Homework 8 Solutions

Abstract

1. Suppose x ∈ Z. Then x is odd if and only if 3x + 5 is even. Solution: Proof. =⇒ Let x be an odd integer. Then x = 2a + 1 for some integer a. So, 3x + 5 = 3(2a + 1) + 5 = 6a + 8 = 2(3a + 4). Since 3a + 4 is an integer, 3x + 5 is even. ⇐= (contrapositive) Suppose that x is an even integer. Then x = 2b for some integer b. So, 3x + 5 = 3(2b) + 5 = 6b + 5 = 2(3b + 2) + 1. Since 3b + 2 is an integer, 3x + 5 is odd. 2. Suppose x, y ∈ R. Then x 3 + x 2 y = y 2 + xy if and only if y = x 2 or y = −x. Solution: Proof. =⇒ Let x, y ∈ R, and suppose that x 3 + x 2 y = y 2 + xy. Then we have x 2 (x + y) = y(x + y). If x + y = 0, then we may divide both sides by x + y to obtain x 2 = y. If x + y = 0, then we obtain 0 = 0, a true statement. But x + y = 0 is the same as y = −x. Therefore, if x 3 + x 2 y = y 2 + xy, then y = x 2 or y = −x.