Let’s define the quantity , and assume Cramer’s conjecture. Then . Let’s denote by the event is -central, and by the event is a couple of twin primes. Then from de Bayes’ theorem, one gets , since if is the half sum of a couple of twin primes, then necessarily is -central. But is equal to the ratio of the number of quantities such that is the half sum of twin primes to the number as defined above, hence . As , one finally gets . Hence the assumption of Cramer’s conjecture implies that there are infinitely many twin primes.
3 réflexions sur “Cramer’s conjecture implies twin prime conjecture”
Could you clarify what you mean by « the number of quantities 2r_{0}(n) such that n is the half sum of twin primes »? Also, how do you deduce P(H(n))\ge\frac{1}{K_{m}\log^{2} n}?
A major issue in here is treating H(n) and U(n) like random events. If you _really_ mean that H(n) is the event that (n-1,n+1) is a twin prime pair, then P(H(n)) is always 0 or 1 – this is a deterministic event and there is no probability involved here. If you want to make it valid, then you have to consider H(n) to be the event « a random number below n is the middle of two twin primes ». Even if this were the case, isn’t 2r_0(n) always 2 for n lying between twin primes, so that r from your previous comment must be equal to 1? That being said, I have no clue why you claim that P(H(n) cap U(n)) = r/G(n).
Indeed, I am convinced that whatever you are trying to prove is wrong. You aren’t using any intrinistic properties of primes here, so the argument should work more generally. But if you consider « paraprimes » to be precisely the numbers 1 mod 4, then K_m = 0 < infinity, so the same argument should show there are infinitely many "paraprimes". But there are none at all!
ideasfornumbertheory 
Could you clarify what you mean by « the number of quantities 2r_{0}(n) such that n is the half sum of twin primes »? Also, how do you deduce P(H(n))\ge\frac{1}{K_{m}\log^{2} n}?
It’s the number of r such that r=2r0+n), n being the half sum of two primes differing by two.
A major issue in here is treating H(n) and U(n) like random events. If you _really_ mean that H(n) is the event that (n-1,n+1) is a twin prime pair, then P(H(n)) is always 0 or 1 – this is a deterministic event and there is no probability involved here. If you want to make it valid, then you have to consider H(n) to be the event « a random number below n is the middle of two twin primes ». Even if this were the case, isn’t 2r_0(n) always 2 for n lying between twin primes, so that r from your previous comment must be equal to 1? That being said, I have no clue why you claim that P(H(n) cap U(n)) = r/G(n).
Indeed, I am convinced that whatever you are trying to prove is wrong. You aren’t using any intrinistic properties of primes here, so the argument should work more generally. But if you consider « paraprimes » to be precisely the numbers 1 mod 4, then K_m = 0 < infinity, so the same argument should show there are infinitely many "paraprimes". But there are none at all!