Re: Dumb arguments, and social stuff

From: James Harris (jstevh_at_msn.com)
Date: 10/16/04 

Date: 16 Oct 2004 05:54:47 -0700

Tim Smith <reply_in_group@mouse-potato.com> wrote in message news:<LsZbd.5724$NX5.2188@newsread3.news.atl.earthlink.net>...
> In article <3c65f87.0410151536.492dbbef@posting.google.com>, James Harris wrote:
> >> > There is no counterexample. That's my point.
> >>
> >> Your "paper" states:
> >>
> >> Therefore, with the factorization
> >> 65x^3 - 12x + 1 = (a_1x + 1)(a_2x + 1)(a_3x + 1)
> >> one of the a's is coprime to 5.
> >>
> >> He shows that 8a^2+4a-45 divides both a and 5, where a is any of your a_1,
> >> a_2, or a_3. That's a counterexample to the claim that one of the a's is
> >> coprime to 5.
> >
> > You're a troll. The a's are algebraic integers.
>
> He shows that none of the a's are coprime to 5, by exhibiting common
> factors. Your paper claims that one of the a's is comprime to 5, implying
> no such common factor exists. Out here in reality, we call that a
> counterexample.

You call it that, but that's not what it is.

Here's the math. If you reply again, follow this argument and if you
wish to challenge it, do so.

Given an algebraic integer x, 1/x cannot be an algebraic integer
unless it is the root of a monic polynomial with integer coefficients.

That requirment is the most important one, so if you have an algebraic
integer x that is not a unit, what do you then know?

If it is irrational, you know that 1/x is not the root of a monic
polynomial with integer coefficients.

That's it. Given classical methods you know nothing else about it!

Now then, if you wish to continue to push your position that W. Dale
Hall gave a counterexample, why don't you explain why x not being a
unit in algebraic integers would tell you more than that 1/x cannot be
an algebraic integer?

James Harris

Relevant Pages

  • Re: JSH: Heart of dispute, number properties
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  • Re: Counting primes, my way versus others
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  • Re: JSH: Neat puzzle, actually
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  • Re: JSH: Neat puzzle, actually
    ... >> The ring of algebraic integers is defined by polynomials such that it ... >it is a root of an irreducible monic polynomial with coefficients ... an algebraic integer if and only if it is the root of SOME monic ... polynomial with integer coefficients; ...
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  • Re: JSH: How easy? Wiless works flaw
    ... > Jim Spriggs wrote: ... root of a monic polynomial with integer coefficients, ...
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