$\begingroup$I suggested the following problem to my friend: prove that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational. The problem seems to have been discussed in this question.
Now, his inital solution was like this: let's take a rational number $r$ and an irrational number $i$. Let's assume
$$a = r^i$$ $$b = \frac{1}{i}$$
So we have
$$a^b = (r^i)^\frac{1}{i} = r$$
which is rational per initial supposition. $b$ is obviously irrational if $i$ is. My friend says that it is also obvious that if $r$ is rational and $i$ is irrational, then $r^i$ is irrational. I quickly objected saying that $r = 1$ is an easy counterexample. To which my friend said, OK, for any positive rational number $r$, other than 1 and for any irrational number $i$ $r^i$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?
Let's stick to real numbers only (i.e. let's forget about complex numbers for now).
$\endgroup$52 Answers
$\begingroup$Consider $2^{\log_2 3}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
$\endgroup$6$\begingroup$A supplement to the answer by Chris above:
Let $r$ be a positive rational number and $i$ a positive irrational number. If $r^{i}$ is rational, then $r^i=\frac{a}{b}$ for $a,b\in \mathbb{Z}$ such that $b\neq 0$. In particular, $i=\log_{r}\left(\frac{a}{b}\right)$. Therefore, Chris Eagle's answer is, in fact, prototypical. (Note also, that if $r=1$, then we get a contradiction as you observed.)
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