Proof of twin prime conjecture

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Oct 2009
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For there not to be infinitely many twin primes, it takes one and only one prime to do away with all the remaining 6n-1 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 6n-1 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 un-eliminated 6n-1 and 6n+1 pairs? It will be absurd to say 5 primes for instance will be needed to collectively eliminate the 6n-1 and 6n+1 pairs because that will mean that the first of the 5 primes did not remove all the 6n-1 and 6n+1 pairs and similarly the second, third and fourth did not remove all the 6n-1 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.
Oh sure, I do agree that if you know there are FINITELY many primes that eliminate all pairs, then it is obvious that there must be one that eliminates the remaining twin primes.

But there are an infinite number of primes. So how do you know that there are finitely many primes that eliminate all pairs?

And, I know we're stupid people. But if you give somebody a proof to get feedback on, and they ask a question, don't reply with saying it's obvious five times. It's just insulting. I don't care, but journals will.
 
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May 2016
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It will be absurd to say 5 primes for instance will be needed to collectively eliminate the 6n-1 and 6n+1 pairs because that will mean that the first of the 5 primes did not remove all the 6n-1 and 6n+1 pairs and similarly the second, third and fourth did not remove all the 6n-1 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.
If each of your five primes removes one fifth of the pairs, then all five are needed, not just the fifth. You have slipped in the assumption that the fifth can do all the eliminations that the first four do.

A cannot lift a beam. B cannot lift the same beam. A and B together cannot lift the same beam. A, B, and C together can lift that beam. That does not prove that C alone can lift the beam. A child can see the flaw in that argument.

So you have not established that if the twin primes conjecture is false, there is a single eliminating prime. Prove it.
 
Oct 2009
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If each of your five primes removes one fifth of the pairs, then all five are needed, not just the fifth. You have slipped in the assumption that the fifth can do all the eliminations that the first four do.

A cannot lift a beam. B cannot lift the same beam. A and B together cannot lift the same beam. A, B, and C together can lift that beam. That does not prove that C alone can lift the beam. A child can see the flaw in that argument.

So you have not established that if the twin primes conjecture is false, there is a single eliminating prime. Prove it.
Exactly.

What he claims in his original post though, is that there is a single prime eliminating all the remaining twin primes ahead. This would be true if there were only 5 primes. However, this seems not what he needs later in the proof. Since later he does need one prime that eliminates EVERY twin prime, not just the remaining ones. So you're right, if he succeeds in proving his original claim that there is one prime that eliminates the remaining ones, that would not be enough.
 
May 2016
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Oh sure, I do agree that if you know there are FINITELY many primes that eliminate all pairs, then it is obvious that there must be one that eliminates the remaining twin primes.
This actually may concede too much. If there is a finite set E of eliminating primes and the number of elements in that set is e > 1, it is indeed obviously true that exactly one element of P must eliminate all the pairs that are not eliminated by the other e - 1 elements. But Mr. Awojobi seems to be making a stronger claim, namely that this true conditional proposition proves that e is not greater than 1.

Of course even if he can prove that stronger conditional proposition, which he has not, this does not address the much more plausible hypothesis that set E is infinite. If there are a finite number of twin primes, then the number of pairs (6n - 1 and 6n + 1) remaining are infinite, and there are an infinite number of primes that potentially can do the necessary eliminating. And in an infinite list, there is no last element.
 
Aug 2010
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I can't see what more I can say to make you see things from what to me is a straight forward point of view. Maybe I will try again. 5, for instance, eliminates all its infinite number of multiples from the infinite number line. Why then can you not all see that if the twin prime conjecture is false then it takes just one prime to eliminate all the un-eliminated, infinite number of 6n-1 and 6n+1 pairs? I repeat again that it will be absurd to say more than 1 prime is required to do this i.e. one might just as well make an absurd comment by saying that more than one prime is required to eliminate all the infinite number of multiples of 5 when it is clear that one prime,5, can do this.
 

skipjack

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Dec 2006
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If it's so straightforward, you should be able to rewrite your proof without using words such as "obvious" and "absurd", and as a series of elementary steps, each of which is a direct consequence of a stated rule (or stated rules) of arithmetic or algebra.
 
Jun 2014
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I can't see what more I can say to make you see things from what to me is a straight forward point of view. Maybe I will try again. 5, for instance, eliminates all its infinite number of multiples from the infinite number line. Why then can you not all see that if the twin prime conjecture is false then it takes just one prime to eliminate all the un-eliminated, infinite number of 6n-1 and 6n+1 pairs? I repeat again that it will be absurd to say more than 1 prime is required to do this i.e. one might just as well make an absurd comment by saying that more than one prime is required to eliminate all the infinite number of multiples of 5 when it is clear that one prime,5, can do this.
You might assume there is a greatest prime that is also a twin prime and derive a contradiction. If such a prime $p$ existed, your method would eventually remove its multiples, but there would still be potential twins at the time you eliminate the multiples of $p$. You would then have to go past $p$ and use other primes to eliminate the remaining potential twin primes. There are infinitely many though and your method cannot halt on any particular prime that eliminates all other potential twin primes from the list. That’s why you fail.
 
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greg1313

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Oct 2008
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This seems to be an ineffective take-off on Zhang's method, which proved there are an infinite number of prime pairs with a prime gap around some number near 70 million.
 

topsquark

Math Team
May 2013
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...Why then can you not all see that if the twin prime conjecture is false then it takes just one prime to eliminate all the un-eliminated, infinite number of 6n-1 and 6n+1 pairs?...
P.R.O.V.E. I.T!!

Your statement here, once again, does not involve a proof.

You are getting exasperated with us. That's a good thing! Provide a proof that doesn't rely on "hand waving" and you'll convince us that we've been wrong about this the whole time.

-Dan
 
Jun 2014
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This actually may concede too much. If there is a finite set E of eliminating primes and the number of elements in that set is e > 1, it is indeed obviously true that exactly one element of P must eliminate all the pairs that are not eliminated by the other e - 1 elements.
It's proven earlier in the thread that there is no finite set E of eliminating primes.

Of course even if he can prove that stronger conditional proposition, which he has not, this does not address the much more plausible hypothesis that set E is infinite.

Yes, E is infinite. So now MrAwojobi, you have to prove that there is an infinite set E of eliminating primes that eliminates all potential twin primes greater than $p$ (where $p$ is the greatest prime that is also a twin prime). Under your method, that is a really tall order. Good luck!
 
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