# Proof of $$\sqrt{2}$$ Irrationality

Suppose to the contrary that $$\sqrt{2}$$ is rational. Then, there must exist two coprime integers $$p$$ and $$q$$ such that$$\sqrt{2} = \frac{p}{q}$$. We can rearrange this equation as follows:

$\sqrt{2} = \frac{p}{q}$ $2q^2 = p^2$

From the above it is clear that $$p^2$$ is even. It follows that $$p$$ is also even, since the square of any odd number is odd. Thus, we can substitute $$p$$ with $$2k$$, where $$k$$ is any integer:

$2q^2 = p^2$ $2q^2 = 4k^2$ $q^2 = 2k^2$

By a similar argument, we can again see that both $$q^2$$ and $$q$$ are even. The fact that both $$p$$ and $$q$$ are even contradicts the initial statement that the two are coprime. Thus, we have reached a contradiction, and $$\sqrt{2}$$ must be irrational.