Re: JSH: Inconsistency with algebraic integers

On May 25, 3:05 pm, jankri...@xxxxxxxxxxx wrote:
Perhaps you (or someone else) could clarify whether you consider the
set of algebraic integers to form a ring or not?

---
J K Hauglandhttp://home.no.net/zamunda

I do. My issue is not with whether or not taking the roots of monic
polynomials with integer coefficients will give you a ring or not, as
I agree that it does.

But I prove that the ring you get is inconsistent, as it allows you to
appear to prove one thing, which can be proven to be false in a field.

I give the argument in my papers which appears to prove one thing,
like that for

x^2 - 6x + 35 = 0

one root has 7 as a factor in the ring of algebraic integers, though
it does not, which can be proven from a field.

I've emphasized that my argument relies on identities, and one
conditional, which is easily shown to be in the ring of algebraic
integers with algebraic integer x.

Factorizations are identities so with

175x^2 - 15 x + 2 = (f(x)+2)(g(x)+1)

you have an identity, where the paper goes through figuring out how
you can solve for f(x) and g(x).

I've done more editing to the paper today as I realized I could focus
on there being only identities used plus one conditional:

r^2 + rs - (2 + 2xt + tQ(x))st + s^2 = (1+2xs+sQ(x))t^2

Now I answered a post from "marcus" in this thread where he wants to
attack factorizations being identities, but I'd hope you'd be a bit
more mathematically sophisticated, and if you are then you have to
admit that my paper uses identities all over the place.

Even when I introduce Q(x) I add and subtract it.

There is just no way logically for identities to introduce conditions.

So there has to be at least one conditional expression among all the
identities as logic requires.

And only that expression can drive conditions, so if it is in the ring
of algebraic integers, it is reasonable to suppose you would be in
that ring, but taking roots of monic polynomials with integer
coefficients you CANNOT have a monic quadratic with integer
coefficients and non-rational roots where just one root has 7 as a
factor.

So the ring of algebraic integers has no choice, and despite the
conditional expression being within it, it cannot allow g(x) to be in
the ring as by its defining rules it cannot have one non-rational root
of a monic quadratic with integer coefficients with 7 as a factor
while the other does not.

Notably you CAN have BOTH roots with 7 as a factor in the ring.

OR you CAN have integer solutions within the ring where one root has 7
as a factor.

But you CANNOT in the ring have one non-rational root with 7 as a
factor while the other does not if they are roots of the same monic
quadratic with integer coefficients.

And that is the arbitrary exclusion, which is analogous to in evens,
the declaration that you only take even numbers keeping 2 from being a
factor of 6 in evens.

The problem is one I've proven multiple ways over years.

The issue here is not about whether or not I can prove that the ring
of algebraic integers is inconsistent allowing you to appear to prove
something that is false. As I can and have.

The problem is that a lot of people built their careers or sense of
self on working through problems wrong, getting wrong answers, using
faux mathematics, and now there is a strong resistance to the truth.

Faux mathematics can be appealing because it allows you to "prove" any
number of things that are not true.

Which is why mathematics is harder than many of you clearly realized
or maybe you'd have stayed out of the field.

As now with a real test of your love of proof, most of you are simply
failing, like that "marcus" guy who has now moved to fighting
identities!

The sad thing is that faux mathematics can make mathematics seem easy.

So if you had the correct mathematics, many of you might not have
managed to even think you were in the field, as it would have been too
difficult for you, while arcane rules and techniques wrapped around
the flaw in algebraic integers might have been difficult to learn, and
gave many of you a sense of intellectual strength from difficulty and
abstruseness, while still being wrong, so you could feel like a
mathematician.

Kind of like actors playing at something, doing a role, but not having
the real ability of the actual.

Mathematics is about proof.

It's not about just having complicated looking stuff, or having to
study weird looking things for hours until someone tells you that you
have a handle on it.

It is about proof.


James Harris

.

Relevant Pages

  • Re: JSH: A simple error
    ... algebraic integers routinely move outside the ring. ... the tautology ... Extreme Mathematics Google group, as is my latest paper which is more ...
    (sci.math)
  • Re: New paper, algebraic integers, Galois Theory
    ... >>You're suggesting that it is possible that one of the roots of the ... existence of common factors requires the context of a ring. ... can find the common factors whose existence you're apparently disputing. ... You have maintained that the ring of algebraic integers is somehow ...
    (sci.math)
  • Re: JSH: A simple error
    ... algebraic integers routinely move outside the ring. ... I start with a modular tautology in terms of variables r, s, ... Extreme Mathematics Google group, as is my latest paper which is more ...
    (sci.math)
  • Re: JSH: Math journals do not just die
    ... Yup, the Southwest Journal of Pure and Applied Mathematics, a minor ... And what electronic journals are very old? ... DECLARED to be in the ring of algebraic integers definitely were not at ... simply enough are roots of monic polynomials with integer coefficients. ...
    (sci.skeptic)
  • JSH Summery #4
    ... the 7 pop up, in only one factor, so that the mathematics clearly shows I'm ... usage of Galois Theory and the theory of ideals. ... where, oddly enough, you are forced to have a ring with fractions!!! ... with the Decker equations, in the ring of algebraic integers, where the goal ...
    (sci.math)