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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.

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    $\begingroup$ According to mathworld, it's still an open problem: mathworld.wolfram.com/e.html $\endgroup$
    – Cocopuffs
    Jun 17, 2012 at 6:39
  • $\begingroup$ The same think is asked in (a part of) this question: math.stackexchange.com/questions/28243/… $\endgroup$ Jun 17, 2012 at 6:41
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    $\begingroup$ 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. $\endgroup$
    – user856
    Jun 17, 2012 at 6:51
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    $\begingroup$ I think the expectation is that much more is true: $\pi$ and $e$ are algebraically independent. See mathoverflow.net/questions/33817/…. $\endgroup$
    – lhf
    Jun 4, 2013 at 2:46
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    $\begingroup$ If it were rational, it would be difficult $\endgroup$
    – PyRulez
    Sep 17, 2015 at 1:22

2 Answers 2

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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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    $\begingroup$ @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 … $\endgroup$ Feb 15, 2017 at 9:37
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    $\begingroup$ 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. $\endgroup$ Feb 15, 2017 at 9:42
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    $\begingroup$ @Ovi I think most people think it's irrational, i.e. they'd be VERY surprised if it turns out to be rational. $\endgroup$ Sep 17, 2018 at 11:55
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    $\begingroup$ Is it strange that I agree with Marcus's comments and also agree with @AkivaWeinberger's comment? $\endgroup$
    – Wildcard
    Sep 17, 2018 at 16:18
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    $\begingroup$ @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. $\endgroup$
    – user76284
    May 13, 2020 at 19:41
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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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  • $\begingroup$ 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. $\endgroup$
    – KalEl
    Sep 21, 2020 at 7:11
  • $\begingroup$ @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. $\endgroup$
    – Nagarajan
    Mar 19 at 1:32

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