Let $x$ be positive, rational, but not an integer. That means $x$ can be written as $\frac{p}{q}$ with $p,q$ coprime, $p,q \neq 0$ and $q \neq 1$. Is $x^x$ always irrational?
I think that this has to be the case, but I can't prove it.
$x^x = (\frac{p}{q})^\frac{p}{q} = \frac{p^\frac{p}{q}}{q^\frac{p}{q}}$
I know that $p^\frac{p}{q}$ and $q^\frac{p}{q}$ can't both be rational, but there are cases where the division of two irrational numbers gives us an rational number. For example $\frac{\sqrt{2}}{\sqrt{2}} = 1$
You have $y=x^x=\left(\cfrac pq\right)^{\frac pq}=\left(\cfrac {p^p}{q^p}\right)^{\frac 1q}$ so that $$y^q=\frac {p^p}{q^p}$$
Hence the rational fraction $y$ is a $p^{th}$ power - say $y=z^p$
Then $z^q=\frac pq$, and $\frac pq$ is a $q^{th}$ power. Now $q$ is an integer which is a $q^{th}$ power of an integer. But for $q\gt 1$ we have $2^q\gt q$.