Re: JSH: Attacking the conclusion

From: Andrzej Kolowski (akolowski_at_hotmail.com)
Date: 06/19/04 

Date: 19 Jun 2004 10:07:13 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0406170813.99f6ad0@posting.google.com>...
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>

On one level, Harris is right.

The conclusion of 'APF' does not follow from the main result
claimed in APF.

Why?

Because the main result is about polynomials of the form

  P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +3m) x^3 - 3(-1 + mf^2) ux^2 + u^3 f),

where f is a non unit, nonzero algebraic integer coprime to 3 and x,
and u is a non unit non zero algebraic integer coprime to f.

Harris's main result about such polynomials is applied to

  65 x^3 - 12 x + 1.

But this polynomial does not have the same form as P(m). In
particular, it has constant term 1 rather than constant term

  u^3 f^3.

Therefore it is *could* be that this 'conclusion' to APF is wrong,
but the 'main result' is right.

Everyone, perhaps including Harris, seems to have overlooked this.
His main result does not apply to the polynomial for which Dale
Hall, among others, provides irrefutable counterarguments.

Has Harris even read his own paper?

Here is what Harris does. In the expression above for P(m),
he lets f = sqrt(5), m = 1, and u = 1. He writes:

[*] (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3(-1 + mf^2) xu^2 + u3 = 65x^3 -12x + 1.

But go back and look at the expression for P(m) above. It is

   P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +3m) x^3 - 3(-1 + mf^2) ux^2 + u^3 f)

        = (m^3 f^6 - 3m^2 f^4 + 3m f^2)x^3 - 3(-1 + mf^2) xu^2 f^2 + u3 f^3.

It does not agree with the left side of [*].

Harris was just plain careless in multiplying through by f^2.

Harris has re-submitted 'APF' to another journal. Does this
version include this careless error and the erroneous conclusion?

OK, let's look at things on another level.

Harris trots out a mantra which says "Proofs cannot duel."

What he wants is not a counterexample to what he claims. He
thinks he has a valid proof. Dale Hall of course thinks he has
a valid counterproof. Harris has not found an error in what Dale
has done. He says Hall is using 'circular reasoning', but he
does not point out where. He says Hall's argument is a 'cheat',
that it is 'specious'.

But he doesn't find an error. Saying it is a 'cheat' or 'specious'
or a 'circular argument' is not equivalent to finding an error.

He himself has a 'proof'. He does not find an error in it either.

The problem is, Hall has not pointed out an error in Harris's
main proof either. Hall only showed that Harris's application of his
main result to 65 x^3 - 12 x + 1 is wrong. As noted above, this
could very well be because Harris was careless in applying his
main conclusion: not because the main conclusion itself was wrong.

So Harris says, "Proofs cannot duel." Hall's proof is a proof.
Harris thinks his proof is a proof. They don't 'duel' because
they are about different things. Both could be right.

Harris is not going to be satisfied until someone convinces him
that there is an error in his 'proof' of his main result. Showing
that a misapplication of his main result is wrong does not do that.
  
So where is Harris's error? IS there an error in his main
result ?

There is.

Contrary to what Harris says, the error has been pointed out before.
DOZENS of times. He notes that P(m) factorizes in a certain way
when m = 0. He then jumps to the conclusion that the same form
of factorization must be true when m <> = 0.

He gives no proof of this. m = 0 is a degenerate case. In that
case a certain associated polynomial is reducible. In general, when
m > 0, it is irreducible. Reducible polynomials factor differently
than irreducible ones. The central result is this: If Q(y) is
a monic polynomial with integer coefficients, and the constant term
is divisible by a prime q, then every root of Q(y) has a nonunit
algebraic integer factor in common with q. Put another way, no
root of Q(y) is coprime to q in the algebraic integers.

  [Harris denies this. He says that, e.g., if Q(y) = y^3 + 12*y^2 - 65,
   then one of the three roots of Q(y) is coprime to 5.]

This fact shows that Harris's main result is wrong.

Here is the precise excerpt from APF (the version that appeared
briefly in the electronic journal) where Harris makes his mistake:

  " ... given g1 = a1 x + uf where with m = 0, g1 gives a factor
    of f IT MUST HAVE THAT SAME FACTOR IN GENERAL, proving that
    two of the a's have a factor that is f."

