Re: JSH: Fool all of the people, all of the time?

From: Brian Chandler (imaginatorium_at_despammed.com)
Date: 12/05/04 

Date: 5 Dec 2004 11:19:53 -0800

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0412050552.40525729@posting.google.com>...
> imaginatorium@despammed.com (Brian Chandler) wrote in message news:<f2c35871.0412042036.542cbcd6@posting.google.com>...
> > jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0412040746.3ff61803@posting.google.com>...
> >
> > [snip]
> >
> > > Good question. But to believe that my results haven't traveled
> > > through mathematical society at this point you have to believe that a
> > > very basic argument, which I know I can explain in about an hour as I
> > > did it in-person at my alma mater Vanderbilt University, which shoots
> > > down one of the underpinnings of algebraic number theory could just
> > > float out there, be argued about by me on Usenet for years, and never
> > > get heard of by leading mathematicians.
> >
> > Impressive bit of sentence construction. Incidentally, though: "argued
> > about by me on Usenet for years" - which bit _is_ this? I know you
> > seem to have been claiming to have found errors in 'core' for years,
> > but I thought they were different arguments. Since as you say
> > yourself, you've been wrong a lot in the past, and the important bit
> > is the current argument, which alone of course is Correct, how long
> > has this bit been going? I'd have thought less than "years"...?
> >
> > [snip]
>
>
> The full timeline is that back in December 1999 I first discovered an
> approach which would lead to the analysis tool of non-polynomial
> factorization.

<snip>

> The earliest arguments in this area go back to December 1999.

OK - I'll give you your "arguing for years".

> > Newsflash!! People are asking questions!! Here's the commonest one:
> >
> > "What does 'properly a unit' mean?"
> >
> > Only you can answer.
> >
>
> That's an old game of trying to cause major arguments over the use of
> some term or other, as I now simply shift from usage that sci.math'ers
> find easily works to provoke confusion.

Sorry, I'm a bit lost here. You're accusing me of "playing a game",
just because I asked what a bit of your argument means? I don't
understand what you mean when you claim that while something-or-other
is not a unit in the ring of algebraic integers, it _is_ "properly a
unit". I guess it means something to you, but I surmise it means
nothing to anyone else. (That means you are speaking a private
language, and the closest to fame you can hope for is the status of a
Voynich manuscript in a half-millennium or two.)

I find all this stuff about peering into polynomials looking for
'factors' a bit confusing, too. I mean, of course in (a) below,
there's a factor of 4 that can be divided off:

(a): 4x^2 + 4x - 4

So if n is an integer - any integer - I know that 4n^2 + 4n -4 is a
multiple of 4. However:

(b): (x+1)(x+2)

Does this polynomial have a factor of 2 in it? I can see a '2', but it
doesn't look like a factor. And again I know that any integer n means
that (n+1)(n+2) is a multiple of 2. So what's going on? Is there any
difference from the case of (a). You seem to reject the idea that
there's really any difference between talking about polynomials *as*
polynomials and evaluations of polynomials, because that's "voodoo
math" - have I got that right?

And what ever should I think about (c) - absolutely no trace of a '3'
anywhere:

(c): x(x+1)(x+2)

Another thing I wonder about: OK, suppose the dam bursts. Suppose
suddenly you are on tv shows and whatnot. The mathematicians are all
disgraced, flung into debtors jails. Are you going to rewrite all the
text books? Who will do it?

FWIW, I have a copy of Herstein's "Topics in [pre-Harrisian] Algebra"
here, and it only mentions algebraic integers very briefly, in the
problems. There's the usual definition, then the first problem reads:

10. If _a_ is any algebraic number, prove that there is a positive
integer _n_ such that _na_ is an algebraic number.

(I can type more if required.) Well, is Herstein's solution to this
problem wrong?

Brian Chandler
http://imaginatorium.org

Relevant Pages

  • Re: JSH: The "Published" paper he dosent what you to know about.
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  • Re: JSH: The "Published" paper he dosent what you to know about.
    ... ADVANCED POLYNOMIAL FACTORIZATION ... Determining the distribution of factors within irrational algebraic integers ... that are themselves polynomials. ... Mathematicians have decided to hold on to flawed ideas that I easily ...
    (sci.math)
  • Re: The actual math, advanced polynomial factorization
    ... > and, remember, x, m, f and u are all independent variables. ... > ALGEBRAIC INTEGERS they do not always have f as a factor. ... Then if w is any factor of r in the ring of algebraic ... My result holds for a family of polynomials. ...
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