January 1st, 2019, 11:37 AM  #1 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4  Proof of twin prime conjecture
PROOF OF THE TWIN PRIME CONJECTURE The twin prime conjecture states that there are an infinite number of twin primes. Twin primes are a pair of primes that differ by 2 e.g. 11 and 13, 17 and 19, 29 and 31. Consider the sieve of Eratosthenes acting on the infinite number line. 2 makes the first move and eliminates all its infinite number of multiples, including itself, (i.e. the even numbers) from the infinite number line. 3 then makes the next move and eliminates all its infinite number of multiples (including itself) that have not already been eliminated by 2, from the infinite number line. At this point of the sieving process, it is not difficult to see that all the infinite number of integers left on the infinite number line (after 1 is removed also) are in pairs of the form 6n1 and 6n+1, where n is some positive integer. Therefore odd primes are of the form 6n1 or 6n+1. So, for particular values of n, twin primes, 6n1 and 6n+1 will exist amongst the infinite integer pairs of the form 6n1 and 6n+1 remaining on the infinite number line. So 5 makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2 and 3 i.e. (in ascending order) 5x5, 5x7, 5x11, 5x13, 5x17, 5x19, 5x23, 5x5x5, 5x29, 5x31, 5x5x7, 5x37, 5x41, 5x43, 5x47, 5x7x7, 5x53, 5x5x11, â€¦ 7 then makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2, 3 and 5 i.e. 7x7, 7x11, 7x13, â€¦ 11 then makes the next move and 13 the next after 11 and the process carries on ad infinitum. So, following this process, the twin primes are simply those pairs of primes, 6n1 and 6n+1 that escape the elimination process. A little thought will show that for the twin prime conjecture to be true, it takes one and only one special prime to eliminate (at some point in its elimination process) the remaining infinite number of 6n1 and 6n+1 pairs of integers ahead of it on the infinite number line (i.e. those that have not already been eliminated by previous primes). Eliminating a pair of 6n1 and 6n+1 integers simply requires that one of them is eliminated i.e. one of them is a multiple of this special prime. It will now be shown that it is impossible for any special prime to achieve this i.e. the twin prime conjecture is true. In this part of the discussion, consider the infinite number line after 2, 3 and their infinite number of multiples have been eliminated (1 is removed also) as mentioned previously. Imagine a prime number hopping (infinitely) like a frog along this infinite number line consisting of integers of the form 6n1 or 6n+1 during its turn in the sieving process, as it eliminates its multiples as described above e.g. 17 takes its first hop and eliminates 17 x 17 = 289 as it lands on it. 17 then takes its second hop and then eliminates 17 x 19 = 323 as it lands on it etc. A little thought will show that each and every one of the infinite number of multiples of 17 with prime factors of 5, 7, 11, 13 and 17 only will have a pair of un  eliminated 6n1 and 6n+1 integers just before it (if the multiple of 17 is of the form 6n1) or after it (if the multiple of 17 is of the form 6n+1). So as 17 is hopping infinitely along the infinite number line described, it will hop over these infinite number of 6n1 and 6n+1 uneliminated pairs. So this will be true for all primes greater than 3. So 37 will hop over un eliminated pairs before or after the infinite number of multiples of 37 with prime factors of 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37 only. So no such special prime exists that can eliminate all the remaining uneliminated 6n1 and 6n+1 pairs ahead of it at some point in its infinite hopping. So the twin prime conjecture is true i.e. there are an infinite number of twin primes. 
January 1st, 2019, 11:58 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 
I think you should go ahead and remove all the whitespace from this so it's easier to understand.

January 1st, 2019, 01:13 PM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 
This means that $\displaystyle \lim_{p\rightarrow \infty } \pi (p) =\lim_{p\rightarrow \infty } \pi (p+2)1$ $\displaystyle \; \;$ for pprime $\displaystyle \pi (p)$ is the number of primes less than $\displaystyle p$ Last edited by idontknow; January 1st, 2019 at 01:15 PM. 
January 1st, 2019, 03:39 PM  #4  
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235  Quote:
 
