## Examples 1.1.5(b):

Prove that if the square of a number is an even integer, then the original
number must also be an even integer. (Try a proof by contradiction).

To prove this we first need to know what exactly an even and odd integer is:
- an integer
*x*is even if*x = 2n*for some integer*n* - an integer
*x*is odd if*x = 2n + 1*for some integer*n*

*x*is a number such that

*x*is even. To start a proof by contradiction we will assume the opposite of what we would like to prove: assume that

^{2}*x*is odd (but

*x*is still even). Then, because

^{2}*x*is odd, we can write it as

*x = 2n + 1*

*x*is

*x*^{2}= (2n + 1)(2n + 1) = 4 n^{2}+ 4n + 1 = 2 (2 n^{2}+ 2n) + 1 = 2k + 1

*k = 2 n*. Therefore

^{2}+ 2n*x*is odd. But that is contrary to our assumption that the square of

^{2}*x*is even. Hence, if the square of

*x*is supposed to be even,

*x*itself must be 'not odd'. But 'not odd' means even. Therefore, the proof is finished.