CHAPTER 2

The Real Numbers

2.1. The Algebraic and Order Properties of RDefinition. A binary operation on a set F is a function B : F ⇥F ! F .

For the binary operations of + and ·, we replace B(a, b) by a + b and a · b,respectively.

Field Axioms of RThe real numbers are a field (as are the rational numbers Q and the complexnumbers C). That is, there are binary operations + and · defined on R 3��(A1) a + b = b + a 8a, b 2 R.

(A2) (a + b) + c = a + (b + c) 8a, b, c 2 R.

(A3) 9 0 2 R 3�� 0 + a = a and a + 0 = a 8a 2 R.

(A4) 8a 2 R,9 � a 2 R 3�� a + (�a) = 0 and (�a) + a = 0.

(M1) ab = ba 8a, b 2 R.

(M2) (ab)c = a(bc) 8a, b, c 2 R.

(M3) 9 1 2 R, 1 6= 0,3�� 1 · a = a and a · 1 = a 8a 2 R.

(M4) 8a 2 R, a 6= 0,9 1

a2 R 3�� a

⇣1

a

⌘= 1 and

⇣1

a

⌘a = 1.

(D) a(b + c) = ab + ac and (b + c)a = ba + ca 8a, b, c 2 R.

11

12 2. THE REAL NUMBERS

Some Properties of R

Theorem (2). If z, a 2 R 3�� z + a = a, then z = 0.

(i.e., the number 0 guaranteed by (A3) is unique.)

Proof. By (A4), 9 � a 2 R 3�� a + (�a) = 0.

Then

z

=A3

z + 0 = z +⇥a + (�a)

⇤=A2

(z + a) + (�a)

= a + (�a) = 0.

⇤

Theorem (3). Let a, b 2 R. Then a + x = b has the unique solutionx = (�a) + b.

(i.e., we are defining b� a.)

Proof.

a +⇥(�a) + b

⇤=A2

⇥a + (�a)

⇤+ b =

A40 + b =

A3b,

so (�a) + b is a solution.

For uniqueness, suppose y is any solution of the equation, i.e., a + y = b. Then

y

=A3

0 + y

=A4

⇥(�a) + a

⇤+ y

=A2

(�a) + (a + y)

= (�a) + b

⇤

2.1. THE ALGEBRAIC AND ORDER PROPERTIES OF R 13

Note.N ✓ Z ✓ Q ✓ R ✓ C| {z }

all fieldsThe last three all satisfy the field axioms, so the field axioms do not characterizeR. Recall that Q is closed under + and ·, i.e., if a, b 2 Q, then a + b 2 Q anda · b 2 Q.

Homework Pages 30-31 #4 (Hint: assume a 6= 0 and prove a = 1), 5 (Hint:1/(ab) = (1/a) · (1/b) if (1/a) · (1/b) does what 1/(ab) is supposed to do), 8b(1st part)

Order Properties of RR is an ordered field, i.e., the following properties are satisfied:

(1) (Trichotomy) For a, b 2 R, exactly one of the following is true: a < b,a = b, or a > b.

(2) (Transitive) For a, b, c 2 R, if a < b and b < c, then a < c.

(3) For a, b, c 2 R, if a < b, then a + c < b + c.

(4) For a, b, c 2 R, if a < b and c > 0, ac < bc.

Some Order Properties

Theorem (4). If a, b 2 R, then a < b () �a > �b.

Proof.

a < b ()a +

⇥(�a) + (�b)

⇤< b +

⇥(�a) + (�b)

⇤()⇥

a + (�a)⇤

+ (�b) < b +⇥(�b) + (�a)

⇤()

0 + (�b) <⇥b + (�b)

⇤+ (�a) ()

� b < 0 + (�a) () �b < �a () �a > �b

⇤

14 2. THE REAL NUMBERS

Theorem (5). If a, b, c 2 R, then a < b and c < 0 =) ac > bc

Proof. c < 0 =)TH4

�c > 0.

