= Solution
Suppose $\sqrt n=a/b$ in lowest terms. Then $a^2=nb^2$. Comparing prime exponents shows that every prime dividing $b$ also divides $a$, unless $b=1$. Hence $\sqrt n=a$ is an integer. Therefore \b[$\sqrt n$ 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:
$$
\sqrt2+(-\sqrt2)=0,\qquad
\sqrt2\,\sqrt2=2,\qquad
(\sqrt2)^2=2.
$$
For $\alpha^\beta$, let $r=(\sqrt2)^{\sqrt2}$. If $r$ is rational, $\alpha=\beta=\sqrt2$ is a counterexample. If $r$ is irrational, take $\alpha=r$ and $\beta=\sqrt2$; then $\alpha^\beta=2$. Thus \b[$\alpha+q$ is the sole expression forced to be irrational].
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