(X) - B{\)B~Ê2\ =

F{\) - B(X)B-iFi =

Then QiW = Br^ {AgX^ + E2\ + F^) =

4

1 2

3 5

1

X +

X +

4

-1 1

4

-1 1= E(\)

= FM

-1 2

-3 5

1 o'

-1X2 +

i?2(X)

3 2

-1X +

4

1 -2

X2 + 3X 2X + 4

-X2 + 1 -X-2See Problem 5.

For the w-square matrix B = [bu] over f, define its characteristic matrix as

X— fell — 6l2 — &13 ... —bin

— bai A— 622 —623 . • . —bin

—631 —632 A. — 633 . • . —banXl-B =

— bnl —bni —bn X-bn

With A{k) as in {1) and B{\) ^ \I-B, (4) and (4') yield

A{k) = Qi{k) ' {\I - B) + Ri (5)

and A(\) = (Xl - B) Qiik) + R2 (5')

in which the remainders Ri and i?2 are free of A. It can be shown, moreover, that

Ri = Ar(B) and R2 = Al{B)

Example 5:

With A(\) = X2 X-1X + 3 X* + 2

and B =1 2

2 3we have X/ — B =

and

A(X) = X + 1 3

3 X + 3(X/-B) +

7 10

12 17= (X/-B)

X + 1 3

3 X + 3+

X-1 -2-2 X-3

7 8

14 17

From Example 3, the remainders are fij = AîB) — 7 10

12 17and R2 = Aj^{B) =

7 8

14 17


= (7,14,9)GF

THE CHARACTERISTIC ROOTS AND VECTORS OF A MATRIXWe return now to further study of a given linear transformation of F„ (f) into itself.

Consider, for example, the transformation of V - V3 {R) given by

~2 2 1I e, ^ (2,2,1)^=131 or ^2 ^ (1,3,1)

_1 2 2J .3-* (1,2,2)

(It is necessary to remind the reader that in our notation the images of the unit vectors£,, e^, £3 of the space are the row vectors of A and that the linear transformation is given by

V-*V: i-*Usince he may find elsewhere the image vectors written as the column vectors of A. Inthis case, the transformation is given by

For the same matrix A, the two transformations are generally different.)

The image of | = (1,2,3) G F is

r2 21"

V = (1,2,3) 1 3 1

[l 2 2_whose only connection with | is through the transformation A. On the other hand, theimage of ^, = (r, 2r, r)GV is 5|,, that is, the image of any vector of the subspace F' c V,spanned by (1,2, 1), is a vector of V\ Similarly, it is easily verified that the image of anyvector of the subspace WcV, spanned by (1,-1,0), is a vector of V^; and the image ofany vector of V^ c V, spanned by (1,0,-1), is a vector of W. Moreover, the image ofany vector {s + 1, -s, -t) of the subspace V* c V, spanned by (1, -1, 0) and (1, 0, -1), isa vector of the subspace generated by itself. We leave for the reader to show that thesame is not true for either the subspace V^ spanned by (1,2,1) and (1,-1 0) or of 7«spanned by (1,2,1) and (1,0,-1).

v

./, ,

We summarize: The linear transformation A of Va(R) carries any vector of the sub-space V\ spanned by (1, 2, 1), into a vector of F> and any vector of the subspace V\ spannedby (1, -1, 0) and (1, 0, -1), into a vector of the subspace generated by itself. We shall callany non-zero vector of V\ also of V*, a characteristic vector (invariant vector or eigenvector)of the transformation.

In general, let a linear transformation of F = VniT) relative to the basis e ,«

,

be given by the n-square matrix A = [a«] over f . Any given non-zero vector |'=

(xi, X2, Xs, .. .,x„) G V is a characteristic vector of A provided iA = A|, i.e.,

(anXl + O21X2 + • • • + anlXn, ai2Xi + 022X2 + • • • + an2Xn, .

.

for some A. e 5^,> \ > ,

n,

We shall now use (6) to solve the following problem: Given A, find all non-zerovectors | such that fA = A| with XGf. After equating corresponding components in(6), the resulting system of equations may be written as follows

(A - aii)xi — a2iX2 — aaiXa - • . . — oma;™ =-ai2Xi + (A - 022)3:2 - 032a;3 - ... - o„2a;n =-aiaXi - 023*2 + (A. - 033)3:3 - • . • - cu^„ = i'^)

(«)

—ainXl —02n3;2 Osn^s — • • • + (A — ann)Xn =

which by Theorem XVIII, Chapter 14, Page 181, has a non-trivial solution if and only ifthe determinant of the coefficient matrix


A, — ttii —021 —aai

— ffll2 A. — ft22 — tt32

—(Il3 —ft23 A. — a33

-ttnl

-ffln2

-Ctn3

—am -Ct2n -asn A, — On

= lAZ-AÎ =:

where A'^ is the transpose of A. Now A./ - A^ = (A/ - A)'' (check this); hence, by Theorem

XXII, Chapter 14, |A/-A''| = |A/-A|, the determinant of the characteristic matrix of A.

For any n-square matrix A over f, ]Xl - A'^| is called the characteristic determinant

of A and its expansion, a polynomial <j,{\) of degree n, is called the characteristic polynomial

of A. The n zeros X^,K^,X^, . . .,A„ of ^(A) are called the characteristic roots (latent roots

or eigenvalues) of A.

Now ^(A) G 7"' [A] and may or may not have all of its zeros in f\ (For example, the

characteristic polynomial of a 2-square matrix over R will have either both or neither of its

zeros in R; that of a 3-square matrix over R will have either one or three zeros in R. One

may then restrict attention solely to the subspaces of Fs (R) associated with the real zeros, if

any, or one may enlarge the space to Fa (C) and find the subspaces associated with all of the

zeros.) For any characteristic root A,, the matrix X.I - A"^ is singular so that the system of

linear equations (7) is linearly dependent and a characteristic vector ^ always exists. Also,

fc| is a characteristic vector associated with A; for every scalar k. Moreover, by Theorem

XVIII, Chapter 14, Page 181, when A./-A^ has rank r, then (7) has ti-r linearly inde-

pendent solutions which span a subspace of dimension n-r. Every non-zero vector of

this subspace is a characteristic vector of A associated with the characteristic root A^.

Example 6: Determine the characteristic roots and associated characteristic vectors of V3(B),

given A =1 2

2 2

1 3

The characteristic polynomial of A is

X-1 1

\\I-Ar\ = = X3 - 6X2 + iix-1 X - 2 -1

-2 -2 X - 3

the characteristic roots are Xi = 1, X2 = 2, X3 = 3; the system of linear equations

^^^ '*' (X - l)a;i + Xi =0

—aci + (X — 2)x2 — X3 =0-2xi - 2X2 + 0^-^)«3 =

«! + ^2 =X3 =0'

X2 = —1, «3 = as a solution. Thus, associated with the characteristic root Xj = 1

is the one-dimensional vector space spanned by Ji = (1,-1,0). Every vector

(k,—k,0), k ¥= 0, of this subspace is a characteristic vector of A.

Xi + X3 =

(a)

When X = Xx = 1, the system (a) reduces to having Xi = 1,

When X = X2 = 2, system (a) reduces to 0'having Xi = 2,

^^Xi + 2a;2

X2 = —1, X3 — —2 as a solution. Thus, associated with the characteristic root

X2 = 2 is the one-dimensional vector space spanned by ^2 = (2, —1, —2), and every

vector (2fe, — fc, —2fc), fc # 0, is a characteristic vector of A.

{a;. -I- a, = , .

_ , havmg Xi = 1,Zxi -I- 0:3 —

X2 — —1, x^ = —2 as a solution. Thus, associated with the characteristic root

X3 = 3 is the one-dimensional vector space spanned by I3 = (1,-1,-2), and every

vector {k,—k,—2k), fc # 0, is a characteristic vector of A.


Example 7: Determine the characteristic roots and associated characteristic vectors of VîR),

2 2 1

given A = 13 1

12 2

The characteristic polynomial is

\\i-An =\-2 -1 -1

-2 X - 3 -2

-1 -1 \ - 2

X3 - 7\2 + IIX - 5 ;

the characteristic roots are Xi = 5, Xj = 1, Xg = 1; and the system of linear equa-tions (7) is

(a)

{\-2)Zi - X2 - «3 =-2xi + (X-3)a;2 - 2xs =

. -«! - X2 + (\-2)X3 =

When X — Xi = 5, the system (o) reduces to*i + ^2 — 3a;3 =*2 ""a — "

havingaî - «3 =

Xi -1, X2 = 2, Xg = 1 as a solution. Thus, associated with Xj = 5 is the one-dimensional vector space spanned by Ji = (1, 2, 1). When x = Xj = 1, the system(o) reduces to xi + x^ + x^^^ having x, = 1, aij = 0, ajg = -1 and x^ = 1, aig = -1,Ks - as linearly independent solutions. Thus, associated with Xj = 1 is the two-dimensional vector space spanned by Ij = (1, 0, —1) and J3 = (1, —1, 0).

The matrix of Example 7 was considered at the beginning of this section. Examples 6and 7, also Problem 6, suggest that associated with each simple characteristic root is aone-dimensional vector space and associated with each characteristic root of multiplicitym > 1 is an w-dimensional vector space. The first is true but (see Problem 7) the secondIS not. We shall not investigate this matter here (the interested reader may consult anybook on matrices); we simply state

If A is a characteristic root of multiplicity w^l of A, then associated with Ais a vector space whose dimension is at least 1 and at most m.


Theorem III. If x^, i^; a^, i^ are distinct characteristic roots and associated characteristicvectors of an «-square matrix, then |, and 1^ are linearly independent.

We leave for the reader to prove

Theorem IV, The diagonal matrix Dteristic roots and e^, e^, .

.

: diag(Aj, Xj AJ has Aj, A^ A„ as charac-, £„ as respective associated characteristic vectors.

SIMILAR MATRICESTwo n-square matrices A and B over f are called similar over J provided there exists

a non-singular matrix P over f such that B - PAP'\In Problems 9 and 10, Page 213, we prove

Theorem V. Two similar matrices have the same characteristic roots,

and

Theorem VI. If ^. is a characteristic vector associated with the characteristic root A^ ofB = PAP \ then |. = ^.P is a characteristic vector associated with thesame characteristic root A. of A.


Let A, an n-square matrix over f having \, \, . . .,\^ as characteristic roots, be

similar to D = diag(Aj, X^, ..., A„) and let P be a non-singular matrix such that PAP'^ = D.

By Theorem IV, e. is a characteristic vector associated with the characteristic root \ of Dand, by Theorem VI, $^ — (^P is a characteristic vector associated with the same charac-

teristic root X.ofA. Now t^P is the ith row vector of P; hence, A has n linearly independent

characteristic vectors €^P which constitute a basis of V„{f).

Conversely, suppose that the set S of all characteristic vectors of an n-square matrix Aspans V„{f). Then we can select a subset {i^,$^, ...,!„} of S which is a basis of Vn(T)-

Since each ^^ is a characteristic vector.

where Aj, A^, are the characteristic roots of A. With P =

PA =

A,

A,

^1

we find

P or

,K) = -DPAP-^ = diag (Aj, A

and A is similar to D. We have proved

Theorem VII. An w-square matrix A over f, having Aj, Ag,

is similar to D = diag(Aj, A^,

. . , A„ as characteristic roots

, AJ if and only if the set S of all charac-

teristic vectors of A spans V„ {f).

Example 8:

2 2 1

For the matrix A = 1 3 1

12 2

"i i i

—1

Then P-i = i i -i 1

i -1 i_

of Example 7, take Ph 1 2 1

?2= 1 -1

?3 1 -1

and

PAP-i =1 2 1 2 2 1

1 -1 • 1 3 1 •

1 -1 12 2

i i f 5

i i -i = 10

i -1 4_1

= diag (\i, X2. >^3)

Not every n-square matrix is similar to a diagonal matrix. In Problem 7, Page 213,

for instance, the condition in Theorem VII is not met since there the set of all characteristic

vectors spans only a two-dimensional subspace of VsiR)-

REAL SYMMETRIC MATRICES

An n-square matrix A - [an] over R is called symmetric provided A^ = A, i.e.,

Oij = an for all i and j. The matrix A of Problem 6, Page 212, is symmetric; the matrices

of Examples 6 and 7 are not.

In Problem 11, Page 214, we prove

Theorem VIII. The characteristic roots of a real symmetric matrix are real.

