Bertrand postulate - MathOverflow

Bertrand postulate

2009-10-26T23:13:23Z

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?

Answer by Michael Lugo for Bertrand postulate

2009-10-26T23:17:42Z

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

Answer by Ricardo for Bertrand postulate

2009-10-27T00:48:29Z

Erdös's proof is also contained in Chapter 2 of Proofs from THE BOOK, by Aigner & Ziegler.

Answer by engelbrekt for Bertrand postulate

2009-10-27T18:13:55Z

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).

Answer by J. H. S. for Bertrand postulate

2009-11-08T04:02:52Z

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.

Answer by Kristal Cantwell for Bertrand postulate

2009-11-08T17:56:09Z

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

Answer by Anweshi for Bertrand postulate

2010-01-02T21:42:00Z

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

Answer by Ben Weiss for Bertrand postulate

2010-01-02T23:04:46Z

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).