Solution
= Solution
If $q=\sqrt2+\sqrt3$ were a <rational number>, then $q^2=5+2\sqrt6$ would make $\sqrt6$ rational. Write $\sqrt6=a/b$ 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
$$
\boxed{\sqrt2+\sqrt3
otin\mathbb Q.}
$$