"Magidin-McKinnon Theorem"
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>I'm curious about the newsgroup reaction to a claim I often seen made
>in posts that Magidin, and I guess some other guy named McKinnon
>deserve credit for the following:
>
>Any polynomial P(x) with integer coefficients of degree n can be
>written as
>
> (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
>
>And in fact not only have many claimed that Magidin and McKinnon
>(though I don't see his name come up as much) should get credit for
>this, but they have been quite heated in their defense, when I've
>challenged their assertion, and heaping in their praise of Magidin.
I'll renew, yet again, my calls for references to results of this sort
so they can be included in our manuscript. Bill Dubuque pointed out
that Kaplansky, in his book on rings, has an exercise asking to prove
that in a GCD-domain (a ring in which any two elements have a greatest
common divisor), the product of primitive polynomials is
primitive. This is a key result in the argument leading to the
statement above, but is not sufficient (that result is true in a
function field, but the analogous result above not true in the
integral closure of the corresponding number ring). Though nothing in
our manuscript is particularly earth-shattering, it involves piecing
together several known results in a way that we think is new.
But if it is not new, then we want to know who first proved. If anyone
has a reference, please send it my way.
Having done that, let me ask you, James:
Do you think the result is correct, or not? You have alternately said
it was "trivial" (though the argument you provided was flawed), and
that it was false. What is it now?
======================================================================
"Yo no estudio para escribir, ni menos para ense~ar,
[...] sino solo para ver si con estudiar ignoro menos."
[My purpose in studying is not to write, much less to
teach, but simply to see whether studying makes
me less ignorant.]
--- _Respuesta a Sor Filotea de la Cruz_
Sor Juana Ines de la Cruz (Juana de Asbaje y Ramirez)
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
David Einstein
> I'm curious about the newsgroup reaction to a claim I often seen made
> in posts that Magidin, and I guess some other guy named McKinnon
> deserve credit for the following:
>
> Any polynomial P(x) with integer coefficients of degree n can be
> written as
>
> (a1 x + b1)...(an x + bn)
>
> where the a's and b's are algebraic integers.
>
> And in fact not only have many claimed that Magidin and McKinnon
> (though I don't see his name come up as much) should get credit for
> this, but they have been quite heated in their defense, when I've
> challenged their assertion, and heaping in their praise of Magidin.
>
To quote one of the top number theorists in the world.
" You keep talking about the "Magidin-McKinnon
Theorem" and now I say let them take credit for that, since I'm now
quite certain it's wrong "
> I'm just curious if sci.math, in general, agrees. James Harris
Tapio Hurme
"David Einstein" <Dei...@world.std.com> wrote in message
news:at57i6$qjd$1...@pcls4.std.com...
> James Harris wrote:
> To quote one of the top number theorists in the world.
>
> " You keep talking about the "Magidin-McKinnon
> Theorem" and now I say let them take credit for that, since I'm now
> quite certain it's wrong "
So this "top number theorist in the world" tells about his/her own opinion,
but not about the fact that he/she can proof. Thus he/she makes the
principal error, because he/she trust in the authority more than in the
proof. Sic! :-)
Tapio
David C. Ullrich
>I'm curious about the newsgroup reaction to a claim I often seen made
>in posts that Magidin, and I guess some other guy named McKinnon
>deserve credit for the following:
>
>Any polynomial P(x) with integer coefficients of degree n can be
>written as
>
> (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
>
>And in fact not only have many claimed that Magidin and McKinnon
>(though I don't see his name come up as much) should get credit for
>this, but they have been quite heated in their defense, when I've
>challenged their assertion, and heaping in their praise of Magidin.
>
>I'm just curious if sci.math, in general, agrees. James Harris
sci.math, in general, is curious about two things:
(i) This week are you claiming the result is true or false?
You've flip-flopped on this many times, so it's impossible to guess
(ii) If you agree it's true, then who do _you_ think proved it?
David C. Ullrich
Arturo Magidin
[.snip.]
>Who do I think proved
>
>> >Any polynomial P(x) with integer coefficients of degree n can be
>> >written as
>> >
>> > (a1 x + b1)...(an x + bn)
>> >
>> >where the a's and b's are algebraic integers.
>?
>
>Gauss. Dedekind.
>
>Which answers both questions.
You are guessing, right? Or do you have actual specific references you
can give me? I would appreciate the latter.
To save you the time of going through the 12 volumes of Gauss'
complete works, the historical essay on Bourbaki's _Commutative
Algebra_ notes that algebraic integers first occur, independently, in
the works of Dirichlet, Hermite, and Eisenstein; and that only
Eisenstein goes as far as to substantially (but not completely) prove
that the sum and product of two algebraic integers is again an
algebraic integer. Eisenstein was a student of Gauss, and all of this
takes place after Gauss did his work, so it seems unlikely that you
will find anything on algebraic integers on any of Gauss' work.
Dedekind could indeed be right; however, I have not been able to
locate anything even remotely similar in his _Theory of Algebraic
Integers_. David Hilbert's book on Algebraic Number Theory, which was
an accumulation, reorganization, and redevelopement of all algebraic
number theory known at the time (including all the work of Dedekind,
of which Hilbert speaks in glowing terms, as well as the work of
Kummer, also highly praised) does not appear to contain the result,
either.
I could have missed it, and I haven't looked through all of Dedekind's
supplement's to Dirichlet's books on Number Theory, so it is possible
that Dedekind proved the result or something similar.
Do you have a reference?
If not, shall we just label what you wrote above as nothing more than
an (un)educated guess, then?
And shall we not draw our own conclusions as to what your motivation
for it might be? Hint: Shakespeare called it the green-eyed monster.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
David C. Ullrich
>David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<dsucvu0cikg6t7lfv...@4ax.com>...
>Who do I think proved
>
>> >Any polynomial P(x) with integer coefficients of degree n can be
>> >written as
>> >
>> > (a1 x + b1)...(an x + bn)
>> >
>> >where the a's and b's are algebraic integers.
>?
>
>Gauss. Dedekind.
>
>Which answers both questions.
Yes, it does answer those questions. It raises two other
questions: (i) if you think it was proved by Gauss and
Dedekind (were they working together on this or what?)
then why have you stated so many times that the result
is false? (ii) can you give any _evidence_ that Gauss
and Dedekind proved this fact?
Nobody expects you will ever reply to (i). But you
should _really_ give an answer to (ii), otherwise
people will assume you're just making it up.
>James Harris
David C. Ullrich
Dik T. Winter
> In retrospect it turned out to be fun, as after others gave the values
> for the f's and g's, and I first disputed then actually checked and
> found them to be correct, I wrote a quick little computer program to
> calculate such values for quadratics for myself.
>
> The problem gets more difficult if you move to cubics, and I'm curious
> as to whether or not anyone has ventured into that domain.
If you had read the posts by Arturo, you would have known.
> From Gauss we have the Fundamental Theorem of Algebra.
>
> From Dedekind we have algebraic integers.
>
> Considering a polynomial of degree n with integer coefficients, let me
> only label the leading or first coefficient as A.
>
> Then from Gauss' FTA I have that there are n roots of the polynomial.
>
> Designating one of those roots as r1, I conjecture that r1 is a ratio
> of algebraic integers.
You need not conjecture it. The roots of such polynomials are algebraic
numbers, and each algebraic number can be expressed as the ratio of two
algebraic integers, or of an algebraic integer and a rational integer,
whichever you want.
> Considering r1 = u1/A, where u1 is assumed to be an algebraic integer,
But why can the denominator be A, the leading coefficient of the
polynomial?
> I substitute for x in my polynomial, and now that it is zero I find
> that I need multiply by a factor of A^{n-1}, and I find myself with a
> monic expression,
I do not understand. Let's start with 5x^2 + 3x + 2. The roots are
[-3 +- sqrt(-31)]/10. So u is [-3 +- sqrt(-31)]/2, which is an
algebraic integer. So I have to multiply something (the polynomial,
I presume) by 1/5, and get x^2 + (3/5)x + (2/5), indeed a monic
expression.
> which I can generalize to r, consider it as a
> polynomial, and get n roots yet again, except now they are algebraic
> integers, as Dedekind defined same to be the roots of a monic
> polynomial with integer coefficients.
But that polynomial does not have integer coefficients. So there is
something wrong here. I think you mean something else...
> Therefore, it is proven that my u1 is an algebraic integer, as then
> are similiar u's for the remaining roots.
But *that* is well known. Given a polynomial an.x^n + ... + a0, with
integer coefficients. Set y/an = x, substitute, multiply with an^(n-1)
and get y^n + a{n-1}.an.y^(n-1) + ... + a0.an^(n-1). So the y's are
algebraic integers, and so your u's are algebraic integers.
> Therefore, the roots of the polynomial are ratios of algebraic
> integers.
That was already known earlier; the roots are algebraic numbers and
algebraic numbers are ratios of algebraic integers.
> Now writing the polynomial as
> A(x-r1)...(x-rn) = P(x)
> I use my previous result to write each r as a ratio of algebraic
> integers, now using the algebraic integers v, and w, with v coprime to
> w,
How do you define co-prime? Do you know your definition works in the
algebraic integers? If so, why?
> I have
> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
> Given that the final coefficient is an integer, I know that
> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
> (w1...wn) as a factor.
Why? How do you know that v1...vn is co-prime to w1...wn? That
each vi is co-prime to each wi is not sufficient. It is true in an UFD,
but the algebraic numbers are *not* an UFD. So the conclusion is not
justified.
> As the last coefficient is another number besides A (labeling it is
> not important) I have that A = w1...wn,
This is handwaving, there is no justification at all.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Virgil
jst...@msn.com (James Harris) wrote:
> Note the key point is the writing of the roots of the polynomial as a
> ratio of algebraic integers. As it is my knowledge that said roots
> are also called algebraic numbers it is then seen that algebraic
> numbers are ratios of algebraic integers.
>
> I thought it worth offering a link for your perusal.
>
> http://www.21stcenturysciencetech.com/articles/Spring02/Gauss_02.html
>
>
> James Harris
It might be worth noting that Lyndon H. LaRouche, Jr., whose
sponsorship is honored in the above link, is as politically and
morally bent as James S, Harris is mathematically bent.
The blind leading the blind?
Arturo Magidin
>did not cover all numbers that I called objects. I won't go into the
>details of the discussions where a problem with my reasoning was
>pointed out, except to remind that a quadratic 5x^2 + 3x + 2 = (f1 x +
>g1)(f2 x + g2) where the f's and g's are algebraic integers played a
>big part.
>
>In retrospect it turned out to be fun, as after others gave the values
>for the f's and g's, and I first disputed then actually checked and
>found them to be correct, I wrote a quick little computer program to
>calculate such values for quadratics for myself.
>
>The problem gets more difficult if you move to cubics, and I'm curious
>as to whether or not anyone has ventured into that domain.
>
>that he should get credit for work done by mathematical greats, I
>consider that now.
>
>From Gauss we have the Fundamental Theorem of Algebra.
>
>From Dedekind we have algebraic integers.
>
>Considering a polynomial of degree n with integer coefficients, let me
>only label the leading or first coefficient as A.
>
>Then from Gauss' FTA I have that there are n roots of the polynomial.
>
>of algebraic integers.
>
>Considering r1 = u1/A,
That's very bad notation. You said "A" was the leading coefficient of
your polynomial.
> where u1 is assumed to be an algebraic integer,
>I substitute for x in my polynomial, and now that it is zero I find
>that I need multiply by a factor of A^{n-1}, and I find myself with a
>monic expression, which I can generalize to r,
Lovely. Now you are cribbing from the post I made two days ago. Not
that the argument I gave was new or mine (it is standard), but you
should really stop stealing one of these days.
>consider it as a
>polynomial, and get n roots yet again, except now they are algebraic
>integers, as Dedekind defined same to be the roots of a monic
>polynomial with integer coefficients.
>
>Therefore, it is proven that my u1 is an algebraic integer, as then
>are similiar u's for the remaining roots.
This is well known: every algebraic number is the ration of two
algebraic integers.
>Therefore, the roots of the polynomial are ratios of algebraic
>integers.
>
>Now writing the polynomial as
>
> A(x-r1)...(x-rn) = P(x)
>
>I use my previous result to write each r as a ratio of algebraic
>integers, now using the algebraic integers v, and w, with v coprime to
>w,
Your "revious result" did not prove that you can choose v and w,
coprime, such that r1 = v/w. So please PROVE it.
>I have
>
> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
>
>Given that the final coefficient is an integer, I know that
>
> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
>
> (w1...wn) as a factor.
This does not follow. Even though v1 is coprime to w1, and v2 to w2,
and v3 to w3, ... and vn to wn, why do you assume that v1*...*vn is
coprime to w1*...*wn?
That is not even true in the INTEGERS. What if v1 = 2, w1 = 3, and v2
= 3, w2 = 2? Then 7*(v1*v2/w1*w2) is an integer, but 7 does not have 6
as a factor.
At this point, your "argument" is again flawed. Just like it was last
time.
Not to mention that big gap over there about writing an algebraic
number as a ration of COPRIME algebraic integers. You can prove that,
right?
>As the last coefficient is another number besides A (labeling it is
>not important) I have that A = w1...wn, so then
>
> w1...wn(x - v1/w1)...(x - vn/wn) = P(x)
>
>and it is then a trivial matter to see that
>
> (w1 x - v1)...(wn x - vn) = P(x).
Unfortunately, as noted above, your conclusion that w1*...*wn divides
A does not follow from your developement up to that point.
>Said result resulting rather easily from the wonderful work of Gauss
>and Dedekind, it seems rather odd that any modern mathematician would
>try and take credit for the result.
As noted above, said result did not in fact follow as you think it
does.
>Note the key point is the writing of the roots of the polynomial as a
>ratio of algebraic integers.
Nope. There is a lot more to it. For one, you need to write them as
ration of COPRIME algebraic integers. Please prove that can be done.
> As it is my knowledge that said roots
>are also called algebraic numbers it is then seen that algebraic
>numbers are ratios of algebraic integers.
Not enough. If you replace Q with a function field K, the algebraic
numbers with the algebraic closure of K, the integers with the
integers of K, and the algebraic integers with the algebraic integers
in the algebraic closure of K, then the Fundamental Theorem of Algebra
holds, as well as the result that every element which is algebraic
over K is a ration of two algebraic integers over K. Yet the analogous
result does not hold.
Arturo Magidin
>In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
[.snip.]
> > I substitute for x in my polynomial, and now that it is zero I find
> > that I need multiply by a factor of A^{n-1}, and I find myself with a
> > monic expression,
>
>I do not understand.
He is, unsuccessfully, copying from my post in which I recounted the
usual proof that any algebraic number can be written as the ration of
an algebraic integer by an integer. If you have
Ax^n + ... + a_0 , and r is a root, multiplying by A^{n-1} shows you
that A*r is an algebraic integer, so r = d/A, where d is an algebraic
integer.
[.snip.]
> > Now writing the polynomial as
> > A(x-r1)...(x-rn) = P(x)
> > I use my previous result to write each r as a ratio of algebraic
> > integers, now using the algebraic integers v, and w, with v coprime to
> > w,
>
>How do you define co-prime? Do you know your definition works in the
>algebraic integers? If so, why?
>
> > I have
> > A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
> > Given that the final coefficient is an integer, I know that
> > A (v1...vn)/(w1...wn) is an integer, which gives me that A has
> > (w1...wn) as a factor.
>
>Why? How do you know that v1...vn is co-prime to w1...wn? That
>each vi is co-prime to each wi is not sufficient. It is true in an
> UFD,
Not even in a UFD. Take v1=2, w1=3; v2 = 3, w2 = 2.
Andrzej Kolowski
> did not cover all numbers that I called objects. I won't go into the
> details of the discussions where a problem with my reasoning was
> pointed out, except to remind that a quadratic 5x^2 + 3x + 2 = (f1 x +
> g1)(f2 x + g2) where the f's and g's are algebraic integers played a
> big part.
>
> In retrospect it turned out to be fun, as after others gave the values
> for the f's and g's, and I first disputed then actually checked and
> found them to be correct, I wrote a quick little computer program to
> calculate such values for quadratics for myself.
>
> The problem gets more difficult if you move to cubics, and I'm curious
> as to whether or not anyone has ventured into that domain.
>
> Now as to the assertion by one Magidin and others on this newsgroup
> that he should get credit for work done by mathematical greats, I
> consider that now.
>
> From Gauss we have the Fundamental Theorem of Algebra.
>
> From Dedekind we have algebraic integers.
>
> Considering a polynomial of degree n with integer coefficients, let me
> only label the leading or first coefficient as A.
>
> Then from Gauss' FTA I have that there are n roots of the polynomial.
>
> Designating one of those roots as r1, I conjecture that r1 is a ratio
> of algebraic integers.
>
No need to conjecture. This is well-known. In fact you can
assume the denominator is a rational integer.
> Considering r1 = u1/A, where u1 is assumed to be an algebraic integer,
> I substitute for x in my polynomial, and now that it is zero I find
> that I need multiply by a factor of A^{n-1}, and I find myself with a
> monic expression, which I can generalize to r, consider it as a
> polynomial, and get n roots yet again, except now they are algebraic
> integers, as Dedekind defined same to be the roots of a monic
> polynomial with integer coefficients.
>
> Therefore, it is proven that my u1 is an algebraic integer, as then
> are similiar u's for the remaining roots.
>
> Therefore, the roots of the polynomial are ratios of algebraic
> integers.
>
> Now writing the polynomial as
>
> A(x-r1)...(x-rn) = P(x)
>
> I use my previous result to write each r as a ratio of algebraic
> integers, now using the algebraic integers v, and w, with v coprime to
> w, I have
>
> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
>
> Given that the final coefficient is an integer, I know that
>
> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
>
> (w1...wn) as a factor.
>
> As the last coefficient is another number besides A (labeling it is
> not important) I have that A = w1...wn, so then
>
What? Doesn't work -
A = a0 * (w1 ... wn) / (v1 ... vn).
> w1...wn(x - v1/w1)...(x - vn/wn) = P(x)
>
No - try this with a specific polynomial, like
3*x^2 - x - 5.
The roots are
r1 = (1 + sqrt(61))/6 and r2 = (1 - sqrt(61))/6.
Thus
v1 = (1 + sqrt(61)), w1 = 6, and
v2 = (1 - sqrt(61)), w2 = 6. But
w1 * w2 * (x - r1) * (x - r2) =
(6*x - (1 + sqrt(61))*(6*x - (1 - sqrt(61)) =
36*x^2 - 12*x - 60 = 12 * P(x).
Thus you do not get back the same polynomial.
You may think, "Well, clearly it's an equivalent polynomial."
True, in the sense that it has the same roots. But for your
application, you need EXACTLY the same polynomial, and this
doesn't do it.
You have badly underestimated how hard this theorem is
to prove.
> and it is then a trivial matter to see that
>
> (w1 x - v1)...(wn x - vn) = P(x).
>
No - it just doesn't work, as shown above.
> Said result resulting rather easily from the wonderful work of Gauss
> and Dedekind, it seems rather odd that any modern mathematician would
> try and take credit for the result.
>
What you obtain by your factorization is a multiple of
P(x), not P(x) itself. It is a trivial result and it is
not equivalent to what Magidin and McKinnon did.
Andrzej
> Note the key point is the writing of the roots of the polynomial as a
> ratio of algebraic integers. As it is my knowledge that said roots
> are also called algebraic numbers it is then seen that algebraic
> numbers are ratios of algebraic integers.
>
Arturo Magidin
[.snip.]
Yet another error:
>Now writing the polynomial as
>
> A(x-r1)...(x-rn) = P(x)
>
>I use my previous result to write each r as a ratio of algebraic
>integers, now using the algebraic integers v, and w, with v coprime to
>w,
As noted, you have not shown it is possible to choose v and w that way
(yes, it ->is<- possible, but (a) you have not shown it; and (b) it is
not an easy result at all; Dedekind calls it both "difficult" and
"deep", for example).
> I have
>
> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
>
>Given that the final coefficient is an integer, I know that
>
> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
>
> (w1...wn) as a factor.
As noted, this does not follow. The fact that for each i vi is coprime
to wi does not imply that v1*...*vn is coprime to w1*...*wn.
One can even come up with a specific example (in case anyone was
wondering).
Let f(x) = 6x^4 - 13 x^2 + 6
The roots are sqrt(2)/sqrt(3), -sqrt(2)/sqrt(3), sqrt(3)/sqrt(2),
-sqrt(3)/sqrt(2) (even written as a quotient of coprime algebraic
integers!)
since
6x^4 - 13x^2 + 6 = 6(x^4 - (13/6)x^2 + 1)
= 6(x^2 - (3/2))(x^2-(2/3))
= 6(x-sqrt(3)/sqrt(2))(x+sqrt(3)/sqrt(2))*
(x-sqrt(2)/sqrt(3))(x-sqrt(3)/sqrt(2)).
Here we have:
v1 = sqrt(3)
w1 = sqrt(2)
v2 = -sqrt(3)
w2 = sqrt(2)
v3 = sqrt(2)
w3 = sqrt(3)
v4 = -sqrt(2)
w4 = sqrt(3)
so that v1*v2*v3*v3 = 6 = w1*w2*w3*w4; it doesn't matter what A is,
A*v1*v2*v3*v3*v4/w1*w2*w3*w4 is always an integer, whether or not
w1*w2*w3*w4 divides A or not.
In fact, since there are no restrictions on the polynomial to begin
with, one can always just pick ANY polynomial f(x) =
a_nx^n+...+a_1x+a_0, then pick g(x) = a_0x^n + a_1x^{n-1} + ... +a_n,
and the product p(x) = f(x)*g(x) is ->guaranteed<- to have
v1*v2*...*vn/w1*w2*...*wn=1, since if r is a root of f(x), then 1/r is
a root of g(x); so if r= u/v, then the corresponding root of g is v/u,
and everything will cancel out.
