For about two years now I've been trying to explain an algebraic
method for analyzing polynomials that relies on non-polynomial
factors. My research is a natural extension in the polynomial domain
of the idea of irrationality in the integer domain. Basically, I look
at the equivalent of irrational factors with polynomials.

Recently Rick Decker, a professor at Hamilton College, apparently
trying to refute my research came up with a quadratic example, which I
like because it's a quadratic, and easier to manipulate than the
cubics I've used before.

If you wish to see his original post here are some headers which also
show that he is indeed at Hamilton College:

Message-ID: <3FF47C4C.6080109@hamilton.edu>
Date: Thu, 01 Jan 2004 15:00:12 -0500
From: Rick Decker <rdecker@hamilton.edu>
Newsgroups: sci.math
Subject: Re: Mathematical consistency, courage

Decker put forward the quadratic

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

where his a's are roots of

a^2 - (x - 1)a + 7(x^2 + x).

The factors (5a_1(x) + 7) and (5a_2(x) + 7) are examples of
non-polynomial factors.

Notice that despite not being polynomials they are algebraic integers
if x is an algebraic integer because a_1(x) and a_2(x) are the two
roots of

a^2 - (x - 1)a + 7(x^2 + x).

To my knowledge mathematicians have never done extensive research with
non-polynomial factors of polynomials, but instead most work in the
area has to do with finding roots to polynomials, or determining if
polynomials with rational coefficients are reducible over rationals.

The extension into the realm of non-polynomial factors has revealed a
problem with a previous understanding of mathematicians in the area of
the ring of algebraic integers. It's easy to prove the problem.

Consider that algebraically, factorizations multiply out in a certain
way demonstrated by

(a+b)(c+d) = ac + ad + bc + bd

and that's just a FACT which is not open to debate.

Looking at

(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)

it is possible then to multiply out the factors on the left to obtain

25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 49 = 7(25x^2 + 30x + 2).

Now subtracting 14 from both sides gives

25 a_1(x) a_2(x) + 35 a_1(x) + 35 a_2(x) + 35 = 7(25x^2 + 30x)

which reveals an imbalance as 35 is the constant left on the left,
while 0 is what's on the right.

A simple linear transformation helps balance the factorization, as
using

a_2(x) = b_2(x) - 1, and making the substitution gives

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2).

Now multiplying out I get

25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x + 2)

and subtracting 14 from both sides gives

25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x)

so as expected the constants 7 and 2 on the left are factors of 7(2)
on the right.

The mathematics is trivially obvious in that regard, but from that
simple result, obvious from basic arithmetic, I have a truly profound
conclusion.

That is, given that with

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)

the 7 and 2 on the left are factors of 7(2) on the right, if I divide
that constant factor 7, visible as a factor of

7(25x^2 + 30x + 2)

from both sides then to remain in the ring of algebraic integers,
there's only one way logically to do it:

(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2

as any other way, like factoring 7 into two other non-unit factors
will require that you have numbers not available in the ring of
algebraic integers.

For instance, assume there exists some value of x within the ring of
algebraic integers where the factors (5a_1(x) + 7) and (5b_2(x) + 2)
each have sqrt(7) as a factor (in fact, x=1 will work), then you'd
have

(5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) =

25x^2 + 30x + 2

which reveals the two factors of 2 on the right, as being sqrt(7) and
2/sqrt(7) on the left.

Remember, algebra gives a rigid format for multiplying out, as I
pointed out before with

(a+b)(c+d) = ac + ad + bc + bd

so there shouldn't be debate here from people who accept algebra.

Notice that works in the field of algebraic numbers, but not in the
ring of algebraic integers.

But it turns out that it's possible to prove that even with the
factorization

(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2

that (5a_1(x)/7 + 1) is not in general an algebraic integer.

What makes this result so surprising is that the necessary conclusion
is that you cannot operate completely in the ring of algebraic
integers if you divide both sides of

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)

by 7, as then *necessarily* from basic algebra, you're forced out of
the ring as you then must have factors of 2 that are outside the ring
of algebraic integers, as I demonstrated with sqrt(7) as *necessarily*
then the factors of 2 are sqrt(7) and 2/sqrt(7) which are not in the
ring of algebraic integers.

Extensions of mathematical knowledge tend to produce surprising
results, and some people can't handle a world where areas once thought
settled are revealed to contain new results. But my hope is that some
of you are actually mathematical researchers versus seeing yourselves
as mere caretakers of dogma.

For more on my research in this area and a broader overview, please
see my blog archives:

http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html


James Harris

For about two years now I've been trying to explain an algebraic
method for analyzing polynomials that relies on non-polynomial
factors.

For about two years now I've been trying to explain an algebraic
method for analyzing polynomials that relies on non-polynomial
factors. My research is a natural extension in the polynomial domain
of the idea of irrationality in the integer domain. Basically, I look
at the equivalent of irrational factors with polynomials.

But it turns out that it's possible to prove that even with the
factorization

(5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2

that (5a_1(x)/7 + 1) is not in general an algebraic integer.

What makes this result so surprising

is that the necessary conclusion
is that you cannot operate completely in the ring of algebraic
integers if you divide both sides of

(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)

by 7,