proof of Goldbach conjecture
in [Math]
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From: Arturo Magidin on 2 Dec 2009 10:57 On Dec 1, 11:34 pm, eestath <stathopoulo...(a)gmail.com> wrote: > I simply tell that: > > Suppose that exist a k such that p+q/=2k > > An i prove that it does not exist such a k. > > I said k>2 so 2k>4 so the two prime numbers are odd.... STOP SENDING ME E-MAIL. I've received no fewer than 7 unwanted missives from you, despite my telling you *not* to send me e-mail. Anything you have to say, say it on the newsgroup. Otherwise, I will report your actions as harassment.
From: T.H. Ray on 2 Dec 2009 01:32 Eestath 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 > Actually, the conjecture states that every even integer greater than 4 can be expressed as the sum of two odd primes. They do not have to be distinct. > 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. > Interesting? Every prime except 2 is odd, and every integer multipled by 2 is even. > 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) > > So k must be devided by q or > k=w*q (w>=1 a positive integer) > Sure, by the fundamental theorem of aithmetic. > 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) > What contradiction? I just can't parse this. > 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. > Hmmm. Then p was either 2, or wasn't a prime at all, as it was supposed to be according to your original statement about "two different primes." Do you understand Arturo's question about identifying p & q? > If p=q Goldbach conjecture is true. > > Thus Goldbach conjecture is true for every number > greater or equal to > 4 > Let's test with real numbers. Take p = 3, q = 5. You want 2^3/5 to be irrational--yes, that works, and will satisfy irrationality for every p and q. Then you say that some integer k such that 2^2k/q will always be rational--When you introduce exponential values of 2, you are simply verifying that the law of exponents in arithmetic (fractional values can be treated the same as integers)holds for every n. It doesn't matter whether we're talking about the 5th root of 2 multiplied by 3 or the qth root of 2 multipled by 2k. What does this have to do with the Goldbach Conjecture? You have to go back, as Magidin said, to what you mean by p & q. In any case, using the real numbers 3 & 5, it's easy to see that if p = q, the exponent is 1. So 2^p/q = 2, for every p and q. So? Tom > Dimitris Stathopoulos > Email: stathopouloscs(a)hotmail.com > > > > > >
From: eestath on 2 Dec 2009 11:59 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
From: Arturo Magidin on 2 Dec 2009 13:02 On Dec 2, 1:04 am, eestath <stathopoulo...(a)gmail.com> wrote: Again: DO NOT reply to me via e-mail, or with copies or your messages, or with messages with no content and only quoting posts. I am not interested in getting any personal e-mail from you. Any e-mail I get from you will be reported as harassment, as I have told you over half a dozen times already. I wrote: >> This is *exactly the same* as saying: >> >> "If for every n>4, the two prime numbers p and q satisfy p=/=q, >> then for every m>4, if m is even then m is the sum of two prime >> numbers. " In reference to the original statement, which was: > Theorem > Golbach conjecture is true for every n>4 if the two prime numbers are > different > > No it is not what i say : Yes, it is *exactly* what you said. If it was not what you *meant*, then the fault is yours. > If p/=q then it does not exist k sutch that p+q/=2k! (i) This is *not* what you said in the "theorem". I am not asking you to explain your argument, I am asking you to explain your *statement*. (ii) The statement above is FALSE. It is *patently* false. It is *idiotically simple* to see it is false. Take p=3 and q=7. Then p=/=q. You claim there is no k such that p+q =/= 2k. Okay: so, if I take k=4, then it is *not* true that p+q is different from 2k. But p+q = 3+7 =10, and 2k = 2(4) = 8. So you are saying that 10 and 8 are equal. So what you are saying is *wrong*. -- Arturo Magidin > you fail to understand that is if NOT if and only if You fail to answer the simple question of explaining what you wrote, and you fail to understand how stupid what you wrote above is. Nowhere did I use an "if and only if". > So what you are saying is simply wrong! What I am saying is that you are not answering my question as to the *meaning* of your "theorem". That is a fact, it is not wrong. What *you* say is trivially wrong. > p and q are prime number that are greater then 2! and correlate to n > (there are all the possible solutions for witch p+q/=2k k>2!) How do they "correlate to n"? Do they go on trips together if they have enough vacation days? "Correlate to n" is not a mathematical answer, it is an evasion. Given n, how do we take p and q? > how the hell did you become a professor! > > i am not polite if you are not polite! You have been abusive and impolite from the very beginning, by repeatedly sending me unwanted e-mail despite my requests that you not do so. Calling you an idiot at this point is not lack of politeness, it is a simple statement of fact. -- Arturo Magidin
From: eestath on 2 Dec 2009 13:22
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! |