January 2nd, 2019, 03:39 PM  #21  
Senior Member Joined: May 2016 From: USA Posts: 1,245 Thanks: 515  Quote:
I think that you are not quite following what his argument boils down to. I think what he originally intended (but did not say) was this syllogism: (1) If there exist unique integers m and u such that for any integer n $\ge$ m > 0 u either (6n  1)/u or (6n + 1)/u is an integer, then the conjecture is false. (2) No such m and u exist. (3) Therefore the conjecture is true. The first proposition is selfevidently true because it would entail that at least one member of each 6n  1 and 6n + 1 pair was composite. I suspect that the second proposition is also true but did not attempt a proof because even if both propositions 1 and 2 are true, proposition 3 does not follow as a consequence. The syllogism is a well known fallacy. I may have misunderstood what he was originally trying to say. In any case, he seems now to have revised his argument. (1a) Only if there exists unique integers m and u such that, for any integer n $\ge$ m, either (6n  1)/u or (6n + 1)/u is an integer will the conjecture be false. (2a) No such m and u exist. (3a) Therefore the conjecture is true. THAT is a valid syllogism. But proposition (1a) is no longer selfevidently true, and no proof has been forthcoming except handwaving about "last" and "final" in the context of infinite sets. In fact, I suspect proposition 1a is false. Last edited by JeffM1; January 2nd, 2019 at 04:12 PM.  
January 2nd, 2019, 05:39 PM  #22 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
I have explained why no prime greater than 3 can eliminate the remaining 6n1 and 6n+1 pairs. I have said that this is because every prime greater than 3 will always hop over an infinite number of uneliminated 6n1 and 6n+1 pairs within the vicinity of a class of multiples of the prime. So it is impossible for any prime to eliminate the remaining 6n1 and 6n+1 pairs.

January 2nd, 2019, 06:34 PM  #23  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,978 Thanks: 788 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
So prove to us why "why no prime greater than 3 can eliminate the remaining 6n1 and 6n+1 pairs." Don't try for a convincing argument, nail it down with the Mathematics. Dan  
January 2nd, 2019, 07:01 PM  #24  
Senior Member Joined: May 2016 From: USA Posts: 1,245 Thanks: 515  Quote:
 
January 2nd, 2019, 08:03 PM  #25  
Senior Member Joined: Jun 2014 From: USA Posts: 430 Thanks: 29  Quote:
Proof: $$\text{Let } x = p_1*p_2*p_3*...*p_i$$ $$\text{Then } x+1 \text{ and } x1 \text{ will fail to be eliminated by the primes } \leq p_i$$ $$\text{The same goes for } 2x+1 \text{ and } 2x1, 3x+1 \text{ and } 3x1, ...$$ That said, I didn't read your OP nor do I intend to try... It's a big messy paragraph that I don't want to try to unravel. I personally have no idea why this doesn't solve the twin prime conjecture, but that's a shortcoming of my own as opposed to a proposed acceptance of your work. I'm sure there is a good explanation (and perhaps it has already been provided for you above). Last edited by AplanisTophet; January 2nd, 2019 at 08:06 PM.  
January 2nd, 2019, 09:02 PM  #26  
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198  Quote:  
January 3rd, 2019, 03:06 AM  #27 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
So it seems not to be in dispute that every prime greater than 3 will always hop over uneliminated 6n1 and 6n+1 pairs. Can you therefore not see that this means that there are infinitely many twin primes because for there not to be infinitely many twin primes, it takes one and only one prime to do away with all the remaining 6n1 and 6n+1 pairs ahead of it. No short prime exists because of the first sentence above.

January 3rd, 2019, 03:09 AM  #28 
Senior Member Joined: Oct 2009 Posts: 684 Thanks: 222  
January 3rd, 2019, 04:20 AM  #29  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,978 Thanks: 788 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Or you could, you know, listen to us and learn something.... Maybe even improve on your own argument. It makes little difference to me. Dan  
January 3rd, 2019, 05:17 AM  #30 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
For there not to be infinitely many twin primes, it takes one and only one prime to do away with all the remaining 6n1 and 6n+1 pairs ahead of it. I will prove this statement which I thought would be quite obvious to those who just do a little bit of thinking about it. Since we are dealing with infinitely many 6n1 and 6n+1 pairs, is it not obvious that if the twin primes are finite then only one prime is needed to eliminate all the infinite number of uneliminated 6n1 and 6n+1 pairs? It will be absurd to say 5 primes for instance will be needed to collectively eliminate the 6n1 and 6n+1 pairs because that will mean that the first of the 5 primes did not remove all the 6n1 and 6n+1 pairs and similarly the second, third and fourth did not remove all the 6n1 and 6n+1 pairs. It will be the fifth prime that will achieve this i.e. one and only one prime will be required to render the twin prime conjecture false. We are in agreement from previous posts that no prime can do this and so the twin prime conjecture is true, no doubt.


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conjecture, prime, proof, twin 
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