In Problem 12, Page 214, we prove

Theorem IX. If Aj, 1^; A^, I2 are distinct characteristic roots and associated charac-

teristic vectors of an n-square real symmetric matrix, then $^ and 1^ are

mutually orthogonal.


Although the proof will not be given here, it can be shown that every real symmetricmatrix A is similar to a diagonal matrix whose diagonal elements are the characteristicroots of A. Then A has n real characteristic roots and n real, mutually orthogonalassociated characteristic vectors, say.

Aj,î; \, I2; • •; \» l„

If, now, we define>?* = Villi. (i = l,2,...,n),

A has n real characteristic roots and n real, mutually orthogonal associated characteristicunit vectors

-^1' Vp ^2' V2; • • • ; K' Vn

Finally, with S ='72

, we have SAS-> = diag(A,,A2, . . .,Aj.

The vectorsVi, V2' -'Vn constitute a basis of V„ (R). Such bases, consisting of mutu-

ally orthogonal unit vectors, are called normal orthogonal or orthonormal bases.

ORTHOGONAL MATRICESThe matrix S, defined in the preceding section, is called an orthogonal matrix. We

develop below a number of its unique properties.

1. Since the row vectors r,. of S are mutually orthogonal unit vectors, i.e., ,,..7/^ =[ 1, when i = j

i ^ , . . it follows readily thatI 0, when i # ;

S'S-^ =

and S'' = S-KVn

• [Vi.Vz' -'Vn] =

Vi •Vi Vi Vi . . r,^ v„

V2 Vi V2 V2 V2- v„

Vn Vi Vn V2 • Vn Vn

2. Since S'-S'" = S^-S = /, the column vectors of S are also mutually orthogonal unitvectors. Thus,

A real matrix H is orthogonal provided H'H^ = H^-H - I.

3. Consider the orthogonal transformation Y - XH of Vn (R), whose matrix H is orthog-onal, and denote by Yi, Y2 respectively the images of arbitrary .STi.Zz G F„(i?). Since

y,.y, = Y,Y^ = (x^H){x^Hy - x^(H'H^)xi = x^xi ^ x^-x^,

an orthogonal transformation preserves inner or dot products of vectors.

4. Since |yj| = {Yi-YiY'^ = {Xi-XiY'^ = \Xi\, an orthogonal transformation preserveslength of vectors.

Y ' Y X^ • IT5. Since cos S' =

| J' =] J = cos e, where ^ e,^' < ,r, we have e' = ^.

|-J:i|'|jr2| |Ai|-|A2|

In particular, if Zi • Z2 = then Yi'Y2 = 0, that is, under an orthogonal transfor-mation the image vectors of mutually orthogonal vectors are mutually orthogonal.


An orthogonal transformation Y = XH (also, the orthogonal matrix H) is called proper

or improper according as \H\ = 1 or |^| = — 1.

Example 9:

For the matrix A of Problem 6, we obtain

,j = |j/|jj| ^ (2/V6, -1/V6. -1/a/6). r,2 = (1/V3, l/VS, 1/V3), ,3 = (0. 1/^2, -1/V^)

Then, with

S = V2

13

2/\/6 -i/Ve -i/Ve

i/Vs i/\/3 l/^/3

1/V2 -1/V2

S-i = SI"

2/\/6 l/v'S

-i/Ve i/Vi i/a/2

-i/Ve i/V^ -xiy/l

and we have S'A-S'^ = diag (9, 3, -3).

The matrix S of Example 9 is improper, i.e., |S| = -1. It can be verified easily that

had the negative of any one of the vectors Vi. "^z- '/s^®" "^®^ *" forming S, the matrix

then would have been proper. Thus, for any real symmetric matrix A, a proper orthogonal

matrix S can always be found such that S-A-S-"^ is a diagonal matrix whose diagonal

elements are the characteristic roots of A.

CONICS AND QUADRIC SURFACES

One of the problems of analytic geometry of the plane and of ordinary space is the

reduction of the equations of conies and quadric surfaces to standard forms which make

apparent the nature of these curves and surfaces.

Relative to rectangular coordinate axes OX and OY, let the equation of a conic be

ax^ + hy^ + 2cxy + 2dx + 2ey + f = (8)

and, relative to rectangular coordinate axes OX,OY and OZ, let the equation of a quadric

surface be given as

ax^ + by^ + cz^ + 2dxy + 2exz + 2fyz + 2gx + 2hy + 2kz + m = (9)

It will be recalled that the necessary reductions are effected by a rotation of the axes to

remove all cross-product terms and a translation of the axes to remove, whenever possible,

terms of degree less than two. It will be our purpose here to outline a standard procedure

for handling both conies and quadric surfaces.

Consider the general conic of equation (8). Its terms of degree two, ax^ + by^ + 2cxy,

may be written in matrix notation as

ax^ + by^ + 2cxy (x,y)a c

c b= X-E-X^

where X - (x.y). Now E is real and symmetric; hence there exists a proper orthogonal

matrix 5 = such that S-E-S-^ = diag(A.j,A2) where \,r,^, X^.r,^ are the charac-

teristic roots and associated characteristic unit vectors of E. Thus there exists a proper

orthogonal transformation X = {x', y')S = X'S such that

X'S-E-S-'X'-" = X'Aj

KX' \x'^ + X^v'-"

in which the cross-product term has as coefficient.

CHAP. 15] MATRIX POLYNOMIALS

Let S = =Jlii '722

; then

(x.y) = X = X'S = {^'.V')îi Via"

= b

209

and we have\x = Vn^' + V^yV = Vii^' + ViiV

This transformation reduces {8) to

V^ + A,r'' + 2(d,„ + e,^X + 2(dv2i + «'722)2/' + / =which is then to be reduced to standard form by a translation.

An alternate procedure for obtaining (8') is as follows:

(i) Obtain the proper orthogonal matrix S.

(ii) Form the associate of (8)

(«')

ax^ + hy^ + 2cxy + 2dxu + 2eyu + fv? = (x, y, u)

where X = (x, y, u).

'S

a c d

c b e

d e f

(iii) Use the transformation X = X'

X S1

1

F

, where X' = {x',y',u'), to obtain

X'-^ =S''

1the associate of (*')•

Example 10: Identify the conic 5x2 _ 2-\/3xy + ly^ + 20\/3 x - 44j/ + 75 =

5 -yfzFor the matrix E =

-v/3 7

as the characteristic roots and associated characteristic unit vectors and form S

5 -Vs 10\^

of terms of degree two, we find 4, (^yfz, ^); 8, (-|, \yfl)

Then X = X'S

1reduces X • F • X^ = X -Vs 7 -22

10^/3 -22 76

-i iVa

^'' = to

"i\/3 i O"

-i iVs

1

5 —v/3 loVs

-a/3 7 -22

10\/3 -22 75

"iVs -i

i iV31

X'T

= 4x'2 + 8j/'2 + 8a;'M' - 32\/3 j/'m' + 75m'2 =

4 4

= («',!/', m') 8 -16^3

4 -16\/3 76

the associate of 4x'2 + 8j/'2 + 8«' - 32^3 2/' + 75 = 4(x' + l)2 + 8(3/'-2V3)2 - 25 = 0.

'x" = a;' + 1Under the translation

'x = i\/3x'-^j/'Using

2\/3

and

this becomes 4x"2 + 8j/"2 = 25. The conic is an ellipse.

1

it follows readily that, in terms of

X' = x'

_ and {

V = ix' + ^y/Zy' [j/' = j/" + 2-v/3

the original coordinate system, the new origin is at 0"(-3V%'2, 5/2) and the new axes 0"X" and0"Y" have respectively the directions of the characteristic unit vectors (^\fl, ^) and (-\, ^^/3 ).

See Problem 14.


Solved Problems

1. Reduce A (A.)

X 2A +

1

X + 2

\' + \ 2X2 + 2X \^ + 2\

X?-2\ 2X2-2X-1 A2 + X-3to normal form.

The greatest common divisor of the elements of A(\) is 1; set /i (X) - 1. Now use K2i(-2)

followed by K12 and then proceed to clear the first row and first column to obtain

A(X) ~1 X X + 2

X2 + X X2 + 2X

2X - 1 X2 - 2X X2 + X - 3

10X2 + X X« + 2X

-X2 - X -X2 - 2X - 1

10X2 + X X2 + 2X

2X - 1 -X2 - X -X2 - 2X - 1

B{\)

The greatest common divisor of the elements of the submatrixX2 + X

-X2-X

X2 + 2X

-X2 - 2X - 1

is 1; set

/2 (X) = 1. On B(X) use ^23(1) »«<! ^23(-l) and then proceed to clear the second row and second

column to obtain

A(X)

1

1

X + 1

-1

-X2 - 2X - 1

101

x + 1 -X* - X

1

1

-X2-X

101

X* + X

= iV(X)

the final step being necessary in order that /a (X) = X^ + X be monic.

2. Reduce (a) A{\)\

A + 1and (b) B{\) =

A»

\^-\A"

to normal form.

(a) The greatest common divisor of the elements of A(X) is 1. We obtain

A(X) = X

X + 1

-X

1

-1

X x +x +

-X2-X

1

1

,^

r

-1 x + 1

-X - 1 x + 1

1

-X2 - X

1

X2 + X

(6) The greatest common divisor of B(\) is X. We obtain

J5(X) =X3

XZ-XX2

X3 X2-XX2-X

X2

X2 X2-X-X3 + X2 X2-X

X2

= N{\)

X X2-X-X3 + X X2-X

X2

X X2-X-X3

x2

X

X*-X3

X2

X

X2

X*-X3

= N{\)


3. Reduce A{\) -k-2 -1 -1

-2 A. - 3 -2

-1 -1 A-2to normal form.

The greatest common divisor of the elements of A{\) is 1. We use K13 followed by Xi(-l) andthen proceed to clear the first row and column to obtain

A(\) ~1 -1 X - 2

2 X - 3 -22 - X -1 -1

10X - 1 2 - 2X

1-X X2-4X + 3

The greatest common divisor of the elements of B{\) is X - 1; then

A(X)

10X - 1 2 - 2X

1-X X2-4X + 3

10X-1

X2 - 6X + 5

1

B(X)

m\)

4. Write A{X) =X + 2 X + 1 X + S

X X -3A.2 + X

A2 + 2A A« + A 3A2 + 5A

as a polynomial in A and compute A(-2),

Ar{C) and Al(C) when C =10 2

-1 -1 -4-1 -2

We obtain A{\) =1 1 1 2 13

-3 X2 + 1 1 1 X +1 1 3 2 15

and A (-2)

Since C =-1 -2

4 1 10

10 2

0-31 1 3

, we have

p -1

111 2 13- 2 111

2 15+ —

0-1 1

-2 -2 -14

2 2

Ar(0 =

Al(0 =

-1 -2

4 1 10

10 2

1 1 ll

+ 1112 15

1 2 2 131 -1 -4 + =1 -2

-1 -2

4 1 10

10 2

-3 +1 1 3

10 2

-1 -1 -4

-1 -2

fl 1 1 2 131 1 1 + =2 15

i— —J L J

1 -1-4 -1 -10

2 4

5 2 8

4 5

-3 -1 -5

5. Given A{X) = and B{X) =A* + A« + 3A2 + A A* + A3 + 2A2 + A + 1

A^ - 2A + 1 2A3 - 3A2 - 2

find Qi{X),Ri{X);Q2{X),R2iX) such that

(a) A{X) = Qi{X)'B{X) + R,{x) and (&) A{x) = B{X) - Q2{X) + R2{X)

We have

A^ + 1 A^ - A'

A^ + A 2A2+1

A(X) = 1 1x* +

1 1

1 2X3 + 3 2

-3X2 +

1 1

-2 X +


B(\) =

B2

1 1

1 2\2 +

-1

1\ +

1 1

1 2and Bo

1

1

2 -1

-1 1

(o) A(X) - AJB-^-\'''B(\) =

CM - CsB-^-\'B{\)

1 2

1 2

1 2

-1 -3

\3 + 2 2

-3X2 +

1 1

-2X +

1

1 -2C(x)

X2 +1

-2 -1X +

1

1 -2= D{\)

D{\) - DJB-^-B(\) == = Bi(X)

and

Qi(X) = (A4X2 + C3X + D2)B-» =

Here, B(X) is a right divisor of A(X).

