mm-1087
===
Subject: Re: Focus on point of dispute, more math
in spite of certain bizarre conventions onetiquette,
anyone can easily refer-back to your original posting,
wherein the sanctified New Math resides. so,
why dont you just come-up with that (single/any) example,
that monsieur Bond requires to convince him?
hed be the first, but the rest of 
usd fall into line!
> You *say* having f=3 wouldnt change it, but why? The rest
of
> the posting doesnt provide an argument. 
Doesnt it bother
you
> to be stuck repeating such a claim without any way to show
> that its actually true besides telling us how weird it
would be
> if it werent?
> Posters repeatedly delete out the math in replies.
--les ducs dEnron!
http://members.tripod.com/~american_almanac
===
Subject: Re: How do you prove this? (Composite functions and
their
inverses)
> below _except_ where I still have questions.
>>Anyway, f : (0, 1) -> [0, 1] : x -> x is one to one and has
no
inverse.
> And then, following my query:
>Sure. (0, 1) is the usual open interval; [0, 1] is the usual
closed
>interval. The point was that neither 0 nor 1 was equal to
f(x) for x in
>(0, 1) so f was not onto and therefor could not have an
inverse. What
>would f^(-1)(1) be?
> This is where Im not following your notation. If (0, 1),
the usual
> open interval, is the domain, and the function is defined by
f(x) =
> x, then I dont see how f(anything) could be 0 or 1.
Exactly the point. There can be no function h : [0, 1] -> (0,
1) such
that f(h(1)) = 1. So, f has no inverse.
>The _real_ nit pick is that f: Domain -> Codomain and
>f: Domain -> Range (Image) way well be different functions.
> I guess I can grasp that, dimly, but I dont understand
what the
> point is. If f(x) cannot have any values outside the open
interval
> (0,1), then what is the benefit to assigning it a codomain
of [0,1]?
In this case, the only advantage is that it is an example of
a one to
one function that is not onto.
It is not unusual to look at a whole bunch of functions; for
example
let C be the set of continuous functions from [0, 1] to [0,
1]. Now, if
I say Let g be in C with g(x) = x/2, it is understood what
the domain
and codomain are and moreover, g is not an invertible element
of C
since there is no h _in_ _C_ such that g(h(a)) = 1.
In Calculus, for example, one frequently sees things like Let
k(x) = sqrt(x). Since the domain is not mentioned, it is
_assumed_ to
be the largest set possible: the non-negative reals. [More
likely, one
just doesnt worry about it.] What is done about the 
codomain
depends
on what comes next - what do you want to with k. Most of the
time the
codomain isnt mentioned because it doesnt 
make any
difference what it
is. However, if you want to use the inverse function theorem
and the
known derivative of s(x) = x^2 to find the derivative of k,
_then_ you
must specify the codomain of s to insure that s^(-1) exists.
> I dont know how I managed to miss this business of onto
functions
> and codomains, but I did. I apologize if my questions are
too
> elementary.
Not at all. Keep Ôem coming if you like. I hope there are some
interested lurkers.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: Math expression *does* say it all
although you call it an uberpoly.,
it is clear taht it is *still* a polynomial, perhaps
with an extra variable, or considered
as a partial differential equation etc.k
the only complex part taht you have yet to deal with,
is the complex *field*.
> The expression
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f)
> is NOT a polynomial, but is what I now call an
uber-polynomial, which
> contains an infinite number of polynomials, so 
its more
like a
> family.
> By using P(m), Im focusing on the variable m, so with
P(m), m=0,
> gives me the constant term with respect to m.
> So in looking at m=0, Im pulling out the constant term
with respect
> to m, from a rather complicated expression, which
represents a more
> complex system that you may have dealt with before, but
still I can do
> some rather basic things to get a handle on it.
--les ducs dEnron!
http://members.tripod.com/~american_almanac
===
Subject: Re: Math expression *does* say it all
In sci.math, James Harris
<jstevh@msn.com>
>> In sci.math, James Harris
>> <jstevh@msn.com>
>> Finally in looking at the expression I call the
uber-polynomial I
>> realize it says it all rather quickly, and directly.
>>
>> 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) =
>>
>> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>>
>> And what Ive proven by checking the constant term P(0) 
is
that
>> Pedant point: If youre referring to the constant term
>> (x^0) of P(m), which is u^3f^3, it is clear that
>> that is *not* equal to P(0), which is 3u^2f^2x + u^3f^3 =
>> f^2(3u^2x+u^3f), unless x=0.
> That is incorrect.
> The expression
> f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
> 3(-1+mf^2 )x u^2 + u^3 f)
> is NOT a polynomial, but is what I now call an
uber-polynomial, which
> contains an infinite number of polynomials, so 
its more
like a
> family.
Its a polynomial with partly-specified 
coefficients, in
terms of x. Or in m. Or in f. Or in u.
Is Dx^3 + Cx^2 + Bx + A a polynomial or uber-polynomial?
Yet set D = f^2(m^3 f^4 - 3m^2 f^2 + 3m), C = 0, B =
-3f^2(-1+mf^2 ),
and A = u^3f^3, and we get your expression.
Id prefer a different expression than P(0), though.
P(m)|0 is a bit odd-looking,
|
P(m)|
|0 is incorrect and unwieldly (and probably munged in
proportional-font newsreaders),
|
P(m)|
|m=0 is correct but redundant.
P(m)_0 or P_0(m) might work; one can then compute
P(m) = P_3(m)x^3 + P_2(m)x^2 + P_1(m)x + P_0(m)
where
P_3(m) = f^2(m^3 f^4 - 3m^2 f^2 + 3m),
P_2(m) = 0,
P_1(m) = -3f^2(-1+mf^2 ),
P_0(m) = u^3f^3.
This should also typeset nicely in TeX.
Of course if youre focusing on m being a variable, rather
than x,
your notation makes some sense although such usage is slightly
unusual.
> By using P(m), Im focusing on the variable m, so with
P(m), m=0,
> gives me the constant term with respect to m.
Well, P(0) is still -3f^2(-1+mf^2 )x + u^3f^3. That x 
doesnt
drop out magically.
> When considering a polynomial the constant term tells you
which part
> of it does not contain the key variable.
> For instance, given P(x) = x^2 + 2x + 1, P(0) = 1, which is
the part
> of the polynomial that does not contain x.
> So in looking at m=0, Im pulling out the constant term
with respect
> to m, from a rather complicated expression, which
represents a more
> complex system that you may have dealt with before, but
still I can do
> some rather basic things to get a handle on it.
I did say it was a Pedant Point.