>As the last coefficient is another number besides A (labeling it is
>not important) I have that A = w1...wn, so then
And this is the other error. This does not follow either, EVEN if we
had that v1*...*vn and w1*...*wn are coprime. The last coefficient
could be a proper multiple of v1*...vn, so that w1*...*wn would have
to be a proper divisor of A.
Multiplying the above polynomial by 7, we would have A=42,
w1*w2*w3*w4=6.
Arturo Magidin
Andrzej Kolowski <akol...@hotmail.com> wrote:
>jst...@msn.com (James Harris) wrote in message news:<3c65f87.02121...@posting.google.com>...
[.snip.]
>> Now writing the polynomial as
>>
>> A(x-r1)...(x-rn) = P(x)
>>
>> I use my previous result to write each r as a ratio of algebraic
>> integers, now using the algebraic integers v, and w, with v coprime to
>> w, I have
>>
>> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
>>
>> Given that the final coefficient is an integer, I know that
>>
>> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
>>
>> (w1...wn) as a factor.
>>
>
>
>> As the last coefficient is another number besides A (labeling it is
>> not important) I have that A = w1...wn, so then
>>
>
> What? Doesn't work -
>
> A = a0 * (w1 ... wn) / (v1 ... vn).
>
>
>> w1...wn(x - v1/w1)...(x - vn/wn) = P(x)
>>
>
>
> No - try this with a specific polynomial, like
>
> 3*x^2 - x - 5.
>
> The roots are
>
> r1 = (1 + sqrt(61))/6 and r2 = (1 - sqrt(61))/6.
Note that James asserted (but did not prove or give any references to
a proof that it could be done) that the expressions should be in
coprime algebraic integers. Are (1+sqrt(61)) and 6 coprime?
Since (1+sqrt(61))(1-sqrt(61)) = -60, and this is a product of
conjugates, both of them cannot be coprime to 2, and so not coprime to
6.
The argument James gives is unsound because he cannot conclude, from
what he has written, that w1*...*wn must divide A. It could easily
divide v1*...*vn, instead. (Not to mention that he cannot conclude
from what he has written that one may always express an algebraic
number as a ration of two algebraic integers which are ->coprime<-).
Arturo Magidin
>> In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
>> > In retrospect it turned out to be fun, as after others gave the values
>> > for the f's and g's, and I first disputed then actually checked and
>> > found them to be correct, I wrote a quick little computer program to
>> > calculate such values for quadratics for myself.
>> >
>> > The problem gets more difficult if you move to cubics, and I'm curious
>> > as to whether or not anyone has ventured into that domain.
>>
>> If you had read the posts by Arturo, you would have known.
>
>
>So why don't you give an *example* where the algebraic integer factors
>for a cubic are actually shown?
Translation: You have no clue how to do it, so you would prefer if
someone else did it for you.
Like it would be terribly hard...
2x^3 - 3 = (ax+b)(ax+wb)(ax+w^2b)
where a = cuberoot(2), b = cuberoot(3), and w = -(1/2) + sqrt(-3)/2,
all algebraic integers.
Or did you mean a nontrivial one? You should really learn to write
what you mean.
The point is that the argument given explains exactly how to go about
finding the coefficients. And that if you knew ->ANYTHING<- about
algebraic integers you would know that while the proof gives an
algorithm, the algorithm is not particularly easy to work out by hand,
even for small values of n. Even for a quadratic polynomial, you will
find it hard to find explicit values if the discriminant is large; you
never did show us your "program" which found values (and given your
insistence that you had counterexamples, I find its existence to
be... questionable), but perhaps you can post it? Perhaps you can
explain how you deal with a polynomial like
5x^2 - 6x + 15 ?
The discriminant is 26 - 300 = -264 = -4(66), so the splitting field
is Q(sqrt{-66}), which has class number 8. So you may need roots of
polynomials of degree as high as 8 before you can find algebraic
integers that will work; if the discriminant were merely -206 times a
square, you might have to go as high as degree 20.
(Some special cases are easy for quadratics: if b=0, or c=1, even
irreducible ones are easy to factor; but other values are far more
complicated)
>And I am curious as I find the area of interest, so don't act like
>there's something sinister in my request.
There's nothing sinister about it. There's just your usual mix of
pride, ignorance, condesension, and contempt for anybody who knows
more than you do.
[.snip.]
>> > From Gauss we have the Fundamental Theorem of Algebra.
>> >
>> > From Dedekind we have algebraic integers.
>> >
>> > Considering a polynomial of degree n with integer coefficients, let me
>> > only label the leading or first coefficient as A.
>> >
>> > Then from Gauss' FTA I have that there are n roots of the polynomial.
>> >
>> > Designating one of those roots as r1, I conjecture that r1 is a ratio
>> > of algebraic integers.
>>
>> You need not conjecture it. The roots of such polynomials are algebraic
>> numbers, and each algebraic number can be expressed as the ratio of two
>> algebraic integers, or of an algebraic integer and a rational integer,
>> whichever you want.
>
>Really? Where's your proof of that?
So, I guess that when you assert that an algebraic number can be
written as a ratio of two algebraic integers, which are actually
coprime, YOU have a proof of ->that<-, right?
Please post it. Otherwise, we might as well just say that your
"argument" does not pass even your own standards of "proof."
In case you were wondering: I gave the standard proof that every
algebraic number can be written as the ratio of an algebraic integer
by a rational integer. If r is a root of
a_nx^n + ... + a_1x + a_0, with a_i integers, then
a_n*r is a root of y^n + a_{n-1}a_n y^{n-1} + .... + a_1a_n^{n-1} +
a_0a_n^n, which is monic with integer coefficients, hence a_n*r is an
algebraic integer. If we write d = a_n*r, then r = d/a_n, and there it
is a ratio of an algebraic integer by a an integer. Since every
integer is an algebraic integer, this also proves that every algebraic
number is the ratio of two algebraic integers.
I eagerly await your FULL proof that any two algebraic integers have a
gcd.
>You know it's amazing to me that you don't realize that's all that's
>necessary to prove what you've been touting for months as the
>"Magidin-McKinnon Theorem", and people who wonder can see that you
>obviously have no shame.
No, it is not enough. And the reason what you write is not enough is
that there are situations in which both theorems hold, but the
factorization result does not hold. Specifically, in function fields.
You also need that any two algebraic integers have a gcd. Please prove it.
>I suggest for those who wonder to do an advanced search at
>groups.google.com for posts where Winter is the author where he talks
>of the "Magidin-McKinnon Theorem", and see just how much of a follower
>he is.
Better to follow to the right place, than lead to a dead end. Guess
where you're going?
>> > Considering r1 = u1/A, where u1 is assumed to be an algebraic integer,
>>
>> But why can the denominator be A, the leading coefficient of the
>> polynomial?
>
>Because I'm considering that possibility, and assuming that u1 is an
>algebraic integer.
Fine. And if that's not happening, then what? What if u1 and A are not
coprime, which you later demand? YOu can't just ASSUME that things
will work out nicely. If you don't ->prove<- that they do, your
argument is unsound.
>Are you seriously asking the newsgroup to wonder if
>
> u1 = r1 A exists?
No. He is asking you why do you think you can assume that if you write
r1 = u1/A, you will get u1 and A coprime, since you later use that.
[.snip.]
>I suggest to those who wonder why I dislike Magidin so much, to
>consider the kind of crap I have to deal with because in arguing with
>him and his acolytes I find them constantly attacking me on dumb
>stuff, and then chortling and partying as if they've done something
>when I get bothered by it.
Actually, the only crap that shows up in those arguments is the crap
you spew. Like the "proof" you have provided of the result, which is
unsound and incomplete, and on which you base your current host of
insults.
[.snip.]
>> > Therefore, the roots of the polynomial are ratios of algebraic
>> > integers.
>>
>> That was already known earlier; the roots are algebraic numbers and
>> algebraic numbers are ratios of algebraic integers.
>
>Really? Where's your proof?
>
>> > Now writing the polynomial as
>> > A(x-r1)...(x-rn) = P(x)
>> > I use my previous result to write each r as a ratio of algebraic
>> > integers, now using the algebraic integers v, and w, with v coprime to
>> > w,
>>
>> How do you define co-prime? Do you know your definition works in the
>> algebraic integers? If so, why?
>
>Dedekind.
Really? Where's your proof?
You know what someone who does not adhere to the standards he demands
of others is called?
HINT: starts with an 'h' and ends with a 'ypocrite'.
Arturo Magidin
>would actually consider u1/3 = r1, so your question is disingenuous.
Quite the contrary, James. You are so blinded by your unthinking hate
that you did not even notice that what I wrote was actually telling
Adrzej that his criticism of what you wrote was invalid, since he is
taking v1 = 1+sqrt(61) and w1 = 6.
But you did not notice that. How come?
>Here it'd give u1 = (1+sqrt(61))/2, and since
> (1+sqrt(61))(1-sqrt(61))/4 = 15
>
>it's clear that u1 is, while an algebraic integer, not coprime to 3.
Right. Which means that your argument as written is incomplete. While
you have said you can write each root as an algebraic integer divided
by the leading coefficient, you haven't shown that you can write them
as quotients of coprime algebraic integers.
>Now in my exposition I go on to consider a ratio of algebraic integers
>v1/w1, which *are* coprime, so you must believe that by giving r1 as
>you have that you can fool people into questioning the existence of v1
>and w1.
>
>Why cheat?
Don't know. Why do you find the need to claim I am wrong, even when I
come in on your side?
Why cheat?
>Dedekind's own work shows that (1+sqrt(61))/2 can be decomposed into
>algebraic integer factors.
This makes no sense. There IS NO UNIQUE DECOMPOSITION in algebraic
integer factors. Any factor can always be decomposed in an infinite
number of essentially distinct ways. There are no irreducibles.
If r is an algebraic integer, then you can write
r = r^{1/2} * r^{1/2} = r^{1/3}*....*r^{1/3}
= r^{1/n} * ... * r^{1/n}
= (1+sqrt(1-r))*(1-sqrt(1-r))
There is no canonical decomposition "into algebraic integer factors."
In face, Dedekind states this QUITE EXPLICITLY:
"I shall therefore end these preliminary considerations of the domain
of ALL [algebraic] integers with the remark that there are absolutely
no numbers in this domain with the character of PRIME
NUMBERS. Because, if a is a nonzero [algebraic] integer, and not a
unit, then we can decompose it in infinitely many ways into factors
which are [algebraic] integers but not units. For example, we have a =
sqrt(a)*sqrt(a), and also a=b_1*b_2 where b_1,b_2 are the two roots b
of the equation b^2 - b + a = 0."
(Richard Dedekind, _Theory of Algebraic Integers_, translated by John
Stillwell, Cambridge University Press; 1996. pp. 106, last paragraph
of Section 14 in Chapter 3; emphasis in the original. Dedekind calls
algebraic integers "integers" and uses "rational integer" or "regular
integer" for the integers, so I added the 'algebraic' above).
> Given that 3 is an algebraic integer as
>well, it can also be decomposed. As you have u1/3, then it's clear
>that u1 and 3 can both be decomposed and shared algebraic integer
>factors canceled off.
Actually, no, it is not clear at all, since the decompositions are not
at all unique, nor are they finite in number.
IN fact, this is precisely WHY Dedekind also says, just prior to the
paragraph I qouted above, that the proof that any two nonzero
algebraic integers have a greatest common divisor which can be
expressed as a linear combination of them both an "important theorem"
and "not at all easy to prove".
If we had factorization as you so simply put it, we would be able to
do this very simply, but we can't.
>Now you want credit for something that follows from the work of Gauss
>and Dedekind, and instead of at this point simply conceding that said
>credit is not due, you fight for it, and depend on snowing the reader.
The fact is, James, you have not managed to produce a correct proof
yourself. The fact is also that the result does not follow "easily"
from the results you claim it does; it ->does<- follow from some
other, very deep theorems, such as the finiteness of the class number
(proven by Dedekind), once you actually go around to doing a fair
amount of extra work.
As to wanting credit. I have made several calls for people to tell me
if the theorem, or a version thereof, has appeared in print
before. You have not been able to produce any reference, nor has
anyone else.
EVEN IF you were right that if follows "easily" (and it does not,
since the two results you mention have perfect analogues for function
fields, but the analogous conclusion is false for function fields), if
nobody had noticed it before, then the first person to notice it gets
credit.
I don't know if we are the first people to notice it. I know that I
haven't found anybody who noticed it before, nor have I found anyone
who has found anyone who has noticed it before.
Rather, it seems to me, that your hatred is ->forcing<- you to find
some way to deny that this could be an original contribution. First
you insisted it had to be false, now you insist it has to be
trivial(which begs the question... if it is ->so<- trivial, just
how dumb are you that you spent MONTHS insisting it was ->obviously<-
false?)
Here I come to point out that Andrzej did not in fact point a false
step in your argument, and you reply accusing me of cheating.
What more evidence is needed that, whatever it is that guides your
pronouncements, it isn't math, it isn't logic, and it isn't intellect?
>As I've seen you at such behavior often, I'm happy to point out to
>those reading in a case where your ego has finally caught you in an
>arena where it's harder for you to win by your usual tactics.
>
>Though eventually you'll probably start claiming Galois Theory proves
>your claims for credit.
Right. My ego. That's why I keep asking if anybody has a
reference. And that's why you have failed TWICE now to provide a
correct proof of the result.
>> Since (1+sqrt(61))(1-sqrt(61)) = -60, and this is a product of
>> conjugates, both of them cannot be coprime to 2, and so not coprime to
>> 6.
>
>However given u1/3, as I pointed out above, it *does* follow, from
>Dedekind's work, that there exists v1/w1, where v1 and w1 are coprime.
Yes, it does follow from Dedekind's work. But the argument you provide
above is simply wrong, as Dedekind himself pointed out.
So even though what you have stated is correct, you are obviously
stating it for the wrong reasons.
>Your statements don't add up Magidin,
Curious. Which statements are false? Is it not true that Andrzej took
v1=1+sqrt(61) and v2 = 1-sqrt(61)? Is it not true that he took
w1=w2=6? Is it not true that this does not satisfy the coprimeness
condition you asserted (but have not proven or given a correct
reference to)? Is it not true that I pointed this out, telling Andrzej
that his objection was not correct?
And is it not true that you came in accusing me of "cheating" for
doing that?
> and only make sense when people
>realize that you're fighting for credit, and apparently think you can
>succeed by sleight-of-hand.
Clarke's First Law says that "Any sufficiently advanced technology is
indistinguishable from magic." It is obvious that to you, any
sufficiently advanced mathematics (read: anything beyond high school)
is indistinguishable from sleight of hand.
>> The argument James gives is unsound because he cannot conclude, from
>> what he has written, that w1*...*wn must divide A. It could easily
>> divide v1*...*vn, instead. (Not to mention that he cannot conclude
>> from what he has written that one may always express an algebraic
>> number as a ration of two algebraic integers which are ->coprime<-).
>
>Which defies what many people learn as kids when they learn about
>adding fractions as I pointed out to Winter,
Hardly defies it. I already posted a polynomial where w1*...*wn
actually is EQUAL to v1*...*vn. Here it is again:
6x^4 - 13x^2 + 6.
Then v1 = sqrt(2), v2 = -sqrt(2), v3=sqrt(3), v4 = -sqrt(3), w1 =
sqrt(3), w2 = sqrt(3), w3 = sqrt(2), and w4 = sqrt(2).
Each root is expressed as a quotient of coprime algebraic integers,
just like you asked. v1 is coprime to w1, v2 to w2, v3 to w3, v4 to
w4. And yet, v1*v2*v3*v3 = w1*w2*w3*w4. So from the fact that
A*v1*v2*v3*v4 is a multiple of w1*w2*w3*w4 you cannot conclude that
w1*w2*w3*w4 must divide A; that conclusion, following from ONLY the
premise that w1*w2*w3*w4 divides A*v1*v2*v3*v4 (which is ALL you have)
requires that w1*w2*w3*w4 be coprime to v1*v2*v3*v4, and it simply is
not.
Which is why your argument is wrong.
And yet again, we see that rather than present actual specific
mathematics, you instead rely on vague pronouncements like "what many
people learn as kids when they learn about adding fractions."
That's not an argument. That's a pathetic attempt at a proof by
intimidation.
It simply won't do as proof. Not here, not anyqhere.
>one of your supporters,
>in a post earlier today by considering how the roots add to give
>integers.
And I pointed out why you are wrong.
>Your problem is that not only isn't the ring of algebraic integers a
>unique factorization domain (UFD), but algebraic integers are
>infinitely decomposable into other algebraic integers, as Dedekind
>himself pointed out.
Funny. That's YOUR Problem. Because, up there, you assumed that there
was ->a<- decomposition that allowed you do "cancel" factors. Yet how
are you going to choose from the infinitely many different
possibilities? How are you going to PROVE that you can always choose
them in such a way that "common factors" ->will<- cancel? And how will
you PROVE that if you try to do it in two different ways, the final
result will nonetheless be the same?
You CANNOT appeal to "what people learn when they learn to add
fractions", because when they learn to add fractions, NONE OF THOSE
PROBLEMS SHOW UP.
>Why you'd want to rob Dedekind is beyond me.
Simple mathematics is beyond you. But, obviously, undeserved hubris is
not.
======================================================================
"[Sir Richard Phillips] had four valuable qualities:
honesty, zeal, ability, and courage. He applied them
all to teaching matters about which he knew nothing;
and gained himself an uncomfortable life and a
ridiculous memory."
--- Augustus de Morgan, _A Budget of Paradoxes_
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
[by the by]
[.snip.]
>
>Here it'd give u1 = (1+sqrt(61))/2, and since
>
> (1+sqrt(61))(1-sqrt(61))/4 = 15
>
>it's clear that u1 is, while an algebraic integer, not coprime to 3.
I'm mighty curious about how you reach this conclusion, since the way
I know it to be true is essentially the argument I use in what you
call my "sleight of hand."
You are trying to see whether (1+sqrt(61))/2 is coprime to 3 or not.
So you check that a MULTIPLE of (1+sqrt(61))/2 is divisible by 3.
Tell me, James. Mathematically: if u*w is a multiple of 3, does that
mean that u and 3 are not coprime?
How do you know that the factor of 3 did not come from (1-sqrt(61))/2?
See, the reason ->I<- know it is because (1+sqrt(61))/2 and
(1-sqrt(61))/2 are conjugate, so one is coprime to 3 if and only if
the other is coprime to 3 (the exact same thing as the ai's in your
polynomial); so if the "factor of 3" came from (1-sqrt(61))/2, I would
know that there is also one in (1+sqrt(61))/2, which is all I need to
show they are not coprime.
Don't tell me it's "obvious". Figuring out whether an algebraic
integer has a factor of another is pretty complicated. For
example. Can you tell me if any two roots of
x^17 - 3857236 x^16 + 9287546891 x^3 - x^2 + 105678490213 x -
105943789312
are coprime or not? Can you tell me which ones have "factors of 2" and
which ones don't?
======================================================================
"Vastus animus"
-- Sallust, Bellum Catilinae
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Virgil
> So why don't you give an *example* where the algebraic integer factors
> for a cubic are actually shown?
>
> And I am curious as I find the area of interest, so don't act like
> there's something sinister in my request.
>
> It is a math newsgroup after all, and it'd be nice if you showed the
> math.
It would have been nice any time in the past few years if you had
shown the math instead of flawed analogies.
If you want others to do this sort of thing, why don't you try doing
it yourself?
A lot of us have expressed an interest in having you show actual
math.
You might even make some friends and influence some people in
positive ways.
Virgil
> > You need not conjecture it. The roots of such polynomials are algebraic
> > numbers, and each algebraic number can be expressed as the ratio of two
> > algebraic integers, or of an algebraic integer and a rational integer,
> > whichever you want.
>
> Really? Where's your proof of that?
Look who's kvetching about proofs.
Harris doesn't give proofs and doesn't accept other's proofs.
Now he bitches when a proof is omitted for a statement that has been
proved here many times.
Virgil
> Why cheat?
Just following your lead.
Arturo Magidin
>> In article <3c65f87.02121...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<H71J3...@cwi.nl>...
>> >> In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
>> >> > In retrospect it turned out to be fun, as after others gave the values
>> >> > for the f's and g's, and I first disputed then actually checked and
>> >> > found them to be correct, I wrote a quick little computer program to
>> >> > calculate such values for quadratics for myself.
>> >> >
>> >> > The problem gets more difficult if you move to cubics, and I'm curious
>> >> > as to whether or not anyone has ventured into that domain.
>> >>
>> >> If you had read the posts by Arturo, you would have known.
>> >
>> >Ah, still invoking the name of your hero I see.
>> >
>> >So why don't you give an *example* where the algebraic integer factors
>> >for a cubic are actually shown?
>>
>> Translation: You have no clue how to do it, so you would prefer if
>> someone else did it for you.
>
>the math, and wouldn't mind seeing someone succeed.
The way you choose to ASK someone is quite telling.
>> Like it would be terribly hard...
>>
>> 2x^3 - 3 = (ax+b)(ax+wb)(ax+w^2b)
>>
>> where a = cuberoot(2), b = cuberoot(3), and w = -(1/2) + sqrt(-3)/2,
>> all algebraic integers.
>
>However I'm a researcher, so trivial results don't mean squat to me.
Interesting. That must explain why you still don't believe that
algebraic integers cannot be roots of irreducible primitive
polynomials with integer coefficients whose leading coefficient is
neither 1 nor -1. It is, after all, rather trivial.
>Oh well, maybe there's someone else out there with ability, and an eye
>for interesting math results.
>
>> Or did you mean a nontrivial one? You should really learn to write
>> what you mean.
>
>Yeah, in the highly charged atmosphere of sci.math where I have to
>always expect attacks from people like you.
All the MORE reason to write what you mean. Why give people the opening
of attacking you for saying obvisouly wrong things, if you don't mean
them? Instead, take the time to write what you ->do<- mean. Maybe that
will be correct.
