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

(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:

[.snip.]

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.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

Well that example works, so now I'll say algebraic integer m, and it
blows up at m^2+1 = 0.

So let me see if I have this right:

(1) You claimed something was impossible under a given set of
conditions.

(2) When I proved that your claim was simply false, you changed the
set of conditions, and now argue that what I said was wrong.

You ****ing dumbass Arturo Magidin, the ring has ALWAYS been algebraic
integers, so having m in the ring of algebraic integers FOLLOWS ANYWAY
you goddamn, ****ing, stupid dumbass.

You are so ****ing stupid to supposedly be a mathematician you **** of
**** Arturo Magidin.

You are **** Arturo Magidin. You are a stupid Magidin piece of ****.

You know what Arturo Magidin? You are a ****ing dumbass.


James Harris

James Harris wrote:

[snip]

You ****ing dumbass Arturo Magidin, the ring has ALWAYS been algebraic
integers, so having m in the ring of algebraic integers FOLLOWS ANYWAY
you goddamn, ****ing, stupid dumbass.

You are so ****ing stupid to supposedly be a mathematician you **** of
**** Arturo Magidin.

You are **** Arturo Magidin. You are a stupid Magidin piece of ****.

You know what Arturo Magidin? You are a ****ing dumbass.

James Harris

Are you related to Dar Kabatoff, by any chance?

--
Proposed slogan for JSH: Veni, vidi, vomito.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com

In alt.math.undergrad James Harris wrote:
(Arturo Magidin) wrote in message ...

So let me see if I have this right:

(1) You claimed something was impossible under a given set of
conditions.

(2) When I proved that your claim was simply false, you changed the
set of conditions, and now argue that what I said was wrong.

You ******* ******* Arturo Magidin, the ring has ALWAYS been algebraic
integers, so having m in the ring of algebraic integers FOLLOWS ANYWAY
you *******, *******, stupid *******.

You are so ******* stupid to supposedly be a mathematician you **** of
**** Arturo Magidin.

You are **** Arturo Magidin. You are a stupid Magidin piece of ****.

You know what Arturo Magidin? You are a ******* *******.


James Harris

Here James once again demonstrates what happens when he sees that he is
wrong -- he loses control and starts foaming at the mouth. Here we also
see, once again, what a low-class, filthy-minded, utterly degenerate
person he is. Beneath all his pretensions of intellect, logic, and
superiority lies a profound recognition (and abhorrence) of his own
inferiority; and his reaction when forced to face the truth about himself
and his depravity is to resort to foul, puerile language as he rants,
raves, and hurls abuse at all he sees as better than himself. He is a
perennial dirty-faced, dirty-minded and squalid little boy throwing mud
at the neatly-dressed people he sees walking by because he knows he is
not fit for their company.

How ashamed his family must be of him.

--
Wayne Brown (HPCC #1104) | "When your tail's in a crack, you improvise
| if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"

(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

"James Harris" wrote in message
m...
"David Moran" wrote in message
...
"James Harris" wrote in message
m...
"David Moran" wrote in message
...
"James Harris" wrote in message
m...
(Nora Baron) wrote in message

deleted

Also incorrect and false. Remember what you said above:

"... consider w_1(m), a factor of a_1 that is a factor of f..."

If w_1(m) is a "factor of f", that can only mean f/w_1(m) is
an algebraic integer, which of course implies that uf/w_1(m)
is an algebraic integer: this is your *assumption* here.

The assumption is itself a contradiction, in that it requires the
ability to define one algebraic integer function by the
multiplicative
inverse of another algebraic integer function.

Basic algebra.
IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I
UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

Unfortunately the people who *do* understand it are doing their best
to confuse everyone else, and you've done your part by basically
heckling me with posts.

If I didn't have mathematicians fighting me, then I could carefully go
through each and every detail to work it out so that everyone who has
a basic grasp of algebra can understand.

But first I need you to help me, by NOT letting posters like "Nora
Baron" and Arturo Magidin get away with muddying the waters.

Other posters need to begin challenging these people who have betrayed
you, after all, they've had many of you nodding along with something
as dumb as xy=2 with *both* x and y algebraic integers where x varies
over the algebraic integers which you get using u=1, f=2, and the
assumption of a w_1(m) as a factor of f that's a varying factor of
a_1, with x=w_1(m), y=uf/w_1(m).

