# The [[Square Root]] of 2 Is Irrational Proving that $\sqrt 2$ is irrational is done by contradiction so we'll start by assuming that it is actually [[Rational Numbers|rational]]. For this proof it's important to point out that [[Squaring a Number Does Not Affect Its Parity]]. $ \begin{align} \sqrt 2 &= \frac pq && \text{square both sides} \\ 2 & =\frac{p^2}{q^2} \\ p^2 &= 2q^2 && \text{⟹ p is even}\\ &&& \text{substitute } p=2n\\ (2n)^2 &= 2q^2 \\ 4n^2 &= 2q^2 \\ 2n^2 &= q^2 && \text{⟹ q is even} \end{align} $ Both p and q cannot be even (as per the definition of [[Rational Numbers]]), and so our original assumption must be incorrect.