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Everyone knows that $\pi$ is an irrational number, and one can refer to this page for the proof that $\pi^{2}$ is also irrational.

What about the highers powers of $\pi$, meaning is $\pi^{n}$ irrational for all $n \in \mathbb{N}$ or does there exists a $m \in \mathbb{N}$ when $\pi^{m}$ is rational.

Glorfindel
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    Not only does everyone know that $\pi$ is irrational, but everyone also knows that $\pi$ is transcendental :-) – Robin Chapman Aug 12 '10 at 20:43
  • @Robin Chapman: Ok. Agreed. Ivan Niven's proof is awesome for the irrationality of $\pi$. –  Aug 12 '10 at 20:45
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    Does anyone have a link to a proof of transcendentality for completeness? – Casebash Aug 12 '10 at 21:49
  • The sketch of it at least: if $\pi$ were algebraic, Lindemann-Weierstrass would imply $\exp(2\pi i)$ is transcendental (proving e is transcendental is another story)... and you can fill in the rest. – J. M. ain't a mathematician Aug 12 '10 at 22:25

1 Answers1

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What Robin hinted at:

If $\pi^{n}$ was rational, then $\pi$ would not be transcendental, as it would be the root of $ax^{n}-b = 0$ for some integers $a,b$.

Aryabhata
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  • @Moron: Ok, then if thats the case then why have they proved the seperate case of $\pi^{2}$ being irrational at planet math. They could have used your argument. –  Aug 12 '10 at 20:54
  • @Chandru: No idea. Perhaps they have it there for historical reasons. – Aryabhata Aug 12 '10 at 20:57
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    There's a (relatively) simple proof that just happens to work for $\pi^2$ (and from which the $\pi$ case is an immediate corollary). – Robin Chapman Aug 12 '10 at 20:57
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    1768 pi irrational, 1882 pi transcendental. proving irrationality is easier than proving transcendence. using the fact that pi is transcendantal seems to be overkill (in my interpretation of OP's question) – yoyo Mar 27 '11 at 20:52
  • @yoyo: The question was, is $\pi^m$ rational for some integer $m \gt 0$. The answer is no. How does it matter how hard it is prove that? – Aryabhata Mar 27 '11 at 21:31
  • By the same method, ${\pi}^r$ is irrational (even transcendental) for all nonzero rational $r.$ A possibly interesting question (which may have an easy answer -- I haven't given it any thought yet) is whether there exists an irrational algebraic real number $x$ such that ${\pi}^x$ is rational, and also whether there exists an explicit irrational algebraic number $y$ (i.e. can be expressed using radicals) such that ${\pi}^y$ is irrational. – Dave L. Renfro Jul 10 '14 at 13:49
  • In my 2nd question, what I actually meant (but didn't think about until it was too late to edit my comment) is that can such a number $y$ can be explicitly exhibited, as opposed to simply proved to exist by some non-constructive method. – Dave L. Renfro Jul 10 '14 at 13:57