Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 7D / i / Solution

If $q=\sqrt{2}+\sqrt{3}$ were a rational number, then $q^2=5+2\sqrt{6}$ would make $\sqrt{6}$ rational. Write $\sqrt{6}=a/b$ in lowest terms. Then $a^2=6b^2$, which is impossible by unique factorization: the exponent of the prime $2$ is even on the left and odd on the right. Therefore