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Here's the classic proof by contradiction that √2 is irrational:

Proof:

1. Assume √2 is rational. Then it can be expressed as a reduced fraction: $\sqrt{2} = \frac{a}{b}$ where a, b ∈ ℤ, b ≠ 0, and gcd(a,b) = 1

2. Square both sides: $2 = \frac{a^2}{b^2}$

3. Multiply both sides by b²: $2b^2 = a^2$

4. This implies a² is even. Since the square of an odd number is odd, a must be even. Let: $a = 2k$ for some integers k

5. Substitute back: $2b^2 = (2k)^2 = 4k^2$ → $b^2 = 2k^2$

6. By similar logic, b² - and thus b - must also be even

7. Contradiction: Both a and b being even violates gcd(a,b) = 1. Therefore, our initial assumption is false. QED.

Key Insight: This demonstrates the power of proof by contradiction and shows how number parity properties reveal fundamental truths about irrational numbers.