### Wednesday, October 13, 2004

## JSH: Now a change

Working out the details in that new paper thread allows me from now on to refer to sections of it when dealing with the obnoxious posters who refuse to accept mathematics.

For instance, when someone claims that W. Dale Hall has a counterexample, I can point to Section 2, and how it shows that finding algebraic integers does not show a contradiction.

If you look over my current draft you can see from beginning to end a fairly basic argument that handles

My case begins with the simple position that constants are constant, and notice from Section 2 and Section 3 that sci.math'ers arguing with me have been relying on unit factors—just not units in the ring of algebraic integers.

For those who don't understand what I mean about constants and unit factors, consider units in the ring of algebraic integers with

P(x) = (x+1)(x+2)

as the constant terms are 1 and 2 of the factors, but you can multiply by u_1, and u_2, where u_1 u_2 = 1, and u_1 and u_2 are algebraic integers to get

P(x) = (u_1 x + u_1)(u_2 x + 2u_2)

and someone might jump up and down and now claim that the constants are not constant.

That is essentially what posters like Dik Winter have done, for months now, except the units are provably not in the ring of algebraic integers, but in a ring that includes the ring of algebraic integers, where two conditions rule:

They are the key requirements.

So, when posters have put forward what they claim are counterexamples, they have simply put forward algebraic integers multiplied by units that act as conversion factors.

They are units that convert non-algebraic integers in a ring defined by the conditions 1. and 2. into an algebraic integer, and it's not hard to prove that such will always exist, as that is basically Section 2. of my draft paper.

I've covered all the bases, and shown what must be true mathematically.

That now leaves the issue of your egos and other psychological issues you may have, but some of you must know that I'm not surprised, as you're not mathematicians.

And make no mistake, I'm emphasizing that point for a purpose, as mathematicians necessarily would care more for what is mathematically true.

Mathematics is a hard discipline. Behaving in one way proves your not a mathematician, without any room for argument, just like a math proof leaves no room for argument.

So later, when you fail as I expect you will, then I will say that you have proven you are not mathematicians, so that other necessary things can be done.

After all, what's mathematically true was always true, and those wrong now, were always wrong.

No one who understand mathematics would hold on to a false position, especially one so easily proven to be false, if they were a true mathematician.

For instance, when someone claims that W. Dale Hall has a counterexample, I can point to Section 2, and how it shows that finding algebraic integers does not show a contradiction.

If you look over my current draft you can see from beginning to end a fairly basic argument that handles

**every**issue raised over years in arguments, and shows how mathematically a fairly straightforward conclusion is necessary to avoid contradiction.My case begins with the simple position that constants are constant, and notice from Section 2 and Section 3 that sci.math'ers arguing with me have been relying on unit factors—just not units in the ring of algebraic integers.

For those who don't understand what I mean about constants and unit factors, consider units in the ring of algebraic integers with

P(x) = (x+1)(x+2)

as the constant terms are 1 and 2 of the factors, but you can multiply by u_1, and u_2, where u_1 u_2 = 1, and u_1 and u_2 are algebraic integers to get

P(x) = (u_1 x + u_1)(u_2 x + 2u_2)

and someone might jump up and down and now claim that the constants are not constant.

That is essentially what posters like Dik Winter have done, for months now, except the units are provably not in the ring of algebraic integers, but in a ring that includes the ring of algebraic integers, where two conditions rule:

- No rational other than 1 or -1 is a unit.
- No non-unit member of the ring is a factor of any two rationals that are coprime to each other.

They are the key requirements.

So, when posters have put forward what they claim are counterexamples, they have simply put forward algebraic integers multiplied by units that act as conversion factors.

They are units that convert non-algebraic integers in a ring defined by the conditions 1. and 2. into an algebraic integer, and it's not hard to prove that such will always exist, as that is basically Section 2. of my draft paper.

I've covered all the bases, and shown what must be true mathematically.

That now leaves the issue of your egos and other psychological issues you may have, but some of you must know that I'm not surprised, as you're not mathematicians.

And make no mistake, I'm emphasizing that point for a purpose, as mathematicians necessarily would care more for what is mathematically true.

Mathematics is a hard discipline. Behaving in one way proves your not a mathematician, without any room for argument, just like a math proof leaves no room for argument.

So later, when you fail as I expect you will, then I will say that you have proven you are not mathematicians, so that other necessary things can be done.

After all, what's mathematically true was always true, and those wrong now, were always wrong.

No one who understand mathematics would hold on to a false position, especially one so easily proven to be false, if they were a true mathematician.