Is there a way of showing there are arbitrarily big gaps between primes by contradiction?

Asked Nov 20 '13 at 12:33

Active Nov 20 '13 at 12:33

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This may be a stupid question. So apologies in advance, if it is,

One proof of this is a straightforward construction.

For any $N$, consider $(N+1)!+2$, $(N+1)!+3$,..., $(N+1)!+N$. All of these numbers are composite.

My question is, is there a proof by contradiction which does not use this argument?

asked Nov 20 '13 at 12:33

Lost1

If the prime gaps were bounded, we would have $\pi(x) \geqslant c\cdot x$ for some $c > 0$. That contradicts the prime number theorem $\pi(x) \sim \frac{x}{\log x}$. – Daniel Fischer Nov 20 '13 at 12:37
@DanielFischer may I ask what $\pi(x)$ is called. this $x/\log(x)$ thing looks vaguely familiar. – Lost1 Nov 20 '13 at 12:42
$\pi(x)$ is the prime counting function. That $\pi(x) \sim \frac{x}{\log x}$ is the celebrated prime number theorem. – Daniel Fischer Nov 20 '13 at 12:48

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