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From this list I came to know that it is hard to conclude $\pi+e$ is an irrational? Can somebody discuss with reference "Why this is hard ?"

Is it still an open problem ? If yes it will be helpful to any student what kind ideas already used but ultimately failed to conclude this.

Bill Dubuque
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    According to mathworld, it's still an open problem: http://mathworld.wolfram.com/e.html – Cocopuffs Jun 17 '12 at 06:39
  • The same think is asked in (a part of) this question: http://math.stackexchange.com/questions/28243/is-there-a-proof-that-pi-times-e-is-irrational – Martin Sleziak Jun 17 '12 at 06:41
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    I don't think this is precisely a duplicate of the other question, as this one asks for references and discussion about why previous techniques are insufficient to resolve the problem. (I've edited the title to match.) This can be more illuminating than a simple yes/no answer, which is what the previous question received. –  Jun 17 '12 at 06:51
  • After Rahul Narain's [edit](http://math.stackexchange.com/posts/159350/revisions) the title of the question corresponds to the body. So it seems that it is a different question - I apologize for being too quick in voting to close. – Martin Sleziak Jun 17 '12 at 06:51
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    I think the expectation is that much more is true: $\pi$ and $e$ are algebraically independent. See http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e. – lhf Jun 04 '13 at 02:46
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    If it were rational, it would be difficult – PyRulez Sep 17 '15 at 01:22
  • It is easy to see that both the sum and the product of two transcendentals can be rational or even integer. But proving that the sum or product of two given transcendentals is rational is an open problem, particularly for the two more famous transcendents $e$ and $\pi$. – Piquito Oct 22 '21 at 15:06

2 Answers2

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"Why is this hard?" I think a different question would be "Why would it be easy?"

But there are some things that are known. It is known that $\pi$ and $e$ are transcendental. Thus $(x-\pi)(x-e) = x^2 - (e + \pi)x + e\pi$ cannot have rational coefficients. So at least one of $e + \pi$ and $e\pi$ is irrational. It's also known that at least one of $e \pi$ and $e^{\pi^2}$ is irrational (see, e.g., this post at MO).

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davidlowryduda
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    Do most mathematicians think it is irrational, or is it viewed as 50/50? – Ovi May 13 '16 at 05:10
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    @Ovi mathematicians don't work like that. They either say, with certainty, "based on our axiom system, X IS", and then it'd be 100/0, or they say "based on our axiom system, X is UNDECIDABLE", or they say "maybe there's a proof for X, but it hasn't been found, so this is something I make NO STATEMENT on". Math is not done via polls, and when working mathematically, you *abhor* the idea of saying "I believe that X is". Either you can prove X, or you cannot make any statement based on a's veracity. You can, of course, take X as a *hypothesis* and build complex things on it, but then you'd … – Marcus Müller Feb 15 '17 at 09:37
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    … **always** remember X is just a hypothesis. Otherwise, you'd be a sloppy mathematician, and that is a *bad* mathematician. Notice that a lot of mathematicians are still social people and you can try to persuade them to pick either X or NOT X in a poll if you don't offer the "NOT YET PROVEN" option, but using the result of that poll would be equivalent to asking healthy people whether they'd rather be struck with either AIDS or cancer, getting e.g. a 70/30 result and then claiming that 70% of people would like to have cancer. – Marcus Müller Feb 15 '17 at 09:42
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    @Ovi I think most people think it's irrational, i.e. they'd be VERY surprised if it turns out to be rational. – Akiva Weinberger Sep 17 '18 at 11:55
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    Is it strange that I agree with Marcus's comments *and also* agree with @AkivaWeinberger's comment? – Wildcard Sep 17 '18 at 16:18
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    @MarcusMüller That's not true at all. Subjective probability guides mathematicians' efforts all the time, to focus them on where they think they're most likely to bear fruit. For instance, most mathematicians believe the Riemann hypothesis is true. That doesn't mean they *know* it's true, just that they think it's more likely to be true than false (in the form of betting odds, if you want). For a formalization of this idea, see [logical induction](https://intelligence.org/files/LogicalInduction.pdf). – user76284 May 13 '20 at 19:41
  • @user76284 exactly my point! So, if you're proving something, and your proof requires the Riemann Hypothesis to be true, you'd very clearly note that. You're not actually contradicting me... – Marcus Müller May 13 '20 at 19:44
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    @MarcusMüller Of course. What I mean is, mathematicians have internal beliefs (reflected as odds or probabilities) about the truth of statements, even before those statements have been proven or disproven. Otherwise, arguably, mathematics would be completely intractable (just a brute-force enumeration of proofs). – user76284 May 13 '20 at 19:46
  • @user76284 or were you referring to my use of the word "abhor"? That might have been a bit strong, but again, I find it a very natural thing (and I see that in any mathematical work) to apply Occam's Razor, i.e. to keep one's axiom system as small as possible in any way – up to a point where working on a hypothesis has the undeniable advantage of letting you continue your work :) – Marcus Müller May 13 '20 at 19:47
  • ah simultaneous commenting, @user76284 :) yes, I think I fully agree with you. – Marcus Müller May 13 '20 at 19:48
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I was actually going to ask the same question... and in particular if the result would follow as the consequence of any hard, still open conjecture. From the MO thread mentioned by lhf (not the same as the one mentioned by mixedmath) I found out that Schanuel's conjecture would imply it.

On the Mathworld page for $e$ there's a bit of info on numerical attempts to (how should I say?) verify that you cannot easily disprove the irrationality:

It is known that $\pi+e$ and $\pi/e$ do not satisfy any polynomial equation of degree $\leq 8$ with integer coefficients of average size $10^9$.

Obtaining this result in 1988 required the use of a Cray-2 supercomputer (at NASA Ames Research Center). I guess one could add that the Ferguson–Forcade algorithm, which was used in this computation, gets a bit of flak on Wikipedia. In fact, the author of this paper, D.H. Bailey, later co-developed the superior PSLQ algorithm. So it is interesting that the problem has advanced computational science too, in a way.

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Fizz
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  • Numerical attempts were done for fun I assume. Otherwise it is obviously a red herring, the numbers are bound to be transcendental. It would be immensely surprising if they are not. – KalEl Sep 21 '20 at 07:11
  • @KalEl Agreed... there would be an amazing connection between e and pi for any of those numbers to be rational. However, then e^(i pi) comes to mind. Its pretty amazing that its rational. – Nagarajan Mar 19 '22 at 01:32