![Section 1.3 x Techniques of Proof: I 8
7. (a) Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then p + q = (2k + 1) + (2r + 1) = 2(k + r + 1), so
p + q is even.
(b) Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then pq = (2k + 1)(2r + 1) = 4kr + 2k + 2r + 1 =
2(2kr + k + r) + 1, so pq is odd.
(c) Here are two approaches. The first mimics part (a) and the second uses parts (a) and (b).
Proof 1: Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then p + 3q = (2k + 1) + 3(2r + 1) =
2(k + 3r + 2), so p + 3q is even.
Proof 2: Suppose p and q are both odd. By part (b), 3q is odd since 3 is odd. So by part (a), p + 3q is even.
(d) Suppose p = 2k + 1 and q = 2r for integers k and r. Then p + q = (2k + 1) + 2r = 2(k + r) + 1, so p + q is odd.
(e) Suppose p = 2k and q = 2r for integers k and r. Then p + q = 2k + 2r = 2(k + r), so p + q is even.
(f) Suppose p = 2k, then pq = 2(kq), so pq is even. A similar argument applies when q is even.
(g) This is the contrapositive of part (f).
(h) Hint in book: look at the contrapositive.
Proof: To prove the contrapositive, suppose p = 2k + 1. Then p2
= (2k + 1)2
= 4k2
+ 4k + 1 =
2(2k2
+ 2k) + 1, so p2
is odd.
(i) To prove the contrapositive, suppose p = 2k. Then p2
= (2k)2
= 4k2
= 2(2k2
), so p2
is even.
8. Suppose f (x1) = f (x2). That is, 4x1 + 7 = 4x2 + 7. Then 4x1 = 4x2, so x1 = x2.
9. Answers in book:
(a) r º ~ s hypothesis
~ s º ~ t contrapositive of hypothesis: 1.3.12(c)
r º ~ t by 1.3.12(l)
(b) ~ t º (~ r ” ~ s) contrapositive of hypothesis: 1.3.12(c)
~ r ” ~ s by 1.3.12(h)
~ s by 1.3.12(j)
(c) r ” ~ r by 1.3.12(d)
~ s º ~ v contrapositive of hypothesis 4 [1.3.12(c)]
r º ~ v hypothesis 1 and 1.3.12(l)
~ r º u hypotheses 2 and 3 and 1.3.12(l)
~ v ” u by 1.3.12(o)
10. (a) ~r hypothesis
~ r º (r ” ~ s) contrapositive of hypothesis: 1.3.12(c)
r ” ~ s by 1.3.12(h)
~ s by lines 1 and 3, and 1.3.12(j)
(b) ~ t hypothesis
~ t º (~ r ” ~ s) contrapositive of hypothesis: 1.3.12(c)
~ r ” ~ s by lines 1 and 2, and 1.3.12(h)
~ s by line 3 and 1.3.12(k)
(c) s º r contrapositive of hypothesis: 1.3.12(c)
t º u hypothesis
s ” t hypothesis
r ” u by 1.3.12(o)](https://image.slidesharecdn.com/analysiswithanintroductiontoproof5theditionlaysolutionsmanual-180120044948/85/Analysis-with-an-introduction-to-proof-5th-edition-lay-solutions-manual-11-320.jpg)
![2
¨ ¸
© ¹
Section 1.3 x Techniques of Proof: I 9
11. Let p: The basketball center is healthy. q: The point guard is hot.
r: The team will win. s: The fans are happy.
t: The coach is a millionaire. u: The college will balance the budget.
The hypotheses are (p ” q) º (r ” s) and (s ” t) º u. The conclusion is p º u.
Proof: p º ( p ” q) from the contrapositive of
1.3.12(k) (p ” q) º (r ” s) hypothesis
(r ” s) º s by 1.3.12(k)
s º (s ” t) from the contrapositive of
1.3.12(k) (s ” t) º u hypothesis
p º u by 1.3.12(m)
Section 1.4 – Techniques of Proof: II
1. (a) True: See the comment before Example 1.4.1.
(b) False: Indirect proofs avoid this.
(c) False: Only the “relevant” steps need to be included.
2. (a) True: See the comment before Practice 1.4.2.
(b) False: The left side of the tautology should be [(p ” ~ q)
º c]. (c) True: See the comment after Practice 1.4.8.
