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Discrete Mathematics with Graph Theory (3rd Edition) 44

# Discrete Mathematics with Graph Theory (3rd Edition) 44 -...

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42 Solutions to Exercises Transitive: If a f"V b and b f"V c, then 3a + b ::::; 4k for some integer k and 3b + c = for some integeri. Since4(k+£) = (3a+b)+(3b+c) = (3a+c) +4b, we see that 3a+c = 4(k+£) -4b is a multiple of 4 and, hence, that a f"V c. (b) 0 = {x E Z I x f"V O} = {x I 3x = 4k for some integer k}. Now if 3x = 4k, k must be a multiple of 3. So 3x = 12£ for some £ E Z and x = 4£. 0 = 4Z. (c) 2 = {x E Z I x f"V 2} = {x I 3x + 2 = 4kforsomeintegerk} = {x I 3x = 4k- 2 for some integer k} Now if 3x = 4k - 2, then 3x = 3k + k - 2 and so k - 2 is a multiple of 3. Therefore, k = 3£ + 2 for some integer £, 3x = 4(3£ + 2) - 2 = 12£ + 6 and x = + 2. S02 = 4Z+2. (d) The quotient set is {4Z, 4Z + 1, 4Z + 2, 4Z + 3}. to. (a) Reflexive: For any a E Z, a f"V a because 3a + 4a = 7a and a is an integer. Symmetric: If a, b E Z and a f"V b, then 3a + 4b = 7n for some integer n. Then 3b + 4a = (7a + 7b) - (3a + 4b) = 7a+ 7b -7n = 7(a + b - n) and a + b- n is an integer. Thus b f"V a. Transitive: Suppose a, b, c E Z with a f"V b and b f"V c. Then 3a + 4b = 7n and 3b + 4c = 7m for some integers n and m.
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