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Two questions came to mind when I was reading the proof for Bertrand's Postulate (there's always a prime between $n$ and $2n$):

(1) Can we change the proof somehow to show that: $\forall x > x_{0}$, there exists a prime $p$ $\in [x, ax]$, for some $a \in (1, 2)$?

(2) Suppose the (1) is true, what is the smallest value of $x_{0}$?

I'm not sure how to prove either of them, any input would be greatly appreciated! And correct me if any of the above statement is wrong. Thank you!

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    @LindsayDuran: You still need to specify where in the sequence of quantifiers that $\exists a\in(1,2)$ appears.2012-09-21

2 Answers 2

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There are better results than Bertrand's Postulate. Pierre Dusart has proven better results, some of which can be found on wikipedia.

Less specifically, for any $k > 1$, $k \in \mathbb{R}$, one can show that $\lim_{n \to \infty} \dfrac{\pi(kn) - \pi(n)}{n/\ln n} = k - 1$ using the prime number theorem, which means that for any $k>1$, there is some $x_0$ such that for $x > x_0$, there is a prime between $x$ and $kx$.

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    @MTurgeon: No, $k$ can be any real number greater than $1$.2012-09-21
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I think you would enjoy the page PRIME GAPS.

My own version of the conjecture of Shanks, actually both a little stronger and a little weaker, is $ p_{n+1} < p_n + 3 \,\left( \log p_n \right)^2, $ for all primes $p_n \geq 2.$ This is true as high as has been checked.

Shanks conjectured that $ \limsup \frac{p_{n+1} - p_n}{\left( \log p_n \right)^2} = 1, $ while Granville later corrected the number $1$ on the right hand side to $2 e^{- \gamma} \approx 1.1229,$ see CRAMER GRANVILLE. There is no hope of proving this, but I enjoy knowing what seems likely as well as what little can be proved.

Here is a table from the third edition (2004) of Unsolved Problems in Number Theory by Richard K. Guy, in which $p = p_n$ is a prime but $n$ is not calculated, then $g = p_{n+1} - p_n,$ and $p = p(g),$ so $p_{n+1} = p + g.$

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