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The history of proving numbers irrational is full of interesting stories, from the ancient proofs for $\sqrt{2}$, to Lambert's irrationality proof for $\pi$, to Roger Apéry's surprise demonstration that $\zeta(3)$ is irrational in 1979.

There are many numbers that seem to be waiting in the wings to have their irrationality status resolved. Famous examples are $\pi+e$, $2^e$, $\pi^{\sqrt 2}$, and the Euler–Mascheroni constant $\gamma$. Correct me if I'm wrong, but wouldn't most mathematicians find it a great deal more surprising if any of these numbers turned out to be rational rather than irrational?

Are there examples of numbers that, while their status was unknown, were "assumed" to be irrational, but eventually shown to be rational?

Amir Sagiv
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I. J. Kennedy
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    One could also ask the same thing with “irrational” bumped up to “transcendental”... – Peter LeFanu Lumsdaine Jul 22 '10 at 16:16
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    BTW, Apéry announced his proof of $\zeta(3)\notin\mathbb Q$ in 1978 (an April conference in Luminy and later the ICM in Helsinki). – Wadim Zudilin Jul 23 '10 at 02:36
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    Imagine some great mathematician publishes a proof for the rationality of $\pi + e$. Would you believe him? :-) – Martin Brandenburg Jul 23 '10 at 07:45
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    If this happens on April 1, ... YES! – Wadim Zudilin Jul 23 '10 at 09:45
  • I think, all things which their result of computation doesn't become math formula(known only value of limit) yet are unknown whether rational or irrational. – Takahiro Waki Jan 05 '17 at 12:43
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    "Roger Apéry's surprise demonstration that $\zeta(3)$ is irrational...." Perhaps it should be clarified that it was not a surprise that $\zeta(3)$ turned out to be irrational; rather, the surprise was that the demonstration was carried out using tools that had been available for 200 years. – Gerry Myerson Aug 01 '19 at 23:06
  • I wonder whether the cosine of the bond angle of the water molecule is an example, in some manner demonstrably equal to exactly $-1/4\text{?}$ $\qquad$ – Michael Hardy Apr 10 '23 at 03:40

14 Answers14

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I don't think Legendre expected this number to be rational, let alone integer...

Gjergji Zaimi
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    (+1) I agree this is a good example! On the other hand, there is something different about Legendre's constant compared to the other examples. With the other examples, one could compute many digits (nowadays about as many as you'd ever care to see), while Legendre's estimate (1.08366) was wildly off by the third digit! If his calculations showed 1.0000023 or .999984, he certainly might have conjectured the exact value of 1. In short, I think there is a difference in the rationality surprise factor for numbers for which we can compute all the digits. – I. J. Kennedy Jul 22 '10 at 17:29
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    In a similar vein, consider the fine structure constant in physics: Initial measurements showed it to be close to 1/137, so you had a bunch of physicists trying to justify why it had to be exactly 1/137, until we had more accurate measurements... – Simon Rose Jul 22 '10 at 17:48
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    I did some numerical computations and Legendre's guess of x/(log x+ 1.08...) nearly eqauls $\pi(x)$ when x=100,000. – Micah Milinovich Jul 22 '10 at 22:02
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    I.J. Computing all the digits is the same as computing the number though, so I don't think that's a really valid counterpoint. – Adam Hughes Feb 28 '11 at 15:09
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    @I. J. Kennedy: The same thing could be said of our guesses 100 years from now: if they had only looked at the first trillion digits, then they would have been led to the correct conjectures. – Jeff Strom Feb 28 '11 at 15:38
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    Great picture at: http://commons.wikimedia.org/wiki/File:Legendre%27s_constant_10_000_000.svg – André Henriques May 12 '11 at 17:16
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    I must say, having the constant 1 named after you is straight ballin' – Benjamin Lindqvist Mar 03 '16 at 12:01
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Another 'opposite' example - a naturally occurring number suspected to be rational but turning out to be irrational - occurs in the study of random polytopes. In 1923, Blaschke asked

What is the expected volume of a tetrahedron with vertices chosen randomly in a unit volume tetrahedron ?

The corresponding answer for a unit line is $\frac{1}{3}$ and for a unit triangle it's $\frac{1}{12}$. Klee made the (very plausible) conjecture that for the tetrahedron the answer is $\frac{1}{60}$ but later Monte Carlo experiments suggested the answer was closer to $\frac{1}{57}$.

