Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $2n$.



share|cite|improve this question
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 – Gerhard Paseman Aug 1 '12 at 23:15
Find a version of the prime number theorem that gives explicit lower and upper bounds... – Qiaochu Yuan Aug 1 '12 at 23:24
The proof for $2n/3n$ is in this paper by el Bacharaoui:… It is an extension of Erdos' proof. – Igor Rivin Aug 2 '12 at 0:34

7 Answers 7

up vote 14 down vote accepted

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, 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.

share|cite|improve this answer
J. J. Sylvester. On arithmetical series. Messenger Math. 21 (1892), pp. 1-19, 87-120, 192. (See p. 4) – Charles Aug 7 '12 at 18:02
Here's an online link to the Sylvester paper:… – Larry Freeman Sep 6 '12 at 1:17

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

share|cite|improve this answer
Could you fix the link to go to the abstract rather than directly to the PDF? – Harry Altman Aug 7 '12 at 8:03
OK, here you go. – Igor Rivin Aug 7 '12 at 13:26
Okay, thank you! – Harry Altman Aug 8 '12 at 13:33

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.

share|cite|improve this answer
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. – user22202 Aug 8 '12 at 5:44

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 $n=1$.

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer
Self-promotion, perhaps? – Fred Kline Jun 26 '13 at 6:55

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

share|cite|improve this answer
This does not answer the question, which asked for different proofs, not simplifications of the known proofs. – Boris Bukh Sep 29 at 20:37
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – András Bátkai Sep 30 at 6:27

protected by S. Carnahan Sep 30 at 2:16

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.