Is an irrational to an irrational rational? I am working on this logical proof for class and trying to either prove or disprove that an irrational to an irrational power is also irrational. Please don't answer the problem for me but I'm completely stuck on how to begin. Any hints would be appreciated.
 A: If an irrational positive real raised to an irrational power were always irrational, then whenever $$\text{irrational}^{x}=\text{rational}\text{,}$$ $x$ would have to be rational. This is equivalent to saying that $\log_{\text{irrational}}(\text{rational})$ would have to be rational. So then for any fixed rational $r$, the map $$\text{irrational}\mapsto\log_{\text{irrational}}({r})=\frac{\ln(r)}{\ln(\text{irrational})}$$ would be a decreasing map from irrationals greater than $1$ to rationals greater than $0$. Can you have a monotonic (and hence one-to-one) map from an uncountable subset of the reals to a countable one?
A: I think I learned this first from MSE but I could not retrieve in what question. I googled for it and found it here, on the blog of Andrej Bauer, 2009.  It is probaby a well-known folklore fact, albeit a lot less known than the standard $\sqrt{2}^{\sqrt{2}}$ based proof.
Here it goes: take $a = \sqrt{2}$ and $b = \log_2{9}= 2\, \log_23$. Then
$$ a^b = \sqrt{2}^{2 \log_2 3} = 2^{\log_2 3} = 3$$ is rational, but it is well known and easy to show that both $\sqrt{2}$ and $\log_2 9$ are irrational. (Both follow from unique factorisation into primes in some sense.)
A: You can disprove the statement by simply finding a situation where an irrational number to the power of an irrational number is a rational number.
For example, consider
$$\left(\sqrt2^\sqrt2\right)^\sqrt2$$
