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There are 3 standard proofs of Bertrand's Postulate:

(1) Chebyshev's original proof

(2) Ramanujan's simplification of Chebyshev's proof

(3) Erdos's proof

I recently learned about the very simple proof that if the Goldbach conjecture is true, then Bertrand's postulate follows (see here).

Does anyone know of any other proofs? There are recent proofs that extend Bertrand's postulate to show that there is always a prime in $2n$/$3n$ and $3n$/$4n$.

I am wondering if there aren't other lesser known proofs that take a different approach to establish the existence of a prime between $n$ and $2n$.

Thanks,

-Larry

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    $\begingroup$ I don't know about different, but Joe Roberts has a text you might like which has Finsler's refinement that gives estimates on pi(2n) -pi(n). It is a calligraphed number theory text. Gerhard "Ask Me About System Design" Paseman, 2012.08.01 $\endgroup$ Aug 1 '12 at 23:15
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    $\begingroup$ Find a version of the prime number theorem that gives explicit lower and upper bounds... $\endgroup$ Aug 1 '12 at 23:24
  • $\begingroup$ The proof for $2n/3n$ is in this paper by el Bacharaoui: m-hikari.com/ijcms-password/ijcms-password13-16-2006/… It is an extension of Erdos' proof. $\endgroup$
    – Igor Rivin
    Aug 2 '12 at 0:34
  • $\begingroup$ Today, I am thinking about Goldbach conjecture and Bertrand's Postulate. I think can using Goldbach conjecture to prove Bertrand's Postulate. But I was born later. $\endgroup$ Jul 5 '18 at 16:44
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Bertrand's Postulate follows as a direct consequence of the following theorem of J.J. Sylvester:

Theorem (Sylvester, 1892): Let $k$ be a positive integer. Then at least one of any $k$ consecutive integers greater than $k$ is divisible by a prime greater than $k$.

(For comparison: Chebyshev's analytic proof dates to 1850; Erdos' elementary proof dates to 1932.)

See Theorem 6 (p. 6) in http://www.math.sc.edu/~filaseta/papers/schurpaper.pdf, from which I quote:

"The theorem implies immediately that for any positive integer $k$, one of $k+1, k+2, \ldots, 2k$ is a prime (since one of these integers must be divisible by a prime $\geq k+1).$"

A copy of the Sylvester paper can be found here.


Edit: An article in the AMM (Aug/Sept, 2013) presents a revised version of Ramanujan's proof of Bertrand's Postulate; in particular, in which the use of Stirling's formula is eliminated. The citation is:

Ramanujan’s Proof of Bertrand’s Postulate. Jaban Meher, M. Ram Murty. The American Mathematical Monthly, Vol. 120, No. 7 (August–September 2013), pp. 650-653. http://www.jstor.org/stable/10.4169/amer.math.monthly.120.07.650.

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  • $\begingroup$ J. J. Sylvester. On arithmetical series. Messenger Math. 21 (1892), pp. 1-19, 87-120, 192. (See p. 4) $\endgroup$
    – Charles
    Aug 7 '12 at 18:02
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    $\begingroup$ Here's an online link to the Sylvester paper: gdz.sub.uni-goettingen.de/dms/load/img/… $\endgroup$ Sep 6 '12 at 1:17
  • $\begingroup$ Can You help me why? "The theorem implies immediately that for any positive integer $k$, one of $k+1, k+2, \ldots, 2k$ is a prime (since one of these integers must be divisible by a prime $\geq k+1).$" $\endgroup$ Jul 5 '18 at 16:53
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    $\begingroup$ @ĐàoThanhOai At least one of the numbers in that list is divisible by a prime $\geq k+1$; but, every number in that list is $\leq 2k$. So the number divisible by a new prime is itself prime: if it had additional prime factors, it would exceed $2k$, which it does not. $\endgroup$ Jul 5 '18 at 17:25
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The proofs for $(2n,3n)$ and $(3n,4n)$ are elementary and very pleasing (based on a quick look.) It is known that there is always a prime between $k$ and $\frac{6k}{5}\ $ for $k \gt 24.$ The proof is more involved but do not use anything analytic. Putting $k=2n$ etc gives a prime in $(4n,\frac{24n}{5})$ (except $n=1,2,6$) and hence in $(4n,5n)$ except for $n=1,2$ and one in $(5n,6n)$ except for $n=1$.

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A stronger version is proved by Jonathan Sondow in this arxiv preprint (which looks like a monthly paper).

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    $\begingroup$ Could you fix the link to go to the abstract rather than directly to the PDF? $\endgroup$ Aug 7 '12 at 8:03
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    $\begingroup$ OK, here you go. $\endgroup$
    – Igor Rivin
    Aug 7 '12 at 13:26
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Nice question! For a point of view from the perspective of Goldbach's conjecture, perhaps one can consider also Theorem 3.7 of "The Hardy-Littlewood Method", 2nd edition, by R.C. Vaughan.

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  • $\begingroup$ This should work by counting prime pairs that sum to $m$ for each even $m$ between $n/2$ and $n$. If there are no primes between $n/2$ and $n$, the sum on the left in the above theorem should be too large. $\endgroup$
    – user22202
    Aug 8 '12 at 5:44
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Chebyshev's proof can also be simplified a bit, I wrote down details in this blog post:

A quite short proof of Bertrand's postulate

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    $\begingroup$ This does not answer the question, which asked for different proofs, not simplifications of the known proofs. $\endgroup$
    – Boris Bukh
    Sep 29 '15 at 20:37
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I found an interesting proof today that demonstrates a stronger form of Bertrand's Postulate. I hadn't seen it before:

Abstract. In this paper we give a stronger form of Bertrand's postulate and use it to prove that every positive integer, except 1, 2, 4, 6, and 9, can be written as the sum of distinct odd primes.

http://www.ams.org/journals/proc/1972-033-02/S0002-9939-1972-0292746-6/S0002-9939-1972-0292746-6.pdf

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Re-extending Chebyshev’s theorem about Bertrand’s conjecture:

http://link.springer.com/article/10.1007%2Fs11253-008-0034-7

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  • $\begingroup$ Self-promotion, perhaps? $\endgroup$
    – Fred Kline
    Jun 26 '13 at 6:55

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