My Work--Objective Review
David C. Ullrich
>Their names are quite familiar to me and to many of you who have kept
>up with the discussions that have raged about my mathematical work:
>
>Ullrich, Magidin, Wayne Brown, Christian Bau, etc.
>
>A group uniform in that they are unrelentingly negative, often take
>cheapshots, like Ullrich, a math professor at a state university, who
>loves the word "idiot" or likening me to a monkey with a typewriter.
>
>Mathematics isn't supposed to be this way. Some might have thought
>there was some dignity in the discipline, some decorum.
>
>From what I've seen people in the math community choose to blame me.
Huh? What do you imagine people in the math community blame you
for? I've never seen anyone "blame" you for anything.
>Wow, I must have some extraordinary level of power to be able to get
>math professors to grovel in the muck, to produce such demeaning
>behavior from career mathematicians and I guess career posters.
Truly remarkable.
>Hmmm...I wonder if I could translate it over to politics. Maybe I
>could get George W. Bush to call me names? Now that could at least be
>useful.
>
>Now it makes sense to me that mathematicians behave without
>professionalism or decorum as I haven't been just sitting and taking
>it as they insult me--I insult back. From what I've seen I have lots
>of reasons to see mathematicians as base liars, who at least are
>public, or cowards who hide in silence, or the worst, the hidden ones
>who knowing my work is correct hide in silence in what must be seen as
>malicious intent.
And in that entire group of mathematicians who know your work
is correct not _one_ of them has decided to come out and say
so. Hard to believe.
Hint: Actually the mathematical community is not as well
organized as you think. There is no secret handshake. When
people look at your work and say it's wrong there's a much
simpler explanation.
>Basically there are two angry groups. I am a harsh force of one.
>Against me is a society of mathematicians.
>
>So far it's been a draw.
It's only a draw in your imagination. Trust me on this - we won
long ago.
>However, my work is not hidden. In response to challenges I've
>*simplified* and explained in depth.
Right. Something else that's not hidden is _hundreds_ of posts
asking you to explain what you mean by this or that, with no reply.
[...]
>
>Given that there are two powerful sides at loggerheads I'm asking for
>an objective and independent review by someone who isn't a jackass.
>
[...]
>
>Now there are a lot of usual suspects who argue with me. They lie.
>They blow smoke. They refuse to acknowledge mathematical truths, but
>will involve me in long threads, with long posts if I let them.
>
>Why should I?
>
>I am a force of one.
>
>If any of you can find an error then present it. Then can someone who
>is objective and invested in either side push forward any claims of
>error that seem cogent?
>
>If so, then I'll address THAT person's post. But I'm sick of the same
>old vicious crew of arrogant assholes. I'm sick of arguing with liars
>who in their posts can go so far as to attack the basis of algebra
>itself.
Yup. Every once in a while someone new joins in. You're happy
to discuss things with the new person for a while. Then he
always finds errors or incomprehensibilities, and at that point you
he becomes a lying asshole.
>My work is out there. I have discussed it in detail in posts. I have
>quite a few pages explaining it. I have worked hard to put the
>information out there.
>
>Now it's time for someone to step up that can be objective.
>
>And yes, I don't think highly of mathematicians, so it might be hard
>to find someone.
GIven that the definition of "objective" seems to be "agrees with
me" yes, that _is_ going to be difficult.
>But consider my point of view; every day that goes by that my work
>isn't acknowledged is another day that the math community is lying.
>
>As far as I'm concerned they're jackasses.
And you wonder why nobody loves you.
>James Harris
George Cox
>
> ... Ullrich, a math professor at a state university, who
> loves the word "idiot"
Well, maybe he knows, or knows of, an idiot and applies the word "idiot"
to him or her?
> or likening me to a monkey with a typewriter.
This I find harder to explain. Perhaps David once met a monkey with a
typewriter and found it to be an unpleasant experience?
GC
Brian Chandler
[ snip ]
> Now it makes sense to me that mathematicians behave without
> professionalism or decorum as I haven't been just sitting and taking
> it as they insult me--I insult back. From what I've seen I have lots
> of reasons to see mathematicians as base liars, who at least are
> public, or cowards who hide in silence, or the worst, the hidden ones
> who knowing my work is correct hide in silence in what must be seen as
> malicious intent.