H H3. Given any H > 0, let G = H /3. Then G is also positive and whenever 2 – G < x < 2 + G, we have 2
3
x 2
3
so
that 6 – H < 3x < 6 + H and 11 – H < 3x + 5 < 11 + H, as required.
4. Given any H > 0, let G = H /5. Then G is also positive and whenever 1 – G < x < 1 + G, we have 1
H x 1
H so
5 5
that 5 – H < 5x < 5 + H and –2 – H < 5x – 7 < –2 + H. Now multiply by –1 and reverse the inequalities: 2 + H >
7 – 5x > 2 – H. This is equivalent to 2 – H < 7 – 5x < 2 + H.
5. Let x = 1. Then for any real number y, we have xy = y.
6. Let x = 0. Then for any real number y, we have xy = x because xy = 0.
7. Given any integer n, we have n
2
+ n
3
= n
2
(1 + n). If n is even, then n
2
is even. If n is odd, then 1 + n is even. In
either case, their product is even. [This uses Exercise 1.3.7(f).]
8. If n is odd, then n = 2m + 1 for some integer m, so n
2
= (2m + 1)
2
= 4m
2
+ 4m + 1 = 4m(m + 1) + 1. If m is even,
then m = 2p for some integer p. But then n2
= 4(2p)(m + 1) + 1 = 8[p(m + 1)] + 1. So n2
= 8k + 1, where k is the
integer p(m + 1). On the other hand, if m is odd, then m + 1 is even and m + 1 = 2q for some integer q. But then,
n2
= 4m(2q) + 1 = 8(mq) + 1. In this case, n2
= 8k + 1, where k is the integer mq. In either case, n2
= 8k + 1 for
some integer k.
9. Answer in book: Let n 2. Then n2 3
n
2
4
§ 3 ·
( 2) 4 3 1, as required. The integer is unique.
© ¹
10. Existence follows from Exercise 3. It is not unique. x = 2 or x = 1/2.
11. Answer in book: Let x be a real number greater than 5 and let y = 3x/(5 x). Then 5 x 0 and 2x ! 0, so y 0.
5
§ 3x ·
¨ ¸
Furthermore,
5y © 5 x ¹ 15x 15x
x, as required.
y 3 § 3x ·
3¨
5 x
¸
3x 3(5 x) 15](https://image.slidesharecdn.com/analysiswithanintroductiontoproof5theditionlaysolutionsmanual-180120044948/85/Analysis-with-an-introduction-to-proof-5th-edition-lay-solutions-manual-12-320.jpg)
![n qn
q .
x
2
q
s
Section 1.4 x Techniques of Proof: II 10
12. Solve the quadratic y2
– 6xy + 9 = 0 to obtain y
organized like Exercise 11.
3x r 3 x2
1. Call one solution y and the other z. The proof is
13. Answer in book: Suppose that x2
x 6 t 0 and x ! 3. It follows that (x 2)(x 3) t 0 and, since x 3 ! 0, it
must be that x 2 t 0. That is, x t 2.
14. Suppose
x
d 3 and x ! 2. Then x – 2 > 0, so we can multiply both sides of the first inequality by x – 2 to
x 2
obtain x d 3(x – 2). Thus x d 3x – 6. That is, x t 3. If x = 2, then
x
is not defined.
x 2
15. Hint in book: Suppose log2 7 is rational and find a contradiction.
a/b
Proof: Suppose log2 7 = a/b, where a and b are integers. We may assume that a > 0 and b > 0. We have 2 = 7,
which implies 2
a
= 7
b
. But the number 2
a
is even and the number 7
b
is odd, a contradiction. Thus log
irrational.
7 must be
16. Suppose |x + 1| d 3 and consider the two cases: x t –1 and x –1. If x ” –1, then x + 1 ” 0 so that |x + 1| = x + 1.
This implies x + 1 ” 3 and x ” 2. So in this case we have –1 ” x ” 2.
On the other hand, if x < –1, then x + 1 < 0 and |x + 1| = – (x + 1). This leads to – (x + 1) ” 3, x + 1 ” – 3, and
x ” –4. In this case we have – 4 ” x < –1.
Combining the two cases, we get – 4 ” x ” 2.
17. (a) This proves the converse, which is not a valid proof of the original implication.
(b) This is a valid proof using the contrapositive.
18. (a) This is a valid proof in the form of Example 1.3.12 (p): [ p º (q ” r)] ” [( p ”
~ q) º r]. (b) This proves the converse, which is not a valid proof of the original implication.
