Properties of the Integers: Mathematical Induction
1. Mathematical Induction
2. Harmonic, Fibonacci, Lucas Numbers
3. Prime Numbers
Chapter 4
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
The Well-Ordering Principle Any nonempty subset of Z+
contains a smallest element. (We often express this by sayingthat Z+ is well ordered.)
This principle serves to distinguish Z+ from Q+ and R+.
Theorem 4.1 Finite Induction Principle or Principle of Mathematical Induction
S(n): an open statement, n a positive integer
(a) If S(1) is true; and /* basis, not necessarily from 1 */(b) If whenever S(k) is true (for some ), then S(k+1) is true; (inductive step)then S(n) is true for all
k Z
n Z
If
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Proof: Let is false}. We wish to provethat so to obtain a contradiction we assume that
Then by the well - ordering principle, has aleast element . Since (1) is true, So > 1, and
consequently -1 Z With -1 (since is the smallest element in ), is true. By condition(b), is also true. A contradiction.
+
F t Z S tF
F Fs S s s
s s F sF S s
S s S s
{ | ( ),
..
.( )
(( ) ) ( )
1
11 1
[ ( ) [ [ ( ) ( )]]], ( ).
S n k n S k S kn n S n
0 00
1
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.1. For any =
Proof: =Induction basis: is true.Induction hypothesis: = is true.Establish the truth of
Note the importance of induction basis. For example, in
Ex. 4.4, if = ( , we also have
n Z i n n
S n i n nS
S k k k kS k
S n i n nS k S k
in
in
ik
in
, ( ) /
( ): ( ) /( )
( ): ( ) /( ).
(
( ): ) /( ) ( ). )
1 2
1 21
1 21
2 21
1
1
1
21
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.5 Find a formula for ( )2 11
ii
n
Ans: Observe and conjecture.
n=1 1 1=12
n=2 1+3 4=22
n=3 1+3+5 9=32
n=4 1+3+5+7 16=42
conjecture: ( )2 11
ii
n
=n2
then prove by induction
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.7 the Harmonic Numbers
H H H
Hn
n H n
n H n H n
n
jj
nn
n
1 2 3
2
1
1 11
21
1
2
1
3
11
2
1
3
1
1
1
, , ,
.
, ( ) .
In general,
Prove N, (a)
(b) If Z then +
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.7 the Harmonic Numbers
For (a), induction hypothesis
Since for all 1 j 2 it follows thatk
S k H k
H
H
j
H H H k k
k
k
k
k k k
k k k k
k k k
k k
kk
( ):
, ,
( ) ( )
2
2
2
2 2 2
1
11
2
1
2
1
2 1
1
2 21
2 1
1
2 21
2
1
2
21
21 1 1 1 1
1
1
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.7 the Harmonic Numbers
For (b), induction hypothesis
.S k H k H k
H H H k H k H
k Hk
k H
k H k
jj
kk
jj
kj
j
kk k k
k k
k
( ): ( )
( )
( )[ ]
( ) ( )
1
1
1
11 1
1 1
1
1
1
11
12 1
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.8 For let R, where |An|=2n and the elements ofAn are listed in ascending order. If R, prove that in order todetermine whether An, we must compare r with no more thann+1 elements in An.
n0 An r
r
Proof: When n=0, A0={a} and only one comparison is needed.
Assume the result is true for some and consider the casefor Ak+1.
k 0
kkkkkk |C||B| where,CBA 21 Let and Bk < Ck
(1) Compare once to determine whether r B r Ck k or
(2) Then compare in Bk or in Ck
1
k+1k+2 Total comparisons
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.10 Prove S(n): n can be written as a sum of 3's and/or 8's(with no regard to order) for any n14.
basis: n=14=3+3+8 OKinduction hypothesis: k can be added up by 3's and/or 8's.
If k=...+8+..., then k+1=...+3+3+3+...Otherwise, there are at least 5 3's in k's summands.k+1=...3+3+3+3+3+1+...=...8+8+...
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Theorem 4.2 Finite Induction Principle--Alternate Form
Let with (a) If are true and(b) If whenever are
true for some where then is also true.then is true for all
n n Z n nS n S n S n
S n S n S n S k
k Z k n S k
S n n n
0 1 0 10 0 1
0 0 1
1
0
11
1
, .( ), ( ), , ( )
( ), ( ), , ( ), , ( )
, , ( )
( ) .
