Proving √2 is irrational

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

Proof (by contradiction): Assume \sqrt{2} \in \mathbb{Q}. This implies there must exist a pair a,b \in \mathbb{N} such that gcd(a,b) = 1 for which \frac{a}{b} = \sqrt{2}. This can be rewritten as 2 \cdot b^{2} = a^{2}, which implies a is even, since a necessary and sufficient that a^{2} = 2 \cdot n (where n \in \mathbb{N}) is that a^{2} (and therefor also a) has 2 as one of its prime factors. This means there must exist an integer k such that 2k = a. If we substitute a for 2k we get \frac{2k}{b} = \sqrt{2} \leftrightarrow \frac{(2k)^{2}}{b^{2}} = 2 \leftrightarrow 2 \cdot b^{2} = 4k^{2} \leftrightarrow b = 2k^{2}. This also implies b is an even integer. But this is a contradiction, because if both a and b are even integers gcd(a,b) > 1. Therefor \sqrt{2} must be irrational. \blacksquare


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