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SOLUTIONS MANUAL — Analysis with an Introduction to Proof, 5th Edition — Steven R. Lay — ISBN 9780137546138

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The Solutions Manual for Analysis with an Introduction to Proof, 5th Edition by Steven R. Lay (ISBN 978-0137546138) provides worked-out solutions and detailed explanations corresponding to all exercises, proofs, hints, and selected problems throughout the text. It follows the official Table of Contents as published by Pearson, ensuring consistency and alignment with the student edition. The manual includes step-by-step solutions for every section: Chapter 1: Logic and Proof with Section 1. Logical Connectives, Section 2. Quantifiers, Section 3. Techniques of Proof: I, Section 4. Techniques of Proof: II; Chapter 2: Sets and Functions covering Section 5. Basic Set Operations, Section 6. Relations, Section 7. Functions, Section 8. Cardinality, Section 9. Axioms for Set Theory (Optional); Chapter 3: The Real Numbers including Section 10. Natural Numbers and Induction, Section 11. Ordered Fields, Section 12. The Completeness Axiom, Section 13. Topology of the Reals, Section 14. Compact Sets, Section 15. Metric Spaces (Optional); Chapter 4: Sequences (Sections 16-19), Chapter 5: Limits and Continuity (Sections 20-24), Chapter 6: Differentiation (Sections 25-28), Chapter 7: Integration (Sections 29-31), Chapter 8: Infinite Series (Sections 32-34), Chapter 9: Sequences and Series of Functions (Sections 35-37), together with the Glossary of Key Terms, References, Hints for Selected Exercises, and a full Index.

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Analysis with an Introduction to Proof
TU
– 5th Edition

SOLUTIONS
V
IA

MANUAL
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AP
Steven R. Lay
PR
Complete Solutions Manual for Instructors and
OV
Students

© Steven R. Lay
ED

All rights reserved. Reproduction or distribution without permission is prohibited.
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©Medexcellence ✅��

, Section 1.1 x Logical Connectives 4


This work is protected by United States copyright laws and is provided solely for
the use of instructors in teaching their courses and assessing student learning.
Dissemination or sale of any part of this work (including on the World Wide Web)
TU
will destroy the integrity of the work and is not permitted. The work and materials
from it should never be made available to students except by instructors using
the accompanying text in their classes. All recipients of this work are expected to
abide by these restrictions and to honor the intended pedagogical purposes and
the needs of other instructors who rely on these materials.
V
IA

Analysis
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with an Introduction to Proof
5th Edition
AP
by Steven R. Lay



Chapter 1 – Logic and Proof
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Solutions to Exercises

Section 1.1 – Logical Connectives
OV
1. (a) False: A statement may be false.
(b) False: A statement cannot be both true and false.
(c) True: See the comment after Practice 1.1.4.
(d) False: See the comment before Example 1.1.3.
(e) False: If the statement is false, then its negation is true.
ED
2. (a) False: p is the antecedent.
(b) True: Practice 1.1.6(a).
(c) False: See the paragraph before Practice 1.1.5.
(d) False: “p whenever q” is “if q, then p”.
(e) False: The negation of p º q is p š ~ q.
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3. Answers in Book: (a) The 3 × 3 identity matrix is not singular.
(b) The function f (x) = sin x is not bounded on .
(c) The function f is not linear or the function g is not linear.
(d) Six is not prime and seven is not odd.
(e) x is in D and f (x) t 5.
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Copyright © 2014 Pearson Education, Inc.

, Section 1.1 x Logical Connectives 5
(f ) (an) is monotone and bounded, but (an) is not convergent.
(g) f is injective, and S is not finite and not denumerable.

4. (a) The function f (x) = x2 – 9 is not continuous at x = 3.
TU
(b) The relation R is not reflexive and not symmetric.
(c) Four and nine are not relatively prime.
(d) x is not in A and x is in B.
(e) x < 7 and f (x) is in C.
(f ) (an) is convergent, but (an) is not monotone or not bounded.
(g) f is continuous and A is open, but f – 1(A) is not open.
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5. Answers in book: (a) Antecedent: M is singular; consequent: M has a zero eigenvalue.
(b) Antecedent: linearity; consequent: continuity.
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(c) Antecedent: a sequence is Cauchy; consequent: it is bounded.
(d) Antecedent: y > 5; consequent: x < 3.

6. (a) Antecedent: it is Cauchy; consequent: a sequence is convergent.
(b) Antecedent: boundedness; consequent: convergence.
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(c) Antecedent: orthogonality; consequent: invertability.
(d) Antecedent: K is closed and bounded; consequent: K is compact.

