Irrationality of $\sqrt{2}$

Suppose, for contradiction, that $\sqrt{2}$ is rational. Then we can write $\sqrt{2} = \frac{p}{q}$ for integers $p,q$ with $q \neq 0$, and by cancelling common factors we may assume $\gcd(p,q)=1$ (the fraction is in lowest terms).

Squaring both sides gives $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$. This means $p^2$ is even. Since the square of an odd number is odd, $p$ itself must be even; write $p = 2k$ for some integer $k$.

Substituting back, $(2k)^2 = 2q^2$, so $4k^2 = 2q^2$, which simplifies to $q^2 = 2k^2$. This means $q^2$ is even, so by the same reasoning $q$ must also be even.

But now both $p$ and $q$ are even, so $2$ divides both of them, contradicting the assumption that $\gcd(p,q)=1$. This contradiction shows that the original assumption — that $\sqrt{2}$ can be written as $\frac{p}{q}$ — must be false. Therefore $\sqrt{2}$ is irrational.