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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Adam Hughes
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Swadhin
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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. – Adam Hughes Sep 03 '14 at 18:04
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There are much tighter bounds for this, like between $n^2$ and $(n+1)^2$. – barak manos Sep 03 '14 at 18:05
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1But a proof for these tighter bounds is much more difficult. – Peter Sep 03 '14 at 18:06
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1It's not absolutely trivial. It's called Bertrand's postulate, – Thomas Andrews Sep 03 '14 at 18:06
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To elaborate on @ThomasAndrews: http://en.wikipedia.org/wiki/Bertrand%27s_postulate... Oooops, beat me by 5 seconds... – barak manos Sep 03 '14 at 18:07
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3@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. – Daniel Fischer Sep 03 '14 at 18:10
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Here is another reference: http://www.math.washington.edu/~mathcircle/circle/2013-14/advanced/mc-13a-w10.pdf – user84413 Sep 03 '14 at 18:14
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@DanielFischer Still open. – Andrés E. Caicedo Sep 03 '14 at 18:16
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@DanielFischer: Here is a question I asked on MO (one of the very few I dared to), which refers exactly to that - http://mathoverflow.net/q/158841/27456. See the two answers to that question. In one of them, the author has replied to a comment of mine, saying that "the $n^2<P<(n+1)^2$ conjecture is almost purely relevant for historical reasons". – barak manos Sep 03 '14 at 18:17
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@AndresCaicedo: Please see the question which I've mentioned in the comment above (well, the accepted answer to that question). – barak manos Sep 03 '14 at 18:19
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@barakmanos I'm familiar with that discussion. – Andrés E. Caicedo Sep 03 '14 at 18:24
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@AndresCaicedo: So do you agree that this conjecture is relevant only for historical reasons? (although it is not in contrast with the fact that it has not been proved). – barak manos Sep 03 '14 at 18:32
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You can access P.29 of "Analytic Number Theory" here: http://books.google.co.uk/books?id=NaACAQAAQBAJ&pg=PA21&lpg=PA21&dq=dekoninck+luca+n+and+2n+prime+number+proof&source=bl&ots=QK2tdj6zyK&sig=OsP3NWuLhkkXcx4BeXu8MVNpJds&hl=en&sa=X&ei=oFgHVOXkG-7H7Aa064HACA&ved=0CCwQ6AEwAQ#v=onepage&q=dekoninck%20luca%20n%20and%202n%20prime%20number%20proof&f=false – user4568 Sep 03 '14 at 18:08
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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.
Jack D'Aurizio
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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. – Mr. Brooks Sep 03 '14 at 21:37
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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. – Jack D'Aurizio Sep 03 '14 at 21:48
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2You're telling me life is hard? That gave me quite a chuckle. Whatever. May you live and prosper as long as I have already. – Mr. Brooks Sep 03 '14 at 21:52