Talk:Bertrand's postulate

Easier intuitive formulation?

I first encountered this postulate as stating that there is no prime number that is more than double the previous prime number. At least for the layman this seems much easier to grasp (assuming it really says the same, which I think it does?) so maybe this could be added to the introduction? Leist (talk) 22:55, 18 April 2021 (UTC)

Untitled

If Chebyshev did indeed use an inequality to prove Bertrand's postulate, it seems highly unlikely he used what is generally known as Chebyshev's inequality. --Henrygb 13:21, 4 May 2004 (UTC)

Generalization

Hi, on reference desk there was a question about 2n<p<3n, and it was answered as true, and that brought me here. My notes from lecture claim that Vinogradov have proven that for every $\epsilon >0$ there exists m such that for all n>m there exists a prime such that $n<p<=(1+\epsilon )n$. But unfortunately, I don't know the source of this interesting generalization. Samohyl Jan 00:36, 30 August 2005 (UTC)

That's a trivial consequence of the prime number theorem. I don't know what Vinogradov has to do with it, either he proved something stronger or the reference is simply bogus. -- EJ 18:24, 10 March 2006 (UTC)

But note that this is not quite a generalization senso stricto of Bertrand's postulate. Bertrand's postulate states "for every n, there is...", not "for all but finitely many n". Since the prime number theorem is about a limit, it does not allow us to decide whether there is some prime between 2n and 3n if n ≥ 100000. —The preceding unsigned comment was added by 136.199.8.42 (talk) 15:00, 2 May 2007 (UTC).

Dubious statement

I've removed this dubious statement as it would prove the twin prime conjecture.

Erdős also proved there always exists at least two prime numbers p with n < p < 2n for all n > 6. Moreover, he showed that one of them must be of the form 4k+1 and the other must be of the form 4k+3.

not if you have different k in each case —Preceding unsigned comment added by 216.198.139.84 (talk) 16:08, 4 August 2010 (UTC)

This is either a hoax, or a corrupt version of a true result by Erdős. Does anybody have a source for such a result? Paul August ☎ 18:22, 12 May 2006 (UTC)

This is only a problem of terminology. In the usual parlance, a "number of the form 4k+1" means exactly the same as an "integer congruent to 1 modulo 4", and in particular, the two k's in the above statement are not assumed to be the same. -- EJ 18:57, 12 May 2006 (UTC)

Thanks EJ, for clearing this up. Do you know of a source for this result? Paul August ☎ 03:58, 13 May 2006 (UTC)

Hmm, not for sure, now that I think about it. MathWorld says so, but the page looks a bit fishy. I can't check the original reference, as our library does not hold J. London Math. Soc. -- EJ 02:21, 14 May 2006 (UTC)

then how do you conclude that there existy prime between 2n to 3n

It's pretty easy to show that there exists a prime between 2n and 3n (and another prime between 3n and 4n), for all integers, n ≥ 11. This is because for all n ≥ 11, (and this is a fact)... that there exists at least 3 primes between n and 2n. Also, if there exists k primes in the interval between n and 2n, then there exists at least k + 2 primes in the interval between 2n and 4n. However, we now know that for all n ≥ 11, there exists at least 5 primes between 2n and 4n. However, it is also true that at least 2 of these primes must be in the interval between 2n and 3n, and at least 2 of these primes must be in the interval between 3n and 4n. However, since there is a fifth prime in the interval between 2n and 4n, one of the intervals, the one between 2n and 3n or the one between 3n and 4n, must contain at least 3 primes. In essence, I believed that I have also proved the Goldbach Conjecture. PhiEaglesfan712 19:34, 23 July 2007 (UTC)

Conjecture

Extending the results by Erdos, the next statement is conjectured.

For any positive integer k, there is a natural number N such that for all n > N, there are at least k primes congruent to 1 modulo 4 and at least k primes congruent to -1 modulo 4 between n and 2n. 218.133.184.93 17:47, 3 February 2007 (UTC)

contradiction?

n >= 2, n < p < 2n or n > 3, n < p < 2n-2

Which was his original claim? The two don't even seem to be the same...

n > 3, n < p < 2n-2 #take n-2 = y gives similar to the first but not identical y >= 2, y-2 < p < 2y

—porges_(talk) 07:02, 12 February 2007 (UTC)

The first, "n >= 2, n < p < 2n", is trivially true for n=2 (p=3) and n=3 (p=5). The only real difference between the two is whether it's assured for n > 3 that p < 2n, or the slightly stronger p < 2n-2 (which eliminates the possibility that p = 2n-1 is needed, as it is for n=2 and n=3). Both versions are proved, and reliable sources mention both versions in the same paragraph. I added MathWorld and Prime Pages references showing this. I don't know which formulation is oldest or most common, but if MathWorld and Prime Pages mention both, then it makes sense that we also do. PrimeHunter 14:32, 12 February 2007 (UTC)

I have removed the contradiction tags. PrimeHunter 10:55, 16 May 2007 (UTC)

final result

I realize that x could be an arbitrary large enough real number but is there any reason not to restrict to integer values (and use n in place of x) so that it looks like Schoenfelds result? Also the final result is an improvement in real sense however for n=2010760 it replaces the 16597 of Schoenfeld with 29.02 which is much worse. The first place it beats Schoenfeld is for n=exp(sqrt(33194)/2) which is roughly 3.6*10^39. For n>10^50 it says that there is a prime between n and (1+1/26509)*n. For now I'll just fix the one x.--Gentlemath (talk) 02:37, 4 March 2010 (UTC)

It follows from the prime number theorem that for any $k>1$, there exists an $n_{0}$ such that there is always a prime between $n$ and $kn$ for all $n>n_{0}$

PNT?

