# 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. $\blacksquare$