>> The point is that the argument given explains exactly how to go about
>> finding the coefficients. And that if you knew ->ANYTHING<- about
>> algebraic integers you would know that while the proof gives an
>> algorithm, the algorithm is not particularly easy to work out by hand,
>> even for small values of n. Even for a quadratic polynomial, you will
>> find it hard to find explicit values if the discriminant is large; you
>> never did show us your "program" which found values (and given your
>> insistence that you had counterexamples, I find its existence to
>> be... questionable), but perhaps you can post it? Perhaps you can
>> explain how you deal with a polynomial like
>>
>> 5x^2 - 6x + 15 ?
>>
>> The discriminant is 26 - 300 = -264 = -4(66), so the splitting field
>> is Q(sqrt{-66}), which has class number 8. So you may need roots of
>> polynomials of degree as high as 8 before you can find algebraic
>> integers that will work; if the discriminant were merely -206 times a
>> square, you might have to go as high as degree 20.
>
>The program simply scanned till it got an answer. Computers don't get
>excited about such things, they just do it.
Oh, it "scanned"? What did it scan? How did it scan them? How did it
write down the solution? It may be impossible to express the
coefficients for the quadratic above in terms of radicals of
integers.
>It wasn't a long program.
Great. Then you can re-do it, post it, shows us how it works.
Or not. Then we can form our own conclusions as to whether such a
program actually ever existed, or if you just figured out you could
come up with one and never actually bothered to find out if it worked.
>> (Some special cases are easy for quadratics: if b=0, or c=1, even
>> irreducible ones are easy to factor; but other values are far more
>> complicated)
>
>Not for a computer. It doesn't care--for quadratics that is.
Excellent. Please post the program. We'll see how it works.
[.snip.]
>> >> > From Gauss we have the Fundamental Theorem of Algebra.
>> >> >
>> >> > From Dedekind we have algebraic integers.
>> >> >
>> >> > Considering a polynomial of degree n with integer coefficients, let me
>> >> > only label the leading or first coefficient as A.
>> >> >
>> >> > Then from Gauss' FTA I have that there are n roots of the polynomial.
>> >> >
>> >> > Designating one of those roots as r1, I conjecture that r1 is a ratio
>> >> > of algebraic integers.
>> >>
>> >> You need not conjecture it. The roots of such polynomials are algebraic
>> >> numbers, and each algebraic number can be expressed as the ratio of two
>> >> algebraic integers, or of an algebraic integer and a rational integer,
>> >> whichever you want.
>> >
>> >Really? Where's your proof of that?
>>
>> So, I guess that when you assert that an algebraic number can be
>> written as a ratio of two algebraic integers, which are actually
>> coprime, YOU have a proof of ->that<-, right?
>
>I have a proof that any algebraic number is a ratio of algebraic
>integers.
Which does not prove that any algebraic number is the ratio of two
COPRIME algebraic integers.
The argument you gave to dervie it from "ratio of algebraic integers"
is false.
->Do<- you have a correct proof that any algebraic integer is the
ratio of two COPRIME algebraic integers?
>Winter just said it was true, and I asked him if he could prove it.
Right. Because, as usual, you can make any demands of others, but no
one can make any of you.
>> Please post it. Otherwise, we might as well just say that your
>> "argument" does not pass even your own standards of "proof."
>
>I already posted the proof for what I actually said, though you tried
>to add to it, and you've replied to what I posted in this thread.
The argument you posted is incorrect. You started from the fact that
every algebraic number is the ratio of algebraic integers. That's
correct. Then you ASSERTED that you could cancel out common
factors. You need to prove that. You attempted to justify it by
mentioning that Dedekind said that every algebraic integer can be
factored in an infinite number of ways. Although that statement is
true, that statement does NOT imply that you can "cancel out" common
factors.
I know you ->hate it<- when I give you actual math, so here it is:
Consider A = C[x,y]/(y^2-x^3+3x-49).
This is an integral domain. Let K be its field of fractions. Let F be
the algebraic closure of K, so every polynomial with coefficients in K
can be written as a product of linear terms (the equivalent of the
Fundamental Theorem of Algebra).
Let M be the collection of all elements of F which are the roots of
monic polynomials with coefficients in A. This is the exact equivalent
of "algebraic integers."
Every element of F can be written as a ratio of two elements of M.
Every element of M can be decomposed in infinitely many ways, since,
for example, if r is a root of a monic polynomial with coefficients in
A, then so is sqrt(r), and the n-th root of r, and so on and so forth,
and also things like 3 - (sqrt(3-r)) and 3+sqrt(3-r), etc.
Both 7+y and x are elements of M. They are NOT coprime. Yet (7+y)/x
CANNOT be written as a ratio of two elements of M which are
coprimie. There is "common factor" that can be cancelled.
Now, there is an example in which ALL the hypothesis you claim to use
to prove the result for algebraic integers hold. And yet the
conclusion does not hold.
Either you are using some hypothesis you have not noticed, and/or your
argument is incorrect.
>> In case you were wondering: I gave the standard proof that every
>> algebraic number can be written as the ratio of an algebraic integer
>> by a rational integer. If r is a root of
>
>I'm not wondering and I don't care, as I've given a simple, short
>proof in this very thread, and you've replied to it.
Yeah. One has to wonder where you got it from. I posted it two days
ago. Did you read ->that<- one?
>> a_nx^n + ... + a_1x + a_0, with a_i integers, then
>>
>> a_n*r is a root of y^n + a_{n-1}a_n y^{n-1} + .... + a_1a_n^{n-1} +
>> a_0a_n^n, which is monic with integer coefficients, hence a_n*r is an
>> algebraic integer. If we write d = a_n*r, then r = d/a_n, and there it
>> is a ratio of an algebraic integer by a an integer. Since every
>> integer is an algebraic integer, this also proves that every algebraic
>> number is the ratio of two algebraic integers.
>>
>> I eagerly await your FULL proof that any two algebraic integers have a
>> gcd.
>
>Hmmm...
I see. No proof, huh?
>> >You know it's amazing to me that you don't realize that's all that's
>> >necessary to prove what you've been touting for months as the
>> >"Magidin-McKinnon Theorem", and people who wonder can see that you
>> >obviously have no shame.
>>
>> No, it is not enough. And the reason what you write is not enough is
>> that there are situations in which both theorems hold, but the
>> factorization result does not hold. Specifically, in function fields.
>>
>> You also need that any two algebraic integers have a gcd. Please prove it.
>
>What's the metric for algebraic integers?
What are you talking about? "metric"?
"gcd" stands for "greatest common divisor".
An element x is a gcd of a and b if and only if it satisfies two
conditions:
(1) x divides a and x divides b; and
(2) if y is any element which divides a and divides b, then y
divides y.
So, please prove that if a and b are algebraic integers, they have an
algebraic integer gcd. You need that for the coprimeness result.
>Besides, I only need the coprimeness result, which I will admit.
Can you prove it?
Have you seen a proof? You've heard us say it is true, but since we
are such dastardly liars, so incompetent as mathematicians, how can
you possibly be sure it is true? We've ->said<- it's in books, but
then, we've said the result about roots of nonmonic primitive
polynomials with integer coefficients is in books.
You believe one, but not the other. Why?
Because one is convenient, and the other is decidedly
->in<-convenient?
How... conveninent.
>However, given that algebraic integers are infinitely decomposable, it
>follows that if I have b/a where b and a are not coprime that I can
>decompose into a product of algebraic integers, and cancel off shared
>factors to get u/v where u and v *are* coprime.
No, it does NOT follow from "infinitely decomposable" at all. The
example above shows a sitaution in which the exact analogue is
infinitely indecomposable, and yet there is no way to cancel off
shared factors. For one thing, you have not shown that "coprime" means
"have no common divisors other than units." THEY ARE NOT EQUIVALENT
(well, they ->are<- for algebraic integers, but to show that you need
to show that any two algebraic integers have a gcd, so you cannot
invoke it in order to prove it).
See, you seem to be confusing two different things. "infinitely
decomposable" means that there infinitely many different ways of
expressing it is a product. It does NOT mean that you can specify how
you want to decompose them and you will always be able to do so.
>> >I suggest for those who wonder to do an advanced search at
>> >groups.google.com for posts where Winter is the author where he talks
>> >of the "Magidin-McKinnon Theorem", and see just how much of a follower
>> >he is.
>>
>> Better to follow to the right place, than lead to a dead end. Guess
>> where you're going?
>
>Hmmm...cryptic.
>
>If people do the search I suggested they will find lots of times that
>Winter glowingly talks of the "Magidin-McKinnon Theorem" over which
>you're fighting now to keep claiming credit.
>
>Now then, what is this that you're claiming credit for?
>
>Given a polynomial P(x) of degree n, there exists the factorization
>
> (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
Polynomial with integer coefficients.
>That's it, and you claim that you should get credit as if Dedekind
>hadn't a clue of such a thing!!!
No. I claim that IF nobody can find ANY reference of ANYBODY else
proving it before we did, THEN we should get credit.
You don't get to say "it's so obvious someone must have thought of
it", particularly not when your "proof" of obviousness is so flawed
that it wouldn't even suffice to prove that if a polynomial with
integer coefficients factors into a product with rational
coefficients, you can find polynomials with integer coefficients that
do the same thing. Your argument is full of holes, misunderstandings,
and at least one false claim.
So. Can you PROVE that somebody else proved it before? Or is the best
you can do whine about how trivial and obvious it is? (so, you never
did answer, if it is so trivial and obvious, just WHY did it take you
months to figure out it was true, and you kept insisting it was
definitely false?)
That's the way it goes: if nobody did it before, it doesn't matter how
easy it is from what others did.
And in fact, this is not as easy as you think it is. Only your
perennial ignorance allows you to miss the subtleties that are involved.
The argument you gave does not work.
>> >> > Considering r1 = u1/A, where u1 is assumed to be an algebraic integer,
>> >>
>> >> But why can the denominator be A, the leading coefficient of the
>> >> polynomial?
>> >
>> >Because I'm considering that possibility, and assuming that u1 is an
>> >algebraic integer.
>>
>> Fine. And if that's not happening, then what? What if u1 and A are not
>> coprime, which you later demand? YOu can't just ASSUME that things
>> will work out nicely. If you don't ->prove<- that they do, your
>> argument is unsound.
>
>I don't assume u1 and A are coprime. I only rely on them both being
>algebraic integers, as I show how trivial it is to prove that roots of
>a polynomial with integer coefficients are ratios of algebraic
>integers.
That is not difficult. The difficulty is proving they are ratio of two
algebraic integers which are COPRIME.
You use ->that<-, much stronger fact. Can you PROVE it?
>Now it seems to me that you're insinuating otherwise in your typical
>type of tricky argument.
Yeah. It is obvious that your inability to understand even your own
"arguments" lead you to all sorts of interesting conclusions.
>However, it also seems to me that you're moving in shallow waters, and
>people can see you thrashing about this time, finally.
Interesting claim. Given that you still have failed to provide a
coherent and correct proof.
>> >Are you seriously asking the newsgroup to wonder if
>> >
>> > u1 = r1 A exists?
>>
>> No. He is asking you why do you think you can assume that if you write
>> r1 = u1/A, you will get u1 and A coprime, since you later use that.
>
>Nope. I later use v1/w1, where v1 and w1 are coprime. I never claim
>that u1 and A are coprime. (Note for the reader: A is an integer, u1
>is an algebraic integer.)
>
>As that is rather clear, it'd help if you explained why you make such
>an accusation.
"Accusation"? I told you what he was asking. You never said, in your
argument, how you went from "the root is u1/A, u1 an algebraic
integer" to "the root is v1/w1, with v1 and w1 coprime". You just SAID
that such v1 and w1 existed. Since the only such algebraic integers
you had exhibited were u1 and A, the ONLY possibility is v1=u1 and
w1=A, unless you explain how to obtain them in so other way.
That's how a proof works: no gaps allowed. No guessing allowed. You
must provide ALL the details.
>> [.snip.]
>>
>> >I suggest to those who wonder why I dislike Magidin so much, to
>> >consider the kind of crap I have to deal with because in arguing with
>> >him and his acolytes I find them constantly attacking me on dumb
>> >stuff, and then chortling and partying as if they've done something
>> >when I get bothered by it.
>>
>> Actually, the only crap that shows up in those arguments is the crap
>> you spew. Like the "proof" you have provided of the result, which is
>> unsound and incomplete, and on which you base your current host of
>> insults.
>
>Well, anyone who looks over what you're trying to claim credit for,
>which I'll give again, should wonder why you're fighting so hard for
>credit in an arena where it looks to me like you don't have a valid
>claim to credit.
>
>Given a polynomial P(x) of degree n, there exists the factorization
>
> (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
>
>Others might be surprised but I see this as something you are
>unethical enough to try and pull off, and expect to get away with it,
>as your endline to me indicates not only your pride in fooling people,
>but in making sure that you actually tell them.
And everyone can see that your "proof" is faulty, and contains
gaps. That you continue to misunderstand and misuse known results.
>> >> > Therefore, the roots of the polynomial are ratios of algebraic
>> >> > integers.
>> >>
>> >> That was already known earlier; the roots are algebraic numbers and
>> >> algebraic numbers are ratios of algebraic integers.
>> >
>> >Really? Where's your proof?
>> >
>> >> > Now writing the polynomial as
>> >> > A(x-r1)...(x-rn) = P(x)
>> >> > I use my previous result to write each r as a ratio of algebraic
>> >> > integers, now using the algebraic integers v, and w, with v coprime to
>> >> > w,
>> >>
>> >> How do you define co-prime? Do you know your definition works in the
>> >> algebraic integers? If so, why?
>> >
>> >Dedekind.
>>
>> Really? Where's your proof?
>
>Dedekind showed that algebraic integers are infinitely decomposable.
Please think about what you are saying.
"infinitely decomposable" means:
there exist an infinite numbers of ->essentially<- distinct ways of
writing an algebraic integer as a product of a finite number of
algebraic integers.
How does ->THAT<- prove that if two algebraic integers have no common
divisor other than units, then they are coprime? How does that show
that given any two algebraic integers a and b, there exist (presumably
unique) algebraic integer u1 and v1 such that u1/v1 = a/b?
It does not.
You are just showing your ignorance again.
>> You know what someone who does not adhere to the standards he demands
>> of others is called?
>>
>> HINT: starts with an 'h' and ends with a 'ypocrite'.
>
>
>Which people should consider when they read your endline which
>follows.
You demand proof. You never provide it. You demand deetails, you never
provide them. And when confronted with details, you bitch and moan
that people are trying to confuse you "with math".
You demand standards of others, but you don't adhere to them.
That makes you a hypocrite.
>For those who can't believe this can go on, they need to remember that
>Magidin *usually* gets away with it.
Because I am *usually* right. I know, you can't even consider the
possibility. But, hey, lots of things you haven't considered possible
are actually true.
>Often people such as him refuse to accept when things change, so I'm
>sure he will continue to believe that he can still fool you for quite
>some time longer.
Well, see, ->YOU<- can put an end to that.
You claim there is a simple proof that exactly two of the ai's are
divisible by f^j. Provide it. No examples. No "it's just like". No
analogies. Just a proof.
That should make it obvious for anyone who is right and who isn't,
right?
You claim that it is clear that the result I have often posted (and
proved on-line) about divisors of the coefficients in the
factorization is false. All you have to do is post a specific
polynomial which meets the hypotehsis but fails to meet the
condition. Then everyone can see who is right and who isn't.
But, no. YOu think it is better to just bitch and moan about how I am
"confusing" everyone and "getting away with" lies and everything.
It's within your grasp to shut me up. JUST POST ACTUAL PROOFS. JUST
POST ACTUAL EXAMPLES.
What is so hard about that?
Maybe that you've tried to find some examples, and coulnd't? Maybe
that whenever you've tried your hand at even simple proofs, you've
failed miserably?
Maybe.
Arturo Magidin
>> >So why don't you give an *example* where the algebraic integer factors
>> >for a cubic are actually shown?
>>
>> Translation: You have no clue how to do it, so you would prefer if
>> someone else did it for you.
>
>Oh I admit that I don't quite see how to do it, but I'm intrigued by
>the math, and wouldn't mind seeing someone succeed.
The manuscript which you so decry tells you exactly how to go about
doing it, if you ever bothered to read it.
It is difficult to do in practice, but doable in principle.
http://www.math.umt.edu/~magidin/preprints/gauss.pdf
That's why I am so interested to see that mythical program of yours
which could do it for quadratics. I cannot think of a way to just
"scan" (scan what?) to find them, although I can figure out how to
find them for any specific example (though it may take considerable
amount of time); I can think of a few things I can program a computer
to do, but cannot imagine a program that can just "do it", as you
claim yours did.
Now, I'm not a computer programmer, so maybe you can show us yours. At
one point John Ramsden posted a few ideas, and only with some very
strong hypothesis (which are not always met) did he come up with some
formulas.
Virgil
> However I'm a researcher, so trivial results don't mean squat to me.
>
> Oh well, maybe there's someone else out there with ability, and an eye
> for interesting math results.
It would appear from your history that even otherwise important
results that do not redound to your own personal fame and fortune
don't mean squat to you.
Virgil
> > >And I am curious as I find the area of interest, so don't act like
> > >there's something sinister in my request.
> >
> > There's nothing sinister about it. There's just your usual mix of
> > pride, ignorance, condesension, and contempt for anybody who knows
> > more than you do.
>
> Nope.
Yup.!
Virgil
> However, given that algebraic integers are infinitely decomposable, it
> follows that if I have b/a where b and a are not coprime that I can
> decompose into a product of algebraic integers, and cancel off shared
> factors to get u/v where u and v *are* coprime.
Given that Harrisonian mathematical reasoning is infinitely
decomposable, it dissolves from its inner contradictions until all
that remains of Harrisonian "proofs" is a Harrisonian wish to have
something true.
Unless it is a result "borrowed" from someone whose mathematical
proofs are less ephemeral.
John
> In article <3c65f87.02121...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<atd8b1$1bvj$1...@agate.berkeley.edu>...
> >> In article <3c65f87.02121...@posting.google.com>,
> >> James Harris <jst...@msn.com> wrote:
> >> >"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<H71J3...@cwi.nl>...
> >> >> In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
> >> >> > In retrospect it turned out to be fun, as after others gave the values
> >> >> > for the f's and g's, and I first disputed then actually checked and
> >> >> > found them to be correct, I wrote a quick little computer program to
> >> >> > calculate such values for quadratics for myself.
> >> >> >
> >> >> > The problem gets more difficult if you move to cubics, and I'm curious
> >> >> > as to whether or not anyone has ventured into that domain.
> >> >>
> >> >> If you had read the posts by Arturo, you would have known.
> >> >
> >> >Ah, still invoking the name of your hero I see.
> >> >
> >> >So why don't you give an *example* where the algebraic integer factors
> >> >for a cubic are actually shown?
> >>
> >> Translation: You have no clue how to do it, so you would prefer if
> >> someone else did it for you.
Isn't it to be expected that *anyone* with academic transcripts
like the following will be more likely to be able to prove this
than JSH?
811 INTRO TOPOLOGY ANAL MATH 202A 4.0 A- 14.8
812 METAMATHEMATICS MATH 225A 4.0 A 16.0
813 GRPS RINGS & FIELDS MATH 250A 4.0 A 16.0
12.0* 46.8*
SPRING SEMESTER 1994
814 INTRO TOPOLOGY ANAL MATH 202B 4.0 A+ 16.0
815 MULTILINEAR ALGEBRA MATH 250B 4.0 A+ 16.0
816 REPRESENTATIONS MATH 252 4.0 A 16.0
12.0* 48.0*
FALL SEMESTER 1994
817 ALGEBRAIC TOPOLOGY MATH 215A 4.0 A+ 16.0
818 RING THEORY MATH 251 4.0 S SF
819 NUMBER THEORY MATH 254A 4.0 A 16.0
8.0* 32.0*
SPRING SEMESTER 1995
820 THRY ALGB STRUCTS MATH 245A 4.0 A+ 16.0
821 NUMBER THEORY MATH 254B 4.0 A 16.0
SPRING SEMESTER 1995
820 THRY ALGB STRUCTS MATH 245A 4.0 A+ 16.0
821 NUMBER THEORY MATH 254B 4.0 A 16.0
822 TEACHING WORKSHOP MATH 300 3.0 S SU
8.0* 32.0*
FALL SEMESTER 1995
823 ELEMENTARY JAPANESE JAPAN 1A 5.0 S SF
824 ALGEBRAIC GEOMETRY MATH 256A 4.0 A+ 16.0
825 INDIVIDUAL RESEARCH MATH 295 4.0 S SU
4.0* 16.0*
SPRING SEMESTER 1996
826 INDIVIDUAL RESEARCH MATH 295 9.0 S SF
0.0* 0.0*
FALL SEMESTER 1996
827 INDIVIDUAL RESEARCH MATH 295 8.0 A 32.0
8.0* 32.0*
SPRING SEMESTER 1997
828 INDIVIDUAL RESEARCH MATH 295 8.0 S SU
0.0* 0.0*
FALL SEMESTER 1997
829 INDIVIDUAL RESEARCH MATH 295 8.0 S SU
0.0* 0.0*
SPRING SEMESTER 1998
830 INDIVIDUAL RESEARCH MATH 295 8.0 S SU
0.0* 0.0*
TOTAL SATISF/UNSATISF ATTM 49.0 PASSED 49.0
SEMESTER CREDITS COMPLETED 101.0 UC GPA 3.977
******************************************************
Undergraduate
UNIVERSIDAD NACIONAL AUTONOMA DE MEXICO
PRIMER SEMESTRE
003 0091 18 CALCULO DIFERENC E INTEGR I MB
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003 0092 18 CALCULO DIFERENC E INTEGR II MB
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003 0008 10 ALGEBRA SUPERIOR II MB
003 0093 18 CALCULO DIFERENC E INTEGR III MB
003 0245 10 GEOMETRIA ANALITICA II MB
003 0625 10 PROBABILIDAD I MB
CUARTO SEMESTRE
003 0006 10 ALGEBRA LINEAL II MB
003 0094 18 CALCULO DIFERENC E INTEGR IV MB
003 0162 10 ECUACIONES DIFERENCIALES I MB
003 0398 10 ESTADISTICA I MB
QUINTO SEMESTRE
003 0001 10 ALGEBRA MODERNA I MB
003 0009 10 ANALISIS MATEMATICO I MB
003 0840 10 VARIABLE COMPLEJA I MB
ASIGNATURAS OPTATIVAS
003 0002 10 ALGEBRA MODERNA II MB
003 0003 10 ALGEBRA MODERNA III MB
003 0004 10 ALGEBRA MODERNA IV MB
003 0010 10 ANALISIS MATEMATICO II MB
003 0011 10 ANALISIS MATEMATICO III MB
003 0118 10 COMPUTACION I MB
003 0242 10 GEOMETRIA ALGEBRAICA I MB
003 0406 10 ESTRUCTURAS DE DATOS MB
003 0422 10 GRAFICAS Y JUEGOS MB
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003 0629 10 PROGRAMACION DE SISTEMAS MB
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Source: <http://www.math.umt.edu/~magidin/applic/transcript.html>
> >Oh I admit that I don't quite see how to do it, but I'm intrigued by
> >the math, and wouldn't mind seeing someone succeed.
