13
$\begingroup$

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

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ 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 $\endgroup$
    – Gerhard Paseman
    Commented Aug 1, 2012 at 23:15
  • 4
    $\begingroup$ Find a version of the prime number theorem that gives explicit lower and upper bounds... $\endgroup$
    – Qiaochu Yuan
    Commented Aug 1, 2012 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
    Commented Aug 2, 2012 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. $\endgroup$
    – Đào Thanh Oai
    Commented Jul 5, 2018 at 16:44

7 Answers 7

Reset to default
19
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ J. J. Sylvester. On arithmetical series. Messenger Math. 21 (1892), pp. 1-19, 87-120, 192. (See p. 4) $\endgroup$
    – Charles
    Commented Aug 7, 2012 at 18:02
  • 1
    $\begingroup$ Here's an online link to the Sylvester paper: gdz.sub.uni-goettingen.de/dms/load/img/… $\endgroup$
    – Larry Freeman
    Commented Sep 6, 2012 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 $\geq k+1).$" $\endgroup$
    – Đào Thanh Oai
    Commented Jul 5, 2018 at 16:53
  • 2
    $\begingroup$ @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. $\endgroup$
    – Benjamin Dickman
    Commented Jul 5, 2018 at 17:25
4
$\begingroup$

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 $n=1$.

Improve this answer
$\endgroup$
3
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Could you fix the link to go to the abstract rather than directly to the PDF? $\endgroup$
    – Harry Altman
    Commented Aug 7, 2012 at 8:03
  • 1
    $\begingroup$ OK, here you go. $\endgroup$
    – Igor Rivin
    Commented Aug 7, 2012 at 13:26
2
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Commented Aug 8, 2012 at 5:44
1
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ This does not answer the question, which asked for different proofs, not simplifications of the known proofs. $\endgroup$
    – Boris Bukh
    Commented Sep 29, 2015 at 20:37
0
$\begingroup$

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

Improve this answer
$\endgroup$
-2
$\begingroup$

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

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ Self-promotion, perhaps? $\endgroup$
    – Fred Daniel Kline
    Commented Jun 26, 2013 at 6:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .