Irrationality of Square Root 2

Theorem: $\sqrt{2}$ is irrational.

Proof (by contradiction):

Assume $\sqrt{2}$ is rational, i.e., $\sqrt{2} = \dfrac{p}{q}$ where $p$ and $q$ are co-prime integers and $q \neq 0$.

Squaring both sides:

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

This means $p^2$ is even, so $p$ is even. Let $p = 2k$:

$$(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$$

So $q^2$ is also even, meaning $q$ is even. But this contradicts our assumption that $p$ and $q$ are co-prime. Hence, $\sqrt{2}$ is irrational. $\blacksquare$