>
> Basically there are two angry groups. I am a harsh force of one.
> Against me is a society of mathematicians.
I don't quite understand this characterisation of "two groups": you
mean the people who keep pointing out errors and obscurities, and the
people who say nothing at all? Seems unclear to me that any of the
first group are angry - there's substantial evidence that mostly
they're just amused. We haven't yet really seen any evidence
whatsoever of the existence of this group of silent observers, nodding
away at your efforts, yet saying nothing - surely they couldn't be
angry either?
[ snip summary of your webpage]
> Now apparently many of you have apparently swallowed the notion that I
> must be wrong just because people keep arguing with me, and *claim*
> I'm wrong, when they can't prove it. Instead they debate the
> definition of "polynomial" or just keep calling me names.
>
> Given that there are two powerful sides at loggerheads I'm asking for
> an objective and independent review by someone who isn't a jackass.
>
> I have NOTHING to hide about my work. It's available 24 hours a day,
> 7 days a week at
>
> http://groups.msn.com/AmateurMath
Right. Well, since you complain a lot about people who (as far as I
can see) are genuinely asking you to explain _exactly_ what you mean
by a polynomial, I've given the following couple of pages of a
(smelly-paper) book, which are also available 24/7 (I hope; I use
pair.com hosting):
"The scope of x", pp 34-35 of Sawyer's "A concrete approach to
abstract algebra":
http://imaginatorium.org/private/sawyer.gif
(136K, 800x600)
In this extract, Sawyer describes two ways of looking at a polynomial;
depending on whether you view the polynomial as a formal object or
whether you view it as simply a generalised form of arithmetic (i.e.
you are always thinking of just its evaluation), you will get
different answers to quite basic questions like "Are these two
polynomials equal?" or "Is this polynomial a factor of that one?"
So you need either: to explain which way you are thinking of your
polynomials, or to explain why Sawyer's carefully delineated
distinction doesn't exist.
Um, actually, have I understood your latest claim correctly? You say
that the algebraic integers are incomplete, because you have x and y,
such that y is a factor of x, but x/y isn't an algebraic integer? So
in your new system of completing the ring of algebraic integers (is
this related to Objects, by the way? I'll called them CAIs for
completed algebraic integers for now) - is it the case that for any
CAIs, r and s, that r/s is also a CAI? So consider two general
polynomials P(x) and Q(x) "in" the ring of CAIs: since you say that
the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is
also a CAI? But if every polynomial divides every other polynomial,
isn't it rather hard to have an irreducible polynomial?
>
> If any of you can find an error then present it. Then can someone who
> is objective and invested in either side push forward any claims of
> error that seem cogent?
Well, I can't understand what an object is. Do you know anyone except
yourself who can? Perhaps a good tack to take would just be a humble
appeal to the silent masses for some good soul to step forward and
explain objects so that at least people like me could understand them
(even if the Nasty Mathematician Mafia say they can't).
Brian Chandler
----------------
geo://Sano.Japan.Planet_3
Jigsaw puzzles from Japan at:
http://imaginatorium.org/shop/
Randy Poe
I think "likens" is the wrong word. James was compared
UNFAVORABLY to the classic statistical model of N monkeys
at N typewriters, on which one can calculate such things
as expected time to produce "Hamlet". The probability
is nonzero that "Hamlet" (or a valid proof of FLT) will
be produced in finite time.
The probability is exactly zero that James will produce
a proof of FLT in finite time.
The monkeys-on-typewriters thing started a very silly
side discussion on the claim that the statistical model
was racist.
- Randy
Brian Quincy Hutchings
imagin...@despammed.com (Brian Chandler) wrote in message news:<f2c35871.03061...@posting.google.com>...
> > http://groups.msn.com/AmateurMath
> http://imaginatorium.org/private/sawyer.gif
> Um, actually, have I understood your latest claim correctly? You say
> that the algebraic integers are incomplete, because you have x and y,
> such that y is a factor of x, but x/y isn't an algebraic integer? So
> in your new system of completing the ring of algebraic integers (is
> this related to Objects, by the way? I'll called them CAIs for
> completed algebraic integers for now) - is it the case that for any
> CAIs, r and s, that r/s is also a CAI? So consider two general
> polynomials P(x) and Q(x) "in" the ring of CAIs: since you say that
> the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is
> also a CAI? But if every polynomial divides every other polynomial,
> isn't it rather hard to have an irreducible polynomial?
