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In many of the sources I have consulted about this, the "probability" that two positive integers chosen at random are relatively prime is calculated as the limit as $n \to \infty$ of the probability that two randomly chosen integers in the set {1,2, ..., $n$} are relatively prime (the limit being $1/\zeta(2)$). My first question is: Is this limit really a probability?

Also, the nonrigorous/heuristic proofs that I have seen of this start by mentioning that "the probability that a prime $p$ divides a positive integer is $1/p$". This makes intuitive sense. I was wondering though: Is there a way of defining a probability measure on the positive integers in such a way that the set {$n \in \mathbf Z_+$ | $p$ divides $n$} has measure $1/p$ (that we can use for a rigorous proof)?

echoone
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3 Answers3

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The answer is yes, there is such a measure, if we weaken the notion of measure suitably. (The "if we weaken" part is a way of saying that this does not really answer the question.)

What is given up is countable additivity. Finite additivity is kept, and, very importantly, we can have translation invariance. (Obviously we cannot have both countable additivity and translation invariance. Since we can't have both, translation-invariance may be the more attractive property.)

There is a translation-invariant finitely additive "measure" on the positive integers that gives measure $1$ to $\mathbb{N}$, and that is defined on all subsets of $\mathbb{N}$. The idea goes back to Banach. One could look under Banach mean, invariant mean, or amenable (group or semigroup). There is a large literature.

The translation invariance guarantees that for any modulus $m$, and any integer $a$, the set of numbers congruent to $a$ modulo $m$ has "measure" $1/m$.

Construction of a Banach mean requires the Axiom of Choice.

Addendum: One can build stronger properties into a Banach mean. Let $A$ be any subset of $\mathbb{N}$, and let $A_n$ be the set of all $x \in A$ such that $x \le n$. If

$$\lim_{n\to\infty} \frac{|A_n|}{n}$$

exists, call that limit $d(A)$. There is a Banach mean that assigns "measure" $d(A)$ to any $A$ for which $d(A)$ exists.

André Nicolas
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  • Presumably one can do a similar construction on ${\bf N}\times{\bf N}$. Does one then find that the set of $(a,b)$ with $\gcd(a,b)=1$ has measure $6/\pi^2$? – Gerry Myerson May 07 '11 at 03:56
  • Cool. I did not anticipate this would lead to talk about amenable groups. I just learned about them recently in connection with the Banach-Tarski paradox and ergodic measures. – echoone May 07 '11 at 04:50
  • @yoyo: Talking here about probability measures, of course. If we allow $\infty$, we can give weight $1$ to every point. Not all points have weight $0$, else countable additivity gives $\mathbb{N}$ weight $0$. And if some point has weight $w>0$, by translation invariance they all do, so $\mathbb{N}$ cannot have weight $1$. – André Nicolas May 07 '11 at 05:13
  • Gerry Myerson: For $\mathbb{N}$ I know how to do it by an ultrapower argument. Will try to dig back into my past to see whether I can do it for $\mathbb{N}\times \mathbb{N}$. Even back then (circa 1980) the stuff for $\mathbb{N}$ wasn't new. – user6312 9 hours ago – André Nicolas May 07 '11 at 14:55
  • @AndréNicolas Did you find how to do it for $\mathbb{N}\times\mathbb{N}$? – Arbuja Apr 17 '17 at 13:19
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There are some probability measures on the set of integers, but quite often they do not correspond to our intuition (they are not translation invariant for example). To measure sets in number theory, we often rely on densities such as the natural density, of which you are speaking, and the analytic density, which is actually better behaved. The good news is when a set has both densities, they are equal.

Joel Cohen
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  • Are these "densities" the same as "Banach means" as in the first answer above? In any case, would it be correct to say that "the density that two random positive integers are coprime is $6/\pi^2$"? – echoone May 07 '11 at 04:48
  • @echoone, I would phrase it this way: the (natural) density of the coprime pairs is $6/\pi^2$, in the same way as I would say the (natural) density of the even numbers is $1/2$ (rather than trying to make sense of "the density that a random positive integer is even is $1/2$"). – Gerry Myerson May 07 '11 at 05:59
  • @ Yoyo : Suppose you have a translation invariant probability measure on $\mathbb{Z}$ (or $\mathbb{N}$). Because of translation invariance, all singletons have the same measure. But from countable additivity, the sum over $\mathbb{Z}$ of this measure must be $1$. Clearly this is impossible.

    That's why you have to either drop translation invariance or countable additivity.

     – Joel Cohen May 07 '11 at 13:39
  • @yoyo: The post dealt with the semigroup $\mathbb{N}$. In that case the $\sigma$-algebra would contain the points. Don't have immediate answer for $\mathbb{Z}$. – André Nicolas May 08 '11 at 16:28
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One may use the honest probability $P(X=n)=n^{-s}/\zeta(s)$ for $s>1$. From this honest probability one may prove that:

1 Events $E_{p}=\{X$ is divisible by $p\}$ are independent.

2 $P(gcd(X,Y)=n)=n^{-2s}/\zeta(2s)$ for independent $X$ and $Y$. If $s\to 1$ for $n=1$ we obtain the claim.

Some other results may be obtained. See ex. 4.2. in Probability with martingales by Williams.

Roah
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