Suppose we order the rational numbers using the diagonal method (used to prove they are countable) using an $n\times n$ grid. Now suppose we count the distinct rational numbers (those points on the grid where gcd(numerator, denominator) = 1. Denote this by $S\_n$.


Does $\lim\_{n\to\infty} n^2/|S\_n|$ exist, and if so what is it?


Some quick computations suggest that it converges to approx 1.64...



does the ratio of the count of rational numbers on an $n\times n$ grid to $n^2$, converge as $n$ tends to infinity
The limit you are looking for is the *reciprocal* of the probability that two random positive integers are coprime. This probability (with the proper intepretation) is the density of square-free numbers:
$$\prod\_{\text{$p$ is prime}}\left(1-\frac{1}{p^2}\right)=\left(\sum\_{n=1}^\infty \frac{1}{n^2}\right)^{-1}=\frac{6}{\pi^2}.$$
Note that
$$\sum\_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}=1.6449340668\dots$$