> geo://Sano.Japan.Planet_3
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
Dik T. Winter
> > or likening me to a monkey with a typewriter.
>
> This I find harder to explain. Perhaps David once met a monkey with a
> typewriter and found it to be an unpleasant experience?
Eh, no. This was an allusion to the knowledge that a million monkeys
at typewriters will at some time produce a work of Shakespeare. Even
I (non-native English speaker) understood it.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
George Cox
Ok, I undertake not to make any more attempts at humour, especially of
the ironic kind.
GC
Virgil
<3c65f87.03061...@posting.google.com>,
jst...@msn.com (James Harris) wrote:
>
> Oh yeah, it'd help if you know mathematics.
>
>
> James Harris
If who knew mathematics?
Nora Baron
> up with the discussions that have raged about my mathematical work:
>
> Ullrich, Magidin, Wayne Brown, Christian Bau, etc.
>
> A group uniform in that they are unrelentingly negative, often take
> loves the word "idiot" or likening me to a monkey with a typewriter.
>
<snip>
> > I have NOTHING to hide about my work. It's available 24 hours a day,
> 7 days a week at
>
> http://groups.msn.com/AmateurMath
>
> and included the paper Advanced Polynomial Factorization, which is
> sweeping in its simplicity, while being the most important math paper
> of the year because it highlights an error in taught mathematics--a
> problem in "core".
>
Let's have a look at it. The main claim, it appears,
is that if P(x) is the polynomial
P(x) = 65*x^3 - 12*x + 1
and P(x) is factored in the form
P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1)
where a1, a2, and a3 are algebraic integers, then
at least one of a1, a2 or a3 is coprime to 5.
Is this correct ?
Let x = 1/u. Then P(x) = 0 implies
P(x) = 65/u^3 - 12/u + 1 = 0.
Multiplying through by u^3 gives
u^3 - 12*u^2 + 65 = 0.
Call this polynomial Q(u). Let r1, r2, and r3 be
roots of Q. Note that a1, a2 and a3 happen to be
the negatives of r1, r2, and r3 - that is, a1 = -r1,
a2 = -r2, a3 = -r3 [the order does not matter].
The polynomial Q(u) is irreducible. Let H be the
field of algebraic numbers; clearly r1, r2, and
r3 are in H. Let F12 be an automorphism of H such
that F12(r1) = r2, and F12 leaves fixed the subfield
of rational numbers. Such exists because of the
irreducibility of Q. Note that F12(a1) = a2, since
a1 = -r1 and a2 = -r2.
Without loss of generality, suppose a1 is coprime
to 5 in the ring of algebraic integers. By definition
this means that there exist algebraic integers r and s
such that
r*a1 + s*5 = 1.
Now apply F12 to both sides:
(1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1).
Note that F12(5) = 5 and F12(1) = 1 because
the automorphism F12 leaves fixed the subfield
of rational numbers. Also F12(a1) = a2.
Let r' = F12(r) and s' = F12(s). Note that
both r' and s' are algebraic integers. Thus eqn (1)
reduces to
r'*a2 + s'*5 = 1.
That is, a2 is also coprime to 5.
Similarly one shows that a3 is coprime to 5.
Therefore if one of a1, a2, or a3 is coprime
to 5, they all are.
But a1*a2*a3 = 65. Thus at least one of a1,
a2, and a3 is NOT coprime to 5. Thus a contradiction.
The claim in the Advanced Polynomial Factorization
paper is therefore wrong.
> Now there are a lot of usual suspects who argue with me. They lie.
> They blow smoke. They refuse to acknowledge mathematical truths, but
> will involve me in long threads, with long posts if I let them.
>
> Why should I?
>
> I am a force of one.
>
> If any of you can find an error then present it.
See above.
> Then can someone who
> is objective and invested in either side push forward any claims of
> error that seem cogent?
>
> If so, then I'll address THAT person's post.
Go for it.
Nora Baron
<snip>
David Collier
> an example to try and help them understand?