>>
>>
>> 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 =
>>
>> (a_1/f x + u)(a_2/f x + u)(a_3 x + uf)
>>
>> where if f is coprime to 3, one of the as is coprime to
f, and here
>> Ive given it the indice 3.
>>
>> That point has been debated a lot, with other posters
claiming that
>> dependent on m, the f^2 separates out differently, as
theyve
>> essentially claimed that its a variable of m.
>> P(m)/f^3 = (f^3m^3 - 3fm^2 + 3m/f)x^3 + (-3u^2fm +
3u^2/f)x + u^3
>> If 3u^2/f and 3m/f are integers, things get interesting.
> Why?
Because P(m)/f^3 becomes a polynomial with integer
coefficients, mostly.
This means there are now *three* factors of f to account for
when looking at the as, not two.
>>
>> It was tonite though in making the post previous to this
one that I
>> noticed that the pattern remains constant, even with
>>
>> P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 -
>>
>> (-1+m 3^2 )x u^2 + u^3 =
>>
>> (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)
>>
>> as then, all that happens, is that with f=3, that last of
the as, a_3
>> now lets f factor through it.
>> P(m)/3^3 = (27m^3 - 9m^2 + m)x^3 + (-9u^2m + u^2)x + u^3
>> = (3^3m^3 - 3^2m^2 + m)x^3 + (-3^2u^2m + u^2)x + u^3
>> = (3^3m^2 - 3^2m + 1)mx^3 + (-3^2u^2m + u^2)x + u^3
>> = (3^3m^2 - 3^2m + 1)mx^3 + (-3^2u^2m + u^2)x + u^3
>> Now m becomes part of the factorization of a_1a_2a_3; the
other
>> part is an irreducible quadratic in m, 3^3m^2 - 3^2m + 1.
>> For what its worth, but its clear m=0 
changes things
slightly.
> Nope. When f=3, the as have a factor that is 3 without
regard to m,
> though for m=0, its trivial for two of them as then they
equal 0, but
> 0 has 3 as a factor, as 3*0 = 0.
Fair enough.
> James Harris
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: My FLT proof, *last* warning to critics
> Saying that my proof is wrong without checking it makes me
angry, and
> can be causing me financial harm. I reserve the right to
seek redress
> for such harm if it is provable that it occurred, and may
do so in a
> court of law.
If James is as good at proving legal theories as he is at
proving
mathematical theories, we have nothing to worry about.
===
Subject: normal subgroup and cosets
Let H be normal in G, a finite, but not necessarily abelian,
group
generated by x and y. Now suppose that there exists a g in
G-H such
that
g <xy^{-1}> subseteq G-H.
( <xy^{-1}> is the cyclic subgroup generated by xy^{-1} ). Is
it true
that
g <xy^{-1}> = x^k <xy^{-1}> or g <xy^{-1}> = y^m <xy^{-1}>
for some k or for some m?
I know that normality lets you write g=x^a y^b for some a and
b, so
then you have
g <xy^{-1}> = x^a y^b <xy^{-1}> = x^{a+b} x^{-b} y^b
<xy^{-1}>,
but I cant seem to get any farther. Can this be proved? Any
hints or
suggestions would be appreciated.
===
Subject: Re: normal subgroup and cosets
Visiting Assistant Professor at the University of Montana.
>Let H be normal in G, a finite, but not necessarily abelian,
group
>generated by x and y.
Im sorry, Im not clear:
Who is generated by x and y? H or G?
>Now suppose that there exists a g in G-H such
>that
>g <xy^{-1}> subseteq G-H.
I assume this is the the set
{g(xy^{-1})^k : k an integer}.
>( <xy^{-1}> is the cyclic subgroup generated by xy^{-1} ).
Is it true
>that
>g <xy^{-1}> = x^k <xy^{-1}> or g <xy^{-1}> = y^m <xy^{-1}>
>for some k or for some m?
So, you are asking if g is congruent to a power of x or a
power of y
modulo xy^{-1}?
>I know that normality lets you write g=x^a y^b for some a
and b,
I dont understand again: are a and b elements of G, or
integers? I
assume the latter...
chose g not belonging to H. If it is G which is generated by
x and y,
then again this is wrong: there is no indication of what H
is, and
there may be elements of G which cannot be written as a power
of x
times a power of y.
So, is it ->perhaps<- the case that G is supposed to be
generated by x
and y, and that H is the cyclic subgroup generated by
xy^{-1}? And we
are assuming that H is normal?
But then for any g in G-H, we would have g<xy^{-1}> a subset
of
G-H... So Im not sure.
Could you clarify?
But, if I assume that this is indeed what you meant, so G is
generated
by x and y, and H=<xy^{-1}>, and H is normal in G, then G/H
is cyclic
(we have identified x and y), so in fact every element is
congruent to
a power of x (or of y), and you would get your conclusion...
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: normal subgroup and cosets
I think the O.P. meant the following:
Let G be 2-generated by x and y, and G not necessarily
abelian. Let H
be an arbitrary normal subgroup of G. Suppose there exists an
element
g in G-H such that
g <xy^{-1}> subseteq G-H.
Then, is it true that, for some k,
g <xy^{-1}> = x^k <xy^{-1}>?
[Notation: <g> = cyclic group generated by g. The group
<xy^{-1}>
isnt H here...H is just an arbitrary normal subgroup.]
> Let H be normal in G, a finite, but not necessarily abelian,
group
> generated by x and y. Now suppose that there exists a g in
G-H such
> that
> g <xy^{-1}> subseteq G-H.
> ( <xy^{-1}> is the cyclic subgroup generated by xy^{-1} ).
Is it true
> that
> g <xy^{-1}> = x^k <xy^{-1}> or g <xy^{-1}> = y^m <xy^{-1}>
> for some k or for some m?
> I know that normality lets you write g=x^a y^b for some a
and b, so
> then you have
> g <xy^{-1}> = x^a y^b <xy^{-1}> = x^{a+b} x^{-b} y^b
<xy^{-1}>,
> but I cant seem to get any farther. Can this be proved?
Any hints or
> suggestions would be appreciated.
===
Subject: Re: normal subgroup and cosets
Now that were on the subject, I would be interested in the
answer to
a possibly-related question:
Let G be 2-generated by x and y, finite, and not necessarily
abelian.
Let H be a normal subgroup of G.
The question: Is it true that, for any g in G-H, the set
g<xy^{-1}>
contains at least one element of H? Or does there exist a g
in G-H
such that g<xy^{-1}> does not contain any element of H?
