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Twin Prime Conjecture Proven? | 5 comments (4 topical, 0 hidden)
[new] TWIN PRIMES CONJECTURE (none / 0) (#2)
by Anonymous Hero on Mon May 16, 2005 at 08:54:54 AM PDT

Why do we need 38 pages to prove something which we can see quite intuitively and demonstrate easily in a page? I challenge anyone to refute the following (bearing in mind that the assumption of randomness is made):- The Twin Primes Conjecture is quite obviously seen to be true if we utilise 2 proven facts and one very reasonable assumption: * The primes are infinite in number * Bertrand's postulate (there is a prime in every interval [x, 2x] * We are prepared to accept the "randomness of the occurrence of primes within intervals". By this I mean that in any randomly selected interval [x, 2x], each integer in the interval has an equal likelihood (probability) of being prime. So lets proceed with our intuitive "proof". Consider the infinite array of intervals formed by the primes and their doubles: [p(m), 2p(m)], [p(m+1), 2p(m+1)], [p(m+2), 2p(m+2)], ... where p(m+k) is the (m+k)th prime such that P(m) < p(m+k), and k = 0,1,2,3, ... By Bertrand's postulate: the probability of any integer (2 + p(m+k) ) being prime is precisely 1/p(m+k) since at least one prime must exist in any interval [ p(m+k), 2p(m+k) ], and every integer in the interval has the equal probability, 1/p(m+k), of being prime. Now let us consider the largest known twin prime to be p(n), say. Let p(m) be some singleton prime such that p(n) < p(m). The probability that no further integer, p(m+k) + 2, is prime is precisely the infinite product (1-1/p(m+1)).(1-1/p(m+2)).(1 - 1/p(m+3))... = 0 Hence the probability that at least one of the intervals [p(m),2p(m)],[p(m+1),2p(m+1)], [p(m+2),2p(m+2)], ... contains a prime integer of form 2 + p(m+k), is exactly 1. Now, call this new twin prime, p(n') and repeat the above argument. Plainly, there will always be shown to be some interval greater than the largest known twin prime (p(n), p(n'), ... etc.) which contains a twin prime of similar form to 2 + p(m+k). Our intuitive proof based on the assumption of randomness is complete.



Twin Prime Conjecture Proven? | 5 comments (4 topical, 0 hidden)
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