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I believe there was an old conjecture that there's always a prime number between N and 2N.

What's the history and how is this proven is the easiest/elementary/deepest ways?

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Is this a joke? This is an old theorem. – Qiaochu Yuan Oct 26 2009 at 23:17
No, not a joke, I'm pretty bad in standard number theory facts and I confused it with a bunch of other primes-related hard statements. I am very red-faced now though :) – Ilya Nikokoshev Oct 26 2009 at 23:20
Well, this was of course quite a stupid question of me because most part of analytic number theory would make no sense if such a bad approximation to pi(x) was not known... – Ilya Nikokoshev Oct 26 2009 at 23:22
Bertrand's Postulate is not a consequence of the Prime Number Theorem, so it doesn't follow that if BP was unproven then even a bad approximation of pi(x) would be beyond reach. What I'm trying to say is, the theorem and the proof are well-known but it's not quite such a stupid question :-) – Alon Amit Oct 27 2009 at 4:45
@Alon, well, it is an asymptotic consequence. – Ilya Nikokoshev Oct 27 2009 at 7:50
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7 Answers

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Proven. This is called Bertrand's postulate. Here is Erdos' elementary proof; the original proof is due to Chebyshev.

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There's also a brief but informative Wikipedia entry: en.wikipedia.org/wiki/Bertrand%27s_postulate – Manny Reyes Oct 26 2009 at 23:19
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Nathan Fine summarized Erdos' accomplishment thus: Chebyshev said, And I say it again, There's always a prime Between $n$ and $2n$. – Gerry Myerson Mar 16 2010 at 21:46
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There are various proofs of Bertrand's postulate. There is quite an easy one available if one treats it together with the proof of the usual (double) Chebyshev bound as a unit. One optimizes the proof of the Chebyshev bound for subsequently proving Bertrand's Postulate by the method of Ramanujan (getting rid of the appeal to Stirling's formula at the same time).

The history of Bertrand's Postulate is set forth in The Development of Prime Number Theory by Wladyslaw Narkiewicz.

A comment to the comment by Michael Lugo. The Prime Number Theorem is considerably harder to prove than Bertrand's Postulate, and getting the PNT in the form of good explicit inequalities is hard work on top of that (such inequalities exist, and are useful for some purposes).

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I am sorry that this failed to come out nicely. I ought to clean it up, but unfortunately I don't know how to about it. engelbrekt – engelbrekt Oct 27 2009 at 18:15
Math Overflow currently has no LaTeX support, it could be best to make a reference to an outside site an post a pdf file there. – Ilya Nikokoshev Oct 27 2009 at 18:43
I cut the unreadable LaTeX. It would be easy for me to make a pdf file, but I have no webpage... – engelbrekt Oct 27 2009 at 19:10
I think there are some wikis with LaTex support... – Ilya Nikokoshev Oct 27 2009 at 19:25
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For large enough x, x+x^.525 contains a prime see:

R. C. Baker, G. Harman and J. Pintz, The difference between consecutive primes, II, Proceedings of the London Mathematical Society 83, (2001), 532–562.

For more on related results on prime gaps see the follwoing

http://en.wikipedia.org/wiki/Prime_gap

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See Chandrasekharan, Analyic Number Theory, for the proof by S. S. Pillai. It is quite easy.

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Note that we can also give a conditional proof of Bertrand's Postulate assuming the veracity of Goldbach's Conjecture. This was the subject matter of a short note that appeared in the sixth issue of volume #112 of the Monthly.

Also, it has to be noted that the reference given by Michael Lugo doesn't contain the original proof of Erdős. The original one is to be found here.

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Thanks for the pointer to the original proof. The reference I gave is where I first saw the proof. (And the original proof was published in German; my German is not so good.) – Michael Lugo Nov 8 2009 at 18:29
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Erdös's proof is also contained in Chapter 2 of Proofs from THE BOOK, by Aigner & Ziegler.

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One remark to relate Bertrand's postulate to the prime number theorem: Chebyshev's work was related to bounding ratios of factorials--in particular $\frac{2n!}{n!n!}.$

His later proof that $C\frac{x}{\log x} < \pi(x) < D\frac{x}{\log x}$ made use of other ratios (in this case $\frac{30n!n!}{15n!5n!3n!}$). In theory one could try and improve the numbers used in the ratio to asymptotically prove the prime number theorem. Jonathan Bober (among others) have worked on this. He has catalogued many different combinations of ratios of factorials (this also ends up tying into G and E functions....but I'm already out of my depth of what I'm capable of explaining).

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