 mersenneforum.org I Think The Twin Prime Conjecture Is True
 Register FAQ Search Today's Posts Mark Forums Read  2019-04-06, 13:58 #12 Batalov   "Serge" Mar 2008 Phi(4,2^7658614+1)/2 27×71 Posts ... or Cain prize.   2019-04-13, 16:10 #13 MathDoggy   Mar 2019 3716 Posts What if I do a proof like this? If the set of twin prime numbers is finite then we can make a list, let S be the list of twin prime numbers, S=P1,P2,P3,P4,PN Now let us construct a number Q such that Q=P1×P2×P3×P4×PN+1 If Q is a twin prime then there exists a larger twin prime then on S If Q is composite then non of the twin primes of S will divide Q. Both of this conclusions yield to a contradiction, therefore there are infinitely many twin prime numbers   2019-04-13, 16:55   #14
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

22×1,061 Posts Quote:
 Originally Posted by MathDoggy What if I do a proof like this? If the set of twin prime numbers is finite then we can make a list, let S be the list of twin prime numbers, S=P1,P2,P3,P4,PN Now let us construct a number Q such that Q=P1×P2×P3×P4×PN+1 If Q is a twin prime then there exists a larger twin prime then on S If Q is composite then non of the twin primes of S will divide Q. Both of this conclusions yield to a contradiction, therefore there are infinitely many twin prime numbers
Where is the contradiction if Q is composite? If Q is composite, then Q is divisible by primes that aren't twin primes. That's hardly contradictory.   2019-04-14, 11:57 #15 MathDoggy   Mar 2019 5510 Posts Instead of the sum of the first n natural and twin primes numbers I correct it into an infinite sum of natural numbers and twin prime numbers   2019-04-14, 12:00 #16 MathDoggy   Mar 2019 5×11 Posts The contradiction of Q being composite is that it would not be divisible by any elemet of S because it always leaves remainder 1   2019-04-14, 18:03   #17
wblipp

"William"
May 2003
New Haven

93616 Posts Quote:
 Originally Posted by MathDoggy The contradiction of Q being composite is that it would not be divisible by any elemet of S because it always leaves remainder 1
But not all primes are in S. Could Q be divisible by one or more of those other primes?   2019-04-15, 13:03 #18 MathDoggy   Mar 2019 5510 Posts Clearly Q is larger than any other twin prime, so it does not equal one of them, since P1,P2,P3,P4...Pn constitute all twin primes Q can not be a twin prime. Thus it most be divisible by at least one of our finitely many twin primes, say Pk ( 1 less than or equal to k less than or equal to n) But when we divide Q by Pk we have a remainder of 1. This is a contradiction so our original assumption that there are finitely many twin prime must be false.   2019-04-15, 13:06   #19
MathDoggy

Mar 2019

3716 Posts Quote:
 Originally Posted by jrsousa2 LOL...post your "proof" on vixra.org/numth/ then. Alongside the proofs of Riemann's Hypothesis and other then-open problems. I'm pretty sure your proof will be deemed correct and you nominated for the next Abel prize.
I don't think so because the proof has a lot of mistakes that I have not been capable of solving   2019-04-15, 13:26   #20
wblipp

"William"
May 2003
New Haven

235810 Posts Quote:
 Originally Posted by MathDoggy Clearly Q is larger than any other twin prime, so it does not equal one of them, since P1,P2,P3,P4...Pn constitute all twin primes Q can not be a twin prime.
I'm with you this far.

Quote:
 Originally Posted by MathDoggy Thus it most be divisible by at least one of our finitely many twin primes,
But I lost you here, There are lots of primes that are not twin primes. Why is it impossible that Q is a product of only these non-twin primes?   2019-04-15, 13:34 #21 MathDoggy   Mar 2019 678 Posts It is impossible because we constructed an arbitrary number Q which is the product of the finite list of the twin primes and adding 1 to the product   2019-04-15, 13:39   #22
MathDoggy

Mar 2019

3716 Posts Quote:
 Originally Posted by MathDoggy Introduction: The twin prime conjecture is a mathematical hypothesis which states that there exists infinitely many prime numbers that differ by 2. Proof by direct method: Let s be the infinite sum of the first n twin prime numbers Let z be the infinite sum of the first n natural numbers Let us assume that there exists a finite amount of twin prime numbers, then by the comparison criterion we will check whether the infinite sum of z diverges or converges because if z diverges then by implication s will also diverge. Now we will prove the divergence of z: z=1+2+3+4+5+6+7.. Now let us take the largest power of 2 which is greater or equal to K, where K is an element of z distinct to 2. Now we have, z=0+1+2+2+2+2+3+3... Obviously this infinite sum diverges then we can conclude that s also diverges therefore there does not exist a finite amount of twin prime numbers. Q.E.D (I have used the same method to prove this conjecture as the Goldbach conjecture) If you want to see the other proof, you can take a look in the Algebraic Number Theory section.
Can somebody point out what parts of the proof are wrong   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Steve One Miscellaneous Math 53 2019-03-18 00:34 Awojobi Miscellaneous Math 40 2019-02-10 16:36 Carl Fischbach Miscellaneous Math 7 2009-06-24 05:52 cipher Twin Prime Search 0 2009-04-15 12:21 Templus Lounge 9 2006-03-14 16:30

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