19. (a) If x
p
and y m , then x y
p m pn qm
, so x + y is rational.
(b) If x
q
p
and y
n
m , then xy
q n qn
pm
, so xy is rational.
(c) Hint in book: Use a proof by contradiction.
Proof: Suppose x
p
, y is irrational, and suppose x y r Then
( ) r p rq ps
,y x y x s q qs
so that y is rational, a contradiction. Thus x + y must be irrational.
20. (a) False. For example, let x = 2 and y = 2. Then x + y = 0.
(b) True. This is the contrapositive of Exercise 7(a).
(c) False. For example, let x = y = 2. Then xy = 2.
(d) True. This is the contrapositive of Exercise 13(b).
21. (a) Let x = 0 and y 2. Then xy = 0.
(b) y z
xy
when x = 0.
(c) If x z 0, then the conclusion is true.](https://image.slidesharecdn.com/analysiswithanintroductiontoproof5theditionlaysolutionsmanual-180120044948/85/Analysis-with-an-introduction-to-proof-5th-edition-lay-solutions-manual-13-320.jpg)
Analysis with an introduction to proof 5th edition lay solutions manual
- 1. Analysis with an Introduction to Proof 5th Edition Lay SOLUTIONS MANUAL Full download at: http://testbanklive.com/download/analysis-with-an-introduction-to- proof-5th-edition-lay-solutions-manual/ Instructor’s Solutions Manual to accompany ANALYSIS with an Introduction to Proof 5th Edition Steven R. Lay Lee University ” &RSULJKW 3HDUVRQ (GXFDWLRQ ,QF Upper Saddle River, New Jersey 07458
- 3. 2 This manual is intended to accompany the 5th edition of Analysis with an Introduction to Proof by Steven R. Lay (Pearson, 2013). It contains solutions to nearly every exercise in the text. Those exercises that have hints (or answers) in the back of the book are numbered in bold print, and the hints are included here for reference. While many of the proofs have been given in full detail, some of the more routine proofs are only outlines. For some of the problems, other approaches may be equally acceptable. This is particularly true for those problems requesting a counterexample. I have not tried to be exhaustive in discussing each exercise, but rather to be suggestive. Let me remind you that the starred exercises are not necessarily the more difficult ones. They are the exercises that are used in some way in subsequent sections. There is a table on page 3 that indicates where starred exercises are used later. The following notations are used throughout this manual: = the set of natural numbers {1, 2, 3, 4, …} = the set of rational numbers = the set of real numbers = “for every” = “there exists” ” = “such that” I have tried to be accurate in the preparation of this manual. Undoubtedly, however, some mistakes will inadvertently slip by. I would appreciate having any errors in this manual or the text brought to my attention. Steven R. Lay slay@leeuniversity.edu
- 4. 3 Table of Starred Exercises Note: The prefix P indicates a practice problem, the prefix E indicates an example, the prefix T refers to a theorem or corollary, and the absence of a prefix before a number indicates an exercise. Starred Exercise Later Use Starred Exercise Later use 2.1.26 T3.4.11 4.3.14 4.4.5 2.2.10 2.4.26 4.4.10 8.2.14 2.3.32 2.5.3 4.4.16 8.3.9 3.1.3 E7.1.7 4.4.17 T8.3.3 3.1.4 7.1.7 5.1.14 6.2.8 3.1.6 E8.1.1 5.1.16 T6.2.9 3.1.7 4.3.10, 4.3.15, E8.1.7, T9.2.9 5.1.18 5.2.14, 5.3.15 3.1.8 P8.1.3 5.1.19 5.2.17 3.1.24 4.1.7f, E5.3.7 5.2.10 T7.2.8 3.1.27 3.3.14 5.2.11 7.2.9b 3.1.30b 3.3.11, E4.1.11, 4.3.14 5.2.13 T5.3.5, T6.1.7, 7.1.13 3.2.6a 4.1.9a, T4.2.1, 6.2.23, 7.2.16, T9.2.9 5.2.16 9.2.15 3.2.6b T6.3.8 5.3.13b T6.2.8, T6.2.10 3.2.6c T4.1.14 6.1.6 6. 2.14, 6.2.19 3.2.7 T8.2.5 6.1.8 7.3.13 3.3.7 T7.2.4, 7.2.3 6.1.17b 6.4.9 3.3.12 7.1.14, T7.2.4 6.2.8 T7.2.1 3.4.15 3.5.12, T4.3.12 6.3.13d 9.3.16 3.4.21 3.5.7 7.1.12 P7.2.5 3.5.8 9.2.15 7.1.13 7.2.5 3.6.12 5.5.9 7.1.16 7.2.17 4.1.6b E4.2.2 7.2.9a P7.3.7 4.1.7f T4.2.7, 4.3.10, E8.1.7 7.2.11 T8.2.13 4.1.9a 5.2.10, 9.2.17 7.2.15 7.3.20 4.1.11 E4.3.4 7.2.20 E7.3.9 4.1.12 5.1.15 8.1.7 E8.2.6 4.1.13 5.1.13 8.1.8 8.2.13 4.1.15b 4.4.11, 4.4.18, 5.3.12 8.1.13a 9.3.8 4.1.16 5.1.15 8.2.12 9.2.7, 9.2.8 4.2.17 E6.4.3 8.2.14 T8.3.4 4.2.18 5.1.14, T9.1.10 9.1.15a 9.2.9
- 5. 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) 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. Analysis with an Introduction to Proof 5th Edition by Steven R. Lay slay@leeuniversity.edu Chapter 1 – Logic and Proof Solutions to Exercises Section 1.1 – Logical Connectives 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. 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 ” ~ q. 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.
