ALL 7 CHAPTERS COVERED
SOLUTIONS MANUAL
, 3
Contents
CHAPTER 1 Sets, Numbers, and Functions . . . . . . . . . . . . . . . . . . . . . . . . 1
CHAPTER 2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
CHAPTER 3 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
CHAPTER 4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
CHAPTER 5 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
CHAPTER 6 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
CHAPTER 7 Commutation of Limit Operations . . . . . . . . . . . . . . . . . . . 163
iii
©2010 Elsevier, Inc.
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C H A P TER 1
Sets, Numbers, and Functions
1.1. No, because T ∧ F is F.
1.2. Yes, because F ∨ T is T. If ∼ (p ∧ q) is T(F), then so is (∼ p∨ ∼ q).
1.3. (a)
p q ∼p ∼p∨q p→q
T T F T T
T F F F F
F T T T T
F F T T T
(b)
p q p→q q → p (p → q) ∧ (q → p) p ↔ q
T T T T T T
T F F T F F
F T T F F F
F F T T T T
(c)
p q ∼p ∼q p→q ∼ q →∼ p (p → q) ↔ (∼ q →∼ p)
T T F F T T T
T F F T F F T
F T T F T T T
F F T T T T T
1.4. (a), (b), (f ) hold.
1.5. A Right Distributive Law is already implied by R2(b), R4: (y + z)x =
x(y + z) = xy + xz = yx + zx. Statements analogous to field axioms
R2(b), R6(b) would fail for 3 × 3 matrices.
1.6. (a) For <Z5 , ⊕, ⊗>, axioms analogous to R1 – R5 are inherited from
R. The additive inverses of 0, 1, 2, 3, 4 are 0, 4, 3, 2, 1, respectively,
Advanced Calculus. DOI: 10.1016/B978-0-12-384694-5.00001-5
Copyright c 2010, Elsevier Inc. All rights reserved. 1
©2010 Elsevier, Inc.
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2 C H A P T E R 1: Sets, Numbers, and Functions
and the multiplicative inverses of 1, 2, 3, 4, are 1, 3, 2, 4, respectively.
Hence, an axiom analogous to R6 holds, so <Z5 , ⊕, ⊗> is a field.
(b) For Z5 suppose that P ⊆ {1, 2, 3, 4} is nonempty; let x ∈ P. Addition
of x to itself a sufficient number of times produces all members of
{1, 2, 3, 4}, so P = {1, 2, 3, 4}. But then −x ∈ P, which is not allowed
by Axiom R7(b), so P = ∅.
(c) If ⊕, ⊗ are defined modularly as with <Z5 , ⊕, ⊗> at the start of Exer-
cise 1.6, then <Z7 , ⊕, ⊗> and <Z11 , ⊕, ⊗> are found to be finite
fields. But the set Z6 does not produce a field where addition and
multiplication are modular because, for example, 2 then has no
multiplicative inverse. CONJECTURE: <Zp , ⊕, ⊗> is a field iff p is
prime.
1.7. (a) If 0, 0′ are distinct additive identities, we interpret “distinct” to mean
that their difference is nonzero. Let 0 + (−0′ ) = c, where c 6= 0, 0′ .
Post-addition of 0′ to both sides gives from Axiom R3(a)
0 + −0′ + 0′ = c + 0′ .
(*)
In the brackets, let 0 be the zero resulting from addition of the num-
ber 0′ to its additive inverse −0′ . On the right-hand side of (*), let
0′ be a zero as in Axiom R5(b). We obtain
0 + 0 = c,
so from R5(b) again we have 0 = c, which is not allowed. The
difficulty can be removed if 0, 0′ are not distinct.
(b) Interpret “1, 1′ being distinct” to mean that 1 · (1′ )−1 is neither 1 nor
1′ . Let 1 · (1′ )−1 = c. Post-multiplication of both sides by 1′ gives
from Axiom R3(b)
1 · (1′ )−1 · 1′ = c · 1′ .
(*)
In the brackets, let 1 be the multiplicative identity resulting from
multiplication of the nonzero number 1′ by its multiplicative inverse
(1′ )−1 . On the right-hand side of (*), apply Axiom R5(b); we obtain
1 = 1 · 1 = c,
which is not allowed. The difficulty can be removed if 1, 1′ are not
distinct.
©2010 Elsevier, Inc.
