TITLE: Using Quotient of Prime Numbers to Approximation Reals
QUESTION [16 upvotes]: We know a positive rational number can be uniquely written as $m/n$ where $m$ and $n$ are coprime positive integers. Particularly, we can pick out those numbers with $m$ and $n$ both prime.
Question 1: Is the collection of all such numbers dense on the positive half of the real line?
Furthermore, we can ask about the efficiency of approximation, more precisely:
Question 2: Suppose we have an inequality $1\le ps-qr\le a$. Fix some $a$, can we find infinitely many solutions where $p$,$s,$,$q$,$r$ are positive primes?

REPLY [2 votes]: I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of looking at Question 1.
Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.
First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$
Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that
$$p\in [rKn, rK(n+1)],$$
$$ q\in [sKn, sK(n+1)].$$
Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$
We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$
Edit: small rewording.