- 6. 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. (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. 5. Answers in book: (a) Antecedent: M is singular; consequent: M has a zero eigenvalue. (b) Antecedent: linearity; consequent: continuity. (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. (c) Antecedent: orthogonality; consequent: invertability. (d) Antecedent: K is closed and bounded; consequent: K is compact. 7 and 8 are routine. 9. Answers in book: (a) T ” 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 ” F) º T is T. (h) (T ” F) º F is F. (i) (T ” F) º F is T. (j) ~(F ” T) is F. 10. (a) T ” 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 ” T) º F is F. (j) ~(F ” T) is T. 11. Answers in book: (a) p ” ~ q; (b) ( p ” q) ” ~ ( p ” q); (c) ~ q º p; (d) ~ p º q; (e) p ” ~ q. 12. (a) n ” ~ m; (b) ~ m ” ~ n or ~ (m ” n); (c) n º m; (d) m º ~ n; (e) ~ (m ” n). 13. (a) and (b) are routine. (c) p ” q. 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. (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. 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.
- 8. 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. (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). 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 (e) True. Let x = 1, or any other real number. 5 such as x = 2. (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. (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. 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. (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. (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. (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. (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.
- 9. Section 1.2 x Quantifiers 7 13. Answer in book: (a) x, f ( x) = f (x); (b) x ” f ( x) z f (x). 14. (a) k ! 0 ” x, f (x + k) = f (x). (b) k ! 0, x ” f (x + k) z f (x). 15. Answer in book: (a) x and y, x d y º f (x) d f ( y). (b) x and y ” x d y and f (x) > f ( y). 16. (a) x and y, x y º f (x) ! f (y). (b) x and y ” x y and f (x) d f ( y). 17. Answer in book: (a) x and y, f (x) = f ( y) º x = y. (b) x and y ” f (x) = f (y) and x z y. 18. (a) y in B x in A ” f (x) = y. (b) y in B ” x in A, f (x) z y. 19. Answer in book: (a) H ! 0, G ! 0 ” x ” D, |x c| G º | f (x) f (c)| H. (b) H ! 0 ” G ! 0, x ” D ” | x c| < G and | f (x) f (c)| t H. 20. (a) H ! 0 G ! 0 ” x and y in S, |x – y| G º | f (x) – f (y)| F. (b) F ! 0 ” E ! 0, x and y in S ” | x – y| E and | f (x) – f ( y)| t F. 21. Answer in book: (a) H ! 0, G ! 0 ” x ” D, 0 |x c| G º | f (x) L| H. (b) H ! 0 ” G ! 0, x ” D ” 0 |x c| G and | f (x) L| t H. 22. Answers will vary. Section 1.3 – Techniques of Proof: I 1. (a) False: p is the hypothesis. (b) False: The contrapositive is ~ q º ~ p. (c) False: The inverse is ~p º ~ q. (d) False: p(n) must be true for all n. (e) True: Example 1.3.1. 2. (a) True: See the comment after Practice 1.3.4. (b) False: It’s called a contradiction. (c) True: See the comment after Practice 1.3.8. (d) True: See the end of Example 1.3.1. (e) False: Must show p(n) is true for all n. 3. Answers in book: (a) If all violets are not blue, then some roses are not red. (b) If A is invertible, then there is no nontrivial solution to Ax = 0. (c) If f (C) is not connected, then either f is not continuous or C is not connected. 4. (a) If some violets are blue, then all roses are red. (b) If A is not invertible, then there exists a nontrivial solution to Ax = 0. (c) If f (C) is connected, then f is continuous and C is connected. 5. (a) If some roses are not red, then no violets are blue. (b) If Ax = 0 has no nontrivial solutions, then A is invertible. (c) If f is not continuous or C is not connected, then f (C) is not connected. 6. For some of these, other answers are possible. (a) Let x = – 4. (b) Let n = 2. (c) If 0 x 1, then x 3 x 2 . In particular, (1/2) 3 = 1/8 < 1/4 = (1/2) 2 . (d) An equilateral triangle. (e) n = 40 or n = 41. (f) 2 is prime, but not odd. (g) 101, 103, etc. (h) 3 5 + 2 = 245 is not prime. (i) Let n = 5 or any odd greater than 3. (j) Let x = 2 and y = 18. (k) Let x = 0. (l) The reciprocal of 1 is not less than 1. (m) Let x = 0. (n) Let x = 1.