Then in 2001, Buchta and Reitzner showed that the answer is actually

$$\frac{13}{720}-\frac{\pi^2}{15015}.$$

Jeremy Rickard
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dke
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    layman here - what is expected volume? Something like average volume? Chosen randomly means I just pick any 4 points inside the volume of a tetrahedron? I tried to google it but didn't find a laymen description. – daniel.sedlacek Sep 16 '14 at 10:14
  • @daniel.sedlacek: http://en.wikipedia.org/wiki/Expected_value – Michael Borgwardt Sep 16 '14 at 11:16
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    @daniel.sedlacek I think you get the gist of it yea. Uniformly pick 4 points inside the tetrahedron. The average volume of doing this repeatedly will near the expected value. – noio Sep 16 '14 at 12:06
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    Not to disparage the authors, but I am really surprised this was only solved so recently. Isnt that just a four-fold nested integral? – John Jiang Oct 17 '15 at 01:23
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    @John Jiang it is (total number of variables is 12), but what then? It is integral of $|P|$ for polynomial function $P$, there is no general recipe how to find their values. See Kontsevich and Zagier "periods". – Fedor Petrov Oct 22 '15 at 08:37
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    @fedor: yes the absolute value here messes things up. I remember that the set of periods is conjectured to properly contain algebraic numbers. It's sad that math is powerless against such a wide swath of innocent looking problems not related to number theory. – John Jiang Oct 23 '15 at 00:21
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A surprising rational number is 32/27. Thomassen showed in 1997 that the closure of the set of all real zeros of all chromatic polynomials of graphs is $\lbrace 0\rbrace \cup \lbrace 1\rbrace \cup [32/27,\infty)$.

Kevin Ventullo
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Richard Stanley
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Consider the hypergeometric function ${}_2F_1(a,b,c;z)$. When $a$, $b$ and $c$ are rational and ${}_2F_1$ is a transcendental function, Siegel sought to prove that--apart from obvious exceptions--the function takes transcendental values at algebraic $z$. But it turns out that there are $a$, $b$ and $c$ for which this is false. For example:

$${}_2F_1(1/3,2/3,5/6;27/32)=8/5$$

$${}_2F_1(1/4,1/2,3/4;80/81)=9/5$$

$${}_2F_1(1/12,5/12,1/2;1323/1331)= 11^{1/4}$$

Adam Hughes
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paul Monsky
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    Aha. This shows a potential loophole in what I described. It is more difficult to notice that at least one of a large family of numbers is rational. – Greg Kuperberg Jul 22 '10 at 22:26
  • Paul, I can add that there are many examples of (not obviously!) algebraic generalised hypergeometric functions, http://mathoverflow.net/questions/27324/what-are-some-naturally-occurring-high-degree-polynomials/27328#27328, whose evaluations at algebraic points are of course algebraic and even sometimes rational. – Wadim Zudilin Jul 23 '10 at 00:15
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    Greg Kuperberg: would it also fall to the same loophole if we conjectured that $ \pi $ and $ e $ are algebraically independent? This is a generalization of the $ \pi + e $ in the question. – Zsbán Ambrus Sep 25 '10 at 16:01
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    How does one prove this sort of equalities? – Gro-Tsen Sep 13 '14 at 18:05
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    See serial Special values of the hypergeometric series by Joyce and Zucker. – Infiniticism Nov 17 '20 at 04:21
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There are reasons that any modern example is likely to resemble the status of Legendre's constant. Most (but not all) interesting numbers admit a polynomial-time algorithm to compute their digits. In fact, there is an interesting semi-review by Borwein and Borwein that shows that most of the usual numbers in calculus (for example, $\exp(\sqrt{2}+\pi)$) have a quasilinear time algorithm on a RAM machine, meaning $\tilde{O}(n) = O(n(\log n)^\alpha)$ time to compute $n$ digits. Once you have $n$ digits, you can use the continued fraction algorithm to find the best rational approximation with at most $n/2-O(1)$ digits in the denominator. The continued fraction algorithm is equivalent to the Euclidean algorithm, which also has a quasilinear time version according to Wikipedia.

Euler's constant has been to computed almost 30 billion digits, using a quasilinear time algorithm due to Brent and McMillan.