(6) A(\) - B{\)'B-ÂiKi =

E{\) - B(\)-B-Ê3\ =

X2 X + 1

1 X-2

x« +1

X2 +1 1'

X +"0 1"

-1 1 -2 -2 1 —2- L J L L -'

0'

X2 +1

X +1

= FM-2 -1 1 -2

_ _ J

EM

-1

-1 -2F(X) - B{\)'B-'Fi =

and

Q2(X) = B-^{Ai\^ + Es\ + F2) -

X +-1

1

-X-1-x + 1 -2\

= fi2(X)

2X2 + X 2X2 + 2

-X2 - X -X2 - 2

6. Find the characteristic roots and associated characteristic vectors of A -

over R.

7 -2 -2-2 14-2 4 1

X-7 2 2

2 X-1 -4

2 -4 X-1X3 _ 9x2 - 9x + 81;

r(X-7)a;, + 2*2 + 2a;3 =

2iKi + (X-l)a!2 - 4^3 =

I 2a;i - 49;2 + (X-l)«3 =

The characteristic polynomial of A is |X/ — A''

the characteristic roots are Xi = 9, X2 = 3. X3 = -3; and the system of linear equations (7) is

(o)

When X = Xi = 9. (a) reduces to {*) + J = \ ^^^'"^ x^ = 2, a,2 = -L ^3= -1 as a solution.

Thus, associated with Xi = 9 is the one-dimensional vector space spanned by ?i = (2, -1, -1).

When X = X2 = 3, (a) reduces to j*' ~ !" I „ having a^j = 1, a;2 = L ^=3 = 1 as a solution.

Thus, associated with X2 = 3 is the one-dimensional vector space spanned by ia - (L 1. !)•

When X = X3 = -3, (a) reduces to{ ^^

+ ^^ I Jhaving o^^ = 0, X2 = 1, X3 = -1 as a solution.

Thus, associated with X3 = -3 is the one-dimensional vector space spanned by J3 = (0, 1, -1).

CHAP. 15] MATRIX POLYNOMIALS

7. Find the characteristic roots and associated characteristic vectors of Aover R.

213

-2 -2-112-1 -1 2

The characteristic polynomial of A is |\/ - A'^\ = \»-3\2 + 4; the

XI 1

2 X-1 1

2 -2 \ - 2

characteristic roots are Xj = -1, x^ = 2, X3 = 2; and the system of linear equations (7) is

{XiCi + aca + Xg =02a;i + (X- 1)0:2 + «3 =02x, - 2*2 + (X-2)a;3 =

When X = X, = -1, the system (a) reduces to J «i " «2 =^^^.^^ ^^ ^^^ ^^^^^ ^^^^

[^

«3 —as â solution. Thus, associated with X, = -1 is the one-dimensional vector space spanned byii — (1> li 0).

When X = Xij = 2, the system (a) reduces to j3«i + *3 = ^

[^», - Wjj = ^ . 3

as â solution. Thus, associated with X2 = 2 is the one-dimensional vector space spanned by

• J^°*^*^** ^^^^ * vector space of dimension one is associated with the double root Xj = 2 whereasm Example 7 a vector space of dimension two was associated with the double root.

8. Prove: If A,,|j; \^,i^ are distinct characteristic roots and associated characteristicvectors of A, then |, and i^ are linearly independent.

Suppose, on the contrary, that £1 and Jj are linearly dependent; then there exist scalars a, and a,not both 0, such that

' ^(i) o,Ji + 02^2 =

Multiplying (i) on the right by A and using {jA = Xj{,, we have

(ii) OjJiA + a2l2A = ajXiii + 02^2^2 =

Now (i) and (ii) hold if and only if

are linearly independent.

1 1

Xj X20. But then Xj = Xg, a contradiction; hence, fj and {2

9. Prove: Two similar matrices have the same characteristic roots.

Let A and B = PAP-^ be the similar matrices; then

XI -B = \I-PAP-i = Px/P-i-PAP-i = P(X/-A)P-i

and |X/-iB| = |P(X/-A)P-i| = |P|.|X/-A|-|P-1| = |x/-A|

Now A and B, having the same characteristic polynomial, must have the same characteristic roots.

10. Prove: If C^ is a characteristic vector associated with the characteristic root \ ofB = PAP-\ then $^ = C^ is a characteristic vector associated with the same charac-teristic root A, of A.

By hypothesis, f,B = Xjf, and BP = (PAP-i)P = PA. Then IjA = SiPA = ifBP = XifjP = Xj{and {j is a characteristic vector associated with the characteristic root Xj of A.

'


11. Prove: The characteristic roots of a real symmetric «-square matrix are real.

Let A be any real symmetric matrix and suppose fc + tfe is a complex characteristic root. Now

(fc + %k)l — A is singular as also is

B = \{h + ik)l-A\'\{h-ik)l-A\ = (h^ + k^)I - 2hA + A^ = (hI~A)^ + kn

Since B is real and singular, there exists a real non-zero vector i such that l,B -- and, hence,

iBjT = ^{(hl - A)^ + kHW = {i(hl - A)}{(hl - A)'r e) + kH(r

= v.v + kH'l. =

where -n= i(hl - A). Now ij . i; - while, since { is real and non-zero, $•? > 0. Hence, k-0

and A has only real characteristic roots.

12. Prove: If X,^,; \>^2 ^^^ distinct characteristic roots and associated characteristic

vectors of an n-square real symmetric matrix A, then |j and 1^ are mutually orthogonal.

By hypothesis, JxA - Xi4i and i2^ - \2l2- Then

iÂil = \iiiil and

'^ - ^ - '^ and

JaA j[ = Xaiz S[

and, taking transposes, iz^ ii M2«[ iiA ii = X2?i (.2

Now £iA ij = XiJijJ XaiiiJ and (Xj - \2)iiil = 0. Since X, - Xa ^ 0, it follows that fil^T _

|i'{2 = and £i and £2 are mutually orthogonal.

13. (a) Show that « = (2,1,3) and p = (1,1,-1) are mutually orthogonal.

(6) Find a vector y which is orthogonal to each,

(c) Use a,p, y to form an orthogonal matrix S such that |S| = 1.

(o) a • j8 = 0; a and p are mutually orthogonal.

(6) y = axp = (-4,5,1).

(c) Take pi= a/\a\ = (2/a/14, l/Vli, 3/-\/l4), Pa = /8/|^| = (l/\/3. 1/^3, -l/'/S

)and P3 = 7/|y| =

(-4/\/42, 5/\/42, l/\/42 ). Then

= 1 and S =Pi

P2

P3

2/V^ 1/-/14 3/VI4

l/\/3 i/Vs -i/Vst

-4/^42 5/V42 l/\/42

14. Identify the quadric surface

3a;2 - 22/2 - 2^ - 4a;3/ - 8xz - 12yz - 8x - I63/ - 342 - 31

For the matrix E3 -2 -4

-2 -2 -6

-4 -6 -1of terms of degree two, take

3,(2/3,-2/3,1/3); 6,(2/3,1/3,-2/3); -9,(1/3,2/3,2/3)

characteristic roots and associated characteristic unit vectors. Then, with S

=

X = (x,y,x,u) = {x',y',z',u')S

1

2/3 -2/3 1/3

with S = 2/3 1/3 -2/3

1/3 2/3 2/3

s 0'

1


reduces X • F 'X-r = X

3 -2 -4 -4-2 -2 -6 -8-4 -6 -1 -17-4 -8 -17 -31

XT

to

2/3 -2/3 1/3 0^3 -2 -4 -4~

2/3 1/3 -2/3 -2 -2 -6 -81/3 2/3 2/3 -4 -6 -1 -17

'

1_ _-4 -8 -17 -31_

3 0-3 A= (x',y',z'.«')

6 6

0-9 -18y

z

-3 6 -18 -31 \ u

2/3 2/3 1/3

-2/3 1/3 2/3

1/3 -2/3 2/3X'T

= 3a;'2 + 6j/'2 - 9z'2 - 6x'u' + 12y'u' - SSz'u' - 31m'2 =

the associate of

3a;'2 + 6j/'2 - Qz'^ - 6x' + 12y' - 36z' - 31 = 3{x' - 1)2 + 6(j/' + 1)2 - 9(z' + 2)2-4 =

Under the translation

hyperboloid of one sheet.

Ix" = x'-l\y" = y' + 1 , this becomes 3x"2 + 6j/"2 - 9z"2 ~ 4; the surface is a

z" = z' + 2

Using (x,y,z) = (x',y',z')S and the equations of the translation, it follows readily that, interms of the original coordinate system, the new origin has coordinates (—2/3, —7/3, —1/3) and thenew axes have the directions of the characteristic unit vectors (2/3,-2/3,1/3), (2/3,1/3,-2/3)(1/3,2/3,2/3).


15. Given A(X) =

(o) A{\) + B{\) =

(b) A(\) - B{\) =

X2 + X X + 1

X2 + 2 Xand B{\)

2X2 + X x2 + 2X + 1

X2 + X + 1 2X

X -X2 + 1

X2 - X + 3

X2 X2 + X

X-1 X

(c) A(X)-B(X) =

(d) B(\)'A{\) =

find

X* + X3 + X2 - 1 X* + 2X3 + 2x2 + \

\* + 3X2 - X X* + X» + 3X2 + 2x

2X* + 2X3 + 2X2 + 2x 2X3 + 2X2

'

2X3 + X 2x2-1

16. In each of the following find Qi(X), fl,(X); QziX), R2M, where fli(x) and RzM are either or of degreeless than that of B(\), such that A(X) = Qi(\) • B{\) + fti(x) and A(X) = B(\) • QzM + RziX).

(a) A(X) = X' - 2X2 + 2X - 2 X* + X -

1

X* + X3 + X - 2 2X2 + X - 1

B(X) = X2 + 1 X

1 X2 + X


(b) AM =

(e) A(\) =

(d) A(\) =

2X2 + 2\ 2\2

\2 + \ + 2 X2 + 2\ - 1

B(X) =

X3 - 2X2 + X - 3 4X2 + 2X + 3

-2X - 2 X3 + 4X2 + 6X + 3

X X

1 X

B(\) = X-2 3

-1 X + 3

X» - X2 + X + 4 2X2 + X X2 + 4X - 2

2X X3 + X2 + 2X - 1 4X2 - 2X + 1

3X2 - 4X - 1 x2 + X X8 - 2X2 + 2X + 2

BM =X-1 1

-1 X + 1 3

2 X-2

Ans. (a) Qi(X)^-3 ^^-M; «,(X)=

2X + 1 4X-1

X2+X-1 -X + 2 X-3 -1

Q2M =

(6) Qi(M =

2(X) =

(e) Qi(X)

Q2M =

(d) Q,(X)

Q2(X)

-2 X2 -

1

X2 1

2X + 2 -2

X + 1 1

X+2 X-1X X + 1

; «2(M = 2X

X

2

1 -1

X2 + 2 X-1X + 1 X2 + X

X2-2 x + 1

X-5 X2 + X + 4

: i?i(x) =

; «2(x) =

Ri(\) =

B2M =

X2 X x + 3

x + 1 X2 + 1 X

x-2 X-1 X2-l

X2 x + 2 X + 4

X-2 X2 X-2X-2 X + 1 X2

«i(X) =

i22(X) =

8 -7

11 -8

-2 4

= 2 -3 -2

-2 3 3

-

6 2

—

8 -2 7

-5 --2 --6

17. Reduce each of the following to its normal form:

X 2X 2X -

1

(a) X2 + 2X 2X2 + 3X 2X2 + X-1

X2 - 2X 3X2 _ 4x 4X2 - 5X + 2

X2 + 1 X3 + X X3 - X2

(6) )X-1 X2 + 1 -2X

X2 X3 X3 - X2 + 1

-X x + 1 x + 2

(c) -X2 X2 + X-1 X2 + 2X-1

X2 + X + 1 -X2 - 2X - 1 -X2 - 3X - 2

(d)

(e)

(/)

X2

X + 1

x + 1

x-1 3 -2

-2 X + 1

3 1 X + 2

X-2 2 3

-3 X + 3 4

2 X + 1

CHAP. 15]

Ana. (a)

(6)

MATRIX POLYNOMIALS 217

1

X

X2

1

X + 1

X3 + 1

(c)

(d)

1010

1

10x + 1

X3 + X2

(e)

(/)

101

X3 + 2X2 + IIX + 20

101

X3 + 2X2 - 5X - 2

18. Find the characteristic roots and associated characteristic vectors of each of the following matricesA over R.

{a)

(b)

1 -2-5 4

2 -1

-8 4

(c)

(d)

3 1

-1 1ie)

2 0-12-2

1-12(If)

1-1012 1

-2 1 -1

1 -2 3 2 4

-2 4 (/) 2 2*- 4 2 3

Aws. (o) 6,(k,-k); -l,(5A;,2fc)

(6) 0,(4*, A:); 6,(2fe,-fc)

(c) 2.(k,k)

(d) 0,(2fc,fc); 5,(ft,-2fc)

where k¥'(i and J 9^0.

