Solutions Manual
Introduction To Real Analysis
By
William F. Trench
( All Chapters Included - 100% Verified Solutions )
1
, Section 1.1 The Real Number System 1
CHAPTER 1
THE REAL NUMBERS
1.1 THE REAL NUMBER SYSTEM
1:1:1. Note that ja bj D max.a; b/ min.a; b/.
(a) a C b C ja bj D a C b C max.a; b/ min.a; b/ D 2 max.a; b/.
(b) a C b ja bj D a C b max.a; b/ C min.a; b/ D 2 min.a; b/.
ˇ ˇ
(c) Let ˛ D aCbC2cCja bjCˇaCb 2cCja bjˇ. From (a), ˛ D 2 Œmax.a; b/ C c C j max.a; b/ cj Ddf
ˇ. From (a) with a and b replaced by max.a; b/ and c, ˇ D 4 max .max.a; b/; c/ D
4 max.a; b; c/.
ˇ ˇ
(d) Let ˛ D aCbC2c ja bj ˇaCb 2c ja bjˇ. From (b), ˛ D 2 Œmin.a; b/ C c j min.a; b/ cj Ddf
ˇ. From (a) with a and b replaced by min.a; b/ and c, ˇ D 4 min .min.a; b/; c/ D
4 min.a; b; c/.
1:1:2. First verify axioms A-E:
Axiom A. See Eqns. (1.1.1) and (1.1.2).
Axiom B. If a D 0 then .a C b/ C c D b C c and a C .b C c/ D b C c, so .a C b/ C
c D a C .b C c/. Similar arguments apply if b D 0 or c D 0. The remaining case is
a D b D c D 1. Since .1 C 1/ C 1 D 0 C 1 D 1 and 1 C .1 C 1/ D 1 C 0 D 1, addition
is associative. Since
0; unless a D b D c D 1;
.ab/c D a.bc/ D
1; if a D b D c D 1;
multiplication is associative.
Axiom C. Since
0; if a D 0;
a.b C c/ D ab C ac D
b C c; if a D 1;
the distributive law holds.
Axiom D. Eqns. (1.1.1) and (1.1.2) imply that 0 and 1 have the required properties.
2
,2 Chapter 1 The Real Numbers
Axiom E. Eqn. (1.1.1) implies that 0 D 0 and 1 D 1; Eqn. (1.1.2) implies that 1=1 D 1.
To see that the field cannot be ordered suppose that 0 < 1. Adding 1 to both sides yields
0 C 1 < 1 C 1, or 1 < 0, a contradiction. On the other hand, if 1 < 0, then 1 C 1 < 0 C 1,
so 0 < 1, also a contradiction.
p p
1:1:3. If 2 is rational we can write 2 D m=n, where either m is odd or n is odd. Then
m2 D 2n2 , so m is even; thus, m D 2m1 where m1 is an integer. Therefore, 4m21 D 2n2 ,
so n2 D 2m21 and n is also even, a contradiction.
p p
1:1:4. If p is rational we can write p D m=n where either m or n is not divisible
2 2
by p. Then m D pn , so m is divisible by p; thus, m D pm1 where m1 is an integer.
Therefore, p 2 m21 D pn2 , so n2 D pm21 and n is also divisible by p, a contradiction.
1:1:6. If S is bounded below then T is bounded above, so T has a unique supremum, by
Theorem 1.1.3. Denote sup T D ˛. Then (i) if x 2 S then x ˛, so x ˛; (ii) if
> 0 there is an x0 2 T such that x0 < ˛ , so x0 > ˛ C . Therefore, there is
an ˛ with properties (i) and (ii). If (i) and (ii) hold with ˛ replaced by ˛1 then ˛1 is a
supremum of T , so ˛1 D ˛ by the uniqueness assertion of Theorem 1.1.3.
1:1:7. (a) If x 2 S , then inf S x sup S , and the transitivity of implies (A), with
equality if and only if S contains exactly one point.
