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Old 2019-04-06, 13:58   #12
Batalov
 
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... or Cain prize.
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Old 2019-04-13, 16:10   #13
MathDoggy
 
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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
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Old 2019-04-13, 16:55   #14
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Quote:
Originally Posted by MathDoggy View Post
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.
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Old 2019-04-14, 11:57   #15
MathDoggy
 
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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
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Old 2019-04-14, 12:00   #16
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The contradiction of Q being composite is that it would not be divisible by any elemet of S because it always leaves remainder 1
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Old 2019-04-14, 18:03   #17
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Originally Posted by MathDoggy View Post
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?
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Old 2019-04-15, 13:03   #18
MathDoggy
 
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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.
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Old 2019-04-15, 13:06   #19
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Originally Posted by jrsousa2 View Post
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
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Old 2019-04-15, 13:26   #20
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Quote:
Originally Posted by MathDoggy View Post
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.

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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?
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Old 2019-04-15, 13:34   #21
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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
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Old 2019-04-15, 13:39   #22
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Originally Posted by MathDoggy View Post
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
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