7 and 8 are routine.
AP
9. Answers in book: (a) T š T is T. (b) F › T is T. (c) F › F is F. (d) T º T is T. (e) F º F is T.
(f) T º F is F. (g) (T š F) º T is T. (h) (T › F) º F is F. (i) (T š F) º F is T. (j) ~ (F › T) is F.

10. (a) T š F is F. (b) F › F is F. (c) F › T is T. (d) T º F is F. (e) F º F is T. (f) F º T is T.
(g) (F › T) º F is F. (h) (T º F) º T is T. (i) (T š T) º F is F. (j) ~ (F š T) is T.
PR
11. Answers in book: (a) p š ~ q; (b) ( p › q) š ~ ( p š q); (c) ~ q º p; (d) ~ p º q; (e) p ” ~ q.

12. (a) n š ~ m; (b) ~ m š ~ n or ~ (m › n); (c) n º m; (d) m º ~ n; (e) ~ (m š n).

13. (a) and (b) are routine. (c) p š q.
OV
14. These truth tables are all straightforward. Note that the tables for (c) through (f ) have 8 rows because there are 3
letters and therefore 23 = 8 possible combinations of T and F.


Section 1.2 - Quantifiers

1. (a) True: See the comment before Example 1.2.1.
ED
(b) False: The negation of a universal statement is an existential statement.
(c) True: See the comment before Example 1.2.1.

2. (a) False: It means there exists at least one.
(b) True: Example 1.2.1.
(c) True: See the comment after Practice 1.2.4.
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3. (a) No pencils are red.
(b) Some chair does not have four legs.
(c) Someone on the basketball team is over 6 feet 4 inches tall.
(d)  x > 2, f (x) z 7.
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Copyright © 2014 Pearson Education, Inc.

, Section 1.2 x Quantifiers 6
(e)  x in A †  y > 2, f ( y) d 0 or f ( y) t f (x).
(f )  x † x > 3 and  H > 0, x2 ” 9 + H.

4. (a) Someone does not like Robert.
TU
(b) No students work part-time.
(c) Some square matrices are triangular.
(d)  x in B, f (x) ” k.
(e)  x † x > 5 and 3 d f (x) d 7.
(f )  x in A † y in B, f (y) ” f (x).
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5. Hints in book: The True/False part of the answers.
(a) True. Let x = 3. (b) True. 4 is less than 7 and anything smaller than 4 will also be less than 7.
(c) True. Let x = 5. (d) False. Choose x z r 5 such as x = 2.
IA
(e) True. Let x = 1, or any other real number.
(f ) True. The square of a real number cannot be negative.
(g) True. Let x = 1, or any real number other than 0. (h) False. Let x = 0.

6. (a) True. Let x = 5. (b) False. Let x = 3. (c) True. Choose x z r 3 such as x = 2.
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(d) False. Let x = 3. (e) False. The square of a real number cannot be negative.
(f ) False. Let x = 1, or any other real number. (g) True. Let x = 1, or any other real number.
(h) True. x – x = x + (– x) and a number plus its additive inverse is zero.

7. Answers in book: (a) You can use (ii) to prove (a) is true. (b) You can use (i) to prove (b) is true.
AP
Additional answers: (c) You can use (ii) to prove (c) is false. (d) You can use (i) to prove (d) is false.

8. The best answer is (c).

9. Hints in book: The True/False part of the answers.
(a) False. For example, let x = 2 and y = 1. Then x > y.
PR
(b) True. For example, let x = 2 and y = 3. Then x ” y.
(c) True. Given any x, let y = x + 1. Then x ” y.
(d) False. Given any y, let x = y + 1. Then x > y.

10. (a) True. Given any x, let y = 0.
(b) False. Let x = 0. Then for all y we have xy = 0 z 1.
OV
(c) False. Let y = 0. Then for all x we have xy = 0 z 1.
(d) True. Given any x, let y = 1. Then xy = x.

11. Hints in book: The True/False part of the answers.
(a) True. Let x = 0. Then given any y, let z = y. (A similar argument works for any x.)
(b) False. Given any x and any y, let z = x + y + 1.
(c) True. Let z = y – x.
ED
(d) False. Let x = 0 and y = 1. (It is a true statement for x z 0.)
(e) True. Let x d 0.
(f ) True. Take z d y. This makes “z ! y ” false so that the implication is true. Or, choose z ! x + y.

12. (a) True. Given x and y, let z = x + y.
(b) False. Let x = 0. Then given any y, let z = y + 1.
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(c) True. Let x = 1. Then given any y, let z = y. (Any x z 0 will work.)
(d) False. Let x = 1 and y = 0. (Any x z 0 will work.)
(e) False. Let x = 2. Given any y, let z = y + 1. Then “z ! y ” is true, but “z ! x + y ” is false.
(f ) True. Given any x and y, either choose z ! x + y or z d y.
??

Copyright © 2014 Pearson Education, Inc.

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