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

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    $\begingroup$ Why are you interested in these particular numbers? $\endgroup$
    – cfranc
    Sep 27, 2010 at 14:08
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    $\begingroup$ 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 $\gamma$. $\endgroup$ Sep 27, 2010 at 17:26
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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Oct 2, 2010 at 16:28
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    $\begingroup$ Pietro said that it would be sad if effort were put into such things (rather than into something more enlightening or useful). I agree. $\endgroup$ May 3, 2013 at 20:29
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    $\begingroup$ @PaulTaylor Don't you think that the rationality of $\pi e$ would be very enlightening and useful? $\endgroup$ May 3, 2013 at 20:50

2 Answers 2


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).

  • 9
    $\begingroup$ You raise a nice question! (Though of course an answer 'yes' would be a lot nicer than 'no'!) $\endgroup$
    – Stefan Kohl
    May 3, 2013 at 20:54
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    $\begingroup$ 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 $\endgroup$ May 3, 2013 at 22:02
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    $\begingroup$ ok, pi^pi^pi^pi is a hell of a lot bigger than I thought it was. Should still work though if you have a good computer and enough time. $\endgroup$
    – user85798
    Apr 21, 2014 at 19:45
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    $\begingroup$ @VladimirReshetnikov Oh, if it actually is an integer then of course this wouldn't work. I'm assuming it's not an integer. (I see no reason why we would get x.00000000000...) $\endgroup$
    – user85798
    Apr 21, 2014 at 20:49
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    $\begingroup$ $\pi^{\pi^{\pi^{\pi}}}$ has over a hundred quadrillion digits. It would take more than two exabytes of storage just to write down the integer part of that number. $\endgroup$ Jan 31, 2015 at 21:07

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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