Mathematical Elegance Math Example 4

Solution

1. 1
   Assume $\sqrt{2} = p/q$ in lowest terms ($\gcd(p,q)=1$). Then $p^2=2q^2$.
2. 2
   So $p^2$ is even $\Rightarrow$ $p$ is even $\Rightarrow$ $p=2k$ $\Rightarrow$ $4k^2=2q^2$ $\Rightarrow$ $q^2=2k^2$ $\Rightarrow$ $q$ is even.
3. 3
   Both $p$ and $q$ are even, contradicting $\gcd(p,q)=1$.
4. 4
   Elegant core: the same argument that makes $p$ even also makes $q$ even — the structure propagates perfectly to the contradiction.