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Is this question well posed? See here for the solution Probability that two random numbers are coprime

I have also seen it in some contests. The question asks to compute $p=\lim p_n$ where $p_n$ is the probability that two random chosen integers less than $n$ are coprime. There is no way to associate a uniform distribution to integers; so I would hesitate to call this limit a probability. So is there any rigorous way to understand this limit as a probability of some event?

See also this post

What's the mean of all real numbers?

where it is mentioned (and I agree) that the mean of reals (or integers) is undefined. But one could in the same way define a uniform distribution for reals or integers with absolute value less than $x>0$ and take the limit of the mean as $x$ goes to infinity. Then the mean of reals would be $0$.

Community
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adrido
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  • How do you define a uniform distribution on the integers less than $x$? On the positive integers less than $x$, sure, but then the limit of the mean is the limit of $(x+1)/2$, which isn't zero. – Gerry Myerson Jan 04 '14 at 03:24
  • fixed it, thanks. – adrido Jan 04 '14 at 03:25
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    But with the coprimality problem, the limit is the same however you take it. With the mean problem, you get a different answer if you take the limit over intervals $[-x,x+1]$, $x\to\infty$. – Gerry Myerson Jan 04 '14 at 03:29
  • That's a good point. but are you saying that if a sequence of pairs of probabilities and events $(p_n,A_n)$ with $p_n=P_n(A_n)$ converge to $(p,A)$ in a way that $p$ does not depend on $A_n$, then $p=P(A)$ for some probability measure $P$? – adrido Jan 04 '14 at 03:51
  • https://math.stackexchange.com/questions/64498/probability-that-two-random-numbers-are-coprime-is-frac6-pi2?noredirect=1&lq=1 – Guy Fsone Feb 02 '22 at 19:52

2 Answers2

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Yes. Define a measure of a set $\Omega$ in the plane to be the limit of number of relatively prime points in $t \Omega$ divided by $t^2,$ as $t$ goes to infinity. It can be shown that for "nice" sets this is a measure, which is a multiple of the Lebesgue measure. The multiple is precisely the limit you talk about ($6/\pi^2$). You can look for a discussion along these lines in this paper if you want to know more.

Igor Rivin
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  • nice paper, thanks! – adrido Jan 04 '14 at 03:57
  • @Igor Rivin : For the purposes of the first question, what reason is there for considering any set $\Omega$ other than, say, $(0,1)^2$? If the answer is in the paper you linked, you can just say so and I'll look at it. – Stefan Smith Jan 04 '14 at 06:32
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    @StefanSmith For your specific question, that is the only set, but it is not a priori clear that that way of defining a "random pair" is any more natural than any other way. – Igor Rivin Jan 04 '14 at 16:26
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Let us start with the following observation:

One integer chosen amongst $'p'$ other integers has one chance to be divisible by $p$

  1. From this we infer that the probability that an integer is divisible by $p$ is $\frac{1}{p}$.
  2. Therefore the probability that two different integers are both simultaneously divisible by a prime $p$ is $\frac{1}{p^2}$
  3. This means that the probability that two different integers are not simultaneously divisible by a prime $p$ is $$1-\frac{1}{p^2}$$
    1. Conclusion: The probability that two different integers are never simultaneously divisible by a prime (meaning that they are co-prime) is then given by
      $$ \color{red}{\prod_{p, prime}\left(1-\frac{1}{p^2} \right) = \left(\prod _{p, prime}\frac {1}{1-p^{-2}}\right)^{-1}=\frac {1}{\zeta (2)}=\frac {6}{\pi ^{2}} \approx 0,607927102 ≈ 61 \%}$$

Where should recall from the Basel problem we have the following Euler identity

$$\frac{\pi^2}{6}=\sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2)=\prod _{p, prime}\frac {1}{1-p^{-2}} $$

By similar Token, The probability that $m$ numbers are co-prime is given by

$$ \color{red}{\prod_{p, prime}\left(1-\frac{1}{p^m} \right) = \left(\prod _{p, prime}\frac {1}{1-p^{-m}}\right)^{-1}=\frac {1}{\zeta (m)}}$$

Here $\zeta$ is the Riemann zeta function. $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$

Guy Fsone
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