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?



Using Quotient of Prime Numbers to Approximation Reals
**Question 1:** The set is dense. 


Suppose that we are given a fixed $x\in\mathbb{R}$. Then let $p$ be a large prime. If $p$ is sufficiently large, then there will be a prime $$q\in\left[px,\ px+\left(px\right)^{0.525}\right]$$ by [the work](http://www.cs.umd.edu/~gasarch/BLOGPAPERS/BakerHarmanPintz.pdf) of Baker, Harman and Pintz on [prime gaps](http://en.wikipedia.org/wiki/Prime_gap). This implies that $$\left|x-\frac{q}{p}\right|\ll\_x p^{-0.475},$$ which becomes arbitrarily small as we take $p\rightarrow\infty $. This proves that for any $\epsilon>0$, there exists $p,q$ such that $\left|x-\frac{q}{p}\right|\leq \epsilon.$


**Question 2:** We can find infinitely many solutions to $$1\leq qp-rs\leq a$$ for primes $p,q,r,s$ and all $a\geq 26$. Under the [Elliott-Halberstam Conjecture](http://en.wikipedia.org/wiki/Elliott%E2%80%93Halberstam_conjecture), we can take $a\geq 6$.


This is a corollary of [the work](http://arxiv.org/abs/math/0506067) of Goldston, Graham, Pintz and Yıldırım on the gaps between almost primes. They prove that if $q\_n$ is the $n^{th}$ almost prime, then $$\liminf\_{n\rightarrow \infty} q\_{n+1}-q\_n \leq 26,$$ and that the upper bound may be reduced to $6$ under the Elliott-Halberstam Conjecture. Since $q\_n=pq$ and $q\_{n+1}=rs$ where $p,q,r,s$ are primes, this yields the above claim.


**Edit:** The [more recent work](http://arxiv.org/abs/math/0609615) of Goldston, Graham, Pintz and Yıldırım show that we can take $a=6$ unconditionally. (Thank you to quid for mentioning this in the comments)


