# disproof of Riemann Hypothesis?

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### Reinhold Burger

Mar 16, 2007, 7:33:32 AM3/16/07
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An unpublished paper by Tribikram Pati, in which he claims to
have disproved the Riemann Hypothesis. I don't have the number
theory background to review it, but I gather the author is a
respected mathematician, possibly retired, from Allahabad, India.

I wondered if any of you could give opinions on it.

Reinhold

### Gerry

Mar 16, 2007, 9:45:02 AM3/16/07
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Hi
interesting, i thought everyone was expecting the opposite.
Did he also include a list of the complex zero's not on the critical
line?
Gerry

### MuTsun Tsai

Mar 16, 2007, 10:35:24 AM3/16/07
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Will, at leaast this does not look like an immediate nonsense.
I'll probably take a look at it if I have time.

### Bob Kolker

Mar 16, 2007, 6:43:54 PM3/16/07
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Reinhold Burger wrote:

Wait until his paper is vetted by the professionals.

Bob Kolker

### David L. Johnson

Mar 16, 2007, 6:43:56 PM3/16/07
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One would assume that would settle the issue, but this paper does not do
that as far as I could tell by glancing at it. The author constructs a
chain of conclusions from the assumption (used essentially throughout)
that the hypothesis is true, ending in something that is supposed to be
obviously false. No construction of other zeros; it fits the word
"disproof" in the title.

--

David L. Johnson

welfare was my business; charity, mercy, forbearance, and
benevolence, were, all, my business. The dealings of my trade were but
a drop of water in the comprehensive ocean of my business!" --Dickens,

### stefanw

Mar 26, 2007, 1:00:17 PM3/26/07
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On 16 Mrz., 13:33, Reinhold Burger <rfbur...@cs.uwaterloo.ca> wrote:
> http://arxiv.org/abs/math.NT/0703367
>
> An unpublished paper by Tribikram Pati, in which he claims to
> have disproved theRiemannHypothesis. I don't have the number

> theory background to review it, but I gather the author is a
> respected mathematician, possibly retired, from Allahabad, India.
>
> I wondered if any of you could give opinions on it.
>
> Reinhold

on p.12, m stands for the multiplicity of a zero of the zeta function,
so
typically m=1. But then the step from the third to the fourth
displayed
something + A < something,
for a positive constant A. Similarly, I do not understand how the
positive
constant B disappears a few lines later. I did not attempt to
investigate,
though, whether this is a major problem or not.

Stefan Wehmeier
ste...@math.upb.de

### tc...@lsa.umich.edu

Mar 26, 2007, 5:00:20 PM3/26/07
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In article <eu8u71$25l$1...@news.ks.uiuc.edu>,

stefanw <ste...@math.upb.de> wrote:
>But then the step from the third to the fourth displayed
>something + A < something,
>for a positive constant A. Similarly, I do not understand how the
>positive >constant B disappears a few lines later.

He explains a couple of pages back that "A" stands for some unspecified
positive constant, which might not be the same constant each time it
appears, even in consecutive lines of a series of inequalities. Sort
of like big-O notation.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences

