If I define the function
$$\Phi(n) := \sum_{k=1}^n \phi(k),$$
where $\phi$ is Euler's totient function,
and I define $Q_n(x)$ to be the number of distinct rational numbers with demoninators $\leq n$ and values $\in (0,x]$,
and I define
$$F(x) := \lim\limits_{n\to\infty} \dfrac{Q_n(x)}{\Phi(n)}$$ on $[0,1]$,
and I finally define $f(x) = F'(x)$.
Does $F(x)$ exist? If so, is it differentiable? If so, what does $f(x)$ look like?
My intuition says that $F$ should be the identity function. Is this true?
Yes, $F(x)=x$ for all $x\in[0,1]$.
Sketch of proof: $\Phi(n)$ counts all primitive lattice points in the triangle with vertices $(0,0)$, $(0,n)$, and $(n,n)$. (Here a lattice point $(a,b)$ is primitive if $\gcd(a,b)=1$. The bijection sends $(a,b)$ to $a/b$.)
You can count primitive lattice points by inclusion-exclusion: if $F(n)$ is the total number of lattice points in the triangle with vertices $(0,0)$, $(0,n)$, and $(n,n)$, then $\Phi(n) = \sum_{d=1}^\infty \mu(d)F(n/d)$.
Similarly, $Q_n(x)$ counts all primitive lattice points in the triangle with vertices $(0,0)$, $(0,n)$, and $(xn,n)$. And thus $Q_n(x)$ can also be calculated using inclusion-exclusion, in terms of the total number of lattice points in scaled versions of this triangle.
Finally, the number of lattice points in a large triangle is basically the same as its area.