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Consider the closed interval $[0,1]$, there is $\frac{2}{3} \in [0,1]$ where $p=2$ and $q=3$. Similarly consider $[2,3]$, one can have $\frac{5}{2} \in [2,3]$ where $p=5$ and $q=2$. Does every interval of the form $[a,b]$, where $a,b \in \mathbb{R}$ contain a rational of this kind. If yes, how can we prove it?

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Roughly speaking this asks whether the quotients of two primes are dense in the positive reals. The answer is yes.

Let $0 < a < b$ and let $q$ be a prime. Then there will a a prime $p$ with $a < p/q\le b$ if and only if $\pi(bq) > \pi(aq)$ where $\pi$ is the prime-counting function. But by the prime number theorem, as $q\to\infty$, $$\frac{\pi(bq)}{\pi(aq)}\sim\frac{b\log(aq)}{a\log(bq)} =\frac{b(\log q+\log a)}{a(\log q+\log b)}\sim\frac ba>1.$$ For all large enough $q$, $\pi(bq)/\pi(aq) > 1$ as required.

Robin Chapman
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    +1 -- although I wonder if we really need the prime number theorem here – balpha Sep 05 '10 at 16:17
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    Right, I was getting at something similar in my comment above. Recently I have been starting to wonder whether we should be giving hints rather than complete answers to questions like this. – Pete L. Clark Sep 05 '10 at 16:20
  • I feel like you don't need the Prime Number Theorem for this. – Matthew Levy Dec 26 '14 at 08:15
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    So given a and b , for a sufficiently large prime q you can always find a prime p between aq and bq. Right, someone remind me, why was this problem rewarded with 50 points? – Cristian Dumitrescu Feb 20 '22 at 07:42
  • Not using the PNT to solve this question is not obvious, see my comments on https://math.stackexchange.com/questions/1013414. – Watson Jun 10 '22 at 05:33