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Goldbach's Conjecture  Proof (better link) 
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#37
Jan1310, 11:26 PM

P: 30

I never suggested that your lack of a counter example means I proved Goldbach!
You attempted an example to counter a particular element of my proof, in an effort to demonstrate the proof as invalid. You were unable to do so. So it is that the proof was not destroyed by your example. Your example has not verified the proof true or false. 


#38
Jan1310, 11:26 PM

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Sloane's A049300 seems relevant, but sadly doesn't list any papers. 


#39
Jan1310, 11:29 PM

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#40
Jan1410, 01:46 AM

P: 7

In the proof, you explicitly use the theorem that there is no string of n consecutive numbers below n^2 so that they are all divisible by at least one of 3,5,...,n, and you must prove this, or it will not be valid.
Furthermore, even if CRGreathouse cannot construct a counterexample (although I think he did construct a counterexample to your original proof), you have given no reason for one not to exist, so the proof is not "complete". Cheers, Rofler 


#41
Jan1410, 08:02 AM

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#42
Jan1410, 08:14 AM

P: 30




#43
Jan1410, 08:16 AM

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#44
Jan1410, 08:48 AM

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In post #29 you agreed that I found a hole. In post #31 you attempt to fix it, but your fix doesn't work. I suggest carefully studying the transition from {3} to {3, 5} to {3, 5, 7} in order to see why this is. For reference, the first mistake in that post is 


#45
Jan1410, 06:54 PM

P: 30

Take my example in post 26 where N=37. There is no reason to extend the iteration beyond sqrt(74). Sundaram's sieve is illustrative... 9, 15, 21, .... 25, 35, 45, .... 49, 63, 77, ... p^2+2px gives all odd composites. 


#46
Jan1410, 08:24 PM

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Your claim is that there are at least (1/sqrt(2N))*((NM)/2)) Goldbach pairs for 2N. This is really the only context in which I care about M. If M is defined in a way that makes it obvious that it is less than sqrt(2N), then I want to know how you came to this statement; if it's defined in a way that makes this statement obvious, I want to know how you can bound it below sqrt(2N). 


#47
Jan1510, 12:42 AM

P: 30

Sorry, I realize now from your post 27 in which you said:
"In fact, out of N consecutive odd numbers, as few as (N  2)/3 might remain." Your N was my (N1)/2, so your (1/3)(N2) becomes (1/3)((N5)/2) for me. It seems I was in fact too generous allowing for M to be the maximum divisor in the process, when 5 will do just fine. I believe then my proof shows there are at least (1/sqrt(2N))*((N5)/2)) Goldbach pairs for 2N. This is greater than 1 for N>=17. 


#48
Jan1510, 08:13 AM

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#49
Jan1510, 08:27 AM

P: 30




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