Answer to Question #123220 in Discrete Mathematics for Kavee

Question

Show that these statements about the integer x are equivalent.
(i) 3x + 2 is even. (ii) x + 5 is odd. (iii) x2 is even.

Answer 1

We prove that all these are equivalent to x being even. If x is even, then x = 2k for some integer k. Therefore 3x + 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1), which is even, because it has been written in the form 2t, where t = 3k + 1. Similarly, x + 5 = 2k + 5 = 2k + 4 + 1 = 2(k + 2) + 1, so x + 5 is odd;  and x² = (2k)² = 2(2k²), so x² is even. For the converses, we will use a proof by contraposition. So assume that x is not even; thus x is odd and we can write x = 2k + 1 for some integer k. Then 3x + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1, which is odd (i.e., not even), because it has been written in the form 2t + 1, where t = 3k + 2. Similarly, x + 5 = 2k + 1 + 5 = 2(k + 3), so x + 5 is even (i.e., not odd). Lastly, x² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, so x² is odd