(e) \,(k,-k.-k); 2, (2fc, -fc, 0); 3,(&,-Jfc,fc)

(/) -l,(fc,2Z,-fc-Z); Z,{2k,k,2k)

(g) 1, (3fc, 2fc, fc); 2,(A;,3fc,A;); -l,(fc,0,fc)

19. For an Jt-square matrix A, show

(a) the constant term of its characteristic polynomial is (-1)«|A|,

(6) the product of its characteristic roots is |A|,

(c) one or more of its characteristic roots is if and only if \A\ = 0.

20. Prove: The characteristic polynomial of an ^-square matrix A is the product of the invariant factorsof X/ — A.

Hint. From P(X) • (X/- A) .S(x) = diag(/i(X)./2(X) /„(X)) obtain

with |P(X)| . ,S(x)| = 1.l^<^)H«(x)|..(x)=A(x)./,(x).../„(x)

^'']!'?^„^f*^*'

°* *^® following real symmetric matrices A, find a proper orthogonal matrix S such thatSAS~^ IS diagonal.

3 -2 -2

(e) -2 8 -2 (g)

-2 -2 3

(o)2 2

2 -1 (c)1 -6

-6 -4

3 -1 -1

-12-10 2

(6)4 -3

-3 -4

A?i«. (a)

(«»)

id)

2/\/5 l/VI

-I/VS 2/\/5

3/\/i0 -1/VlO

1/VlO 3/VlO

1 -2-2 4

(/)

2-5-5 -1 3

3-6

4-2 4

(h) -2 1 -2

4-2 4

(c)

(d)

'3/\/i3 -2/\/l3

2/Vl3 3/\/l3

i/\^ -2/V5

2/\/5 l/\/5

(e)

(/)

l/3\/2 -4/3-v/2 l/3^/2

l/\/2 -I/V2

2/3 1/3 2/3

5/^/42 -4/\/42 -l/Vil

l/\/i4 2/\/l4 -3/a/14

1/V^ 1/v^ i/v/3


io)

'2/V^ -i/Ve -i/Ve

l/\/2 -l/\/2

1/V3 i/Vs 1/V3_

W2/3 -1/3 2/3

l/\/2 -l/^f2

\IZyf2 4/3V2 1/3V2_

22. Identify the following conies:

(a) 4*2 + 24xj/ + llj/2 + 16x + 42j/ + 15 =

(6) 9a!2 - 12x1/ + 4j/2 + 8VT3 x + 12\/l3 y + 52 =

(c) 3x2 4. 2xj/ + 3j/2 + 4V2x + 12V2j/ -4 =

Ans. (a) Hyperbola, (6) Parabola, (c) Ellipse

23. Identify the following quadric surfaces:

(a) 3x2 + 8j,2 + 3z2 _ 4a;j, _ 4a;g - 4y« - 4x - 2j/ - 4« + 12 =

(6) 2x2 _ 3,2 _ 6«2 - 10x1/ + 6yz + 60x - 74i/ + 42« + 107 =

(c) 4«2 + 1/2 + «2 - 4x1/ - 4xz + 2i/« - 61/ + 6« + 2 =

(d) 2xy + 2x« + 2i/« + 1 =

Am8. (o) Elliptic paraboloid, (6) Hyperboloid of two sheets, (c) Parabolic cylinder

24. Let A have Xi, Xj, . . ., X„ as characteristic roots and let S be such that

S'A-S-^ = diag(Xi,X2. •••.^n) = ^

Show that SAT^S-^ = D when S = S'K Thus any matrix A similar to a diagonal matrix is similar

to its transpose A'''.

25. Prove: If Q is orthonormal, then Q'' = Q~*.

26. Prove: Every real 2-square matrix A for which |A| < is similar to a diagonal matrix.

27. Prove by direct substitution that A — a 6

c dis a zero of its characteristic polynomial.

28. Under what conditions will the real matrix A =

(o) equal characteristic roots,

(5) the characteristic roots ±1.

a b

c dhave

chapter 16

Linear Algebras

LINEAR ALGEBRAA set ^ having binary operations of addition and multiplication, together with scalar

multiplication by elements of a field f, is called a linear algebra over f provided

(i) Under addition and scalar multiplication, „C is a vector space j(l{f) over f.(ii) Multiplication is associative.

(iii) Multiplication is both left and right distributive relative to addition.

(iv) Ji has a multiplicative identity (unity) element.

(v) (ka)p = a{kp) = k{a- p) for all a,p&jci and k&f.Example 1

:(a) The field C of complex numbers is a linear algebra of dimension (order) 2 over

the field R of real numbers since (see Chapter 13) C(R) is a vector space ofdimension 2 and satisfies the postulates (ii)-(v).

(6) In general, if ^ is any field having y as a subfield, then ^{J) is a lin.ar algebraover f.

Example 2: Clearly, the algebra of all linear transformations of the vector space V„(7) is alinear algebra of order n^. Hence the isomorphic algebra M„(f) of all ?v-squarematrices over f is also a linear algebra.

AN ISOMORPHISMThe linear algebra of Example 2 plays a role here similar to that of the symmetric

group S„ in group theory. In Chapter 9 it viras shown that every abstract group oforder n is isomorphic to a subgroup of S„. We shall now show that every linear algebraof order n over f is isomorphic to some subalgebra of Mn{f).

Let .^ be a linear algebra of order n over f having {xuXi,Xz, . . .,Xn} as basis. Witheach a&Jl, associate the mapping

Ta-. xTa = X'a, X S ^By (iii), xTa + yT„ = x'a + y-a = {x + y)a = (x + y)Toc

and by (v), {kx) r„ = {kx)a = &(«•«) = k{x r„)

for any x,y & Ji and k&f. Hence Ta is a linear transformation of the vector spaceJi{f). Moreover, the linear transformations r„ and T» associated with the distinct elementsa and ^ of ^ are distinct. For, if a¥'P, then wa # u- p, where u is the unity of jr,

implies r„ ¥- Tp.

Now, by (iii) and (v),

xTa + xTe = X'a + X-p = xia + i3)= xTa^p

{xTa)T» = {x-a)p = X(a-p) = xTa-p

and {kx) Ta = (kx)a = k{X'a) = X-ka = X Tka

219

220 LINEAR ALGEBRAS [CHAP. 16

Thus the mapping a ^ r« is an isomorphism of =C onto a subalgebra of the algebra of all

linear transformations of the vector space JiiT). Since this, in turn, is isomorphic to a

subalgebra of Mn{T), we have proved

Theorem I. Every linear algebra of order n over f is- isomorphic to a subalgebra of M„(f).

Example 3: Consider the linear algebra Q[\/2] of order 3 with basis {1, \/2, VJ }. For any

element a = aj 1 + Oj v'Z + ag VJ of Qly/2], we have

I'a - Oj 1 + O2V2 + OsVi

V^ • a = 2as 1 + OiV2 + a^y/Z

y/Z-a = 2a2 1 + 203V2 + o,\/4

Then the mapping

ai 1 + 02\/2 + a3\/4

«! O2 "^3

2a2 2a3 a^

is an isomorphism of the linear algebra ^[^2"] onto the algebra of all matrices of

r 8 t

M„(Q) of the form 2t r 8

28 2t r See also Problem 1.

Solved Problems

1. Show that ^ = {tti • 1 + aaa + as/S : Oi^R} with multiplication defined such that

1 is the unity, = 0'H-0-a + 0-/8 is the additive identity, and

(«)

• a |8

a a P

P

and (b)

• a p

a a

/8

are linear algebras over R.

We may simply verify in each case that the postulates (i)-(v) are satisfied. Instead, we prefer to

show that in each case 4. is isomorphic to a certain subalgebra of MîR).

{a) For any a = a^'l + a^ + a^p, we have

1 • a — ail + Oja + 03/8

a' a — («! + 02)0 + tta/S

P'a — aiP _

Hence ^ is isomorphic to the algebra of all matrices MîB) of the form

and is a linear algebra over R.

Ol di 0,3

aj + 02 tts

ai

(6) For any a — Oi'l + a^ + 03 we have

1 • a — ail + aja + 03)8

a • a = (ai + a2)a

P'a = ai/8

Hence .C is isomorphic to the algebra of all matrices MgiR) of the form

tti a2 a3

ai + a2

ai

CHAP. 16] LINEAR ALGEBRAS 221


2. Verify that each of the following, with addition and multiplication defined as on R, is a linear algebraover Q.

(a) Q[V3] = {al + 6^3 : a,b 6 Q}

(6) ^ = {al + bVs + cVB" + dVl5: a,b,c,d G Q}

3. Show that the linear algebra ^ = QlVt], where t G N is not a perfect square, is isomorphic to the

a balgebra of all matrices of M2(Q) of the form

tb a

4. Show that the linear algebra C over R is isomorphic to the algebra of all matrices of MziR) of the

a bform

—6 a

5. Show that each of the following is a linear algebra over R. Obtain the set of matrices isomorphicto each,

(a) Jl = {al + ba + ca^ : a,b,c G R}, where G = {a, o?, c? = 1} is the cyclic group of order 3.

(b) J^ = {ajl + a2X + Ogj/ : Oj e R}, with multiplication • defined so that 1 is the unity, -

X y

0'l + 0'a; + 0'j/ is the additive identity, and x

y

1 y

y

(c) Si — {ai + oji + ttgi + a4k : Oj G fi} with multiplication table

• i J k

i -1 k -}

i -k -1 i

k i —i -1

a b e

ns. (a) cabb e a

(b)

Oj O2 O3

O2 Oj 03

(oi + 02)

(c)

Oj O2 *3 tt4

-O3 O4 Oj —02

-O4 —03 02 tti

chapter 17

Boolean Algebras

BOOLEAN ALGEBRAA set <B, on which binary operations U and D are defined, is called a Boolean algebra

provided the following postulates hold:

(i) U and n are commutative,

(ii) 'S contains an identity element with respect to U and an identity element 1

with respect to n.

(iii) Each operation is distributive with respect to the other, i.e., for all a, 6, c G "B

a U (6 n c) = (a U 6) n (a U c)

and a n (6 U c) = (a n 6) U (a n c)

(iv) For each aG'B there exists an a' G'=B such that

a U a' = 1 and ana' =

The more familiar symbols + and • are frequently used here instead of U and n.We

use the latter since if the empty set be now denoted by and the universal set U be now

denoted by 1, it is clear that the identities 1.9-1.9', 1.4-1.4', 1.10-1.10', 1.7-1.7', proved valid

in Chapter I for the algebra of all subsets of a given set, are precisely the postulates

(i)-(iv) for a Boolean algebra. Our first task then will be to prove, without recourse to

subsets of a given set, that the identities 1.1, 1.2-1.2', 1.5-1.5', 1.6-1.6', 1.8-1.8', 1.11-1.11'

of Chapter 1 are also valid for any Boolean algebra, that is, that these identities are

consequences of the postulates (i)-(iv). It will be noted that there is complete symmetry

in the postulates with respect to the operations U and n and also in the identities of (iv).

Hence there follows for any Boolean algebra the

Principle of Duality. Any theorem deducible from the postulates (i)-(iv) of a

Boolean algebra remains valid when the operation symbols U and n and the identity

elements and 1 are interchanged throughout.

As a consequence of the duality principle it is necessary to prove only one of each pair of

dual statements.

Example 1 : Prove: For every aG'B,

aU a = a and a n a = a (1)

(See 1.6-1.6', Chapter 1, Page 5.)

Using in turn (ii), (iv), (iii), (iv), (ii):

a U o = (o u o) n 1 = (o U a) n (a U a') = a U (a n a') - o U = a

Example 2: Prove: For every o e "B,

a U 1 = 1 and a n = (2)


Using in turn (ii), (iv), (iii), (ii), (iv):

a n = U (a n 0) = (a n a') U (o n 0) = o n (a' U 0) = a n a' =

222

CHAP. 17] BOOLEAN ALGEBRAS 223

Example 3: Prove: For every a,b & 'B,

aU (a n b) = a and a n {a U b) = a (s)

Using in turn (ii), (iii), (2), (ii):

au (an b) = (a n 1) u (o n 6) = o n (1 u 6) = o n 1 = a


BOOLEAN FUNCTIONSLet « = {a, 6, c, . .

