Proof that √2 is Irrational

Theorem

$\sqrt{2}$ is irrational — it cannot be expressed as $\frac{p}{q}$ where $p, q$ are integers with no common factors.

Proof by contradiction

Assume $\sqrt{2} = \frac{p}{q}$ in lowest terms (i.e., $\gcd(p, q) = 1$).

Squaring both sides: $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$.

Since $p^2$ is even, $p$ must be even (the square of an odd number is odd). Write $p = 2r$.

Substituting: $(2r)^2 = 2q^2 \Rightarrow 4r^2 = 2q^2 \Rightarrow q^2 = 2r^2$.

So $q^2$ is even, meaning $q$ is also even.

But if both $p$ and $q$ are even, they share factor 2 — contradicting our assumption that $\gcd(p, q) = 1$. $\blacksquare$