>
> The way you choose to ASK someone is quite telling.
Only an ingrate--and a sadistic one at that--would do his best to
humiliate and degrade an amateur whose assertions and gaps--and yes,
whose MISTAKES--have caused him to think productively.
> >> Like it would be terribly hard...
> >>
> >> 2x^3 - 3 = (ax+b)(ax+wb)(ax+w^2b)
> >>
> >> where a = cuberoot(2), b = cuberoot(3), and w = -(1/2) + sqrt(-3)/2,
> >> all algebraic integers.
> >
> >However I'm a researcher, so trivial results don't mean squat to me.
>
> Interesting. That must explain why you still don't believe that
> algebraic integers cannot be roots of irreducible primitive
> polynomials with integer coefficients whose leading coefficient is
> neither 1 nor -1. It is, after all, rather trivial.
> >Oh well, maybe there's someone else out there with ability, and an eye
> >for interesting math results.
> >
> >> Or did you mean a nontrivial one? You should really learn to write
> >> what you mean.
> >
> >Yeah, in the highly charged atmosphere of sci.math where I have to
> >always expect attacks from people like you.
>
> All the MORE reason to write what you mean. Why give people the opening
> of attacking you for saying obvisouly wrong things, if you don't mean
> them? Instead, take the time to write what you ->do<- mean. Maybe that
> will be correct.
GOOD professors are aware of the ladders they have climbed that others
have not. And they don't humiliate others--in public or otherwise--for
not having climbed them.
And if it doesn't work, will you rub his nose in it?
--John
David Kastrup
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<ate2n4$1m08$1...@agate.berkeley.edu>...
> > In article <3c65f87.02121...@posting.google.com>,
> > James Harris <jst...@msn.com> wrote:
> > >>
> > >> Translation: You have no clue how to do it, so you would prefer if
> > >> someone else did it for you.
>
> Isn't it to be expected that *anyone* with academic transcripts
> like the following will be more likely to be able to prove this
> than JSH?
Yes. That is why James could ask politely if he wants to know
something instead of trying to get the information he wants by
insinuating that nobody else could come up with it.
People ask question and get answers here _all_ the time. It is James
special style to not ask, but _taunt_ for information. So you may
expect that this style gets its comments.
> > >Oh I admit that I don't quite see how to do it, but I'm intrigued by
> > >the math, and wouldn't mind seeing someone succeed.
> >
> > The way you choose to ASK someone is quite telling.
>
> Only an ingrate--and a sadistic one at that--would do his best to
> humiliate and degrade an amateur whose assertions and gaps--and yes,
> whose MISTAKES--have caused him to think productively.
Uh, could you explicate how the above quote from Magidin about James'
style of trying to elicit information is humiliating and degrading? I
am afraid I don't get it.
> > All the MORE reason to write what you mean. Why give people the
> > opening of attacking you for saying obvisouly wrong things, if you
> > don't mean them? Instead, take the time to write what you ->do<-
> > mean. Maybe that will be correct.
>
> GOOD professors are aware of the ladders they have climbed that others
> have not. And they don't humiliate others--in public or otherwise--for
> not having climbed them.
And good students don't call their teachers names and "challenge" them
as a prerequisite to asking information.
> > >Not for a computer. It doesn't care--for quadratics that is.
> >
> >
> > Excellent. Please post the program. We'll see how it works.
>
> And if it doesn't work, will you rub his nose in it?
According to your taxonomy, pretty certainly. Wrong things that get
posted here get pointed out as a rule.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<ate2n4$1m08$1...@agate.berkeley.edu>...
>> In article <3c65f87.02121...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<atd8b1$1bvj$1...@agate.berkeley.edu>...
>> >> In article <3c65f87.02121...@posting.google.com>,
>> >> James Harris <jst...@msn.com> wrote:
>> >> >"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<H71J3...@cwi.nl>...
>> >> >> In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
>> >> >> > In retrospect it turned out to be fun, as after others gave the values
>> >> >> > for the f's and g's, and I first disputed then actually checked and
>> >> >> > found them to be correct, I wrote a quick little computer program to
>> >> >> > calculate such values for quadratics for myself.
>> >> >> >
>> >> >> > The problem gets more difficult if you move to cubics, and I'm curious
>> >> >> > as to whether or not anyone has ventured into that domain.
>> >> >>
>> >> >> If you had read the posts by Arturo, you would have known.
>> >> >
>> >> >Ah, still invoking the name of your hero I see.
>> >> >
>> >> >So why don't you give an *example* where the algebraic integer factors
>> >> >for a cubic are actually shown?
>> >>
>> >> Translation: You have no clue how to do it, so you would prefer if
>> >> someone else did it for you.
>
>Isn't it to be expected that *anyone* with academic transcripts
>like the following will be more likely to be able to prove this
>than JSH?
[...]
Of course it is. That's why James's insistence that he's one of
the top number theorists in the world, his continued undocumented
claims of proofs of all sorts of things, his insistence that all
the experts are _wrong_ on various topics, etc, is so, um,
let's say "curious".
You seem to think you're making some sort of point in James'
favor by exhibiting Magidin's credentials as a competent
mathematician. But James continually insists that he's a
total incompetent, that it's hard to see how he got a
degree.
>Source: <http://www.math.umt.edu/~magidin/applic/transcript.html>
>
>> >Oh I admit that I don't quite see how to do it, but I'm intrigued by
>> >the math, and wouldn't mind seeing someone succeed.
>>
>> The way you choose to ASK someone is quite telling.
>
>Only an ingrate--and a sadistic one at that--would do his best to
>humiliate and degrade an amateur whose assertions and gaps--and yes,
>whose MISTAKES--have caused him to think productively.
"amateur"? James is one of the world's top number theorists!
"productively"? There's _no_ evidence that his "research" on
FLT has produced anything useful.
It certainly has not produced a proof of FLT. Your comments
would make more sense if James said "oops" when people pointed
out his errors. Except then your comments would not exist in
the first place, because people would not have the attitude
they do towards him. He doesn't say oops when a mistake is
pointed out, he invariably calls the person pointing out the
error a lying incompetent.
[...]
>
>GOOD professors are aware of the ladders they have climbed that others
>have not. And they don't humiliate others--in public or otherwise--for
>not having climbed them.
James humiliates himself.
"GOOD professors" indeed. If I spoke to a student in class the
way I sometimes speak to James I'd get in big trouble. But that
doesn't come up, because I _don't_ speak to students in class
that way. This has something to do with the fact that when I'm
teaching class the students do not call me a lying incompetent
when I say something they don't understand.
[...]
>>
>>
>> Excellent. Please post the program. We'll see how it works.
>
>And if it doesn't work, will you rub his nose in it?
If he doesn't I will. If James doesn't like having his nose
rubbed in things there's a simple solution: He could stop
posting things that are utterly wrong; stop insisting he
undestands the math better than the mathematicians while
_exhibiting_ abysmal ignorance.
>--John
David C. Ullrich
Moufang Loop
>
> ...
>
> Just look at the automorphisms.
>
When James uses a bit of technical mathematese like "automorphism" does
he know what he's talking about?
ML
Virgil
john_...@yahoo.com (John) wrote:
> Only an ingrate--and a sadistic one at that--would do his best to
> humiliate and degrade an amateur whose assertions and gaps--and yes,
> whose MISTAKES--have caused him to think productively.
Only and ingrate-- and a sadistic one at that-- would reject the
many opportunities to learn what he needs to know to achieve his
ambitions that have been offered to him br Magidin et al.
JSH's mistakes, without the help of Magidin et al, would have
forever remained unchanged. Any productivity JSH may have achieved
is due to tuition by others which he uses but ungratefully denies
using.
Any improvements in JSH's claim towards a proof of FLT must be
credited to others pointing out what would not work. JSH has not
knowledge enough to find those faults unaided.
Then he attacks anyone who attempts to help him.
Then he attacks anyone who objects to his attacks.
Any humiliation and degradation which has fallen on him has been
brough upon him by his own selfishness and blind ambition.
Virgil
> Well, as I explained above I know that (1+sqrt(61))/2 cannot be a
> ratio of the leading coefficient and some other algebraic integer
> where the leading coefficient is coprime to that leading coefficient.
The only way I can see for the leading coefficient to be coprime to
the leading coefficient is for it to be a unit. In which case, it is
not divisible by 3 or any other non-unit.
But perhaps,James, in his infinite wisdom, can wiggle his way out of
this contradiction.
John
No less curious than the sci.math abuse of JSH that leads him to
make such claims.
>
> You seem to think you're making some sort of point in James'
> favor by exhibiting Magidin's credentials as a competent
> mathematician. But James continually insists that he's a
> total incompetent, that it's hard to see how he got a
> degree.
>
> >Source: <http://www.math.umt.edu/~magidin/applic/transcript.html>
> >
> >> >Oh I admit that I don't quite see how to do it, but I'm intrigued by
> >> >the math, and wouldn't mind seeing someone succeed.
> >>
> >> The way you choose to ASK someone is quite telling.
> >
> >Only an ingrate--and a sadistic one at that--would do his best to
> >humiliate and degrade an amateur whose assertions and gaps--and yes,
> >whose MISTAKES--have caused him to think productively.
>
> "amateur"?
Yes, amateur.
> James is one of the world's top number theorists!
James elevates himself because you debase him. Ever heard of
Newton's Third Law?
>
> "productively"? There's _no_ evidence that his "research" on
> FLT has produced anything useful.
Without JSH, would there have been a "Magidin-MacKinnon" theorem?
Without JSH, would Magidin have produced the mathematics that he
has during the course of these discussions? If not, hasn't JSH
helped Magidin along the road to publication? Or is this beyond
the ken of a JSH-basher like you?
>
> It certainly has not produced a proof of FLT.
Granted that this assertion has no bearing on whether the
polemics with JSH have been mathematically productive,
wasn't your purpose in asserting this to mislead and
misdirect?
> Your comments
> would make more sense if James said "oops" when people pointed
> out his errors.
His errors are used--not by everyone, but by many (including
Magidin)--as a pretext for debasing him. And to this JSH's
response should be "Oops"?
> Except then your comments would not exist in
> the first place, because people would not have the attitude
> they do towards him. He doesn't say oops when a mistake is
> pointed out, he invariably calls the person pointing out the
> error a lying incompetent.
>
> [...]
> >
> >GOOD professors are aware of the ladders they have climbed that others
> >have not. And they don't humiliate others--in public or otherwise--for
> >not having climbed them.
>
> James humiliates himself.
Claims his chief debaser.
>
> "GOOD professors" indeed. If I spoke to a student in class the
> way I sometimes speak to James I'd get in big trouble.
Indeed you have.
> But that
> doesn't come up, because I _don't_ speak to students in class
> that way. This has something to do with the fact that when I'm
> teaching class the students do not call me a lying incompetent
> when I say something they don't understand.
"Kicker la tête d'un élève du premier rang c'est depuis toujours
le rêve de Ullrich."
(To kick in the head of a student in the first row has always
been Ullrich's dream.)
>
> [...]
> >>
> >>
> >> Excellent. Please post the program. We'll see how it works.
> >
> >And if it doesn't work, will you rub his nose in it?
>
> If he doesn't I will.
>
--John
Virgil
>
> You don't even understand the mathematics being discussed.
>
> But you still talk a lot though.
Do you often talk to yourself like this?
You must be talking to yourself, since everyone else posting about
the math under discussion understands it a good deal better than you
do.
Virgil
> I've said that Magidin is either incompetent or a liar.
You have said a lot of other things that also proved not to be true.
>
> Given his continued claim for credit for the "Magidin-McKinnon
> Theorem" I'm leaning towards the latter.
Since Magidin has never claimed to be the ->origina<-l prover of
that theorem, on what do you base your statement that he calims
credit for it?
He has claimed to have a proof of the theorem, and that proof has
been carefully vetted by mathematicians of a great deal more talent
than you have.
But Magidin has repeatedly said that he does not know whether there
are earlier proofs extant, and has repatedly asked for references to
such earlier proofs if they exist.
This is hardly the same as your reiterated claims of having a short
proof of FLT in the face of a multitude of objections to various
parts of that claimed proof.
>
> As if you know Ullrich, though you've supported Magidin's claims of
> credit, I've yet to see you post mathematics, as usual, and I suspect
> that as usual you don't have a clue what the math is actually about.
As you habitually refuse to read carefully enough to understand any
mathematics not in support of your claimed proof, why should anyone
bother with posting such math.
That Magidin still posts any math relevant to your claim, after
having seen how you treat such posts is a wonder.
One day soon, he will realize the futility of such posts, and you
will have no one to goad into providing for you the mathematics you
cannot provide for yourself.
Much as I enjoy his postings, I could wish that his postings in your
threads cease, and that he expend his talents on those more
deserving of them.
Virgil
> > That's why James's insistence that he's one of
> > the top number theorists in the world, his continued undocumented
> > claims of proofs of all sorts of things, his insistence that all
> > the experts are _wrong_ on various topics, etc, is so, um,
> > let's say "curious".
>
> No less curious than the sci.math abuse of JSH that leads him to
> make such claims.
You have a severe case of cart-before-horse-itis.
Nat Silver
>...and I suspect that as usual you don't have a
> clue what the math is actually about.
> Especially as there's algebra in it, and
> you've admitted not being good at algebra,
> now haven't you?...
He said that he knew no algebra.
You may not be aware of the fact
that in math-speak, one regularly
communicates using understatement.
A logician may say stuff like: Which is
on top, the numerator or the denominator?
None of these remarks is meant to be taken literally.
Why do mathematicians speak like this?
For one, when correcting someone
it's gentler to remark, "You have to
be a little careful here" rather than
saying "How could you have gone
wrong with known results? Your
Areas 1, 2, and 3 are garbage,
you idiot."
For another, after having been humbled
so many times themselves, mathematicians
have a lot of empathy for someone else's
struggles.
Finally, besides being charming, understatement
is funny, sometimes in a subtle way. After all,
mathematicians see themselves as underdogs,
which lends itself to self-deprecating humor.
Clearly, Ullrich's "admission" falls into this
category, understatement. It means that he
is not an expert in the field, period. However,
it does not mean he does not know a whole
lot of algebra. You are being neither charming
nor funny. Seizing upon Ullrich's understatement
to insist he knows no algebra is dumb and boring.
John
Proofs are part of the process of *exactification*, and exact-
ification is a skill which is acquired through training and
practise. However what *gets* exactified are intuitions--and
those who exactify sometimes exactify intuitions other than
their own.
--John
**********************************************************************
Intuition: The ability to understand or produce new ideas in-
stantly and without prior rational elaboration. Syn insight,
vision. Thus "intuition" is opposed to "rational" and in partic-
ular to "exact" and "formal." However, the intuitive and the
formal are only the extremes of a broad gamut. Moreover, intu-
itions never come out of the blue, but culminate in processes of
learning and search. And, if promising, they can often be exacti-
fied. This shows that intuition is often only the first stage in
the process of concept formation. Moreover, the practice of
reason strengthens intuition: the experienced scholar develops an
intuitive "feel"--though never an infallible one. (Mario Bunge,
_Dictionary of Philosophy_, pp. 144-145)
Exactification: The transformation of an imprecise idea or
intuitive idea into an exact one. Example 1: The subject-predicate
relation, mysterious in Aristotelian logic, can be analyzed thus:
"b is a P" is the value of the function P at b, that is, Pb. The
fuzzy old concept of copula, "is," has been absorbed in the *pred-
icate*, which in turn has been conceived of as a function from
individuals to propositions. Example 2: The aura of mystery about
*emergence* evaporates once this definition is adopted: "An
emergent property is a property of a system as a whole, such that
none of its components possess it." Example 3: *Operational
definitions* turn out to be indicators or criteria, as in "litmus
paper is an acidity indicator." Example 4: The *value* of an item
is the degree to which it satisifes a need, and the disvalue of an
item is the degree to which it generates a need. Caution: Since it
is easier to exactify a simple idea than a complex one, there is a
risk involved in exactification, namely that of trivialization. In
other words, if exactification is given absolute priority, then
relevance, depth, or even truth are likely to be underrated.
(loc cit., p. 87)
Zachary Turner
"Nat Silver" <mat...@worldnet.att.net> wrote in message
news:W4PK9.62245$hK4.5...@bgtnsc05-news.ops.worldnet.att.net...
> James Harris wrote:
>
> >...and I suspect that as usual you don't have a
> > clue what the math is actually about.
> > Especially as there's algebra in it, and
> > you've admitted not being good at algebra,
> > now haven't you?...
>
> He said that he knew no algebra.
>
> You may not be aware of the fact
> that in math-speak, one regularly
> communicates using understatement.
> A logician may say stuff like: Which is
> on top, the numerator or the denominator?
I suppose he also thinks that a pediatric cardiologist should either be well
versed in gynecology or fired.
Arturo Magidin
>> In article <3c65f87.02121...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [by the by]
>>
>> [.snip.]
>>
>> >
>> >Here it'd give u1 = (1+sqrt(61))/2, and since
>> >
>> > (1+sqrt(61))(1-sqrt(61))/4 = 15
>> >
>> >it's clear that u1 is, while an algebraic integer, not coprime to 3.
>>
>> I'm mighty curious about how you reach this conclusion, since the way
>> I know it to be true is essentially the argument I use in what you
>> call my "sleight of hand."
>
But what?
>Besides it's easy enough to see what I did, as I found an algebraic
>integer to multiply u1 by and get an integer, which I found had a
>factor of 3.
Right. But then, you don't accept that the root of a nonmonic cannot
be an algebraic integer. How do you know the original root was not
already an algebraic integer?
>> You are trying to see whether (1+sqrt(61))/2 is coprime to 3 or not.
>
>Nope. I knew it wasn't, as it couldn't be.
How did you know that?
>It couldn't be because I'm using the leading coefficient in my ratio
>of algebraic integers.
You know that 3*r is an algebraic integer, where r is a root. Tell me,
how did you know that r itself was not an algebraic integer already?
You are assuming your conclusions. You may be able to do that in
physics to do what I've heard refered to as "spooning" your results
(you assume you have found the answer, and you work backwards to see
what the premises should be), but that doesn't work in mathematics.
>Obviously, all the roots can't have the full first coefficient as what
>you can call their denominator, so there must be shared factors.
>
>It's quite simple.
"Obviously?"
Explain. I dind't say ALL of them did. I asked how you knew it wasn't
the other one.
>> So you check that a MULTIPLE of (1+sqrt(61))/2 is divisible by 3.
>
>I checked a *product* and found it to have a factor of 3.
Which is what I said, James. You want to know if a given number, x,
was coprime to 3 or not. Wha tyou did was multiply x by another number
y, of which you also did not know if it was coprime to 3 or not, and
concluded that xy was a multiple of 3.
That does NOT prove that x is not coprime to 3.
>> Tell me, James. Mathematically: if u*w is a multiple of 3, does that
>> mean that u and 3 are not coprime?
>
>It means that u, w, or uw has non unit algebraic integer factors of 3.
Right! So, you have u=(1+sqrt(61))/2; w = (1-sqrt(61))/2, and you have
concluded from the fact that u*w is a multiple of 3 that u is not
coprime to 3.
But as you have noted, this conclusion is UNWARRANTED. The best you
can conclude is that EITHER u or w will not be coprime to 3. How do
you know it wasn't just w?
>> How do you know that the factor of 3 did not come from (1-sqrt(61))/2?
>
>Well, as I explained above I know that (1+sqrt(61))/2 cannot be a
>ratio of the leading coefficient and some other algebraic integer
>where the leading coefficient is coprime to that leading coefficient.
No, you did not explain. You just said it was "obvious". Please give a
full explanation.
>And, if you want independent confirmation, Galois Theory should tell
>you that, if *you* are not sure.
You can stop bluffing, James. ->I<- know it. Everyone knows I know
it. I'm trying to show to you that you have simply asserted something
correct for the wrong reasons, yet again.
>Just look at the automorphisms.
Oh, more bluff! If you thought this was correct, then you wouldn't be
arguing over the proof I've posted on your polynomial.
Stop bluffing. Just give the proof.
>> See, the reason ->I<- know it is because (1+sqrt(61))/2 and
>> (1-sqrt(61))/2 are conjugate, so one is coprime to 3 if and only if
>> the other is coprime to 3 (the exact same thing as the ai's in your
>> polynomial); so if the "factor of 3" came from (1-sqrt(61))/2, I would
>> know that there is also one in (1+sqrt(61))/2, which is all I need to
>> show they are not coprime.
>
>Prove it then.
Yet more bluffs on your part. I've proven it countless times
already. I'm not going to do it here for your benefit.
If you want to see a proof, ASK for it.
Say "please".
Didn't your mother tell you that you get more flies with honey than
with vinegar?
>I'm just curious to see what you'll write down as a proof.
The same as I always have. What you always claim is "nonsense."
The obvious problem is that you don't know what a proof is. You don't
know how to write one. You don't know how to read one.
