probability that two positive integers are relatively prime

The probability that two positive integers chosen randomly arerelatively prime is

$$\frac{6}{{\pi }^2}=0.60792710185\mathrm{\dots }.$$

At first glance this “naked” result is beautiful, but nosuitable definition is there: there isn’t a probability spacedefined. Indeed, the word “probability” here is an abuse oflanguage.So, now, let’s write the mathematical statement.

For each $n\in {\mathbb{Z}}^{+}$, let $S_n$ be the set $\{1,2,\mathrm{\dots },n\}\times \{1,2,\mathrm{\dots },n\}$ and define${\mathrm{\Sigma }}_n$ to be the powerset of $S_n$. Define$\mu :{\mathrm{\Sigma }}_n\to ℝ$ by $\mu (E)=|E|/|S_n|$. This makes$(S_n,{\mathrm{\Sigma }}_n,\mu )$ into a probability space.

We wish to consider the event of some $(x,y)\in S_n$ also beingin the set$A_n=\{(a,b)\in S_n:\gcd (a,b)=1\}$.The probability of this event is

$$P((x,y)\in A_n)={\int }_{S_n}{\chi }_{A_n}d\mu =\frac{|A_n|}{|S_n|}.$$

Our statement is thus the following. For each $n\in {\mathbb{Z}}^{+}$,select random integers $x_n$ and $y_n$ with $1\le x_n,y_n\le n$.Then the limit ${lim}_{n\to \mathrm{\infty }}P((x_n,y_n)\in A_n)$ exists and

$$\underset{n\to \mathrm{\infty }}{lim}P((x_n,y_n)\in A_n)=\frac{6}{{\pi }^2}.$$

In other words, as $n$ gets large, the fraction of $|S_n|$ consisting of relatively prime pairs of positive integers tends to $6/{\pi }^2$.