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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.

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    $\begingroup$ Did you look at the mathoverflow question linked to in the question you posted? Seems to do a pretty good job of answering.... $\endgroup$ – lulu Apr 22 '17 at 11:23
  • $\begingroup$ Thanks then this afternoon I will try to understand the answer of Mark Sapir, and other comments in Math Overflow. With my Question I am asking what should be a discussion of simple cases $r^{\sqrt{2}}$ being irrational, and of course using the theorems from the literature about transcendent numbers. Many thanks for your attention @lulu $\endgroup$ – user243301 Apr 22 '17 at 11:33
  • $\begingroup$ IMHO, if there's no restriction on $x$ other than a real number, things get easy. Let $b$ be an irrational number (infinitely many such numbers) in the interval $(0,1)$ such that $r^{\sqrt(2)}=b$, you can express $r$ with the logarithm function. $\endgroup$ – Huang Apr 22 '17 at 11:54
  • $\begingroup$ Many thanks also for your contribution @Huang $\endgroup$ – user243301 Apr 22 '17 at 11:56
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    $\begingroup$ You can choose every algebraic irrational number $r$ with $0<r<1$. As pointed out below the Gelfond-Schneider-theorem guarantees that $r^{\sqrt{2}}$ will be transcendental, hence irrational. You could , for example, choose the golden-ratio-number $\frac{\sqrt{5}-1}{2}$ $\endgroup$ – Peter Apr 22 '17 at 15:15
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The Gelfond-Schneider-theorem mentioned by Jack allows a more general choice for $r$ :

For every algebraic irrational $r$ with $0<r<1$, the number $r^{\sqrt{2}}$ is transcendental, hence irrational.

A particular cute choice is the golden-ratio-number $\phi=\frac{\sqrt{5}-1}{2}$

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$\sqrt{2}^{\sqrt{2}}$ is a trascendental number from the Gelfond-Schneider theorem, hence its reciprocal $\left(\frac{1}{\sqrt{2}}\right)^{\sqrt{2}}$ is clearly an irrational number.

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  • $\begingroup$ Thank you for clarifying the example. $\endgroup$ – user243301 Apr 22 '17 at 14:05

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