I stumbled across this proof, and thought it was pretty neat, so I wanted to sketch up my own proof:

**Proof (by contradiction)**: Assume . This implies there must exist a pair such that for which . This can be rewritten as , which implies is *even*, since a necessary and sufficient that (where ) is that (and therefor also ) has 2 as one of its prime factors. This means there must exist an integer such that . If we substitute for we get . This also implies is an even integer. But this is a contradiction, because if both and are even integers . Therefor must be irrational.

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