As a result, for any such number it's difficult to be surprised. You would need a mathematical coincidence that the number is rational, but with a denominator that is out of reach for modern computers. (This was Brent and MacMillian's stated motivation in the case of Euler's constant.) I think that it would be fairly newsworthy if it happened. On the other hand, if you can only compute the digits very slowly, then your situation resembles Legendre's.

I got e-mail asking for a reference to the paper of Borwein and Borwein. The paper is On the complexity of familiar functions and numbers. To summarize the relevant part of this survey paper, any value or inverse value of an elementary function in the sense of calculus, including also hypergeometric functions as primitives, can be computed in quasilinear time. So can the gamma or zeta function evaluated at a rational number.

Greg Kuperberg
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    Brun's constant, on the other hand... – Victor Protsak Jul 23 '10 at 00:03
  • Greg, but the OP asks for a constant which has already a status of being rational, right? – Wadim Zudilin Jul 23 '10 at 00:11
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    I can compute Brun's constant for prime triplets of the form $(p,p+2,p+4)$. To no one's surprise, it is rational. – Greg Kuperberg Jul 23 '10 at 00:24
  • @Wadim, my point is an argument why it isn't easy to find an example of what the OP wants, unless it is hard to compute like Legendre's constant, or one of a large family of numbers as in Siegel's problem. – Greg Kuperberg Jul 23 '10 at 00:55
  • Greg, your point is taken. I have another article of Jon and Peter in mind which provides several examples of very irrational rationals. – Wadim Zudilin Jul 23 '10 at 01:19
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    I won't put more answers to this question but the article with many pretty irrational rationals and quite rational irrationals is [J.M. Borwein and P.B. Borwein, Strange Series and High Precision Fraud, Amer. Math. Monthly 99 (1992) 622-–640] (http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P56.pdf). Enjoy! – Wadim Zudilin Jul 23 '10 at 01:33
  • At first I was confused by the answer, but I guess you meant “There are reasons that any modern example is unlikely to resemble the status of Legendre's constant.” – Tsuyoshi Ito Sep 25 '10 at 02:35
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    But isn't it true that a randomly chosen rational number will almost surely have a denominator out of reach for modern computers? Why should we expect that most rational constants will have small denominators? – Mike Shulman Sep 25 '10 at 06:02
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    @Tsuyoshi I think that I meant likely, in the sense that Legendre's constant was extremely difficult to compute in his time. – Greg Kuperberg Sep 27 '10 at 13:38
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    @Mike There is no standard sense of a "random" rational number, since it is an infinite set without one preferred normalized measure. (It's the same issue as choosing a "random" real number or a "random" integer.) If you did imagine a normalized distribution on rational numbers, then of course the median size of the denominator could be as large as you like, but in a motivated choice it would typically be small. – Greg Kuperberg Sep 27 '10 at 14:14
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Hmmm, I am upset to be not not in time for the question (a short night sleep was necessary!).

Let me comment on a quite opposite to the question

Are there examples of numbers that, while their status was unknown, were "assumed" to be irrational, but eventually shown to be rational?

There is one famous constant, the One-Ninth Constant, which for a very long time was expected to be a rational number, namely, $1/9$. It was only in the 1980s when A. Gonchar and E. Rakhmanov found an explicit formula for it through the elliptic integrals and Nesterenko's 1996 theorem on the algebraic independence of modular functions resulted in the transcendence of this constant. There is a nice chapter on this constant in Steven Finch's Mathematical Constants, Cambridge University Press 2003 (§4.5, pp. 259--262), although the transcendence is not mentioned there.

Wadim Zudilin
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    In this direction there is an interesting example in the Cohen--Lenstra heuristics (Springer LNM 1068 pp. 52--53) of a nonconstant complex analytic function which is 1 at all integers. When Cohen plotted its real values, they all seemed to be 1, which is impossible. It turns out the function differs from 1 on the real line by at most 10^(-38). – KConrad Jul 23 '10 at 02:39
  • Keith, see in this direction my last (for the moment!) comment to Greg's post. – Wadim Zudilin Jul 23 '10 at 02:41
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    Of course, the ur-example of a number suspected to be rational but not so would be $\sqrt{2}$. But that's ancient history. – Harrison Brown Sep 29 '10 at 04:55
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This certainly doesn't answer the question, but I can't help but mention Conway's constant:

http://mathworld.wolfram.com/ConwaysConstant.html

It relates to Pete's comment about "bumping" it up a notch, in that it gives an example of a number that I think any reasonable person would conjecture to be transcendental, but turns out to be algebraic (of degree 71, of all things). And algebraic numbers are sort of finitely far from being rational, so...