Then a(�c) < b(�c) =) �ac < �bc =)TH4

ac > bc. ⇤

Theorem (2.1.9). If a 2 R 3�� 0 a < ✏ 8✏ > 0, then a = 0.

Proof. Suppose a > 0.

Since 0 <1

2< 1, 0 < 1

2a < a.

Let ✏0 = 12a.

Then 0 < ✏0 < a, contradicting our hypothesis. Thus a = 0. ⇤

Problem (Page 31 #18). Let a, b 2 R, and suppose a b + ✏

(or a� ✏ b) 8✏ > 0. Then a b.

Proof. By way of contradiction, suppose b < a.

[Need to find an ✏ that gives a contradiction.]

Let ✏0 =1

2(a� b). Then

a� ✏0 = a� 1

2(a� b) =

1

2a +

1

2b >

1

2b +

1

2b = b,

contradicting our hypothesis. Thus a b. ⇤

2.1. THE ALGEBRAIC AND ORDER PROPERTIES OF R 15

Theorem (Arithmetic-Geometric Mean Inequality).

Suppose a, b > 0. Then pab 1

2(a + b)

with equality holding () a = b.Proof.

(1) Suppose a 6= b. Thenp

a > 0,p

b > 0, andp

a 6=p

b.

Thusp

a�p

b 6= 0 =)�pa�

pb�2

> 0 =)a� 2

pap

b + b > 0 =)�2p

ap

b > �(a + b) =)p

ab <1

2(a + b)

(2) If a = b,p

ab =p

a2 = |a| = a =1

2(2a) =

1

2(a + a) =

1

2(a + b).

(3) Ifp

ab =1

2(a + b),

ab =1

4

�a + b

�2=)

4ab = a2 + 2ab + b2 =)0 = a2 � 2ab + b2 =)

0 = (a� b)2 =)0 = a� b =)

a = b.

⇤

16 2. THE REAL NUMBERS

Theorem (Bernoulli’s Inequality). If x > �1, then

(1 + x)n � 1 + nx 8n 2 N.Proof.

[We use MI to prove this.]

Let S ✓ N for which (1 + x)n � 1 + nx.

1 2 S since (1 + x)1 = 1 + 1 · x.

Suppose k 2 S, i.e., (1 + x)k � 1 + kx. Then

(1 + x)k+1

= (1 + x)k(1 + x)

� (1 + kx)(1 + x)

= 1 + (k + 1)x + kx2

� 1 + (k + 1)x

Thus S = N by MI . ⇤

Note. We now have

N ✓ Z ✓ Q ✓ R| {z }ordered fields

✓ C.

2.1. THE ALGEBRAIC AND ORDER PROPERTIES OF R 17

Problem (Page 30 #16d). Find all x 2 R 3�� 1

x< x2.

Solution.1

x< x2 ()

x2 � 1

x> 0 ()

1

x(x3 � 1) > 0 ()n1

x> 0 and x3 � 1 > 0

oorn1

x< 0 and x3 � 1 < 0

o()n

x > 0 and x3 > 1o

ornx < 0 and x3 < 1

o()n

x > 0 and x > 1o

ornx < 0 and x < 1

o()

x > 1 or x < 0.

⇤

Homework Page 31 # 13 (Hint: for =), suppose, WLOG, a 6= 0, then reacha contradiction), 16c, 20

18 2. THE REAL NUMBERS

2.2. Absolute Value and the Real Line

Definition.

8a 2 R, |a| =

(a if a � 0

�a if a < 0

Theorem (2.2.2).

(a) |ab| = |a||b| 8a, b 2 R.

(b) |a|2 = a2 8a 2 R.

(c) If c � 0, then |a| c () �c a c.

(c’) If c > 0, then |a| < c () �c < a < c.

(d) �|a| a |a| 8a 2 R.Theorem (2.2.3 — Triangle Inequality).

8a, b 2 R, |a + b| |a| + |b|.Proof.

