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The rationals are clearly dense in the real number system, i.e. for every pair a < b of real numbers there exists a rational number p/q s.t. a < p/q < b. I conjecture the same to be true with p and q both primes. Any idea of how one could prove it? It should depend on some strong result on the distribution of prime numbers.

edited Oct 23 2009 at 17:22

asked Oct 23 2009 at 16:58

Gian Maria Dall'Ara
1,408●6●19

Frank Thorne · Accepted Answer · 2009-10-23 17:07:06Z UTC

Yes. Take q sufficiently big and fixed (in terms of a and b). Then the question is, is there some prime p between qa and qb? Use the prime number theorem to estimate pi(qb) - pi(qa) > 0, where q is chosen to be big enough so that the main term is bigger than the error terms. QED.

Qiaochu Yuan · Answer 2 · 2009-10-23 17:14:45Z UTC

3

You don't need the prime number theorem. Suppose the result is false, i.e. for fixed a and b there are only finitely many q such that there is a prime between qa and qb. Then the nth prime p_n grows at least as fast as (b/a)^n; in particular, sum 1/p_n converges, which we know to be false (and which is totally elementary). (This argument is slightly thorny to make rigorous, but it's just a matter of handling constants.)

answered Oct 23 2009 at 17:14

Qiaochu Yuan
38.2k●5●116●301

	It's enough to assume (by contradiction) that for any prime q, there are no primes between qa and qb. Then the conclusion follows without any complication if a=1, but I can't still figure out how you manage the case a=!1. – Gian Maria Dall'Ara Oct 23 2009 at 18:29
	Hmm. I might have to take this one back; this works fine if q is allowed to be arbitrary but for q prime the intervals [qa, qb] might skip all the intervals where primes are close together. – Qiaochu Yuan Oct 23 2009 at 18:52

Density of a subset of the reals

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