Discrete Mathematics – 8th Edition – Richard Johnsonbaugh – Complete Solutions Manual with Step-by-Step Answers

SOLUTIONS for
Discrete Mathematics, 8th
edition
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Author (s): Richard Johnsonbaugh
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, Solutions to Selected Exercises
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Section 1.1
2. {2, 4} 3. {7, 10} 5. {2, 3, 5, 6, 8, 9} 6. {1, 3, 5, 7, 9, 10}
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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.
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38. Since 32 − 1 > 3, 3 ∈
/ B. Since 3 ∈ A, A 6= B. 41. Equal 42. Not equal

45. Let x ∈ A. Then x = 1, 2. If x = 1,

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

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


53.

U
A B




54.
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U
A B
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56.


A B U
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C



57.
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U
A
B
C
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59.
U
A B
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C



62. 32 63. 105 65. 51

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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, SOLUTIONS 3


Math
'$
'$
CompSci

4n n 7n

&%
&%

Thus, the total number of students is
4n + n + 7n = 12n.

The proportion taking a mathematics course is
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5n 5
= ,
12n 12
which is greater than one-third.
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69. {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}

70. {(1, 1), (1, 2), (2, 1), (2, 2)} 73. {(1, a, a), (2, a, a)}
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74. {(1, 1, 1), (1, 2, 1), (2, 1, 1), (2, 2, 1), (1, 1, 2), (1, 2, 2), (2, 1, 2), (2, 2, 2)}

77. Vertical lines (parallel) spaced one unit apart extending infinitely to the left and right.

79. Consider all points on a horizontal line one unit apart. Now copy these points by moving the horizontal
line n units straight up and straight down for all integers n > 0. The set of all points obtained in this
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way is the set Z × Z.

80. Ordinary 3-space

82. Take the lines described in the instructions for this set of exercises and copy them by moving n units out
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and back for all n > 0. The set of all points obtained in this way is the set R × Z × Z.

84. {1, 2}
{1}, {2}

85. {a, b, c}
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{a, b}, {c}
{a, c}, {b}
{b, c}, {a}
{a}, {b}, {c}
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88. False 89. True 91. False 92. True
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94. ∅, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d},
{a, c, d}, {b, c, d}, {a, b, c, d}. All except {a, b, c, d} are proper subsets.

95. 210 = 1024; 210 − 1 = 1023 98. B ⊆ A 99. A = U

102. The symmetric difference of two sets consists of the elements in one or the other but not both.

103. A 4 A = ∅, A 4 A = U , U 4 A = A, ∅ 4 A = A

105. The set of primes

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