Instructor’s Solutions Manual — Discrete Mathematics, 8th Edition — Richard Johnsonbaugh — ISBN 9780321964687

Discrete Mathematics – 8th Edition
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INSTRUCTOR’S
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SOLUTIONS
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MANUAL
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Richard Johnsonbaugh
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Comprehensive Solutions Manual for Instructors

and Students
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© Richard Johnsonbaugh

All rights reserved. Reproduction or distribution without permission is prohibited.
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©STUDYSTREAM

, Instructor’s Solutions Manual, Discrete Mathematics, 8e
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Contents

Chapter 1 1
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Chapter 2 14

Chapter 3 39
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Chapter 4 56

Chapter 5 77

Chapter 6 85
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Chapter 7 115

Chapter 8 136
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Chapter 9 158

Chapter 10 182

Chapter 11 187
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Chapter 12 196

Appendices 208
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Solutions to Selected Exercises
Section 1.1
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2. {2, 4} 3. {7, 10} 5. {2, 3, 5, 6, 8, 9} 6. {1, 3, 5, 7, 9, 10}

8. A 9. ∅ 11. B 12. {1, 4} 14. {1}
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15. {2, 3, 4, 5, 6, 7, 8, 9, 10} 18. {n ∈ Z+ | n ≥ 6} 19. {2n − 1 | n ∈ Z+ }

21. {n ∈ Z+ | n ≤ 5 or n = 2m, m ≥ 3} 22. {2n | n ≥ 3} 24. {1, 3, 5}

25. {n ∈ Z+ | n ≤ 5 or n = 2m + 1, m ≥ 3} 27. {n ∈ Z+ | n ≥ 6 or n = 2 or n = 4}

29. 1 30. 3
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33. We find that B = {2, 3}. Since A and B have the same elements, they are equal.

34. Let x ∈ A. Then x = 1, 2, 3. If x = 1, since 1 ∈ Z+ and 12 < 10, then x ∈ B. If x = 2, since 2 ∈ Z+ and
22 < 10, then x ∈ B. If x = 3, since 3 ∈ Z+ and 32 < 10, then x ∈ B. Thus if x ∈ A, then x ∈ B.
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Now suppose that x ∈ B. Then x ∈ Z+ and x2 < 10. If x ≥ 4, then x2 > 10 and, for these values of x,
x∈/ B. Therefore x = 1, 2, 3. For each of these values, x2 < 10 and x is indeed in B. Also, for each of
the values x = 1, 2, 3, x ∈ A. Thus if x ∈ B, then x ∈ A. Therefore A = B.

37. Since (−1)3 − 2(−1)2 − (−1) + 2 = 0, −1 ∈ B. Since −1 ∈
/ A, A 6= B.

38. Since 32 − 1 > 3, 3 ∈
/ B. Since 3 ∈ A, A 6= B. 41. Equal 42. Not equal
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45. Let x ∈ A. Then x = 1, 2. If x = 1,

x3 − 6x2 + 11x = 13 − 6 · 12 + 11 · 1 = 6.

Thus x ∈ B. If x = 2,
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x3 − 6x2 + 11x = 23 − 6 · 22 + 11 · 2 = 6.
Again x ∈ B. Therefore A ⊆ B.

46. Let x ∈ A. Then x = (1, 1) or x = (1, 2). In either case, x ∈ B. Therefore A ⊆ B.
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49. Since (−1)3 − 2(−1)2 − (−1) + 2 = 0, −1 ∈ A. However, −1 ∈
/ B. Therefore A is not a subset of B.

50. Consider 4, which is in A. If 4 ∈ B, then 4 ∈ A and 4 + m = 8 for some m ∈ C. However, the only value
of m for which 4 + m = 8 is m = 4 and 4 ∈ / C. Therefore 4 ∈
/ B. Since 4 ∈ A and 4 ∈
/ B, A is not a
subset of B.

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, 2 SOLUTIONS


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62. 32 63. 105 65. 51
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67. Suppose that n students are taking both a mathematics course and a computer science course. Then
4n students are taking a mathematics course, but not a computer science course, and 7n students are
taking a computer science course, but not a mathematics course. The following Venn diagram depicts
the situation:

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