Friday, December 27, 2002
JSH: Finally, something you can *see*
Posting my challenge helped me to see something as in response to some replies I went ahead and had the math software calculate out the answers. Then I looked at the answer for the a's and b's and realized that you can actually see that only two of the a's can have factors of 2!!!
Now I'm thinking about explaining that without totally giving away my challenge as I want those who spend so much time attacking my creative output to show that they actually know enough mathematics that their opinion is worth your time.
Basically the factorization of 2x^3 - 3x - 2, gives answers that luckily enough for me easily show that only two of the a's have a factor of 2, where that factor is sqrt(2).
Now understanding that evidence is going to require that some ground is covered, which I'll do in this post, while I leave my challenge up, so I won't give all the answers here.
And yes, the factorization of 2x^3 - 3x - 2 shows a lot, luckily for me.
Ok, so let's talk about cuberoots.
To understand my evidence you need to accept that given something like
(1 + Sqrt[2])^(1/3) = 1 + s
where s is an algebraic integer, s has non unit factors of 2.
That's it. If that doesn't bother you, then I can show you easily.
Again, you need to accept that with
(1 + Sqrt[2])^(1/3) = 1 + s,
s must have non unit algebraic integer factors of 2.
I'm going to leave it at that, but I have to say I'm very excited.
After so many battles to FINALLY be in a position to demonstrate to the rest of you with what hopefully is a simple enough example is making me optimistic.
So now what happens is I wait, as I look to see responses to my challenge, and to what I've said here about 1+s.
Oh, and it is rather interesting that numbers had a surprise for all of us, as even I hadn't considered that with an algebraic integer, say w, where
w = sqrt(2)x
that x doesn't have to be an algebraic integer.
Whew!!! It's nice to be in a position to demonstrate. I'm not naive enough though to think it'll be easy, so I'm going to be structured in how I present.
First the challenge, and the s thing. Then the evidence.
But, if you're one of those people who can simply figure it all out yourself, then you can jump ahead and see for yourself.
Just factor 2x^3 - 3x - 2, and look at the a's and b's.
That is look at the factorization
2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)
with the a's and b's algebraic integers.
Today is a rather big one for me because I can answer critics of my FLT proof with a factorization of a polynomial. Though I feel a need to point out that the mathematical argument supporting my results is solid.
I don't know if any of you can imagine what it's been like to have a correct FLT proof, while people have been claiming otherwise.
It is exciting though to have pushed number theory with a demonstration of numbers which are units and algebraic integers, while their inverses are units but not algebraic integers.
Some of you may be wondering what's the ring then. Well, it's what I call an object ring, and I figure some of you may think you have room to criticize me because you don't understand the mathematics.
I suggest you pause.
You're at the edge of human understanding of number. It can take a while to soak in, or you may simply back away from the intensity, as you're in extreme mathematics now. Not everyone goes for the extreme.
I'm expecting some challenges to my post explaining the factorization of 2x^3 - 3x - 2, and some of you may wish to see portions of it expanded.
That's fine.
Ok, so now you can finally see with an actual polynomial a factorization where only two of the a's have a factor of f.
The numbers are rather interesting in a way that can bend your mind. It kind of reminds me of when I was a student contemplating quantum mechanics or Einstein's theory of relativity in that way.
I guess it only makes sense that number theory should get as strange as some of the hardest to understand concepts in modern physics.
However, no matter how quirky it seems, or how disconcerting it may be to have your intuition totally lost, remember that it's mathematical logic that counts.
If you wish to dispute my results, find an error in the mathematical argument!!!
Now I'm thinking about explaining that without totally giving away my challenge as I want those who spend so much time attacking my creative output to show that they actually know enough mathematics that their opinion is worth your time.
Basically the factorization of 2x^3 - 3x - 2, gives answers that luckily enough for me easily show that only two of the a's have a factor of 2, where that factor is sqrt(2).
Now understanding that evidence is going to require that some ground is covered, which I'll do in this post, while I leave my challenge up, so I won't give all the answers here.
And yes, the factorization of 2x^3 - 3x - 2 shows a lot, luckily for me.
Ok, so let's talk about cuberoots.
To understand my evidence you need to accept that given something like
(1 + Sqrt[2])^(1/3) = 1 + s
where s is an algebraic integer, s has non unit factors of 2.
That's it. If that doesn't bother you, then I can show you easily.
Again, you need to accept that with
(1 + Sqrt[2])^(1/3) = 1 + s,
s must have non unit algebraic integer factors of 2.
I'm going to leave it at that, but I have to say I'm very excited.
After so many battles to FINALLY be in a position to demonstrate to the rest of you with what hopefully is a simple enough example is making me optimistic.
So now what happens is I wait, as I look to see responses to my challenge, and to what I've said here about 1+s.
Oh, and it is rather interesting that numbers had a surprise for all of us, as even I hadn't considered that with an algebraic integer, say w, where
w = sqrt(2)x
that x doesn't have to be an algebraic integer.
Whew!!! It's nice to be in a position to demonstrate. I'm not naive enough though to think it'll be easy, so I'm going to be structured in how I present.
First the challenge, and the s thing. Then the evidence.
But, if you're one of those people who can simply figure it all out yourself, then you can jump ahead and see for yourself.
Just factor 2x^3 - 3x - 2, and look at the a's and b's.
That is look at the factorization
2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)
with the a's and b's algebraic integers.
Today is a rather big one for me because I can answer critics of my FLT proof with a factorization of a polynomial. Though I feel a need to point out that the mathematical argument supporting my results is solid.
I don't know if any of you can imagine what it's been like to have a correct FLT proof, while people have been claiming otherwise.
It is exciting though to have pushed number theory with a demonstration of numbers which are units and algebraic integers, while their inverses are units but not algebraic integers.
Some of you may be wondering what's the ring then. Well, it's what I call an object ring, and I figure some of you may think you have room to criticize me because you don't understand the mathematics.
I suggest you pause.
You're at the edge of human understanding of number. It can take a while to soak in, or you may simply back away from the intensity, as you're in extreme mathematics now. Not everyone goes for the extreme.
I'm expecting some challenges to my post explaining the factorization of 2x^3 - 3x - 2, and some of you may wish to see portions of it expanded.
That's fine.
Ok, so now you can finally see with an actual polynomial a factorization where only two of the a's have a factor of f.
The numbers are rather interesting in a way that can bend your mind. It kind of reminds me of when I was a student contemplating quantum mechanics or Einstein's theory of relativity in that way.
I guess it only makes sense that number theory should get as strange as some of the hardest to understand concepts in modern physics.
However, no matter how quirky it seems, or how disconcerting it may be to have your intuition totally lost, remember that it's mathematical logic that counts.
If you wish to dispute my results, find an error in the mathematical argument!!!
# posted by James Harris @ 16:43 
