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This is a really natural question for which I know a stunning solution. So I admit I have a solution, however I would like to see if anybody will come up with something different. The question is

What is the probability that two numbers randomly chosen are coprime?

More formally, calculate the limit as $n\to\infty$ of the probability that two randomly chosen numbers, both less than $n$ are coprime.

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    $\begingroup$ I think you will find this disappointing, but this problem is no secret: see here, here, and here. And I clicked only the first 3-4 links. ;) Actually, to be fair, some work has to be done to make the arguments in the wikipedia page rigorous, but that isn't the stunning part anyway. That said, let's see how many different proofs exist. $\endgroup$ – Srivatsan Sep 14 '11 at 14:46
  • $\begingroup$ ah.. ok :( i thought it was something less known. Well, seems i was wrong... it is not stunning anymore now :).. btw thanks for the references $\endgroup$ – uforoboa Sep 14 '11 at 14:48
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    $\begingroup$ @user15453 I didn't mean that this is not a stunning problem. It is very much so, and it's good to see that you find it so fascinating. The point that I wanted to convey was simply: it's more well-known that you seem to imagine. (That will be especially true in this site than (say) among your friends, since many people here do a lot more mathematics.) And I didn't mean to discourage you from learning such interesting things. On the contrary, I hope this will turn up lot of good answers (like, Eric's answer below). $\endgroup$ – Srivatsan Sep 14 '11 at 15:52
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Lets look at the function $$S(x)=\sum_{\begin{array}{c} m,n\leq x\\ \gcd(m,n)=1\end{array}}1.$$

Then notice that $$S(x)=\sum_{m,n\leq x}\sum_{d|\gcd(m,n)}\mu(d)=\sum_{d\leq x}\sum_{r,s\leq\frac{x}{d}}\mu(d)= \sum_{d\leq x}\mu(d)\left[\frac{x}{d}\right]^{2}$$

From here it is straight forward to see that in the limit, $$\frac{S(x)}{x^2}\rightarrow\sum_{n=1}^\infty \frac{\mu(n)}{n^2}=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}.$$

However, there are still some interesting questions here. How fast does in converge, and what are the secondary terms? It turns out we can easily relate this to the summatory totient function, which has a rich history. See these two math stack exchange posts: Totient function, Asymptotic formula. What follows below is a modification of my answer on the second post.

The History Of The Error Term

In 1874, Mertens proved that $$S(x)=\frac{6}{\pi^{2}}x^{2}+O\left(x\log x\right).$$ Throughout we use $E(x)=S(x)-\frac{6}{\pi^2}x^2$ for the error function.

The best unconditional result is given by Walfisz 1963: $$E(x)\ll x\left(\log x\right)^{\frac{2}{3}}\left(\log\log x\right)^{\frac{4}{3}}.$$

In 1930, Chowla and Pillai showed this cannot be improved much more, and that $E(x)$ is not $$o\left(x\log\log\log x\right).$$

In particular, they showed that $\sum_{n\leq x}E(n)\sim\frac{3}{\pi^{2}}x^{2}$ so that $E(n)\asymp n$ on average. In 1950, Erdos and Shapiro proved that there exists $c$ such that for infinitely many positive integers $N,M$ we have $$E(N)>cN\log\log\log\log N\ \ \text{and}\ \ E(M)<-cM\log\log\log\log M, $$

or more concisely

$$E(x)=\Omega_{\pm}\left(x\log\log\log\log x\right).$$

In 1987 Montgomery improved this to

$$E(x)=\Omega_{\pm}\left(x\sqrt{\log\log x}\right).$$

Hope you enjoyed that,

Added: At some point, I wrote a long blog post about this, complete with a proof of Montgomery's lower bound.

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Let us start with a very simple 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 $\frac{1}{p}$. This is having one chance over $p-$chances to be divisible by $p$.
  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 therefore given by
      $$ \color{blue}{\prod_{p, prime}\left(1-\frac{1}{p^2} \right) = \left(\prod _{p, prime}\frac {1}{1-p^{-2}}\right)^{-1}=\frac {1}{\zeta (2)}=\frac {6}{\pi ^{2}} \approx 0,607927102 ≈ 61 \%}$$

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. $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$

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    $\begingroup$ Quite a bump, but +1 for the easier explanation, and the extension to m integers. $\endgroup$ – theREALyumdub Jan 26 '18 at 22:09
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    $\begingroup$ @theREALyumdub in fact this is the cesaro theorem we just learn this in class. And I love it $\endgroup$ – Guy Fsone Jan 26 '18 at 22:11

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