- 11. Section 1.3 x Techniques of Proof: I 8 7. (a) Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then p + q = (2k + 1) + (2r + 1) = 2(k + r + 1), so p + q is even. (b) Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then pq = (2k + 1)(2r + 1) = 4kr + 2k + 2r + 1 = 2(2kr + k + r) + 1, so pq is odd. (c) Here are two approaches. The first mimics part (a) and the second uses parts (a) and (b). Proof 1: Suppose p = 2k + 1 and q = 2r + 1 for integers k and r. Then p + 3q = (2k + 1) + 3(2r + 1) = 2(k + 3r + 2), so p + 3q is even. Proof 2: Suppose p and q are both odd. By part (b), 3q is odd since 3 is odd. So by part (a), p + 3q is even. (d) Suppose p = 2k + 1 and q = 2r for integers k and r. Then p + q = (2k + 1) + 2r = 2(k + r) + 1, so p + q is odd. (e) Suppose p = 2k and q = 2r for integers k and r. Then p + q = 2k + 2r = 2(k + r), so p + q is even. (f) Suppose p = 2k, then pq = 2(kq), so pq is even. A similar argument applies when q is even. (g) This is the contrapositive of part (f). (h) Hint in book: look at the contrapositive. Proof: To prove the contrapositive, suppose p = 2k + 1. Then p2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1, so p2 is odd. (i) To prove the contrapositive, suppose p = 2k. Then p2 = (2k)2 = 4k2 = 2(2k2 ), so p2 is even. 8. Suppose f (x1) = f (x2). That is, 4x1 + 7 = 4x2 + 7. Then 4x1 = 4x2, so x1 = x2. 9. Answers in book: (a) r º ~ s hypothesis ~ s º ~ t contrapositive of hypothesis: 1.3.12(c) r º ~ t by 1.3.12(l) (b) ~ t º (~ r ” ~ s) contrapositive of hypothesis: 1.3.12(c) ~ r ” ~ s by 1.3.12(h) ~ s by 1.3.12(j) (c) r ” ~ r by 1.3.12(d) ~ s º ~ v contrapositive of hypothesis 4 [1.3.12(c)] r º ~ v hypothesis 1 and 1.3.12(l) ~ r º u hypotheses 2 and 3 and 1.3.12(l) ~ v ” u by 1.3.12(o) 10. (a) ~r hypothesis ~ r º (r ” ~ s) contrapositive of hypothesis: 1.3.12(c) r ” ~ s by 1.3.12(h) ~ s by lines 1 and 3, and 1.3.12(j) (b) ~ t hypothesis ~ t º (~ r ” ~ s) contrapositive of hypothesis: 1.3.12(c) ~ r ” ~ s by lines 1 and 2, and 1.3.12(h) ~ s by line 3 and 1.3.12(k) (c) s º r contrapositive of hypothesis: 1.3.12(c) t º u hypothesis s ” t hypothesis r ” u by 1.3.12(o)
- 12. 2 ¨ ¸ © ¹ Section 1.3 x Techniques of Proof: I 9 11. Let p: The basketball center is healthy. q: The point guard is hot. r: The team will win. s: The fans are happy. t: The coach is a millionaire. u: The college will balance the budget. The hypotheses are (p ” q) º (r ” s) and (s ” t) º u. The conclusion is p º u. Proof: p º ( p ” q) from the contrapositive of 1.3.12(k) (p ” q) º (r ” s) hypothesis (r ” s) º s by 1.3.12(k) s º (s ” t) from the contrapositive of 1.3.12(k) (s ” t) º u hypothesis p º u by 1.3.12(m) Section 1.4 – Techniques of Proof: II 1. (a) True: See the comment before Example 1.4.1. (b) False: Indirect proofs avoid this. (c) False: Only the “relevant” steps need to be included. 2. (a) True: See the comment before Practice 1.4.2. (b) False: The left side of the tautology should be [(p ” ~ q) º c]. (c) True: See the comment after Practice 1.4.8. H H3. Given any H > 0, let G = H /3. Then G is also positive and whenever 2 – G < x < 2 + G, we have 2 3 x 2 3 so that 6 – H < 3x < 6 + H and 11 – H < 3x + 5 < 11 + H, as required. 