Physics Banter - View Single Post - Core error argument objection refuted, short

**James Harris**

(James Harris) wrote in message om...
"Dik T. Winter" wrote in message ...
In article (James Harris) writes:
(Nora Baron) wrote in message . com...
...
However, consider w_1(m), a factor of a_1 that is a factor of f, as
well as a function that varies with m, then it follows that

a_1 x + uf has w_1(m) as a factor,

so dividing through by w_1(m) gives

a_1 x/w_1(m) + uf/w_1(m)

but then uf/w_1(m) cannot in general be an algebraic integer as it's
not representable as a polynomial with a finite number of terms if
w_1(m) varies with m.

A totally off-the-wall, unjustified statement, and, as it
so happens, incorrect. But for now, if you want to claim
it is true, the shoe is on your foot: try to prove it.

I've introduced r(m), to handle the result of uf/w_1(m).

So the poster is requesting that I prove that

r(m) w_1(m) = uf, does not exist over the ring of algebraic integers.

I've concluded that using numbers for u and f, as they are
*independent* of m, helps, so let u=2, f=13.

Then you have r(m) w_1(m) = 26, and let x=r(m), y=w_1(m), so you have

xy=26

and now I'll chat further.

Then the question is, does their exist a multiplicative inverse in the
ring of algebraic integers for 26/w_1(m) for *all* algebraic integers
m?

The simple answer is that if w_1(m) varies with m, then it must vary
over an infinite number of algebraic integer values as m varies over
algebraic integers.

Why? Please prove that.

It has to do with continuity and slope. If you could have w_1(m) and
w_1(m') equal when m does not equal m' then at that point you'd have
infinite slope or a discontinuity.

OOPS! What I said was STUPID!!!

What's interesting about that error is that you can see replies to it
in this thread.

I want you all to *focus* on the replies. Read them carefully.

Oh yeah, so how do you prove that a varying function in algebraic
integers has to have an infinite number of results?

Anybody? Anybody?

James Harris