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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?

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    $\begingroup$ One could also ask the same thing with “irrational” bumped up to “transcendental”... $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jul 22, 2010 at 16:16
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    $\begingroup$ BTW, Apéry announced his proof of $\zeta(3)\notin\mathbb Q$ in 1978 (an April conference in Luminy and later the ICM in Helsinki). $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 2:36
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    $\begingroup$ Imagine some great mathematician publishes a proof for the rationality of $\pi + e$. Would you believe him? :-) $\endgroup$
    – Martin Brandenburg
    Commented Jul 23, 2010 at 7:45
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    $\begingroup$ If this happens on April 1, ... YES! $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 9:45
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    $\begingroup$ "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. $\endgroup$
    – Gerry Myerson
    Commented Aug 1, 2019 at 23:06

14 Answers 14

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

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    $\begingroup$ (+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. $\endgroup$
    – I. J. Kennedy
    Commented Jul 22, 2010 at 17:29
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    $\begingroup$ 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... $\endgroup$
    – Simon Rose
    Commented Jul 22, 2010 at 17:48
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    $\begingroup$ I did some numerical computations and Legendre's guess of x/(log x+ 1.08...) nearly eqauls $\pi(x)$ when x=100,000. $\endgroup$
    – Micah Milinovich
    Commented Jul 22, 2010 at 22:02
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    $\begingroup$ Great picture at: commons.wikimedia.org/wiki/… $\endgroup$
    – André Henriques
    Commented May 12, 2011 at 17:16
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    $\begingroup$ I must say, having the constant 1 named after you is straight ballin' $\endgroup$
    – Benjamin Lindqvist
    Commented Mar 3, 2016 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

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

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    $\begingroup$ 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. $\endgroup$
    – daniel.sedlacek
    Commented Sep 16, 2014 at 10:14
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    $\begingroup$ @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. $\endgroup$
    – noio
    Commented Sep 16, 2014 at 12:06
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    $\begingroup$ Not to disparage the authors, but I am really surprised this was only solved so recently. Isnt that just a four-fold nested integral? $\endgroup$
    – John Jiang
    Commented Oct 17, 2015 at 1:23
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    $\begingroup$ @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". $\endgroup$
    – Fedor Petrov
    Commented Oct 22, 2015 at 8:37
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    $\begingroup$ @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. $\endgroup$
    – John Jiang
    Commented Oct 23, 2015 at 0:21
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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 $\lbrace 0\rbrace \cup \lbrace 1\rbrace \cup [32/27,\infty)$.

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    $\begingroup$ Is there a rational explanation? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Sep 25, 2010 at 20:52
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    $\begingroup$ I'm not surprised that 32/27 is rational. $\endgroup$
    – KConrad
    Commented Mar 1, 2011 at 3:53
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    $\begingroup$ I too always suspected it. $\endgroup$
    – R.P.
    Commented Nov 5, 2012 at 1:25
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    $\begingroup$ The tricky part was done by Jackson (1993), see dx.doi.org/10.1017/S0963548300000705 $\endgroup$
    – didest
    Commented Apr 7, 2014 at 4:29
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    $\begingroup$ @TomášZato, re, as @‍KConrad wittily points out, the surprise is not, as a quick reading suggests, that the number is rational, but rather that this obviously rational number occurs as an answer to this particular question. $\endgroup$
    – LSpice
    Commented Dec 26, 2022 at 20:10
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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:

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

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

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

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    $\begingroup$ 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. $\endgroup$
    – Greg Kuperberg
    Commented Jul 22, 2010 at 22:26
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    $\begingroup$ Paul, I can add that there are many examples of (not obviously!) algebraic generalised hypergeometric functions, mathoverflow.net/questions/27324/…, whose evaluations at algebraic points are of course algebraic and even sometimes rational. $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 0:15
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    $\begingroup$ 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. $\endgroup$
    – Zsbán Ambrus
    Commented Sep 25, 2010 at 16:01
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    $\begingroup$ How does one prove this sort of equalities? $\endgroup$
    – Gro-Tsen
    Commented Sep 13, 2014 at 18:05
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    $\begingroup$ See serial Special values of the hypergeometric series by Joyce and Zucker. $\endgroup$
    – Infiniticism
    Commented Nov 17, 2020 at 4:21
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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.

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    $\begingroup$ Brun's constant, on the other hand... $\endgroup$
    – Victor Protsak
    Commented Jul 23, 2010 at 0:03
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    $\begingroup$ 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. $\endgroup$
    – Greg Kuperberg
    Commented Jul 23, 2010 at 0:24
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    $\begingroup$ 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] (cecm.sfu.ca/personal/pborwein/PAPERS/P56.pdf). Enjoy! $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 1:33
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    $\begingroup$ @Tsuyoshi I think that I meant likely, in the sense that Legendre's constant was extremely difficult to compute in his time. $\endgroup$
    – Greg Kuperberg
    Commented Sep 27, 2010 at 13:38
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    $\begingroup$ @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. $\endgroup$
    – Greg Kuperberg
    Commented Sep 27, 2010 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.

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    $\begingroup$ 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). $\endgroup$
    – KConrad
    Commented Jul 23, 2010 at 2:39
  • $\begingroup$ Keith, see in this direction my last (for the moment!) comment to Greg's post. $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 2:41
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    $\begingroup$ Of course, the ur-example of a number suspected to be rational but not so would be $\sqrt{2}$. But that's ancient history. $\endgroup$
    – Harrison Brown
    Commented Sep 29, 2010 at 4:55
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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...

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    $\begingroup$ 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. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 3, 2010 at 1:14
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    $\begingroup$ 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. $\endgroup$
    – Harrison Brown
    Commented Sep 29, 2010 at 4:49
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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.

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    $\begingroup$ That's not a fair example because no one knew that $\zeta(2)$ involved $\pi^2$ until Euler computed it. $\endgroup$
    – lhf
    Commented Jul 23, 2010 at 0:56
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    $\begingroup$ 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"... $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 1:15
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    $\begingroup$ 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 3*zeta(2)/pi^2 = 1/2 or 6*zeta(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. $\endgroup$
    – KConrad
    Commented Jul 23, 2010 at 2:47
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    $\begingroup$ 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. $\endgroup$
    – Kevin Ventullo
    Commented Jul 23, 2010 at 3:08
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    $\begingroup$ 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. $\endgroup$
    – Charles
    Commented Jul 23, 2010 at 3:39
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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.

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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?

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  • $\begingroup$ Refudiate?..... $\endgroup$
    – LSpice
    Commented Dec 26, 2022 at 20:10
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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.

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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 $=1$.

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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.

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    $\begingroup$ 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. $\endgroup$
    – Wadim Zudilin
    Commented Jul 23, 2010 at 0:18
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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).

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

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    $\begingroup$ 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. $\endgroup$
    – Trixie Wolf
    Commented Oct 12, 2022 at 23:13
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    $\begingroup$ 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 $\int_0^1 r\cdot(2r)\,dr=\frac23.$ $\endgroup$
    – Thomas
    Commented Jan 4, 2023 at 18:32

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