[We wish to use Theorem 2.2.2(c)]

By Theorem 2.2.2(d),

�|a| a |a| and � |b| b |b|.Then

��|a| + |b|

�= �|a|� |b| a + b |a| + |b| =)

|a + b| |a| + |b|by Theorem 2.2.2(c) ⇤

2.2. ABSOLUTE VALUE AND THE REAL LINE 19

Corollary (2.2.4). If a, b 2 R, then

(a)��|a|� |b|

�� |a� b|(b) |a� b| |a| + |b|

Note. These are also referred to as triangle inequalities.Proof. [We use a “smuggling” technique.]

(a)

|a| = |a� b + b| |a� b| + |b| =)|a|� |b| |a� b|.

|b| = |b� a + a| |b� a| + |a| =)|b|� |a| |b� a| =)�|a� b| |a|� |b|.

Thus�|a� b| |a|� |b| |a� b| =)��|a|� |b|

�� |a� b|by Theorem 2.2.2.(c)

(b) Just replace b by (�b) in the triangle inequality. ⇤

Corollary (2.2.5). 8a1, a2, . . . , an 2 R,

|a1 + a2 + · · · + an| |a1| + |a2| + · · · + |an|.

20 2. THE REAL NUMBERS

Problem (Page 36 #10a). Find all x 2 R 3�� |x� 1| > |x + 1|.Solution.

First, considering the values of x that make one of the absolute values 0,

x < �1 or � 1 x < 1 or x � 1.

If x < �1,

|x� 1| > |x + 1| =) �x + 1 > �x� 1 =) 1 > �1,

so x is a solution.

If �1 x < 1,

|x� 1| > |x + 1| =) �x + 1 > x + 1 =) 0 > 2x =) x < 0,

so �1 x < 0 are solutions.

If x > 1,|x� 1| > |x + 1| =) x� 1 > x + 1 =) �1 > 1,

which is impossible.

Therefore, {x : x < 0} is the solution set. ⇤

Recall. |a� b| give the distance from a to b on the number line.

Definition (2.2.7). Let a 2 R and ✏ > 0. Then the ✏-neighborhood of ais the set

V✏(a) = {x 2 R : |x� a| < ✏}.Corollary.

x 2 V✏(a) () �✏ < x� a < e () a� ✏ < x < a + ✏.

2.2. ABSOLUTE VALUE AND THE REAL LINE 21

Problem (Page 36 # 17). If a, b 2 R and a 6= b, then 9 ✏-neighborhoodsU of a and V of b 3�� U \ V = ;.

Proof.

WLOG (without loss of generality), suppose a < b. Now

a < a +1

3(b� a) < a +

2

3(b� a) = b� 1

3(b� a) < b.

ChooseU = V1

3(b�a)(a) and V = V13(b�a)(b)

ThenU \ V = ;.

⇤

Homework Page 36 # 12, 13

22 2. THE REAL NUMBERS

2.3. The Completeness Property of RGOAL — to characterize the real numbers

Definition (2.3.1). Let ; 6= S ✓ R.

(a) S is bounded above if 9 u 2 R 3�� s u 8s 2 S.

Then u is an upper bound (u.b.) of S.

(b) S is bounded below if 9 w 2 R 3�� w s 8s 2 S.

Then w is a lower bound (l.b.) of S.

(c) S is bounded if it is both bounded above and below.

Otherwise it is unbounded.

Corollary.

(a) ⌫ 2 R is not an u.b. of S if

9 s0 2 S 3�� v < s0.

(b) z 2 R is not an l.b. of S if

9 s00 2 S 3�� s00 < z.Example.

(1) S = {x 2 R : x � 5}.— S is not bounded above since, if v were an upper bound,

max{5, v + 1} > v and max{5, v + 1} 2 S,

a contradiction.

— S is bounded below by any w 5.

2.3. THE COMPLETENESS PROPERTY OF R 23

(2) (from Page 39 #4) S4 =n

1 � (�1)n

n: n 2 N

o. Find a lower bound and

upper bound for S4.