When confronted with one, you panic and you claim it is "nonsense" and
"babbling". When you try to read one, you get so hopelessly confused
that you must say that obviously the ->object<- of the writing is to
confuse people.
When you try to write one, you never succeed. You have never written
down a correct proof that you didn't copy from someone. Even what is
correct in your webpage is not properly justified to call it a
"proof".
You simply have no idea what a proof is.
You don't want to learn any algebra, you claim, so as not to be
contaminated with the ideas that dind't work out for FLT. Fine.
Could you please read a book on mathematics and how to prove stuff,
then?
I might suggest Paul Halmos' _Naive Set Theory_.
Try to familiarize yourself with what a proof is.
>> Don't tell me it's "obvious". Figuring out whether an algebraic
>> integer has a factor of another is pretty complicated. For
>> example. Can you tell me if any two roots of
>
>It's not that complicated, but I'm sure you'd like people to believe
>it is, so that they'll believe your claim of credit.
Well, see, the problem is that if you are right that this is all a
dastardly plot to confuse people, then I'm afraid you've been the
perfect accomplice.
See, all you have to do to show to the world what a liar I am is to
explain your algorithm and show your program. Then everyone will see
that you know how to do it, and that when I said it was in general
hard I was full of it.
When you say I am just lying and trying to confuse people with the
proof that an irreducible polynomial with integer coefficients,
primitive, with leading coefficient which is not 1 or -1 cannot have
algebraic integer roots, all you would have to do is produce a
polynomial and the roots.
When you say that my proof about divisors of the coefficients in a
polynomial is false, all you would have to do is produce a
counterexample.
But no. Instead, you rant and rave and whine about how I'm trying to
"corner the math". When you thought the result you now say is
"trivial" was false, you were complaining about how I thought the
"coefficients have far more control over roots than they do". Well,
apparently, I wasn't.
See, instead of ranting and whining about how everyone believes me and
nobody believes you, all you had to do was to was produce an actual
example.
Why haven't you?
Are you trying to ->help<- me "confuse" and "lie"?
Obviously, if I am indeed so successful at "lies", I have to thank you
for your unflagging support and help.
But you know what? They aren't lies. They aren't attempts to
confuse. The only one confused is you, but you are too proud and to
ignorant to admit it.
>> x^17 - 3857236 x^16 + 9287546891 x^3 - x^2 + 105678490213 x -
>> 105943789312
>>
>> are coprime or not? Can you tell me which ones have "factors of 2" and
>> which ones don't?
>
>Here "factors of 2" meaning having non unit algebraic integer factors
>of 2, versus having a factor that is 2.
>
>Well, going from a quadratic to an irreducible polynomial of degree 17
>is an interesting move Magidin.
Not really. YOu said it was "obvious". I want to see you do it. You
already failed miserably with a degree 2, maybe your ideas only work
for higher degrees?
>The answer depends on *each* of the coefficients and there are 18 to
>deal with, and I'm not one to call that easy.
Weird. You said it was "easy" to see whether or not it was.
>However, you claim to have shown the answer through Galois Theory,
>while managing to ignore coefficients other than the first, or leading
>coefficient, and the last.
Again, you are misunderstanding what I wrote. Which is hardly
surprising, since you simply DO NOT UNDERSTAND MATHEMATICAL
STATEMENTS.
What I said, each and every time, is that if you factor it into linear
terms with algebraic integer coefficients, and you look at the leading
coefficients, then ONE of them is not coprime to 2 IF AND ONLY IF ALL
off them are not coprime to 2.
Tell me again. How does that say whether or not one of them is or is
not coprime to 2?
> When I called you on that with cubics, you
>just gave the coefficient equations.
Again, get it straight: you didn't "call me". You only made evident to
everyone your own ignorance and incapacity to understand a simple
mathematical implication.
>So I'm calling you on it with your own example.
Oh, really? I ->said<- it was hard to figure it out. So how will you
"call" me? Only if you can produce them easily.
>I note for the reader that Magidin switched to a monic, though non
>monics were being discussed before.
That meakes it even easier: all the roots are ALREADY algebraic
integers. Surely you know what the factorization will be in that case,
right?
And here's what you will find:
For every integer m, and every positive integer n>0, m^{1/n} is
coprime to one of the roots if and only if it is coprime to ALL the
roots.
>A quick web search gave the phrase "Vastus animus" as meaning that
>someone's mind is a vast wasteland.
>
>Well, at least Magidin's no longer laughing at you.
You seem confused again. The laughter you hear is directed at you.
======================================================================
"Taum autem eras excors, ut tota in oratione tua tecum ipse pugnares,
non modo non cohaerentia inter se diceres, sed maxime disiuncta ataque
contraria, ut non tanta mecum quanta tivi trecum esset contentio."
-- Cicero, Philippicae II
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
David Kastrup <David....@t-online.de> wrote:
>john_...@yahoo.com (John) writes:
>
>> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<ate2n4$1m08$1...@agate.berkeley.edu>...
>>
>> Only an ingrate--and a sadistic one at that--would do his best to
>> humiliate and degrade an amateur whose assertions and gaps--and yes,
>> whose MISTAKES--have caused him to think productively.
>
>Uh, could you explicate how the above quote from Magidin about James'
>style of trying to elicit information is humiliating and degrading? I
>am afraid I don't get it.
He might also try explaining what exactly has been "productive." The
manuscript James is now challenging did NOT come from James, but from
someone else. What I have been posting and proving in response to
James' is hardly "productive;" it's all very well known stuff. It's
baby algebraic number theory. I certainly haven't gotten anything
"productive" out of them, except perhaps stress relief.
>> > All the MORE reason to write what you mean. Why give people the
>> > opening of attacking you for saying obvisouly wrong things, if you
>> > don't mean them? Instead, take the time to write what you ->do<-
>> > mean. Maybe that will be correct.
>>
>> GOOD professors are aware of the ladders they have climbed that others
>> have not. And they don't humiliate others--in public or otherwise--for
>> not having climbed them.
>
>And good students don't call their teachers names and "challenge" them
>as a prerequisite to asking information.
Oh, I agree that a professor should never behave with a student the
way I behave with James.
But then, if a student ever behaved even one tenth as badly as James
does, he would not be in any classroom or office hours: he would have
been expelled already.
Virgil
On the basis of this definition, JSH may have some weak glimmerings
of intuition about math, but his main intuition is about how to con
others into doing the math he cannot do himself.
>
> Exactification: The transformation of an imprecise idea or
> intuitive idea into an exact one. Example 1: The subject-predicate
> relation, mysterious in Aristotelian logic, can be analyzed thus:
> "b is a P" is the value of the function P at b, that is, Pb. The
> fuzzy old concept of copula, "is," has been absorbed in the *pred-
> icate*, which in turn has been conceived of as a function from
> individuals to propositions. Example 2: The aura of mystery about
> *emergence* evaporates once this definition is adopted: "An
> emergent property is a property of a system as a whole, such that
> none of its components possess it." Example 3: *Operational
> definitions* turn out to be indicators or criteria, as in "litmus
> paper is an acidity indicator." Example 4: The *value* of an item
> is the degree to which it satisifes a need, and the disvalue of an
> item is the degree to which it generates a need. Caution: Since it
> is easier to exactify a simple idea than a complex one, there is a
> risk involved in exactification, namely that of trivialization. In
> other words, if exactification is given absolute priority, then
> relevance, depth, or even truth are likely to be underrated.
> (loc cit., p. 87)
JSH's exactification, so far as it exists at all, seems to consist
mainly of incorporating others' works into his own without offering
any credit to the others.
John
Do *all* sadists blame their victims?
In news:<c37480a7.02121...@posting.google.com>
I wrote:
> >Isn't it to be expected that *anyone* with academic transcripts
> >like the following will be more likely to be able to prove this
> >than JSH?
> [...]
>
> Of course it is.
> That's why James's insistence that he's one of
> the top number theorists in the world, his continued undocumented
> claims of proofs of all sorts of things, his insistence that all
> the experts are _wrong_ on various topics, etc, is so, um,
> let's say "curious".
No less curious than the sci.math abuse of JSH that leads him to
make such claims.
>
John
What's mainly wrong with Ullrich is that after
heaping abuse on JSH, encouraging others
to do the same (He does it & he's a professor,
so why shouldn't I?), he has the gall to deny
that he & a few others are the ones who keep
this flame-war going. Coming from college
professors, such conduct is unconscionable.
--John
Nat Silver
> > James Harris wrote:
> > For another, after having been humbled
> > so many times themselves, mathematicians
> > have a lot of empathy for someone else's
> > struggles.
> Ah, so you mean Ullrich's demonstration of his empathy??!!!
Maybe you'd be surprised to know that the rest of us
can identify with some of the frustrations that stem
from difficulties with math.
> > Mathematicians see themselves as underdogs,
> > which lends itself to self-deprecating humor.
> But how can you be underdogs getting paid for "pure math"?
Pure math begets applied math.
> It seems to me that money for nothing, couldn't be better
> unless you got the "chicks for free".
Not everything reduces to partying.
Maybe you should move to Las Vegas.
> > Clearly, Ullrich's "admission" falls into this
> > category, understatement. It means that he
> > is not an expert in the field, period. However,
> > it does not mean he does not know a whole
> > lot of algebra. You are being neither charming
> > nor funny. Seizing upon Ullrich's understatement
> > to insist he knows no algebra is dumb and boring.
> Um, remember, I don't necessarily depend on one source.
Ullrich has shown, in YOUR threads, that he knows some algebra.
> And I definitely don't depend on someone like yourself, who has
> demonstrated your propensity for "math-speak" in quite a few places.
You don't listen very well or maybe you're too math challenged to
take advantage of the expert help you have had over the years.
> Thanks for providing humor to the newsgroup though, and I'm sure some
> of them learned some valuable information from your post besides new
> math terminology!!!
Even a poor soul, such as yourself, knows your quest for FLT is over,
and you've been beating a dead horse. That's why you have shifted
focus to your sieve program that alternates with rehashing FLT and
other stuff.
Virgil
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message
> news:<ate38o$1m6q$1...@agate.berkeley.edu>...
> > In article <3c65f87.02121...@posting.google.com>,
> > James Harris <jst...@msn.com> wrote:
> >
> > >> >So why don't you give an *example* where the algebraic integer factors
> > >> >for a cubic are actually shown?
> > >>
> > >> Translation: You have no clue how to do it, so you would prefer if
> > >> someone else did it for you.
> > >
> > >Oh I admit that I don't quite see how to do it, but I'm intrigued by
> > >the math, and wouldn't mind seeing someone succeed.
> >
> > The manuscript which you so decry tells you exactly how to go about
> > doing it, if you ever bothered to read it.
> >
> > It is difficult to do in practice, but doable in principle.
> >
> > http://www.math.umt.edu/~magidin/preprints/gauss.pdf
>
> Oh yeah right, you put in URL's and see I don't delete them out.
>
> And it's *not* difficult for quadratics, as my pc barely indicates
> effort.
But your latest effort with quadratics would seem to indicate that,
at least for you, it is difficult.
>
> But then computers are so amazingly powerful and wonderful and
> *logical*.
>
> Thank God my computer never lies to me.
If what you posted was what it told you about that quadratic that it
analysed for you , then it lied to you quite thoroughly. Or perhaps
you lied to it when you told it how to analyse quadratics.
>
> I wish human beings were more like computers in that way.
You don't seem to get on with your computer any better than with
those mathematicians of whom you are so jealous.
>
> > That's why I am so interested to see that mythical program of yours
> > which could do it for quadratics. I cannot think of a way to just
> > "scan" (scan what?) to find them, although I can figure out how to
> > find them for any specific example (though it may take considerable
> > amount of time); I can think of a few things I can program a computer
> > to do, but cannot imagine a program that can just "do it", as you
> > claim yours did.
>
> Well I'm a problem solver, so it's not a big deal.
You haven't solved any problems, except in your own mind, whose
solution has appeared here, now, have you?
>
> In fact, I lost the other program, which was really basic anyway.
>
> Well, no big deal, I just threw together another one.
>
> What I have now is nicer in that it actually puts out the numbers, so
> I can just copy them off.
>
> Less effort for me that way you see.
> Well then I think you should take some computer programming courses
> because I've written programs that run rings around you, like this
> quadratic thing.
If the combined result of your math and your quadratic analyser
program are any indication of your talents, you would have trouble
with a "Hello World" program. The programs we have seen wouldn't run
rings around a rosie, much less around any real person.
>
> My computer doesn't actually have to make much of an effort to solve
> what you claim is a hard problem.
On the other hand, that lack of effort is matched by a total lack of
results.
>
> Um, maybe I shouldn't brag too much too quickly, as I'm still fiddling
> with it, so it might still need some tweaking, though I have posted
> what it's saying now.
Perhaps the best tweak would be to junk it and start over.
>
> Please feel free to post more quadratics to throw at it.
Can it do anyting with x^2 - 1 = 0, or is that still too difficult
for it?
>
> And I think turnabout is definitely fairplay as so often people have
> gotten on my case about taking math course, so you should learn
> computer programming!!!
Before you recommend it so highly, so should you.
>
> Did I mention I took a class in computer programming at Duke
> University when I was a kid?
Too bad it didn't take.
>
> We kids labeled gifted were taught by an IBM researcher from IBM's
> facility in Durham, but don't hold that against me please, you know,
> being taught by someone from IBM.
Then it must be that some sort of premature senility has deprived
you of those gifts, since you don't seem able to demonstrate that
you still have them.
>
>
> James Harris
Virgil
> > Why cheat?
>
> I didn't cheat.
>
> Why do you lie?
Probably for better reasons and a with good deal less frequency than
you lie. he doesn't have to cover up his cheats with lies the ay you
do.
By the way
Virgil
> > >> Note that James asserted (but did not prove or give any references to
> > >> a proof that it could be done) that the expressions should be in
> > >> coprime algebraic integers. Are (1+sqrt(61)) and 6 coprime?
> > >
> > >However, from my earlier exposition it is clear that at this point one
> > >would actually consider u1/3 = r1, so your question is disingenuous.
We may all note that the question is also unanswered.
Because James can't answer it, he attempts to sleaze his way around
the question by attcking the questioner.
> >
> > Quite the contrary, James. You are so blinded by your unthinking hate
> > that you did not even notice that what I wrote was actually telling
> > Adrzej that his criticism of what you wrote was invalid, since he is
> > taking v1 = 1+sqrt(61) and w1 = 6.
>
> Nope, as from my earlier exposition it is clear that at this point one
> would actually consider u1/3 = r1, so your question is disingenuous.
James seem to think that his calling a question disingenuous
constitutes an answer to the question, rather than underlining James
ignorance of the answer.
>
> > But you did not notice that. How come?
>
> I answered above.
You evaded above.
>
> > >Here it'd give u1 = (1+sqrt(61))/2, and since
> > > (1+sqrt(61))(1-sqrt(61))/4 = 15
> > >
> > >it's clear that u1 is, while an algebraic integer, not coprime to 3.
> >
> > Right. Which means that your argument as written is incomplete. While
> > you have said you can write each root as an algebraic integer divided
> > by the leading coefficient, you haven't shown that you can write them
> > as quotients of coprime algebraic integers.
>
> Dedekind.
What makes you think so?
>
> > >Now in my exposition I go on to consider a ratio of algebraic integers
> > >v1/w1, which *are* coprime, so you must believe that by giving r1 as
> > >you have that you can fool people into questioning the existence of v1
> > >and w1.
> > >
> > >Why cheat?
> >
> > Don't know. Why do you find the need to claim I am wrong, even when I
> > come in on your side?
> >
> > Why cheat?
>
> I didn't cheat.
Lie.
>
> Why do you lie?
Attack the person when all legitmate arguments fail.
>
> > >Dedekind's own work shows that (1+sqrt(61))/2 can be decomposed into
> > >algebraic integer factors.
> >
> > This makes no sense. There IS NO UNIQUE DECOMPOSITION in algebraic
> > integer factors. Any factor can always be decomposed in an infinite
> > number of essentially distinct ways. There are no irreducibles.
> >
> > If r is an algebraic integer, then you can write
> >
> > r = r^{1/2} * r^{1/2} = r^{1/3}*....*r^{1/3}
> > = r^{1/n} * ... * r^{1/n}
> > = (1+sqrt(1-r))*(1-sqrt(1-r))
> >
> >
> > There is no canonical decomposition "into algebraic integer factors."
> > In face, Dedekind states this QUITE EXPLICITLY:
> >
> > "I shall therefore end these preliminary considerations of the domain
> > of ALL [algebraic] integers with the remark that there are absolutely
> > no numbers in this domain with the character of PRIME
> > NUMBERS. Because, if a is a nonzero [algebraic] integer, and not a
> > unit, then we can decompose it in infinitely many ways into factors
> > which are [algebraic] integers but not units. For example, we have a =
> > sqrt(a)*sqrt(a), and also a=b_1*b_2 where b_1,b_2 are the two roots b
> > of the equation b^2 - b + a = 0."
> >
> > (Richard Dedekind, _Theory of Algebraic Integers_, translated by John
> > Stillwell, Cambridge University Press; 1996. pp. 106, last paragraph
> > of Section 14 in Chapter 3; emphasis in the original. Dedekind calls
> > algebraic integers "integers" and uses "rational integer" or "regular
> > integer" for the integers, so I added the 'algebraic' above).
>
> <deleted>
>
> Your quote shows that Dedekind knew, as I've said, that an algebraic
> integer can be decomposed into algebraic integer factors.
In infintely many ways, but in no unique way.
>
> And for the reader I give my statement again, so they don't have to
> scroll back up unless they want to, as I said:
>
> >>Dedekind's own work shows that (1+sqrt(61))/2 can be decomposed into
> >>algebraic integer factors.
>
> Now your trying to make it seem like I said something wrong just
> because there are an *infinity* of such factors is irrational.
>
> And as I've said before, nothing you've come up with was unknown to
> Dedekind, so why would anyone *dare* call the "Magidin-McKinnon
> Theorem", something like that given P(x) a polynomial of degree n,
> there exists
>
> (a1 x + b1)...(an x + bn)
>
> where the a's and b's are algebraic integers?
You have yet to show that Dedekind, or anyone else, proved the M&M
theorem before M&M did. You have not even shown that anyone else
stated that theorem before M&M proved it. What is your basis, other
than petty jealosy, for wanting to deny M&M the credit?
If you have any evidence, in terms of specific references, that
anyone preceded M&M in either statement or proof declare them now or
forever hold your peace. The fact that you don't like it does not
make it false. The world does not wag to your will.
>
> Now I can understand why you're fighting for credit, as my
> understanding is that there's a lot of pressure to publish, which is
> why so much crap gets published and why some mathematicians need this
> "pure math" business, so that it's easier to publish.
What is pushing you to publish, Jimmy, my boy? As long as you keep
publishing the garbage you publish here, you are hardly in a
position to carp about what others publish anywhere.
Clean up your own act first.
>
> But don't you think you could have done better than pick on Gauss and
> Dedekind?
It is you, Jimmy boy, who are picking on Gauss and Ddekind, to try
and justify your dirty work with senselessly bringing up there names
in inappropriate contexts. Nobody else had bothered with them in the
least.
>
> Is it really *that* bad in the math world today?
The worst of that world is all those flakes outside it who imagine
that all sorts of non-existent things are going on inside it.
>
>
> James Harris
Virgil
john_...@yahoo.com (John) wrote:
> Virgil <vmh...@attbi.com> wrote in message
> news:<vmhjr2-1C925E....@netnews.attbi.com>...
> > In article <c37480a7.02121...@posting.google.com>,
> > john_...@yahoo.com (John) wrote:
> >
> > > > That's why James's insistence that he's one of
> > > > the top number theorists in the world, his continued undocumented
> > > > claims of proofs of all sorts of things, his insistence that all
> > > > the experts are _wrong_ on various topics, etc, is so, um,
> > > > let's say "curious".
> > >
> > > No less curious than the sci.math abuse of JSH that leads him to
> > > make such claims.
> >
> > You have a severe case of cart-before-horse-itis.
>
> Do *all* sadists blame their victims?
I don't know. If you blame me then some do.
It works both ways, we only debase him to the extent that he overly
elevates himself, so maybe it about comes out even.
>
> >
> > "productively"? There's _no_ evidence that his "research" on
> > FLT has produced anything useful.
>
> Without JSH, would there have been a "Magidin-MacKinnon" theorem?
> Without JSH, would Magidin have produced the mathematics that he
> has during the course of these discussions? If not, hasn't JSH
> helped Magidin along the road to publication? Or is this beyond
> the ken of a JSH-basher like you?
That is analogous to crediting diseases with creating their cures.
It is true, but only in the most perverted sense.
>
> >
> > It certainly has not produced a proof of FLT.
>
> Granted that this assertion has no bearing on whether the
> polemics with JSH have been mathematically productive,
> wasn't your purpose in asserting this to mislead and
> misdirect?
Probably a good deal less than it was your purpose to mislead and
misdirect about something of which you seem to have only a miniscule
grasp.
Virgil
> What's mainly wrong with Ullrich is that after
> heaping abuse on JSH, encouraging others
> to do the same (He does it & he's a professor,
> so why shouldn't I?), he has the gall to deny
> that he & a few others are the ones who keep
> this flame-war going. Coming from college
> professors, such conduct is unconscionable.
>
> --John
Harris has it in his power to end this merely by finding somewhere
else to post his unsustainable claims and personal attacks.
Regarding flame wars, it takes two to tango.
And Harris does most of the leading. The rest of us, including
Ullrich, mostly follow.
Harris is a most artful dodger, and a troll of unmatched talents, at
least in sci.math. You needn't feel sorry for him, as he can carry
his own end without help, and would probably prefer to do so with no
one else on his side diluting his fun.
John
> "Nat Silver" <mat...@worldnet.att.net> wrote in message news:<W4PK9.62245$hK4.5...@bgtnsc05-news.ops.worldnet.att.net>...
> >
> > >...and I suspect that as usual you don't have a
> > > clue what the math is actually about.
> > > Especially as there's algebra in it, and
> > > you've admitted not being good at algebra,
> > > now haven't you?...
