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

**David C. Ullrich**

On 19 Oct 2003 07:11:11 -0700, (James Harris) wrote:

"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.

Whee! In another post you just said

"The function has to be continuous given the other equations which not
surprisingly the poster deleted. These people are dumb as rocks."

and I considered asking whether you could _define_ the word
"continuous". Decided not to bother, because continuity is one
of those things that a person _could_ have a fairly reasonable
understanding of even if he were unable to give the precise
definition.

But here you show you have _no_ clue regarding "continuity
and slope", which is funnier than your cluelessness regarding
algebraic number theory, because continuity and slope
were things you were supposed to learn about when you
got that famous degree in physics.

Hint: "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" is funny. Like say f(m) = 42; f is constant.
It follows from what you say here that a _constant_ function
has "infinite slope or a discontinuity". Wow.

I'm not interested in explaining basic mathematics but at least
hopefully I can give other readers some sense of the frustration I've
had to handle dealing with posters who are so mathematically ignorant,
while fanatically replying to my posts negatively.

Yeah. So mathematically ignorant that they don't realize that
constant functions are discontinuous (or that integers are
irrational, that sqrt(i) is not a complex number, etc... it's really
remarkable how ignorant we all are. guffaw)

But w_1(m) must vary from 0 to infinity if it varies with m.

Why? Please prove that.

It turns out that you need the the absolute value, like r(m)r*(m), and
with it it's possible to show that for an algebraic integer function
that varies as m varies--a continuous function--as m varies over all
of algebraic integer r(m)r*(m) must vary from 0 to positive or
negative infinity.

So if you had any doubts about how low mathematicians could go,
consider that after all, they're trying to convince that you can have
one algebraic integer function defined by the multiplicative inverse
of another algebraic integer function over all algebraic integer m,
like with f=7, 14/w_1(m) = r(m).

It's like saying that you can have xy=2, where x and y are integers,
or algebraic integers, where x varies over the ring, and y remains in
it, though of course, I can just show that for x=5 that doesn't work.

Not at all. The range of w_1(m) may be severely limited, while the
range of x is not limited. It is similar to saying that you can have
xy = 2, where x ranges over the algebraic integer factors of 2, and so y
also ranges over the algebraic integer factors of 2. 5 is now not
a counterexample because it is not an algebraic integer factor of 2.

Yeah you can *say* just about anything, but mathematically that
statement is bull****, and for me to have made the discoveries I've
made and be stuck because a lot of posters can get away with bull****
is just ****ing me off.

James Harris

************************

David C. Ullrich