. } be a Boolean algebra. By a constant we shall mean any symbol,as and 1, which represents a specified element of "B; by a variable we shall mean a symbolwhich represents an arbitrary element of ^. If in the expression x' U {y n z) we replaceU by + and n by • to obtain x' + yz, it seems natural to call x' and y n z monomialsand the entire expression x' U {y n z) a polynomial.

Any expression as xUx'.an b', [a n (b u c')] u {a' n b' n c) consisting of combina-tions by U and n of a finite number of elements of a Boolean algebra « will be called aBoolean function. The number of variables in any function is the number of distinctletters appearing without regard to whether the letter is primed or unprimed. Thus x u x'is a function of one variable x while a n b' is a function of two variables a and b.

'

In ordinary algebra every integral function of several variables can always be expressedas a polynomial (including 0) but cannot always be expressed as a product of linear factorsIn Boolean algebra, on the contrary. Boolean functions can generally be expressed inpolynomial form (including and 1), i.e. a union of distinct intersections, and in factoredform, i.e. an intersection of distinct unions.

Example 4:

Simplify:

(a) (xny)u [{X u V') n y]'. (6) [(x u y') n (x n y' n z)']'. (c) {[(*' n y')' u z] n (x u z)}'

(a) {xny)u [(X U y') n y]' ^ (x n y) U [{x U y')' u y'] = {x n y) U [(x' n y) U y']

= {xny)U [{x' U y') n{yu y')] = (x n j/) u [(x' U y') n 1]

= (x n J/) u (x' u y') = {xny)u{xnyy = i

(b) [(x U y') n(xny' n «)']' = (x U y')' U [(x ny' n z)']'

= (x' ny)u {x ny' n z), a union of intersections.

[(x u y') nixny'n «)']' = (x' n y) u (x n y' n z)

= (x' u x) n (x' U y') n (x' V z)n (xuy)n(yu y') n(yuz)= in (x' u y') n (x' u z) n (x u {/) n 1 n (1/ u 2)

= {xuy)n(yuz)n (x' u z) n (x' u y'), an intersection of unions.

(c) {[(«' n y'Y \jz]n(xu z)}' = [(x' n y')' u z]' u (x u «)'

= (X' ny'n z') u (x' n z') = x' n z' (by Example 3)

See also Problem 5.

Since (see Problem 15, Page 234) there exists a Boolean algebra having only theelements and 1, any identity may be verified by assigning in all possible ways the valuesand 1 to the variables.

Example 5: To verify the proposed identity (see Example 4(o))

(x n 1/) U [(x U y') n y]' - 1

we form the following Table 17-1.

224 BOOLEAN ALGEBRAS [CHAP. 17

X y a = X n V xUy' 6 == {xUy') ny au b'

1 1 1 1 1 1

1 1 1

11

1 1

Table 17-1

NORMAL FORMSThe Boolean function in three variables of Example 4(b) when expressed as a union

of intersections (x' n y) U {x n y' n z) contains one term in which only two of the vari-

ables are present. In the next section we shall see that there is good reason at times to

replace this expression by a less simple one in which each term present involves all of the

variables. Since the variable z is missing in the first term of the above expression, we

obtain the required form, called the canonical form or the disjunctive normal form of the

given function, as follows

(«' ny)u(xny' nz) = {x' ny nl)u {x ny n z)

= [(x' ny)r\{zu z')] u{xny' nz)

= («' n 2/ n 2) u {X' ny nz')u (xn y' n z)

See also Problem 6.

It is easy to show that the canonical form of a Boolean function in three variables can

contain at most 2" distinct terms. For, if x,y,z are the variables, a term is obtained by

selecting x or x', y or y', z or z' and forming their intersection. In general, the canonical

form of a Boolean function in n variables can contain at most 2" distinct terms. The

canonical form containing all of these 2" terms is called the complete canonical form or

complete disjunctive normal form in n variables.

The complete canonical form in n variables is identically 1. This is shown for the

case n = 3 in Problem 7, Page 231, while the general case can be proved by mduction. It

follows immediately that the complement F' of a Boolean function F expressed m canonical

form is the union of all terms of the complete canonical form which do not appear in the

canonical form of F. For example, if F = {x n y) U (x' n y) V {x' n y'), then F' = (x n y').


Theorem I. If. in the complete canonical form in n variables, each variable is assigned

arbitrarily the value or 1, then just one term will have the value 1 while

all other terms will have the value 0.

and

Theorem II. Two Boolean functions are equal if and only if their respective canonical

forms are identical, i.e., consist of the same terms.

The Boolean function in three variables of Example 4(6), when expressed as an inter-

section of unions in which each union contains all of the variables, is

{xuy)n{yLiz)n {x' u 2) n {x' u y')

= [{X u 2/) u (z n z')] n [{y uz)u(xn x')] n [(x' u z) u (2/ n y')] n [{x' u y) u (2 n 2')]

= (x u 3/ u 2) n (a; u 2/ u z')n (x'uyuz)n {x' u 2/' u z) n (x' u 2/' u 2')

This expression is called the dual canonical form or the conjunctive normal form of the

function. Note that it is not the dual of the canonical form of that function.


The dual of each statement concerning the canonical form of a Boolean function is avalid statement concerning the dual canonical form of that function. (Note that the dualof term is factor.) The dual canonical form of a Boolean function in n variables can containat most 2" distinct factors. The dual canonical form containing all of these factors iscalled the complete dual canonical form or the complete conjunctive normal form in nvariables; its value is identically 0. The complement F' of a Boolean function F expressedin dual canonical form is the intersection of all factors of the complete dual canonical formwhich do not appear in the dual canonical form of F. Also, we have

Theorem I'. If, in the complete dual canonical form in n variables, each variable is

assigned arbitrarily the value or 1, then just one factor will have thevalue while all other factors will have the value 1.

Two Boolean functions are equal if and only if their respective dual canonicalforms are identical, i.e., consist of the same factors.

In the next section we will use these theorems to determine the Boolean function whenits values for all possible assignments of the values and 1 to the variables are given.

and

Theorem II'

X V z F(x,y,z)

1 1 1 1

1 1

1 1 1

1 1

1

1

1 1

1

CHANGING THE FORM OF A BOOLEAN FUNCTION

Denote by F{x,y,z) the Boolean function whosevalues for all possible assignments of or 1 to thevariables is given by Table 17-2.

The proof of Theorem I suggests that the termsappearing in the canonical form of F(x,y,z) areprecisely those of the complete canonical form in

three variables which have the value 1 wheneverF{x, y,z) = 1. For example, the first row of the tableestablishes x ny (Iz as a term while the third rowyields x n y' n z as another. Thus,

Table 17-2

F{x,y,z) = (a n 2/ n «) u (a; n 2/' n «) u («' n J/' n «) u {x' n r n z')

- (a n z) u («' n y)

Similarly, the factors appearing in the dual canonical form of F{x,y,z) are preciselythose of the complete dual canonical form which have the value whenever F{x, y, z) = 0.

We have

F(x,y,z) = (x' Uy'Uz)n(xUy'U z') n (x' U y U z) n (x U y' U z)

= {X' U z) n {X U y') See Problem 10.

If a Boolean function F is given in either the canonical or dual canonical form, thechange to the other form may be readily made by using successively the two rules forfinding the complement. The order in their use is immaterial; however, at times one orderwill require less computing than the other.

Example 6: Find the dual canonical form for

F = (x ny n z')u (xny' n z)u (x ny' n z')u (x' ny nz)u (x' n y' n z')

Here F' = (x n y n z) u {x' n y' n z) v {x' n y n z')


(the union of all terms of the complete canonical form not appearing in F) and

F - (F'Y = (a; U ^ U z') n (« U J/' U «) n («' U j/' U z') (by Problem 4)


The procedure for changing from canonical form to dual canonical form and vice versa

may also be used advantageously in simplifying certain Boolean functions.

Example 7: Simplify F = [(y n z') u (y' n z)] n [{x' n y) u {x' n z) u {x n y' n z')]

Set Fi = (yn z') U (y' n z) and F^ = («' n y) u (x' n z) u (x n y' n z').

Then Fi = (x n y n z') V (x' n y n z') u {x n y' n z) u (x' n j/' n z)

f[ - (x' n J/ n z) u («' n J/' n z') u (« n J/ n z) u (x n J/' n z')

and Fi = (« u i/' U z') n (a; u j/ u z) n (»' u v' u z') n (a;' u y u z)

Also Fg = («' n y n z) u (a;' n J/ n z') u (x' n i/' n z) u (x n j/' n z')

F'i = (x' n 1/' n z') u (« n 1/ n z) u (a; n 1/' n z) u (x n ^ n z')

and Ft = (x U j/ U z) n (x' u y' U z') n (x' U 3/ U z') n (x' u y' U z)

Then F - F^nFi= {x u y V z) n {x u y' V z') n (x' u y u z) n {x' u y U z')

n (x' u 3/' u z) n (x' u 3/' u z')

Now F' = (x u 3/' u z) n (x u 3/ u z')

and F = {x' n y n z') u (x' n y' n z) = x' n [(3/ n 2') u (y' n z)]

ORDER RELATION IN A BOOLEAN ALGEBRA

Let f7 = {a,b,c} and S = {0,A,B,C,D,E,F, U) where A = {a}, B = {b}, C = {c},

D = {a,b}, E = {a,c}, F = {6,c}. The relation C, defined in Chapter 1, when applied

to S satisfies the following laws:

For any X,Y,Z &S,

(a) XqX(6) If Z c y and Y cX, then X=Y.(c) If X C y and Y qZ, then XqZ.(d) If X c y and X c Z, then X c (y n Z).

(e) If X c y, then X c (y U Z).

(/) XqY if and only if Y' C X'.

(y) X c y if and only if X U y = y or the equivalent X n y == 0.

The first three laws assure (see Chapter 2) that C effects a partial ordering in S illus-

trated byjj"*"^

}F

Fig. 17-1


We shall now define the relation C (read, "under") in any Boolean algebra 'B by

a Qb if and only if o U 6 = & or the equivalent a n b' =

for every a,b G'B. (Note that this is merely a restatement of (g) in terms of the elementsof <B.) There follows readily

(oi) a ca(6i) If a c 6 and b Ca, then a — b.

(ci) If a C 6 and b Cc, then a c c.

so that c defines a partial order in 'B. We leave for the reader to prove

(di) If a c 6 and a c c, then a c (& n c).

(ei) If acb, then a C (6 U c) for any c G'B.

ifi) acb if and only if 6' c a'.


Theorem III, For every a,b G'B, a U 6 is the least upper bound and a D 6 is the greatestlower bound of a and b.


Theorem IV. c 6 C 1, for every b G'B.

ALGEBRA OF ELECTRICAL NETWORKSThe algebra of electrical networks is an interesting and highly important example of

the Boolean algebra (see Problem 15) of the two elements and 1. The discussion herewill be limited to the simplest kind of networks, that is, a network consisting only ofswitches. The simplest such network consists of a wire containing a single switch r:

When the switch is closed so that current flows through the wire, we assign the value 1to r; when the switch is open so that no current flows through the wire, we assign thevalue to r. Also, we shall assign the value 1 or to any network according as currentdoes or does not flow through it. In this simple case, the network has value 1 if and onlyif r has value 1 and the network has value if and only if r has value 0.

Consider now a network consisting of two switches r and s. [When connected in series:

-• 8 •-

Fig. 17-2

it is clear that the network has value 1 if and only if both r and s

have value 1, while the network has value in all other assign-ments of or 1 to r and s. Hence this network can be repre-sented by the function F = F(r,s) which satisfies Table 17-12We find easily F - r n si\ [When connected in parallel:

r 8 F

1 1 1

1

1

Table 17-1 Z.

Fig. 17-3


it is clear that the network has value 1 if and only if at least

one of r and s has value 1, and the network has value if and

only if both r and s have value 0. This network can be repre-

sented by the function F = F{r,s) which satisfies Table 17-43

We find easily F = r U s!J For the various networks consisting

of three switches, see Problem 13.

