Bertrand's Postulate asserts that there is a prime between $n$ and $2n$.

Is this the best such upper bound on prime gaps known today, or have stronger estimates been proved? I mean results of the kind:

  • there will always be a prime between $n$ and $2n-2$, or
  • there will always be a prime between $n$ and $cn$ with $1<c<2$?

Was any such improvement proved rigorously, or is Bertrand's Postulate still the best we have?

  • 3
    $\begingroup$ You can read Wikipedia. There is no reason to ask such a question here. $\endgroup$ Sep 27, 2014 at 4:39
  • $\begingroup$ There is a nice proof of this from the book "Proofs from the book", which is elementary and you might like. $\endgroup$ Sep 27, 2014 at 5:11

1 Answer 1


Yes. See theorem 3 of:

O. Ramar´e, Y. Saouter, Short effective intervals containing primes, J. Number Theory 98 (2003), no. 1.

A related statement states for all $\epsilon > 0$, for sufficiently large $n$, there is a prime between $n, (1+ \epsilon)n$ which is a trivial corollary of the PNT.

For your first question, both of those statements have been proven.

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    $\begingroup$ "sufficiently large" makes this a non-generalisation of Bertrands postulate. $\endgroup$ Sep 27, 2014 at 4:42
  • $\begingroup$ What about my second question? All known primes gaps have been observed to be smaller than n or n-2, with maybe the exception of the largest known prime. $\endgroup$ Sep 27, 2014 at 5:10
  • $\begingroup$ Have you looked at the reference? And what do you mean by "with maybe the exception of the largest known prime"? $\endgroup$ Sep 27, 2014 at 5:27
  • $\begingroup$ @MayankPandey From the largest known prime to the next prime, what is the distance? Let k be the largest known prime. Let u be the next consecutive prime, which is currently unknown. Will u be only slightly less than 2k? $\endgroup$ Sep 27, 2014 at 6:28
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    $\begingroup$ @JeffreyYoung: Far from it. The next prime is at most 1.000000000000000025k (Dusart 2010), and probably a lot closer than that. Apparently the result of Baker-Harman-Pintz could be made effective and that would give a much better gap, more like $(1+10^{-9148214})k$, but we don't have that yet. And far better is conjectured, more like $(1+3.11\cdot10^{-17425155})k$. $\endgroup$
    – Charles
    Sep 29, 2014 at 14:23

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