# Using Quotient of Prime Numbers to Approximation Reals

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?

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?

-
For question 1, a negative answer would imply unusually large gaps between primes, in particular that there are no primes between x and x + x^(2/3) for too many x. For question 2, it is likely true even for a=1, given that the number of divisors function is 4 pretty often, even for consecutive numbers. Gerhard "Ask Me About System Design" Paseman, 2012.12.25 –  Gerhard Paseman Dec 25 '12 at 9:07
A related question mathoverflow.net/questions/53736/… –  S.C. Dec 26 '12 at 8:13

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 of Baker, Harman and Pintz on prime gaps. 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, we can take $a\geq 6$.

This is a corollary of the work 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 of Goldston, Graham, Pintz and Yıldırım show that we can take $a=6$ unconditionally. (Thank you to quid for mentioning this in the comments)

-
Nice answer. Two minor comments: in the second display it seems a minus in the exponent is missing. Further, I believe that by subsequent work of the same authors 6 (instead of 26) is known unconditionally. –  quid Dec 25 '12 at 15:48
@quid: Thank you for your comments. I found the paper, it is also by Goldston, Graham, Pintz and Yıldırım. –  Eric Naslund Dec 25 '12 at 17:49

Here is a simpler solution to question 1. By the prime number theorem, the $n$th prime $p_n$ admits the asymptotic estimate $p_n \sim n\log n$. It follows for any real number $x>0$ that $p_{[nx]}/p_n \rightarrow x$ as $n \rightarrow \infty$. The rate of convergence for this explicit ratio of primes tending to $x$ is pretty slow, however. For instance, taking $x = \pi$, I checked with PARI that $p_{[n\pi]}/p_n$ is $3.642$ when $n = 1000$ and $3.517$ when $n=10000$.

-

Question 1 comes up a lot. For example, it was discussed on sci.math back in 1990. It was also discussed in print in the following article:

Quotients of primes, by David Hobby and D. M. Silberger, Amer. Math Monthly 100 (1993), 50–52.

More recently it has shown up on Yahoo Answers and math.stackexchange.com.

-

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.

-