My Math Forum Proof of twin prime conjecture

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January 2nd, 2019, 02:39 PM   #21
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Quote:
 Originally Posted by Micrm@ss 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!
@Micrm

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 self-evidently 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 self-evidently true, and no proof has been forthcoming except hand-waving 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 03:12 PM.

 January 2nd, 2019, 04:39 PM #22 Banned Camp   Joined: Aug 2010 Posts: 170 Thanks: 4 I have explained why no prime greater than 3 can eliminate the remaining 6n-1 and 6n+1 pairs. I have said that this is because every prime greater than 3 will always hop over an infinite number of un-eliminated 6n-1 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 6n-1 and 6n+1 pairs.
January 2nd, 2019, 05:34 PM   #23
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Quote:
 Originally Posted by MrAwojobi I have explained why no prime greater than 3 can eliminate the remaining 6n-1 and 6n+1 pairs.
You've explained the criterion but you haven't offered an actual proof of anything. Of course the 6n -1 and 6n + 1 pairs can be eliminated if you use primes of a sufficient size by using Eratosthene's sieve, but the sieve doesn't even cancel out all numbers not of the form above. So there is a gap in your concept that needs to be worked on.

So prove to us why "why no prime greater than 3 can eliminate the remaining 6n-1 and 6n+1 pairs." Don't try for a convincing argument, nail it down with the Mathematics.

-Dan

January 2nd, 2019, 06:01 PM   #24
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Quote:
 Originally Posted by MrAwojobi I have explained why no prime greater than 3 can eliminate the remaining 6n-1 and 6n+1 pairs. I have said that this is because every prime greater than 3 will always hop over an infinite number of un-eliminated 6n-1 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 6n-1 and 6n+1 pairs.
You have not explained why the elimination must be done by a single prime.

January 2nd, 2019, 07:03 PM   #25
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Quote:
 Originally Posted by MrAwojobi I have explained why no prime greater than 3 can eliminate the remaining 6n-1 and 6n+1 pairs. I have said that this is because every prime greater than 3 will always hop over an infinite number of un-eliminated 6n-1 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 6n-1 and 6n+1 pairs.
Every prime greater than 3 will "hop over" (when sequencially eliminating its multiples on a number line after having eliminated all the multiples of previous primes under a standard ordering, and where "hop over" implies "fail to eliminate") an infinite number of 6n-1 and 6n+1 pairs, yes. I do understand you perfectly and I know that many others (myself included) have noticed this fact over the years.

Proof:

$$\text{Let } x = p_1*p_2*p_3*...*p_i$$
$$\text{Then } x+1 \text{ and } x-1 \text{ will fail to be eliminated by the primes } \leq p_i$$
$$\text{The same goes for } 2x+1 \text{ and } 2x-1, 3x+1 \text{ and } 3x-1, ...$$

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 07:06 PM.

January 2nd, 2019, 08:02 PM   #26
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Quote:
 Originally Posted by Collag3n Twin prime conjecture proof
but but... 2019 isn't prime!

 January 3rd, 2019, 02:06 AM #27 Banned Camp   Joined: Aug 2010 Posts: 170 Thanks: 4 So it seems not to be in dispute that every prime greater than 3 will always hop over un-eliminated 6n-1 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 6n-1 and 6n+1 pairs ahead of it. No short prime exists because of the first sentence above.
January 3rd, 2019, 02:09 AM   #28
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Quote:
 Originally Posted by MrAwojobi 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.
Prove it

January 3rd, 2019, 03:20 AM   #29
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Quote:
 Originally Posted by MrAwojobi So it seems not to be in dispute that every prime greater than 3 will always hop over un-eliminated 6n-1 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 6n-1 and 6n+1 pairs ahead of it. No short prime exists because of the first sentence above.
Hey, if you want to try to go to the Journals and present this then they'll laugh themselves silly.

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, 04:17 AM #30 Banned Camp   Joined: Aug 2010 Posts: 170 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 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.

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