Solution. 0 <1

n 1 and �1 �1

n< 0 8n 2 N =)

�1 (�1)n

n 1 8n 2 N =)

�1 �(�1)n

n 1 8n 2 N =)

0 1� (�1)n

n 2 8n 2 N.

Thus 0 is a lower bound and 2 is an upper bound of S4. ⇤

Definition (2.3.2). Let ; 6= S ✓ R.

(a) If S is bounded above, then u is a supremum (or least upper bound) of S,written u = sup S, if

(1) u is an upper bound of S;

(2) if v is any upper bound of S, u v.

(b) If S is bounded below, then w is an infimum (or greatest lower bound) ofS, written w = inf S, if

(1) w is a lower bound of S;

(2) if t is any lower bound of S, t w.

24 2. THE REAL NUMBERS

Lemma (2.3.3). Suppose ; 6= S ✓ R.

(a) u = sup S ()(1) s u 8s 2 S,

(2) if v < u, Then 9 s0 2 S 3�� v < s0.

(b) w = inf S ()(1) w s 8s 2 S,

(2) if w < z, then 9 s00 2 S 3�� s00 < z.

Lemma (2.3.4).

(Property S) Let ; 6= S ✓ R. u = sup S ()(1) u is an u.b. for S;

(2) 8✏ > 0,9 s✏ 2 S 3�� u� ✏ < s✏.

(Property I) Let ; 6= S ✓ R. w = inf S ()(1) w is a l.b. for S;

(2) 8✏ > 0,9 s✏ 2 S 3�� w + ✏ > s✏.

Proof. (of Property S)

(=)) Assume u = sup S. Then, by definition, u is an u.b. for S.

Let ✏ > 0 be given. Then u� ✏ < u, so by Lemma 2.3.3

9 s0 2 S 3�� u� ✏ < s0. Let s✏ = s0.

((=) (1) u is an u.b. for S =) s u 8s 2 S.

(2) Suppose v < u.

Let ✏ = u� v. Then

9 s✏ 2 S 3�� u� (u� v) < s✏ =) v < s✏. Let s0 = s✏ =) v < s0.

Then u = sup S by Lemma 2.3.3. ⇤

2.3. THE COMPLETENESS PROPERTY OF R 25

Example.

(1) If ; 6= S ✓ R is a finite set,

sup S is the largest element of S and

inf S is the least element of S.

(2) S =�x 2 R : 2 < x 5

.

(a) sup S = 5 2 S.Proof.

(1) 8x 2 S, x 5 =) 5 is an u.b. of S.

(2) Let ✏ > 0 be given. 5� ✏ < 5 2 S. Let s✏ = 5.

Then 5 = sup S by Property S. ⇤

(b) inf S = 2 /2 S.Proof.

(1) 8 x 2 S, 2 < x =) 2 x =) 2 is a l.b. of S.

(2) Let ✏ > 0 be given.

If ✏ > 3, 2 + ✏ > 2 + 3 = 5 2 S, so let s✏ = 5.

If ✏ 3,

2 < 2 +✏

2 2 +

3

2=

7

2 5, so 2 +

✏

22 S.

Then 2 + ✏ > 2 +✏

22 S, so let s✏ = 2 +

✏

2.

Thus 2 = inf S by Property I . ⇤

Axiom (Completeness Property of R or Supremum Property of R).

Every non-empty set of real numbers that has an upper bound also has asupremum in R.

Note. Thus R is a complete ordered field, while Q is not.

26 2. THE REAL NUMBERS

Theorem (Infimum Property of R).

Every nonempty set of real numbers that is bounded below has an infimumin R.

Proof. Suppose ; 6= S ✓ R is bounded below. Then

; 6= S0 =�� s : s 2 S

is bounded above. This is true since

w a lower bound of S =) w s 8s 2 S =)�s �w 8s 2 S =) �w is an upper bound of S0.

By the Completeness Property, u = sup S0 exists. We claim �u = inf S.

(1) u = sup S0 =) �s u 8s 2 S =) �u s 8s 2 S =)�u is a lower bound of S.

