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In its page on Bertrand's postulate Wikipedia says

It follows from the prime number theorem that for any real $\varepsilon >0$ there is an $n_0 > 0$ such that for all $n > n_0$ there is a prime $p $ such that $n < p < ( 1 + ε )n$ .

I need to quote this result for $\varepsilon = 1/2$. I'd like a reference more formal than Wikipedia.

hardmath
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Ethan Bolker
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    Why don't you cite the reference that very source that Wikipedia used: G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, p. 494. – fleablood Jul 16 '18 at 00:00
  • @fleablood Of course. I missed that reference. Post as an answer, or I can delete the question as too narrow. – Ethan Bolker Jul 16 '18 at 00:04

2 Answers2

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The ariticle on wikipededia cites that with a foot note (as of the time I write this) #8 refering to: G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, p. 494.

from Wikipedia

It follows from the prime number theorem that for any real $ {\displaystyle \varepsilon >0}$ there is a ${\displaystyle n_{0}>0} $ such that for all ${\displaystyle n>n_{0}} $ there is a prime $ {\displaystyle p}$ such ${\displaystyle n<p<(1+\varepsilon )n}$. It can be shown, for instance, that

${\displaystyle \lim _{n\to \infty }{\frac {\pi ((1+\varepsilon )n)-\pi > (n)}{n/\log n}}=\varepsilon ,} $[8]

[8]G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, p. 494.

fleablood
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  • This is pretty unreadable. I doubt that Hardy and Wright says “for any real $\epsilon > 0 > \epsilon > 0\epsilon > 0$, for example (and it doesn’t get better). Can you edit the quote so it doesn’t look like a copy/paste from some non-LaTeX source? Even an image of the relevant Hardy and Wright page would be better than this. – Steve Kass Jul 16 '18 at 00:34
  • It is a copy and paste from a non-LaTeX source! I thought the was clear from context. – fleablood Jul 16 '18 at 02:19
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tersely worded, but Dusart gives, for $x \geq 396738,$ a prime between $x$ and $$ x \left( 1 + \frac{1}{25 \log^2 x} \right) $$

I think it was on the arxiv, let me find it.

Yes, this is Proposition 6.8.

For your fixed target $3x/2,$ you may fill in the smaller numbers with the table of maximal prime gaps at https://en.wikipedia.org/wiki/Prime_gap#Numerical_results . I went through the thing once by machine, for $p \geq 11$ and $p < 4 \cdot 10^{18},$ we get gap $g < \log^2 p.$

Will Jagy
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