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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?

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  • $\begingroup$ 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 $\endgroup$ Commented Dec 25, 2012 at 9:07
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    $\begingroup$ A related question mathoverflow.net/questions/53736/… $\endgroup$
    – C.S.
    Commented Dec 26, 2012 at 8:13
  • $\begingroup$ The first question is answered in Quotients of Primes by David Hobby and D. M. Silberger. See also math.stackexchange.com/questions/1013414/…. $\endgroup$
    – Watson
    Commented Aug 25, 2016 at 8:26

4 Answers 4

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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)

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    $\begingroup$ 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. $\endgroup$
    – user9072
    Commented Dec 25, 2012 at 15:48
  • $\begingroup$ @quid: Thank you for your comments. I found the paper, it is also by Goldston, Graham, Pintz and Yıldırım. $\endgroup$ Commented Dec 25, 2012 at 17:49
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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 $n=10000$.

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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.

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    $\begingroup$ Tim, you posed Question 1 yourself in 1990 during that course we took using Serre's Course in Arithmetic. I was presenting material from the section on Dirichlet's theorem, and with that on my mind I noticed that if Dirichlet's theorem admits a stronger form where $\gcd(a,b) = 1 \Rightarrow an+b$ and $bn+a$ are both prime (together) for infinitely many $n$ then from $(an+b)/(bn+a) \to a/b$ as $n \to \infty$ we'd get the ratios of primes to be dense in $\mathbf Q_{>0}$ and thus also dense in $\mathbf R_{>0}$. The next year I found the right context for that in the Bateman-Horn conjecture. $\endgroup$
    – KConrad
    Commented Dec 20, 2020 at 0:43
  • $\begingroup$ @KConrad : Yes, I remember asking that question back then, but I no longer recall if I came up with the question myself or heard it from someone else first. $\endgroup$ Commented Dec 20, 2020 at 1:05
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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.

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