Statement For every n &gt; 1 there is always at least one prime p such that n &lt; p &lt; 2 n.I am curious to know that if I replace that 2 n by 2 n − ϵ, ( ϵ &gt; 0) then what is the inf ( ϵ ) so that the inequality still holds, meaning there is always a prime between n and 2 n − ϵ

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Statement For every   n  &amp;gt;  1 there is always at least one prime   p such that   n  &amp;lt;  p  &amp;lt;  2  n.I am curious to know that if I replace that   2  n by   2  n  −  ϵ, (  ϵ  &amp;gt;  0) then what is the   inf  (  ϵ  ) so that the inequality still holds, meaning there is always a prime between   n and   2  n  −  ϵ

Gornil2 · Accepted Answer

Three related points are worthy of mention, showing that epsilon can be close to n.There is a result of Finsler that approximates how many primes lie between n and 2n, which is of order o(n/log(n)) as is to be expected by the Prime Number Theorem.Literature on prime gaps will tell you the exponent delta such that there is (for sufficiently large n) at least one prime in the interval (n , n + n^delta). I think delta is less than 11/20.Observed data suggests that n^delta can be replaced by something much smaller: for n between something like 3 and 10^14 , some function like 2(log(n))^2 works

Ayaan Barr · Accepted Answer

Bertrand&#039;s postulate isif   n  &amp;gt;  3 is an integer, then there always exists at least one prime number p with   n  &amp;lt;  p  &amp;lt;  2  n  −  2.Thus   ξ  &amp;lt;  2 for   n  &amp;gt;  3. What if   n  ≤  3?For   n  =  3  ,  3  &amp;lt;  5  &amp;lt;  6  −  ξ  ⇒  ξ  &amp;lt;  1For   n  =  2  ,  2  &amp;lt;  3  &amp;lt;  4  −  ξ  ⇒  ξ  &amp;lt;  1Hence we have   0  &amp;lt;  ξ  &amp;lt;  1, if ε is a constant.

Statement For every n > 1 there is always at least one prime p such that n

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