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What is known about irrationality of $\pi e$, $\pi^\pi$ and $e^{\pi^2}$?

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Why are you interested in these particular numbers? –  cfranc Sep 27 '10 at 14:08
No doubt that a proof of irrationality of one of these numbers would be a monument of the human intelligence... But isn't a bit sad, such a big effort to prove something that everybody would believe true? What I would really like to see is a proof of rationality of at least one of these combinations of $\pi$ $e$ and $\gamma$. –  Pietro Majer Sep 27 '10 at 17:26
Pietro, why would it be sad to prove something people believe? It happens all the time! More often than not (but not always) long-standing conjectures which are solved turn out to be true in the way that they were conjectured. –  KConrad Oct 2 '10 at 16:28
Pietro said that it would be sad if effort were put into such things (rather than into something more enlightening or useful). I agree. –  Paul Taylor May 3 '13 at 20:29
@PaulTaylor Don't you think that the rationality of $\pi e$ would be very enlightening and useful? –  Oksana Gimmel May 3 '13 at 20:50

2 Answers 2

up vote 20 down vote accepted

I believe most such questions are still very far from being resolved.

Apparently, it is not even known if $\pi^{\pi^{\pi^\pi}}$ is an integer (let alone irrational).

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You raise a nice question! (Though of course an answer 'yes' would be a lot nicer than 'no'!) –  Stefan Kohl May 3 '13 at 20:54
@Oksana Gimmel: very interesting! Can you suggest any references for reading on that last bit? (It’s rather difficult to search about!) –  Peter LeFanu Lumsdaine May 3 '13 at 21:42
It is mentioned on the Russian Wikipedia page Open mathematical problems. A very similar question was discussed at math.stackexchange.com/questions/13050/eee79-and-ultrafinitism –  Oksana Gimmel May 3 '13 at 22:02
Surely it can be proven that pi^pi^pi^pi is not an integer? Just bound it between two consecutive integers... It would be inelegant as factorial but it would work ? –  LTS Apr 21 at 14:36
@Oliver But what if it is actually an integer? –  Vladimir Reshetnikov Apr 21 at 18:09

Brownawell and Waldschmidt do have results in these directions which do not rely on Schanuel's Conjecture. The references are

M. Waldschmidt, "Solution du Huitième Problème de Schneider," J. Number Theory 5 (1973), 191-202.

W. D. Brownawell, "The algebraic independence of certain numbers related by the exponential function," J. Number Theory 6 (1974), 23-31.

The two papers independently prove results along the following lines. (The following version is taken from Brownawell.) Let $\alpha$, $\beta$, and $\gamma$ be nonzero complex numbers with $\alpha$ and $\beta$ both irrational. If $e^\gamma$ and $e^{\alpha\gamma}$ are both algebraic numbers, then at least two of the numbers $$\alpha, \beta, \gamma, e^{\beta\gamma}, e^{\alpha\beta\gamma}$$ are algebraically independent over $\mathbb{Q}$.

This theorem has several interesting consequences:

  • Taking $\alpha=\beta=e^{-1}, \gamma=e^2$, we see that at least one of $e^e$ and $e^{e^2}$ must be transcendental. This was conjectured by Schneider.

  • Taking $\alpha=\beta=\gamma$, we see that given any nonzero complex number $\alpha$, at least one of the numbers $e^{\alpha}, e^{\alpha^2}, e^{\alpha^3}$ must be transcendental.

  • Taking $\alpha = \beta = i/\pi, \gamma=\pi^2$, we see that at least one of the following holds: (i) $e^{\pi^2}$ is transcendental, or (ii) $e$ and $\pi$ are algebraically independent.

So as a partial answer to this question, at least one of $e\pi$ and $e^{\pi^2}$ is transcendental.

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