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Are there examples that we know a number is a rational number but we do not know what is its numerator and denominator?

In order to say clearly, this number should given by a certain formula, such as $\sum_{i=1}^\infty f(n)$ (or $\int_0^\infty f(x)dx$)where $f(n)$ is a certain function so that we can calculate $f(n)$ for any given integer number $n$ (or real number $x$). Hence we avoid the answers like "the least even integer $N$ which makes Goldbach conjecture not true" or "the age when I get married".

Thanks in advance!

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To say we don't know the numerator and denominator is equivalent to saying we don't know the value. Could one have a convergent series and not know the limit ? –  Tom Collinge Feb 22 '14 at 14:19
There exists an integer-valued computable total function $f$ such that $\sum_{n=1}^{\infty} f(n)$ is the least integer $k$ such that the number of primes $\le k$ is greater than $\operatorname{li} k$. We know this series converges: the Skewes number is an upper bound for it. So it is not as easy to exclude these kinds of examples as you might think... –  Zhen Lin Feb 22 '14 at 14:22
@TomCollinge: I think the OP is asking for an example on the lines of various examples of numbers which are irrational. Instead now the number is rational, although an exact value is unknown. –  Shahab Feb 22 '14 at 14:24
Bad example of an answer to the question: nearest rational approximation of $\pi$ within $10^{-{10^{10}}}.$ –  Jeff Snider Feb 22 '14 at 14:32
@JeffSnider There is not a nearest rational approxination in that range (you can always find a better one) –  chubakueno Feb 22 '14 at 14:38

2 Answers 2

Take any Turing machine $M$, and let $h_M$ be $\frac{1}{N}$ iff $M$ halts after exactly $N$ steps, and let $h_M := 0$ if $M$ never halts. It is clear that $h_M$ is a rational number, given $M$ we can compute $h_M$ as a real number (i.e. find better and better approximations to it), but there is no general procedure to write $h_M$ as a quotient of natural numbers (because that would mean solving the Halting problem).

Edit: I might point out that there is a topological counterpart to this argument, namely the observation that the $id : \mathbb{Q}_e \to \mathbb{Q}_d$ is discontinuous, where $\mathbb{Q}_e$ are the rationals with the subspace topology inherited from $\mathbb{R}$, while $\mathbb{Q}_d$ are the rationals with the discrete topology.

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I would call this "a rational representation of the answer to a question we cannot answer," which the question tried (poorly) to exclude. –  Jeff Snider Feb 22 '14 at 18:09
The crucial difference between my examples and those ruled out in the question is that here is IS possible to write down the number as a REAL number. –  Arno Feb 22 '14 at 18:15
I don't think this is a good example at all. Determining $h_M$ does not require a solution to the Halting Problem. The theorem about the Halting Problem says there is no method that can calculate $h_M$ for all $M$. It says nothing about whether one can calculate $h_M$ for one particular $M$, as you are doing here. There certainly exist Turing machines for which solving the Halting problem is easy, or even trivial. –  MJD Feb 26 '14 at 4:59
It is not a single example, but a parameterized family of examples. While one can know some of them as fractions, one cannot know all of them as fractions - note the phrase "general procedure". That is the best one can get in this setting anyway, as any fraction is of course knowable in principle - if one only knew it. –  Arno Feb 26 '14 at 9:42

Let $E$ be an elliptic curve over $\mathbb Q$ of (algebraic) rank $0$. Let $L(E,s)$ be its associated $L$-function. Then the quantity $L(E,1)/\Omega_E$, where $\Omega_E$ is the period of $E$, is known to be rational, even if we don't know its numerator and denominator.

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