### stefanw

Mar 27, 2007, 12:30:03 PM3/27/07
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On 26 Mrz., 23:00, t...@lsa.umich.edu wrote:
< f(n,t)$(the left "<" sign is never used in the sequel). Here t is bound with delta via relation$t=t_0+\delta$,$t_0$being a zero of zeta between n-1 and n+1, and we may consider$t_0$uniquely determined by n. When choosing$A^*$numbers t and delta are not yet defined, delta is introduced later as$\delta=A^*/\log(n+1)$. Thus we are choosing delta as satisfying inequality of type$\delta<f(delta)$. In order to define it well, we should eliminate dependence on delta in f. It is easy to do for several factors using estimates$n-1<t_0+\delta<n+1$; but then there remains$|\zeta(1/2+it_0+i\delta)|^{\tilde A}$which we cannot estimate by n only. Thus existence of such delta needs proof; and the rest of the paper shows this delta cannot exist. So here is the contradiction and not in the RH. ### lange...@hotmail.com unread, May 23, 2007, 11:30:01 AM5/23/07 to Your results looks fine. But the main definition of delta is not "introduced later" - after (18), I think you point on (36) and on that the author give attantion to (11) and (20) - but on (10) and (11). So the main idea of the author was to show, that we need that delta to proof RH, but there is no delta, so we can not proof RH after all (it is not a disproof but a 'not proveable' - statement). ### Julia Kuznetsova unread, May 28, 2007, 12:30:03 PM5/28/07 to On May 23, 7:30 pm, lange_kl...@hotmail.com wrote: > Your results looks fine. But the main definition of delta is not > "introduced later" - after (18), I think you point on (36) and on that > the author give attantion to (11) and (20) - but on (10) and (11). Delta is$A^*/log n$, so it is defined by A*, i.e. in (20). In (11) it is only announced: "... where A* is a ... number to be further specified". I don't mean (36) because it is deduced correctly from (20) and below. > the main idea of the author was to show, that we need that delta to > proof RH, but there is no delta, so we can not proof RH after all (it > is not a disproof but a 'not proveable' - statement). No, not so. He wants to show that if we proved RH then there would exist such a delta (what would lead to a contradiction). ### lange...@hotmail.com unread, May 30, 2007, 9:30:03 AM5/30/07 to Yes, Pati by himself didn't uses the strategy showing that RH is not proveable under the axioms of standard analysis. But indirectly - his "disproof" shows - that the only way for standard analysis handle the RH may lead to the conclusion, that I give before. But you are right: He himself claims a contradicition found on RH. My point was: This main idea, I talked about, was unintended given by the author's paper. But its path opened for that. ### tc...@lsa.umich.edu unread, May 30, 2007, 10:30:00 AM5/30/07 to In article <f3ju8r$cmf\$1...@news.ks.uiuc.edu>, <lange...@hotmail.com> wrote:
>Yes, Pati by himself didn't uses the strategy showing that RH is not
>proveable under the axioms of standard analysis. But indirectly - his
>"disproof" shows - that the only way for standard analysis handle the
>RH may lead to the conclusion, that I give before.

Let me see if I understand. You claim that the author wanted to show
something like:

1. RH implies the existence of delta.
2. In fact, delta does not exist.
3. Therefore, RH is false.

However, the author did not show this. But are you claiming that the
author *did* in fact show, unintentionally, that

1'. If RH were provable using standard axioms, then delta would exist.
2'. In fact, delta does not exist.
3'. Therefore, RH is not provable using standard axioms?

It would be remarkable for a paper that makes no explicit mention of
axioms or related concepts from mathematical logic to show 1'-2'-3'
without showing 1-2-3. Without looking at the paper in detail, I would
suspect that Kuznetsova's interpretation is correct, that the author
has shown neither 1 nor 1'.

### lange...@hotmail.com

Jun 11, 2007, 10:00:08 AM6/11/07
to
Bernd Krötz is showing, that step by step Pati's paper is correct, but
the logical strategy behind them failed. With or without RH the

http://guests.mpim-bonn.mpg.de/kroetz/RH.pdf

So, and that is my point:

I mean that this error of Patis strategy leads us to the idea, that RH
is not proveable/ or disproveable in boundarys of standard analysis
axioms.

### Julia Kuznetsova

Jun 12, 2007, 3:10:53 PM6/12/07
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> Bernd Krötz is showing, that step by step Pati's paper is correct, but
> the logical strategy behind them failed.

I have read the text at the link. He says the same what I did.
Calculations are correct but it doesn't mean "Pati's paper is
correct".
If I choose a number x such that 0 < x < sin x and carry lots of
correct calculations with it -- how do you think, maybe I'll get

> I mean that this error of Patis strategy leads us to the idea, that RH
> is not proveable/ or disproveable in boundarys of standard analysis
> axioms.

It has nothing to do with axioms. It's a mere logical error of one
definite mathematician. This happens from time to time... Erratum
humanum est.