The clue to the proof of Riemann hypothesis

[robert80]

This could be the way to proof. remember, this is not a proof.

today I found a clue to solution to Riemann hypothesis:

Let it be Riemann zeta function :ζ(s)

The proof that all the non trivial zeroes lie on the critical strip when s = 1/2 + it

let us suppose there are other zeroes having the real part between 0 < Re < 1/2

since to sattisfy the convergence of Re between 0 and 1/2 we know that THE VERY SAME AMOUNT OF ZEROS would accour symetrically to Re = 1/2

so the zeroes would accure on Re(1/2 - x) and they would occur on Re = (1/2 + x) because of convergence condition.

since the 2 values of Riemann Zeta functions of applying to it 2 different s1 and s2 s1 = 1/2 - X +b*i and s2 = 1/2 + X +b*i never equal, that means they couldn't have the same zeroes symetrically distributed to the Re = 1/2

---------------> the zeroes occur at Re=(1/2) if they exist.

The end.

 

[robert80]

If this is of any help to somebody, I would be glad, if not I appologize. I will take now some days off of Math problems. And yes, there is a great possibility, I am wrong... But I feel really passionate about Math and I think I will go to study theoretical Math. Greetings to all.

 

[robert80]

Revised :

In order to sattisfy DIVERGENCE criteria on interval 1> Re > 0 the zeroes are distributed simetrically to the Re = 1/2. Sorry I mixed the terms ...

 

[robert80]

REVISED:This could be the way to proof. remember, this is not a proof.

today I found a clue to solution to Riemann hypothesis:

Let it be Riemann zeta function :ζ(s)

The proof that all the non trivial zeroes lie on the critical strip when s = 1/2 + it

let us suppose there are other zeroes having the real part between 0 < Re < 1/2

since to sattisfy the divergence criteria of Re between 0 and 1 we know that THE VERY SAME AMOUNT OF ZEROS would accour on the right of Re=1/2 symetrically to Re = 1/2

so the zeroes would accure on Re(1/2 - x) and they would occur on Re = (1/2 + x) because of divergence condition.

since the 2 values of Riemann Zeta functions of applying to it 2 different s1 and s2 s1 = 1/2 - X +b*i and s2 = 1/2 + X +b*i never equal, that means they couldn't have the same zeroes symetrically distributed to the Re = 1/2

---------------> the zeroes occur at Re=(1/2) if they exist.

 

[blue_raver22]

I appreciate the effort and enthusiasm put into finding a clue to the solution of the Riemann hypothesis. However, it is important to note that this is not a proof and more evidence and rigorous mathematical analysis would be needed to fully establish the validity of this approach. Additionally, the Riemann hypothesis is a highly complex and longstanding problem in mathematics, and it is unlikely that a single clue or solution will be able to fully address it. It is important to continue to approach the problem from multiple angles and continue to build upon previous research in order to potentially reach a proof.