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For example, let $n=20$.

The lower range [$0$,$\frac{n}{2}$] has primes $p$ = $2,3,5,7$.

The upper range [$\frac{n}{2}$,$n$] has primes $p=$ $11,13,17,19$.

For any even number, does there exist at least one prime $p$ in [$\frac{n}{2}$,$n$]?

It seems true but is there a proof?

Thanks!

vengy
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    See https://en.wikipedia.org/wiki/Bertrand%27s_postulate – rogerl Nov 16 '21 at 16:12
  • Related: https://math.stackexchange.com/questions/2852944/strengthening-bertrands-postulate-using-the-prime-number-theorem – Ethan Bolker Nov 16 '21 at 16:17
  • Very interesting! Never heard about Bertrands Theorem. So at least there's a prime $p$ in that range $[\frac{n}{2},n]$. Thanks. – vengy Nov 16 '21 at 16:20
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    @vengy This result can be extended : For every $\epsilon>0$, there is a positive integer $n_0$, such that for every $n\ge n_0$, then range $[n,n(1+\epsilon)]$ contains at least one prime number. – Peter Nov 16 '21 at 16:44

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