>
> Are you confused by the profound and extraordinary possibilities of
> P(x)=x+1 in the ring of integers?
I certainly am. It's very confusing to write "P(x) in the ring of
integers". Taken literally, that ought to mean that P(x) is an integer;
but it's not - you could think of it as a function from the set of
integers to itself, perhaps, but that's not the same as being an
integer.
Surely what you mean to say is "P(x) in the ring of polynomials
with integer coefficients".
If you can't be bothered to write out "the ring of polynomials with
integer coefficients", I'm sure we'd be happy with "Z[x]".
C. Bond
> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03061...@posting.google.com>...
> > > I have NOTHING to hide about my work. It's available 24 hours a day,
> > 7 days a week at
> >
> > http://groups.msn.com/AmateurMath
> >
> > and included the paper Advanced Polynomial Factorization, which is
> > sweeping in its simplicity, while being the most important math paper
> > of the year because it highlights an error in taught mathematics--a
> > problem in "core".
> >
>
> Let's have a look at it. The main claim, it appears,
> is that if P(x) is the polynomial
>
> P(x) = 65*x^3 - 12*x + 1
>
> and P(x) is factored in the form
>
> P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1)
>
> where a1, a2, and a3 are algebraic integers, then
> at least one of a1, a2 or a3 is coprime to 5.
>
> Is this correct ?
[snip analysis of claim]
> Therefore if one of a1, a2, or a3 is coprime
> to 5, they all are.
>
> But a1*a2*a3 = 65. Thus at least one of a1,
> a2, and a3 is NOT coprime to 5. Thus a contradiction.
>
> The claim in the Advanced Polynomial Factorization
> paper is therefore wrong.
>
> > If any of you can find an error then present it.
>
> See above.
>
> > Then can someone who
> > is objective and invested in either side push forward any claims of
> > error that seem cogent?
> >
> > If so, then I'll address THAT person's post.
>
> Go for it.
>
> Nora Baron
I predict that either (1) James will ignore your post altogether, in spite of the fact that he actively
solicited such responses, or (2) he will completely fail to grasp what you posted and, as an alternative to
constructive dialog, will accuse you of lying, cheating, blowing smoke, 'blocking' his contribution, etc.
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
Ktulwxwatcher
mathematics, as is everyone else is that reads your stuff so if so many people
say it's wrong, get a clue.
David Moran
Joona I Palaste
by Ullrich et al. are wrong, please say where you think there are
wrong. That's all. I don't want to read these rants that have nothing
to do with mathematics, and I don't think anyone else wants to either.
--
/-- Joona Palaste (pal...@cc.helsinki.fi) ---------------------------\
| Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
| http://www.helsinki.fi/~palaste W++ B OP+ |
\----------------------------------------- Finland rules! ------------/
"You could take his life and..."
- Mirja Tolsa
Arturo Magidin
James Harris <jst...@msn.com> wrote:
[.snip.]
>How can they ask that when I gave P(x)=x+1 in the ring of integers as
>an example to try and help them understand?
You call yourself a writer; the sentence above does not parse, let
alone make mathematical sense.
Mathematical formulas are part of the language. They are read out loud
together with the rest of the sentence. What you wrote above is:
"How can they ask that[,] when I gave capital P of x is equal to x
plus one in the ring of integers as an example to try and help them
understand?"
>Are you confused by the profound and extraordinary possibilities of
>P(x)=x+1 in the ring of integers?
Again, this does not parse.
Setting that aside, you are still confusing polynomials, functions,
and their values. Polynomials, qua polynomials, are not functions,
though they can be used to define functions. However, their properties
as functions do not always translate to properties as polynomials: the
constant function 2 from the integers to the integers divides the
function x|->(x^2+x) in the ring of all functions from the integers to
the integers, but the polynomial 2 does not divide the polynomial
x^2+x in the ring of all polynomials with integer coefficients.
And neither the polynomial function nor the function are
->integers<-.
As to your preference of examples by way of explanation, when will it
get through that when people ask for a ->definition<-, and example
will ->never<- do by itself?
[.snip.]
>Why don't you show just how well you understand Sawyer by explaining
>with
>
> P(x) = x + 1
>
>in the ring of integers?
