# Bertrand's Postulate

Statement For every $n > 1$ there is always at least one prime $p$ such that $n < p < 2n$.

I am curious to know that if I replace that $2n$ by $2n-\epsilon$, ($\epsilon>0$) then what is the $\inf (\epsilon)$ so that the inequality still holds, meaning there is always a prime between $n$ and $2n-\epsilon$

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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.

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if n > 3 is an integer, then there always exists at least one prime number p with n < p < 2n − 2.

Thus ε < 2 for n > 3. What if n ≤ 3?

• For n = 3, 3 < 5 < 6 - ε ⇒ ε < 1
• For n = 2, 2 < 3 < 4 - ε ⇒ ε < 1

Hence we have 0 < ε < 1, if ε is a constant.

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now the question may be more interesting. I actually wanted to find the least postive $\epsilon$ such that the condition remains true. –  anonymous Aug 11 '10 at 16:21
@Chandru1: Any ε between 0 and 1 will do, so the infimum of all possible positive ε is 0. This is not surprising. Did you want the supremum instead? (The supremum is 1 of you want it to hold for n=2 or 3, and infinity if you only want sufficiently large n.) –  ShreevatsaR Aug 11 '10 at 16:44
@ShreevatsaR : Hi i got it. –  anonymous Aug 11 '10 at 16:50
if $n>60$, then $\varepsilon=\frac{2n}{3}$.