It's posters like Arturo Magidin and "Nora Baron" who've helped make
mathematicians look like fools and patsies, willing to lie and
challenge mathematics itself either willingly, or from incompetence.


James Harris

It's you who are subjecting yourself to the heckling. If you want the
heckling to stop, either stop posting, or stop the personal attacks. I
can
think of a much better way to do what you're doing, but will you listen
to
me? Probably not.

David Moran

Readers here's what I was replying to from this *same* poster, which
I'm copying down from this post.

IF THIS IS BASIC ALGEBRA, WHY CAN'T I UNDERSTAND YOUR ARGUMENT? I
UNDERSTAND
ALGEBRA PERFECTLY AND NONE OF THIS MAKES SENSE TO ME.

So I reply thinking that maybe the guy is serious, and you can see
what he said in return, and notice his arrogance and lack of social
graces.

It's like I'm dealing with children with VERY bad manners.


James Harris

Hey, I'm not arrogant to anyone who doesn't deserve it. You need to grow up,
James. Your name calling makes me think I'm reading something from my 7 year
old brother. I can be nice, or I can be mean; you decide.

David Moran

In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:
(Arturo Magidin) wrote in message ...
In article ,
James Harris wrote:

[.snip.]

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.


w_1(m) = 13^{1/(m^2+1)}*2^{|m|/(m^2+1)}

I'm not saying that's the function in question (it is not), but there
is a function that "actually varies with m such that 26/w_1(m) is an
algebraic integer for *all* integer m." So the argument that no such
function can exist is simply bogus.

Well that example works, so now I'll say algebraic integer m, and it
blows up at m^2+1 = 0.

So let me see if I have this right:

(1) You claimed something was impossible under a given set of
conditions.

(2) When I proved that your claim was simply false, you changed the
set of conditions, and now argue that what I said was wrong.

You ****ing dumbass Arturo Magidin, the ring has ALWAYS been algebraic
integers, so having m in the ring of algebraic integers FOLLOWS ANYWAY
you goddamn, ****ing, stupid dumbass.


You asked for a function that satisfied certain, specific, explicitly
given properties, claiming such a function was impossible.

I gave a function that satisfied EACH AND EVERY ONE of the properties
you listed.

Then you changed the properties you wanted. I noted the change, and
proceeded to give you a function that satisified all the NEW
properties, AND MORE (being defined over all the algebraic numbers, not
just the algebraic integers; you can restrict it to the algebraic
integers if you want). But you deleted it. Here it is again:

-- Begin Insert --

Here's an example that works for EVERY algebraic NUMBER m:

Step 1. Given an algebraic number m, let f(x) be the unique monic
polynomial with rational coefficients, irreducible over Q, which has m
as a root. Write it as:

f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0.

Note that a_0 must be different from 0.

Step 2. Write a_0, a rational number, as a_0 = r/s, with r and s
integers, r and s relatively prime.

Step 3. If r=1 or -1, let q = 1.

Step 4. If r is not equal to 1 or -1, then let q be the largest
rational prime that divides r.

Step 5. Let w(m) be a root of the polynomial

x^4 + 13qx^3 + qx + 13.


Then:

(a) w(m) is an algebraic integer.

(b) w(m) divides 13 in the ring of algebraic integers, since the
product of all the roots is 13, and every root is an algebraic
integer.

(c) w(m) is not constant: it takes different values at different
integers.

(d) w(m) takes each value a countably infinite number of times, so
w(m) cannot be given by a polynomial.


-- End Insert --

I gave a variant elsewhere defined for all complex numbers m as well.

Still think such a function is impossible?

================================================== ====================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A great
many people are staggered to this extent, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
================================================== ====================

Arturo Magidin

are you saying that the algebraic integers are continuous
"on hte real line," or just incredibly dense?... or,
just taht there are an infinity of them?

(James Harris) wrote in message om...

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?

--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR "@" http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81

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

Ah, you assume a continuous function. Why?

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.

Assuming a continuous function again. Why?

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.

You are producing the bull****. Pray produce a proof that w_1(m) is a
continuous function and we can talk further.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

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