Discrete Mathematics with Graph Theory (3rd Edition) 9

Discrete - Section 0.2 7 23 By Exercise 22 there exists an integer k such that n = 4 k 1 o r n = 4 k 3 C ase 1 n = 4 k 1 I f k is even there exists

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Section 0.2 7 23. By Exercise 22, there exists an integer k such that n = 4k + 1 or n = 4k + 3. Case 1: n = 4k + 1. If k is even, there exists an integer m such that k = 2m, so n = 4(2m) + 1 = 8m + 1, and the desired conclusion is true. If k is odd, there exists an integer m such that k = 2m + 1, so n = 4(2m + 1) + 1 = 8m + 5, and the desired conclusion is true. Case 2: n = 4k + 3. If k is even, there exists an integer m such that k = 2m, so n = 4(2m) + 3 = 8m + 3, and the desired conclusion is true. If k is odd, there exists an integer m such that k = 2m + 1, so n = 4(2m + 1) + 3 = 8m + 7, and the desired conclusion is true. In all cases, the desired conclusion is true. 24. [BB] If the statement is false, then there does exist a smallest positive real number r. Since ~r is positive and smaller than r, we have reached an absurdity. So the statement must be true. 25. We give a proof by contradiction. If the result is false, then both a > Vn and b > Vn. (Note that the negation of an "or" statement is an "and" statement.) But then
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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