Recently I have found out about a proof through a video.

This proof shows that an irrational number can be raised to a irrational power to get and irrational number, but this proof only requires one case, so I wanted to see if I could prove it for all irrational square roots. This is what I cam up with:

First take the square root of a positive integer, x. $\sqrt x$ Then apply the power for the square root of x: $\sqrt x^{\sqrt x}$. Some numbers might evaluate rationally here, if so stop. If not this can be continued to $(\sqrt x^{\sqrt x})^{\sqrt x} = \sqrt x^{x}$. Here if x is even the equation can be written as $\sqrt x^{2n} = x^{n}$ where n is a positive integer, and the equation will evaluate to a whole number [Note 1]. If x is odd this can be continued to $(\sqrt x^{x})^{\sqrt 2} = (\sqrt x^{\sqrt 2x})$ [Note 2]. Then continue to raise it to the power of the square root of two once more. $(\sqrt x^{x\sqrt 2})^{\sqrt 2}$ [Note 3]. This evaluates to $\sqrt x^{2x} = x^x$ which must be a whole number!

Thank you for reading. Is this proof correct? Hopefully, I made no mistakes and everything here is right. I assume this has been proven before. If so, can I get a link and was my proof nice compared to the other one?

[Note 1]: You cannot say $\sqrt x^{\sqrt x}$ could have evaluated to a whole number invalidating this proof. This case was already covered earlier.

[Note 2]: Because x is a positive integer multiplying it by a irrational number will make it irrational. The proof is still valid.

[Note 3] $\sqrt 2^ \sqrt 2$ is irrational


1 Answer 1


If you are trying to prove that for all integers $x$, there is some irrational number $y$ such that $\sqrt{x}^{y} \in \mathbb{Z}$, then you have not done that. You have successfully proven that $\left(\sqrt{x}^{x\sqrt{2}}\right)^\sqrt{2}$ is an integer, but this expression simplifies to $\sqrt{x}^{2x}$, and since $2x$ is rational, you have not proven the theorem.

In fact, the original theorem does not even say that. The original theorem states that $\left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2}=\sqrt{2}^2$ is rational. This does not give an irrational number $y$ such that $\sqrt{2}^y$ is rational, because all we are doing is raising $\sqrt{2}$ to the second power.

In fact, I don't think that this theorem can even be proven in this way. If you want a way to prove it, you can just note: $$\sqrt{x}^{2 \log_x(2)}=2$$ And both the base and exponent are irrational whenever $x$ is not a power of $2$. If $x$ is a power of $2$, then replace (the second and third) $2$ with $3$.


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