Because if it is a polynomial, it is not an integer (except for
constant polynomials, which sometimes are considered to be integers as
well). If it is an integer, then at best you can think of it as a
constant polynomial. But you are clearly not doing either. So P(x) is
->not<- "in the ring of integers."
At least not as the words are understood by the entire world bar James
Harris.
>> So you need either: to explain which way you are thinking of your
>> polynomials, or to explain why Sawyer's carefully delineated
>> distinction doesn't exist.
>
>More than once I've talked about polynomials as a family of values.
Which is incorrect. Perhaps you should talk about the polynomial
function associated to the polynomial instead...
[.snip.]
>>But if every polynomial divides every other polynomial,
>> isn't it rather hard to have an irreducible polynomial?
>
>You don't know much mathematics do you?
>
>Irreducibility is typically expressed over Q, or the field of
>rationals, though it can also be about Z, the integers. And I learned
>that from arguing with people about it.
You don't know much mathematics, do you?
Irreducibility of a polynomial is ALWAYS expressed relative to a
particular ring of polynomials. Only when it is understood from
context can this be dropped. "Irreducible over Q" means that the
polynomial is irreducible, when considered as an element of the ring
of polynomials with rational coefficients, Q[x].
[.snip.]
>Even with current math teaching I doubt anyone would talk of
>irreducibility over the ring of algebraic integers.
You are incorrect; it makes perfect sense to talk about irreducibility
over the ring of algebraic integers. It means that the polynomial,
when considered as an element of A[x], where A is the ring of all
algebraic integers, is irreducible. As it happens, the only
irreducible polynomials are the linear polynomials which are
primitive; that is, polynomials ax+b, where a and b are algebraic
integers, a nonzero, and a and b are coprime (there exist r,s
algebraic integers such that ar+bs=1).
[.snip.]
>Well you can't help. You betrayed that you don't know even basic
>mathematics.
You are being very careful with your rare exotic flower, aren't you?
>Oh yeah, it'd help if you know mathematics.
It would help is ->YOU<- knew mathematics. But you do not. Certainly,
not the kind of mathematics you are trying to talk about.
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Nora Baron
Yes, looks like prediction (1) is on target, but of course
prediction (2) might come true eventually. You realize that
neither of these predictions puts you up there in a class with
Jeanne Dixon or other renowned prophets.
A third possibility is that he may say, "You did not find an
error in my paper. Therefore if you are right and I am right,
there is a contradiction in 'core' mathematics, which is what
I said in the first place." The problems with that argument
are (1) sight unseen, it is far more likely that there is an
error in his argument than that there is a heretofore over-
looked contradiction in 'core' math, whatever that is - if
you see a strange light in a dark wooded area, you don't assume
as a first explanation that it is extraterrestrials out hunting
for mushrooms; and (2) his paper, particularly the latter
part, is quite unreadable.
A fourth possibility is that my argument is wrong and he
will find an error in it. I don't totally rule that out.
However I agree that the chance that he even understands my
argument is pretty slim.
Nora B.
Robert J. Kolker
James Harris wrote:
> Now given my original post, which called for an objective review of
> the actual paper, which is available 24 hours a day, one might
> understand why I see Nora as a yet another jackass.
>
Blovating jackass!
Bob Kolker
Brian Chandler
> > can see) are genuinely asking you to explain _exactly_ what you mean
> > by a polynomial, I've given the following couple of pages of a
> > (smelly-paper) book, which are also available 24/7 (I hope; I use
> > pair.com hosting):
>
> an example to try and help them understand?
I think the point is that they all know that "polynomial" refers to
something with a combination of letters and numbers, with plus signs
in the middle, and bits raised to different powers.
> Are you confused by the profound and extraordinary possibilities of
> P(x)=x+1 in the ring of integers?
Yes, I suppose, because like everyone else I don't know what you mean
by "in the ring of integers".
> > "The scope of x", pp 34-35 of Sawyer's "A concrete approach to
> > abstract algebra":
> > http://imaginatorium.org/private/sawyer.gif
> > (136K, 800x600)
> >
> > In this extract, Sawyer describes two ways of looking at a polynomial;
> > depending on whether you view the polynomial as a formal object or
> > whether you view it as simply a generalised form of arithmetic (i.e.
