Let $\mathbb{Q}[n]$ be the set of rational numbers with denominator $\le n$ and for any $X\subseteq \mathbb{Q}$, let $X[n]=X\cap \mathbb{Q}[n]$.

Is there a set of rational numbers, X, such that for any interval Y of rationals: $$\underset{n\to \infty }{\mathop{\lim }}\,\frac{card(X[n]\cap Y)}{card(\mathbb{Q}[n]\cap Y)} = 1/2 ?$$

  • $\begingroup$ It's confusing to use lower-case letters to denote sets; usually sets are denoted by upper-case letters. $\endgroup$ Nov 1, 2012 at 6:19
  • $\begingroup$ The rationals with odd denominator in reduced form? $\endgroup$
    – ronno
    Nov 1, 2012 at 6:38
  • $\begingroup$ @ronno, do you mean odd numerator? That would've been my guess. $\endgroup$ Nov 1, 2012 at 6:53
  • $\begingroup$ I think that the proportion of fractions with even numerator and denominator bounded by $N$ is aproximately one third, and the same for even denominator. $\endgroup$ Nov 1, 2012 at 10:21
  • $\begingroup$ A good candidate for your set $X$ with a very simple description is the set of reduced fractions $a/b$ with $$ a\cdot b \equiv 0 \pmod{3} $$ $\endgroup$ Nov 1, 2012 at 11:22

3 Answers 3


Yes. Let $\phi(n)$ be the number of rationals in $[0,1)$ with denominator (when in reduced form) exactly equal to $n$. This is the Euler totient function (OEIS:A000010), and while it jumps around, on average it grows linearly, such that $\sum_{k=1}^{n}\phi(k)\sim 3n^2/\pi^2$, and clearly $\phi(n) \le n$. Define the indicator variable $a_n$ recursively: $a_1=1$, and $a_n=1$ for $n>1$ if $\sum_{k=1}^{n-1}a_{k}\phi(k) < \frac{1}{2} \sum_{k=1}^{n-1}\phi(k);$ otherwise $a_n=0$. Then the set consisting of all rational numbers with denominator $n$ such that $a_n=1$ satisfies the given condition. For each new denominator $n$, we are including rationals with that denominator if the density so far is less than $1/2$. Because the contribution to the density from rationals with denominator $n$ decreases to zero (as $1/n$), the density converges to $1/2$ as $n\rightarrow\infty$. The generalization to fractions other than $1/2$ is obvious.

  • $\begingroup$ I think there something missing in the right hand side of the inequality $\sum_{k=1}^{n-1} a_k\phi(k) < 1/2$, I suppose you mean $<\frac{1}{2}\sum_{k=1}^{n-1} \phi(k)$? On the other hand, even if it is probably true, I can't see how to prove that this construction works for all the intervals $Y$. $\endgroup$ Nov 1, 2012 at 10:34
  • $\begingroup$ @EstebanCrespi: You're right about the missing term, thanks. It works for any interval because the number of rationals with denominator $n$ in an interval of length $\lambda$ is $\lambda \phi(n) + o(n)$ as $n \rightarrow \infty$... that is, the corrections grow much more slowly than $n$. $\endgroup$
    – mjqxxxx
    Nov 1, 2012 at 12:20

The probabilistic method is well suited to such settings...

Order the positive rationals as $\{q_n\mid n\geqslant1\}$ in any way you prefer (for example, ordering the reduced fractions $p/q$ first along the increasing values of $p+q$ and then along the increasing values of $p$ in case of equal $p+q$).

Let $(\xi_n)_{n\geqslant1}$ denote an i.i.d. sequence of Bernoulli random variables such that, for every $n$, $\mathbb P(\xi_n=1)=\mathbb P(\xi_n=0)=\frac12$. Let $X\subseteq\mathbb Q_+$ denote the random set $X=\{q_n\mid n\in N\}$ where $N=\{n\geqslant1\mid\xi_n=1\}$.

Then the law of large numbers asserts that $X$ is almost surely a solution, for every fixed interval $Y=(a,b)$ with $0\leqslant a\lt b$ rationals. There are countably many such intervals $Y$ hence $X$ is almost surely a solution, simultaneously for all of them.

To get a suitable random subset of $\mathbb Q$, do the same thing, independently, with $\{-q_n\mid n\geqslant1\}$ (or start from any ordering of $\mathbb Q$).

