Thursday, July 02, 2009


JSH: Why choice rules

On my math blog Penny Hassett had a comment that got me to thinking and after pondering at her suggestion what others believe to be true in their disagreements with me about a certain serious mathematical issue, I concluded that infinity could help to clear the air.

So I have simple examples yet again to start (believe me it gets really hard later so please pay close attention to the easy part):

7(x^2 + 3x + 2) = (7x + 7)(x + 2)

and let the ring be the ring of algebraic integers. And notice that reflects CHOICE. My choice for a simple example as the 7 can be factored an INFINITY of ways!!!

After all,

7(x^2 + 3x + 2) = (x + 1)(7x + 14)

is ALSO just as valid mathematically, and in fact there are an infinity of possible variations all equally valid.

Key here also to notice is that there are an infinity of choices FOR EVERY x, and remember the ring is the ring of algebraic integers.

So far easy and there should be no disagreement with the above!

Now to the hard mathematics.

Now we'll TRY to assume the ring is still the ring of algebraic integers and now a harder example:

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


7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1

and the a's are roots of

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

Notice the b's are chosen such that when x=0, both the b's equal 0, which is to say, they are normalized.

Now just like before you have human choice, and you can imagine moving that 7 around easily enough:

7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)

which is just one more human choice out of infinity.

And there are an INFINITY of possibilities for EVERY x. Every x has an infinite number of choices.

So how does the mathematics choose?

It doesn't. I did. I chose.

But that's where the fights start and the arguing and the dead math journal comes into the picture as what follows from mathematical logic and the rigidity of infinity is a conclusion that is devastating emotionally to thousands of mathematicians around the world which is that there is a core error: the ring of algebraic integers gives a false result.

Because now you have that one of the a's has 7 as a factor by the choice argument.

But the a's are roots of

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

and at x=1, that gives

a^2 - 6a + 35 = 0

and provably in the ring of algebraic integers NEITHER of the roots can have 7 as a factor and there is shown the direct contradiction.

Notice the ring of algebraic integers does not fail gracefully and its error is not fixable as your alternative is to introduce some mystical force to choose out of infinity, like, maybe God?

(Posters have been remarkably vague in past arguments about how a particular factor of 7 arises, often simply repeating what is taught in math classes about ways to find the factors at any particular x, but remember there are an infinity of choices for EVERY x, so the choice problem gives you infinity at every point. So who chooses? God?)

But not even God can choose for you here if you go with mathematical logic. Of course, you can like so many human beings before you simply choose to believe in what you've been taught, even though it is very, very wrong.

The mystical nature of the disagreement with me is remarkable but I think it's natural for human beings to turn to spirits or mystical forces in the face of something difficult to understand. That's what your ancestors did.

But then you're no longer really mathematicians.

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