# Why is it hard to prove whether $\pi+e$ is an irrational number?

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.

• According to mathworld, it's still an open problem: mathworld.wolfram.com/e.html – Cocopuffs Jun 17 '12 at 6:39
• The same think is asked in (a part of) this question: math.stackexchange.com/questions/28243/… – Martin Sleziak Jun 17 '12 at 6:41
• 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. – user856 Jun 17 '12 at 6:51
• After Rahul Narain's edit 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 6:51
• I think the expectation is that much more is true: $\pi$ and $e$ are algebraically independent. See mathoverflow.net/questions/33817/…. – lhf Jun 4 '13 at 2:46

"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).

• @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 â€¦ – Marcus Müller Feb 15 '17 at 9:37
• â€¦ 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 9:42
• @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
• Is it strange that I agree with Marcus's comments and also agree with @AkivaWeinberger's comment? – Wildcard Sep 17 '18 at 16:18
• @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. – user76284 May 13 at 19:41

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 Fergusonâ€“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.

• 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 at 7:11