4. Given any H > 0, let G = H /5. Then G is also positive and whenever 1 – G < x < 1 + G, we have 1 H x 1 H so 5 5 that 5 – H < 5x < 5 + H and –2 – H < 5x – 7 < –2 + H. Now multiply by –1 and reverse the inequalities: 2 + H > 7 – 5x > 2 – H. This is equivalent to 2 – H < 7 – 5x < 2 + H. 5. Let x = 1. Then for any real number y, we have xy = y. 6. Let x = 0. Then for any real number y, we have xy = x because xy = 0. 7. Given any integer n, we have n 2 + n 3 = n 2 (1 + n). If n is even, then n 2 is even. If n is odd, then 1 + n is even. In either case, their product is even. [This uses Exercise 1.3.7(f).] 8. If n is odd, then n = 2m + 1 for some integer m, so n 2 = (2m + 1) 2 = 4m 2 + 4m + 1 = 4m(m + 1) + 1. If m is even, then m = 2p for some integer p. But then n2 = 4(2p)(m + 1) + 1 = 8[p(m + 1)] + 1. So n2 = 8k + 1, where k is the integer p(m + 1). On the other hand, if m is odd, then m + 1 is even and m + 1 = 2q for some integer q. But then, n2 = 4m(2q) + 1 = 8(mq) + 1. In this case, n2 = 8k + 1, where k is the integer mq. In either case, n2 = 8k + 1 for some integer k. 9. Answer in book: Let n 2. Then n2 3 n 2 4 § 3 · ( 2) 4 3 1, as required. The integer is unique. © ¹ 10. Existence follows from Exercise 3. It is not unique. x = 2 or x = 1/2. 11. Answer in book: Let x be a real number greater than 5 and let y = 3x/(5 x). Then 5 x 0 and 2x ! 0, so y 0. 5 § 3x · ¨ ¸ Furthermore, 5y © 5 x ¹ 15x 15x x, as required. y 3 § 3x · 3¨ 5 x ¸ 3x 3(5 x) 15
- 13. n qn q . x 2 q s Section 1.4 x Techniques of Proof: II 10 12. Solve the quadratic y2 – 6xy + 9 = 0 to obtain y organized like Exercise 11. 3x r 3 x2 1. Call one solution y and the other z. The proof is 13. Answer in book: Suppose that x2 x 6 t 0 and x ! 3. It follows that (x 2)(x 3) t 0 and, since x 3 ! 0, it must be that x 2 t 0. That is, x t 2. 14. Suppose x d 3 and x ! 2. Then x – 2 > 0, so we can multiply both sides of the first inequality by x – 2 to x 2 obtain x d 3(x – 2). Thus x d 3x – 6. That is, x t 3. If x = 2, then x is not defined. x 2 15. Hint in book: Suppose log2 7 is rational and find a contradiction. a/b Proof: Suppose log2 7 = a/b, where a and b are integers. We may assume that a > 0 and b > 0. We have 2 = 7, which implies 2 a = 7 b . But the number 2 a is even and the number 7 b is odd, a contradiction. Thus log irrational. 7 must be 16. Suppose |x + 1| d 3 and consider the two cases: x t –1 and x –1. If x ” –1, then x + 1 ” 0 so that |x + 1| = x + 1. This implies x + 1 ” 3 and x ” 2. So in this case we have –1 ” x ” 2. On the other hand, if x < –1, then x + 1 < 0 and |x + 1| = – (x + 1). This leads to – (x + 1) ” 3, x + 1 ” – 3, and x ” –4. In this case we have – 4 ” x < –1. Combining the two cases, we get – 4 ” x ” 2. 17. (a) This proves the converse, which is not a valid proof of the original implication. (b) This is a valid proof using the contrapositive. 18. (a) This is a valid proof in the form of Example 1.3.12 (p): [ p º (q ” r)] ” [( p ” ~ q) º r]. (b) This proves the converse, which is not a valid proof of the original implication. 19. (a) If x p and y m , then x y p m pn qm , so x + y is rational. (b) If x q p and y n m , then xy q n qn pm , so xy is rational. (c) Hint in book: Use a proof by contradiction. Proof: Suppose x p , y is irrational, and suppose x y r Then ( ) r p rq ps ,y x y x s q qs so that y is rational, a contradiction. Thus x + y must be irrational. 