>Let H be normal in G, a finite, but not necessarily abelian,
group
>generated by x and y.
> Im sorry, Im not clear:
> Who is generated by x and y? H or G?
>Now suppose that there exists a g in G-H such
>that
>g <xy^{-1}> subseteq G-H.
> I assume this is the the set
> {g(xy^{-1})^k : k an integer}.
>( <xy^{-1}> is the cyclic subgroup generated by xy^{-1} ).
Is it true
>that
>g <xy^{-1}> = x^k <xy^{-1}> or g <xy^{-1}> = y^m <xy^{-1}>
>for some k or for some m?
> So, you are asking if g is congruent to a power of x or a
power of y
> modulo xy^{-1}?
>I know that normality lets you write g=x^a y^b for some a
and b,
> I dont understand again: are a and b elements of G, or
integers? I
> assume the latter...
> chose g not belonging to H. If it is G which is generated
by x and y,
> then again this is wrong: there is no indication of what H
is, and
> there may be elements of G which cannot be written as a
power of x
> times a power of y.
> So, is it ->perhaps<- the case that G is supposed to be
generated by x
> and y, and that H is the cyclic subgroup generated by
xy^{-1}? And we
> are assuming that H is normal?
> But then for any g in G-H, we would have g<xy^{-1}> a
subset of
> G-H... So Im not sure.
> Could you clarify?
> But, if I assume that this is indeed what you meant, so G
is generated
> by x and y, and H=<xy^{-1}>, and H is normal in G, then G/H
is cyclic
> (we have identified x and y), so in fact every element is
congruent to
> a power of x (or of y), and you would get your conclusion...
> Its not denial. Im just very selective 
about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
> Arturo Magidin
> magidin@math.berkeley.edu
===
Subject: Re: normal subgroup and cosets
>Now that were on the subject, I would be interested in the
answer to
>a possibly-related question:
>Let G be 2-generated by x and y, finite, and not necessarily
abelian.
>Let H be a normal subgroup of G.
>The question: Is it true that, for any g in G-H, the set
g<xy^{-1}>
>contains at least one element of H? Or does there exist a g
in G-H
>such that g<xy^{-1}> does not contain any element of H?
No. For example, the permutations x = (12) and y = (23)
generate G = S_3,
and x y^(-1) = (123) is in the normal subgroup H = A_3. Thus
<xy^{-1}>
is contained in H. If g is any permutation not in H,
g<xy^(-1)> does not
intersect H.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: normal subgroup and cosets
> Now that were on the subject, I would be interested in 
the
answer to
> a possibly-related question:
> Let G be 2-generated by x and y, finite, and not necessarily
abelian.
> Let H be a normal subgroup of G.
> The question: Is it true that, for any g in G-H, the set
g<xy^{-1}>
> contains at least one element of H? Or does there exist a g
in G-H
> such that g<xy^{-1}> does not contain any element of H?
If I understand correctly what youre asking, the answers 
are
trivially no and yes, respectively. Let G = <x> be the
cyclic group of order 2, y = x, H = {1} and g = x. Then
g<xy^{-1}> = {x} contains no element of H.
Perhaps you want to refine your question.
--
Jim Heckman
===
Subject: Re: normal subgroup and cosets
Visiting Assistant Professor at the University of Montana.
>I think the O.P. meant the following:
>Let G be 2-generated by x and y, and G not necessarily
abelian. Let H
>be an arbitrary normal subgroup of G. Suppose there exists
an element
>g in G-H such that
>g <xy^{-1}> subseteq G-H.
>Then, is it true that, for some k,
>g <xy^{-1}> = x^k <xy^{-1}>?
>[Notation: <g> = cyclic group generated by g. The group
<xy^{-1}>
>isnt H here...H is just an arbitrary normal subgroup.]
The answer is no.
Let G be the group
G = <x,y,c | yx=xyc, yc=cy, xc=cx, x^3=y^3=c^3=e>
Note that since c=y^{-1}x^{-1}xy=[y,x], the group is
generated by x
and y.
Every element of G may be written uniquely in the form
x^a*y^b*[y,x]^c
with a,b,c in {0,1,2}.
Let H={e} be the trivial group. Let g=[y,x].
xy^{-1}=xy^2. The subgroup <xy^{-1}> consists exactly of the
three
elements
{ e, xy^2, x^2y[y,x]^2}
So [y,x]<xy^{-1}> = {[y,x], xy^2[y,x], x^2y} which is
trivially
contained in G-H. Since it does not contain either a power of
x or a
power of y, it cannot be equal to either x^k<xy^{-1}> or
y^k<xy^{-1}>
for any integer k.
Its not denial. Im just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: probability density function
About probability variable X,Y
Bivariate probability density function f(x,y) is given :
f(x,y)=(1/2)xy if 0=<x=<y=<2, otherwise 0
Question : Find probability density function of Z=Y-X.
If anyone can help me, please post reply...
===
Subject: Re: probability density function
>About probability variable X,Y
>Bivariate probability density function f(x,y) is given :
>f(x,y)=(1/2)xy if 0=<x=<y=<2, otherwise 0
>Question : Find probability density function of Z=Y-X.
>If anyone can help me, please post reply...
I admit up front that I dont know the answer. But since no
one else
has answered, maybe my thinking out loud will help get you
started.
Visualize f(x,y). The original density is a curved surface
rising
above a triangular base. The edges of the base are the three
lines
x=0, y=x, y=2; and the height above any (x,y) is xy/2. The
total
probability is the double integral
INT[0, 2, INT[0, y, xy/2 dx] dy]
which equals 1 (as it should).
To answer your question, I think you need to begin by looking
at
what z = y-x looks like in the region where the probability
function
is nonzero. Since z=y-x, y=x+z. The defined region must have
0 <= z <= 2 as well as 0 <= x <= y <= 2.
z = 2 represents point (x,y) = (0,2)
z = 1 represents line segment y=x+1 between (0,1) and (1,2);
z = 0 represents line segment y=x between (0,0) and (2,2);
and so forth. More generally, any z is represented y the line
segment y = x+z between the points (0,z) and (2-z,2).
So far Im pretty confident.
Now I _think_ that the probability density for each z value
is the
area of the vertical slice that cuts the base in the
indicated line
segment (or point, in the case of z=2). That area can be
found by a
line integral. Re-expressing the path y=x+z for any
particular z as
parametric equations, we have
x = t, y = t+z; x = y = 1 where 0 <= t <= 
2-z.
