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Discrete Mathematics with Graph Theory (3rd Edition) 77

# Discrete Mathematics with Graph Theory (3rd Edition) 77 -...

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Section 4.1 2. (a) [BB] True. If a and b are real numbers, certainly a - b is a real number. (b) False. If a = 1, b = 2, then a - b = -1 but b - a = 1. 75 (c) False. Ifa = 1,b = 2,c = 3, then (a-b) -c = -1-3 = -4, while a- (b-c) = 1- (-1) = 2. 3. Let a = 3, b = 4, c = -2. Then 3 :5 4 but 3( -2) = -6 1:. -8 = 4( -2). 4. (a) [BB] q = 29, r = 7; (b) [BB] q = -30, r = 10; (c) [BB] q = -29, r = 7; (d) [BB] q = 30, r = 10. 5. (a) q = 278, r = 4; (d) q = 279, r = 15; 6. (a) q = -316, r = 21; (d) q = 20, r = 2984; (b) q = -279, r = 15; (e) q = 0, r = 19; (b) q = 29, r = 972; (e) q = 55, r = 134; (c) q = -278, r = 4; (0 q = -1, r = 5267. (c) q = 70, r = 613; (0 555,555,123 = 555,555,555 - 432 = 5(111,111,111) - 432 = 4(111,111,111) + 111,111,111 - 432 = 4(111,111,111) + 111,110,679 So q = 4; r = 111,110,679 (g) q = 2,104,000; r = 0 7. [BB] Write a = 3q + r with r = 0,1,2. Case 1: r = O. In this case, a = 3q, so a 2 = 9q2 = 3k with k = 3q2. Case 2: r = 1. Here, a = 3q + 1, so a 2 = 9q2 + 6q + 1 = 3k + 1 with k = 3q2 + 2q. Case 3: r = 2. In this case, a = 3q + 2 and a 2 = 9q2 + 12q + 4 = 3k + 1 with k = 3q2 + 4q + 1. S. (a) [BB] The domain of f is Z; its range is Z as well, because given q E Z, we have q = f(qn). (b) [BB] f is not one-to-one since f(O) = f(l).
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