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
Ex. 4.11 Prove S(n): n can be written as a sum of 3's and/or 8's(with no regard to order) for any n14.
Assume the truth of the statements:
S S S k S k k Zk n k k
k k S kS k
( ), ( ), , ( ), ( ) ,. ,
( ) . ( )( ).
14 15 116 1 17
1 2 3 21
for some where And now if then and The truth of implies the truth of
Chapter 4 Properties of the Integers: Mathematical Induction
4.1 The Well-Ordering Principle: Mathematical Induction
12
12222323
12212323
11
1012
122
2373
3
222
1211
2
2
2110
k
kkk
kkkkk
nn
nn
nnn
)()(
)()(ppp
.kn Consider true. all are )k(S),k(S
,),(S),(S Assume .p :)n(S :Proof
.p that Show.n whereZn for
,ppp and ,p,p Let
(recurrence relation)
Chapter 4 Properties of the Integers: Mathematical Induction
4.2 Recursive Definitions
0,2,4,6,8,10,12,...: bn=2n
1,2,3,6,11,20,37,68,125,...:an=?
explicit definition
a a a an n n n 1 2 3
implicit recursivedefinition
Examples:
factorial: (n+1)!=(n+1)(n!)
Harmonic Number: H Hnn n 11
1
Chapter 4 Properties of the Integers: Mathematical Induction
4.2 Recursive Definitions
Ex. 4.16 The Fibonacci Numbers
(1) and
(2) for with
F F
F F F n Z nn n n
0 1
1 2
0 1
2
, ;
,
Prove that (a)
for all N.
n Z F F F
b F n
ii
nn n
n
n
2
01
5
3( )
Chapter 4 Properties of the Integers: Mathematical Induction
4.2 Recursive Definitions
Ex. 4.16 The Fibonacci Numbers
( ,
( )
(
a) Induction hypothesis:
b)
F F F
F F F F F F
F F F F F
F F F
ii
kk k
ii
ki
i
kk k k k
k k k k k
k k k
k k
k k k
2
01
2
0
1 2
01
21 1
2
1 1 1 2
1 1
1
1 1 1
5
3
5
3
5
3
8
3
5
3
24
9
5
3
25
9
5
3
1k
Please check out the induction basis for yourselves first.
Chapter 4 Properties of the Integers: Mathematical Induction
4.2 Recursive Definitions
Ex. 4.17 The Lucas Number
( ) , ;
.
.
1 2 1
2
1
0 1
1 2
02
1 1
and
(2) for Z with
Prove
(a) For N,
(b) Z
+
+
L L
L L L n n
n L L
n L F F
n n n
ii
nn
n n n
Chapter 4 Properties of the Integers: Mathematical Induction
4.2 Recursive Definitions
Ex. 4.17 The Lucas Number
(
, .
.
a) Assume
Then =
(b) Assume for the integers= 1,2,3, . . . , where
Then
L L
L L L L
L F Fn k k k
L L L F FF F F F
ii
kk
ii
kk k k
n n n
k k k k kk k k k
02
0
12 1 3
1 1
1 1 1 12 2
1
1 1
1 2
Please check out the induction basis for yourselves first.
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
Def. 4.1 For integers a,b, we say b divides a, and we write b|a,if there is an integer n, such that a=bn. We say b is a divisorof a or a is a multiple of b.
Theorem 4.3. For any Z (a) 1| (b) (c) (d) Z(e) If , for some Z, and divides two ofthe three integers, then divides the remaining integer.(f) Z.(g) For let Z. If divides each then
where Z for
a b ca a a b b a a ba b b c a c a b a bx xx y z x y z a
aa b a c a bx cy x y
i n c a ca c x c x c x x i
i i
n n i
, ,| [( | ) ( | )]
[( | ) ( | )] | | | ,, ,
[( | ) ( | )] |( ), ,, ,
| ( ), all
0
111 1 2 2 n.
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
proof of (f):If and then for some
Z. So If follows that
a b a c b am c anm n bx cy am x an y a mx ny
a bx cy
| | , ,, ( ) ( ) ( ).
|( ).
Ex. 4.21 Let a,b in Z so that 2a+3b is a multiple of 17. Prove that17 divides 9a+5b.