To realize any density $\alpha$ in $(0,1)$ instead of $\frac12$, use an i.i.d. sequence $(\xi^\alpha_n)_{n\geqslant1}$ of Bernoulli random variables such that, for every $n$, $\mathbb P(\xi^\alpha_n=1)=\alpha$ and $\mathbb P(\xi^\alpha_n=0)=1-\alpha$. This has the additional advantage to show that one can realize each set $X_\alpha$ with density $\alpha$ in a way such that $X_\alpha\subseteq X_\beta$ (and $X_\alpha\subset X_\beta$ almost surely) for every $\alpha\lt\beta$.


As I mentioned in a comment above there is a nice set $X$ containing half the rationals with a very simple description: the set of reduced fractions $a/b$ with $\operatorname{gcd}(ab,3)=1$.

To see it, we can map each reduced fraction $a/b$ to the pair $(x,y) \equiv (a,b) \pmod{3}$. There are 8 possible pairs $(x,y)$: $$ (0,1),(0,2),(1,0),(2,0),(1,1),(1,2),(2,1),(2,2) $$ if we can prove that reduced fractions in an interval $Y$ are uniformly distributed between these 8 pairs then we have the result.

It is clearly enough to prove this for an interval $Y = [0,\lambda)$ for some real $\lambda > 0$. Lets fix a pair $(x,y)$ and call $A_{(x,y)}(n)$ the number of fractions $a/b$ with $b\le n$, $\operatorname{gcd}(a,b)=1$, $a < \lambda b$, and $(a,b)\equiv(x,y\pmod{3}$ ie the horrid sum;

$$ A_{(x,y)}(n) = \sum_{b\le n}\sum_{\begin{matrix}\operatorname{gcd}(a,b)=1\\(a,b)\equiv (x,y) \pmod{3}\\a \le \lambda b\end{matrix}} 1 $$

Now we use the Möbius identity: $$ \sum_{d \vert \operatorname{gcd}(a,b)}\mu(d) = \begin{cases}1 \quad&\text{if }\operatorname{gcd}(a,b)=1\\0&\text{otherwise}\end{cases} $$ to get

$$ A_{(x,y)}(n) = \sum_{b\le n}\sum_{\begin{matrix}(a,b)\equiv (x,y) \pmod{3}\\a \le \lambda b\end{matrix}} \sum_{d\vert \operatorname{gcd}(a,b)} \mu(d) $$

We can invert this sums puting $b = kd, a = td$ and then $t < \lambda k$ so we have

$$ A_{(x,y)}(n) = \sum_{d\le n} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}}\sum_{\begin{matrix}t<\lambda k \\td\equiv x\pmod{3}\end{matrix}} 1 $$

In the outer sum we can limit ouselves to integers $d$ coprime with 3, (if $3 \vert d$ as $(x,y)\not\equiv (0,0)\pmod{3}$ there is nothing to sum) so we have:

$$ \begin{align} A_{(x,y)}(n) &= \sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}}\sum_{\begin{matrix}t<\lambda k \\td\equiv x\pmod{3}\end{matrix}} 1\\ &=\sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}} \mu(d) \sum_{\begin{matrix}kd\le n\\kd\equiv y\pmod{3}\end{matrix}} \left(\frac{\lambda k}3 + O(1)\right) \\ &=\sum_{\begin{matrix}d\le n\\d\not\equiv 0\pmod{3}\end{matrix}} \mu(d) \left( \frac{ \lambda n^2}{18d^2} + O(n/d) \right)\\ &=\frac{\lambda n^2}{18}\sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} + O(n) + O(n\log n) \end{align} $$

We can evaluate the constant in the last equation, call $$ B = \sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} $$ then $$ B - \frac{B}{9} = \sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(d)}{d^2} +\sum_{\begin{matrix}d=1\\d\not\equiv 0\pmod{3}\end{matrix}}^\infty \frac{\mu(3d)}{(3d)^2} = \sum_{d=1}^\infty \frac{\mu(d)}{d^2} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}$$ and so: $$ B = \frac{27}{4\pi^2} $$ and finally $$ A_{(x,y)}(n) = \frac{3\lambda }{8\pi^2} n^2 + O(n\log n) $$ so each pair $(x,y)$ have ultimately the same proportion $1/8$ of all reduced fraction in the interval, as was to be shown.

Note: I suppose that with some additional work it can be proven that for a given modulus $m$ the reduced fractions $a/b$ are uniformly distributed when reducing mod $m$ between the pairs $(x,y)$ such that $\operatorname{gcd}(x,y,m)=1$. And then we could find sets containing any given proportion of the rationals with the same procedure.

  • $\begingroup$ Wow. Thank you. That's the kind of explicit solution I was hoping for - though, I must admit, it will take me a while to work through and understand the proof. (The proof certainly has enough detail; I'm just weak on the number theory.) $\endgroup$
    – fmkatz
    Nov 3, 2012 at 22:27

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