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).
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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

Now that we have a precise definition, the actual proof is easy: Suppose x is a number such that x² is even. To start a proof by contradiction we will assume the opposite of what we would like to prove: assume that x is odd (but x² is still even). Then, because x is odd, we can write it as

x = 2n + 1

But then the square of x is

x² = (2n + 1)(2n + 1) = 4 n² + 4n + 1 = 2 (2 n² + 2n) + 1 = 2k + 1

with k = 2 n² + 2n. Therefore x² is odd. But that is contrary to our assumption that the square of 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.

Interactive Real Analysis, ver. 1.9.5
(c) 1994-2007, Bert G. Wachsmuth
Page last modified: Mar 28, 2007