Examples 1.1.5(b):
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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
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
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