Cam McLeman
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    Given that Conway's constant exists, I don't think it's unreasonable to conjecture that it's algebraic. Natural combinatorial sequences with exponential growth rate tend to count regular languages. – Qiaochu Yuan Aug 03 '10 at 01:14
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    I'm pretty sure that the algebraicity of Conway's constant follows immediately from the cosmological theorem (although this is based on dim memories of my proving this fact to myself when I was in high school, so take it with a couple grains of salt.) Whether it's reasonable to conjecture that there is a cosmological-type theorem, I don't know. – Harrison Brown Sep 29 '10 at 04:49
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It was not known for a long time that the number $$ \frac{\zeta(2)}{\pi^2} =\frac1{\pi^2}\sum_{n=1}\frac1{n^2} $$ is rational, $1/6$. Euler showed this in his solution of the Basel problem. Related examples include $\zeta(2)^2/\zeta(4)$ and, more generally, $\zeta(2k)/\pi^{2k}$ for integer $k$. I mention this historical fact because of several attempts on MO to find a "closed form" evaluation of $\zeta(3)$ (mostly of the form $\zeta(3)/\pi^3\overset?\in\mathbb Q$, which is numerically confirmed to be doubtfully true).

EDIT. I do understand that not everybody feels this post to be in a (magic) "spirit" of the OP. But I do not understand your downvotes here. Why don't you downvote when somebody puts a problem on finding a "closed form" for $\zeta(3)$? Or when somebody "proves" that $\log2$ is a rational multiple of $\pi^2$? Anyway, I do not remove this post but put it in the community wiki mode, as it might be used, together with this answer and comments therein, as a reference to later silly questions about zeta values.

Community
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Wadim Zudilin
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    That's not a fair example because no one knew that $\zeta(2)$ involved $\pi^2$ until Euler computed it. – lhf Jul 23 '10 at 00:56
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    What I'm saying is that nobody expected that $\zeta(2)/\pi^2$ could be rational. I can't really catch what is counted by "not fair"... – Wadim Zudilin Jul 23 '10 at 01:15
  • And a lot of thanks to an anonymous downvoter! Euler would have appreciated such evaluation of one of his famous results. :-) – Wadim Zudilin Jul 23 '10 at 01:23
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    What lhf presumably means about this example not being fair is that you are rigging the terms to force a rational value. To emphasize the point, you could just as well have said that for a long time nobody knew 3zeta(2)/pi^2 = 1/2 or 6zeta(2)/pi^2 = 1, and these are not the way people usually think about this zeta evaluation. Moreover, if anybody had computed zeta(2)/pi^2 before anyone knew an exact formula for zeta(2), the estimate would be .16666... so the natural guess is that the ratio is 1/6. Thus this example does not seem to be in the spirit of the question. – KConrad Jul 23 '10 at 02:47
  • Keith, thanks for explaining the criticism. I take it. But not your conclusion of "this example does not seem to be in the spirit of the question", as it is really subjective. For example, I don't count Marty's response as "following the spirit", but Marty and some others do. – Wadim Zudilin Jul 23 '10 at 02:54
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    Perhaps it's not a perfect example, but I think it's at least in the spirit of the question. The idea is that we had some description of a number which was believed to be the ONLY description of that number, but some clever person comes along and shows it's actually equal to something we're already familiar with. – Kevin Ventullo Jul 23 '10 at 03:08
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    Thanks, Kevin, this is exactly my feeling about the question. At least, I don't see myself alone in my subjectivity. – Wadim Zudilin Jul 23 '10 at 03:18
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    I think a key difference here is that no one was asking about $\zeta(2)/\pi^2$, so you could have chosen any first- or second-tier mathematical constant to divide or multiply $\zeta(2)$ by. So instead of a single number which is surprisingly rational, it's a large finite class of numbers, one of which turns out to be rational. $\zeta(2)/e$, $\zeta(2)/(\pi+e)$, $\zeta(2)e^\gamma$, etc. – Charles Jul 23 '10 at 03:39
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    I think this is in "the spirit" of the question. For one thing it shows that $\zeta (2)$ is a period, and periods are countable which means they are special and a random number is not a period with probability one. (Just like someone above suggested that substituting rational with algebraic is also in the spirit of the question...) – Gjergji Zaimi Aug 03 '10 at 01:17
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    @ Wadim Zudilin: that 'somebody' who "proved" that $log(2)$ is a rational multiple of $\pi^2$ was me. In the future, I'll try to ask such non-research questions to a professor (as soon as I have one) or math stackexchange. – Max Muller Sep 24 '10 at 15:33
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    +1. I read this question and answers with months of delay, but for what it is worth, I think this one is the best answer. This example illustrates the point made in the question that it is often more surprising, more intersting and more fruitful to prove that some number is rational rather than the opposite. – Joël Dec 13 '10 at 02:23
  • Could you clarify what you mean by "doubtfully true"? Does that mean "unlikely to be true", or "possibly true but there is some doubt", or what? – tparker Jan 20 '22 at 01:18
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A couple of other numbers that have no business being rational: the minimum density of a letter in an infinite ternary squarefree word is 883/3215. The maximum density is 255/653. See https://en.wikipedia.org/wiki/Square-free_word.

