January 3rd, 2019, 09:23 AM  #41 
Newbie Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 
Take $N^+$ and instead of sieving all multiples of some prime $p$, you sieve all positive exponent of 2 of these special numbers which I'll call tetaprimes: $a\cdot 2^0$ where $a$ is odd (and obviously the tetaprime is too). After sieving, the question is: "Is there an infinite number of tetatwin" (tetatwin are just 2 successive numbers, 1 odd and 1 even, or viceversa). First pass, you sieve all tetamultiples of $1\cdot 2^0$: $1\cdot 2^1$, $1\cdot 2^2$, $1\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of tetatwins. second pass, you sieve all tetamultiples of $3\cdot 2^0$: $3\cdot 2^1$, $3\cdot 2^2$, $3\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of tetatwins. Third pass, you sieve all tetamultiples of $5\cdot 2^0$: $5\cdot 2^1$, $5\cdot 2^2$, $5\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of tetatwins. After $n$ passes, you sieve all tetamultiples of $(2n1)\cdot 2^0$: $(2n1)\cdot 2^1$, $(2n1)\cdot 2^2$, $(2n1)\cdot 2^3$, ... In the new infinite set, there are still an infinite numbers of tetatwins. Your answer: There are infinitely many tetatwins. My answer: There are none. 
January 3rd, 2019, 09:24 AM  #42  
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326  Quote:
Why can't it be, for example, that $5$ eliminates one of $6N \pm 1$ $7$ eliminates one of $6(N+k)\pm 1$ for $k>0$ $11$ eliminates one of $6(N+k')\pm 1$ for $k'>k$ $13$ eliminates one of $6(N+k'')\pm 1$ for $k''>k'$ etc. Why is this scenario impossible?  
January 3rd, 2019, 09:30 AM  #43 
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326  
January 3rd, 2019, 09:31 AM  #44 
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326  
January 3rd, 2019, 09:37 AM  #45  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
If such a proof exists, why is Mr. Owojobi so intent on proving that if the twin primes conjecture is false, such a set has exactly one member?  
January 3rd, 2019, 10:30 AM  #46  
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Quote:
Quote:
MrAwojobi is failing to consider the possibility of an infinite E. That's why he's looking for the "one member" that eliminates all twin primes not eliminated by previous primes.  
January 3rd, 2019, 11:06 AM  #47 
Banned Camp Joined: Aug 2010 Posts: 170 Thanks: 4 
You are all making a very simple explanation of the truth of the twin prime conjecture over complicated probably because the maths gurus like Field medalist, Terence Tao have made the twin prime conjecture a big deal when really it is not. My proof is just as logical and as simple as proving the infinitude of primes. My proof uses the sieve of Eratosthenes, which by the way is the algorithmic generator of the primes, to show the infinitude of twin primes. I have shown that the sieve of Eratosthenes cannot 'hit' all the 6n1 and 6n+1 pairs using a simple proof. This sieve operates by giving every prime greater than 3 opportunities to eliminate 6n1 and 6n+1 pairs. It is agreed that each prime in its infinite hopping will hop over an infinite number of uneliminated 6n1and 6n+1 pairs and therefore no one prime can eliminate all infinite number of 6n1 and 6n+1 pairs. This means that the twin primes will never run out.

January 3rd, 2019, 11:07 AM  #48  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
And his failure to consider that E was infinite was exposed when he said "last" and "final" way back when. So his uniqueness argument is not merely unproven, but unprovable and unnecessary. ROFL  
January 3rd, 2019, 11:13 AM  #49  
Senior Member Joined: Oct 2009 Posts: 850 Thanks: 326  Quote:
 
January 3rd, 2019, 11:14 AM  #50 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  It does not follow that what one prime cannot do an infinite set of primes cannot do. It is not as though we are going to run out of primes to work with.


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conjecture, prime, proof, twin 
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