Using more switches, various networks of a more complicated

nature may be devised; for example.

r s F

1 1 1

1 1

1 1

Table 17-i 3

Fig, 17-4

The corresponding function for this network consists of the intersection of three factors:

(rUs)ntn{uUvDw)

So far all switches in a network have been tacitly assumed to act independently of

one another. Two or more switches may, however, be connected so that (o) they open

and close simultaneously or (6) the closing (opening) of one switch will open (close) all of

the others. In case (a), we shall denote all switches by the same letter; in case (6), we

shall denote some one of the switches by, say, r and all others by r'. In this case, any

primed letter has value when the unprimed letter has value 1 and vice versa. Thus the

network

-• r' •-

Fig. 17-5

consists of three pairs of switches: one pair, each of which is denoted by t, open and close

simultaneously and two other pairs, denoted by r, r' and s, s', such that in each pair the

closing of either switch opens the other. The corresponding function is basically a union

of two terms each involving the three variables. For the upper wire we have r n s' n t

and for the lower {tU s)n r'. Thus, the function corresponding to the network is

F = (r n s' n <) u [{t u s) n v]

and the table giving the values (closure properties) of the function is

r 8 t rn »' n « (( u 8) n r' F

1 1 1

1 1

1 1 1 1

1 1 1 1

1

1 1 1

1 1 1

Table 17-5


It is clear that current will flow through the network only in the following cases:

(1) r and t are closed, s is open; (2) s and t are closed, r is open; (3) s is closed, r and t

are open; (4) t is closed, r and s are open.

In further analysis of series-parallel networks it will be helpful to know that the

algebra of such networks is truly a Boolean algebra. In terms of a given network, theproblem is this: Suppose F is the (switching) function associated with the network andsuppose that by means of the laws of Boolean algebra this function is changed in form to

G associated with a different network. Are the two networks interchangeable; in other

words, will they have the same closure properties (table)? To settle the matter, first

consider Tables 17-3 and 17-4 together with their associated networks and functions

r n s and r U s respectively. In the course of forming these tables, we have verified thatthe postulates (i), (ii), (iv) for a Boolean algebra hold also for network algebra. For thecase of postulate (iii), consider the networks

Fig. 17-6

corresponding to the Boolean identity a U (6 n c) = (a U 6) n (a U c).

from the table of closure properties

It is then clear

a 6 c o u (6 n c) (a U 6) n (a u c)

1 1 1 1

1 1 1

1 1 1

1 1 1

1 1

1

1

Table 17-6

that the networks are interchangeable.

We leave for the reader to consider the case of the Boolean identity

a n (6 U c) = (a n 6) U (a n c)

and conclude that network algebra is a Boolean algebra.

SIMPLIFICATION OF NETWORKSSuppose now that the first three and last columns of Table 17-5 are given and we are

asked to devise a network having the given closure properties. Using the rows in whichF = 1, we obtain

F = (rn s' n t) u (r' n s n t) u (r' n s n «') u (r' n s' n «) = (r' n s) u (s' n t)

Since this is not the function (network) fromwhich the table was originally computed, the

network of Fig. 17-5 is unnecessarily complexand can be replaced by the simpler network, ' • «' • • *

shown in Fig. 17-7. Fig. 17-7


Note. The fact that one of the switches of Fig. 17-7 is denoted by r' when there is no switch

denoted by r has no significance here. If the reader has any doubt about this, let himinterchange r and r' in Fig. 17-5 and obtain F = (r n s) U (s' n t) with diagram

Fig. 17-8

See Problem 14.

Solved Problems

1. Prove: U and n are associative, i.e., for every a,b,c G'B

(a U 6) U c = a U (6 U c) and (a n 6) n c = a n (b n c) (4)


Let X = {a n b) n c and y = a n (b n e). We are to prove x = y. Nov^ using (iii) and (3),

aU X = a U [(o n 6) n c] = [a U (a n 6)] n (a U c)

= o n (a U c) = a = o U [a n (5 n c)] = aV y

and a' u X = a' U [{a n b) n e] = [a' u (o n b)] n (a' u c) = [{a' V a) n («' u 6)] n (a' u c)

= [1 n (a' u b)] n (a' U e) = (a' U 6) n (o' u e) = a' U (6 n c)

= (o' U a) n [a' U (b n c)] = a' U [a n (b n c)] = a' U y

Hence (o u «) n (a' u x) = (a u j/) n (a' U j/)

(a n a') u « = (on a') u y

and X = y

We leave for the reader to show that as a consequence parentheses may be inserted in

o, U 02 u • u «„ and ai n a2n •• n a„ at will.

2. Prove: For each a e "S, the element a' defined in (iv) is unique.

Suppose the contrary, i.e., suppose for any aG<B there are two elements a', a" G "B such that

a U a' = 1 a n a' =and

au a" - 1 o n a" =

Then a' = 1 n a' = (a U a") no' = (on o') U (o" n o')

= (o n a") u (o" n o') = o" n (o u o') - o" n 1 = a"

and a' is unique.

3. Prove: For every a,b G'B

(a U by = o' n 6' and (a n 6)' = a' U 6' (5)

(See 1.11-1.11', Chapter 1, Page 5.)

Since by Problem 2 there exists for every a; e <B a unique x' such that x U x' = 1 and

a; n a;' = 0, we need only verify that

(o u 6) u (o' n b') = [(o u 5) u o'] n [(o u b) u 6'] = [(o U a') U 6] n [<x u (6 u 6')]

= (1 u 6) n (a u 1) = in 1 = 1


and (we leave it for the reader) (a U 6) n (a' n 6') =Using the results of Problem 2, it follows readily that

(aj U 02 U • • • u a„)' = a[n a^n •• n a'^

»n^ (oi n 02 n • • • n aJ' = o.' u a; u • • • u o'

4. Prove: (a')' = a for every aG'B. (See 1.1, Chapter 1, Page 5.)

{a'Y = 1 n (a')' = (a U a') n (a')' = [a n (o')'] U [a' n {a')'] = [a n (a')'] U

= u [o n (a')'] = (an a') u [a n {a')'] = o n [a' u (a')'] = a n 1 = a

5. Simplify: [x U {x' U 3/)'] n [a; U {y n «')']•

[« U (»' u 1/)'] n [a; U (j/' n z'Y] = [a; U (* n 2/')] n[xu(yuz)] = a; n [a; U (j/ U «)] = a;

6. Obtain the canonical form of [x U {x' U y)'] n [x U (1/' n 2')'].

Using the identity of Problem 5,

[aj u (x' u y)'] n [a; u (j/' n 2')'] = x = x n (y u y') n (a u z')

= (« n 3/ n z) u (a; n J/' n z) u (a; n J/ n «') u (a; n J/' n «')

7. Prove: The complete canonical form in 3 variables is identically 1.

First, we shov that the complete canonical form in 2 variables

(a; n J/) u (a n y') u (x' n y) \j (x' n y') = [(x n y) u (x n y')] u [(x' n 3/) u (a;' n j/')]

= [a; n (1/ U y')] U [x' n (j/ U J/')]

„^ ^= (« U a;') n (j/ U 3/') = 1 n 1 = 1

Then the canonical form in 3 variables

[(a; n 2/ n z) u (a; n J/ n z')] u [(x n 1/' n z) u (« n 3/' n z')]

u [(x' n 3/ n z) u (a;' n y n z')] u [(a;' n 3/' n z) u (a;' n 3/' n z')]

= [(« n 3/) U (a; n 3/') U (a;' n 3/) U (a;' n 3/')] n (z U z') = 1 n 1 = 1

8. Prove: If, in the complete canonical form in n variables, each variable is assignedarbitrarily the value or 1, then just one term will have the value 1 while all otherswill have the value 0.

Let the values be assigned to the variables a:i,a;2, . . .,a;„. The term whose value is 1 containsxi if x^ has the value 1 assigned or a:; if x^ has the value assigned, a;2 if x^ has the value 1 orx.^ if X2 has the value 0, . . ., a;„ if a;„ has the value 1 or x'^ if x„ has the value 0. Every otherterm of the complete canonical form will then have as at least one factor and, hence, has as value.

9. Prove: Two Boolean functions are equal if and only if their respective canonical formsare identical, i.e. consist of the same terms.

Clearly two functions are equal if their canonical forms consist of the same terms. Conversely,if the two functions are equal, they must have the same value for each of the 2" possible assignmentsof or 1 to the variables. Moreover, each of the 2" assignments for which the function has thevalue 1 determines a term of the canonical form of that function. Hence the two normal forms containthe same terms.


10. Find the Boolean function F defined by

X V z F

1 1 1

1 1 1

1 1 1

1 1

1 1

1 1

1

1

Table 17-7

It is clear that the canonical form of F will consist of 5 terms while the dual canonical form

will consist of 3 factors. We use the latter form. Then

F — (x' u y' u z') c\ (x<J y' u z') n {xu y u z')

= (y' U z') n (a; U J/ U z') = [y' n (at U y)\ \J z' ^ (x n y') U z'

11. Find the canonical form for F = {x VJ y \J z) r\ (x' U 3/' U z).

Here F' = («' n j/' n z') u (« n j/ n z')

(by the identity of Problem 3) and

F = (F'Y = (a; n J/ n z) u (a; n J/' n z) u (a; n J/' n z') u (a;' n 1/ n z) u (x' n j/' n z) u (a;' n 1/ n 2')

(the union of all terms of the complete canonical form not appearing in F').

12. Prove: For every a,b G'B, a U 6 is the least upper bound and a n b is the greatest

lower bound of a and b.

That a U 6 is an upper bound of a and b follows from

oU(ou6) - aU b - bU(aU6)

Let c be any other upper bound of a and 6. Then o C c and b Qc so that a U c = c and

5 U c = c. Now (au6)uc = oU(6uc) = aUc = c

Thus, (a U b) C c and a u b is the least upper bound as required.

Similarly, a. n 6 is a lower bound of a and 6 since

(o n b) U a = a and (a n b) u 6 — b

Let c be any other lower bound of a and b. Then c C a and c c b so that c u a = o and c U b = b.

c u (o n b) = (c u a) n (c u b) = o n bNow

Thus, c c (a n b) and o n b is the greatest lower bound as required.

13. Discuss the possible networks consisting of three switches r,&,t.

There are four cases:

(i) The switches are connected in series. The diagram is

* •-

and the function is r n s n t.


(ii) The switches are connected in parallel. The diagram is

and the function is r U s u t.

(iii) The series-parallel combination

with function r n (g U t).

(iv) The series-parallel combination

-• 8 •-

-• t •-

-• r •-

with function r u (« n t).

14. If possible, replace the network

-• r 9- -• t •-

t' '

Fig. 17-9(0)

by a simpler one.

The Boolean function for the given network is

F = (rn t) u {« n («' u t) n [r' u (s n t')]}

= (rn t) u {« n [(«' u t) n (r' u t')]}

= (rn t) u [r' n (« n *)] = (r u «) n t

The simpler network is

Fig. 17-9(6)



u n 1

1 1

u a b c d

a a b e d

b b b d d

c c d e d

d d d d d

n a b c d

a a a a a

b a b a b

c a a c e

d a b c d

15. Show that the set {0, 1} together with the operations as defined in 1 and

is a Boolean algebra. Ill16. Show that the set {a,b,e,d} together with the operations defined in

and

is a Boolean algebra.

17. Show that the Boolean algebra of Problem 16 is isomorphic to the algebra of all subsets of a set of

two elements.

18. Why is there no Boolean algebra having just three distinct elements?

19. Let S be a subset of N, and for any a, 6 S S define a U b and a n b to be respectively the least

common multiple and greatest common divisor of a and 6. Show

(a) <5 is a Boolean algebra when S = {1,2,3,6,7,14,21,42}.

(6) S is not a Boolean algebra when S = {1,2,3,4,6,8,12,24}.

20. Show that a U {a n b) = a n {a U b) without using Example 3, Page 223. State the dual of the

identity and prove it.

21. Prove: For every a, 5 G 'B, a U (o' n 6) = a U 6. State the dual and prove it.

22. Obtain the identities of Example 1, Page 222, by taking 6 = o in the identities of Problem 21.

23. Obtain as in Problem 22 the identities of Example 2, Page 222.

24. Prove: 0' = 1 and 1' = 0. (See 1.2-1.2', Chapter 1, Page 5.)

Hint. Take a = and 6 = 1 in the identity of Problem 21.

25. Prove: (a n 6') U (6 n a') - (a U 6) n (a' U 6'). Write the dual.

26. Prove: (o u 6) n (6 u c) n (c u o) = (a n 6) u (6 n c) u (c n a). What is its dual?

27. Prove: If aU x = b U x and aU x' - b U x', then a = b.

Hint. Consider (a U a) n (a U x') - (6 U «) n (6 U «')•

28. State the dual of Problem 27 and prove it.

29. Prove: If an6 = anc and a u 6 = a U c for any a, 6, c G "B, then h = c.

30. Simplify: (a) (a U 6) n a' n 6'

(6) (o n 6 n c) u a' u 6' u c'

(c) (o n 6) u [c n (a' u b')\

(d) [a u (a' n b)] n [5 u (6 n c)]

Ans. (a) 0, (b) 1, (c) (a n b) U c, (d) b, (e) x' n y, (f) o! n b', (jr) a U b

(e) [(«' n J/')' u z] n (a U J/')'

(/) (a u b') n (a' u b) n (»' u V)

(g) [{a u b) n (c u b')] u [b n (a' u c')]

31.