> >
> > He said that he knew no algebra.
> >
> > You may not be aware of the fact
> > that in math-speak, one regularly
> > communicates using understatement.
> > A logician may say stuff like: Which is
> > on top, the numerator or the denominator?
>
>
> Like the phrase "pure math", which means stuff mathematicians do
> that's useless but they get paid for it anyway, often at taxpayer's
> expense.
>
> Ah, yes, it's so nice of you to provide what I take as the common term
> from within your group.
>
> > None of these remarks is meant to be taken literally.
> > Why do mathematicians speak like this?
> > For one, when correcting someone
> > it's gentler to remark, "You have to
> > be a little careful here" rather than
> > saying "How could you have gone
> > wrong with known results? Your
> > Areas 1, 2, and 3 are garbage,
> > you idiot."
>
>
> I'm beginning to think that possibly he heard it often, possibly as a
> child for some reason, which might help explain why he uses it so much
> now.
>
> But possibly you're still using "math-speak", no?
>
> > For another, after having been humbled
> > so many times themselves, mathematicians
> > have a lot of empathy for someone else's
> > struggles.
>
>
> practice as mathematicians like you clearly do!!!
>
> > Finally, besides being charming, understatement
> > is funny, sometimes in a subtle way. After all,
> > mathematicians see themselves as underdogs,
> > which lends itself to self-deprecating humor.
>
>
> got the "chicks for free".
>
> > category, understatement. It means that he
> > is not an expert in the field, period. However,
> > it does not mean he does not know a whole
> > lot of algebra. You are being neither charming
> > nor funny. Seizing upon Ullrich's understatement
> > to insist he knows no algebra is dumb and boring.
>
>
> demonstrated your propensity for "math-speak" in quite a few places.
>
> of them learned some valuable information from your post besides new
> math terminology!!!
>
>
The test of decency on sci.math is not whether someone is for me or
against me. The test is whether someone can criticize my
mathematics, or my responses to others' postings, without debasing
my person. Such criticism I can accept with equanimity, or reject
without malice.
Based on this, there is much to be admired in Nat Silver's
behaviour, which differs from the flaming of Ullrich, Kastrup
and Varney as day differs from night. As for Magidin, I have
no basis for judging his mathematics--unlike Ullrich, his responses
appear to be of the sort one would expect from a mathematician--but
the way he comes on to you is execrable, as is his failure to recognize
the (undeniable!) fact that you have caused him to think (as he has
caused you to).
As for math-speak, its users should be willing to explain it when
asked to, graciously and without putting on airs.
--John
David Kastrup
> Virgil <vmh...@attbi.com> wrote in message news:<vmhjr2-1C925E....@netnews.attbi.com>...
> > In article <c37480a7.02121...@posting.google.com>,
> > john_...@yahoo.com (John) wrote:
> >
> > > > That's why James's insistence that he's one of
> > > > the top number theorists in the world, his continued undocumented
> > > > claims of proofs of all sorts of things, his insistence that all
> > > > the experts are _wrong_ on various topics, etc, is so, um,
> > > > let's say "curious".
> > >
> > > No less curious than the sci.math abuse of JSH that leads him to
> > > make such claims.
> >
> > You have a severe case of cart-before-horse-itis.
>
> Do *all* sadists blame their victims?
Well, you do all the time, and with obvious relish. Your behavior is
not any different than that which you presume to be chastising, you
obviously enjoy it, and use "justice" or something just as a front.
And it is quite obviously much more a front than love of mathematics
serves as a front for habitual JSH-bashers.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
David C. Ullrich
>David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<msfmvu8tpef25pgtv...@4ax.com>...
>> On 13 Dec 2002 23:46:22 -0800, john_...@yahoo.com (John) wrote:
>>
>> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<ate2n4$1m08$1...@agate.berkeley.edu>...
>> >> In article <3c65f87.02121...@posting.google.com>,
>> >> James Harris <jst...@msn.com> wrote:
>> >> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<atd8b1$1bvj$1...@agate.berkeley.edu>...
>> >> >> In article <3c65f87.02121...@posting.google.com>,
>> >> >> James Harris <jst...@msn.com> wrote:
>> >> >> >"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message news:<H71J3...@cwi.nl>...
>> >> >> >> In article <3c65f87.02121...@posting.google.com> jst...@msn.com (James Harris) writes:
>> >> >> >> > In retrospect it turned out to be fun, as after others gave the values
>> >> >> >> > for the f's and g's, and I first disputed then actually checked and
>> >> >> >> > found them to be correct, I wrote a quick little computer program to
>> >> >> >> > calculate such values for quadratics for myself.
>> >> >> >> >
>> >> >> >> > The problem gets more difficult if you move to cubics, and I'm curious
>> >> >> >> > as to whether or not anyone has ventured into that domain.
>> >> >> >>
>> >> >> >> If you had read the posts by Arturo, you would have known.
>> >> >> >
>> >> >> >Ah, still invoking the name of your hero I see.
>> >> >> >
>> >> >> >So why don't you give an *example* where the algebraic integer factors
>> >> >> >for a cubic are actually shown?
>> >> >>
>> >> >> Translation: You have no clue how to do it, so you would prefer if
>> >> >> someone else did it for you.
>> >
>> >Isn't it to be expected that *anyone* with academic transcripts
>> >like the following will be more likely to be able to prove this
>> >than JSH?
>> [...]
>>
>> Of course it is. That's why James's insistence that he's one of
>> the top number theorists in the world, his continued undocumented
>> claims of proofs of all sorts of things, his insistence that all
>> the experts are _wrong_ on various topics, etc, is so, um,
>> let's say "curious".
>
>You don't even understand the mathematics being discussed.
The last time you said that I asked for an example of a bit
of math needed to understand the Proof that you knew but
you thought I didn't know. No reply.
(Are you still claiming that the Proof is accessible to
high-school students? I'm pretty sure I understand most
of the math in the standard high-school curriculum. No,
I don't know much about things like Galois theory. The
difference between me and you, well _a_ difference, is
that I _know_ I don't know much about Galois theory so
I refrain from talking about it as though I did.)
>But you still talk a lot though.
>
>> You seem to think you're making some sort of point in James'
>> favor by exhibiting Magidin's credentials as a competent
>> mathematician. But James continually insists that he's a
>> total incompetent, that it's hard to see how he got a
>> degree.
>
><deleted>
>
>I've said that Magidin is either incompetent or a liar.
>
>Given his continued claim for credit for the "Magidin-McKinnon
>Theorem" I'm leaning towards the latter.
And given the fact that you haven't given any reason for anyone
to think the result was proved by anyone else, the way you
talk about who proved that result is a good example of the
reason people find you so obnoxious.
(No, your proof that it's immediate from elementary things
like the Fundamental Theorem of Algebra is full of holes.)
>As if you know Ullrich, though you've supported Magidin's claims of
>credit, I've yet to see you post mathematics, as usual, and I suspect
>that as usual you don't have a clue what the math is actually about.
>
>Especially as there's algebra in it, and you've admitted not being
>good at algebra, now haven't you?
No. I've admitted to _knowing_ "no" algebra, where "no algebra" means
no algebra by comparison with the amount that an actual algebraist
knows. I've also "admitted" to knowing much more algebra than you
do.
Never said anything about not being good at it. Those are two
entirely different things. You keep saying that I've admitted
to not being good at algebra. But you never explain how it is
that I _was_ able to prove that result about algebraic integers
not being roots of monic polynomials. You know, the one that
as far as we can tell you're still saying is wrong...
>You don't even *know* the math Ullrich, so why make your pretensions
>of knowledge?
>
>
>James Harris
David C. Ullrich
David C. Ullrich
>David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<msfmvu8tpef25pgtv...@4ax.com>...
>> On 13 Dec 2002 23:46:22 -0800, john_...@yahoo.com (John) wrote:
[...]
>
>>
>> "GOOD professors" indeed. If I spoke to a student in class the
>> way I sometimes speak to James I'd get in big trouble.
>
>Indeed you have.
What? Which student, in which class?
>> But that
>> doesn't come up, because I _don't_ speak to students in class
>> that way. This has something to do with the fact that when I'm
>> teaching class the students do not call me a lying incompetent
>> when I say something they don't understand.
>
>"Kicker la tête d'un élève du premier rang c'est depuis toujours
>le rêve de Ullrich."
>(To kick in the head of a student in the first row has always
>been Ullrich's dream.)
You know there _is_ such a thing as libel... of course the
truth is a defense. You should really answer the question above.
>> [...]
>> >>
>> >>
>> >> Excellent. Please post the program. We'll see how it works.
>> >
>> >And if it doesn't work, will you rub his nose in it?
>>
>> If he doesn't I will.
>>
>
>--John
David C. Ullrich
David C. Ullrich
<David....@t-online.de> wrote:
>john_...@yahoo.com (John) writes:
>
>> Virgil <vmh...@attbi.com> wrote in message news:<vmhjr2-1C925E....@netnews.attbi.com>...
>> > In article <c37480a7.02121...@posting.google.com>,
>> > john_...@yahoo.com (John) wrote:
>> >
>> > > > That's why James's insistence that he's one of
>> > > > the top number theorists in the world, his continued undocumented
>> > > > claims of proofs of all sorts of things, his insistence that all
>> > > > the experts are _wrong_ on various topics, etc, is so, um,
>> > > > let's say "curious".
>> > >
>> > > No less curious than the sci.math abuse of JSH that leads him to
>> > > make such claims.
>> >
>> > You have a severe case of cart-before-horse-itis.
>>
>> Do *all* sadists blame their victims?
>
>Well, you do all the time, and with obvious relish.
I wish you hadn't said that. The fun part of all this has been
the idea that he doesn't seem to realize that what you just
said is obvious to everyone - now you've gone and given it away.
(No, John, I'm not just making this up in reply to Kastrup -
more or less what he said is what I had in mind when I suggested
that what you were actually accomplishing here is not what I
suspected you _thought_ you were accomplishing.)
>Your behavior is
>not any different than that which you presume to be chastising, you
>obviously enjoy it, and use "justice" or something just as a front.
>And it is quite obviously much more a front than love of mathematics
>serves as a front for habitual JSH-bashers.
>
>--
>David Kastrup, Kriemhildstr. 15, 44793 Bochum
David C. Ullrich
David Kastrup
> On 15 Dec 2002 11:41:21 +0100, David Kastrup
> <David....@t-online.de> wrote:
>
> >john_...@yahoo.com (John) writes:
> >
> >> Virgil <vmh...@attbi.com> wrote in message news:<vmhjr2-1C925E....@netnews.attbi.com>...
> >> > In article <c37480a7.02121...@posting.google.com>,
> >> > john_...@yahoo.com (John) wrote:
> >> >
> >> > > > That's why James's insistence that he's one of
> >> > > > the top number theorists in the world, his continued undocumented
> >> > > > claims of proofs of all sorts of things, his insistence that all
> >> > > > the experts are _wrong_ on various topics, etc, is so, um,
> >> > > > let's say "curious".
> >> > >
> >> > > No less curious than the sci.math abuse of JSH that leads him to
> >> > > make such claims.
> >> >
> >> > You have a severe case of cart-before-horse-itis.
> >>
> >> Do *all* sadists blame their victims?
> >
> >Well, you do all the time, and with obvious relish.
>
> I wish you hadn't said that. The fun part of all this has been
> the idea that he doesn't seem to realize that what you just
> said is obvious to everyone - now you've gone and given it away.
You don't actually suppose he would get the point? I think there is
pretty little danger of that. Just wait and see.
Zachary Turner
"James Harris" <jst...@msn.com> wrote in message
news:3c65f87.02121...@posting.google.com...news:<vmhjr2-6D38AC....@netnews.attbi.com>...
> > You haven't solved any problems, except in your own mind, whose
> > solution has appeared here, now, have you?
>
>
> Hmmm...Virgil are you actually claiming to have checked?
>
> Or Virgil, are you lying to the readers of sci.math?
>
> Here's what I have, in case you missed it.
>
> The question was about the following quadratic, the a's and b's are
> algebraic integers.
>
> 3x^2 - x - 5 = (a1 x + b1)(a2 x + b2).
>
> The program I wrote yesterday is now giving:
>
> Solution one:
>
> Factored number is 3
>
> (13+/-sqrt(61))/2)^{1/3}
>
> Solution two:
>
> Factored number is 5
>
> (45+/-5sqrt(61))/2)^{1/3}
>
> Where the a's are Solution one, and the b's are Solution two.
>
> I'm thinking I'll fiddle with the program some more today.
>
> Note: It takes the coefficients of the quadratic as input.
>
>
> James Harris
He probably _has_ checked. I certainly did. It is utterly trivial to
verify that (a1 x + b1)(a2 x + b2) is nowhere near 3x^2 - x - 5. How in
God's name can you have an odd number divided by an even number , and
expect to get an integer when you take the cube root. You've always
wondered why people so vehemently call you stupid and idiotic. Well, this
is why. Even if we give you that proving FLT may be quite difficult and
require other people to look over it and find errors (after all, even Wiles
submitted a proof that he thought was correct originally), something like
this is absolute nonsense. How can you honestly expect people to take you
seriously when you say that
3x^2 - x - 5 = (a1 x + b1)(a2 x + b2)
where
a1 = (13+sqrt(61))/2)^{1/3}
a2 = (13-sqrt(61))/2)^{1/3}
b1 = (45+5sqrt(61))/2)^{1/3}
b2 = (45-5sqrt(61))/2)^{1/3}
That's all I really want to know. _How_ can you expect people to take you
seriously when the above demonstrates you can't even do 7th grade
arithmetic. Yet somehow you're supposed to understand algebraic number
theory? In the 7 years you've been on this futile attempt to prove the
world without knowing what water is, you could have spent a mere 4 years
actually _learning_ something, and the other 3 years producing something of
value (By the way, since you have trouble with simple arithmetic, 4 + 3 =
7). Instead, you're _still_ at square one. You haven't even left the start
line yet. The gun went off 7 years ago and you're still standing around
twiddling your thumbs even though everyone in the crowd is yelling RUN, RUN.
Gus Gassmann
Zachary Turner wrote:
Aside from anything else, not even the parentheses balance. I'm guessing that
this
is supposed to be a1 = ((13+sqrt(61))/2)^{1/3} etc., but this is just a guess.
Zachary Turner
"Gus Gassmann" <hgas...@mgmt.dal.ca> wrote in message
news:3DFCD0AF...@mgmt.dal.ca...
Well it was a guess for me too, and I accidentally interpreted it
incorrectly. Anyway, with this newfound enlightenment, I do get 3 and -5
for a1*a2 and b1*b2, but I don't get -1 for a1 b2 + a2 b1, although I
punched this into my calculator only once as I'm in a rush to get out the
door, and I unlike certain other people can admit if there may be an error
:)
Virgil
> The test of decency on sci.math is not whether someone is for me or
> against me. The test is whether someone can criticize my
> mathematics, or my responses to others' postings, without debasing
> my person. Such criticism I can accept with equanimity, or reject
> without malice.
Another test of decency is whether the person criticized, when
criticized in a decent way, accepts that criticism with equanimity
or rejects it without malice.
You hero, James S, Harris, differs from this definition of decent as
day differs from night. He calls virtually everyone who disagrees
with him vile names and insults them in myriad ways, regardless of
the decency with which those disagreements have been expressed. He
denies any error as long as he can, and when he makes any
corrections he claims that the corrected version was what he meant
all along. He deliberately misquotes corrections in order to deny
that he, the great JSH, has made any error.
If is is a question of decency, if ever JSH will try it, the rest of
us might follow, but we only follow his lead in this matter of
decency.
> Based on this, there is much to be admired in Nat Silver's
> behaviour, which differs from the flaming of Ullrich, Kastrup
> and Varney as day differs from night.
And, if you read back far enough, you will find that the flaming
between JSH and those who flame him now built up gradually with JSH
offering at least as much provocation as those you now denigrate.
> As for Magidin, I have
> no basis for judging his mathematics--unlike Ullrich, his responses
> appear to be of the sort one would expect from a mathematician--but
> the way he comes on to you is execrable, as is his failure to recognize
> the (undeniable!) fact that you have caused him to think
And not denied. Magidin has said several times that some points of
considerable interest have arisen from this extended thread.
> (as he has
> caused you to).
>
> As for math-speak, its users should be willing to explain it when
> asked to, graciously and without putting on airs.
Its users have done so politely when such requests were polite, and
have done so even for some impolite requests, or, on occasion, for
no requests at all.
>
> --John
You really should give up the silly notion that your hero is a hero.
He lacks all of the hero's traditional virtues, such as modesty,
honesty, etc. His "noble" quest is not in search of honor, truth or
beauty, but personal aggrandizement, specifically fame and fortune,
as he has often stated brazenly and openly in sci.math.
He is really kind of a bastard mixture the worst of Don Quixote and
the worst of Til Eulenspiegel, ignorantly charging windmills in the
futile hope of pilfering from them fame and glory (and wealth).
Virgil
> > If what you posted was what it told you about that quadratic that it
> > analysed for you , then it lied to you quite thoroughly. Or perhaps
> > you lied to it when you told it how to analyse quadratics.
>
> Computers don't lie, as there is no *intent* to deceive.
>
> Computer don't have intent, not yet.
>
> But humans lie as you so clearly demonstrate in yet another post.
Ah. I knew you would get aroung to accusing me of lying, since you
eventually accuse anyone not pandering to your own falsehoods of
lying.
I am proud to have joined the honored band of those whose virtue has
been established by your disapproval.
Virgil
> > You don't seem to get on with your computer any better than with
> > those mathematicians of whom you are so jealous.
>
> I'm not jealous.
>
> I have discovered a short proof of Fermat's Last Theorem, and a prime
> counting function which offers paths to new research on primes, and
> mathematicians are treating me rather oddly.
>
> It seems more likely that *they* are jealous, as they refuse to even
> behave with more positives than indifference to the prime counting
> function, and that's in the areas of *primes*, which some people might
> have foolishly thought were more important to mathematicians than
> petty emotional concerns.
>
>
> > >
> > > > That's why I am so interested to see that mythical program of yours
> > > > which could do it for quadratics. I cannot think of a way to just
> > > > "scan" (scan what?) to find them, although I can figure out how to
> > > > find them for any specific example (though it may take considerable
> > > > amount of time); I can think of a few things I can program a computer
> > > > to do, but cannot imagine a program that can just "do it", as you
> > > > claim yours did.
> > >
> > > Well I'm a problem solver, so it's not a big deal.
> >
> > You haven't solved any problems, except in your own mind, whose
> > solution has appeared here, now, have you?
>
> <deleted>
>
> Hmmm...Virgil are you actually claiming to have checked?
I was asking you whether you have actually solved and problems and
then published their solutions here. If you were more cognizant of
the meanings of language, you would realize that I was not making
any claim at all. Learn to read what is actually written, not what
you would like to find written.
This habit of yours of reading between the lines things that are not
there is very counterproductive. You are unlikely to make much
progress until you rid yourself of it.
> Or Virgil, are you lying to the readers of sci.math?
No. I do not have to lie, as I am not trying to prove how good I am
as you are.
>
> Here's what I have, in case you missed it.
>
> The question was about the following quadratic, the a's and b's are
> algebraic integers.
>
> 3x^2 - x - 5 = (a1 x + b1)(a2 x + b2).
>
> The program I wrote yesterday is now giving:
>
> Solution one:
>
> Factored number is 3
>
> (13+/-sqrt(61))/2)^{1/3}
>
> Solution two:
>
> Factored number is 5
>
> (45+/-5sqrt(61))/2)^{1/3}
>
> Where the a's are Solution one, and the b's are Solution two.
>
> I'm thinking I'll fiddle with the program some more today.
>
> Note: It takes the coefficients of the quadratic as input.
>
>
> James Harris
First of all, both your solutions have more right parentheses, ")",
than left parenteses, "(", so they are not even well-formed
expressions.
Assuming that this is mere oversight on your part and that you meant
to have another left parenthesis at the left of each expression:
The product of your b's is
((45+5sqrt(61))/2)^{1/3}*((45-5sqrt(61))/2)^{1/3}
= ((2025 - 305)/2)^{1/3}
= 125^{1/3} = 5, but the product of the b's should be -5, not 5.
Also, a1*b2 + a2*b1 = -1, but with your a's and b's, there is no
combination any where close to -1.
Do you ever bother to check your work at all, or do you just post
it, untested and unproofread, in hopes that someone who knows
arithmetic, much less algebra, will do it for you?
Even sixth graders learn to check their work. Can you imagine the
physicists who built the first A-bomb being so cavalier as not to
check their arithmetic before thying out the first bomb?
If these last two programs are typical of the way James Harris works
as a programmer, may Dedkind and Gauss, to whom James seems to pray,
save us all from ever using any software that James Harris has had a
hand in developing.
John
> On 14 Dec 2002 13:47:34 -0800, john_...@yahoo.com (John) wrote:
>
> >David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<msfmvu8tpef25pgtv...@4ax.com>...
> >> On 13 Dec 2002 23:46:22 -0800, john_...@yahoo.com (John) wrote:
> [...]
>
> >>
> >> "GOOD professors" indeed. If I spoke to a student in class the
> >> way I sometimes speak to James I'd get in big trouble.
> >
> >Indeed you have.
>
> What? Which student, in which class?
Has a tenured professor like you ever got in trouble at Oklahoma State
University for any reason?
On sci.math, tenure is no defense for a sadistic bully.
>
> >> But that
> >> doesn't come up, because I _don't_ speak to students in class
> >> that way. This has something to do with the fact that when I'm
> >> teaching class the students do not call me a lying incompetent
> >> when I say something they don't understand.
> >
> >"Kicker la tête d'un élève du premier rang c'est depuis toujours
> >le rêve de Ullrich."
> >(To kick in the head of a student in the first row has always
> >been Ullrich's dream.)
>
> You know there _is_ such a thing as libel... of course the
> truth is a defense. You should really answer the question above.
