proof of hardy-littlewood 2nd conjecture
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From: jbriggs444 on 5 May 2010 07:55 On May 4, 4:31 pm, "christian.bau" <christian....(a)cbau.wanadoo.co.uk> wrote: > On May 4, 5:24 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > > > > > > On May 4, 10:51 am, master1729 <tommy1...(a)gmail.com> wrote: > > > > gnasher729 wrote : > > > > > On May 1, 10:19 pm, master1729 <tommy1...(a)gmail.com> > > > > wrote: > > > > > proof of hardy-littlewood 2nd conjecture. > > > > > > consider the interval (a,b). > > > > > > we need to prove that the number of primes in that > > > > interval cannot be larger than pi(b - a). > > > > > > thus pi(a,b) <= pi(b - a). > > > > > Please post your proof again after adding a > > > > definition of pi (a, b). > > > > pi(a,b) is the number of primes in the interval (a,b). > > > So, for instance, pi(17,17) = 1 and pi(0) = 0? > > And you need to prove that the former is smaller than the latter? > > (17, 17) is an empty set. With the definition above, pi (17 17) = 0.- Hide quoted text - > > - Show quoted text - True -- if you adopt the common notational convention that parentheses denote open intervals in the reals. I was trying to let the notation slide. If you insist on the "open interval" interpretation, then: pi(16.9,17.1) = 1 and pi(0.2) = 0
From: master1729 on 5 May 2010 14:42 proof of hardy-littlewood 2nd conjecture. all variables are positive integers. all intervals are closed intervals. first we will restate the 2nd h-l conjecture. consider the closed interval (a,b). we need to prove that the number of primes in that interval cannot be larger than pi(b - a). the primes in the interval(a,b) are denoted by pi(a,b) thus pi(a,b) <= pi(b - a). needs to be proven. let the interval lenght be c = b - a by definition. we only need to prove it for large c , since for small c , it has already been checked using modular arithmetic. to find the primes in interval (a,b) we need to 'sieve all primes' up to sqrt(c). by sieving all primes , i mean sieving all multiples of the prime and 1* the prime in the erastotenes way. e.g. for 7 , we sieve the 7th element(1*7) , the 14th element (2*7) , ... and note we dont really sieve numbers , but the nth element of interval. ordinals thus. to prove the conjecture , we need to consider the worst case scenario. for clarity i will give examples , but it can be restated formally. 7 < sqrt(c). for every interval (a,b) , 7 sieves it C or C - 1 times. c =/= C for clarity. thus the most times 7 can sieve , compared to the least time differs by one time , since C - ( C - 1 ) = 1. thus an interval of length c is "sieved by a prime" p =< sqrt(c) at worst one time less than the interval (0,c). thus we add +1 prime ( since not sieved ) for every p < sqrt(c). since that is the " worst case scenario for hardy-littlewood conj. " this gives us a formula for the maximum number of unsieved numbers in the interval c , since we simply do sum (n=p) f(n)=1 going from 2 till sqrt(c). this is simply pi(sqrt(c)). thus an interval c appears to have at most pi(c) + pi(sqrt(c)) primes. however note that pi(c) requires a correction too. since 2,3,5,7,... are no longer counted as primes , we need to remove these. (they are not sieved) let me explain. we sieved 1*2 1*3 1*5 1*7 th element in the interval. but the primes in the interval (0,c) do not sieve those , since 2, 3, 5, 7,.. are primes. therefore we need a correction. and we need to remove up to sqrt(c) , thus we arrive at - pi(sqrt(c)) as the correction. then we end up with : sup = maximum value max is the same as sup. c = b - a max pi(a,b) = pi(c) + pi(sqrt(c)) - pi(sqrt(c)) which follows from the above. which reduces to : max pi(a,b) = pi(c) thus pi(a,b) <= pi(b - a). which is equivalent to hardy-littlewood 2nd conjecture. For those who think we need to consider when sieving 2 or more primes sieves the same or different ordinals in the (a,b) makes any difference , this is not the case. ( because then you are sieving the composites , not the primes. ) Q.E.D. the master tommy1729
From: master1729 on 6 May 2010 06:14 no response. surely you must understand parts of it ??
From: ThinkTank on 6 May 2010 07:58 > no response. > > surely you must understand parts of it ?? Could you please state it a bit more formally? I'm not one to be picky about formality but this, in my opinion, is informal to the point where it becomes ambiguous. Could you restate this a bit more formally? If not, could you at least make it a bit less ambiguous? Don't get me wrong, I want to be able to understand what you are trying to say here, and I am not at all trying to say your proof is wrong. I simply need a little more clarification.
From: A on 6 May 2010 12:39
On May 6, 10:14 am, master1729 <tommy1...(a)gmail.com> wrote: > no response. > > surely you must understand parts of it ?? I just looked over your "proof." It looks as though you have overlooked that any number which is the product of more than one prime will be sieved more than once. This means your formula for the maximum number of primes in the interval (a,b), which you obtain by taking b-a and subtracting 1 for each number sieved from the interval (a,b) for each prime less than sqrt(b-a), this formula is wrong--there can be more than that many primes in the interval (a,b). |