Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
24  
One could also ask the same thing with “irrational” bumped up to “transcendental”... –  Peter LeFanu Lumsdaine Jul 22 '10 at 16:16
9  
Imagine some great mathematician publishes a proof for the rationality of $\pi + e$. Would you believe him? :-) –  Martin Brandenburg Jul 23 '10 at 7:45
13  
If this happens on April 1, ... YES! –  Wadim Zudilin Jul 23 '10 at 9:45
6  
Isn't this the kind of question that ought to be community wiki? –  Rasmus Bentmann Jul 23 '10 at 19:09
5  
community wiki??? –  Sergei Tropanets Aug 3 '10 at 15:01

11 Answers 11

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

share|improve this answer
32  
(+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
25  
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
6  
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
2  
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
8  
Great picture at: commons.wikimedia.org/wiki/… –  André Henriques May 12 '11 at 17:16

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

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

share|improve this answer
    
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 2 days ago
    
@daniel.sedlacek: en.wikipedia.org/wiki/Expected_value –  Michael Borgwardt 2 days ago
    
@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 2 days ago

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:

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

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

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

share|improve this answer
6  
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, mathoverflow.net/questions/27324/…, whose evaluations at algebraic points are of course algebraic and even sometimes rational. –  Wadim Zudilin Jul 23 '10 at 0:15
    
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
1  
How does one prove this sort of equalities? –  Gro-Tsen Sep 13 at 18:05

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 $\lbrace 0\rbrace \cup \lbrace 1\rbrace \cup [32/27,\infty)$.

share|improve this answer
48  
Is there a rational explanation? –  Mariano Suárez-Alvarez Sep 25 '10 at 20:52
41  
I'm not surprised that 32/27 is rational. –  KConrad Mar 1 '11 at 3:53
7  
I too always suspected it. –  René Nov 5 '12 at 1:25
2  
The tricky part was done by Jackson (1993), see dx.doi.org/10.1017/S0963548300000705 –  Diego de Estrada Apr 7 at 4:29

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.

share|improve this answer
25  
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 2:39
    
Keith, see in this direction my last (for the moment!) comment to Greg's post. –  Wadim Zudilin Jul 23 '10 at 2:41
12  
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 4:55

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.

share|improve this answer
1  
Brun's constant, on the other hand... –  Victor Protsak Jul 23 '10 at 0:03
    
Greg, but the OP asks for a constant which has already a status of being rational, right? –  Wadim Zudilin Jul 23 '10 at 0:11
3  
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 0:24
4  
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! –  Wadim Zudilin Jul 23 '10 at 1:33
2  
@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

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...

share|improve this answer
10  
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 3 '10 at 1:14
    
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 4:49

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.

share|improve this answer
17  
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 0:56
5  
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 1:15
24  
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. –  KConrad Jul 23 '10 at 2:47
8  
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 3:08
6  
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 3:39

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?

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
    
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 0:18

protected by Felipe Voloch Jun 25 at 1:03

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.