A little renewed Disproof of Riemann hypothesis

[choe]

hmm...

 

[CRGreathouse]

Are you missing some pages? This pdf begins with "proposition)" rather than, say, a title or abstract.

 

[epkid08]

Am I missing something, why aren't you using

\zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n\geq1}\frac{(-1)^{n-1}}{n^{s}}

as your definition of zeta?

The definition of zeta in your proof is invalid for \text{Re<s>} &lt; 1</s>.

 

[Ashwin_Kumar]

Your zeta function is not well-defined. In fact, it is wrongly defined. That mistake at the beginning completely. Also note that e to the power of pi times imaginary is -1, and by adding in the other values, you get -(u) in which u is all the other values.
The zeta function definition in the papers is a serious problem because you must not forget that the zeta function is a complex plane, and your shortened definition totally misses the actual definition.

 

[zetafunction]

Am I missing something, why aren't you using

\zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n\geq1}\frac{(-1)^{n-1}}{n^{s}}

as your definition of zeta?

The definition of zeta in your proof is invalid for \text{Re<s>} &lt; 1</s>.

but Eta function is defined for s >0 i think this sum is still valid for s >0

 

[amitjohar]

Disproof means showing a counter example, where the real part of the non-trivial zero is not half. If there is no counter example, then it cannot be accepted as disproof.

 

[micromass]

Disproof means showing a counter example, where the real part of the non-trivial zero is not half. If there is no counter example, then it cannot be accepted as disproof.

That's nonsense. A disproof doesn't need to give a counterexample, as the proof doesn't need to be constructive. One could proof this by contradiction by assuming that the Riemann-hypothesis is true and then obtaining a contradiction.

 

[amitjohar]

Well not for Riemann Hypothesis. What I meant to say is that in the case of RH, the only way to disprove is to provide a counter-example. All the disproofs without any counter-example will never be right in the case of RH. All such RH disproofs, attempted by numerous people were scrapped. Therefore it is reasonable to conclude that if a disproof is shown in RH, it is only going to be through counter-example. Otherwise it simply cannot work. Now in other cases, disproof without counter example can work. But definitely not for RH

 

[amitjohar]

hmm...

Your first definition of zeta(s) is correct and it is correct that s = a + bi. But you also state it is valid for every a such that 0 < a < 1. This may not be true. a is the real part of s. You claim that zeta(s) is valid for every positive a less than 1. Zeta(s) converges only when a > 1. See http://en.wikipedia.org/wiki/Riemann_zeta_function and the first sentence states that. If you claim that it is valid for 0< Re(s) < 1, then show its reference or source.

 

[micromass]

Well not for Riemann Hypothesis. What I meant to say is that in the case of RH, the only way to disprove is to provide a counter-example. All the disproofs without any counter-example will never be right in the case of RH. All such RH disproofs, attempted by numerous people were scrapped. Therefore it is reasonable to conclude that if a disproof is shown in RH, it is only going to be through counter-example. Otherwise it simply cannot work. Now in other cases, disproof without counter example can work. But definitely not for RH

I don't see why not. If there is a proof that a counterexample exists, but does not provide one, then it's a good proof!
It's not because all such disproofs did not work, that there might not be a disproof that will work.

In fact, there are many proofs that assert the existence of something, but that never actually give the value. For example x^5-x+1=0 has been proven to have a solution, but nobody can actually find the solution! It's not unlikely that the disproof of the Riemann-zeta function will be a similar situation.

 

[amitjohar]

I agree with you but the main problem lies that RH may not really be false. If it is true, then there is no point in trying to create a disproof. There is no reason to think that the RH is wrong. Most mathematicians think it is true. RH maybe unprovable because of Godel's theorem. Godel's theorem states that for any given mathematical system, there are mathematical truths that cannot be proven within that system. The Riemann Hypothesis maybe one of these. If Godel theorem holds, then RH can never be proven true and it can also never be proven false.

 

[micromass]

I agree with you but the main problem lies that RH may not really be false. If it is true, then there is no point in trying to create a disproof. There is no reason to think that the RH is wrong. Most mathematicians think it is true.

What mathematicians think or guess is irrelevant. It must be proven true. As long as its not proven true, a proof or disproof must be sought. Looking for a disproof or looking for a proof requires quite the same methods actually...
There is no reason to think RH is true or wrong.

RH maybe unprovable because of Godel's theorem. Godel's theorem states that for any given mathematical system, there are mathematical truths that cannot be proven within that system. The Riemann Hypothesis maybe one of these. If Godel theorem holds, then RH can never be proven true and it can also never be proven false.