January 1st, 2019, 04:49 PM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,252 Thanks: 519  Quote:
If there exists a positive integer u that divides one of every pair of integers of the form 6n + 1 and 6n  1 > v, there are no twin primes > v. So I think he means that if such a u exists, the twin prime conjecture is false rather than true. He then tries to prove that no such u exists. He then concludes that because no such u exists, the twin primes conjecture is true. This is fallacious. (Of course, I do not guarantee that I understand his word salad.) $\{\alpha \implies \neg \ \beta\} \not \implies \{\neg \ \alpha \implies \beta\}.$ The true statement of "If an animal is a monkey, that animal is not a whale" does not entail the truth of the obviously false statement of "If an animal is not a monkey, that animal is a whale." Last edited by JeffM1; January 1st, 2019 at 04:54 PM.  
January 1st, 2019, 05:04 PM  #6  
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
In all seriousness, congrats on the revolutionary proof. Who would have thought there was an upper bound on the number of primes eliminated in each iteration of the sieve of Eratosthenes. Fields medal inc.  
January 2nd, 2019, 05:43 AM  #7 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
If you took time to follow my presentation, I would have thought that it is quite obvious that it takes just one prime to eliminate all the uneliminated 6n1 and 6n+1 primes for the twin prime conjecture not to be true. I have shown that no such prime exists because every prime greater than 3 will always hop over an infinite number of uneliminated 6n1 and 6n+1 prime pairs in the vicinity of its infinite number of multiples made up of all the primes less than the prime and the prime only.

January 2nd, 2019, 05:48 AM  #8  
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235  Quote:
Basically, your proof of the twinprime conjecture is that "it's obvious". Ok...  
January 2nd, 2019, 05:55 AM  #9 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
PROOF OF THE TWIN PRIME CONJECTURE The twin prime conjecture states that there are an infinite number of twin primes. Twin primes are a pair of primes that differ by 2 e.g. 11 and 13, 17 and 19, 29 and 31. Consider the sieve of Eratosthenes acting on the infinite number line. 2 makes the first move and eliminates all its infinite number of multiples, including itself, (i.e. the even numbers) from the infinite number line. 3 then makes the next move and eliminates all its infinite number of multiples (including itself) that have not already been eliminated by 2, from the infinite number line. At this point of the sieving process, it is not difficult to see that all the infinite number of integers left on the infinite number line (after 1 is removed also) are in pairs of the form 6n1 and 6n+1, where n is some positive integer. Therefore odd primes are of the form 6n1 or 6n+1. So, for particular values of n, twin primes, 6n1 and 6n+1 will exist amongst the infinite integer pairs of the form 6n1 and 6n+1 remaining on the infinite number line. So 5 makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2 and 3 i.e. (in ascending order) 5x5, 5x7, 5x11, 5x13, 5x17, 5x19, 5x23, 5x5x5, 5x29, 5x31, 5x5x7, 5x37, 5x41, 5x43, 5x47, 5x7x7, 5x53, 5x5x11, â€¦ 7 then makes the next move and eliminates all its infinite number of multiples (apart from itself) that have not already been eliminated by 2, 3 and 5 i.e. 7x7, 7x11, 7x13, â€¦ 11 then makes the next move and 13 the next after 11 and the process carries on ad infinitum. So, following this process, the twin primes are simply those pairs of primes, 6n1 and 6n+1 that escape the elimination process. A little thought will show that for the twin prime conjecture to be false, it takes one and only one special prime to eliminate (at some point in its elimination process) the remaining infinite number of 6n1 and 6n+1 pairs of integers ahead of it on the infinite number line (i.e. those that have not already been eliminated by previous primes). Eliminating a pair of 6n1 and 6n+1 integers simply requires that one of them is eliminated i.e. one of them is a multiple of this special prime. It will now be shown that it is impossible for any special prime to achieve this i.e. the twin prime conjecture is true. In this part of the discussion, consider the infinite number line after 2, 3 and their infinite number of multiples have been eliminated (1 is removed also) as mentioned previously. Imagine a prime number hopping (infinitely) like a frog along this infinite number line consisting of integers of the form 6n1 or 6n+1 during its turn in the sieving process, as it eliminates its multiples as described above e.g. 17 takes its first hop and eliminates 17 x 17 = 289 as it lands on it. 17 then takes its second hop and then eliminates 17 x 19 = 323 as it lands on it etc. A little thought will show that each and every one of the infinite number of multiples of 17 with prime factors of 5, 7, 11, 13 and 17 only will have a pair of un  eliminated 6n1 and 6n+1 integers just before it (if the multiple of 17 is of the form 6n1) or after it (if the multiple of 17 is of the form 6n+1). So as 17 is hopping infinitely along the infinite number line described, it will hop over these infinite number of 6n1 and 6n+1 uneliminated pairs. So this will be true for all primes greater than 3. So 37 will hop over un eliminated pairs before or after the infinite number of multiples of 37 with prime factors of 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37 only. So no such special prime exists that can eliminate all the remaining uneliminated 6n1 and 6n+1 pairs ahead of it at some point in its infinite hopping. So the twin prime conjecture is true i.e. there are an infinite number of twin primes. 
January 2nd, 2019, 05:57 AM  #10 
Senior Member Joined: Aug 2010 Posts: 169 Thanks: 4 
I have corrected the mistake i.e. 'false' instead of 'true'.


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conjecture, prime, proof, twin 
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