(b) There are three cases: (i) If S is bounded below and unbounded above, then inf S D ˛
(finite) and sup S D 1 from (13); (ii) If S is unbounded below and bounded above, then
inf S D 1 from (14) and sup S D ˇ (finite); (iii) If S is unbounded below and above
then sup S D 1 from (13) and inf S D 1 from (14). In all three cases (12) implies (A).
1:1:8. Let a D inf T and b D sup S . We first show that a D b. If a < b then a <
.a C b/=2 < b. Since b D sup S , there is an s0 2 S such that s0 > .a C b/=2. Since
a D inf T , there is a t0 2 T such that t0 < .a C b/=2. Therefore, t0 < s0 , a contradiction.
Hence a b. If a > b there is an x such that b < x < a. From the definitions of a and
b, x 62 T and x 62 S , a contradiction. Hence a D b. Let ˇ D a.D b/. Since a and b are
uniquely, defined so is ˇ. If x > ˇ then x … S (since ˇ D sup S ), so x 2 T . If x < ˇ then
x … T (since ˇ D inf T ), so x 2 S .
1:1:9. Every real number is in either S or T . T is nonempty because U is bounded. S is
nonempty because if u 2 U and x < u then x 2 S . If s 2 S there is a u 2 U such that
u > s, since s is not an upper bound of U . If t 2 T then t u, since t is an upper bound
of U . Since u > s and t u, t > s. Therefore, S and T satisfy the conditions imposed
in Exercise 1.1.8, so there is a number ˇ such that every number greater than ˇ is an upper
bound of U and no number less that ˇ is an upper bound of U . However, ˇ is also an upper
bound of U (if not, there would be a u0 2 U such that u0 > ˇ, which is impossible, since
if u0 > ˇ then every number in .ˇ; u0 / is an upper bound of U ). Therefore, ˇ D sup U .
1:1:10. (a) Let ˛ D sup S and ˇ D sup T . If x D s C t then x ˛ C ˇ. If > 0 choose s0
in S and t0 in T such that s0 > ˛ =2 and t0 > ˇ =2. Then x0 D s0 C t0 > ˛ C ˇ
and ˛ C ˇ D sup.S C T / by Theorem 1.1.3. This proves (A). The proof of (B) by mean
of Theorem 1.1.8 is similar.
(b) For (A), suppose that S is not bounded above. Then sup S D sup.S C T / D 1. Since
sup T > 1 if T is nonempty, (A) holds. The proof of (B) is similar.
3
, Section 1.1 The Real Number System 3
1:1:11. Apply Exercise 1.1.10 with S and T replaced by S and T .
1:1:12. If a D 0 then T D fbg, so inf T D sup T D b.
Now suppose that a ¤ 0 and let ˛ D inf S and ˇ D sup S . From the definitions of ˛ and
ˇ,
˛ s and s ˇ for all s 2 S; .A/
and if > 0 is given there are elements s1 and s2 of S such that
s1 < ˛ C =jaj and s2 > ˇ =jaj: .B/
C ASE 2. If a > 0, multiplying the inequalities in (A) by a shows that
a˛ as and as aˇ for all s 2 S:
Therefore,
a˛ C b as C b for all s 2 S; .C/
as C b aˇ C b for all s 2 S: .D/
Multiplying the inequalities in (B) by a shows that
as1 < a˛ C and as2 > aˇ ;
since jaj D a. Therefore,
as1 C b < .a˛ C b/ C ; .E/
as2 C b > .aˇ C b/ : .F/
Now (C) and (E) imply that a˛ C b D inf T , while (D) and (F) imply that aˇ C b D sup T .
C ASE 3. Suppose that a < 0. Multiplying the inequalities in (A) by a shows that
a˛ as and as aˇ for all s 2 S:
Therefore,
a˛ C b as C b for all s 2 S; .G/
as C b aˇ C b for all s 2 S: .H/
Multiplying the inequalities in (B) by a shows that
as1 > a˛ and as2 < aˇ C ;
since jaj D a. Therefore,
as1 C b > .a˛ C b/ ; .I/
as2 C b < .aˇ C b/ C : .J/
Now (G) and (I) imply that a˛ C b D sup T , while (H) and (J) imply that aˇ C b D inf T .
