# Interactive Real Analysis - part of MathCS.org

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## 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
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.
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