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This is just a thought: if gaps between prime numbers can be arbitrarily large then it should be possible to find infinitely many gaps, such that the product $m=\prod_{n=1}^{N}Pn<P_{N+1}^{2}$, where Pn is the n-th prime. Then m-1, m+1 would have to be twin primes. My question is, what is wrong with this argument:).

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  • $\begingroup$ The gaps are large, but they are way out there. $\endgroup$ – Umberto P. Oct 22 '18 at 13:27
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The problem with your claim is that it is not possible to find infinitely many inequalities of the sort you claim -- and in particular it does not follow from the previous sentence. Although gaps get arbitrarily large, their size grows much slower than what you need for your argument. The gaps that occur near the $N$th prime will be vastly smaller than needed to make $\prod_{n=1}^N P_n$ less than $P_{N+1}^2$.

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  • $\begingroup$ Thanks for the answer. I got the point. I must say this is counter intuitive for me as you can approx $\pi(x)=x/(\ln(x) - 1)$ then the ratio of prime numbers to all integers should go to 0 as $\lim_{x\to\infty}\frac{\frac{x}{(\ln(x) - 1)}}{x}$, which for me means primes are infinitely rare out there. $\endgroup$ – Rafx Oct 24 '18 at 11:13
  • $\begingroup$ They are "infinitely rare", but they get rarer slowly. The product $\prod_{n=1}^N P_n$, for $N > 5$, will be at least $2 \times 3 \times 5 \times 7 \times 11 = 2310$ times larger than $P_N$, and Bertrand's postulate tells us that $P_{N+1} < 2P_N$. $\endgroup$ – Mees de Vries Oct 24 '18 at 11:20

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