4
Introduction To Real Analysis
By
William F. Trench
( All Chapters Included - 100% Verified Solutions )
1
, Section 1.1 The Real Number System 1
CHAPTER 1
THE REAL NUMBERS
1.1 THE REAL NUMBER SYSTEM
1:1:1. Note that ja bj D max.a; b/ min.a; b/.
(a) a C b C ja bj D a C b C max.a; b/ min.a; b/ D 2 max.a; b/.
(b) a C b ja bj D a C b max.a; b/ C min.a; b/ D 2 min.a; b/.
ˇ ˇ
(c) Let ˛ D aCbC2cCja bjCˇaCb 2cCja bjˇ. From (a), ˛ D 2 Œmax.a; b/ C c C j max.a; b/ cj Ddf
ˇ. From (a) with a and b replaced by max.a; b/ and c, ˇ D 4 max .max.a; b/; c/ D
4 max.a; b; c/.
ˇ ˇ
(d) Let ˛ D aCbC2c ja bj ˇaCb 2c ja bjˇ. From (b), ˛ D 2 Œmin.a; b/ C c j min.a; b/ cj Ddf
ˇ. From (a) with a and b replaced by min.a; b/ and c, ˇ D 4 min .min.a; b/; c/ D
4 min.a; b; c/.
1:1:2. First verify axioms A-E:
Axiom A. See Eqns. (1.1.1) and (1.1.2).
Axiom B. If a D 0 then .a C b/ C c D b C c and a C .b C c/ D b C c, so .a C b/ C
c D a C .b C c/. Similar arguments apply if b D 0 or c D 0. The remaining case is
a D b D c D 1. Since .1 C 1/ C 1 D 0 C 1 D 1 and 1 C .1 C 1/ D 1 C 0 D 1, addition
is associative. Since
0; unless a D b D c D 1;
.ab/c D a.bc/ D
1; if a D b D c D 1;
multiplication is associative.
Axiom C. Since
0; if a D 0;
a.b C c/ D ab C ac D
b C c; if a D 1;
the distributive law holds.
Axiom D. Eqns. (1.1.1) and (1.1.2) imply that 0 and 1 have the required properties.
2
,2 Chapter 1 The Real Numbers
Axiom E. Eqn. (1.1.1) implies that 0 D 0 and 1 D 1; Eqn. (1.1.2) implies that 1=1 D 1.
To see that the field cannot be ordered suppose that 0 < 1. Adding 1 to both sides yields
0 C 1 < 1 C 1, or 1 < 0, a contradiction. On the other hand, if 1 < 0, then 1 C 1 < 0 C 1,
so 0 < 1, also a contradiction.
p p
1:1:3. If 2 is rational we can write 2 D m=n, where either m is odd or n is odd. Then
m2 D 2n2 , so m is even; thus, m D 2m1 where m1 is an integer. Therefore, 4m21 D 2n2 ,
so n2 D 2m21 and n is also even, a contradiction.
p p
1:1:4. If p is rational we can write p D m=n where either m or n is not divisible
2 2
by p. Then m D pn , so m is divisible by p; thus, m D pm1 where m1 is an integer.
Therefore, p 2 m21 D pn2 , so n2 D pm21 and n is also divisible by p, a contradiction.
1:1:6. If S is bounded below then T is bounded above, so T has a unique supremum, by
Theorem 1.1.3. Denote sup T D ˛. Then (i) if x 2 S then x ˛, so x ˛; (ii) if
> 0 there is an x0 2 T such that x0 < ˛ , so x0 > ˛ C . Therefore, there is
an ˛ with properties (i) and (ii). If (i) and (ii) hold with ˛ replaced by ˛1 then ˛1 is a
supremum of T , so ˛1 D ˛ by the uniqueness assertion of Theorem 1.1.3.
1:1:7. (a) If x 2 S , then inf S x sup S , and the transitivity of implies (A), with
equality if and only if S contains exactly one point.
