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We are interested in the following statement:

For each $n>1$ and $x>2$ there is at least one prime $p$ satisfying $x<p<n x$.

For $n=2$ we get precisely the Bertrand's postulate which is true. As corollaries, the statements for arbitrary $n\geqslant 2$ are true. However, I am interested maybe there exist independent proofs (prossibly short and elementary) for some of the cases $n>2$.

Thanks for your help.

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A recent preprint by Verma on ArXiv has a retooled approach of Erdos which may be of interest. It was posted less than 3 days ago. – The Masked Avenger Jan 25 '14 at 16:38
up vote 6 down vote accepted

Chebyshev used the divisibility of the middle binomial coefficient ${{2n}\choose{n}}$ to set upper and lower bounds on the number of primes in the form

$\frac{an}{\log(n)} > \pi(n) > \frac{bn}{\log(n)}$

for some constants $a$ and $b$. If $a$ is not too small and $b$ is not too large then the proofs can be elementary and quite short. See e.g. for proofs with $a = 6 \log(2)$ and $b = \log(2)/2$

There must then be a prime between $x$ and $nx$ if

$ \frac{ax}{\log(x)} < \frac{bnx}{\log(nx)} $

which is equivalent to

$a \log(nx) < b n \log(x)$

so for the $a/b = 12$ as above you get

$12 \log(n) < (n-12) \log(x) $

$ x > \exp\left({\frac{12 \log(n)}{n-12}}\right)$.

So, for any value of $n > 12$ this gives a lower bound for $x$ above which there is always a prime between $x$ and $xn$; for smaller $x$ it can be checked by hand.

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I think Erdős's proof of Bertrand's postulate (in the pdf file that you linked) is shorter and simpler. – GH from MO Jan 25 '14 at 15:00
Erdős's proof is another example of the same idea isn't it? But it is tighter and gets down to n=2 – Philip Gibbs Jan 25 '14 at 15:28
Yes. This is why I think there is no really good answer to the original question. Erdős developed Chebyshev's idea further, and it is hard to do better than him! – GH from MO Jan 25 '14 at 16:02

Pal Erdös' version can be found in "proofs from the book". This was rather elementary.

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This statement is false. For example, if $n=1.5$ and $x=3$, then there is no prime $p$ satisfying $x<p<nx$.

On the other hand, it is easy to derive from the prime number theorem that, for a given $n>1$, there is a prime $p$ satisfying $x<p<nx$ for any sufficiently large $x$.

In fact much more is true. By a result of Baker-Harman-Pintz (Proc. London Math. Soc. 83 (2001), 532-562), there is a prime $x<p<x+x^{0.526}$ for any sufficiently large $x$.

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Thanks. I am familiar with these results. But I would like to find a direct proof for cases where $n>2$, for example, $n=3,4,5,$ etc. Are there any references regarding such problems? – Lukas Jan 25 '14 at 13:09
There is a simple direct proof for $n=2$. Why would you want to prove a weaker result? – GH from MO Jan 25 '14 at 13:14

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