A rational number is any number that can be expressed $\frac{P}{Q}$, where $P$ and $Q$ are integers, and $Q$ is not 0.
Let’s assume $\sqrt{2}$ is rational. This means it can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $\gcd(p, q) = 1$ (i.e. $\frac{p}{q}$ is in its simplest form). Therefore:
$$ \sqrt{2} = \frac{p}{q} $$
$$ 2 = \frac{p^2}{q^2} $$
$$ 2q^2 = p^2 $$
Therefore $p^2$ is even since it is 2 times $q^2$ (which is an integer). This means that $p$ must be even. Here’s why:
Assume $p$ is odd. That means $p = 2k+1$ for some integer $k$. Therefore $p^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2+k)+1$ which is odd. But we know that $p^2$ is even, so $p$ cannot equal $2k+1$ (i.e. $p$ cannot be odd and must be even).
So since $p$ is even, it can be written as $p = 2k$ for some integer $k$. Let’s substitute this into $2q^2 = p^2$:
$$ 2q^2=(2k)^2 $$
$$ 2q^2=4k^2 $$
$$ q^2=2k^2 $$
So $q^2$ is even and so $q$ must also be even (by the same argument we used for $p$). Therefore both $p$ and $q$ are even and so $\gcd(p, q) \ge 2$. But this contradicts our original assumption that $\gcd(p,q)=1$, so $\sqrt{2}$ must be irrational.
Consider $\alpha = \sqrt{2}^{\sqrt{2}}$. Either $\alpha$ is rational or irrational.
Case 1: $\alpha$ is rational
Let $p = q = \sqrt{2}$. Therefore both $p$ and $q$ are irrational. But $p^q = \sqrt{2}^{\sqrt{2}} = \alpha$, so if $\alpha$ is rational, then we have irrational numbers $p$ and $q$ such that $p^q$ is rational. Case 2: $\alpha$ is irrational
Let $p = \alpha = \sqrt{2}^{\sqrt{2}}, q = \sqrt{2}$. So $p^q = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2}\sqrt{2}} = (\sqrt{2})^2 = 2$.