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Discrete Mathematics with Graph Theory (3rd Edition) 8

# Discrete Mathematics with Graph Theory (3rd Edition) 8 - 6...

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6 Solutions to Exercises 17. n 2 - n + 5 = n(n - 1) + 5. Now either n - 1 or n is even, since these integers are consecutive. So n( n - 1) is even. Since the sum of an even integer and the odd integer 5 is odd, the result follows. 18. [BB] 2X2 - 4x + 3 = 2(x 2 - 2x) + 3 = 2[(x _1)2 - 1] + 3 = 2(x - 1)2 + 1 is the sum of 1 and a nonnegative number. So it is at least 1 and hence positive. 19. For a 2 - b 2 to be odd, it is necessary and sufficient for one of a or b to be even while the other is odd. Here's why. Case i: a, b even. In this case, a = 2n and b = 2m for some integers m and n, so a 2 - b 2 = 4n 2 - 4m 2 = 4(n 2 - m 2 ) is even. Case ii: a, b odd. In this case, a = 2n + 1 and b = 2m + 1 for some integers m and n, so a 2 - b 2 = (4n 2 + 4n + 1) - (4m 2 + 4m + 1) = 4(n 2 + n - m 2 - m) is even. Case iii: a even, b odd. In this case, a = 2n and b = 2m+1 for some integers n and m, so a 2 _b 2 = 4n 2 - (4m2+4m+ 1) = 4(n 2 - m 2 - m) - 1 is odd. Case iv: a odd, b even. This is similar to Case iii, and the result follows. 20. [BB] ( ----+) To prove this direction, we establish the contrapositive, that is, we prove that
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