Friday, September 17, 2004
Some math, algebraic integers
Luckily for me the math that explains my work is actually quite simple in many ways. All that is required is that you accept algebra. That might sound weird, but pay attention and you'll understand by the end.
Now with rationals you can have a simple case like
2x^2 + 5x - 3 = 0
where the roots are x=1/2 and x=-3, if I did my algebra correctly.
Now notice you have an integer paired with a fraction, which, of course, is NOT an integer.
Now consider some numbers q_1 and q_2, which I won't say a lot about now, though in a bit I'll put them in a particular ring, where you get a polynomial with
x = q_1/2 and x=-3q_2 as the roots.
That polynomial looks like
2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0
and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.
It turns out that q_2 can't be an algebraic integer.
Mathematicians at this point make an assumption, which is that q_1 MUST have non-unit algebraic integer factors in common with 2.
Well, yeah, but the implication is wrong.
That might sound complicated or weird, but think about the rational example, where you have roots 1/2 and -3, and notice that -3 does not have any non-unit factors in common with 2, but somehow, someway if you have irrational factors, for some reason 3q_2 MUST have non-unit factors in common with 2?
It begs the imagination that despite the infinity of possibility that 3q_2 is so constrained, and in fact, it's not.
Mathematicians made a mistake.
It turns out that q_1 and q_2 CAN be in a ring where 1 and -1 are the only integers that are units, and to be more specific just in case some are still confused, where 1 is the only NATURAL that is a unit.
I figured out a rather basic way to prove that using some very simple algebra.
To date, no one had shown an error in my proof, though the paper in which I go through the proof did have at least on minor error that has been pointed out by a sci.math poster.
What I need is fairness. I can go through the details of that paper, point-by-point, step-by-step and show how it follows from basic algebra.
But some posters cheat, and they are the same posters who succeeded with that email campaign against my paper.
These people don't play. They are very good at manipulating the bulk of you, and they have an agenda at hiding the mathematical truth.
My challenge to you is to stand up for yourselves and actually THINK.
Time will tell if any of you have what it takes to be a real mathematician.
Here it will require that you focus on the math, and accept algebra, no matter what tricks posters throw at you.
So far, most of you have failed…but I have a little bit of faith left.
There may be hope yet.
Now with rationals you can have a simple case like
2x^2 + 5x - 3 = 0
where the roots are x=1/2 and x=-3, if I did my algebra correctly.
Now notice you have an integer paired with a fraction, which, of course, is NOT an integer.
Now consider some numbers q_1 and q_2, which I won't say a lot about now, though in a bit I'll put them in a particular ring, where you get a polynomial with
x = q_1/2 and x=-3q_2 as the roots.
That polynomial looks like
2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0
and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.
It turns out that q_2 can't be an algebraic integer.
Mathematicians at this point make an assumption, which is that q_1 MUST have non-unit algebraic integer factors in common with 2.
Well, yeah, but the implication is wrong.
That might sound complicated or weird, but think about the rational example, where you have roots 1/2 and -3, and notice that -3 does not have any non-unit factors in common with 2, but somehow, someway if you have irrational factors, for some reason 3q_2 MUST have non-unit factors in common with 2?
It begs the imagination that despite the infinity of possibility that 3q_2 is so constrained, and in fact, it's not.
Mathematicians made a mistake.
It turns out that q_1 and q_2 CAN be in a ring where 1 and -1 are the only integers that are units, and to be more specific just in case some are still confused, where 1 is the only NATURAL that is a unit.
I figured out a rather basic way to prove that using some very simple algebra.
To date, no one had shown an error in my proof, though the paper in which I go through the proof did have at least on minor error that has been pointed out by a sci.math poster.
What I need is fairness. I can go through the details of that paper, point-by-point, step-by-step and show how it follows from basic algebra.
But some posters cheat, and they are the same posters who succeeded with that email campaign against my paper.
These people don't play. They are very good at manipulating the bulk of you, and they have an agenda at hiding the mathematical truth.
My challenge to you is to stand up for yourselves and actually THINK.
Time will tell if any of you have what it takes to be a real mathematician.
Here it will require that you focus on the math, and accept algebra, no matter what tricks posters throw at you.
So far, most of you have failed…but I have a little bit of faith left.
There may be hope yet.
# posted by James Harris @ 19:39 