>Gulp<
Arturo Magidin
>> In article <3c65f87.02121...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> >> >So why don't you give an *example* where the algebraic integer factors
>> >> >for a cubic are actually shown?
>> >>
>> >> Translation: You have no clue how to do it, so you would prefer if
>> >> someone else did it for you.
>> >
>> >the math, and wouldn't mind seeing someone succeed.
>>
>> doing it, if you ever bothered to read it.
>>
>> It is difficult to do in practice, but doable in principle.
>>
>> http://www.math.umt.edu/~magidin/preprints/gauss.pdf
>
>Oh yeah right, you put in URL's and see I don't delete them out.
Delete it if you want!
The question is: did you bother to read it?
>And it's *not* difficult for quadratics, as my pc barely indicates
>effort.
It is indeed very easy to provide wrong answers. You have provided two
so far from your "program." The first was so obviously wrong it took
about 2 nanoseconds to note that 9-sqrt(61) times 9+sqrt(61) cannot
equal -5. The second was a bit more difficult, but it was pretty
obvious the cross term was off.
So, yes, it is "*not* difficutl"... as long as you don't care whether
the answer is correct or not.
>But then computers are so amazingly powerful and wonderful and
>*logical*.
>
>Thank God my computer never lies to me.
Too bad you lie to your computer. You told it that if it this this and
that the correct answer would come out. It dutifully did everything
you told it to do... and came up with the wrong answer.
>I wish human beings were more like computers in that way.
>
>> That's why I am so interested to see that mythical program of yours
>> which could do it for quadratics. I cannot think of a way to just
>> "scan" (scan what?) to find them, although I can figure out how to
>> find them for any specific example (though it may take considerable
>> amount of time); I can think of a few things I can program a computer
>> to do, but cannot imagine a program that can just "do it", as you
>> claim yours did.
>
>Well I'm a problem solver, so it's not a big deal.
Interesting. YOu have failed to solve a single problem so far as I can tell.
>In fact, I lost the other program, which was really basic anyway.
>
>Well, no big deal, I just threw together another one.
>
>What I have now is nicer in that it actually puts out the numbers, so
>I can just copy them off.
>
>Less effort for me that way you see.
Indeed. You don't even have to check if it is correct.
>> Now, I'm not a computer programmer, so maybe you can show us yours. At
>> one point John Ramsden posted a few ideas, and only with some very
>> strong hypothesis (which are not always met) did he come up with some
>> formulas.
>
>Don't you mean Dave Rusin?
No, I mean John Ramsden. Articles
<3ca388b9...@news.demon.co.uk>,
<3ca46943...@news.demon.co.uk>,
<3ca5fa44...@news.demon.co.uk>, and
<3ca76abb...@news.demon.co.uk>.
Oh, wait. You're still claiming they are one and the same, right?
>Well then I think you should take some computer programming courses
>because I've written programs that run rings around you, like this
>quadratic thing.
I would suggest that before you claim that something "run[s] rings
around" someone, you might want to make sure that the program actually
comes up with the correct answers.
Yours does not, as is painfully obvious.
>My computer doesn't actually have to make much of an effort to solve
>what you claim is a hard problem.
Too bad the solution is wrong, yes?
>Um, maybe I shouldn't brag too much too quickly, as I'm still fiddling
>with it, so it might still need some tweaking, though I have posted
>what it's saying now.
And what it says is wrong.
>Please feel free to post more quadratics to throw at it.
I'll just wait until you find the correct answer for the one that is
giving you so much trouble. Is that okay for you?
>And I think turnabout is definitely fairplay as so often people have
>gotten on my case about taking math course, so you should learn
>computer programming!!!
Here's the difference: I don't claim to be one of the top computer
programmers in the world.
But see, I know enough about computer programming to know the
folloiwng:
1. In order to write a program that does X, one must have an algorithm
that does X, so one can program it into the computer.
2. IN this case, "X" is "given three integers (a,b,c), with a nonzero,
find four algebraic integers a1, a2, b1, b2, such that
a1*a2 = a
b1*b2 = c
a1*b2 + a2*b1 = -b.
3. Whenever you have attempted to say how to do this, you have said
things which are either incorrect, or nonsensical.
4. Hence, it is reasonable to assume that you do not in fact have an
algorithm to do X.
5. I cannot think of an algorithm that would do X, even in principle,
for the general case, simply by "scanning", as you claim yours
does.
6. While I know an algorithm to solve the problem, the algorithm is
difficult. In particular, it requires knowledge of algebraic
integers that you ->obviously<- lack in abudance. As such, I think
it is realistic to believe you do not know a correct way of solving
the problem, even in principle.
7. Which means, you cannot write a computer program to solve the
problem.
>Did I mention I took a class in computer programming at Duke
>University when I was a kid?
>
>We kids labeled gifted were taught by an IBM researcher from IBM's
>facility in Durham, but don't hold that against me please, you know,
>being taught by someone from IBM.
Irrelevant. Especially given that you are trying to program your
computer to do things you yourself have no idea how to do.
======================================================================
"Certo scio, occisam saepe sapere plus multo suem."
-- Plautus, Miles Gloriosus
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Virgil
> > >> "GOOD professors" indeed. If I spoke to a student in class the
> > >> way I sometimes speak to James I'd get in big trouble.
> > >
> > >Indeed you have.
> >
> > What? Which student, in which class?
>
> Has a tenured professor like you ever got in trouble at Oklahoma State
> University for any reason?
>
> On sci.math, tenure is no defense for a sadistic bully.
On sci.math, nobody has tenure, so your remarks are idiotic.
Nobody in sci.math can be bullied without his consent.
Anyone in sci.math can killfile anyone else if they dislike their
posts.
If Ullrich's posts cause Harris so much pain, Harris must be
compulsively masochistic to keep on reading them.
If Ullrich's posts cause you so much pain, you must also be
compulsively masochistic to keep on reading them.
If you can't manage a killfile, maybe you ought to try therapy.
Arturo Magidin
Zachary Turner <_NOzturner...@hotmail.com> wrote:
>He probably _has_ checked. I certainly did. It is utterly trivial to
>verify that (a1 x + b1)(a2 x + b2) is nowhere near 3x^2 - x - 5. How in
>God's name can you have an odd number divided by an even number , and
>expect to get an integer when you take the cube root.
Ehr, it's not an odd number. It's a number of the form
13+sqrt(61). Since 61 is congruent to 1 modulo 4, the number
(13 + sqrt(61))/2
is indeed an algebraic integer.
As to getting an integer, a1*a2 = ((169 -61)/4)^{1/3} = 27^{1/3} = 3,
so that one is right.
LIkewise, b1*b2 = ((2025-1525)/4)^{1/3} = 125^{1/3} = 5, so ->that<-
one is right incorrect, since it should be -5.
The real mess is a1*b2 + b1*a2.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
>> >> In article <a1fa83d9.02121...@posting.google.com>,
>> >> Andrzej Kolowski <akol...@hotmail.com> wrote:
>> >> >jst...@msn.com (James Harris) wrote in message news:<3c65f87.02121...@posting.google.com>...
>> >>
>> >>
>> >> >> Now writing the polynomial as
>> >> >>
>> >> >> A(x-r1)...(x-rn) = P(x)
>> >> >>
>> >> >> I use my previous result to write each r as a ratio of algebraic
>> >> >> integers, now using the algebraic integers v, and w, with v coprime to
>> >> >> w, I have
>> >> >>
>> >> >> A(x-r1)...(x-rn) = A(x - v1/w1)...(x - vn/wn) = P(x).
>> >> >>
>> >> >> Given that the final coefficient is an integer, I know that
>> >> >>
>> >> >> A (v1...vn)/(w1...wn) is an integer, which gives me that A has
>> >> >>
>> >> >> (w1...wn) as a factor.
>> >> >>
>> >> >
>> >> >
>> >> >> As the last coefficient is another number besides A (labeling it is
>> >> >> not important) I have that A = w1...wn, so then
>> >> >>
>> >> >
>> >> > What? Doesn't work -
>> >> >
>> >> > A = a0 * (w1 ... wn) / (v1 ... vn).
>> >> >
>> >> >
>> >> >> w1...wn(x - v1/w1)...(x - vn/wn) = P(x)
>> >> >>
>> >> >
>> >> >
>> >> > No - try this with a specific polynomial, like
>> >> >
>> >> > 3*x^2 - x - 5.
>> >> >
>> >> > The roots are
>> >> >
>> >> > r1 = (1 + sqrt(61))/6 and r2 = (1 - sqrt(61))/6.
>> >> Note that James asserted (but did not prove or give any references to
>> >> a proof that it could be done) that the expressions should be in
>> >> coprime algebraic integers. Are (1+sqrt(61)) and 6 coprime?
>> >
>> >However, from my earlier exposition it is clear that at this point one
>> >would actually consider u1/3 = r1, so your question is disingenuous.
>>
>> that you did not even notice that what I wrote was actually telling
>> Adrzej that his criticism of what you wrote was invalid, since he is
>> taking v1 = 1+sqrt(61) and w1 = 6.
>
>Nope, as from my earlier exposition it is clear that at this point one
>would actually consider u1/3 = r1, so your question is disingenuous.
You ->still<- don't get it, do you?
You said "u1/w1 with u1,w1 coprime."
Andrzej said "I'm letting u1 = 1+sqrt(61), and w1 = 6".
I pointed out to him, by asking the question, that HIS choice of u1
and w1 DO NOT SATISFY THE REQUIREMENTS YOU HAD WRITTEN DOWN. Since
they do not satisfy the requirements, any criticism of your argument
based on taking the values u1=1+sqrt(61) and w1=6 is incorrect.
In short, ->I<- was pointing out the criticism was wrong.
But again, you are so hateful and so ignorant that you don't get that
I was actually defending your argument at that point.
>> But you did not notice that. How come?
>
>I answered above.
Yes: your answer makes it clear you hav eno idea what was going on.
>> >Here it'd give u1 = (1+sqrt(61))/2, and since
>> > (1+sqrt(61))(1-sqrt(61))/4 = 15
>> >
>> >it's clear that u1 is, while an algebraic integer, not coprime to 3.
>>
>> Right. Which means that your argument as written is incomplete. While
>> you have said you can write each root as an algebraic integer divided
>> by the leading coefficient, you haven't shown that you can write them
>> as quotients of coprime algebraic integers.
>
>Dedekind.
This is not a proof. This is not even a complete sentence.
You've said you believe it follows from "infinite decomposibility",
proven by Dedekind, but in fact it does not.
You don't even have ot "take my word" for it, you can just see what
Dedekind himself has to say, since he mentions both things in the same
paragraph:
"A deeper investigation will enable us to see that two nonzero
[algebraic] integers a and b have a *greatest common divisor*, which
can be put in the form aa'+bb' where a' and b' are [algebraic]
integers. This important theorem is not at all easy to prove with
the help of the principles developed thus far, but we shall later
(section 30) be able to derive it quite simply from the theory of
ideals. I shall therefore end these preliminary considerations of
the domain fo *all* [algebraic] integers with the remark that there
are absolutely no numbers in this domain with the character of
*prime numbers*. Because, if a is a nonzero [algebraic] integer, and
not a unit, then we can decompose it in infinitely many ways into
factors which are [algebraic] integers but not units. For example,
we have a = sqrt(a)*sqrt(a), and also a=b_1*b_2, where b_1,b_2 are
the two roots b of the equation b^2-b+a = 0. It follows from Theorem
2 of Section 13 that sqrt(a), b_1, b_2 are [algebraic] integers as
well as a."
-- Richard Dedekind, _Theory of Algebraic Integers_, translated
by John Stillwell. Cambridge Mathematical Library,
1996. Emphasis in the original. Quotation taken from the last
paragraph of Section 14, Chapter 3, pp. 106. The promised
proof ccurs in the final paragraph of Section 30, pp. 152.
You'll see that Dedekind says the result "is not at all easy to prove"
with what he has developed thus far, which includes the infinite
decomposibility. In fact, Dedekind is being a bit facetious, as it is
not just "not at all easy", but actually quite hard to prove, as the
fact that it takes him almost 50 pages to develop the result shows.
IN case you are wondering, "every quotient of algebraic integers can
be written as a ratio of two coprime algebraic integers" is equivalent
to "any two algebraic integers have a greatest common divisor which
can be written as a linear combination of them."
Fist, suppose every quotient can be written as a ration of coprime
algebraic integers. Given a and b, nto both zero, say b not zero, then
write a/b = u/v with u and v coprime. Then va=ub. Since v and u are
copriime, but v divides ub, it follows that v divides b; likewise, u
divides a. So a = u*d, b = v*d'. Cancelling, we have that d=d', so d
is the greatest common divisor of a and b. Since in fact u and v are
coprime, we can write 1 = ru+sv, so d = rdu + sdv = ra+sb, showing
that any two algebraic integers have a gcd which can be written as a
linear combination of them.
Conversely, if any two have a gcd, then take a/b. Let d =
gcd(a,b). Then a=d*u, b=d*v, so a/b = u/v. And since d was the gcd of
a and b, u and v have no common divisor other than units. Thus, the
gcd is 1, and 1 = ru+sv for some algebraic integers r and s, which
means that u and v are indeed coprime.
So what you are claiming follows easily is something that Dedekind
claims is "not at all easy to prove."
Was Dedekind wrong? Or in the habit of saying that easy stuff is
actually hard?
Or pehaps... ->YOU<- are incorrect in your assertion that it follows easily?
>> >Now in my exposition I go on to consider a ratio of algebraic integers
>> >v1/w1, which *are* coprime, so you must believe that by giving r1 as
>> >you have that you can fool people into questioning the existence of v1
>> >and w1.
>> >
>> >Why cheat?
>>
>> Don't know. Why do you find the need to claim I am wrong, even when I
>> come in on your side?
>>
>> Why cheat?
>
>I didn't cheat.
>
>Why do you lie?
I did not lie. Nothing I have written is false. Hence, nothing I have
written could I have written ->knowing<- it was false.
INdeed. And as I note above, he ALSO says that what you claim follows
from this does not follow from this quite as easily as you think it does.
>And for the reader I give my statement again, so they don't have to
>scroll back up unless they want to, as I said:
>
>>>Dedekind's own work shows that (1+sqrt(61))/2 can be decomposed into
>>>algebraic integer factors.
>
>Now your trying to make it seem like I said something wrong just
>because there are an *infinity* of such factors is irrational.
No. YOu still don't understand what I said:
From the fact that there are infinitely many different factorizations
IT DOES NOT FOLLOW that you can write a/b as u/v with u and v
coprime. Dedekind himself notes it above.
>And as I've said before, nothing you've come up with was unknown to
>Dedekind, so why would anyone *dare* call the "Magidin-McKinnon
>Theorem",
If he never put the results together, then that's it. It's not his.
As for you, your argument isn't even correct enough to show that if a
polynomial with integer coefficints can be written as a product of
polynomials with RATIONAL coefficients, then youc an also write it as
a product of polynomials of the same degrees with integer
coefficients. And before you say it is "obvious", take into account
that in order to prove only about half of what is needed for that
result, namely "Gauss' Lemma", Gauss takes the time and a full page in
order to prove that lemma, in the _Disquisitiones_. This is Gauss, the
guy who was famous for never publishing a proof unless he felt it was
both elegant, clear, and necessary.
>something like that given P(x) a polynomial of degree n,
>there exists
>
> (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers?
Unless he wrote it down, or somebody else did, then whoever notices it
first gets credit.
I realize you are suffering an accute case of sour grapes. Here you
are, after seven years of struggling you have NOTHING worth a damn
that you did yourself, and you think I am a liar and an incompetent
(although you have ever failed to justify either charge), so how dare
I come up with something correct, and ->possibly<- original?
>Now I can understand why you're fighting for credit, as my
>understanding is that there's a lot of pressure to publish, which is
>why so much crap gets published and why some mathematicians need this
>"pure math" business, so that it's easier to publish.
>
>But don't you think you could have done better than pick on Gauss and
>Dedekind?
>
>Is it really *that* bad in the math world today?
If anyone took your word for ->anything<-, then the math world would
indeed be bad. But here you are, an obvious incompetent, comparing
himself to Gauss and Dedekind, and so hatefull, so small, and so
pitiful, that you cannot stand for ->anyone<- to do anything right,
since you obviously cannot.
Sour grapes, and hate. That's all you have.
Either produce an ->actual<- statement from Dedekind which shows he
knew the Theorem, or shut up. The actual theorem. Not all the
necessary ingredients, not all the necessary partial results, but the
actual theorem. Knowing all the partial results is not worth squat if
you don't put them together.
======================================================================
"Quod minimum specimen in te ingenii?"
-- Cicero, In Pisonem
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
David Kastrup
> > > It seems to me that money for nothing, couldn't be better
> > > unless you got the "chicks for free".
> >
> > Not everything reduces to partying.
> > Maybe you should move to Las Vegas.
>
> Taxpayers expect results for their money. Oh wait, we are kind of
> used to having our money thrown away, but it still galls us.
>
> Now to me, mathematicians getting paid to do nothing useful IS
> partying, whether it's in Vegas or not!!!
>
> At least in Vegas you know that's what's going on.
What is your position on music and art teachers? Or English
literature?
Arturo Magidin
[.snip.]
>Let's consider the facts:
>
>1. I said that I had written a program that finds the roots of
>quadratics as algebraic integers.
>
>2. I noted that my original program wasn't available so *yesterday* I
>went at rewriting it.
>
>3. I posted some output from what I was doing, and the attacks began.
Let's consider some other facts:
I. You claimed the program was "easy". You in fact called me all sorts of
names, including liar, partly on the grounds that you could write such
a program "easily."
II. You failed.
Because, you ->DO<- agree that your "program" is giving you wrong
answers, right?
III. Rather than risk a wrong program output, the easiest solution
would have been to post your algorithm for solving the problem. You
must have one, otherwise you would not be able to write down a program
to solve the problem. Posting the algorithm has the added advantage
that any silly mistakes (forgotten semi-colon or parenthesis,
incorrect copying when printing results, etc) would not matter at all.
Why have you not posted the algorithm? Why, instead, have you posted
two solutions which are ->obviously<- wrong?
I can think of two reasons: 1. you are too lazy to check your own
work; you post it so others can check it for you and you needn't
bother doing it. Or 2. You are too incompetent to check your own work;
you post it because having others check it for you is the only way you
can be sure it is correct.
Which one is it?
As to people "relishing" the assault, not caring about the
consequences, you are just again showing you are a hypocrite.
You have never thought even once (let alone twice) about calling
people liars and incompetents, even when you were wrong. And you have
NEVER apologized for calling someone a lying incompetent, simply for
pointing out your mathematical errors, even when you had to admit they
were indeed errors.
You ->certainly<- relish in the assault, and you ->certainly<- don't
care about the consequences of yoru assaults. So don't come whining
like a child just because someone else does it to you. If you can't
take it, you shouldn't spend so much time dishing it out.
Zachary Turner
news:<Y05L9.161382$Gc.52...@twister.austin.rr.com>...
> posting that's typical on this newsgroup.
>
> Let's consider the facts:
>
> 1. I said that I had written a program that finds the roots of
> quadratics as algebraic integers.
>
> 2. I noted that my original program wasn't available so *yesterday* I
> went at rewriting it.
>
> 3. I posted some output from what I was doing, and the attacks began.
<snip>
Let's consider some other facts.
1) It takes under a minute for people who can do arithmetic to multiply the
two numbers you said constitute the factorization.
2) The numbers don't constitute a factorization
3) You said the program was easy
4) It is hilariously wrong
5) You're lying again,
How do I know 5? Well, because you said the data was unchecked. Either it
was checked or it was unchecked, you can't have it both ways. If it was
checked and you deliberately put out false data, then you're lying by saying
it was unchecked. If it was unchecked, then you're lying by saying you
deliberately put out false data.
Either way, it's obvious who the liar is.
Just out of curiosity, exactly what is the mathematical significance of
incorrect output? You say you're just going along posting output as you
compile, maybe you could go ahead and some screen shots from your machine as
well. You know, just for the hell of it. Not necessarily while this
program is running, just some screenshots. It's equally as relevant as
incorrect output to the problem at hand.
Arturo Magidin
Zachary Turner <_NOzturner...@hotmail.com> wrote:
[.snip.]
>5) You're lying again,
>
>How do I know 5? Well, because you said the data was unchecked. Either it
>was checked or it was unchecked, you can't have it both ways. If it was
>checked and you deliberately put out false data, then you're lying by saying
>it was unchecked. If it was unchecked, then you're lying by saying you
>deliberately put out false data.
>
>Either way, it's obvious who the liar is.
Ehr, I don't think James said he was deliberately putting out false
data. He said that some ->might<- think he is deliberately putting out
false data as a test (in fact, in the past he has claimed that as an
explanation for wrong answers), but that in this case he is still
"fiddling" with the program.
Arturo Magidin
[.snip.]
>Virgil <vmh...@attbi.com> wrote in message news:<vmhjr2-994D85....@netnews.attbi.com>...
>> In article <c37480a7.02121...@posting.google.com>,
>What seems to escape you Virgil, though I seriously doubt it actually
>escapes you, is that you can't behave badly and blame someone else
>saying it's their fault, and expect to be taken seriously by adults.
[.snip.]
>So I know the reality versus what you're both saying, and Magidin
>knows when to nice things up.
>
>In the past I found him to be one of the more creatively vicious
>people.
>
>Just go back a few years and read what he was saying to me then.
Are you trying to claim that you can behave badly towards me, because
I'm to blame for that?
What about that lovely thing you said before about being taken
seriously by adults?
I guess, as usual, you don't actually mean it to apply to ->you<-,
just to others.
[.snip.]
Zachary Turner
> > john_...@yahoo.com (John) wrote:
> >
> > > The test of decency on sci.math is not whether someone is for me or
> > > against me. The test is whether someone can criticize my
> > > mathematics, or my responses to others' postings, without debasing
> > > my person. Such criticism I can accept with equanimity, or reject
> > > without malice.
> >
> > Another test of decency is whether the person criticized, when
> > criticized in a decent way, accepts that criticism with equanimity
> > or rejects it without malice.