The line integral for any particular z, which is the
probability
density for that z, is
f(z) = INT[ 0, 2-z, (xy/2) * sqrt(2), dt ]
f(z) = INT[ 0, 2-z, (t^2+zt), dt ] / sqrt(2)
f(z) = [(1/3)t^3 + (1/2)zt^2] / sqrt(2), where t = 2-z
f(z) = (1/6)t^2 * [2t + 3z] / sqrt(2), where t = 2-z
f(z) = 1/(6*sqrt(2)) * (z-2)^2 * (z+4)
Now, IF Im right, the integral of this from 0 to 2 will be 
1.
[pause to tap keys of TI-89] Alas, its not. 
Its sqrt(2).
So I must have made some mistake, but in checking over my
work I
cant find it. Can someone else point it out?
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com
Fortunately, I live in the United States of America, where we
are
gradually coming to understand that nothing we do is ever our
fault, especially if it is really stupid. --Dave Barry
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
Visiting Assistant Professor at the University of Montana.
>
> [.snip.]
>
>>Essentially objections to how f^2 divides off now come down
to
>>claiming that the ws are functions of m, but consider 
that
w_1 w_2
=
>>1, when m=0, if f is coprime to 3.
>>
>>Now Im focusing on what has been revealed to be an area 
of
confusion.
>> Apparently some people believe that when I divide off f^2
that it
can
>>divide off as a *function* of m, so that m=0 might be a
special
case.
>>Im now starting the argument to address that belief by
noting
again
>>that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is,
when f
>>doesnt have 3 as a factor.
>>
>>But that was an arbitrary choice, so let f=3.
>>
>
> f = 3 is irrelevant to what you want. There is
>no reason to consider it. As you will see below, it
>is a red herring and it does not show what you want.
>
> I have not been following all the details of Jamess most
recent
> presentation. But it ->seems<- to me that he is falling
back to an
old
> mistake yet again.
>>
>>What Ive seen is that Arturo Magidin engages in making
posts which
>>apparently are meant to persuade.
>> What Ive seen is that you avoid any mathematical 
argument
that
>> demolishes your claims, no matter how basic that argument
is, and
>> instead post the same thing over and over and over, as
well as posts
>> like this one, where you engage in ad hominem.
>Youre the one who deleted out the mathematical equations
from my post
>in your reply. If there were a ßaw in my argument, why not
point it
>out in place?
I did: you continue to assume that what happens when m=0 will
also
happen when m is not zero.
I will also note that YOU delete my mathematical arguments
ROUTINELY. So dont come crying when people do the same to
you.
Simply put: I have a proof (nay, dozens of proofs) that you
are
wrong. If we start using YOUR standards of evidence, then it
is up to
YOU to find fault with my proofs, not up to me to 
find fault
with
yours.
Since, in addition, what I say agrees with standard results
that are
proven and reproven and checked by hundreds of people every
year for
the past 100 years, surely the burden of proof is on you.
>> My post was not meant to persuade anyone. It was meant as
a comment
>> of what my IMPRESSION of your argument is. If you had
something to
>> correct to that impression, I missed it. So I assume that
you are now
>> endorsing my impression as correct, and acknowledge the
basic logical
>> ßaw in your approach?
>There is no logical ßaw in my approach, and you have not
demonstrated
>one.
Keep telling yourself that. Maybe some day you will believe
it. The
fact that your approach leads to patently false statements,
which can
be verified with direct numeric examples, should be a clue
that your
logic is anything but ßawless.
>Instead you deleted out the mathematics and simply made
commentary,
>which clearly was meant to persuade.
Yeah, I truly apologize for not being an abrasive libelous
git who
goes out of his way to insult people. Instead, I present
complete
arguments in a corteous manner. How dare I!?
>You have demonstrated a need to try and convince others that
Im
>wrong.
No, I have demonstrated that you are wrong. You have
demonstrate a
deep hatred of me.
>It explains why you continually post as you do, often
deleting out key
>mathematical content.
I deleted no correct mathematical content.
[.snip.]
>> My interest is in what the actual
>>mathematical truth is, though it is of secondary interest
that I
>>convince others.
>> And that, I think, is something that not even you believe.
Your
>> interest has ALWAYS been to convince others. Your entire
behavior, as
>> well as your words, prove it.
>Again, you simply assert as if people should trust you.
No. I state my OPINION. You see that I think? What is it, you
think,
it represents?
>Why dont you just say the moon is made out of green 
cheese?
Because thats false. Whereas what I have stated about
mathematics is
true.
>Why argue in such a manner if you have the *math* to back up
your
>case?
Sigh. I see it went through your head.
>> I dont think that even YOU believe that your primary
interest is
>> mathematical truth. If it were, instead of replying to
THIS post, you
>> would replied to
>Oh, so its a sandbagging.
You dont know the meaning of the term, do you?
I pointed out, simply, that you ignored mathematical content
and
instead decided to reply to a post that was explicitly
identified as
tangential. I am free to do that, surely.
> Well, lets see what youve got.
===
>> Subject: Re: Constant factors and polynomials
>> [...]
> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
>
>any as exist, within the ring of algebraic integers, such
that
>sqrt(5) is a factor of them in that ring.
>> [...]
>The problem is that neither a_1, a_2, nor a_3 have ANY
non-unit
>factors in common with 5 in the ring of algebraic integers.
>
> And thats false. I am pretty sure that Dale produced
explicit common
> factors; but in any case, your claim here is certainly
false, since
> their product is not coprime to 65.
>>
>>Ive proven it true that neither a_1, a_2 nor a_3 have ANY
non-unit
>>factors in common with 5 in the ring of algebraic integers.
>> This is paraphrased from the post
>> Take the polynomial 65x^3 - 12x + 1, and factor it as
>> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1)
>Ok.
>> with a1, a2, a3 (necessarily) algebraic integers (in fact,
minus the
>> roots of x^3 - 12x^2 + 65). Let z be any root of that
polynomial. That
>> is, z will be either -a1, -a2, or -a3. It is trivial that
any common
>> factor between z and 5 will be a common factor of the
corresponding ai.
>Hmmm...so you call the following trivial. Interesting.
No, I did not call the following trivial. What I said was
trivial is
that, since z=-a1, if there is a common factor between z and
5, then
there is a common factor between a1 and 5. Surely you agree
that it is
a trivial claim? That any common factor between two numbers,
x and y,
is also a common factor between the numbers -x and y?
>In any event,
>Ill destroy your claim, and lets see if you 
admit the
truth.
Why do you need to argue that way, if you had math on your
side?