32.

33.

Prove: (o U b) n (a' U c) = (a' n b) U (a n c) U (b n c)

= (a u b) n (a' u c) n (b u c) = (a n c) u (a' n b)

Find, by inspection, the complement of each of the following in two ways:

(o) (xny)u(xn y') (e) (x U y' u z) n {x u y u z') n (x u j/' u z')

(b) {xny' r\z)u (x' ny n z') (d) (x u j/' u z) n (x' u j/ u z)

Express each of the following in both canonical form and dual canonical form in three variables:

(o) x' U y', (b) (a; n y') U (x' n y), (c) (x U y) r\ (x' U z'), (d) x n z, (e) x n (y' U z)


Partial Ans.

(a) (xny' nz)u (xny' n z') u (x' n y m) u (x' n j/' n «) u (»' n j/ n 2') u (x' r\y' r\ z')

(b) {xuyuz)n(xuyu z') n (»' Uy' uz)n (x' Uy' u z')

(c) {x n y n z') V (x n y' n z') u {x' n y n z) o (x' n y n z')

(d) (xuyuz)r\(xuy'uz)n(xuyu z') n (x u y' u z') n {x' u y u z) n (x' Uy' u z)

(e) (xu yu z) n {xu y' u z) n {xu y u z') n (x u y' u z') n {x' u j/'.u z)

34. Express each of the following in both canonical and dual canonical form in the minimum number ofvariables:

(a) a U (x' n y) (d) (xnynz)U[(xUy)n{xU z)]

(6) [xn{yu z)] u[xn(yu z')] (e) {x U y) n (x U z') n (x' U y') n («' U z)

(c) (xuyi>z)n[(xny)[j {x' n z)] (/) (xny)v{xn z') u («' n z)

Partial Ans.

(a) {x ny)u (x n y') u (x' n v) (d) (a; u j/ u 2) n (» u j/ u 2') n(x\Jy' vz)(5) (a; u J/) n (x u J/') (e) (x n j/' n 2) u (x' n j/ n 2')

(c) (x n J/ n 2) u (x n J/ n z') u (x' n j/ n 2) u (x' n j/' n 2) (/) (x u v u z) n (x' u j/ u 2') n (x u 1/' u z)

35. Write the term of the complete canonical form in x, y, 2 having the value 1 when:(a) X = 2 = 0, 3/ = 1; (6) X = 1/ = 1, 2 = 0; (c) x = 0, 1/ = 2 = 1.

Ans. (a) x' n J/ n z', (6) x n j/ n z'

36. Write the term of the complete canonical form in x, y, z, w having the value 1 when:

(a) X =^ y = 1, z = w = 0; (b) x = y = w - 0, z - l; (c) x = 0, y = z - w = 1.

Ans. (a) X n y n z' n w', (e) x' n y n z n w

37. Write the factor of the complete dual canonical form in x,y,z having the value when:

(a) X = 2 = 0, J/ = 1; (6) X = 3/ = 1, 2 = 0; (c) x = 0, j/ = z = 1.

Ans. (a) X u v' u 2, (6) x' u 3/' u 2

38. Write the factor of the complete dual canonical form in x, y, z, w having the value when:

{a) x = y-l, z = w = 0; (b) x = y = w = 0, z = l; (e) x = 0, y = z = w = 1.

Ans. (a) x' U 3/' u 2 U w, (c) x U 3/' U 2' U w'

39. Write the function in three variables whose value is 1

(a) if and only if two of the variables are 1 and the other is 0,

(6) if and only if more than one variable is 1.

Ans. (a) (x n 3/ n 2') u (x n 3/' n 2) u (x' n 3/ n 2), (6) [x n (3/ u 2)] u (j/ n 2)

40. Write the function in three variables whose value is

(a) if and only if two of the variables are and the other is 1,

(6) if and only if more than one variable is 0.

Ans. The duals in Problem 39.

41. Obtain in simplest form the Boolean functions Fi,F2 Fg defined as follows:

X V 2 Fi F2 Fs ^4 F, n Fr ^8

1 1 1 1 1 1 1 1

1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1

Ans. Fi - (xu y') n 2', /^g = x' u (3/' n 2) u (3/ n 2'), F^ = (x u z) n [y' u (x n z)], F7 =


42. Show that F^ and Fg of Problem 41 can be found by inspection.

43. Prove: (dj) It a Qh and o c c, then o C (6 n c).

(ei) If a C 6 then a C (6 U c) for any c G ^.

(/,) a C 6 if and only if V c a'.

44. Prove: If a, 6 G "B such that a C 6 then, for any ee:<=B, a U (6 n c) = 6 n (a U c).

45. Prove: For every 6 G 'B, C 6 C 1-

46. Construct a diagram similar to Fig. 17-1 for the Boolean algebra of all subsets of B = {a, 6, c,d}.

47. Diagram the networks represented by a U (a' n 6) and a u 6 and show by tables that they have

the same closure properties.

48. Diagram the networks (i) (o U 6) n o' n 6' and (ii) (a n 5 n c) U (o' U 6' u c'). Construct tables

of closure properties for each. What do you conclude?

49. Diagram each of the following networks

(i) (a u h') n (a' u 6) n (a' u V) (iii) [(a u b) n (c u h')] u \h n (a' U c')]

(ii) (a n 6) U [c n (a' U h')] (iv) (a n 6 n e) u a' u 6' u c'

Using the results obtained in Problem 30, diagram the simpler network for each.

Partial Ans.

(i) -c:>{::>-c:> --• 6'*-

b •-

(ii)Ca • • 6 >

——• c •

50. Diagram each of the networks (r U «') n (r' U s) and (r n s) U (r' n s') and show that they have

the same closure properties.

51. Diagram the simplest network having the closure properties of each of Fj-Fg in Problem 41.

Partial Ans.

-{::>- ^^' -c;>-c:;

52. Simplify:

(i)

ify: !-«•—.

t » s • ' •

r «—• 8 •—• t

—• t •—s' t' •-

(ii) (iv)

1— V •—• z •—

I

I—• I/'•—• 2' •—

'

z'm-

To afford practice and also to check the results, it is suggested that (iii) and (iv) be solved by forming

the table of closure properties and also by the procedure of Example 7, Page 226.

Partial Ans. (i) and (">-c:>


53. Simplify:

(a)

r •- c

I q •—I I—* r •-J

p •— —

I

'—• r »—

(b)

-• P •—• 9 •—• r •-

HI p •—• g •—• 8

-• P •—• q •—• r' •-

(c)

-• P •—• 9' •—• s

• p •—• r' •—• s'•—

• g •—• s' »

-• 9 •—• 8 •-

Ana. (a)

(c)

(6)

54. Simplify:

--c::]

-• g •-

I—• g' •—• 8 •—

I

-• a • • b'

-• a' m • b

e •——• d'

c' 9 • d •-

c •——• d' •-

c' •—• d •-

55. Show that the network of Problem 50 will permit a light over a stairway to be turned on or off eitherat the bottom or top of the stairway.

56. From his garage M may enter either of two rooms. Obtain the network which will permit M to turnon or off the light in the garage either from the garage or one of the rooms irrespective of the positionof the other two switches.

Ana.

I—• »' •—• t' •—

I

57. Design a network which will permit independent control of a light from any one of four switches.

INDEXAbelian group, 82Absolute value, 42

of complex numbers, 77Addition,

of complex numbers, 75of integers, 39of linear transformations, 151of matrices, 164of matrix polynomials, 200of natural numbers, 30of pol3Tiomials, 125of rational numbers, 60of real numbers, 67, 68of subspaces, 148ot vectors, 143, 144

Additive inverse, 40, 60, 68, 75, 101, 125Algebra,

linear, 219of linear transformations, 151of matrices, 167of residue classes, 53

Algebraic system, 22Alternating group, 84Amplitude of a complex number, 77Angle between vectors, 149Angle of a complex number, 77Anti-symmetric relation, 17Archimedean property,

of rational numbers, 62of real numbers, 69, 71

Argument of a complex number, 77Associate, 115

Associative law,

for Boolean algebras, 230for groups, 82for linear algebras, 219for matrices, 166for permutations, 23for polynomials, 125for rings, 101for the complex numbers, 75for the integers, 42for the natural numbers, 30, 31for the rational numbers, 60, 61for the real numbers, 67, 69, 71for union and intersection of sets, 5genera], 19

Augmented matrix, 178

Basis,

normal orthogonal, 207of a vector space, 147orthonormal, 207

Binary,

operation, 18relation, 15

Boolean,

algebra, 222

Boolean (cont.)

function, 223polynomial, 223ring, 112

Cancellation law,

for groups, 83for the integers, 41, 42for the natural numbers, 30, 31for the rational numbers, 60for the real numbers, 69, 71

Canonical forms,

for Boolean polynomials, 224for matrices, 171, 199

Cap, 3

Cayley's theorem, 86Characteristic,

determinant, 204of an integral domain, 115of a ring, 103

polynomial, 204roots, 203

vectors, 203Closure, 19

Codomain, 6

Coefficient, 125Column,

equivalent, 170rank, 172, 199

transformation, 170, 198vector, 164

Common divisor (see also Greatest commondivisor), 49, 117

Commutative,

group, 82

operation, 19

ring, 103

Commutative law,

for Boolean algebras, 222for fields, 117, 118for groups, 82for matrices, 166for polynomials, 125for rings, 101, 103for the complex numbers, 75for the integers, 42for the natural numbers, 30, 31for the rational numbers, 60, 61for the real numbers, 67, 69, 71for union and intersection of sets, 5general, 19

Complement of a set, 2Completeness property, 70Complex number, 75Complex plane, 76Components,