20. (a) False. For example, let x = 2 and y = 2. Then x + y = 0. (b) True. This is the contrapositive of Exercise 7(a). (c) False. For example, let x = y = 2. Then xy = 2. (d) True. This is the contrapositive of Exercise 13(b). 21. (a) Let x = 0 and y 2. Then xy = 0. (b) y z xy when x = 0. (c) If x z 0, then the conclusion is true.
- 14. ” Section 1.4 x Techniques of Proof: II 11 22. It is false as stated, and a counterexample is x = 2. Since x is negative, it’s square root is not real, and hence not irrational. (Every irrational number is a real number.) If x t 0, then the result is true, and can be established by looking at the contrapositive. (In order to use the contrapositive, you have to know that the negation of “ x is irrational” is “ x is rational.” This is only true if you know that x is a real number.) 23. Hint in book: Find a counterexample. Solution: 62 + 82 = 102 or ( 2)2 + 02 = 22 . 24. If a, b, c are consecutive odd integers, then a = 2k + 1, b = 2k + 3, and c = 2k + 5 for some integer k. Suppose a2 + b2 = c2 . Then (2k + 1)2 + (2k + 3)2 = (2k + 5)2 . Whence 4k2 – 4k – 15 = 0 and k = 5/2 or k = 3/2. This contradicts k being an integer. 25. It is true. We may label the numbers as n, n + 1, n + 2, n + 3, and n + 4. We have the sum S = n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 10 = 5(n + 2), so that S is divisible by 5. 26. It is true. We may label the numbers as n, n + 1, n + 2, and n + 3. We have the sum S = n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3). Since 2n + 3 is odd, S is not divisible by 4. 27. Hint in book: Consider two cases depending on whether n is odd or even. Proof: If n is odd, then n2 and 3n are both odd, so n2 3n is even. Thus n2 3n 8 is even. If n is even, then so are n2 , 3n, and n2 3n 8. (This uses Exercise 1.3.7.) 28. False. Let n = 1. Then n2 4n 8 = 13. Any other odd n will also work. 2 229. Hint in book: Let z = 2 . If z is rational, we are done. If z is irrational, look at z . 2 2 ( 2 2) 2 Solution: We have z 2 § 2 · 2 2 2.¨ ¸ © ¹ 30. Counterexample: let x = 1/3 and y = –8. Then (1/3)– 8 = 38 is a positive integer and (–8)1/3 is a negative integer. 31. Suppose x ! 0. Then (x + 1)2 = x2 + 2x + 1 ! x2 + 1, so the first inequality holds. For the second inequality, consider the difference 2(x2 +1) (x +1)2 . We have 2(x2 +1) (x +1)2 = 2x2 + 2 x2 2x 1 = x2 2x + 1 = (x 1)2 t 0, for all x, so the second inequality also holds. Analysis with an Introduction to Proof 5th Edition Lay SOLUTIONS MANUAL Full download at: http://testbanklive.com/download/analysis-with-an-introduction-to-proof-5th- edition-lay-solutions-manual/ analysis with an introduction to proof 5th pdf analysis with an introduction to proof 5th edition pdf free analysis with an introduction to proof 5th edition solutions manual pdf analysis with an introduction to proof 5th edition download analysis with an introduction to proof 4th edition analysis with an introduction to proof 4e by steven lay pdf analysis with an introduction to proof solutions 5th analysis with an introduction to proof pdf download