For all integers n, if n2 is odd, then n is odd. Use a proof by contraposition, as in Lemma 1.1. Let n be an integer. Suppose that n is even, i. e., n =

Let's assume that the statement "if n^2 is odd, then is odd" is false. That would mean "n^2 is odd" leads to "n is even"

Suppose n is even. That means n = 2k where k is any integer.

Square both sides

n = 2k

n^2 = (2k) ^2

n^2 = 4k^2

n^2 = 2 * (2k^2)

The expression 2 (2k^2) is in the form 2m where m is an integer (m = 2k^2) which shows us that n^2 is also even.

So this contradicts the initial statement which forces n to be odd.