Disproof of Riemann Hypothesis

[Ashwin_Kumar]

Ok, I think that would help micro mass. Don't know much about the riemann sphere myself. And dimension, that is an incredulously silly mistake you made there.

 

[dimension10]

As I can see you have made a mistake in the 2nd line of the third page

sin(\frac{\pi }{4}+\frac{\pi it}{2})=sin(\pi(\frac{2it+1}{4}))\neq sin(\pi(\frac{4it+1}{4}))

I think you are correct. So I would need to cut that short again. I have attached that here.

 

[micromass]

You could delete your entire document altogether, because the very first step is not valid. Or you haven't given a reason why it is valid.

Also "Euler and others showed..." should have a reference to where you found the result.

 

[Hurkyl]

On the Riemann sphere (or equivalently, the projective complex numbers), \infty \cdot \infty = \infty. But \infty + \infty is undefined as mm said.



Sometimes, other compactifications of the complex numbers are used. One common one is to let both the real and imaginary parts be extended real numbers. We can extend functions by continuity to get identities such as (for complex x):

1. e^{x - \infty} =0 [/itex ]<br /> [*]\log (x + i \infty) = \infty + i \frac{\pi}{2}

<br /> And contour integrals are often defined to use such points as endpoints. e.g. to write the inverse Laplace transform as an integral.

 

[Ashwin_Kumar]

So technically this is no longer a disproof nor a proof.

 

[Ashwin_Kumar]

I think the paper should be scrapped because it makes no sense anymore. It no longer disproves the hypothesis. Your first steps are incorrect, once again I feel that the gamma function is wrong. And your definition of the theta function, once again, does not seem to make sense. And because the paper ends with no definite result, it is meaningless.

 

[dimension10]

You could delete your entire document altogether, because the very first step is not valid. Or you haven't given a reason why it is valid.

Also "Euler and others showed..." should have a reference to where you found the result.

I just substituted the theta function and the Euler-Mascheroni constant.

So technically this is no longer a disproof nor a proof.

Yes.

I think the paper should be scrapped because it makes no sense anymore. It no longer disproves the hypothesis. Your first steps are incorrect,

I just substituted the theta function and the Euler Mascheroni Constant.

once again I feel that the gamma function is wrong.

Where is it wrong?

And your definition of the theta function, once again, does not seem to make sense.

Go to http://en.wikipedia.org/wiki/Theta_function. That is the definition of theta function I used. The general one, not the special case.

And because the paper ends with no definite result, it is meaningless.

Well but it seems to be much easier do prove the end result rather than the original Riemann Hypoothesis.

 

[amitjohar]

The only way to disprove Riemann Hypothesis is to give a counter example where the non-trivial zero of the Riemman Zeta Function does not have half as its real part. Its that simple. The disproof should not be more than a line or 2 at most. It is the proof that is expected to be very lengthy. To date the first 10 trillion non-trivial zeros of the zeta function are confirmed to be on the critical line. What you have done is really not a disproof.

 

[dimension10]

The only way to disprove Riemann Hypothesis is to give a counter example where the non-trivial zero of the Riemman Zeta Function does not have half as its real part. Its that simple. The disproof should not be more than a line or 2 at most. It is the proof that is expected to be very lengthy. To date the first 10 trillion non-trivial zeros of the zeta function are confirmed to be on the critical line. What you have done is really not a disproof.

That's why its an extension.

 

[amitjohar]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

 

[pmsrw3]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

Ah, the classic reductio ad praemium method of proof...

 

[micromass]

I believe that the Riemann Hypothesis is true. There is a one million dollar prize for anyone who can prove the Riemann Hypothesis. But there is no prize for the disproof of Riemman Hypothesis. This shows that RH is likely true. Why else is no prize offered for disproof of RH?

A prize is offered for the solution of the Riemann Hypothesis. So even a counterexample could get you 1000000$.

 

[pmsrw3]

A prize is offered for the solution of the Riemann Hypothesis. So even a counterexample could get you 1000000$.

Well, now we have a problem. By the method of proof advanced by amitjohar, we now have shown that the Riemann hypothesis is both true and false.

 

[micromass]

Well, now we have a problem. By the method of proof advanced by amitjohar, we now have shown that the Riemann hypothesis is both true and false.

Aha! So we have shown mathematics to be inconsistent!

 

[pmsrw3]

Aha! So we have shown mathematics to be inconsistent!

Isn't there a prize for that?