Saturday, March 08, 2003


JSH: Magidin, Galois Theory, credibility

One of the more annoying things I faced months ago were assertions from Arturo Magidin that he'd used Galois Theory to prove that with

2x^3 - 3x - 2 = (a1 x + b1)(a2 x + b2)(a3 x + b3)

two of the a's couldn't have a factor of sqrt(2).

So I argued with him, and there were two threads that went over 200 posts I think it was, and I decided that was getting nowhere.

This time he's left Galois Theory out and I'm not surprised.

But just so he doesn't try to bring it back in, I can focus on

x^3 + 3x - 2

so that he has less room to confuse and note that only one of its roots is not coprime to 2.

Now I can prove that, while mathematicians don't have the techniques, unless they use mine, to tell you anything about distribution.

That's because Galois Theory can't distinguish between the roots.

Now sure you can give the roots with a lot of operators, which doesn't tell you anything. Believe me, I know as most of my mistakes in posts have been in trying to find some neat trick to break through Galois Theory with conventional techniques.

And you can't.

So now that I know I can't do it using Galois Theory or anything conventional, then I know that Magidin can't either, and his past statements were false.

How I manage to get through though is actually simple and an analogy comes quickly.

Consider if you were looking for factors of 3 in algebraic integers, having just been told the intriguing fact that 3 is not prime in that ring, when before you'd just been playing around with integers.

Now you want to look at a factor of 3. Well you can introduce the variable x, and use the polynomial

x^2 + 2x + 1 = 3

so now you have

x^2 + 2x - 2 = 0,

and can solve for x and plug a solution into the factorization

(x+1)(x+1) = 3

and then you can look at factors of 3.

So to look at factors in something like x^3 + 3x - 2, I introduce an additional variable that I call "m", so that I can factor the polynomial which is a step beyond, and I know it blows the minds of many of you. But the mathematical logic is direct. And Magidin's claims that "m" must be some kind of constant as if that changes anything don't matter anymore with my argument than they would with my
example above. Sure, x there can be considered to be a constant, with a particular x solution, but it's also a variable, and I can still factor x^2 + 2x + 1.

Now that analogy is direct, and I found myself fascinated that any of you would believe Magidin making a variable is not a variable argument, but I guessed you just wanted to believe.

So, I'm using advanced mathematics that has concepts just a slight step beyond what you're used to seeing. Which puts me in the position that others have been in, like those who had to argue about sqrt(-1), and convince people that it wasn't imaginary. The name still stuck though.

[A reply to Arturo Magidin, who raised objections to James' ideas and added “I shall await your replies”.]

Well, you're going to be waiting a while.

Come on Magidin, it's not as if I haven't told you that I post for a purpose.

Now all I have to do is get my paper complete, and send it to a journal.

I post drafts as usual and you can comment on those drafts, but there's no way I'm going to read through a ton of stuff.

So if you don't make very succinct replies then I'm not going to bother reading through your posts.

Don't overrate your importance. As far as I'm concerned your distinguishing characteristic is a fascinating ability to convince others of false math.

It's not as if you've added much. You've just often been very confusing.

Why don't you take a break? I could be throwing up threads for a while and replying to myself quite a bit. Though I think I finally got the gist of it down.

This page is powered by Blogger. Isn't yours?