"Dik T. Winter" wrote in message ...
In article (James Harris) writes:
(James Harris) wrote in message om...
"Dik T. Winter" wrote in message ...
...
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!!!

Yes.

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

Anybody? Anybody?

The simple answer is: you can't. For instance the function (assume x an
algebraic integer):
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise,
is a perfectly well-defined varying function in the function that has
only two possible results.

Wow that poster is astoundingly stupid as if you're *in the ring of
algebraic integers* how can you have x/2 NOT be an algebraic integer?

Trick question: Remember, the "/" operator is NOT defined in general
in the ring, and by convention you can only use it when a given
numerator has the denominator as a factor.

Therefore, to use x/2 it must be that x is even, or the condition is
undefined in the ring.


James Harris

In article (James Harris) writes:
"Dik T. Winter" wrote in message ...
In article (James Harris) writes:
(James Harris) wrote in message om...
"Dik T. Winter" wrote in message ...
...
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!!!

Yes.

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

Anybody? Anybody?

The simple answer is: you can't. For instance the function (assume x an
algebraic integer):
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise,
is a perfectly well-defined varying function in the function that has
only two possible results.

Wow that poster is astoundingly stupid as if you're *in the ring of
algebraic integers* how can you have x/2 NOT be an algebraic integer?

That poster is astoundingly stupid as if you're *in the ring of
algebraic integers* when the only thing I said is that *x is an
algebraic integer*. And it is easily shown that f is a function
from the algebraic integers to the algebraic integers (which you
call as being in the algebraic integers).

Therefore, to use x/2 it must be that x is even, or the condition is
undefined in the ring.

Nonsense.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

On 20 Oct 2003 03:07:10 -0700, (James Harris) wrote:

"Dik T. Winter" wrote in message ...
In article (James Harris) writes:
(James Harris) wrote in message om...
"Dik T. Winter" wrote in message ...
...
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!!!

Yes.

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

Anybody? Anybody?

The simple answer is: you can't. For instance the function (assume x an
algebraic integer):
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise,
is a perfectly well-defined varying function in the function that has
only two possible results.

Wow that poster is astoundingly stupid as if you're *in the ring of
algebraic integers* how can you have x/2 NOT be an algebraic integer?

If x=2, x is an algebraic integer but x/2 is not.

If x=5, x is an algebraic integer, but x/2 is not.

If x = sqrt(2), x is an algebraic integer, but x/2 is not.


Trick question: Remember, the "/" operator is NOT defined in general
in the ring, and by convention you can only use it when a given
numerator has the denominator as a factor.

Therefore, to use x/2 it must be that x is even, or the condition is
undefined in the ring.

Unusually idiotic even for you. But if you insist...

Let x/2 represent the division of x by 2 in the complex numbers. For
every algebraic integer x, let
f(x) = 2 when x/2 is an algebraic integer,
f(x) = 1 otherwise.

I hope even you realize that you can take an x from the algebraic
integers, calculate the number x/2, and check whether it too is an
algebraic integer.

- Randy

James Harris wrote:

I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

....

My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.


When do you anticipate that you'll succeed, James? I imagine a sort
of steely-eyed General Patton response: "As long as it takes, damn it!"
That's the Army spirit . . .

But what if ten years from now you're still presenting them with an
argument whose truth is so blatantly obvious to you that you begin
to wonder if they even understand the words you're using? You write
"number"; they read "plumber".

Maybe this is just fine with you. Sometimes it seems that you enjoy
the struggle so much -- your intellect shining out into unquenchable
darkness -- that deep down you no longer even want victory.

Or maybe you realize that the path you've chosen is fraught with the
possibility of failure, not because of an immovable obstacle ahead,
but, more insidiously, because it could be like a dream path which
extends ever onward toward a receding, illusory goal. (But with so
much drama along the way!)

You believe that the path you're on is the only path which leads to
your goal. But look! Come this way! You protest that it's not a
path at all, that it is mere wandering in the woods, and that, worst
of all, it's going almost backwards from your goal! No promising
vistas. No bright sun. And for God's sake, take off that Walkman!
You didn't even realize, did you? What were you listening to?
Magidin Ullrich Overdrive? Now tie this rope around your waist.
We're going down the cliff. That smoke comes from a village I hope
to reach before nightfall.

But that's a lot of silly talk to an Army man, right? Dig the trench.
Brace the cannon. Call artillery for coordinates.

What will you do in twenty years if you're still fighting the same
battle? How long is too long, James?

James Harris wrote:

I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.

Perhaps because there is a standing counter-example.


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.

It would, however, be a *function* which represents different algebraic
integers for various values of m.


My guess is that some may be assuming that f is replaceable by some
function of m, but in fact,

No. The a's are functions of m. f is a constant. Since we have been
very clear about where the problem is, you are showing low comprehension
of what people are saying to you.

its independent of the value of m, so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer, for all algebraic integers m, if w_1(m) varies with
m.

You have completely failed to understand where the problem lies. You
have not refuted the counter-example or the counter-arguments. You have
not even addressed them.