of a complex number, 75of a vector, 143

239

240 INDEX

Composite, 49

Composition series, 89 ^Cong:ruence modulo m, 52

Conies, 208

Conjugate complex number, 76

Conjunctive normal form, 224

Correspondence, one-to-one, 8

Coset, 86

Countable, 8

Cubic polynomial, 139

Cup, 3

Cut, 66

Cycle, 23

Cyclic group, 84, 86

Cyclic notation ifor permutations, 23

Decimal representation of rational numbers, 62

Dedekind cut, 66

Degree, 125

De Molvre's theorem, 77, 79

De Morgan's laws, 5

Density property,

of rational numbers, 62

of real numbers, 69, 71

Denumerable, 8

Dependence, linear, 146

Determinant, 182

Diagonal matrix, 171

Diagram,for partial ordering, 17

Venn, 3

Difference of sets, 4

Dihedral group, 92

Dimension of vector space, 147

Disjoint,

cycles, 24

subsets, 3

Disjunctive normal form, 224

Distributive law,

for Boolean algebras, 222

for linear algebras, 219

for matrices, 166

for polynomials, 126

for rings, 101

for the complex numbers, 76

for the integers, 42

for the natural numbers, 31

for the rational numbers, 60, 61

for the real numbers, 67, 69, 71

for union and intersection of sets, 6

for vector spaces, 144

general, 20

left, right, 20

Division, 61, 69, 76

algorithm, 50, 117, 128, 201

ring, 117

Divisor, 49, 115, 128

Divisors of zero, 103

Domain,integral, 114, 127

of a mapping, 6

ordered integral, 116

Dot product, 149

Dual canonical form, 224

Echelon matrix, 171

Eigenvalue, 204

Eigenvector, 203

Electrical networks, 227

Element,

first, last, 18

identity, 19

maximal, minimal, 18

of a set, 1

Elementary,

matrix, 173

transformation, 169, 198

Empty set, 2

Equations (see also Polynomial),

homogeneous linear, 181

non-homogeneous linear, 179

systems of linear, 178

Equivalence,

class, 16

relation, 16

Equivalent matrices, 170, 200

Euclidean ring, 107

Even permutation, 24, 27

Expansion of determinant, 182

Exponents,

in a group, 83

integral, 43

natural numbers, 33

real, 70

Factor, 49, 128

group, 88

theorem, 128

Field, 118

skew, 117

Finite,

dimension, 147

set, 8

Form, polynomial, 124

Four-group, 97

Fractions, 60

Function (see also Mapping, Transformation), 7

Boolean, 223

Fundamental Theorem^ of Algebra, 129

Gaussian integer, 110

Generator,

of a cyclic group, 84

of a vector space, 145

Greatest common divisor, 49, 117, 131

Greatest lower bound, 70

Group, 82

Abelian, 82

alternating, 84

cyclic, 84

dihedral, 92

Klein 4-, 97

octic, 92

permutation, 84

quotient, 88

symmetric, 84

INDEX 241

Highest common factor (see Greatest commondivisor), 49

Homogeneous linear equations, 181

Homomorphism,between groups, 84

between rings, 103between vector spaces, 160

Ideal (left, right), 104

maximal, 106

prime, 106

principal, 105

proper, 104

Identity,

element, 19

mapping, 9

matrix, 166

uniqueness of, 20Image, 6

Imaginary numbers, 75

pure, 76

Imaginary part (of a complex number), 76Imaginary unit, 76

Improper,

ideal, 104

orthogonal transformation, 208subgroup, 83

subring, 102

subset, 2

Inclusion,

for Boolean algebras, 226

for sets, 2

Independence, linear, 146

Indeterminate, 124Index of subgroup, 87Induction, 31, 37Inequality,

Schwarz, 149

triangle, 149

Infinite dimensional vector space, 147Infinite set, 8Inner product, 149

Integers, 38

Gaussian, 110

negative, 40

positive (see also Natural numbers), 39Integral domain, 114, 127Intersection,

of sets, 3

of subgroups, 84

of subspaces, 148

Invariant,

subgroup, 87

subring, 104

vector, 203Inverse,

additive, 40, 60, 68, 75, 101, 125, 166in a field, 118

multiplicative, 60, 68, 76of a mapping, 9

of a matrix, 176, 176of an element, 20uniqueness of, 20

Irrational number, 69

Irred]îble polynomial, 128Isomorphism, 21

between groups, 86between rings, 103between vector spaces, 150

Jordan-Holder theorem, 90

Kernel of homomorphism, 88

Lagrange's theorem, 87Lambda matrix, 198

normal form, 199Laws of exponents (see also Exponents), 33Leading coefficient, 125Least common multiple, 59Least upper bound, 70Left,

coset, 86

ideal, 104

Length of a vector, 143, 149Linear,

algebra, 219

combination, 49, 145

congruence, 64

dependence, 146

equations, 178form, 178

independence, 146

transformation, 149Lower bound, 70

Mapping, 6

one-to-one, 8

Mathematical induction, 31, 37Matrix, 165, 167

augmented, 178column rank, 172, 199diagonal, 171

elementary, 173

identity, 166

lambda, 198

non-singular, 172orthogonal, 207over f, 166

product, 166

rank, 172

row rank, 172

scalar product, 164singular, 172

sum, 164

symmetric, 206triangular, 171

zero, 166

Matrix polynomial, 198Maximal ideal, 106Minimum polynomial, 133, 177Modulus of a complex number, 77Monic polynomial, 126Multiples, 33, 43

Multiplication,

of complex numbers, 75

242 INDEX

Multiplication (cont.)

of integers, 39

of linear transformations, 151

of matrices, 165

of matrix polynomials, 200

of natural numbers, 31

of polynomials, 125

of rational numbers, 60

of real numbers, 67

Multiplicative inverse, 60, 68, 75

Multiplicity of root, 129

Natural numbers, 30

Negative integers, 40

Networks, electrical, 227

Non-singular,

matrix, 172

transformation, 151

Norm, 119

Normal divisor, 87

Normal form,

conjunctive, disjunctive, 224

of matrix over f, 171

of X-matrix, 199

Normal orthogonal basis, 207

Normal subgroup, 87

Null set, 2

Numbers,complex, 75

irrational, 69

natural, 30

prime, 49

rational, 60

real, 65

Odd permutation, 24, 27

One-to-one mapping, 8

Onto (mapping), 6

Operations, 18

binary, 19

well-defined, 20

Order,

of a group, 83

of a group element, 83

relations, 32, 40, 61, 226

Ordered integral domain, 116

Ordered pair, 5

Orthogonal,

matrix, 207

normal orthogonal basis, 207

transformation, 207

vectors, 149

Orthonormal basis, 207

Partial ordering, 17

Partition, 17

Peano postulates, 30

Permutation, 22

even, 24, 27

group, 84

odd, 24, 27

Perpendicular vectors, 149

Polar form, 76

Polynomial, 124

Boolean, 223

degree of a, 125

domain C\x\, 129

irreducible, 128

prime, 128

ring of, 125

roots of, 127

zero of a, 127

Positive cut, 67

Positive integers (see also Natural numbers), 39

Powers (see also Exponents), 33, 43

Prime, 49

factors, 52

field, 118

ideal, 106

integer, 49

polynomial, 128

relatively prime integer, 52

Primitive roots of unity, 78

Principal ideal, 105

ring, 106

Product,

dot, 149

inner, 149

of cosets, 88

of linear transformations, 151

of mappings, 8

of matrices, 165

of matrix polynomials, 200

of polynomials, 125

of subgroups, 89

scalar, 149, 151, 164

set, 5

Proper,

ideal, 104

orthogonal transformation, 208

subgroup, 83

subring, 102

subset, 2

Quadric surfaces, 208

Quartic polynomial, 140

Quaternion group, 100, 111, 196

Quotient,

group, 88

ring, 106

Range of a mapping, 6

Rank,of a linear transformation, 151

of a matrix, 172, 199

Rational number, 60

Real number, 65

Real part (of a complex number), 76

Real symmetric matrix, 206

Reflexive relation, 15

Regular permutation group, 93

Relation, 15

equivalence, 16

Relatively prime integers, 52

Remainder theorem, 128

Residue classes, 53

INDEX 243

Right,

coset, 86

ideal, 104

Ring, 101

Boolean, 112

commutative, 103

division, 117

Euclidean, 107

principal ideal, 106

quotient, 106

Roots,

characteristic, 204

latent, 204

of cubic polynomials, 139

of polynomials, 127of quartic polynomials, 140of unity, 78

Row,equivalent, 170

rank, 172, 199

transformation, 170, 198

vector, 164

Scalar, 143

multiplication, 143

product, 149, 164

Schwarz inequality, 149

Set, 1

denumerable, 8

finite, 8

infinite, 8

Similar matrices, 205

Simple,

group, 88

ring, 104

zero, 129

Simultaneous linear equations, 178

Singular,

matrix, 172

transformation, 151

Skew field, 117

Span, 145

Square, octic group of a, 92Subalgebra, 167

Subdomain, 115

Subfield, 118

Subgroup, 83

invariant, 87

normal, 87

proper, 83

Subring, 102

invariant, 104

proper, 102

Subset, 2

Subspace, 144

Subtraction, 41, 61, 69, 76Successor, 30

Sum (see Addition)

Symmetric,group, 84

Symmetric (cont.)

matrix, 206relation, 16

Systems of linear equations, 178

homogeneous, 181

non-homogeneous, 179

Total matrix algebra, 167

Transformation,

group of, 152

linear, 149

singular, 151

orthogonal, 207

Transitive relation, 16Transpose, 183

Transposition, 24

Triangle inequality, 149Triangular matrix, 171

Trichotomy law, 32, 40, 61, 68Trigonometric representation of complex

numbers, 76, 77

Two-sided ideal, 104

Union, 3

Unique factorization theorem, 52, 117, 132Uniqueness,

of identity, 20

ot inverses, 20

Unit, 115

Unity element, 19

Universal set, 2

Upper bound, 70

Value, absolute, 42

of complex numbers, 77

Vector(s), 143

characteristic, 203column, 164

invariant, 203

length of, 143, 149

orthogonal, 149

row, 164

unit, 147

Vector space, 144

basis of a, 147, 207Venn diagram, 3

Well-defined,

operation, 20

set, 1

Well-ordered set, 18

Zero, 40, 60

divisors of, 103

matrix, 166

of a cubic polynomial, 139of a polynomial, 127of a quartic polynomial, 140polynomial, 125vector, 144

Index of Symbols

A Set, 1

Matrix, 164

A' Complement of set A, 3

AT Transpose of matrix A, 183

\A\ Determinant of matrix A, 182

[a] Equivalence class, 16

\a\ Absolute value of a, 42

[«ii] Matrix, 165

a]b a is a divisor of 6, 49

(a,b) Greatest common divisor, 50

S Boolean algebra, 222

C Set of all complex numbers, 75

Dedekind cut, 66

ID Integral domain, 114

Eij Basis matrix for c^„(f), 167

X- IdU, J.XO

Set of all polynomials in x

with coefficients in f, 127

G Group, 82

g.l.b. Greatest lower bound, 70

/ Set of all integers, 1, 38

/+ Set of all positive integers, 39

I- Set of all negative integers, 40

Il(m) Integers modulo m, 63

i Complex unit yf—l, 76

s Ideal, 104

-c Linear algebra, 219

l.u.b. Least upper bound, 70

^niT) Total matrix algebra (the set

N(a) Norm, 119

Zero matrix, 166

Q Set of all rational numbers, 1, 60

Q + Set of all positive rational numbers, 61

Q~ Set of all negative rational numbers, 61, 67

R Set of all real numbers, 1, 65

% Ring, 101

^[a] Set of all polynomials in xwith coefficients in %, 125

S„ Set of all permutations on n symbols, 22, 84

ef Division ring, 117

T Linear transformation, 149

M Identity element of a group, 82

Unity element of a ring with unity, 103

y> V„(f) Vector space, 144

N

nX n matrices over y), 167

Set of all natural numbers, 1, 30

z Complex number, 75

Zero element of a ring, 102

z Conjugate of complex number z, 75

a.fi Mappings, 6

a{x),p(x) Polynomials, 124

e Is in, belongs to, 1

«{ Unit vectors, 147

i.'j Vectors, 143

r Zero vector, 143

{a,h,e} Set, 1

c,C Is contained in, 2

The empty set, 2

n Intersection, 3

u Union, 3

-> Mapping, 6

«-» One-to-one mapping, 8

= Congruent, 52

245

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BESCRIPTIVE GEOMETRYIndwdins 17S SOLVED PROBLEMS

By MINOR C. HAWK. M-.?=' <if

ENQINEERING MECHANICSInslwdina .tin SOlVlD PROBLEMS

iy W. G. M(LE AN, H.S. in E,E,, M.S.,VroHnirs.' uS Miihonld, iofo^nhh* Ci^JfrS*

and E, W. NILSON, B.5. in W,.E., M. Adm, t,,

STRENGTH OF MATERIAlS

indudtng fl30 SOLVED PROBLEMS

By WHLfAWi A. MASH, Pli.D-,

."ro^rtpo- o' E»3, MKôniii, UflificMj)), ftf fibfjtfo

ELECTRIC CIRCUITSindudlng 550 SOLVliQ PROBLEMS

By JOSEPH A, ^PMlNISTER, W.S.E,E.

MECHANICAL VtBRATiaNSirtduding 2IS SOLVED PR08LEMS

By WILliAWi V^. SETO, B,S. in M.E., M.S.,

FLUJC MiCHANfCS and KYDRAUUCSifictuding 47S SOLVED PROBLEMS

By RANALD V. GUfS; a,$,, M.S. in C.t.,

REINFORCED CONCRETE OESJGNnoiuding iCa SOLVED PROBLEMS

By N. J. EVESARD, M5CF, Ph.D.,Fnd^. ai cHfl. Mffth, & Sjrjc. J^KtJi<grofl Stpra CoJ.

nnd J. L. TANMER lil, MSCE,IorJin,'fal ConijHro.ni', Tnxas indumiaz ii]C,

MACHINE OESiSNiniiudins lit) SOLVED PlOaUMS

By HALL, HOLOWENKO. LA^GHLINH'-friiir; 't''p^r. :tn-cii L-in . "..tlIlb Viiiârsit^

BASiC ENBINEERING EQUATIONSImiudJng 1 400 BA:>aC EOUATIONS

By W.F.HUGHES, E W GAyiOilD.Jh.O,tLint.rnjI.nlMr-h Fug,. Ciiri»Eî»f.ni( s(r»{h,

ELEMENTARY AL&EGRAInrluSing 2700 SOLVED paOftLEMS

By BAfNETT lirCH, Ph.D., £,'oDJif(.r. TkH. H.i-

PLANE GEOMETRYIndtfding 3S0 SOLVED PK{}B,LEMS

By BAlfNFIT RICH, Ph.D.,

TEST STEMS IN EOUCATItlN

Imibdtng 310DTEST ITEjMS

ByG,J.WDULY.I>1i,D,. L. f. WALTON, Ki.D.,

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