Proof that √2 is Irrational: Interactive Walkthrough

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 $\gcd(p, q) = 1$. $\blacksquare$

The card below walks the same argument one line at a time, and checks every line on a pair you choose. Drag $p/q$ along the number line or set $p$ and $q$: a ✓ marks a line your pair makes true. However you choose, the first line (coprime) and the eighth (both even) never hold together, and $p^2 - 2q^2$ is never zero. “Next near miss” jumps through the fractions that come closest, each missing by exactly one.

Nikolay Eremeev, Senior Data Infrastructure Engineer. Writes about what he builds in production. CV ·hi@nikolay-eremeev.com