I added caps to highlight Harris's erroneous leap of logic.

Thus we have:

   1. The main result of APF is wrong.

   2. The exact spot where Harris's reasoning goes bad
       has been identified.

Moreover, Rick Decker and Keith Ramsay provided examples back
in January-February which showed that Harris's reasoning is
incorrect. Harris started 20 or more threads in sci.math, frantically
trying to squirm out of Decker's argument. He failed.

Decker's example showed that the central argument of APF was wrong.

That's why Harris was so worried about it. If he couldn't refute
Decker, he knew that APF should be withdrawn.

He could not refute Decker or Ramsay, no more than he was able to
refute Hall.

So he went quiet for a long time.

But somehow now his solipsism has prevailed. APF "passed peer review",
although it is clear now that there was never any peer review,
and that the editor of the electronic journal to which he submitted
it accidentally put it in the wrong electronic pile.

But the damage was done. For a day or so, APF was published.
People here wrote the editor saying it was obviously, clearly wrong.
The editor hastily withdrew it and gave no explanation: a cowardly
act, no matter what the motivation.

But this stupid mistake gave Harris all he needed. "It passed
peer review." Therefore someone must have thought it was right.
The editor must have been lying when he said it was a 'technical mistake.'
He was beaten into withdrawing the paper and then lying by sci.math
"bullies". This gave Harris the excuse to ignore all the counter-arguments:
Decker's, Ramsay's, Hall's, Magidin's, whoever. Never mind that
he could not refute any of these. Some anonymous reviewer, he thought,
gave it the stamp of approval. That outweighs everything: Galois
theory, basic algebraic number theory, even obvious arithmetic.

  Andrzej
  

> It's a cheat. It's a dodge. It's specious.
>
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
>
> In every case posters rely on a circular argument which basically
> relies on the ring of algebraic integers not having the problem I've
> proven it has.
>
> Here's basically how it goes:
>
> There are numbers that should properly be considered factors of 1.
>
> However, in the ring of algebraic integers, these numbers are NOT
> factors of 1.
>
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
>
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
>
> They are factors of 1 in a ring made up of numbers such that only 1
> and -1 are integers units, which includes algebraic integers, and
> other numbers besides them.
>
> That's it. You have a ring where only 1 and -1 are integer units,
> where that's the principal defining characteristic and you can show
> that some irrational units in that ring, are not algebraic integers.
>
> Simple.
>
> But, when attacking my work, posters like Hall, Magidin, and Decker
> assert that because these numbers are not units in the ring of
> algebraic integers i.e. factors of 1 in that ring, then I must be
> wrong.
>
> But I've proven that I'm right using rather basic algebra.
>
> That algebra shows that there is a ring of numbers where -1 and 1 are
> the only integer units that is indeed LARGER than the ring of
> algebraic integers.
>
> That's it. That's the amazing conclusion that mathematicians have
> been running from for so long.
>
> It's actually kind of a neat fact.
>
>
> James Harris

Relevant Pages

  • Re: Questioning assumptions
    ... James Harris wrote: ... > show how it is that posters could find algebraic integers that they ... > Like one BIG assumption is that if a number is not a unit in the ring ... > really mathematicians because they couldn't see the truth. ...
    (sci.math)
  • Re: JSH: Attacking the conclusion
    ... The conclusion of 'APF' does not follow from the main result ... Everyone, perhaps including Harris, seems to have overlooked this. ... root of Qis coprime to q in the algebraic integers. ... > However, in the ring of algebraic integers, these numbers are NOT ...
    (sci.math)
  • Re: From sign conventions to Galois Theory
    ... > flaws of the ring of algebraic integers. ... > You should be aware, by now, that just as the algebra of polynomials ... Harris constructs ... he writes down is going to work because he has this special 'gift' ...
    (sci.math)
  • Re: Dumb arguments, and social stuff
    ... >> contradiction, so he refuses to believe his own eyes. ... > conjugates of a minimal polynomial are added to the ring. ... > actually wrote about algebraic integers, ... Harris recently is ambiguous regarding the Galois argument. ...
    (sci.math)
  • Re: Dumb arguments, and social stuff
    ... Yet another overlong post from "Nora Baron". ... that they do in the ring of algebraic integers. ... Instead I say that one of the a's is coprime to f in the ring. ... James Harris ...
    (sci.math)