# What is the probability of finding a number of the form $\frac{n}{n+1}$ in $\Bbb{Q}\cap [0,1]$?

Based on this question, I want to ask:

Suppose $q$ is drawn uniformly from $[0,1]$. What is the probability that the reduced form of $q$ has the form $\frac{n}{n+1}$ conditioned on $q\in\Bbb{Q}$?

(I'm reminded that one can't have a probability distribution on a countably infinite discrete set; but I don't recall offhand how to prove that there are no Borel probability measures on any countably infinite set. So I think the modification above will make sense but am happy to contemplate reasons why it doesn't and/or alternative probability measures to use on $\Bbb{Q}\cap [0,1]$.)

Here's my quick answer attempt. It will become evident that it's flawed but I don't have clarity on how exactly to fix the argument -- mostly I don't have good intuition for the function $$n\mapsto \#\{\mbox{integers k<n that are coprime to n}\}.$$

I'm going to write $$\Bbb{Q}\cap [0,1] = \bigcup_{n=0}^\infty Q_n$$ where $$Q_n = \bigg\{ \frac{k}{n+1}\ \bigg|\ k=0,1,\ldots,n+1 \bigg\}.$$ For each $n$ we have $|Q_n| = n+1$. As there is precisely one element of $Q_n$ of the form $\frac{n}{n+1}$ we have for each $N$ the probability of randomly drawing such an element from $\cup_{n=0}^N Q_n$ is no less than $$\frac{N+1}{\sum_{j=0}^N Q_j} = \frac{N+1}{1+2+\cdots + N+1} = \frac{2(N+1)}{(N+1)(N+2)} = \frac{2}{N+2}$$ However: This goes to zero, so: great, the probability is no less than zero

Notice that for composite $n$, many elements of $Q_n$ are not in reduced form, so I'm overcounting a lot of fractions.

Am I overcounting enough fractions to push the probability that $q = \frac{n}{n+1}$ above zero?

• You can't have a uniform probability distribution on a countably infinite set. – paw88789 Jul 6 '17 at 17:06
• The probability is zero. Consider how many numbers are of the form $\frac nk$ with $k>n+1$. – Simply Beautiful Art Jul 6 '17 at 17:10
• I'd say it is "not even zero" – Hagen von Eitzen Jul 6 '17 at 17:11
• Can you condition on a null event? – Ranc Jul 6 '17 at 17:30
• As has been said in the comments above : there is no uniform probability on a countable set. Therefore the answer will completely depend on the probability measure. It could be any $x\in [0;1]$, depending on the measure you're willing to put. According to the measurable sets you allow, it is possibly not defined. So the answer is "as you wish" – Max Jul 6 '17 at 17:36

I don't think that the conditioning solves the problem of defining a uniform probability measure over the rationals.

If, for$X\sim U[0,1]$, we were allowed to condition on the (zero probability) event $A \equiv X \in \mathbb{Q}$ (I don't think we can) then, for any $q\in \mathbb{Q}$ we'd should be able to calculate $P(X=q \mid A)=\alpha$ ; but again, this cannot work, because of additivity: no value of $\alpha$ can give us $\sum_{q\in \mathbb{Q}} P(X=q \mid A) =1$

We'd need some non-uniform (and not simple) measure, for example.

So, I'm afraid it's hard to make sense of the question.

For every $n$ such that

$$x_n=\frac n{n+1}$$

There exists infinitely many rationals of the form

$$y_n=\frac nk,k>n+1$$

Thus, the 'probability' of choosing a number of the form $x_n$ is $0$.

Note a probability of $0$ does not mean it is impossible for you to choose a number of the form of $x_n$, but rather that it is improbable.

A classic example is the probability you'd choose $1$ out of all the real numbers in $[0,1]$.

• Sorry to be dense - does this imply an answer to my revised question, conditioning $q = \frac{n}{n+1}$ on $q\in\Bbb{Q}$? – Neal Jul 6 '17 at 17:46
• @Neal I'm unsure what you mean. Are you taking $n\in\mathbb Z\setminus-1$? Either way, yes, the density of your $q$ is zero in $\mathbb Q$. – Simply Beautiful Art Jul 6 '17 at 17:48
• Let $R = \{\frac{n}{n+1}\ |\ n = 0, 1, 2, \ldots\}$ and $Q = \Bbb{Q}\cap[0,1]$. My revised question is: what is $\operatorname{Pr}[R|Q]$ with respect to the Lebesgue measure on $[0,1]$? – Neal Jul 6 '17 at 17:54
• Pretty sure it's still zero. – Simply Beautiful Art Jul 6 '17 at 18:10
• Does this mean that the probability of each one specific rational is $0$? Because if so, then this is not a (conditional) probability distribution. (The (conditional) probabilities don't sum to $1$.) – paw88789 Jul 6 '17 at 22:23