### Monday, November 30, 2009

## JSH: Ranting aside, a remarkable error

When people refuse to acknowledge even the most basic mathematics it can be amazingly hard to get your point across where with this particular error, a hundred years plus of research built on the error gives some people (unfortunately) a lot of motivation to just fail as mathematicians in the face of it.

In a lot of ways it's a trivial demonstration, given

7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

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

the mathematics does not allow that 7 to be controlled by the factorization of what it is multiplying, so you have with a simple example:

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

a CHOICE.

That's because it could be: 7(x^2 + 3x + 2) = (x + 1)(7x + 14).

Now I've pointed this out many times over the years as I've repeatedly explained this error!

No person thinking rationally would suppose that 7 is being told how to multiply by mysterious forces. I used to say, years ago, that the tail does not wag the dog.

So, ok, a non-standard factorization reveals a bizarre problem with the ring of algebraic integers. The construct creatively FORCES the roots of monic polynomial with integer coefficients when x is an integer i.e.

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

to be part of a factorization:

7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

so yes you can solve for the a's with the quadratic formula:

a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2,

a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

And you can make the substitutions and see that it all works. If Dedekind had been given this demonstration over a hundred years ago, you would not be learning the ring of algebraic integers today, except maybe as an oddity: a fairly useless ring which has a fatal flaw.

But a hundred years of error built up around this error.

It takes away the usefulness of Galois Theory. It removes the ring of algebraic integers as a useful ring.

Ok, since I found this error students have continued to be indoctrinated into it. That is a lot of human waste.

Professors who keep feeding it to students may in some cases be functionally insane due to the enormity of the problem.

Or they may just be cynically determined not to face consequences of it. Who knows.

But if you learn it, you are training your brain useless knowledge which does not work.

The math done with it is pure—purely wrong.

So yeah, they'll teach you style. It's style WITHOUT substance.

They force style because the math itself is wrong. Style crap also helps to keep out outsiders like me!

Rigor in error is still error. You are just rigorously wrong.

The problem is a remarkable error. It has taken over a hundred years to be revealed. Puzzle over it. Take your time.

Ask yourself: why can't I divide that 7 off?

There is only one answer because it's mathematics.

Remember that word? "Mathematics"

There is an answer. The right one.

In a lot of ways it's a trivial demonstration, given

7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

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

the mathematics does not allow that 7 to be controlled by the factorization of what it is multiplying, so you have with a simple example:

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

a CHOICE.

That's because it could be: 7(x^2 + 3x + 2) = (x + 1)(7x + 14).

Now I've pointed this out many times over the years as I've repeatedly explained this error!

No person thinking rationally would suppose that 7 is being told how to multiply by mysterious forces. I used to say, years ago, that the tail does not wag the dog.

So, ok, a non-standard factorization reveals a bizarre problem with the ring of algebraic integers. The construct creatively FORCES the roots of monic polynomial with integer coefficients when x is an integer i.e.

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

to be part of a factorization:

7(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

so yes you can solve for the a's with the quadratic formula:

a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2,

a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

And you can make the substitutions and see that it all works. If Dedekind had been given this demonstration over a hundred years ago, you would not be learning the ring of algebraic integers today, except maybe as an oddity: a fairly useless ring which has a fatal flaw.

But a hundred years of error built up around this error.

It takes away the usefulness of Galois Theory. It removes the ring of algebraic integers as a useful ring.

Ok, since I found this error students have continued to be indoctrinated into it. That is a lot of human waste.

Professors who keep feeding it to students may in some cases be functionally insane due to the enormity of the problem.

Or they may just be cynically determined not to face consequences of it. Who knows.

But if you learn it, you are training your brain useless knowledge which does not work.

The math done with it is pure—purely wrong.

So yeah, they'll teach you style. It's style WITHOUT substance.

They force style because the math itself is wrong. Style crap also helps to keep out outsiders like me!

Rigor in error is still error. You are just rigorously wrong.

The problem is a remarkable error. It has taken over a hundred years to be revealed. Puzzle over it. Take your time.

Ask yourself: why can't I divide that 7 off?

There is only one answer because it's mathematics.

Remember that word? "Mathematics"

There is an answer. The right one.