To prove that √2 is irrational we assume √2 = p/q, where p and q are integers with no common factor. The contradiction appears because:

Squaring gives $p^2=2q^2$, so $p^2$ is even and hence $p$ is even. Writing $p=2m$ gives $4m^2=2q^2$, i.e. $q^2=2m^2$, so $q$ is even too. But then $p$ and $q$ share the factor 2, contradicting the assumption that they have no common factor — so no such fraction exists.