(2) Let ✏ > 0 be given. By Property S, 9 (�s✏) 2 S0, where s✏ 2 S,

3�� u� ✏ < �s✏ =) �u + ✏ > s✏.

Thus �u = inf S = � sup�� s : s 2 S

by Property I. ⇤

2.3. THE COMPLETENESS PROPERTY OF R 27

Problem (Page 39 #4). Let S4 =

⇢1� (�1)n

n: n 2 N

�.

Find inf S4 and sup S4.Solution.

We showed earlier that 0 is a lower bound of S4 and 2 is an upper bound since

0 1� (�1)n

n 2 8n 2 N.

Now S4 =

⇢2,

1

2,4

3,3

4,6

5,5

6,8

7,7

8, . . .

�.

We claim inf S4 =1

2.

(1) If n is odd,

1� (�1)n

n= 1� �1

n= 1 +

1

n=

n + 1

n> 1 >

1

2.

If n is even,1

2n � 1, so

1� (�1)n

n= 1� 1

n=

n� 1

n�

n� 12n

n=

12n

n=

1

2.

Thus1

2is a lower bound for S4.

(2) Given ✏ > 0,1

22 S4 and

1

2+ ✏ >

1

2, so choose s✏ =

1

2.

By Property I ,1

2= inf S4.

[Finding and proving sup S4 is Homework]

⇤

Homework

Pages 39-40 #2, 4 (prove sup S4 =?), 5bc.

28 2. THE REAL NUMBERS

2.4. Applications of the Supremum Property

Two questions

Is there a largest natural number?

Is N bounded above in R?

Theorem (2.4.3 —Archimedean Property). If x 2 R, then 9 nx 2 N 3��x < nx.

Proof. Suppose n x 8n 2 N. [We are using contradiction.]

Then x is an u.b. of N, so

N has a supremum. Let u = sup N.

By Property S, 9m 2 N 3�� u� 1 < m.

Then u < m + 1 2 N,

contradicting that u is an upper bound of N.

Thus 9 nx 2 N 3�� x < nx. ⇤

Note. The next 3 corollaries can also be referred to as Archimedean.

Corollary (2.4.4). If S =n1

n: n 2 N

o, inf S = 0.

Proof. Clearly, 0 is a lower bound of S, so inf S exists.

Let w = inf S, so w � 0.

8✏ > 0,9 n 2 N 3�� 1

✏< n (Archimedean) =) 1

n< ✏. Thus

0 w 1

n< ✏ =)

w = 0 by Theorem 2.1.9. Thus inf S = 0. ⇤

2.4. APPLICATIONS OF THE SUPREMUM PROPERTY 29

Corollary (2.4.5). If t > 0, then 9 nt 2 N 3�� 0 <1

nt< t.

Proof. Since infn1

n: n 2 N

o= 0 and t > 0,

t is not a lower bound ofn1

n: n 2 N

o, so

9 nt 2 N 3�� 0 <1

nt< t. ⇤

Corollary (2.4.6). If y > 0, then 9 ny 2 N 3�� ny � 1 y < ny.

Proof. Let Ey = {m 2 N : y < m}.

By the Archimedean property, Ey 6= ;.By Well-Ordering, Ey has a least element, say ny.

Then ny � 1 62 Ey, so

ny � 1 y < ny. ⇤

30 2. THE REAL NUMBERS

Theorem (Density Theorem). If x, y 2 R 3�� x < y, then 9 r 2Q 3�� x < r < y.

Proof. WLOG, assume x > 0.⇥Suppose the theorem is true for x > 0.

Then, if x 0, the Archmedean Property says that

9n 2 N 3�� �x < n, so x + n > 0.

Since x + n < y + n,9r 2 Q 3�� x + n < r < y + n =)x < r � n < y and r � n 2 Q.

⇤Now 0 < x < y =) y � x > 0 =) (by Corollary 2.4.5)

9n 2 N 3�� 1

n< y � x =) nx + 1 < ny.

