Why is the square root of 2 an irrational number?

The classic proof shows that $\sqrt{2}$ cannot be written as a fraction of whole numbers. Assume the opposite: $$\sqrt{2}=\frac{p}{q},$$ where $p$ and $q$ have no common factor. Squaring both sides gives: $$2=\frac{p^2}{q^2},$$ so: $$p^2=2q^2.$$ That means $p^2$ is even, so $p$ must be even. Write $p=2k$. Substitute: $$4k^2=2q^2,$$ so: $$q^2=2k^2.$$ Now $q^2$ is even, so $q$ is even too. But now both $p$ and $q$ are even, which means they have a common factor of 2. That contradicts our assumption that the fraction was already simplified. So $\sqrt{2}$ is irrational. The geometric reason this feels surprising is that $\sqrt{2}$ is just the diagonal of a $1$ by $1$ square. A very simple length can still fail to fit into the fraction system.