How to prove that √2 is irrational by contradiction?

These are excellent questions that get to the heart of the proof. **Why $p^2$ even $\Rightarrow$ $p$ even?** This relies on the fundamental property of integers: if an integer is odd, its square is odd. Let $p = 2k+1$, then $p^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$, which is odd. By contrapositive: if $p^2$ is even, then $p$ cannot be odd, so $p$ is even. **Why doesn't this work for $\sqrt{4}$?** If we try the same proof for $\sqrt{4} = p/q$ in lowest terms: $$4q^2 = p^2$$ Here $p^2$ is divisible by 4, which implies $p$ is even, say $p = 2k$. Then: $$4q^2 = 4k^2 \quad \Rightarrow \quad q^2 = k^2$$ So $q = k$, meaning $p/q = 2k/k = 2$, which is already in lowest terms when $k = 1$. The proof does **not** force a contradiction because the factor of 4 cancels completely. The critical difference is that $2$ is not a perfect square, so the prime factorisation argument forces $p$ and $q$ to share a factor of 2. For $\sqrt{4} = 2$, the equation $4q^2 = p^2$ is satisfied by $p = 2q$ without contradiction.