20190326, 19:58  #1 
Mar 2019
37_{16} Posts 
I Think The Twin Prime Conjecture Is True
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. Last fiddled with by MathDoggy on 20190326 at 19:59 
20190326, 20:43  #2 
Einyen
Dec 2003
Denmark
3·17·59 Posts 
Why do you think you can prove the twin prime conjecture so easily, when all other mathematicians could not prove it for centuries? Does it seem likely or unlikely that it will be very complicated to prove?
Last fiddled with by ATH on 20190326 at 20:43 
20190326, 20:55  #3 
"Curtis"
Feb 2005
Riverside, CA
31·149 Posts 
1. What does it mean to take an infinite sum of the first n objects?
2. What "implication" links divergence of z to divergence of s? You use one word in place of actual proof of anything. What "comparison criterion" are you talking about? This isn't even slightly an outline of a proof of anything, let alone a proof. You slapped some math words (improperly, see #1 above) around the unsubstantiated claim that "because natural numbers are an infinite list, twin primes are an infinite list." Using words like divergence and comparison, and assuming the list of twins is finite do nothing to support your claims; you may as well have said "they're infinite because they are a type of natural number." Last fiddled with by VBCurtis on 20190326 at 20:56 
20190326, 21:35  #4 
"Luke Richards"
Jan 2018
Birmingham, UK
2^{5}·3^{2} Posts 

20190326, 21:56  #5 
Aug 2006
3×1,987 Posts 
Where does your proof fail if you replace "twin" with "even"?

20190326, 22:24  #6  
Mar 2019
5×11 Posts 
Quote:
Answer to number 2: The comparison criterion that I was talking about is a test to determine whether an infinite series diverges or not. It says that if you have two infinite series A and B that satisfy the following you can test if A or B converges or diverges: A is less or equal to B A and B > 0 If A diverges then B diverges If B converges then A converges And there is the implication I was talking about 

20190326, 22:28  #7 
Mar 2019
5·11 Posts 

20190326, 22:30  #8 
Mar 2019
5×11 Posts 

20190326, 23:02  #9  
Aug 2006
3×1,987 Posts 
Quote:
3 + 5 + 7 + 11 + 13 + 17 + 19 + 29 + 31 + ... and compare it to the positive integers: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + ... and notice that the former is always larger than the latter. This would indeed show that the former diverges, if it was also known that it had infinitely many terms. But this is what we wanted to prove in the first place, so we're stuck. 

20190403, 16:33  #10 
"Luke Richards"
Jan 2018
Birmingham, UK
2^{5}×3^{2} Posts 

20190406, 05:44  #11 
Dec 2018
Miami
29 Posts 
LOL...post your "proof" on vixra.org/numth/ then. Alongside the proofs of Riemann's Hypothesis and other thenopen problems.
I'm pretty sure your proof will be deemed correct and you nominated for the next Abel prize. Last fiddled with by jrsousa2 on 20190406 at 05:45 
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