# Between $n$ and $2n$ there is always a prime number. [duplicate]

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.

## marked as duplicate by Eric Naslund, Daniel Fischer♦, user133281, user98602, SemiclassicalSep 10 '14 at 21:51

• 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 3 '14 at 18:04
• There are much tighter bounds for this, like between $n^2$ and $(n+1)^2$. – barak manos Sep 3 '14 at 18:05
• But a proof for these tighter bounds is much more difficult. – Peter Sep 3 '14 at 18:06
• It's not absolutely trivial. It's called Bertrand's postulate, – Thomas Andrews Sep 3 '14 at 18:06
• @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 3 '14 at 18:10

• 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 3 '14 at 21:48