What is the ratio of all distinct fractions to all distinct pairs of naturals? I've been thinking of this lately. Clearly, $|\mathbb{N}| = |\mathbb{N}\times\mathbb{N}|$ and the rationals are equal in count to the integers, which is equal in count to the number of integer pairs, from a number theory point of view.
But what if we take limits? Clearly
$$\lim_{n\rightarrow\infty} \frac{|[n]\times[n]|}{|[n]|} = \infty$$where $[n]={1,2,3,\dots,n}$, but what is
$$\lim_{n\rightarrow\infty} \frac{|[n]\times[n]|}{|\mathbb{F}_n|}$$
where $\mathbb{F}_n$ is the set of all fully-reduced and distinct fractions with numerator and denominator less than or equal to $n$?
 A: You are counting the proportion of all pairs of positive integers to coprime pairs of positive integers.  This proportion equals $\pi^2/6$ (the value of $\zeta(2)$). 
The computation can be found in many places; wikipedia gives a sketch.
A: If we restrict to choosing only $\frac{a}{b}$ where $a\le b$, then we may count as $$\frac{n+{n\choose 2}}{|G_n|}=\frac{n^2+n}{2|G_n|}$$
where $G_n$ represent the Farey sequence of order $n$.   Using the approximation $|G_n|\approx \frac{3n^2}{\pi^2}$ lets us reproduce Matt E's result.
Followup: There are ${n\choose 2}$ with $a<b$, and $n$ with $a=b$. Although this restricts the fractions counted to $[0,1]$, these are asymptotically half of the fractions counted by the OP's $F_n$.
A: Clearly numerator is clearly $n^2$.
For the denominator see, $|F_n|=\displaystyle\sum_{p,q\leq n,(p,q)=1}1=\displaystyle\sum_{p,q \leq n}\displaystyle\sum_{d|(p,q)}\mu(d)=\displaystyle\sum_{d\leq n}\displaystyle\sum_{r,s\leq n/d}\mu(d)=\displaystyle\sum_{d\leq n}\mu(d)[\frac{n}{d}]^2$
$=2\displaystyle\sum_{d\leq n}\varphi(d) -1=\frac{6n^2}{\pi^2}+O(n\log n)$
Hence, $\displaystyle\lim_{n\to \infty}\frac{n^2}{|F_n|}=\frac{6}{\pi^2}$
