Is $\text{rational}^{\text{irrational}}$ rational or irrational? Is the number $\text{rational}^{\text{irrational}}$ rational or irrational?
For example $2^{\sqrt{2}}$: is it rational or irrational?
I tried using a logarithm but it didn't work. It seems by superficial studying that it will be irrational. But what is the proof?
 A: This is essentially the Gelfond-Schneider Theorem which says:
$$
\text{If } a,b \text{ are algebraic}, a \neq 0,1 \text{ and } b \in \mathbb{R} \setminus \mathbb{Q} \text{ then } a^b \text{ is transcendental.} 
$$
Now every transcendental number is also irrational, and every rational number is algebraic. $\sqrt{2}$ is also algebraic so in this case, yes $2^\sqrt{2}$ is irrational and transcendental.
But (I believe) a non-zero algebraic number to the power of a transcendental number can either be rational or irrational, so a rational to the power of an irrational may be rational, only if the irrational is also transcendental. For concrete examples note that $\log_2 (3)$ is irrational, but $2^{\log_2 (3)} = 3$, however I suspect $2^\pi$ is irrational.
Examples of algebraic numbers are say $\sqrt{n}$ or $\sqrt{3 + \sqrt{2}}$ or anything that is the root of a polynomial (that is in fact the definition), whereas transcendental numbers are more like $\pi$ or $e$.
Hopefully this was helpful for you, but in that particular example, yes  $2^\sqrt{2}$ is irrational.
A: Given any two rational numbers, $a,b$, note that $a^{\log_a(b)} = b$ is rational
but $\log_a(b)$ is generally not rational.
For example $2^{\log_2(3)} = 3$ but $\log_2(3)$ is not rational.
So a rational number taken to an irrational power can easily be rational.
However, if $a\neq 0,1$ is rational (in fact even algebraic) and $b$ is an irrational algebraic number, then it is known that $a^b$ is actually transcendental. This is called the Gelfond–Schneider theorem.