Proof: We observe that 17|Also Hence 17|
( ) |( )( ).|( ). [( ) ( )]
|( ).
2 3 17 4 2 317 17 17 17 17 8 12
17 9 5
a b a ba b a b a b
a b
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
For n in Z+ where n>1, n is a prime number if n has only two divisors, 1 and n. Otherwise n is a composite.
Lemma 4.1. If n in Z+ and n is composite, then there is aprime p such that p|n.Proof: If not, let S be the set of all composite integers that have noprime divisors. If S is not empty, then by the well-orderingprinciple, S has a least element m. But if m is composite,m=m1m2 with 1<m1<m and 1<m2<m. Since m1 is not in S, m1 is a prime or divisible by a prime, which means m is also divisibleby a prime, a contradiction.
lemma, theorem, corollary
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
Theorem 4.4 (Euclid) There are infinitely many primes.
Proof: If not, let be the finite set of primes,and let Since for all
cannot be a prime. Hence is a composite. So by Lemma 4.1 there is a prime and Since
and it follows that a contradiction.
p p pB p p p B p i k
B Bp p B p B
p p p p p
k
k i
j j j
j k j
1 2
1 2
1 2
1 1
1
, , ,. ,
| . |
| , | ,
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
Theorem 4.5 (The Division Algorithm)
.0 , with
uniqueexist then 0, with , If
brrqbaZq,r
therebZa, b
a: dividendb: divisorq: quotientr: remainder
Chapter 4 Properties of the Integers: Mathematical Induction
4.3 The Division Algorithm: Prime Numbers
Ex. 4.28 If Z and is composite, then there existsa prime such that and Proof: Since is composite, we can write where1 < and 1 < We claim that one of the integers
must be less than or equal to If not, then and give the contradiction
Without loss of generality, assume If is a prime, the result follows. Otherwise, by Lemma4.1, there exists a prime where So and
+n np p n p n
n n n nn n n n
n n nn n n n n n n
n n n n nn
p n p n p np n
| .,
., .
. .
| . |.
1 2
1 2
1 2
1 2 1 2
1
1
1 1
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Def. 4.2 For a, b in Z, a position integer c is said to be a commondivisor of a and b if c|a and c|b.
Def 4.3 Greatest Common Divisor
For any common divisor d of a and b, we have d|c. Then c is a greatest common divisor of a and b.
Theorem 4.6 For any a,b in Z+, there exists a unique c in Z+
that is the greatest common divisor of a and b.
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Proof: Given , Z let = { + | , ,+ > }. Since , by the Well - Ordering principle
has a least element . We claim that is a greatestcommon divisor of , . Since , = + for some
, . Consequently, if and | and | , then| + , so | . If | , = + , < < . Then = -
= - ( + ) = ( - ) + (- ) , so , contradictory to is the least in . Consequen
+a b S as bt s t Zas bt SS c c
a b c S c ax byx y Z d Z d a d bd ax by d c c a a qc r r c r a qc
a q ax by qx a qy b r Sc S
,0
01
tly, | and | . If both and satisfy, then and which implies
1
2 1 2 1
c a c b cc c c c c c c| | , .2 1 2
(existence and uniqueness)
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
From Theorem 4.6, gcd(a,b) is the smallest positive integer wecan write as a linear combination of a and b.
Integers a and b are called relative prime when gcd(a,b)=1.That is, when there exist x,y in Z with ax+by=1.
Ex. 4.30 Since gcd(42,70)=14, we can find x,y in Z with 42x+70y=14, or 3x+5y=1. By inspection, x=2, y=-1 is asolution. But, general solution? x=2-5k, y=-1+3k orx=2+5k, y=-1-3k..
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Theorem 4.7 Euclidean Algorithm
If , Z we apply the division algorithm as follows:
Then , the last nonzero remainder, equals gcd(
+a ba q b r r bb q r r r rr q r r r r
r q r r r r
r q r r r rr q r r r rr q r
r
i i i i i i
k k k k k k
k k k k k k
k k k
k
,,,,
,
,,
1 1 1
2 1 2 2 1
1 3 2 3 3 2
2 1 2 2 1
3 1 2 3 1 2
2 1 1
1 1
000
0
00
a b, ).
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Ex. 4.31 gcd(250,111)=?
