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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. ⇤
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