Math Forum / Mathematics / General Topics / July 2009

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)

where

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.

James Harris