Re: My paper, and the cheaters
From: James Harris (jstevh_at_msn.com)
Date: 09/19/04
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Date: 19 Sep 2004 05:29:20 -0700
norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0409181058.2437d01@posting.google.com>...
> jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409171443.51d8ff25@posting.google.com>...
> > norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0409170732.566a5e55@posting.google.com>...
> > > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409161404.365eaa1e@posting.google.com>...
> > > > I've made some pointed criticisms of math society but if you actually
> > > > pay attention you'll realize that the evidence against math society is
> > > > nastier than I go into detail about usually.
> > > >
> > > > Supposedly a correct math result is what's important, but despite my
> > > > having a quite correct paper, which was to be published, sci.math'ers
> > > > managed to get it yanked IN ONE DAY with some hostile emails.
> > > >
> > > > The paper has been by a lot of editors and mathematicians and to date,
> > > > no major error has been found, which I say because there was one minor
> > > > error that actually got pointed out by a sci.math poster.
> > > >
> > >
> > > Please do not ever again claim to be an honest person.
> > > Your "proof" contradicts known mathematics. At least 4
> > > counterproofs have been given. Dale Hall's counterproof
> > > can be checked with simple arithmetic. The exact spot
> > > in your "proof" where you make your major error has been pointed
> > > out repeatedly. You know all this and yet you continue to
> > > deny it.
> >
> > That's not true.
> >
> > In fact, I don't disagree with much of what you say.
> >
> > And I'd like you to admit that in a post, so that I can move on to
> > what I actually *DO* say.
> >
> > First though, I need you to read what I have below and acnowledge that
> > you see the agreement on a key point.
> >
> >
> > >
> > > Below is one of the simplest counterproofs. Feel free
> > > to point out errors (assuming you are still interested in
> > > math as opposed to social studies). This is, incidentally,
> > > for your future reference, what real proof is supposed to
> > > look like.
> > >
> > > ================================================================
> > >
> > > James Harris has written a paper called "Advanced Polynomial
> > > Factorization" in which he claims the following: if the
> > > polynomial
> > >
> > > 65*x^3 - 12*x + 1
> > >
> > > is factored in the form
> > >
> > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1),
> > >
> > > where a1, a2 and a3 are algebraic integers, then exactly
> > > two of a1, a2, and a3 are divisible in the algebraic integers
> > > by sqrt(5), and the third one is coprime to 5.
> > >
> > > It should be noted that -a1, -a2, and -a3 are roots of
> > >
> > > x^3 + 12*x^2 - 65.
> > >
> > > That Harris's claim is false can be seen from the following:
> > >
> > > ==========================================================================
> > >
> > > Assume m(x) is an irreducible monic polynomial with integer coefficients
> > > and algebraic integers a and b are two roots of m(x).
> > >
> > > Theorem. If p is a nonzero rational integer and a is divisible by
> > > sqrt(p) in the algebraic integers, then b is also divisible
> > > by sqrt(p).
> >
> > That last should say, in the algebraic integers.
> >
>
> Agreed.
>
>
> > Theorem accepted. Proof not needed here so I deleted it out to focus
> > on what's important. It is in fact true that if 'a' is divisible by
> > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p)
> > ***in the ring of algebraic integers***.
> >
> > Concede agreement "Nora Baron" and then I'll explain the rest.
> >
>
> Of course I "concede". That's what I proved.
>
> The question now becomes, what did you claim to be true in
> "Advanced Polynomial Factorization."
>
> Just below is the full text of APF exactly as it was submitted
> to the SWJPAM:
>
That's unnecessary. No sense piling on a lot of information for
readers, eh?
Now then at issue is the factorization
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
and we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor
of sqrt(5) in the ring of algebraic integers.
Do you understand that?
If so then more progress can be made.
Now you should be able to understand that and no longer make posts
accusing me of being dishonest and then making statements that I don't
actually disagree with, as if I do.
I'll repeat it so that there's no room for confusion:
Given the factorization
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor of
sqrt(5) in the ring of algebraic integers.
Now is that settled?
James Harris
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