Rationality of a power with irrational exponent [duplicate]

Question

The following fact is known:

If $a$ and $b$ are both irrational numbers, then $a^b$ can be a rational number.

Proof. Suppose $a^b$ is always irrational. Then $\sqrt{2}^\sqrt{2}$ is an irrational number, which in turn implies that $\left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2}$ must also be irrational. However: $$ \left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2} = \sqrt{2}^{\sqrt{2}\times\sqrt{2}} = 2 $$

---

Now assume $a$ is a rational number and $b$ is irrational. I wonder if $a^b$ can be a rational number. Any way to prove, or a reference maybe?

Answer 1

Let $\log_a$ be the logarithm to the base $a$. $$a^{\log_a c}=c$$ If $$a^\frac{q}{p}=c$$ then $$a^q=c^p$$ and each prime that divides $a$ must divide $c$ and each prime that divides $c$ must divide $a$. So if this is not the case then $\log_ac$ must be irrational.

Answer 2

$2^b = 3$ where $b = \log_2(3)$ is irrational. In fact, if $2^b = c$ where $b$ and $c$ are rational, $b$ must be an integer.

Answer 3

You can use the Gelfond-Schneider theorem to prove that $a^b$ is not only irrational, but transcendental. The theorem states that:

If $a$ and $b$ are algebraic numbers with $a ≠ 0,1$ and $b$ non-rational, then any value of $a^b$ is a transcendental number.

This not only gives rise to trivial results such that $2^{\sqrt 2}$ is irrational, but to some more interesting ones as well. For instance, by taking $a=-1$ and $b=-i$, where $i=\sqrt {-1}$, we can obtain the following result:

$$ (-1)^{-i} = \left( e^{i \pi} \right)^{-i} = e^\pi, $$

showing that $e^\pi$ is transcendental.