proof of Goldbach conjecture
in [Math]
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From: Arturo Magidin on 2 Dec 2009 13:27 On Dec 2, 12:22 pm, eestath <stathopoulo...(a)gmail.com> wrote: > what i am going to do with you Arturo! > > i have a lot of fun with you:) > > you are funny :) And you are an incompetent boob. So? > > try to post somewhere you can understand what is going on > > i found the proof in 10 minutes and you are asking silly questions 2 > days! And you are failing to answer for three days. What does that tell us? > > if someone understand what i am saying please explain to him! It is *your* responsibility to explain what you are trying to say, nobody else's. If you cannot explain what your "theorem" says, then you don't have a statement, let alone a proof. > with you is a waste of time...ask a coworker or someone smarter than > you! > > you undergoing these 5 stages: > DENIAL > ANGER > BARGAINING > DIPRESSION > ACCEPTANCE! I thought you were just an incompetent boob. I see now that, despite your laughable claims over e-mail of being "smart", you are just an idiot and a troll. -- Arturo Magidin
From: Aatu Koskensilta on 2 Dec 2009 13:35 Arturo Magidin <magidin(a)member.ams.org> writes: > I thought you were just an incompetent boob. I see now that, despite > your laughable claims over e-mail of being "smart", you are just an > idiot and a troll. I thought the exhortation from eestath I received in e-mail: i hope that you understand that i am quite smart! was pretty cute, in a pathetic sort of way. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: T.H. Ray on 2 Dec 2009 04:03 eestath wrote > p and q are all the possible prime numbers for witch > p+q/=2k and k>2 > 2^(2k/q) > could be rational or irrational from the assumption > we know that it > could not be irrational > so there must be a k witch > is devisable by q > i stated that k>2 because p and q are odd for every > k>2 > it cannot be even prime (2 only) because simply 2k > for every k>2 is > the sum of two odd primes > > > Dude, "2k for every k > 2 is the sum of two odd primes" is what you are supposed to be trying to prove. Substituting the symbol k for p + q (odd primes) does not change the identity, unless you think that n + n differs from 2(n). What you've said is that because p + q distinct primes and 2k integers are both even, Goldbach's Conjecture follows. But that _is_ Goldbach's Conjecture (as well as the law of parity, which is independent of the conjecture). Whatever sense I can make of the exponential expression at all in your context, 2k/q cannot differ from 1 when p = q, as I mentioned before. Are we also trying to rewrite the exponent law? Tom
From: Tonico on 2 Dec 2009 15:07 On Dec 2, 8:22 pm, eestath <stathopoulo...(a)gmail.com> wrote: > what i am going to do with you Arturo! > > i have a lot of fun with you:) > > you are funny :) > > try to post somewhere you can understand what is going on > > i found the proof in 10 minutes and you are asking silly questions 2 > days! > > if someone understand what i am saying please explain to him! > > with you is a waste of time...ask a coworker or someone smarter than > you! > > you undergoing these 5 stages: > DENIAL > ANGER > BARGAINING > DIPRESSION > ACCEPTANCE! It was just a matter of time: after so many trolls, cranks, or simply stupid individuals, the group has finally met a deranged person: eestath Perhaps it's funny, perhaps it's sad, or perhaps it's neither, but it is new...at least for me. Tonio
From: eestath on 2 Dec 2009 22:42
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 can anyone find a k sutch that p+q/=2k simply NOT |