 Solution

Suppose $\sqrt{n}=a/b$ in lowest terms. Then $a^2=n{b}^2$. Comparing prime exponents shows that every prime dividing $b$ also divides $a$, unless $b=1$. Hence $\sqrt{n}=a$ is an integer. Therefore $\sqrt{n}$ is either an integer or irrational.

Only $\alpha +q$ must be irrational: if it were rational, subtracting rational $q$ would make $\alpha$ rational. None of the others must be irrational:For ${\alpha }^{\beta }$, let $r=(\sqrt{2}{)}^{\sqrt{2}}$. If $r$ is rational, $\alpha =\beta =\sqrt{2}$ is a counterexample. If $r$ is irrational, take $\alpha =r$ and $\beta =\sqrt{2}$; then ${\alpha }^{\beta }=2$. Thus $\alpha +q$ is the sole expression forced to be irrational.