Re: JSH: Keep it simple



jstevh@xxxxxxx wrote: 
The arguing is kind of fun, but let's not lose sight of the actual
challenge here, and why I'm making it.

With

a^3 + 3(-1+xf^2)a^2 - f^2(x^3 f^4 - 3x^2 f^2 + 3x) = 0

I have proven that two of the roots have f as a factor.


You've done no such thing. In recent articles, it appears you're
trying to claim two contradictory things:

	1. None of the roots is coprime to f in the ring of
	   algebraic integers (in the special case below):

	" For example, with x=1 and f=7 you get the result

		a^3 + 144a^2 - 110593 = 0

	  and that cubic's roots must follow my theory in
	  that only two have 7 as a factor while one is coprime
	  to 7, though not in the ring of algebraic integers."


	2. Your result does hold  in the ring of algebraic
	   integers (although the text I use to support my
	   claim is a bit vague):

	" The ring of algebraic integers is built using the
	  arbitrary rule that you take roots of monic
	  polynomials with integer coefficients.

	" My research shows that builds a ring with special
	  properties which allow the appearance of
	  contradiction, but Dedekind makes a claim to have
	  proven something true about the ring of algebraic
	  integers, which my research proves cannot be true,
	  where he said he relied on the theory of ideals."

Further, your insistence on rational solutions to this family
of polynomials (parametrized by f and x) relies on the specific
polynomial being reducible (the polynomial is monic, and any
rational root of a monic polynomial over Z is in Z: thus, the
polynomial factors over Z into [at least] a linear and a
quadratic factor). To claim that any coprimeness result here
contradicts the results of the theory of ideals is ludicrous:
there is no connection (nor can there be) between the roots
of different irreducible factors of a reducible polynomial.

In other words, you have an integral root, and a pair of
roots of the remaining quadratic factor. You find that the
integral root is divisible by something that is coprime to
the roots of the quadratic factor. That doesn't contradict
anything.

The only possible contradiction between what you want to
claim as your "new mathematics" and standard algebra is
between something that the standard theory actually does
say, and something *different* that your theory says, given
the same data. All this stuff with integer solutions of
an integral polynomial equation is totally irrelevant for
this purpose.

Come back with an integer polynomial that is irreducible over
Z, and an integer that is coprime to some, but not all, of the
roots of that polynomial, in the ring of algebraic integers.

The problem is that people don't accept the proof as acceptance of it
leads down to the conclusion that the theory of ideals is false, so
they challenge the proof and claim it's false.


BTW, claiming that "the theory of ideals is false" is silly,
considering the depth of your ignorance of the area. Cite one
theorem from this theory and refute it.

One theorem.

If the whole theory is false, as you seem to need to believe, then
it should be child's play, even for a know-nothing, to find one
statement at variance with the facts.

That's human weakness displayed--a proof is ignored for social reasons.

Ok, to counter human weakness I just put up a demonstration which shows
the mathematics following the proof.

Few of you can fail to understand that if all these integers follow the
proof then maybe human beings should as well, and if you investigate
the mathematics, you find it's simple and it makes sense that it passed
formal peer review and was published.

My results ARE correct and proven.

But there are people who lie about them.


And you can believe me, because I never lie and I'm always right.

I just found a way to show you directly and clearly that they do lie.

And it's something you can see as I've focused on integers.


As I said above: any claim that the integer root of a reducible
cubic is different from the roots of the remaining quadratic, by
virtue of divisibility properties wrt the standard integers, cannot
show a flaw in the theory of ideals [or with any part of ordinary
algebra], because algebra makes no claims connecting those two
things.

Perhaps you don't get the idea: to show a theory is "false", it
is necessary to demonstrate a claim that the theory actually makes,
and show it's at variance with the facts. Your claim regarding
rational roots of this polynomial cannot do that, since the standard
theory makes no claims regarding common factors (or absence thereof)
among such roots.

Some of you may have noticed that only a few actual integer cases have
so far popped up, as posters are probably dealing with a lot more, but
refusing to put them up, as all of them support me.

That's what I want those of you to focus on who don't have the math
software to do the searches yourself and see the result--some of these
people replying to me may be looking at thousands of results by now
with the mathematics behaving according to my theory, but refuse to put
that information up.

So they are lying to you.  Lying by omission because they have the
tools and the information, but they don't like the result.

Just remember, here an integer counterexample doesn't leave me room.

But integer examples proving my point start shutting down room for
disagreement, so the posters who, as I've told you over the years, lie
to you, will I'm sure not give you as many examples as they find.

If you have the math software, I suggest you run the searches yourself.

Hey, it's a brutal world.  People lie.  But the numbers don't lie.


Back to your typical accusatory style, I see.

Check.  I push you to the mathematics as it supports me.  Check the
numbers.


How about coming up with a claim that actually addresses something
that the standard theory says?


James Harris


Hey, I'm being objective and calm, can you do the same?

If so, stop calling us liars.

Dale.
.

Relevant Pages

  • Re: Resolving the as
    ... It has also been shown RIGOROUSLY that none of the roots ... > so obviously two of the a's have f as a factor, and one is coprime to ... For IRREDUCIBLE monic polynomials, ... with all algebraic integer factors of the constant term. ...
    (sci.math)
  • Re: Resolving the as
    ... James Harris wrote: ... > so the a's are the negatives of the three roots of the cubic ... without exception that "one a is coprime to f". ... assert that polynomials with rational roots should be treated ...
    (sci.math)
  • Re: Some math, algebraic integers
    ... > where A and B are integers, iff the quadratic is reducible, then roots ... > of it CAN be coprime to a prime factor of B, ... It is a property of the polynomials. ... must s_1 and s_2 be algebraic integers? ...
    (sci.math)
  • Re: Orthogonal polynomials (was Chebyshv, etc.)
    ... Legendre, Chebyshev, Hermite, etc.) have n real roots in the ... This general property of orthogonal polynomials is proved as ... you can simply ignore any zeros ... If alpha is a real root of phi_k, ...
    (sci.math)
  • Re: New paper, algebraic integers, Galois Theory
    ... > Now consider the case that m, f, and u are algebraic integers, then I ... > something about the factors of roots of monic polynomials with integer ... Note that this claim does not require Galois Theory, ...
    (sci.math)