If we know for sure that there exists a prime between n and 2n for all n, why do we need the prime number theorem to show that there exists a prime between n and kn for all n>n0? I'm not sure I understand the following math so I won't remove that line altogether, but I'm replacing "from the PNT" to "trivially". .froth. (talk) 00:36, 28 May 2010 (UTC)

Oh, I guess k can be real, not just an integer. So I added real. .froth. (talk) 00:39, 28 May 2010 (UTC)

Proof of consequence?

The article states that Bertrand's postulate implies "[t]he number 1 is the only integer which is a harmonic number" but the connection is not immediately obvious to me. Can a brief explanation be added? Dcoetzee 20:47, 26 January 2012 (UTC)

Better results

The following is in Better results:

"It follows from the prime number theorem" ... "which implies that π(kn) − π(n) goes to infinity (and in particular is greater than 1 for sufficiently large n)."

What I am wondering about is the source reference. What is it? — Preceding unsigned comment added by Reddwarf2956 (talk • contribs) 01:35, 26 October 2012 (UTC)

Added one.—Emil J. 15:12, 16 July 2013 (UTC)

Better than noting Goldbach's conjecture implies a stronger form... Roderick MacPhee (talk) 01:01, 14 August 2024 (UTC)

Short proof

(n²-a²) is an odd prime

=> (n+a) is prime and (n-a) is 1

=> a<n while (n-a) = 1

=> n < (n+a) < 2n . — Preceding unsigned comment added by 84.135.22.28 (talk)

You apparently assume there is a prime of form (n²-a²) for any n. But there is only such a prime if 2n-1 is prime with a = n-1. Moreover, one version of Bertrand's postulate states n < p < 2n-2, so the prime of form (n²-a²) = 2n-1 is too large even when it exists. PrimeHunter (talk) 09:41, 6 June 2014 (UTC)

Sylvester's theorem

Can you add more on Sylvester's theorem? It's barely mentioned here, even though it seems to be even more powerful than Bertrand's postulate. There's a whole page on proofs of Bertrand's postulate, but nothing on this more powerful theorem. --AndreRD (talk) 13:33, 8 October 2015 (UTC)

arxiv is not a reliable source

arxiv is not a reliable publication; it is not reviewed by MathSciNet nor By Zbl; it contains many incorrect papers; moreover Kyle Balliet is credited with zero publication on MathSciNet and zero publication on Zbl: this is suspect. We have to wait this paper is published in a serious, peer-reviewed publication, before we mention it here. I also note that Andy Loo, being credited with this only one minor publication (which is only indexed, and not reviewed, on MathSciNet), will certainly not deserve a page on Wikipedia in the near future: so I removed the red link. Sapphorain (talk) 11:43, 26 November 2015 (UTC)

Bertrand's postulate and the binary conjecture

The binary conjecture is sufficient for Bertrand's postulate, whereas Bertrand's postulate is necessary for the binary conjecture. Also, the binary conjecture and the ternary conjecture are equivalent. If one is true, so is the other. 2605:E000:6116:7D00:4CD6:5569:EA6F:731C (talk) 14:40, 4 October 2017 (UTC)

It's unclear what you refer to but I guess it's about the binary and ternary variants of Goldbach's conjecture. They have not been proved to be equivalent. It's trivial but not worth mentioning that Bertrand's postulate would follow from Goldbach's binary conjeture. The former was proved in 1852 and the latter is still open so it's useless that it would prove something which was already proved 165 years ago. PrimeHunter (talk) 15:46, 4 October 2017 (UTC)

Any odd number M≥7 can be written as M=3+E, where E≥4 is an even number. If E=Pa+Pb is true, then M=3+Pa+Pb is also true. Any even number E≥4 can be written as E=M-3, M≥7 is an odd number. If M=3+Pa+Pb is true, then E=Pa+Pb is also true. 184.146.30.85 (talk) 13:04, 5 October 2017 (UTC)

? And ? True, but where is this supposed to lead? Sapphorain (talk) 16:33, 5 October 2017 (UTC)

Then the binary conjecture cannot descend indefinitely. Please let's stop the talk. Thanks. 70.31.160.220 (talk) 18:41, 6 October 2017 (UTC)

Sure, let's stop the talk. The pleasure is all mine. Sapphorain (talk) 22:38, 6 October 2017 (UTC)

Any three odd primes are allowed by the ternary conjecture. If we assume the conjecture is true then we don't know whether M=3+Pa+Pb. PrimeHunter (talk) 21:35, 5 October 2017 (UTC)

M=3+Pa+Pb is true if E=Pa+Pb is true. Conversely, E=Pa+Pb is true if M=3+Pa+Pb is true. Please let's stop the talk. Thanks. 70.31.160.220 (talk) 18:41, 6 October 2017 (UTC)

Earliest statement?

I just thought of this 'conjecture' myself, assumed it was well-known, and looked online to see if there was a proof. So I'm not surprised to see that it is well-known, but I'm very surprised to see that the first statement of the conjecture is said to be as late as 1845. Surely Fermat or Euler or Gauss or someone must have noticed it before then? 86.132.140.196 (talk) 19:13, 28 September 2018 (UTC)

There is no point arguing one way or the other, as this conjecture was not published before by anybody else. Sapphorain (talk) 21:25, 28 September 2018 (UTC)