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.