January 2nd, 2019, 05:00 AM  #11 
Senior Member Joined: Oct 2009 Posts: 784 Thanks: 280 
So you're not going to explain the missing step?

January 2nd, 2019, 06:35 AM  #12 
Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 
So it seems what you want to be convinced about is why it takes 1 prime to invalidate the twin prime conjecture and not maybe 2 or more primes. Since we are talking about a sieving process which carries on ad infinitum, can you not see that if the twin prime conjecture is false then it will take just 1 prime to put an end to the infinite number of uneliminated 6n1 and 6n+1 ahead of it.

January 2nd, 2019, 07:10 AM  #13  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
At great length, he explains the sieve of Eratosthenes and relies on the theorem that every prime greater than 3 can be expressed as 6n  1 or 6n + 1, where n is a positive integers. Now what he is going on about is this. He considers the infinite set of pairs of numbers {(6 * 1  1, 6 * 1 + 1), (6 * 2  1, 6 * 2 + 1), ...). Let's label these pairs by the factor of 6. He then MEANS to assert that if there exists a positive integer u that divides either 6v  1 or 6v + 1 for every integer v $\ge$ w, a positive integer, then the twin prime hypothesis is NOT true. Because every prime > 3 is in his set of pairs, every twin prime from (5, 7) is in his set. If u divides one member of each set of pairs from from the wth pair on, that member is not prime and so that pair is not a pair of twin primes. In fact, under that supposition, the number of twin primes is w  1, a finite number. Therefore, if such a u exists, the hypothesis of an infinitude of twin primes is false. This is a perfectly valid conditional argument. Of course, in his original presentation, he muffed it by saying that if u exists, the hypothesis of the infinitude of twin primes is true. He then presents an argument that such a u does not exist. I have not bothered to try to penetrate his prose in order to evaluate that argument because, as he originally intended to proceed, he was arguing $\{\{\alpha \implies \neg \ \beta\} \text { and } \neg \ \alpha\} \implies \beta.$ This of course is a fallacy as I showed by a childlike example in my previous post.  
January 2nd, 2019, 07:20 AM  #14 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  
January 2nd, 2019, 07:30 AM  #15  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
I agree that if just one dividing prime exists, the conjecture is false. But there could be an infinite set of such dividing primes, some of which do not divide into every pair but which collectively do. In which case, the conjecture is false even though no one dividing prime will do the trick.  
January 2nd, 2019, 08:25 AM  #16 
Senior Member Joined: Oct 2009 Posts: 784 Thanks: 280  
January 2nd, 2019, 08:53 AM  #17 
Newbie Joined: Sep 2017 From: Belgium Posts: 18 Thanks: 7  
January 2nd, 2019, 11:12 AM  #18 
Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 
Where is the logic when you say say that more than one prime can collectively get rid of the 6n1 and 6n+1 uneliminated pairs. How will this happen except for the last of these primes you talk about to do the final elimination. This is what I mean when I say that it takes one prime to do the final elimination if the twin prime conjecture is false.

January 2nd, 2019, 12:30 PM  #19  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
 
January 2nd, 2019, 12:57 PM  #20 
Senior Member Joined: Oct 2009 Posts: 784 Thanks: 280  So you are assuming it is false now? So the logic of your proof is: Assume twin prime conjecture is false. Then there is one prime doing the last elimination. If there is such a prime, then the conjecture is false. Therefore it's true. Great logic! 

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conjecture, prime, proof, twin 
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