20190204, 14:35  #12 
Feb 2017
Nowhere
110110000100_{2} Posts 
OP added to list...

20190204, 21:25  #13 
Feb 2019
7×13 Posts 
So you are in agreement that since one prime cannot eliminate all the pairs then this is why the twin prime conjecture is true. If you say more than one prime can eliminate all of them if the twin prime conjecture is false, explain how this can happen. The last of such primes will do the final elimination and this is why I say it will take one and only one prime to achieve this.

20190204, 22:05  #14  
"Curtis"
Feb 2005
Riverside, CA
3^{2}×479 Posts 
Quote:
You say "why is it difficult to see that...." Well, you haven't proven it; I don't think it's true, and neither do some others who posted in this thread. So, you should go about proving your claim rather than asking why we can't see it. When you realize you can't prove it, you will likely decide to more loudly proclaim that we cannot appreciate your genius rather than admitting you made an unprovable assumption, but that's what cranks do. 

20190204, 22:24  #15 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}·3·79 Posts 
No, I am in no such agreement. You need to read up on countableinfinite sets and how they all have a proven onetoone relationship together. If one or any other finite set of primes (say a gazillion of them) and their multiples can not "eliminate" all the twin integers, that is not sufficient proof that an infinite number of primes and their multiples can't ("eliminate" them) either.

20190204, 23:40  #16 
Nov 2008
2×3^{3}×43 Posts 
Bringing back an argument I've used before.
Proof that there are infinitely many primes of the form n^{2}1: Suppose there are finitely many primes of this form. Then there must be a prime that eliminates the last possibilities for primes of the form n^{2}1. Call this prime p. Let N = p#, the product of all primes up to p. Then none of the primes up to p divide N^{2}1, so it has not been eliminated, contradicting the claim that primes up to p eliminate all possibilities for primes of this form. Therefore there are infinitely many primes of the form n^{2}1. This "proof" works in exactly the same way as the claimed proof of the twin prime conjecture. But it is clearly nonsense, as n^{2}1 = (n+1)(n1). 
20190205, 15:15  #18 
Feb 2019
7×13 Posts 
For the twin prime conjecture to be false, there must come a time in the Eratosthenes sieving process whereby there will no longer be anymore 6n1 and 6n+1 pairs left in the infinitely long number line. The sieve of Eratosthenes 'creates' the primes and twin primes i.e. the integers that escape its 'wrath'. I have shown that every prime greater than 3 will keep hopping over 6n1 and 6n+1 pairs infinitely often. This clearly shows that these pairs will NEVER run out. For the twin prime conjecture to be false, they must run out at some point, but they can’t. Even though some of them get eliminated later by larger primes when it is their turn in the sieving process, just some and not all remaining get eliminated. It will take one and only one prime to finish the job if the twin prime conjecture is false. This means that if the twin prime conjecture is false, one of the 6n1 or 6n+1 infinitely many pairs remaining will be a multiple of this special prime before this special prime eliminates them. I have shown that the mechanics of the sieve of Eratosthenes will never let this happen. No complicated mathematics is needed to understand this. All these answers some of you are giving do not address the basic concepts in my proof. My proof is just as basic as Euler’s proof of the infinitude of primes. If any of you can come up with a basic concept like I have done to show how else all the 6n1 and 6n+1 pairs can be eliminated (i.e. if possible) without just one ‘hawkish’ prime doing the job in ‘one fell swoop’, then let your concept be known without using hard mathematics that isn’t needed. Give examples and explain clearly and concisely like I do instead of just dismissing my proof and talking about hard mathematics concepts that are not even needed to see what I am trying to show.

20190205, 16:30  #19 
Aug 2006
31·191 Posts 
You spend a lot of time telling us what you think you told us, but not a lot of time actually making what you told us precise. This is a fairly typical defensive reaction for cranks when the know, deep down, that they haven't actually solved the problem at hand. Underwood Dudley talks about this in his various books* and papers (one is available online). If you want to be taken seriously you must make your argument much more clear and rigorous.
* "Numerology: Or, What Pythagoras Wrought"; "Mathematical Cranks"; "A Budget of Trisections". There may be other books. Of the authors who have written on the topic, Dudley does the best job (IMO) on dissecting the psychology of the pseudomathematician or crank. 
20190205, 18:40  #20  
"Curtis"
Feb 2005
Riverside, CA
3^{2}×479 Posts 
Quote:
Your concept of infinite sets is, simply, mistaken. Once you fix that, you'll be able to agree that the above is a problem with your reasoning. For the conjecture to be false, every candidate pair larger than the last twinprime pair must have a factor. Your above statement is equivalent to saying "every candidate pair must have a factor smaller than some number N." This bound N is what you must prove, and claiming it exists is far far far from a proof (not least because such a bound almost surely does not exist). As another poster wrote, your argument would also prove the infinitude of primes of the form N^2  1. That set also has no time in the infinitelist sieving process where every candidate is eliminated, yet it's quite clear (and, importantly, provable) that N = 2 is the only prime. 

20190206, 13:39  #21 
Aug 2006
1011100100001_{2} Posts 
If what we write sounds somewhat polished, it's because this proof is very familiar to us. It's by far the most common 'proof' of the twin prime conjecture; I've read about three dozen versions of it over the years. You can see that I'm still working on a good response to convince proofwriters that what they have is not sound. There are many possible angles, but it's not completely straightforward. Perhaps when you see why it doesn't work you can give us advice on what convinced you and help us show the next person why.

20190207, 14:51  #22 
Feb 2019
5B_{16} Posts 
I have given one possibility in which all the 6n1 and 6n+1 pairs remaining can be eliminated all at once by only one prime just as 5, for example, can eliminate all its infinite number of multiples all at once. Of course, I have shown why this cannot be possible. None of you, as yet, have come up with another way in which all the pairs remaining can be eliminated all at once or otherwise if the twin prime conjecture was false. I insist that 'otherwise' is not even plausible and 'all at once' is the only plausible way these pairs could be eliminated if the twin prime conjecture was false. These pairs never run out as the sieve of Eratosthenes operates on the number line in perpetuity. Don't forget that this sieve is the creator of the primes and I have simplified its operation by disregarding the multiples of 2 and 3 and removed repetitive elimination i.e. a number is eliminated only once.

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