My hope is that posters who have been so successful in convincing
others that my argument is flawed will post concessions.

Hope springs eternal, does it not?


--
Will Twentyman
email: wtwentyman at copper dot net

James Harris wrote:

(Peter van Rossum) wrote in message ...

In article ,
James Harris wrote:

I've noted a problem in algebraic number theory with the inclusiveness
of the definition of algebraic integers as roots of monic polynomials
with integer coefficients.

Various posters have argued that in fact there is no problem, but
here's a short refutation of their primary objection.

First I have

P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)

and the factorization

P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)

where the a's are given by the following cubic:

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).

If the a_i's depend on m, why don't you write it properly?

P(m) = ( a_1(m) x + u f ) ( a_2(m) x + u f ) ( a_3(m) x + u f )

That's a style issue. I see it as a gesture of futility and anguish
at the reality of this easy refutation.

Mathematicians are such babies.

It's not a style issue, it's a clarity issue. If you make your
arguments clear by using precise notation, it becomes easier to follow
your arguments and it may be easier to see where your mistakes are.

My finding that only two of the a's have f as a factor without regard
to the value of m has been vigorously disputed.

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,

i.e. w_1(m) | a_1(m) x + u f for all algebraic numbers m

Mathematicians in trying to deny the reality here have been hoping on
some w_1(m) that can give a variable factor of f for a_1. I'm just
following that idea through to the necessary conclusion.

Good. Now if you'd accept that it exists, you could admit your error
and we could all move on. It has been shown in raw numbers to be the case.

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.

I hope you realize that the functions a_i are also not given by polynomials
in m (i.e., there does not exists a polynomial A_i(M) in A[M], the

That's a rather stupid lie given that I put the polynomial in this
post.

Again it is

a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)

and its roots are a_1, a_2 and a_3.

The a's being *roots of a polynomial* in m does not make *them*
polynomials in m.


But mathematicians have been persistent in showing a rather pathetic
and stupid refusal to accept rather basic mathematics.

I find them disgusting.


polynomial ring in one variable over the algebraic numbers, such that
A_i(m) = A_i(m) for all algebraic numbers m).
For the function w_1, well, the only thing you've mentioned about it is
that w_1(m) | a_1(m) and w_1(m) | f (for all algebraic numbers m,
I think).
So how could one expect a_1(m) x / w_1(m) + u f / w_1(m), or u f / w_1(m)
to have any particular form at all? Of course, for u f / w_1(m)
to be an algebraic integer for all algebraic integers m, it doesn't
have to be expressible as a polynomial in m over the algebraic integers.


The poster is babbling, possibly in shock. Such weakness from the
math world is telling. Mathematicians don't know how to really think,
but have gotten away with faking it.


With all you assumptions, it is of course trivial that u f / w_1(m)
is an algebraic integer (since f is divisible by w_1(m) in the algebraic
integers).


It's not possible. I'll put in values to help those readers confused
by symbols (though algebra is BASED on symbols) by letting u=2, f=13,
then you have

26/w_1(m)

and you don't have to be a rocket scientist to know that no function
w_1(m) that actually varies with m can exist such that 26/w_1(m) is an
algebraic integer for *all* integer m.

Sure there is. w_1(m) = 13 if m0, w_1(m)=2 if m = 0. The actual
w_1(m) will be rather more complicated, and may not have a nice
representation, but that doesn't mean it doesn't exist.


It's a show of how broken math society is that Peter van Rossum would
dare to make that stupid assertion.


My guess is that some may be assuming that f is replaceable by some
function of m, but in fact, its independent of the value of m,

Everybody understands that, I'm sure.


so it's
like 1/(x+1) which is also not representable by a polynomial if x is
an algebraic integer not equal to 0 or -2.

But I wonder who understands this. I definitely don't.

(By the way, if x *is* an algebraic integer unequal to -1, then
1/(x+1) *is* expressible as a polynomial over the algebraic integers -
a constant polynomial. But you probably mean that there is no
polynomial F(X) over the algebraic integers such that F(x) = 1/(x+1)
for all algebraic integers unequal to -1. I still wonder what 0 and
-2 have to do with it - maybe just a mistake.)


I was thinking about x being an integer, though in algebraic integers,
any x such that x=u_1 - 1, where u_1 is a unit in algebraic integers
will work.


So the objection is refuted by the impossibility of uf/w_1(m) being an
algebraic integer, for all algebraic integers m, if w_1(m) varies with
m.

Can you repeat the definition of algebraic integer again for
the newsgroup and tell us how you conclude that u f / w_1(m)
is not an algebraic integer?


Mathematicians are pathetic liars.

I'll use u=2, f=13 again, now then, NO function in algebraic integers
exists such taht 26/w_1(m) is an algebraic integer for all integers m,
if w_1(m) varies with m.

I just wrote one earlier. Nice and simple.


It's just not possible, but it takes a mathematician to lie about it.



--
Will Twentyman
email: wtwentyman at copper dot net