Richard Stanley
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Bernstein's constant doesn't strictly fit the parameters of the question, but it's notable as a more recent "Legendre-type" example. Bernstein conjectured that his constant was exactly $\frac{1}{2\sqrt{\pi}}$ in 1914; it wasn't until the '80s that it became possible to compute enough digits to refu(dia)te the conjecture.

Although perhaps it wasn't surprising -- I have no idea whether Bernstein's conjecture was generally believed; can anyone shed light on that?

Elizabeth S. Q. Goodman
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Harrison Brown
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Here is another example of a number that was thought to be rational until it was proved to be irrational. Erdős conjectured that not much more integers are representable as a sum of two squareful numbers than as a sum of two squares. More precisely, he conjectured that up to $x$ the number of integers in the first set is $x/(\log x)^{1/2+o(1)}$. Blomer proved that the exponent $1/2$ is wrong, the correct value is $1-2^{-1/3}$. He also showed that the same estimate is valid for sums of a square and a squareful number. See J. London Math. Soc. (2) 71 (2005), 69-84.

GH from MO
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Reviewer’s account of remarks of M. Duflo in On the Plancherel formula for almost algebraic real Lie groups (1984, p. 158), further confirmed in (1988, p. 328):

I find it amusing when the author points out that some constants entering into the formula for semisimple groups, for which there are very explicit but complicated formulas in the literature, are actually all $=1$.

Francois Ziegler
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It's hazardous to guess the reactions of most mathematicians. But I imagine a very large number of mathematicians would be surprised if Schanuel's conjecture turned out to be false. And this conjecture implies the irrationality of 3/4 of your examples, I think.

As for the Euler–Mascheroni constant, I have never thought about it.

Marty
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    Marty, as with Greg's answer you do not provide an example of a constant which is known to be irrational. Schanuel's conjecture is a quite deep observation, with many numerical and functional confirmations. I am unaware of an expert in this area who thinks it might be wrong. – Wadim Zudilin Jul 23 '10 at 00:18
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Another constant that has "no business being rational" I think, although a bit elementary: Choose a point at random in the unit disk $D=\left\{x^2+y^2\leq 1\right\}$. Then the expected value $E$ of its distance to the origin is a rational number! (click below for solution).

$E=\frac{2}{3}$

Chris McDaniel
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    I think "choose a point at random in the unit disc" is not well-defined. For example, every point falls on a line intersecting the origin, save the origin itself. If I choose a point by selecting an angle [0, pi) at "random", then a point on that line at "random", the answer is 1/2 not 2/3. Points and numbers cannot be random, only processes. Processes are not as straightforward as you'd think for shapes in R^2 as they are for one-dimensional metric spaces. – Trixie Wolf Oct 12 '22 at 23:13
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    This doesn't seem that surprising. If $R$ is the distance, $P(R<r)=\pi r^2/\pi=r^2$ for $r\in[0,1].$ Then the expected value is $\int_0^1 r\cdot(2r),dr=\frac23.$ – Thomas Jan 04 '23 at 18:32