Proof that √2 is irrational

Claim. √2 is irrational.

Method. Proof by contradiction. A rational number is a number that can be written as a/b with integers a, b, b ≠ 0. We assume √2 is rational and derive a contradiction.

Proof

1. Suppose √2 is rational. Then √2 = a/b where a and b are integers, b ≠ 0, and a/b is in lowest terms, so gcd(a, b) = 1.
2. Squaring both sides: 2 = a²/b², so a² = 2b².
3. Hence a² is even. If a were odd, a² would be odd. So a is even; write a = 2k for some integer k.
4. Substitute: (2k)² = 2b², that is 4k² = 2b², so b² = 2k².
5. Hence b² is even, and by the same argument as step 3, b is even.
6. Now both a and b are even, so 2 divides both. This contradicts gcd(a, b) = 1.
7. The assumption that √2 is rational is false.

Therefore √2 is irrational. ∎

Note: the fact used in step 3 is "if a² is even then a is even". It is proved by contraposition: if a = 2m + 1, then a² = 4m² + 4m + 1 = 2(2m² + 2m) + 1, which is odd.