SOLUTIONS MANUAL
, 3
Contents
CHAPTER 1 Sets, Numbers, and Functions . . . . . . . . . . . . . . . . . . . . . . . . 1
CHAPTER 2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
CHAPTER 3 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
CHAPTER 4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
CHAPTER 5 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
CHAPTER 6 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
CHAPTER 7 Commutation of Limit Operations . . . . . . . . . . . . . . . . . . . 163
iii
©2010 Elsevier, Inc.
, 5
C H A P TER 1
Sets, Numbers, and Functions
1.1. No, because T ∧ F is F.
1.2. Yes, because F ∨ T is T. If ∼ (p ∧ q) is T(F), then so is (∼ p∨ ∼ q).
1.3. (a)
p q ∼p ∼p∨q p→q
T T F T T
T F F F F
F T T T T
F F T T T
(b)
p q p→q q → p (p → q) ∧ (q → p) p ↔ q
T T T T T T
T F F T F F
F T T F F F
F F T T T T
(c)
p q ∼p ∼q p→q ∼ q →∼ p (p → q) ↔ (∼ q →∼ p)
T T F F T T T
T F F T F F T
F T T F T T T
F F T T T T T
1.4. (a), (b), (f ) hold.
1.5. A Right Distributive Law is already implied by R2(b), R4: (y + z)x =
x(y + z) = xy + xz = yx + zx. Statements analogous to field axioms
R2(b), R6(b) would fail for 3 × 3 matrices.
1.6. (a) For <Z5 , ⊕, ⊗>, axioms analogous to R1 – R5 are inherited from
R. The additive inverses of 0, 1, 2, 3, 4 are 0, 4, 3, 2, 1, respectively,
Advanced Calculus. DOI: 10.1016/B978-0-12-384694-5.00001-5
Copyright c 2010, Elsevier Inc. All rights reserved. 1
©2010 Elsevier, Inc.
, 6
2 C H A P T E R 1: Sets, Numbers, and Functions
and the multiplicative inverses of 1, 2, 3, 4, are 1, 3, 2, 4, respectively.
Hence, an axiom analogous to R6 holds, so <Z5 , ⊕, ⊗> is a field.
(b) For Z5 suppose that P ⊆ {1, 2, 3, 4} is nonempty; let x ∈ P. Addition
of x to itself a sufficient number of times produces all members of
{1, 2, 3, 4}, so P = {1, 2, 3, 4}. But then −x ∈ P, which is not allowed
by Axiom R7(b), so P = ∅.
(c) If ⊕, ⊗ are defined modularly as with <Z5 , ⊕, ⊗> at the start of Exer-
cise 1.6, then <Z7 , ⊕, ⊗> and <Z11 , ⊕, ⊗> are found to be finite
fields. But the set Z6 does not produce a field where addition and
multiplication are modular because, for example, 2 then has no
multiplicative inverse. CONJECTURE: <Zp , ⊕, ⊗> is a field iff p is
prime.
1.7. (a) If 0, 0′ are distinct additive identities, we interpret “distinct” to mean
that their difference is nonzero. Let 0 + (−0′ ) = c, where c 6= 0, 0′ .
Post-addition of 0′ to both sides gives from Axiom R3(a)
0 + −0′ + 0′ = c + 0′ .
(*)
In the brackets, let 0 be the zero resulting from addition of the num-
ber 0′ to its additive inverse −0′ . On the right-hand side of (*), let
0′ be a zero as in Axiom R5(b). We obtain
0 + 0 = c,
so from R5(b) again we have 0 = c, which is not allowed. The
difficulty can be removed if 0, 0′ are not distinct.
(b) Interpret “1, 1′ being distinct” to mean that 1 · (1′ )−1 is neither 1 nor
1′ . Let 1 · (1′ )−1 = c. Post-multiplication of both sides by 1′ gives
from Axiom R3(b)
1 · (1′ )−1 · 1′ = c · 1′ .
(*)
In the brackets, let 1 be the multiplicative identity resulting from
multiplication of the nonzero number 1′ by its multiplicative inverse
(1′ )−1 . On the right-hand side of (*), apply Axiom R5(b); we obtain
1 = 1 · 1 = c,
which is not allowed. The difficulty can be removed if 1, 1′ are not
distinct.
©2010 Elsevier, Inc.