>> Take the following three polynomials
>> q(x) = 8 x^2 - 76 x - 185
>> r(x) = 8 x^2 - 4 x - 45
>> s(x) = 4 x^2 - 37 x - 104
>> Then any product of q(z), r(z), s(z) and integers will be
an algebraic
>> integer, necessarily.
>> We have that
>> q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325
>> Note that
>> (64 x + 128)*(x^3 - 12 x^2 + 65)
>> = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320
>> so (65z + 128)(z^3 - 12z^2 + 65) = 64z^4 - 650z^3 -
1536z^2 +
>> 4160z+8320.
>> But z^3-12z^2+65 = 0. So 64z^4 - 650z^3 - 1536z^2
+4160z+8320 = 0.
>> Therefore,
>> q(z)*r(z) = 64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325
>> = (64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8320) + 5
>> = 5.
>Im not interested in checking the math preferring to 
assume
its
>correct.
Oh, so you have no concern for the truth, then? (-;
>At this point then the poster is saying that
> q(z) r(z) = 5.
>> Likewise, take
>> r(z)*s(z) = 32 z^4 - 312 z^3 - 864 z^2 + 2081 z + 4680
>> This time, note that
>> (32 x + 72)*(x^3 - 12 x^2 + 65)
>> = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680
>> So (32z+72)*(z^3-12z^2+65)
>> = 32z^4 - 312z^3 - 864z^2 + 2080z + 4680.
>> But since z^3-12z^2+65 is equal to 0, it follows that
>> 32z^4 - 312z^3 - 864z^2 + 2080z + 4680=0. Therefore,
>> r(z)*s(z) = 32z^4 - 312z^3 - 864z^2 + 2081z + 4680
>> = (32z^4 - 312z^3 - 864z^2 + 2080z + 4680) + z
>> = z.
>As before, Im not interested in checking, preferring to
assume the
>operations were correct, and I note that at this point the
poster is
>saying that
> r(z) s(z) = z.
>> Therefore, r(z), which is an algebraic integer, is a
common factor of
>> 5 and of z. It only remains to show that it is not a unit.
>Its worth considering at this time what the intermediate
results are,
>assuming as before:
> q(z) r(z) = 5
> r(z) s(z) = z.
Note that both q(z), r(z), and s(z) are algebraic integers. I
have
produced three algebraic integers, two of which multiply to
be equal
to 5, two which multiply to be equal to z.
>Switching the latter around I have
> q(z) r(z) = 5
> z = r(z) s(z).
>Multiplying them together gives
> q(z) z = 5 s(z).
You did more than multiply together, but oh well. You also
r(z) after multiplying.
>Ok, now lets go back to the posters argument.
>> The claim is that r(z) is a root of f(x) = x^3 - 969 x^2 +
315 x +5.
>> If this is so, then r(z) is not a unit in the ring of all
algebraic
>> integers, since this is a monic irreducible polynomial
with integer
>> coefficients whose constant term is neither 1 nor -1.
>That is true. However, Arturo Magidin is relying on the
*definition*
>of unit as being an factor of 1 or -1 in the ring of
algebraic
>integers.
What exactly was I supposed to rely on? Thats what a unit 
is.
And, no, I am relying on a THEOREM. Which I have proven in the
newsgroup many times:
THEOREM. Let f(x) be a monic polynomial with integer
coefficients,
irreducible over Q. Then the roots of f(x), which are
algebraic
integers, are units in the ring of algebraic integers if and
only if
the constant term of f(x) is either 1 or -1.
To this theorem, the only objection you have raised is that
it uses
the meaning of units in the ring of algebraic integers...
>As Ive proven, algebraic integers are ßawed in that you 
can
have
> abc = 5,
>where neither Ôa, 
Ôb, nor Ôc has a 
non-unit factor in
common with
No, you have not proven this. Your argument is ßawed, AS THE
CALCULATIONS ABOVE SHOW. You have found no mistake above, and
so you
must agree with its conclusions: YOUR ASSERTIONS ARE FALSE.
You have not demolished anything, you have found no mistakes.
All
you have done is come to the end, found no errors, and so
declared
that it is wrong, simply because you say so.
Thats not math.
> *in the ring of algebraic integers* as youre pushed out 
of
the
>ring, when you consider factors they share with 5.
Nope: all of q(z), r(z), and s(z) are in the ring of algebraic
integers. I have been pushed nowhere. I have explicitly
exhibited
both the factor in common, and the cofactor, and they are all
algebraic integers.
What you are doing now is called sophistry and doubletalk.
> Two of them do
>share non-unit factors with 5 in a higher ring, where one is
a unit in
>that ring, which doesnt have the problem that the ring of
algebraic
>integers does.
You have found no errors, nor any reason to assert that any
of q(z),
r(z), or s(z) are not algebraic integers. They are all
algebraic
integers. The multiplications were made explicit. You are
simply
arguing in a vacuum.
>The wacky thing about the ring of algebraic integer is that
in that
>case Ôa, Ôb and 
Ôc are off in some kind of weird zone
where they
>cant be called factors of 5, in the ring of algebraic
integers, and
>neither can any of them be called units.
>So in citing that result, Arturo Magidin is using the very
error that
>Ive pointed out in a central point in his argument.
You have pointed out no error. Your only objection to the
result is
that it uses the meaning of the term. Thats nonsense. 
Thats
like
saying that one cannot use the fact that even integer means
multiple of 2 because thats the definition of 
even.
>Sneaky, eh?
No, not sneaky. Just a lot of desperate nonsense coming from
you.
>> As Dale noted, we have
>> f(r(z)) = (r(z))^3 - 969 (r(z))^2 + 315 (r(z)) + 5
>> = (8 z^2 - 4 z - 45)^3 - 969 (8 z^2 - 4 z - 45)^2
>> + 315 (8 z^2 - 4 z - 45) + 5
>> = 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3
>> + 731136 z^2 - 374400 z - 2067520.
>> Letting w(z) = 512 z^3 + 5376 z^2 - 5760 z - 31808
>> we have that
>> p(z)*w(z) = 512 z^6 - 768 z^5 - 70272 z^4 + 70592 z^3
>> + 731136 z^2 - 374400 z - 2067520
>> and therefore, f(r(z)) = p(z)*w(z). But we know that
p(z)=0, so
>> f(r(z))=0. This proves that r(z) is not a unit, and yet is
a common
>> factor of z and 5.
>Well youve proven that r(z) is not a unit, but that 
doesnt
prove the
>positive, as the ring of algebraic integers is screwed up,
as Ive
>shown.
WHAT?
You agree that r(z) is both an algebraic integer, and not a
unit.
