proof of Goldbach conjecture
in [Math]
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From: eestath on 3 Dec 2009 02:43 There is a story about a Turkish reformer who wanted to discourage women from wearing the veil. Instead of attempting to forbid it directly (the mathematicians approach), he issued a decree that all prostitutes must wear veils. This indirect trick proved the workable, effective way to his objective and shows the sort of thinking which chessplayers are often rather good at.
From: eestath on 3 Dec 2009 02:45 There is a story about a Turkish reformer who wanted to discourage women from wearing the veil. Instead of attempting to forbid it directly (the mathematicians approach), he issued a decree that all prostitutes must wear veils. This indirect trick proved the workable, effective way to his objective and shows the sort of thinking which chessplayers are often rather good at.
From: Ostap S. B. M. Bender Jr. on 3 Dec 2009 03:00 On Nov 30, 10:09 pm, eestath <stathopoulo...(a)gmail.com> wrote: > Goldbach conjecture states that every even integer greater then 4 is > the sum of two primes > > Proof > > Theorem > > Golbach conjecture is true for every n>4 if the two prime numbers are > different > > Proof > /= (different from) > Suppose that p/=q and that exist a k (positive integer) such that p+q/ > =2k then we have: > Is interesting to notice that p and q are odd. > How about p=2, q = 5 and k = 2009? p /= q and p+q /= 2k, so you are all set. > > 2^(p+q)=2^(2k)=> > 2*2^(p/q)/=2^(2k/q) > > Theorem 2^(p/q) is irrational number if q does not devide p ( in this > case is allways irrational because p and q are different prime > numbers) > Is 3/5 REALLY as irrational as you think (or as you are)? > > So 2*2(p/q) is irrational > > In order 2^(2k/q) to be different from 2*2^(p/q), 2^(2k/q) must be > rational (it cannot be irrational from the assumption) > > So k must be devided by q or > k=w*q (w>=1 a positive integer) > > We have a contradiction > > if k= w*q then we have: > > p+q/=2k=> > p/=2k-q=> > p/=2*w*q -q=> > p/=q*(2w-1) > > q is odd from the assumption and 2w-1 is odd (the product of odd > numbers is always odd) > So p must be different from odd which contradicts the assumption. > Q.E.D. > > If p=q Goldbach conjecture is true. > > Thus Goldbach conjecture is true for every number greater or equal to > 4 > You boggle minds.
From: Ostap S. B. M. Bender Jr. on 3 Dec 2009 03:03 On Dec 1, 12:03 am, Henry <s...(a)btinternet.com> wrote: > On 1 Dec, 06:09, eestath <stathopoulo...(a)gmail.com> wrote: > > > > > Goldbach conjecture states that every even integer greater then 4 is > > the sum of two primes > > > Proof > > > Theorem > > > Golbach conjecture is true for every n>4 if the two prime numbers are > > different > > > Proof > > /= (different from) > > Suppose that p/=q and that exist a k (positive integer) such that p+q/ > > =2k then we have: > > Is interesting to notice that p and q are odd. > > > 2^(p+q)=2^(2k)=> > > 2*2^(p/q)/=2^(2k/q) > > > Theorem 2^(p/q) is irrational number if q does not devide p ( in this > > case is allways irrational because p and q are different prime > > numbers) > > > So 2*2(p/q) is irrational > > > In order 2^(2k/q) to be different from 2*2^(p/q), 2^(2k/q) must be > > rational (it cannot be irrational from the assumption) > > Why not? there is more than one irrational number > Who's counting? > > > So k must be devided by q or > > k=w*q (w>=1 a positive integer) > > > We have a contradiction > > > if k= w*q then we have: > > > p+q/=2k=> > > p/=2k-q=> > > p/=2*w*q -q=> > > p/=q*(2w-1) > > > q is odd from the assumption and 2w-1 is odd (the product of odd > > numbers is always odd) > > So p must be different from odd which contradicts the assumption. > > Why not? There is more than one odd number. > These are minor technicalities. Just do everything modulo 2. :-) > > > Q.E.D. > > > If p=q Goldbach conjecture is true. > > > Thus Goldbach conjecture is true for every number greater or equal to > > 4 > > > Dimitris Stathopoulos > > Email: stathopoulo...(a)hotmail.com > >
From: eestath on 3 Dec 2009 03:10
i sipmply claim: for all k>2: p+q=2k if p and q are odd and p/=q! for all k>1: p+q=2k if p=q! thus i cover all the cases p=2 contradicts assumption 2^(3/5) is irrational! |