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Consider the Riemann $\zeta$ function. We know that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$ (in particular is transcendental). We also know that $\zeta(3)$ is irrational, and we expect $\zeta(n)$ to be irrational (if not even transcendental) for every $n\in\mathbb{N}$, or at least I would be amazed if - say - $\zeta(5)$ turned out to be rational.

Are there known instances of rational numbers appearing where we would not have expected them?

Of course the notion of "expectation" here is very subjective, so this is a soft question just out of curiosity, since I have the feeling that usually (in my limited experience always) complicated expressions yield irrational ($\mathbb{C}-\mathbb{Q}$) numbers.

As a non-example, we have the series $\sum_{n=1}^\infty 2^{-n}=1$. A series is complicated enough (compared to say a finite arithmetic expression), but of course here we have the explicit formula for geometric series so the resulting $1$ is not really a big surprise.

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    $\begingroup$ Related: mathoverflow.net/questions/32967/… $\endgroup$ – user3002473 Apr 12 '17 at 4:00
  • $\begingroup$ I don't know if that's really that much of a non example, it was pretty unexpected at the time to essentially say an unending sum of numbers produces a finite answer when series like the harmonic series diverge despite however one might have wanted it to converge $\endgroup$ – Triatticus Apr 12 '17 at 7:31
  • $\begingroup$ @user3002473. I like the "Legendre constant" example in your link, and the comment to it that having the constant $1$ named after you is extremely cool. $\endgroup$ – DanielWainfleet Apr 12 '17 at 19:46
  • $\begingroup$ @Triatticus Mh, I only partially agree. The first time I encountered series I was surprised how an infinite sum could converge to a finite quantity; now, assuming said series converges, I am actually still amazed that even if all the summands are rational, the series yields am irrational number (consider for instance $e=\sum 1/n!$). In a sense, I am never too surprised to see rational numbers coming out of such series - even though heuristically it might not be reasonable - I just find it counter-intuitive :) $\endgroup$ – Angelo Rendina Apr 13 '17 at 23:04
  • $\begingroup$ Lots of irrational numbers come out of series though, it's not all that surprising, for instance the sum of terms $1/n^2$ yields an irrational number, specifically $\pi^2/6$ $\endgroup$ – Triatticus Apr 14 '17 at 0:37
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Legrendre's Constant is the best example of this that I know. Legendre was interested in the expression

$$\lim_{n\to\infty}\left(\ln(x)-\frac{n}{\pi(n)}\right)$$ due to its relation to the (then unproven) Prime Number Theorem. In particular, if the limit exists then the prime number theorem is true. Legrendre estimated the value to be $\sim 1.08$ but the exact value turned out to be $1$! Although I don't know of any specific conjectures that it is irrational, the fact that it turns out to be exactly $1$ was highly surprising to me and I see no reason to expect it to be rational.

I would imagine that there are many examples of this in probability theory from before the Komogorov Zero-One Law was established.

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    $\begingroup$ Nice one. Ignoring all the background, I would not expect a rational by just looking at that limit! $\endgroup$ – Angelo Rendina Apr 13 '17 at 23:06
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The average distance between two randomly chosen points in the Sierpinski triangle (of side $1$) is

$$\frac{466}{885}$$

(where "distance" means the length of the shortest path between the points that lies within the Sierpinski triangle).

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  • $\begingroup$ This is mindblowing and "obviously wrong." Do you have a proof? $\endgroup$ – Stella Biderman Apr 13 '17 at 23:06
  • $\begingroup$ @StellaBiderman There's one here, and it involves solving the Towers of Hanoi puzzle! $\endgroup$ – Oscar Cunningham Apr 14 '17 at 7:48
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(1). If $f:[0,1]\to \mathbb R$ is continuous with $f(0)=f(1)$ and $n\in \mathbb N$ then $\exists x\in [0,1-1/n]\;(f(x)=f(x+1/n)\;).$ (The theorem of the horizontal chord.) But if $y\in (0,1)$ and $y$ is not the reciprocal of a natural number then there exists a continuous $f:[0,1]\to \mathbb R$ with $f(0)=f(1)$ such that $\forall x\in [0,1-y]\;(f(x)\ne f(x+y)\;).$

(2). In any triangle, the orthocenter $O$, the barycenter $B$, and the circumcenter $C$ not only are collinear but $OB:BC=1:2$ (Euler, 1765). If this was "obvious" or "expected" it would likely have been known about 20 centuries earlier in Greece.

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    $\begingroup$ It's unclear to me how (1) answers the question. Can you reformulate it to be clearer? $\endgroup$ – Stella Biderman Apr 12 '17 at 21:37
  • $\begingroup$ @StellaBiderman. Lengths of horizontal chords of the graph of $f$ are guaranteed to occur only for a certain subset of $\mathbb Q.$ $\endgroup$ – DanielWainfleet Apr 12 '17 at 21:51
  • $\begingroup$ I still don't understand that comment. What is the number that is expected to be irrational but turns out to be rational? $\endgroup$ – Stella Biderman Apr 13 '17 at 23:06
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The exponents for 2 dimensional percolation. Here you consider a two dimensional lattice where each site is going to be made black or white with some probability (this is called site percolation), or alternatively, you delete bonds between sites with some probability (this is called bond percolation). Then, what you find is that there exists a critical probability above which there will be an infinite cluster of the same color in case of site percolation, in case of bond percolation there will be an infinity large connected set of sites when the probability of choosing bonds between sites is larger than the critical probability (which is known to be exactly $\frac{1}{2}$).

One can then ask questions like how the probability that points a distance $R$ apart are on the same cluster behaves for large $R$ when the system is at the percolation threshold. This is then described by a power law, the exponent can be calculated and turns out to be a rational number. One can also consider exponents associated with the approach to the percolation threshold. E.g. the size of the largest cluster sizes will grow as you approach the critical percolation probability $p_c$ as $(p - p_c)^{-\nu}$, and one can show that $\nu = \frac{4}{3}$.

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