(b) There are three cases: (i) If S is bounded below and unbounded above, then inf S D ˛
(finite) and sup S D 1 from (13); (ii) If S is unbounded below and bounded above, then
inf S D 1 from (14) and sup S D ˇ (finite); (iii) If S is unbounded below and above
then sup S D 1 from (13) and inf S D 1 from (14). In all three cases (12) implies (A).
1:1:8. Let a D inf T and b D sup S . We first show that a D b. If a < b then a <
.a C b/=2 < b. Since b D sup S , there is an s0 2 S such that s0 > .a C b/=2. Since
a D inf T , there is a t0 2 T such that t0 < .a C b/=2. Therefore, t0 < s0 , a contradiction.
Hence a b. If a > b there is an x such that b < x < a. From the definitions of a and
b, x 62 T and x 62 S , a contradiction. Hence a D b. Let ˇ D a.D b/. Since a and b are
uniquely, defined so is ˇ. If x > ˇ then x … S (since ˇ D sup S ), so x 2 T . If x < ˇ then
x … T (since ˇ D inf T ), so x 2 S .
1:1:9. Every real number is in either S or T . T is nonempty because U is bounded. S is
nonempty because if u 2 U and x < u then x 2 S . If s 2 S there is a u 2 U such that
u > s, since s is not an upper bound of U . If t 2 T then t u, since t is an upper bound
of U . Since u > s and t u, t > s. Therefore, S and T satisfy the conditions imposed
in Exercise 1.1.8, so there is a number ˇ such that every number greater than ˇ is an upper
bound of U and no number less that ˇ is an upper bound of U . However, ˇ is also an upper
bound of U (if not, there would be a u0 2 U such that u0 > ˇ, which is impossible, since
if u0 > ˇ then every number in .ˇ; u0 / is an upper bound of U ). Therefore, ˇ D sup U .
1:1:10. (a) Let ˛ D sup S and ˇ D sup T . If x D s C t then x ˛ C ˇ. If > 0 choose s0
in S and t0 in T such that s0 > ˛ =2 and t0 > ˇ =2. Then x0 D s0 C t0 > ˛ C ˇ
and ˛ C ˇ D sup.S C T / by Theorem 1.1.3. This proves (A). The proof of (B) by mean
of Theorem 1.1.8 is similar.
(b) For (A), suppose that S is not bounded above. Then sup S D sup.S C T / D 1. Since
sup T > 1 if T is nonempty, (A) holds. The proof of (B) is similar.
3
, Section 1.1 The Real Number System 3
1:1:11. Apply Exercise 1.1.10 with S and T replaced by S and T .
1:1:12. If a D 0 then T D fbg, so inf T D sup T D b.
Now suppose that a ¤ 0 and let ˛ D inf S and ˇ D sup S . From the definitions of ˛ and
ˇ,
˛ s and s ˇ for all s 2 S; .A/
and if > 0 is given there are elements s1 and s2 of S such that
s1 < ˛ C =jaj and s2 > ˇ =jaj: .B/
C ASE 2. If a > 0, multiplying the inequalities in (A) by a shows that
a˛ as and as aˇ for all s 2 S:
Therefore,
a˛ C b as C b for all s 2 S; .C/
as C b aˇ C b for all s 2 S: .D/
Multiplying the inequalities in (B) by a shows that
as1 < a˛ C and as2 > aˇ ;
since jaj D a. Therefore,
as1 C b < .a˛ C b/ C ; .E/
as2 C b > .aˇ C b/ : .F/
Now (C) and (E) imply that a˛ C b D inf T , while (D) and (F) imply that aˇ C b D sup T .
C ASE 3. Suppose that a < 0. Multiplying the inequalities in (A) by a shows that
a˛ as and as aˇ for all s 2 S:
Therefore,
a˛ C b as C b for all s 2 S; .G/
as C b aˇ C b for all s 2 S: .H/
Multiplying the inequalities in (B) by a shows that
as1 > a˛ and as2 < aˇ C ;
since jaj D a. Therefore,
as1 C b > .a˛ C b/ ; .I/
as2 C b < .aˇ C b/ C : .J/
Now (G) and (I) imply that a˛ C b D sup T , while (H) and (J) imply that aˇ C b D inf T .
4
