Irrationality of $ \pi e, \pi^{\pi}$ and $e^{\pi^2}$

Question

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

I recently asked a similar question - http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e — muad, Sep 27 '10 at 14:47
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 02 '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 03 '13 at 20:29
@PaulTaylor Don't you think that the rationality of $\pi e$ would be very enlightening and useful? — Oksana Gimmel, May 03 '13 at 20:50
Irrationality proofs generally aren't useful in any practical sense, but they can certainly be enlightening. — Henry Cohn, May 03 '13 at 21:19
A similar question: http://math.stackexchange.com/questions/386207/what-is-the-role-of-mathematical-intuition-and-common-sense-in-questions-of-irra — Vladimir Reshetnikov, May 09 '13 at 03:37
If an oracle told you that exactly one of these is rational, which would you bet on? — Steven Landsburg, Mar 20 '24 at 04:11
If someone proves that $\pi^\pi$ is irrational, the interesting statement is not "$\pi^\pi$ is irrational" but "humans (or computers) can prove that $\pi^\pi$ is irrational". The first statement is expected, the second statement is unexpected and absolutely amazing. — Hjalmar Rosengren, Mar 20 '24 at 06:15

Matt Papanikolas · Answer 1 · 2010-09-27T15:26:55.803

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.

score 53 · Accepted Answer · answered May 03 '13 at 19:58

53

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).

answered May 03 '13 at 19:58

Oksana Gimmel

1,679

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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 03 '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 03 '13 at 21:42
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It is mentioned on the Russian Wikipedia page Open mathematical problems. A very similar question was discussed at http://math.stackexchange.com/questions/13050/eee79-and-ultrafinitism – Oksana Gimmel May 03 '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 ? – user85798 Apr 21 '14 at 14:36
@Oliver But what if it is actually an integer? – Vladimir Reshetnikov Apr 21 '14 at 18:09
@VladimirReshetnikov Then we would fail, and be unable to bound it between two consecutive integers. But that (probably) won't happen.. just get a computer to find a good enough upper and lower bound, which would count as a proof. – user85798 Apr 21 '14 at 19:14
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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. – user85798 Apr 21 '14 at 19:45
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@Oliver Even if we had such a computer, and got a result like 435643…85685.00000000000…, how do we know if it is the exact integer, or we just need higher precision to discover non-zero fractional part? – Vladimir Reshetnikov Apr 21 '14 at 20:11
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@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...) – user85798 Apr 21 '14 at 20:49
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$\pi^{\pi^{\pi^{\pi}}}$ 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. – Daniel McLaury Jan 31 '15 at 21:07
@VladimirReshetnikov In that case you can run two programs in parallel: one obtaining higher and higher precision of the said number, and the other enumerate all valid proofs and see if one is a proof that the said number is an integer. I bet the first program terminate first. – Fan Zheng Nov 13 '15 at 15:33
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@FanZheng It's still possible that none of the programs terminates, because of incompleteness (there are true statements without proofs). – Vladimir Reshetnikov Nov 13 '15 at 17:02
@VladimirReshetnikov OK, after putting it in wolframalpha, I realized like LTS above that it is much larger than I had thought... so your scenario may well be true, at least in practice. – Fan Zheng Nov 13 '15 at 18:04
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as well as in theory, which I of course believe. To put it more concretely, you need to compute at least $10^{17}$ digits of $\pi$ in order to tell if that number is an integer, and the best approximation of $\pi$ that I'm aware of gives $10^{12}$ digits. There's still someway to go. – Fan Zheng Nov 13 '15 at 18:30

Irrationality of $ \pi e, \pi^{\pi}$ and $e^{\pi^2}$

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