Proof that twin primes are infinite.

[Z.10.46]

 

Hello,

I have π(x)~x/ln(x), the larger the value of x, the more true this relationship becomes.
Suppose I have two very large twin prime numbers, then I would have n/ln(n)~N so E(n/ln(n))=N and (n+2)/ln(n+2)~N+1 so E((n+2)/ln(n+2))=N+1 with Using E to denote the floor function.
So, if there exists an infinity of twin primes, the equation E((n+2)/ln(n+2))=1+E(n/ln(n)) would have a solution where n is very large. If this equation does not have a solution with a sufficiently large n, then there are a finite number of twin primes that are less than a specific value of n.

This equation is true only if I have infinity = 1 + infinity,So, twin prime numbers are infinite.

 

For your information: When proving the infinitude of prime numbers, it is also demonstrated through this type of equation: q = p1 * p2 * ..{pi}.. * pn + 1/pi. It is stated that the only possibility for q to be an integer based on this equation: infinity = infinity + 1/Pi. Thus, we conclude from this that prime numbers are infinite.
And similarly, to demonstrate an infinity of twin primes, I also use the equation infinity = 1 + infinity.

[uncool]

On 8/16/2023 at 5:48 PM, Z.10.46 said:

then I would have n/ln(n)~N so E(n/ln(n))=N

that's...not how it works at all

[Z.10.46]

yes, but there's nothing preventing transitioning from an approximation to an equality, for instance, if π~3 then I can state that π=3+a with a=π-3, thus If the symbol ~ gives you trouble, then I can even avoid using the integer part to make it an equals sign: if n/ln(n)~N and (n+2)/ln(n+2)~N+1, then n/ln(n)+a=N and (n+2)/ln(n+2)+b=N+1, with a and b being real numbers .

 So, if there exists an infinity of twin primes, the equation (n+2)/ln(n+2)+b=1+n/ln(n)+a would have a solution where n is very large. If this equation does not have a solution for very large n, then there exists a finite number of twin primes that are less than a specific n..

Edited August 18, 2023 by Z.10.46

[Z.10.46]

For your information, even if 'a' and 'b' tend towards infinity, they would still be negligible compared to the infinity of 'n/ln(n)' and 'n+2/ln(n+2)', similar to, for example, 'n' or '-n' being negligible compared to 'n^2'. That's why the ratio π(x)/(x/ln(x)) tends towards 1, even as 'a' and 'b' tend towards infinity.

To be more mathematically rigorous, I choose 'a' and 'b' as sequences 'a_k' and 'b_k' with 'n/ln(n)+a_k = N' and '(n+2)/ln(n+2)+b_k = N+1', so '1+n/ln(n)+a_k = (n+2)/ln(n+2)+b_k' Even as 'a_k' and 'b_k' tend towards infinity, the equation remains true, thus indicating the existence of an infinite number of  twin prime numbers. If there were not an infinite of  twin prime numbers,This equation would admit only one solution for a specific small value of 'n'.

 

Edited August 19, 2023 by Z.10.46