Rational number to the power of irrational number = irrational number. True?

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).

Solution 1:

Consider $2^{\log_2 3}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

Solution 2:

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.)