After I've read this question in this Mathematics Stack Exchange, and I tried some failed calculations, I've typed in Wolfram Alpha online calculator the code (sqrt(2)/2)^sqrt(2) to get as output that also this number is transcendent, and thus irrational.
Before this calculation with Wolfram Alpha, as I am saying was a curiosity that I was asking myself if it is possible deduce or are known some cases for which one can state that $r^{\sqrt{2}}$ is irrational, when $$0<r<1$$ is a real number. The only calculations that I did were, on assumption that $p$ and $q$ are positive integers with $\gcd(p,q)=1$, from $$r^{\sqrt{2}}=\frac{p}{q}$$ that $$\sqrt{2}\log r=\log p-\log q.\tag{1}$$ taking logarithms. And additionally if we presume, by contradiction, that $\sqrt{2}$ is a rational number, I can write the condition $$\frac{P}{Q}\log r=\log p-\log q,\tag{2}$$ where $P$ and $Q$ are positive integers satisfying $\gcd(P,Q)=1$.
But $(1)$ neither $(2)$ don't say nothing to me.
Question. Imagine that a friend ask me for a reasoning to get examples of irrational numbers of the form $$r^{\sqrt{2}},$$ when the real number $r$ runs on the set $0<r<1$. What is the reasoning that I should be tell my friend? If we want to create simple examples of irrational number of the form $r^{\sqrt{2}}$, what are simple requirements/conditions that need to be met those real numbers $0<r<1$? Of course if you need theorems of the kind of Gelfond-Schneider's theorem, or a different approach you can combine with these statements in your discussion to get some examples using a mathematical reasoning. If you know literature you can reference it. Thanks in advance.