You also agree that r(z), s(z), and q(z) are algebraic
integers.
You also agree that r(z)*q(z) = 5, so you agree that r(z) is
a NON
UNIT FACTOR OF 5 (in the ring of algebraic integers), and that
r(z)*s(z)=z, so that r(z) is a NON UNIT FACTOR OF z (in the
ring of
algebraic integers).
Therefore, you agree that r(z) is a common non-unit factor of
z and 5.
Since any common factor of z and 5 is a common factor of -z
and 5,
r(z) is a common factor of -z and 5, not a unit, and in the
ring of
algebraic integers.
>That is, Arturo Magidin has NOT proven that r(z) has
non-unit factors
>in common with 5, in the ring of algebraic integers.
r(z) ITSELF is a factor of 5, not a unit, in the ring of
algebraic
integers. You are confused. The point was to prove that ->z<-
and
->5<- have a common, non-unit factor in the ring of algebraic
integers. r(z) is that common factor.
>Here the trick is to get you to assume that r(z) not being a
unit,
I did not assume r(z) is not a unit. I proved that r(z) is
not a unit
(well, really, Dale proved it; I merely copied his
calculations, after
checking them).
>the ring of algebraic integers, proves that it has non-unit
factors in
>common with 5, which is NOT proven. And it cannot be proven.
You are confused again. r(z) ITSELF is the common factor of z
and 5.
>And thats why I say you cant attack a proof 
with a proof,
as proofs
>dont duel.
>Obviously, I can challenge any argument, which supposedly
refutes my
>own mathematical argument by presuming the very thing that I
say the
>argument refutes!!!
>It seems to me that Arturo Magidin is more interested in
*persuading*
>others rather than getting to the truth.
I have gotten to the truth. The truth is you are confused and
you are
grasping at straws.
>Possibly he hoped that his exposition here would look good
enough to
>convince without anyone paying attention to all of the facts.
>Maybe now you see why I keep saying that math is not a
fashion show.
>> Letting z be -a1, -a2, and -a3, in turn, you obtain common
factors of
>> EACH of a1, a2, a3 with 5 in the ring of all algebraic
integers, which
>> are not units.
>Sure, if you assume that the problem Ive outlined 
doesnt
exist.
I have explicitly exhibited both the factor and the cofactor.
Your
imaginary problem is a non sequitur here.
[.snip.]
>>authoritative or looks mathematical, but he rarely actually
proves
>>anything relevant to his central objections.
>> I let the record speak for itself. You have demonstrated
yourself
>> incapable of recognizing either objections, relevancy, or
proofs. I am
>> content with what my posts demonstrate, and what tone they
take. Your
>> comments are neither accurate, correct, nor relevant. They
are nothing
>> but thinly veiled personal attacks.
>Oh, so now you want readers to go the the record as if 
youre
>incapable of giving the information needed to make a
judgement in a
>post.
The record is several years of posts. If you feel up to the
task of
putting it all in a post, please do so.
I shall now delete the rest of your ad hominem and your
nonsense.
[.snip.]
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 mans 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 critize. A
great
many people are staggered to this extend, 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
magidin@math.berkeley.edu
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
it seems to be a big wash. so,
why dont you try some other problems,
such as in construtive geometry,
to get a better handle on your math?
on the other hand,
you could try reading shakespeare, as
in reading your part with others in a play;
that is a key to the English language,
from he who Classified it.
> You have proven no such thing.
> If Ôa, Ôb and 
Ôc are algebraic integers,
> then ab = 5/c, so Ôc divides 
Ô5 *in
> the ring of algebraic integers*. The same goes
> for Ôb and Ôc. 
Hence, each of them is a factor of 5 in
the ring
> of algebraic integers.
> Is 5/c one the numbers you claim *should be* an algebraic
integer, but
> which is outside the ring of algebraic
> integers? If that is your claim, it is false.
> Let a, b, and c be roots of the monic polynomial
> with integer coefficients: x^3-B x^2+Cx-5. Then a monic
> polynomial with integer coefficients whose roots are ab, bc,
and
> ab is: x^3-Cx^2+5*Bx-25.
> Therefore ab, bc, and ca *are* algebraic integers. Therefore
> 5/a, 5/b and 5/c are algebraic integers.
--les ducs dEnron!
http://members.tripod.com/~american_almanac
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
> James,
>
> You sometimes get into an argument as to whether something
is a
> function of something else. It might help if you made it
clear at the
> beginning that P is a function of f, m, u and x. And that
the b_i and
> w_i are functions of f and m. You might then be able to
show,
> subsequently, that something that looked at though it
depended on
> something else, in fact does not.
>
> For instance:
>
> Let f, m, u and x be algebraic integers.
>
> Consider the function:
>
> P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x
u^2 + u^3 f)
>
> Since the set of algebraic integers is an integral domain,
P(f,m,u,x)
> is necessarily an algebraic integer.
>
> Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer?
>
> It is possible to find functions (of f and m) b_1, b_2, b_3,
w_1, w_2,
> and w_3 such that
>
> P(f,m,u,x)/f^2 =
> (b_1(f,m) x + u w_1(f,m))*
> (b_2(f,m) x + u w_2(f,m))*
> (b_3(f,m) x + u w_3(f,m))
>
> Clearly these functions are not uniquely determined.
>
> Aside: Can the functions b_1, b_2, b_3, w_1, w_2, and w_3
all be
> chosen such that, if f and m are algebraic integers,
b_1(f,m) b_2(f,m)
> b_3(f,m) w_1(f,m) w_2(f,m) and w_3(f,m) are all algebraic
integers?
>
> If m=0, we have
>
> P(f,0,u,x)/f^2 =
> (b_1(f,0) x + u w_1(f,0))*
> (b_2(f,0) x + u w_2(f,0))*
> (b_3(f,0) x + u w_3(f,0))
>
> etc
>
> Math Fan
> This does two things which James does not seem to like.
> 1) It makes the formulas messier looking.
> 2) It forces him to clearly define what he is talking about.
> I made similar suggestions to him a month or two ago and
was informed
> that he had no interest in issues of style. After several
posts back
> and forth it became clear that he doesnt understand the
difference
> between style and clarity, nor does he seem to appreciate
that clarity
> often comes at the expense of having simple looking
expressions.
Maybe if people responding to him just adopted the P(f,m,u,x)
b_i(f,m)
w_i(f,m) terminology that would help. It would make the whole
discussion a little more understandable, to me anyway.
> The fact that this may be part of why his papers dont get
accepted
> doesnt seem to be something he understands.
Yes, to get published he would probably have to adopt the
P(f,m,u,x)
format eventually.
