Thursday, March 09, 2006


JSH: Putting it all together

What continually impresses me about the mathematics which proves my case is that it is so simple, and the proof relies on such basic logical principles—yet people can argue about them, and most mathematicians can ignore the results!

The mathematical story starts with the distributive property, commonly expressed as

a*(b+c) = a*b + a*c

where my brilliant and clever idea (if I do say so myself) means that you have the distributive property with functions, so you need to consider something like

a*(f(x) + b) = a*f(x) + a*b

and my dramatic, awe inspiring, and controversial position, over which mathematicians on this newsgroup have argued for years, and a paper got published and retracted by a journal which later keeled over and died, is that the value of f(x) <gasp> is irrelevant to the operation of the distributive property.

Yup, that's it. Sorry if you're disappointed given the build-up, but my position which is so controversial, over which so much discussion has gone on for years is that the value of f(x) has no meaning on

a*(f(x) + b) = a*f(x) + a*b

as if you accept that as true then when f(x)=0 is not a special case where you can have behavior one way, and then when x doesn't equal 0, have it go another.

If you think that can't be it that people couldn't argue against that position for years, I'm going to quickly make it more complicated for you by giving you the result valid over the complex plane that given

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

where all the functions go to 0 at x=0, it must be true then, by the distributive property that

A(x) = 7A'(x) and B'(x) = B(x)

where I use the convention that A is the first function and B the second, though if you hate ordering conventions, you might argue that possibly

B(x) = 7A'(x).

And if you think that doesn't follow from the distributive property, think again.

After all, you can just check—get this—at x=0, as all the functions go to 0, and find that 7 must have multiplied through the first as you just have

7(0 + 1)(0 + 1) = (0 + 7)(0 + 1)

and if the value of the functions is irrelevant to the distributive property, then you know that 7 multiplied through the first for ALL x.

You see, the distributive property doesn't care what value x is…which is the crucial point that sci.math'ers have argued against for years, claiming that x=0 is a "special case".

What can you do in the face of irrational beliefs?

If some person says that 2+2=5, and you hold up two fingers, and then hold up two more and they say, yup, five fingers, what can you do?

You may think you can do something, but I'm here to tell you that if people wish to just argue away something and deny, they will just deny, just like posters deny that they're arguing against the distributive property when they do just that repeatedly.

So why would they? What's so important that people would argue about it for YEARS anyway?

Well, if you accept the distributive property and accept that it works for functions too without caring about what value they have then with a few simple ideas I call non-polynomial factorization you can prove that with the quadratic generator:

a^2 -(1+fx)a + (f^2 x^2 + 2fx) = 0

given algebraic integers f and x, it must be true that f is a factor of only one of the roots, but then you can also prove that in the ring of algebraic integers, if the roots are non-rational then f cannot be a factor of EITHER of the roots, so you have this nifty apparent contradiction.

Posters arguing with me say that I'm wrong, but remember, my results rely on the distributive property, so the way mathematics works, they are attacking the distributive property as that's the linchpin of the result—the keystone.

If you think, hey, maybe the distributive property DOES fail, then, um, you're in that 2+2=5 crowd, and you may as well move along, nothing for you here.

For the rest of you, the resolution to the apparent contradiction is that the ring of algebraic integers is missing some numbers—those numbers that would allow 7 to be a factor of one of the roots—and is doing it in a special way.

My favorite analogy when trying to introduce people to the problem with the ring of algebraic integers is to consider evens as a ring, as notice 2 and 6 are then coprime—because 3 is NOT even.

But now we're getting into heavy duty territory, as I'm challenging standard usage of Galois Theory, and it turns out, also the ideal theory, as the position that the ring of algebraic integers IS complete and cannot have the problem I describe, relies on, guess what, ideal theory!!!

Didn't know that?

Yup. The position that the ring of algebraic integers cannot have the problem I claim it does, relies on ideal theory. And ideal theory is THE basis for that position, so that it is just, well, it's just ideal theory!

But I can explain to you why I think ideal theory doesn't work very quickly, as, you see, it dosn't take into consideration convergent infinite series.

You see, if you append ½ to the ring of integers, you get the field of reals. If you append ½ and i to the ring of integers, you get the field of complex numbers.

It's like a switch. If you break the rules that give you meaningful coprimeness, then you break them completely. You can't break them halfway!!!

So what are those rules? Not to worry, I figured them out:
  1. 1 and -1 are the only rationals that are units in the ring.

  2. Given a member m of the ring there must exist a non-zero member n such that mn is an integer, and if mn is not a factor of m, then n cannot be a unit in the ring.

Those are THE RULES that are necessary for coprimeness to meaningfully hold in a ring, so that you can talk about factors and say things like 2 is coprime to 3.

If you break those rules, then you lose out on integer-like behavior.

If you append ½ to the ring of integers, you break those rules, so ideal theory fails and specifically, convergent infinite series step in, and you get the field of reals.

Denial is a potent thing. Mathematicians get taught things are a certain way, and I see a lot of posters talking about definitions. They seem to think these definitions are what mathematics is.

But mathematics is about what follows logically from the basic axioms.

Human beings can define away but if the definitions aren't in line with the basic axioms and what follows logically, then the definitions DO NOT HOLD mathematically.

It's like, define 2+2=5, so?

So there, you can see why mathematicians would argue against the distributive property for years as my ideas are revolutionary.

I can go from the distributive property to overthrowing ideal theory, and do it in a couple of pages using mostly quadratics and basic algebra.

No other achievement is like it in human history, resistance to it is unlike any other in human history.

Mathematicians do not want to admit being wrong, so I get to argue with people fighting the distributive property!

Remember the logical chain and how quickly it got hard for some of you, how I went from

a*(b+c) = a*b + a*c


a*(f(x) + b) = a*f(x) + a*b

with the DRAMATIC CLAIM that the value of f(x) doesn't matter to the distributive property, to pushing on you that given

7(A'(x) + 1)(B'(x) + 1) = (A(x) + 7)(B(x) + 1)

where all the functions go to 0 at x=0, it must be true—by the distributive property—that

A(x) = 7A'(x) and B'(x) = B(x)

and next thing you know I'm talking about the failure of ideal theory!!!

That's mathematics.

The logic flows from axioms by logical steps to a conclusion which then must be true.

Human beings though, well, they're quirky. When they get wrapped up in some idea, they can be very ornery about letting it go, even when it's been mathematically proven to be false.

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