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

From: James Harris (jstevh_at_msn.com)
Date: 12/06/04 

Date: 5 Dec 2004 19:13:41 -0800

imaginatorium@despammed.com (Brian Chandler) wrote in message news:<f2c35871.0412051119.5b676b91@posting.google.com>...
> 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".
>

So you will accept what is true. How wonderful for you.
 
> > > 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.)

You take a phrase out of context, and make a big deal out of it,
asking for some explanation, when in context when used the explanation
was in what I said.

It's an old tactic that sci.math'ers have used for years.

Pick a phrase, take it out of context, make a big deal out of it.

And, you know? Lots of times I HAVE used certain phrase differently
from standard usage, and misused terms I didn't understand or simply
just didn't care a lot to get exactly right in informal discussions.

Time after time, posters have reacted as if these posts are such
important communications that perfection is a requirement.

I reserve the right to at times just babble. It's not a big deal.

This is Usenet.
 
> 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?

Well here's a good chance to show how sci.math'ers routinely try to
mislead.

With integer you have the any integer either is even or has a residue
of 1.

That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2.

So trivially you can put up something like (x+1)(x+2) and know it must
be even.

However, in the ring of algebraic integers, no such relations exist.

That is, there is no non-unit in the ring of algebraic integers such
that EVERY algebraic integer is either divisible by that number or has
the same residue.

You don't even have the even case with algebraic integers as in the
ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2
as a factor.

Now the example (x+1)(x+2) in the ring of integers is childish
mathematics at the most basic level, but posters have routinely used
that example for years to try and claim that it shows how my research
can be wrong.

You people don't even try hard.
 
> And what ever should I think about (c) - absolutely no trace of a '3'
> anywhere:
>
> (c): x(x+1)(x+2)
>

Yet another childish example based on any *integer* x, either being
divisible by 3 or having a residue of 1 or 2.

That childish game can be played on and on with integers, but it has
no meaning outside of the ring of integers, and it doesn't invalidate
my research findings.

> 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?

Mathematicians will NOT all be disgraced and flung into jail.

If there are some who end up prosecuted it'd most likely be a choice
few.

And hey, not all mathematicians work in the area of algebraic number
theory. Number theorists are the ones who are really going to take
the hit.

How might a prosecution work?

Well, consider a federal prosecution that considers whether or not
some number theorist knowingly continued both to teach what I'd shown
to be false, and to receive federal funds for their research.

So they'd have tax dollars for two things: teaching and research

The charge would be fraud. Evidence might include emails of that
mathematician and conversations had with colleagues who would be
pulled into grand juries and later into court to testify under oath.

Likely punishment? A fine. Blockage from receipt of any more federal
funds. Censure from their university, and probably removal from
teaching position.
 
> 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?
>

Amazingly enough, most people actually care about what's TRUE.

More than likely there will be a flood of mathematicians into
algebraic number theory, as it will be an opened up field with major
opportunities for advancement.

Think about it. In many areas a young mathematician can work for
years and get nowhere. In algebraic number theory, they will have the
opportunity to be a significant figure in the remaking of an entire
discipline.

When that finally gets out, it will be the hottest field in
mathematics.

James Harris