Math Fan
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
>
>
>>A mathematical proof begins with a truth, and proceeds by
logical
>>steps to a conclusion which then must be true.
Thats important which is why I keep emphasizing it.
>>Ive pulled a detailed exposition of a short argument that
quickly
>>shows a problem with algebraic integers. It starts after the
>>reference.
>>
>>
>>Now heres a math proof. Those who doubt that fact can
believe its a
>>claim of proof, but its verified to be a 
proof by tracing
the
>>argument out.
>>
>>In this case, I begin with an expression. The expression
exists, so
>>that is the truth from which you start.
>>
>>Consider, in the ring of algebraic integers,
>>
>> 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).
>>
>>That is, I have the identity which defines P(m) in terms of
various
>>symbols, and its all in the ring of algebraic integers,
which means
>>that the symbols can only represent numbers that are
algebraic
>>integers.
>>
>>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
>>
>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
>
> James,
>
> You sometimes get into an argument as to whether something
is a
> function of something else. It might help if you made it
clear at the
> beginning that P is a function of f, m, u and x. And that
the b_i and
> w_i are functions of f and m. You might then be able to
show,
> subsequently, that something that looked at though it
depended on
> something else, in fact does not.
Thats not a useful view and it can lead people to false
conclusions,
like the ones this poster made. See below...
> For instance:
>
> Let f, m, u and x be algebraic integers.
>
> Consider the function:
>
> P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x
u^2 + u^3 f)
>
> Since the set of algebraic integers is an integral domain,
P(f,m,u,x)
> is necessarily an algebraic integer.
>
> Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer?
>
> It is possible to find functions (of f and m) b_1, b_2, b_3,
w_1, w_2,
> and w_3 such that
>
> P(f,m,u,x)/f^2 =
> (b_1(f,m) x + u w_1(f,m))*
> (b_2(f,m) x + u w_2(f,m))*
> (b_3(f,m) x + u w_3(f,m))
>
> Clearly these functions are not uniquely determined.
Whats important here for readers is the *assertion* that 
the
ws are
functions.
Here there is the assertion that they are functions of f and
m.
Remember w_1 w_2 w_3 = f.
Also remember that when f=3, w_1, w_2, and w_3 are provably
*constants*, and in fact in that case w_1 = w_2 = 1, and w_3
= 3.
P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 -
(-1+mf^2 )x u^2 + u^3 =
(b_1/w_1 x + u)(b_2/w_2 x + u)(b_3/w_3 x + u).
So why does the poster assert that the ws are functions of 
m?
Possibly because the reality is embarrassing to
mathematicians as it
leads to the conclusion of a problem with the ring of
algebraic
integers.
> Aside: Can the functions b_1, b_2, b_3, w_1, w_2, and w_3
all be
> chosen such that, if f and m are algebraic integers,
b_1(f,m) b_2(f,m)
> b_3(f,m) w_1(f,m) w_2(f,m) and w_3(f,m) are all algebraic
integers?
>
> If m=0, we have
>
> P(f,0,u,x)/f^2 =
> (b_1(f,0) x + u w_1(f,0))*
> (b_2(f,0) x + u w_2(f,0))*
> (b_3(f,0) x + u w_3(f,0))
>
> etc
>
> Math Fan
> This does two things which James does not seem to like.
> 1) It makes the formulas messier looking.
> 2) It forces him to clearly define what he is talking about.
> I made similar suggestions to him a month or two ago and
was informed
> that he had no interest in issues of style. After several
posts back
> and forth it became clear that he doesnt understand the
difference
> between style and clarity, nor does he seem to appreciate
that clarity
> often comes at the expense of having simple looking
expressions.
Notice the insulting tone, and I want to emphasize to the
newsgroups
that these posters have a mission, which is to *convince*
others that
there isnt a problem in mathematics.
Rather than get to the truth, they continually send up
information
meant to convince other readers, and often engage in personal
attacks.
It probably IS personal to them as they work to protect math
society
by trying to keep the world from knowing about this strange,
esoteric
error with algebraic integers.
Remember above you saw an *assertion* that the ws are
functions of m,
that I destroyed simply by letting f=3.
> The fact that this may be part of why his papers dont get
accepted
> doesnt seem to be something he understands.
These posters are working a campaign to convince readers that
my work
isnt important, and there are errors, when in actuality I
keep
catching them in errors or making assertions that are false.
Mathematicians have your trust already, remember?
James Harris
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
>
>
>>A mathematical proof begins with a truth, and proceeds by
logical
>>steps to a conclusion which then must be true.
> Thats important which is why I keep emphasizing it.
>>Ive pulled a detailed exposition of a short argument that
quickly
>>shows a problem with algebraic integers. It starts after the
>>reference.
>>
>>
>>Now heres a math proof. Those who doubt that fact can
believe its
a
>>claim of proof, but its verified to be a 
proof by tracing
the
>>argument out.
>>
>>In this case, I begin with an expression. The expression
exists, so
>>that is the truth from which you start.
>>
>>Consider, in the ring of algebraic integers,
>>
>> 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).
>>
>>That is, I have the identity which defines P(m) in terms of
various
>>symbols, and its all in the ring of algebraic integers,
which means
>>that the symbols can only represent numbers that are
algebraic
>>integers.
>>
>>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
>>
>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
>
> James,
>
> You sometimes get into an argument as to whether something
is a
> function of something else. It might help if you made it
clear at the
> beginning that P is a function of f, m, u and x. And that
the b_i and
> w_i are functions of f and m. You might then be able to
show,
> subsequently, that something that looked at though it
depended on
> something else, in fact does not.
> Thats not a useful view and it can lead people to false
conclusions,
> like the ones this poster made. See below...
> For instance:
>
> Let f, m, u and x be algebraic integers.
>
> Consider the function:
>
> P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x
u^2 + u^3
f)
>
> Since the set of algebraic integers is an integral domain,
P(f,m,u,x)
> is necessarily an algebraic integer.
>
> Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer?
>
> It is possible to find functions (of f and m) b_1, b_2, b_3,
w_1,
w_2,
> and w_3 such that
>
> P(f,m,u,x)/f^2 =
> (b_1(f,m) x + u w_1(f,m))*
> (b_2(f,m) x + u w_2(f,m))*
> (b_3(f,m) x + u w_3(f,m))
>
> Clearly these functions are not uniquely determined.
> Whats important here for readers is the *assertion* that
the ws are
> functions.
> Here there is the assertion that they are functions of f
and m.