>
> escapes you, is that you can't behave badly and blame someone else
> saying it's their fault, and expect to be taken seriously by adults.
frequent poster to this thread.
>
> Now, say you think you give me valid criticism, and you want me to
> behave in a way that you see as without malice, does that give you the
> right to then behave badly if I don't do what you want?
>
> Let's keep the facts known here:
>
> I'm someone who's been trying to post about my math ideas on a math
> newsgroup.
>
> That's the key fact to keep in mind.
Nobody has ever denied that. They also point out that none of your math
ideas have ever been _correct_, and in fact have tried endlessly to help you
spot your mistakes so that you might in fact come up with something that
_is_ correct.
>
> Personal responsibility Virgil, you need to think about that phrase.
As should you. You should think about how YOU coming up with incorrect
mathematics means YOU'RE responsible for the incorrect mathematics.
>
> Now as to your charge, what I know is that years ago my posting about
> looking for a short, simple proof of FLT attracted derision, often at
> best.
>
> The newsgroup is actually *nicer* in a lot of ways than it was back
> then, amazingly enough.
Now the "short FLT proof" actually contains a few correct statements. So it
isn't quite as egregiously wrong as it used to be.
Nat Silver
> What most people call math is a series of techniques that they learn.
That may be true, but so what?
> However, those techniques were discovered, which is what many people
> often forget, when they see things neat and tidy in a textbook.
It's like baking a cake.
No one knows what a mess the kitchen was.
> Those *discoverers* are the ones who faced down the major
> frustrations.
What about it? If you haven't published a paper, it's hard to
understand that it's one's baby out there in print.
> > Pure math begets applied math.
> That's the party line.
It's fact. In 1948, Hardy acknowledged
that number theory was useless. Techniques
of factoring large numbers into primes are now
applied. Building a math community and
culture steeped in "useless" research paves
the way for applied advances.
In an very oversimplified example, we have seen the dangers
of not supporting research. U.S. corporations dropped the
ball between 1945 and 1960 when Japan and Germany were
out of commission.They didn't keep the lead in innovative
production of automobiles, televisions, etc., partly because
they discontinued funding and valuing research. They
cashed in, and the U.S. became a service economy.
> However, hasn't it occurred to anyone that if you have people telling
> you upfront that they only do "pure math" that there's lots of room
> for abuse?
Wake up! In every field, every job there is that kind of abuse.
As far as I can tell, there's less of it in academia. Many
academics work very hard. The tenure process kind
of insures that a lot are with the program.
> Answer me a few questions, and no math-speak please:
> Did Newton do only pure math?
Obviously, not.
> Do you consider him a mathematician?
Sure.
> Did Gauss do only pure math?
Nope.
> Do you consider him a mathematician?
Sure.
The above two are superstars,
on everybody's all-time top 5 list.
> And to throw in someone from our modern times,
> did John Nash do only pure math?
Nope.
> Do you consider him a mathematician?
Yup. Yup. Yup.
> Now here's the *really* important question,
> do you personally know any mathematicians
> who only do pure math?
Yes. There are many.
I don't know him other than having been in the
same room with him, but probably Wiles falls
into this category.
> Taxpayers expect results for their money. Oh wait, we are kind of
> used to having our money thrown away, but it still galls us.
Baloney. Taxpayers' money is wasted by political kickbacks,
for housing and military and other government contracts.
And if not, it's wasted within the contracts themselves.
> Now to me, mathematicians getting paid to do nothing useful IS
> partying, whether it's in Vegas or not!!!
Are you jealous that they have a better life-style than yourself?
Well, they have earned it by getting their union card, a Ph.D.
Then they nurtured their careers by doing research, published
in refereed journals, and/or they made themselves useful to
the university in other ways like obtaining funding through
grants or by doing administrative work.
> At least in Vegas you know that's what's going on.
> > Ullrich has shown, in YOUR threads, that he knows some algebra.
> I see you wish to set the bar very low.
He proved something. How does that compare with your end?
> To me that's just more of the problem as I see *modern*
> mathematicians as having very low standards, and I think
> that's because they've been corrupted by a welfare system.
Buwahahaha! Members of mathematics departments have to
work extremely hard, because the bulk of their clientele are
students in service courses, which are extremely difficult to
teach and cause burnout. So, part of their job is to service
the university. It may bring to mind an analogy of work of
the most distasteful kind.
> And I think you have a better idea of just how I've used the
> newsgroup and to what purpose.
Dunno.
> ...the short FLT Proof is indeed a proof, and my prime
> counting research is of interest.
I don't believe you have a proof.
It's nice to see you working in your area,
which I take to be programming.
> Whatever my reasons are though, you of all people should accept my
> right to post.
I never said anything about that but doubt that it's
constructive for anybody when you post on FLT.
Virgil
> "Zachary Turner" <_NOzturner...@hotmail.com> wrote in message
> news:<Y05L9.161382$Gc.52...@twister.austin.rr.com>...
> I left the entire post in so people can see the kind of *vicious*
> posting that's typical on this newsgroup.
>
> Let's consider the facts:
>
> 1. I said that I had written a program that finds the roots of
> quadratics as algebraic integers.
>
> 2. I noted that my original program wasn't available so *yesterday* I
> went at rewriting it.
>
> 3. I posted some output from what I was doing, and the attacks began.
>
James repeatedly claims to have a perfect short proof of Fermat's
Last Theorem, a problem that withstood solution by almost all of the
worlds best mathematicians for over 300 years and then he shows that
he can't tell when his own simple arithmetic is wrong.
And then he can't understand why people treat him with distain.
> If you look over the posts you see an intolerance for anything but
> perfection even when people are repeatedly told that I'm just throwing
> something together and haven't really checked it yet.
Nobody is forcing you to publish your mistakes. And, why the hell do
you publish things that you haven't checked unless you are trying to
get others to do that checking for you? That makes you a lazy slob
in addition to your many other faults.
If you are that lazy, you deserve the treatment you get.
>
> Some of you may feel that's fair. You may suppose that this is some
> formal place where no one should post ideas, or half-thought out
> notions, or output from programs they're still working on.
>
> If you think that way, I think you're at the wrong newsgroup.
Or perhaps you are. If you posted something as a work in progress
AND asked for help in perfecting it, as others have done, you might
receive the gentler treatment that they have earned.
>
> Some of you may consider that I *deliberately* put out wrong data to
> prove a point, but in fact, I am, as I said, fiddling with a program,
> and putting out the data as I go along.
Well you fiddle with the truth, so fiddling things is nothing new
for you.
>
> However, now that some people have reacted viciously I want you to
> consider what a chill such people can produce on talk of current
> research.
If it cools your research to the point where you actually do some
trivial checks of your own work before publishing, than the benefits
will have far outweighed any costs.
>
> Do you all really suppose that top mathematicians haven't heard of
> sci.math?
>
> Do you suppose any of them are stupid enough to get within range of
> people like Zachary Turner?
And most of them probably shy away from James S. Harris, King Troll,
Like Plague.
>
> The Zachary Turner's of the world are a loud and demanding lot, who
> feel quite gratified in attacking discoverers, whereas, discoverers
> are usually fun loving, adventurers who like to actually solve
> problems, rather than worry about people who wish to attack.
If you are trying to pass yourself off as fun loving, your sense of
fun would entertain the Marquis de Sade.
If you are trying to pass yourself off as an adventurer, the name
has a pejorative meaning that fits you to perfection.
If you are trying to pass yourself off as a problem solver, you
performance on your current problem suggests that you couldn't find
your way out of a paper bag without help.
>
> The fact of the matter is that whether 99% of you reading the
> newsgroup are basically decent people it doesn't take more than a
> couple of Zachary Turner's to ruin the newsgroup experience for a
> discoverer.
And it doesn't take more that one "discoverer" of your caliber to
pollute the whole group.
>
> And I'm an adult, so I have more tools for handling such people and
> not taking their attacks to heart, but what about a child?
You behaviour so far contradicts your calim to adulthood.
>
> Can you imagine some 12 year old, possible great mathematician to be,
> who makes the mistake of posting on this newsgroup?
If you got on his case, he might be soured forever, but if he just
lurks, he might get a few laughs from the transparency of Emperor
Harris new clothes.
>
> Do we know it hasn't already happened?
>
> Possibly humanity already paid a price that it will never be aware of
> because that poor child simply retreated from the assault, and quit
> mathematics.
Yes, and you should hang your head in shame at having been the
possible cause.
>
> Some of you are unaware of the impact of words, and you're way too
> tolerant of the Zachary Turner's.
>
> These people *relish* the assault, or unthinkingly don't care about
> the damage.
Speaking for yourself again?
>
> It is my firm belief that children and many young adults, if they wish
> to be researchers of merit, should be kept away from Usenet and the
> Internet.
>
> They should NOT under any circumstances be put in a position where
> they can get feedback from a Zachary Turner.
Or from:
> James Harris
Virgil
> Virgil <vmh...@attbi.com> wrote in message
> news:<vmhjr2-994D85....@netnews.attbi.com>...
> > In article <c37480a7.02121...@posting.google.com>,
> > john_...@yahoo.com (John) wrote:
> >
> > > The test of decency on sci.math is not whether someone is for me or
> > > against me. The test is whether someone can criticize my
> > > mathematics, or my responses to others' postings, without debasing
> > > my person. Such criticism I can accept with equanimity, or reject
> > > without malice.
> >
> > Another test of decency is whether the person criticized, when
> > criticized in a decent way, accepts that criticism with equanimity
> > or rejects it without malice.
>
> What seems to escape you Virgil, though I seriously doubt it actually
> escapes you, is that you can't behave badly and blame someone else
> saying it's their fault, and expect to be taken seriously by adults.
You complain of my behaviour as if I have done anything that you
haven't done or wouldn't do, if you could,
Some of the things I do, I'm not sure you could do, such as checking
simple arithmetic before posting.
>
> Now, say you think you give me valid criticism, and you want me to
> behave in a way that you see as without malice, does that give you the
> right to then behave badly if I don't do what you want?
When I find a post of yours without malice towards others in it, I
either do not reply to it or reply without malice towards you.
>
> Let's keep the facts known here:
>
> I'm someone who's been trying to post about my math ideas on a math
> newsgroup.
You also post a lot of malicious stuff that has nothing to do with
mathematics, but is strictly ad hominem. Lets keep that fact in mind
as well.
>
> That's the key fact to keep in mind.
>
> > You hero, James S, Harris, differs from this definition of decent as
> > day differs from night. He calls virtually everyone who disagrees
> > with him vile names and insults them in myriad ways, regardless of
> > the decency with which those disagreements have been expressed. He
> > denies any error as long as he can, and when he makes any
> > corrections he claims that the corrected version was what he meant
> > all along. He deliberately misquotes corrections in order to deny
> > that he, the great JSH, has made any error.
>
> Um, maybe you're trying to be amusing or deliberately presenting a
> caricature without expecting anyone to take you seriously, but let's
> keep in mind the central fact which is that I'm someone posting my
> math ideas on a math newsgroup.
I call em as I see em. The only recent post of yours with no malice
in it was your apology for attacking a person whom you had
mistakenly believed had attacked you. That way a gracious and
appropriate act on your part. I did not, and will not, attack you
for that.
>
> Granted I've also talked about more than math at times, but I've tried
> to keep it math related as I see mathematics as part of everyday life.
>
When you misquote someone and then use that misquotation as the
basis of a claim that that person has lied, is that how you see
mathematics as a part of everyday life?
When you repeat many times that what is called the M&M theorem is
false, then switch to claiming that Dedkind and/or Gauss proved it
first, but without any references to such proof, then switch to
claiming that it is too trivial to of interest (accompanying this
claim with a multiply flawed argument), is that how you see
mathematics as a part of everyday life?
You must have a life of incredible peculiarity, if these deceits,
possibly self deceits, are an everyday part of it.
> > If is is a question of decency, if ever JSH will try it, the rest of
> > us might follow, but we only follow his lead in this matter of
> > decency.
>
> And it looks like you're trying to abrogate personal responsibility.
When you call a man a liar or publicly denigrate him in other ways,
particularly when he is innocent of your charges, you give him the
right to counter your own false statements. You have given the right
of counterattack in this way to virtually everyone except John
Correy.
>
> Sorry Virgil, but *you* make the choice how to behave on the
> newsgroup.
True, but I have the right to counter falsehoods, and have no moral
qualms about doing so.
>
> I don't sit there and type up your posts and send them--you do.
But you do sit there and post all sorts of falsehood, to which I
sometimes reply to establish the truth.
>
> > > Based on this, there is much to be admired in Nat Silver's
> > > behaviour, which differs from the flaming of Ullrich, Kastrup
> > > and Varney as day differs from night.
> >
> > And, if you read back far enough, you will find that the flaming
> > between JSH and those who flame him now built up gradually with JSH
> > offering at least as much provocation as those you now denigrate.
>
> There are standards for behavior.
Which you would impose on others provided when you not follow them
yourself.
>
> Personal responsibility Virgil, you need to think about that phrase.
I think about it often, and the way you ignore it. If I could get
->you<- to accept it and live by it, I would regard that as
miraculous.
>
> Now as to your charge, what I know is that years ago my posting about
> looking for a short, simple proof of FLT attracted derision, often at
> best.
>
> The newsgroup is actually *nicer* in a lot of ways than it was back
> then, amazingly enough.
>
> What is not talked about is what I've picked up as a feeling among a
> lot of people in mathematics that even talking about a short proof of
> FLT is an affront to them.
>
> They simply hate the subject. Period.
A short proof would be lovely. But, so far, there isn't one.
>
> > > As for Magidin, I have
> > > no basis for judging his mathematics--unlike Ullrich, his responses
> > > appear to be of the sort one would expect from a mathematician--but
> > > the way he comes on to you is execrable, as is his failure to recognize
> > > the (undeniable!) fact that you have caused him to think
> >
> > And not denied. Magidin has said several times that some points of
> > considerable interest have arisen from this extended thread.
>
> And I have to remind that I've posted for years, and I've seen quite a
> few posts from Magidin.
>
> So I know the reality versus what you're both saying, and Magidin
> knows when to nice things up.
>
> In the past I found him to be one of the more creatively vicious
> people.
I bow to your premier experience at being creatively vicious.
But you have certainly given him sufficient cause to justify
whatever he has said about you.
>
> Just go back a few years and read what he was saying to me then.
>
> > > (as he has
> > > caused you to).
> > >
> > > As for math-speak, its users should be willing to explain it when
> > > asked to, graciously and without putting on airs.
> >
> > Its users have done so politely when such requests were polite, and
> > have done so even for some impolite requests, or, on occasion, for
> > no requests at all.
>
> Come on, let's be adults here. These are posts not spin talks on
> cable news.
>
> Ullrich admitted to having problems with algebra.
>
> Now everyone wants to interpret that statement in different ways.
>
> So now it's "math-speak" and means that he isn't an algebra expert.
>
> Ok, I'll take it to mean that he's not an algebra expert and *still*
> be bothered by his problems when discussing my work, especially when
> he keeps attacking it when he clearly doesn't have a clue, and has
> made posts which to me are *bragging* about not caring about the
> actual math content of my work.
As the errors that he finds in your work seem to be generally
confirmed by others, perhaps it is you who are short a clue.
Ullrich clearly regards your bating each other as an amusing game,
which, given the provocation you have given him, makes him
remarkably mild tempered.
>
> There's been a propaganda game for a while now, where in the past they
> used to just call my work "junk", but it's amazing to me how many
> people just accept the propaganda.
You two treat each other with about equal levels of disrespect,
although I judge him to have the better imagination and sense of
humor. The two of you about cancel each other out, in genreal
insults. Ullrich's math is a good deal better, even in algebra, and
when he errs, he admits it and apologizes, a behaviour you could
benefit from by imitating more often.
>
> So someone like Ullrich comes out and admits that he doesn't actually
> check my work, but then attacks it, claiming it's wrong, and no one
> seems to care, except maybe John here.
You may well be right. I read the interchanges between you two more
for amusement than edification.
>
> > >
> > > --John
> >
> > You really should give up the silly notion that your hero is a hero.
>
> I don't think that John is saying I'm his hero or that I'm a hero.
>
> He's simply someone speaking out bothered by a failure to behave with
> common decency.
Then straighten up.
>
> Now most of you can act without standards in your posts on this
> newsgroup and no one can do anything about it.
>
> But Ullrich works at a public university and there are standards that
> citizens of this country have a right to hold him to!
So you would deny Ullrich his civil rights because you don't like
what he says! That is much worse behaviour than anything Ullrich has
done. That makes you a sort of fascist.
>
> And John has the right to express his own outrage at behavior that he
> sees as wrong.
And we have an equal right to expess ourselves about him.
>
> > He lacks all of the hero's traditional virtues, such as modesty,
> > honesty, etc. His "noble" quest is not in search of honor, truth or
> > beauty, but personal aggrandizement, specifically fame and fortune,
> > as he has often stated brazenly and openly in sci.math.
>
> I think that paragraphs says more about you than me,
It is not just my opinion, I cribbed most of the ideas from other
posters, and just put it in my own words. If it says something about
the authors, one of the things it says is that they argee about you.
> as it stands to
> reason that someone finding a short proof of FLT would get some kind
> of reward from recognition.
Two bits and a ride on the ferris wheel is too much for what has
been presented so far. A good deal more if a complete short proof is
ever presented in readable form.
>
> I mean, it's not like that's a strange concept in our modern world.
>
> > He is really kind of a bastard mixture the worst of Don Quixote and
> > the worst of Til Eulenspiegel, ignorantly charging windmills in the
> > futile hope of pilfering from them fame and glory (and wealth).
>
> Pilfering?
Swindling? Getting without earning? I can provide more alternative
expressions at need.
>
> Yet let's be clear about the facts. I post about math on a math
> newsgroup.
That has not been the case in the past. You have posted a lot of
stuff with no math at all. You have posted a lot with some math
content and some squabbling about other things, and you post
occasionally something that is primarily mathematical in content.
Much of the math content of your posts is merely unsnipped carryover
from prior posts so that the amount of original math appearing is a
miniscule proportion of the total postings.
>
>
> James Harris
John
What's the matter, dude? Am I crampin' yer style?
--John
John
> > heaping abuse on JSH, encouraging others
> > to do the same (He does it & he's a professor,
> > so why shouldn't I?), he has the gall to deny
> > that he & a few others are the ones who keep
> > this flame-war going. Coming from college
> > professors, such conduct is unconscionable.
> >
> > --John
>
> Harris has it in his power to end this merely by finding somewhere
> else to post his unsustainable claims and personal attacks.
>
> Regarding flame wars, it takes two to tango.
>
> And Harris does most of the leading. The rest of us, including
> Ullrich, mostly follow.
>
> Harris is a most artful dodger, and a troll of unmatched talents, at
> least in sci.math. You needn't feel sorry for him, as he can carry
> his own end without help, and would probably prefer to do so with no
> one else on his side diluting his fun.
If you had been relentlessly stalked and mobbed by pathological
sadists, with no better use for mathematics than to jack
people up, you might sing a different song.
--John
John
> > > the top number theorists in the world, his continued undocumented
> > > claims of proofs of all sorts of things, his insistence that all
> > > the experts are _wrong_ on various topics, etc, is so, um,
> > > let's say "curious".
> >
> > No less curious than the sci.math abuse of JSH that leads him to
> > make such claims.
>
> You have a severe case of cart-before-horse-itis.
Are you in charge of writing the sci.math version of the
Protocols of the Elders of Zion--in which JSH wil
be calumnied as a drinker of the blood of mathie babies,
not to mention as a harasser of Ullrich's employer (guess
what folks: Ullrich invited JSH to contact his employer,
and when he didn't do it fast enough rebuked JSH for
being all words and no action), and a terrorizer of
clean-living, God-fearing mathies like Ullrich, Uncle
Al, Kastrup, Varney and Wayne Brown--who never ever
have done anything to JSH other than criticize his math?
--John
David Kastrup
> Virgil <vmh...@attbi.com> wrote in message news:<vmhjr2-1C925E....@netnews.attbi.com>...
>
> > You have a severe case of cart-before-horse-itis.
>
> Are you in charge of writing the sci.math version of the
> Protocols of the Elders of Zion--in which JSH wil
> be calumnied as a drinker of the blood of mathie babies,
> not to mention as a harasser of Ullrich's employer (guess
> what folks: Ullrich invited JSH to contact his employer,
> and when he didn't do it fast enough rebuked JSH for
> being all words and no action), and a terrorizer of
> clean-living, God-fearing mathies like Ullrich, Uncle
> Al, Kastrup, Varney and Wayne Brown--who never ever
> have done anything to JSH other than criticize his math?
Pot, kettle, black.
David C Ullrich
[...]
>
>That's false, as the short FLT Proof is indeed a proof,
Nope.
>and my prime
>counting research is of interest.
Primarily to you.
>Whatever my reasons are though, you of all people should accept my
>right to post.
You keep saying this, implying that people _have_ been questioning
your right to post. I don't recall anyone ever saying you _didn't_
have that right.
In all the time I've been participating in sci.math I've only seen
_one_ instance of someone actually taking steps in an
(admitted or otherwise, in this case _admitted_) attempt to
stop someone from posting. That person is _you_.
>James Harris
David C Ullrich
>David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<qi7pvuce8rpgnuhot...@4ax.com>...
>> On 14 Dec 2002 13:47:34 -0800, john_...@yahoo.com (John) wrote:
>>
>> >David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<msfmvu8tpef25pgtv...@4ax.com>...
>> >> On 13 Dec 2002 23:46:22 -0800, john_...@yahoo.com (John) wrote:
>> [...]
>>
>> >>
>> >> "GOOD professors" indeed. If I spoke to a student in class the
>> >> way I sometimes speak to James I'd get in big trouble.
>> >
>> >Indeed you have.
>>
>> What? Which student, in which class?
>
>Has a tenured professor like you ever got in trouble at Oklahoma State
>University for any reason?
I don't see how that answers my question. You _really_ should
come up with an answer.