### Friday, October 31, 2008

## Defend algebraic integers, simple challenge

I have found a problem with the established view of something called the ring of algebraic integers, but rather than acknowledge the issue the mathematical community has tried to hide from it, so here is some very basic algebra with a simple request for those defending the classical view:

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

I'll demonstrate how I do it.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

My hope is that a simple demonstration of their inability to fulfill a basic request will help some of you see the truth here as it is EXTREMELY important. The problem being highlighted here came into the mathematical field in the late 1800's so there is a devastating impact that has snowballed for over a hundred years.

Proving the problem is easier than getting people to accept that it exists.

For more information please Google: non-polynomial factorization

It is a term I invented to describe the type of factorizations shown above, as they are not polynomial factorizations. An example of a polynomial factorization is

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

I advanced the field with non-polynomial factorizations, found an astounding error, and mathematicians ran from it, and even destroyed a mathematical journal to try and keep it hidden. Google: SWJPAM

Given

175x^2 - 15x + 2 = (5b_1(x) + 2)(5b_2(x)+ 1)

Step through getting the factorization:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0.

I'll demonstrate how I do it.

Multiply both sides by 7 in a particular way which is my controversial way:

7*(175x^2 - 15x + 2) = (5b_1(x) + 2)(5*7*b_2(x)+ 7)

and now use the substitutions b_1(x) = a_1(x) + 1, and 7*b_2(x) = a_2(x), to get

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

and you're done. Note that b_1(x) and b_2(x) are given by this method. That is also a requirement for any answers from the mathematical community: give b_1(x) and b_2(x).

My hope is that a simple demonstration of their inability to fulfill a basic request will help some of you see the truth here as it is EXTREMELY important. The problem being highlighted here came into the mathematical field in the late 1800's so there is a devastating impact that has snowballed for over a hundred years.

Proving the problem is easier than getting people to accept that it exists.

For more information please Google: non-polynomial factorization

It is a term I invented to describe the type of factorizations shown above, as they are not polynomial factorizations. An example of a polynomial factorization is

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

I advanced the field with non-polynomial factorizations, found an astounding error, and mathematicians ran from it, and even destroyed a mathematical journal to try and keep it hidden. Google: SWJPAM