If we define (as I did earlier):
P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x u^2
+ u^3 f)
we can give values to f,m and x and get a cubic polynomial in
u. We
can then find the zeros of that polynomial.
Specifically, the zeros of P(4,9,u,1) are (roughly)
-19.02218070771525900968938
23.33415993093670440580763
102.93802077677855460388174
the zeros os P(4,10,u,1) are (roughly)
-21.15053567454956316464579
25.94497362387374822705549
114.45556205067581493759030
Do you consider the b_i and w_i to be functions of m?
> Remember w_1 w_2 w_3 = f.
> Also remember that when f=3, w_1, w_2, and w_3 are provably
> *constants*, and in fact in that case w_1 = w_2 = 1, and
w_3 = 3.
> P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 -
> (-1+mf^2 )x u^2 + u^3 =
> (b_1/w_1 x + u)(b_2/w_2 x + u)(b_3/w_3 x + u).
> So why does the poster assert that the ws are functions 
of
m?
> Possibly because the reality is embarrassing to
mathematicians as it
> leads to the conclusion of a problem with the ring of
algebraic
> integers.
> Aside: Can the functions b_1, b_2, b_3, w_1, w_2, and w_3
all be
> chosen such that, if f and m are algebraic integers,
b_1(f,m)
b_2(f,m)
> b_3(f,m) w_1(f,m) w_2(f,m) and w_3(f,m) are all algebraic
integers?
>
> If m=0, we have
>
> P(f,0,u,x)/f^2 =
> (b_1(f,0) x + u w_1(f,0))*
> (b_2(f,0) x + u w_2(f,0))*
> (b_3(f,0) x + u w_3(f,0))
>
> etc
>
> Math Fan
>
> This does two things which James does not seem to like.
>
> 1) It makes the formulas messier looking.
>
> 2) It forces him to clearly define what he is talking about.
>
> I made similar suggestions to him a month or two ago and
was informed
> that he had no interest in issues of style. After several
posts
back
> and forth it became clear that he doesnt understand the
difference
> between style and clarity, nor does he seem to appreciate
that clarity
> often comes at the expense of having simple looking
expressions.
> Notice the insulting tone, and I want to emphasize to the
newsgroups
> that these posters have a mission, which is to *convince*
others that
> there isnt a problem in mathematics.
> Rather than get to the truth, they continually send up
information
> meant to convince other readers, and often engage in
personal attacks.
> It probably IS personal to them as they work to protect
math society
> by trying to keep the world from knowing about this
strange, esoteric
> error with algebraic integers.
> Remember above you saw an *assertion* that the ws are
functions of m,
> that I destroyed simply by letting f=3.
> The fact that this may be part of why his papers dont get
accepted
> doesnt seem to be something he understands.
> These posters are working a campaign to convince readers
that my work
> isnt important, and there are errors, when in actuality I
keep
> catching them in errors or making assertions that are false.
> Mathematicians have your trust already, remember?
> James Harris
Math Fan
===
Subject: Re: Problem with Algebraic Integers: Detailed
Exposition
>
>
>>A mathematical proof begins with a truth, and proceeds by
logical
>>steps to a conclusion which then must be true.
>
> Thats important which is why I keep emphasizing it.
>
>
>>Ive pulled a detailed exposition of a short argument that
quickly
>>shows a problem with algebraic integers. It starts after the
>>reference.
>>
>>
>>Now heres a math proof. Those who doubt that fact can
believe its
a
>>claim of proof, but its verified to be a 
proof by tracing
the
>>argument out.
>>
>>In this case, I begin with an expression. The expression
exists,
so
>>that is the truth from which you start.
>>
>>Consider, in the ring of algebraic integers,
>>
>> 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).
>>
>>That is, I have the identity which defines P(m) in terms of
various
>>symbols, and its all in the ring of algebraic integers,
which
means
>>that the symbols can only represent numbers that are
algebraic
>>integers.
>>
>>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization
>>
>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)
>
>
> James,
>
> You sometimes get into an argument as to whether something
is a
> function of something else. It might help if you made it
clear at
the
> beginning that P is a function of f, m, u and x. And that
the b_i
and
> w_i are functions of f and m. You might then be able to
show,
> subsequently, that something that looked at though it
depended on
> something else, in fact does not.
>
> Thats not a useful view and it can lead people to false
conclusions,
> like the ones this poster made. See below...
>
> For instance:
>
> Let f, m, u and x be algebraic integers.
>
> Consider the function:
>
> P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x
u^2 + u^3
f)
>
> Since the set of algebraic integers is an integral domain,
P(f,m,u,x)
> is necessarily an algebraic integer.
>
> Aside: Is P(f,m,u,x)/f^2 necessarily an algebraic integer?
>
> It is possible to find functions (of f and m) b_1, b_2, b_3,
w_1,
w_2,
> and w_3 such that
>
> P(f,m,u,x)/f^2 =
> (b_1(f,m) x + u w_1(f,m))*
> (b_2(f,m) x + u w_2(f,m))*
> (b_3(f,m) x + u w_3(f,m))
>
> Clearly these functions are not uniquely determined.
>
> Whats important here for readers is the *assertion* that
the ws are
> functions.
>
> Here there is the assertion that they are functions of f
and m.
> If we define (as I did earlier):
> P(f,m,u,x)=f^2((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3(-1+mf^2)x
u^2 + u^3 f)
> we can give values to f,m and x and get a cubic polynomial
in u. We
> can then find the zeros of that polynomial.
> Specifically, the zeros of P(4,9,u,1) are (roughly)
> -19.02218070771525900968938
> 23.33415993093670440580763
> 102.93802077677855460388174
> the zeros os P(4,10,u,1) are (roughly)
> -21.15053567454956316464579
> 25.94497362387374822705549
> 114.45556205067581493759030
So?
> Do you consider the b_i and w_i to be functions of m?
The bs can be considered to be functions of m, but the 
ws
cannot
reasonably be considered to be functions or m, or even
dependent on
it.
In fact, assertions otherwise just dont make sense. 
Consider
the
numbers you used above for your second example, which are
f=4, m=10,
u=u, x=1, with P(m).
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), so
P(10) = 16((1000(256) - 3(100)(16) + 30)1^3 -
3(-1+10(16) )(1)u^2 + 4u^3), so
P(10) = 4019680 - 7632 u^2 + 64 u^3.
Now why would you ever believe that 16 divides off from that
expression as a function of m?
Why dont you do the calculation for P(9) yourself and ask
yourself
why you thought the approximations you give above are
significant.
What Ill suggest to you is that its not 
about mathematical
logic.
James Harris