Godel's theorem is true. And Godel's theorem has nothing to do with this here. Godel says that there exists statements that can be proven nor disproven. But it is not known if the Riemann hypothesis is such a statement. This must be proven.

 

[lostcauses10x]

Interesting paper. Strangely it may do more to imply the hypothesis than dispute it.
The Riemann zeta function is a strange thing in its factoring,or more precise the factors and how it forms into the graph.
It may even be the proof, and or disproof will be shown by how the results of such are positioned to the complex plane.

 

[atomthick]

I agree with you but the main problem lies that RH may not really be false. If it is true, then there is no point in trying to create a disproof. There is no reason to think that the RH is wrong. Most mathematicians think it is true. RH maybe unprovable because of Godel's theorem. Godel's theorem states that for any given mathematical system, there are mathematical truths that cannot be proven within that system. The Riemann Hypothesis maybe one of these. If Godel theorem holds, then RH can never be proven true and it can also never be proven false.

That's interesting because if you prove that RH is not provable due to Godel's theorem that means RH is actually true because you can prove it to be false by searching for a root not on the critical line (of course there are an infinite number of them but nonetheless there is a way to prove it to be false).

 

[micromass]

That's interesting because if you prove that RH is not provable due to Godel's theorem that means RH is actually true because you can prove it to be false by searching for a root not on the critical line (of course there are an infinite number of them but nonetheless there is a way to prove it to be false).

No, if Godel's theorem is unprovable then it is unprovable. There is no way to make sure whether it is true or false in that case. You just need more axioms to be able to prove it.

 

[HallsofIvy]

Well not for Riemann Hypothesis. What I meant to say is that in the case of RH, the only way to disprove is to provide a counter-example. All the disproofs without any counter-example will never be right in the case of RH. All such RH disproofs, attempted by numerous people were scrapped. Therefore it is reasonable to conclude that if a disproof is shown in RH, it is only going to be through counter-example. Otherwise it simply cannot work. Now in other cases, disproof without counter example can work. But definitely not for RH

What terrible "logic"! Your argument is that such proofs have not worked, so it is "reasonable to conclude" that they can't work, which then becomes, it is impossible!

 

[zhentil]

I think that since ALL disproofs have been scrapped, RH is true by induction.

 

[ramsey2879]

I think that since ALL disproofs have been scrapped, RH is true by induction.

This is not how proof by induction works, but you probably already knew that. I liked your post though, especially where you put ALL in caps.

 

[lostcauses10x]

been a bit bored so... The first trivial zero is at -2, the patern from -2 is every even negative intiger continuing from -2.
So -2 - (-2)= 0
The value of the Reiman zeta function at 0 = -1/2
The offset this value designates is due to the pole at the intiger of 1.
1+(-1/2)=1/2
This is the line the non trivial zeros will occur at.

 

[lostcauses10x]

been a bit bored so...


The first trivial zero is at -2, the pattern from -2 is every even negative intiger continuing from -2.
So -2 - (-2)= 0
The value of the Reiman zeta function at 0 = -1/2
The offset this value designates is due to the pole at the integer of 1.
1+(-1/2)=1/2
This is the line the non trivial zeros will occur at.

Even with me misspellings, I would have though some one would say some thing. Yet I did expect silence.
https://www.physicsforums.com/Nexus/editor/menupop.png
Most will simple take this as nonsense, and maybe rightfully so.

So far in all my reading of the RH proofs papers disproofs proofs of unprovable etc.
I find that no one has given any reason for the why the line of real part 1/2 exists other than brute force calculation of the non trivial zeros.
I do find the ideas that suggest it might be true, but still no reason as to why the line exists other than it does.

So as many folks that have examined the Riemann zeta function it seems to me the part that makes the line exist is built into the complex number system.
So a simple evaluation of the complex number system was done simply by substituting -1 times the square root of 1 for i; and then calculating an equivalent value for each position. Intriguing enough there exists a situation in the complex number system that can and does create the situation of the why of the line directly in relation the the Riemann zeta function.
It can be applied to the Riemann zeta and show the why of the video of the convergence to real part 1/2 and the decrease monotonically.

That is all I will give for the time. What I wrote in the last post may or may not be so nonsensical: or as most might think, I just might be crazy... or not.
An offset does exist.

Well for one I am not trying to prove the thing, just come up with a starting point of why it
(the line) is.