√2 is Irrational

Square it, factor it, contradict yourself — the classic proof by descent.

The idea

$\sqrt{2}$ is irrational: no fraction $p/q$ of integers equals it.

The claim covers every fraction at once, so no amount of checking individual fractions can establish it. We argue instead by contradiction: assume the opposite, derive from that assumption something impossible, and conclude that the assumption was false, so the original claim holds.

The opposite here is that $\sqrt{2} = p/q$ for some integers $p$ and $q$. We may further assume the fraction is in lowest terms, meaning $\gcd(p, q) = 1$. This loses nothing, because cancelling brings any fraction to that form — so ruling out a lowest-terms fraction rules out every fraction.

From the assumption the argument is ordinary algebra: square both sides, clear the denominator, and examine what the resulting equation forces $p$ and $q$ to be. If it forces something incompatible with $\gcd(p, q) = 1$, the assumed fraction cannot exist.

Ways to work on it

• Walkthrough. We prove √2 ℚ from scratch — five steps.
• Practice. Same proof, but for a different prime p.
• Hardest. Why the argument works for primes but fails for perfect squares.

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