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This question already has an answer here:

Between $n$ and $2n$ there is always a prime number.

I was thinking of this and looked it up on the google to find that this is true. Now, I am wondering what is the proof for it? Does any elementary proof exist for it?

Thank you.

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marked as duplicate by Eric Naslund, Daniel Fischer, user133281, user98602, Semiclassical Sep 10 '14 at 21:51

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  • $\begingroup$ Yes, it's quite direct. Most basic analytic number theory books will have a proof. DeKonick and Luca have it on p.29, for example. $\endgroup$ – Adam Hughes Sep 3 '14 at 18:04
  • $\begingroup$ There are much tighter bounds for this, like between $n^2$ and $(n+1)^2$. $\endgroup$ – barak manos Sep 3 '14 at 18:05
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    $\begingroup$ But a proof for these tighter bounds is much more difficult. $\endgroup$ – Peter Sep 3 '14 at 18:06
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    $\begingroup$ It's not absolutely trivial. It's called Bertrand's postulate, $\endgroup$ – Thomas Andrews Sep 3 '14 at 18:06
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    $\begingroup$ @barakmanos Between $n^2$ and $(n+1)^2$ [that's Legendre's conjecture, iirc?] is as far as I know still open. If it's proved, that would be very recent. $\endgroup$ – Daniel Fischer Sep 3 '14 at 18:10
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Yes, many elementary proofs are known. Apart from the the original proof of Chebyshev's theorem, have a look at the Erdos' elementary proof - you can find it also in Proofs from THE BOOK.

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  • $\begingroup$ The latter is a 6-page PDF. It might seem short compared to some other proofs of various theorems I've seen, but it would still take me a couple of hours of careful reading. $\endgroup$ – Mr. Brooks Sep 3 '14 at 21:37
  • $\begingroup$ Life is hard, little my padawan! :D Out of joke, the main idea is pretty straighforward to grasp: any prime between $n$ and $2n$ divide the numerator of $\binom{2n}{n}=\frac{2n(2n-1)\cdot\ldots\cdot(n+1)}{n!}$ but not the denominator, hence we can say something about the distribution of prime numbers by studying how fast the central binomials grow. $\endgroup$ – Jack D'Aurizio Sep 3 '14 at 21:48
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    $\begingroup$ You're telling me life is hard? That gave me quite a chuckle. Whatever. May you live and prosper as long as I have already. $\endgroup$ – Mr. Brooks Sep 3 '14 at 21:52

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