Applying Corollary 2.4.6 to nx > 0,

9 m 2 N 3�� m� 1 nx < m.

Then m nx + 1 < ny =)nx < m < ny =)x <

m

n< y. Let r =

m

n. ⇤

Corollary (2.4.9). If x, y 2 R with x < y, then 9 an irrational z 2R 3�� x < z < y.

Proof. x < y =) xp2

<yp2.

By density, 9 r 2 Q 3�� xp2

< r <yp2

and r 6= 0 (why?).

Then x p

2r < y. Note thatp

2r is irrational as the product of a rationalwith an irrational. Let z =

p2r. ⇤

2.5. INTERVALS 31

Homework

Pages 44-46 #1 (Use Property S. You are likely to need the ArchimedeanProperty in Part (2)), 2 (Just find and prove sup S - same hint as for #1).

Page 46 #19 (Look at proof of Corollary 2.4.9 as a model).

2.5. Intervals

Theorem (2.5.1 — Characterization of Intervals). If S is a subset of Rthat contains at least two points and has the property

if x, y 2 S and x < y, then [x, y] ✓ S,

then S is an interval.

(a, b) = {x 2 R : a < x < b}[a, b] = {x 2 R : a x b}(a, b] = {x 2 R : a < x b}(a,1) = {x 2 R : x > a}(�1, b] = {x 2 R : x b}

I = [0, 1] is called the unit interval.

Definition. A sequence of intervals In, n 2 N, is nested if the followingchain of inclusions holds:

I1 ◆ I2 ◆ I3 ◆ · · · ◆ In ◆ In+1 ◆ · · ·Example.

(1) In =h0,

1

n

i8n 2 N.

1\n=1

In = {0}.

32 2. THE REAL NUMBERS

(2) Jn =⇣0,

1

n

⌘8n 2 N.

1\n=1

Jn = ;.

(3) Kn = (n,1) 8n 2 N.1\

n=1

Kn = ;.

Theorem (2.5.2 — Nested Intervals Property). If In = [an, bn], n 2 N,is a nested sequence of closed, bounded intervals, then 9 ⇠ 2 R 3�� ⇠ 2In 8 n 2 N.

Proof. By nesting, In ✓ I1 8 n 2 N, so an b1 8 n 2 N. Thus

; 6= {an : n 2 N} = A

is bounded above.

Let ⇠ = sup A, so an ⇠ 8 n 2 N .

[To show ⇠ bn 8 n 2 N.]

Let n 2 N be given (so bn is arbitrary, but fixed).

[To show bn is a u.b. of A, so then ⇠ bn.]

(1) Suppose n k. Then In � Ik, so ak bk bn.

(2) Suppose n > k. Then Ik ◆ In, so ak an bn.

Thus, 8 k 2 N, ak bn =) bn is an u.b. of A =) ⇠ bn. ⇤

2.5. INTERVALS 33

Theorem (2.5.3). If In = [an, bn], n 2 N, is a nested sequence of closed,bounded intervals 3��

inf{bn � an : n 2 N} = 0,

then 9 a unique ⇠ 2 R 3�� ⇠ 2 In 8 n 2 N.

Proof. Let ⌘ = inf{bn : n 2 N}. Using an argument similar to that of theprevious theorem, we have an ⌘ 8 n 2 N, so ⇠ = sup{an : n 2 N} ⌘.

Thus, x 2 In 8 n 2 N () ⇠ x ⌘.

Let ✏ > 0 be given. [To show ⌘ � ⇠ = 0.]

By Property I , since inf{bn � an : n 2 N} = 0,

9 m 2 N 3�� 0 + ✏ > bm � am.

Then 0 ⌘ � ⇠ bm � am < ✏.

Thus 0 ⌘ � ⇠ < ✏ 8 ✏ > 0.

By Theorem 2.1.9, ⌘ � ✏ = 0,

so ⌘ = ⇠ is the only point belonging to In 8n 2 N. ⇤