The actual math, advanced polynomial factorization

From: James Harris (jstevh_at_msn.com)
Date: 09/25/04 

Date: 25 Sep 2004 13:31:44 -0700

The following factorization is useful:

f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) =

    (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

Note that x, m, f, and u are all independent variables. It's a lot of
variables, yes, but they're independent of each other.

Also note that a_1, a_2 and a_3 are dependent variables.

Now divide both sides by f^2 to get

(m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =

    (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

and all the arguing for over TWO YEARS NOW has been over what that
means.

Let m=0, and you get

(0*f^4 - 3* 0*f^2 + 3*0) x^3 - 3(-1 + 0*f^2 )x u^2 + u^3 f =

    (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

which is

-3(-1)x u^2 + u^3 f = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

which is

u^2(3x + uf) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

and it's clear that two of the a's have to equal 0, while one equals
3.

But that means that the f^2 divides through two of the factors

(a_1 x + uf), (a_2 x + uf), or (a_3 x + uf)

and, remember, x, m, f and u are all independent variables.

Therefore, two of the factors have f as a factor in general without
regard to the value of m, but it turns out that IN THE RING OF
ALGEBRAIC INTEGERS they do not always have f as a factor. Notice that
"not always", it's important.

Some posters rather than accept that x, m, f and u are independent
variables so their value doesn't matter to each other, claim my result
only holds if m=0, as a special case, but if variables are independent
of each other, how can a particular value be a special case?

So why fight the result? Well Evariste Galois came up with the theory
for how you can *represent* roots of irrational polynomials.

What I've shown is an analysis tool for actually figuring out factors
with irrationals, but no one had tools of this power before, and with
the results of Galois they made a leap: They decided that the work of
Galois was not just about representation, but also about actual
factors.

Now the algebra supporting my case is simple enough, but I've been
arguing for over two years on the same points and posters have
routinely rejected basic algebra, like that independent variables are
independent, while making up their own kind of weird mathematics I've
at times called voodoo mathematics.

Let me show you how it works. Like they'll say there exists algebraic
integer functions w_1(m), w_2(m), and w_3(m), where

w_1(m) w_2(m) w_3(m) = f, and

a_1 = b_1 w_1(m), a_2 = b_2 w_2(m), and a_3 = b_3 w_3(m)

where the b's are algebraic integers.

Now with

(m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =

    (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2

you then have

(m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f =

  (b_1 x + u w_2(m)w_3(m))(b_2 + uw_1(m)w_3(m))(b_3 x + uw_1(m)w_2(m))

where all might seem well and good, like why can't you just have these
functions that somehow relate INDEPENDENT VARIABLES but I've said
that before and the sci.math'ers make fun of me.

There's just one small little problem with that setup, f^2 was
multiplied times both sides before, but what if I play a trick on the
voodoo math?

Let's say that rather than multiply both sides by f^2, I multiply
times 13^2?

You see, the point of have some constant multiple like f^2 is that if
you divide it off it's gone, like if I have

4(x^2 + 3x + 2) = 4(x+1)(x+2)

and divide 4 off, it's gone. But not in the voodoo math of sci.math
posters...if the factorization is irrational. They always come back
to say, yeah, if you're talking about polynomial factors like in my
simple example, then yeah, that's how constant multiples behave!!!

It gets weirder. My result holds for a family of polynomials.

To see that let's make some adjustments:

f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and multiplying through by f^2 gives

(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x u^2 f^2 + u^3 f^3

where notice in the last two terms you have u and f paired, so let
y=uf, so you get

(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3

and the factorization now is

(m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - 3(-1+mf^2 )x y^2 + y^3 =

    (a_1 x + y)(a_2 x + y)(a_3 x + y)

and my result from before still applies because the a's are not
dependent on the value of y, while still having the same values as
before, so now I can change y, as yet another independent variable, so
let's put some numbers in.

Let f=sqrt(2), m=1, y=1, which gives

(4 - 3(2) + 3)(2)x^3 - 3(-1 + 2)x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x
+ 1)

which is

2x^3 - 3x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)

and you can just look at it and see that x=1, works as a root of the
polynomial 2x^3 - 3x + 1.

That leaves room for only two of the a's to have sqrt(2) as a factor,
and in this case you can see it work IN the ring of algebraic
integers.

Notice that if you wish you can try your own numbers.

Mathematically the results are always the same without regard to how
reducible over rationals the primary cubic may be.

That's how math works.

If Galois Theory wasn't such a big deal there wouldn't have been all
the arguing, which has gone on for over two years now.

So, mathematically, my case is solid and absolute, but math society is
protecting its wrong beliefs about roots of polynomials irreducible
over Q, and to do that the people engaging in that activity cannot be
real mathematicians.

They're social animals. A real mathematician would just accept the
math as correct and disdain the earlier false assumption as it's just
wrong.

Mathematicians don't defend wrong beliefs about mathematics, as they
actually love mathematics itself.

So, if you fight a correct result, to hold on to a false belief, you
prove you're not really a mathematician, no matter what people call
you or you call yourself.

That's the problem I think with the modern math world as somehow the
belief has taken hold that you have a social position, not a calling,
and that as long as society calls you a mathematician, then you are,
no matter how false mathematically what you believe is.

James Harris

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