250 111222 28 84 27 27 1
*2
*3
*1
gcd(250,111)=1, what is x, and y suchthat 250x+111y=1?
1=28-1(27)=28-1[111-3(28)]=(-1)111+4(28)=(-1)111+4[250-2(111)]=4(250)+(-9)(111)
x=4+111k, y=-9-250k
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Ex. 4.32 For any n in Z+, prove that 8n+3 and 5n+2 are relativeprime.
8n+3 5n+25n+23n+1 3n+1 2n+12n+1n 2n 1
gcd(8n+3,5n+2)=1. But we could alsoarrived at this conclusion if we hadnoticed that(8n+3)(-5)+(5n+2)(8)=1
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Ex. 4.34. Griffin has two unmarked containers. One containerholds 17 ounces and the other holds 55 ounces. Explain how Griffin can use his two containers to have exactly one ounce.
gcd(17,55)=1, 1=13(17)-4(55), Consequently, Griffin mustfill his smaller container 13 times and empty the contentsinto the larger container.
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Ex. 4.35. Diophantine equation
Find nonnegative integer solutions for 6x+10y=104.
Ans: 3x+5y=52, gcd(3,5)=1, 1=3(2)+5(-1), 52=3(104)+5(-52)x=104-5k, y=-52+3k, 104-5k 0 and -52+3k 0. Sok=18,19,20.
Theorem 4.8. If a,b,c in Z+, the Diophantine equcation ax+by=c has an integer solution if and only if gcd(a,b) divides c.
Chapter 4 Properties of the Integers: Mathematical Induction
4.4 The Greatest Common Divisor: the Euclidean Algorithm
Def 4.4. For a,b,c in Z+, c is called a common multiple of a,b ifc is a multiple of both a and b. Furthermore, c is the least commonmultiple if it is the smallest of all positive integers that are commonmultiple of a,b. We denote c by lcm(a,b).
Theorem 4.10 For a,b in Z+, ab=lcm(a,b)gcd(a,b)
Chapter 4 Properties of the Integers: Mathematical Induction
4.5 The Fundamental Theorem of Arithmetic
Lemma 4.2 If , Z and is a prime, then | | or | .
Lemma 4.3 Let Z for all 1 . If is a primeand | then | for some 1 .
+a b pp ab p a p b
a i n pp a a a p a i n
in i
1 2 ,
Ex 4.38 Show that 2 is irrational.
Proof: If not, let 2 where , Z and
gcd( , ) = 1. Then 2 =
which contradicts theearlier claim that gcd( , ) = 1.
+
a b a b
a b a b b a a
a a c a c b c b c
b b a ba b
/ ,
/ |
|
| | gcd( , ) ,
2 2 2 2 2
2 2 2 2 2 2
2
2 2
2 2 4 2 4 2
2 2 2
Chapter 4 Properties of the Integers: Mathematical Induction
4.5 The Fundamental Theorem of Arithmetic
Theorem 4.11 (The fundamental theorem of arithmetic) Everyinteger n>1 can be written as a product of primes uniquely, upto the order of the primes.
Ex. 4.41 For n in Z+, we want to count the number of positivedivisors of n.
By Theorem 4.11, = where foreach 1 , is a prime and If | , then
= where 0 for all 1 . So by the rule of product, the number of positive divisors of n is ( 1
n p p pi k p e m n
m p p p f ei k
e e e
e eke
i if f
kf
i i
k
k
k
1 2
1 2
2
1 2
1 2
0
1 1 1
,.
)( ) ( ).
Chapter 4 Properties of the Integers: Mathematical Induction
4.5 The Fundamental Theorem of Arithmetic
Ex. 4.43 Can we find three consecutive positive integers whoseproduct is a perfect square, that is, do there exist m,n in Z+ with(m)(m+1)(m+2)=n2?
Sol: Suppose m,n do exist. Since gcd(m,m+1)=gcd(m+1,m+2)=1,so for any prime p, if p|(m+1), then p|m and p|m+2. Furthermore,if p|(m+1), then p|n2. Since n2 is a perfect square, the exponentson p in the prime factorizations of both m+1 and n2 must be thesame even integer. So m+1 is a perfect square. And m(m+2) mustbe a perfect square too. But m2<m(m+2)<m2+2m+1<(m+1)2. Som(m+2) cannot be a perfect square. m,n do not exist.