Talk:Bertrand's postulate

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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)

Contents 1 generalization 2 Dubious statement 3 Conjecture 4 contradiction?

[edit] 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 ε > 0 there exists m such that for all n>m there exists a prime such that n < p < = (1 + ε)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)

[edit] 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.

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

[edit] 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)

[edit] 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)