> > you are always thinking of just its evaluation), you will get
> > different answers to quite basic questions like "Are these two
> > polynomials equal?" or "Is this polynomial a factor of that one?"
>
> Hmmm...I think you're trying to blow smoke.
I'm paraphrasing Sawyer: are you saying he's "blowing smoke"? (You
could at least have accused me of blowing smoke rings - that might
have been mildly funny.)
> Why don't you show just how well you understand Sawyer by explaining
> with
>
> P(x) = x + 1
>
> in the ring of integers?
>
> > So you need either: to explain which way you are thinking of your
> > polynomials, or to explain why Sawyer's carefully delineated
> > distinction doesn't exist.
>
> More than once I've talked about polynomials as a family of values.
OK: this sounds a little like the idea of a polynomial function: for
each integer, n say, there's another integer m, got by evaluating this
P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the
'bright pupils' tend to see introductory algebra.
> But at times I also speak of them as objects in their own right.
OK: this sounds a bit like the formal view of a polynomial. How Sawyer
suggests the 'bright pupils' tend to see introductory algebra.
(**Of course**, he's not saying that one view is right or wrong, or
weak or powerful, just that it turns out that different beginning
students grapple with this differently.)
Well, forgive me for choosing a different example, but I'd like to
take
>
> > Um, actually, have I understood your latest claim correctly? You say
> > that the algebraic integers are incomplete, because you have x and y,
> > such that y is a factor of x, but x/y isn't an algebraic integer? So
> > in your new system of completing the ring of algebraic integers (is
> > this related to Objects, by the way? I'll called them CAIs for
> > completed algebraic integers for now) - is it the case that for any
> > CAIs, r and s, that r/s is also a CAI? So consider two general
> > polynomials P(x) and Q(x) "in" the ring of CAIs: since you say that
> > the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is
> > also a CAI?
>
> No.
>
> >But if every polynomial divides every other polynomial,
> > isn't it rather hard to have an irreducible polynomial?
>
> You don't know much mathematics do you?
>
> Irreducibility is typically expressed over Q, or the field of
> rationals, though it can also be about Z, the integers. And I learned
> that from arguing with people about it.
>
> Even with current math teaching I doubt anyone would talk of
> irreducibility over the ring of algebraic integers.
>
> Damn. So you're just another jackass.
>
> > >
> > > If any of you can find an error then present it. Then can someone who
> > > is objective and invested in either side push forward any claims of
> > > error that seem cogent?
> >
> > Well, I can't understand what an object is. Do you know anyone except
> > yourself who can? Perhaps a good tack to take would just be a humble
> > appeal to the silent masses for some good soul to step forward and
> > explain objects so that at least people like me could understand them
> > (even if the Nasty Mathematician Mafia say they can't).
>
> Well you can't help. You betrayed that you don't know even basic
> mathematics.
>
> Oh well, is there anyone out there who can be objective?
>
> Oh yeah, it'd help if you know mathematics.
>
>
> James Harris
Brian Chandler
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03061...@posting.google.com>...
> imagin...@despammed.com (Brian Chandler) wrote in message news:<f2c35871.03061...@posting.google.com>...
> > Right. Well, since you complain a lot about people who (as far as I
> > can see) are genuinely asking you to explain _exactly_ what you mean
> > by a polynomial, I've given the following couple of pages of a
> > (smelly-paper) book, which are also available 24/7 (I hope; I use
> > pair.com hosting):
>
> How can they ask that when I gave P(x)=x+1 in the ring of integers as
> an example to try and help them understand?
I think the point is that they all know that "polynomial" refers to
something with a combination of letters and numbers, with plus signs
in the middle, and bits raised to different powers.
> Are you confused by the profound and extraordinary possibilities of
> P(x)=x+1 in the ring of integers?
Yes, I suppose, because like everyone else I don't know what you mean
by "in the ring of integers".
> > "The scope of x", pp 34-35 of Sawyer's "A concrete approach to
> > abstract algebra":
> > http://imaginatorium.org/private/sawyer.gif
> > (136K, 800x600)
> >
> > In this extract, Sawyer describes two ways of looking at a polynomial;
> > depending on whether you view the polynomial as a formal object or
> > whether you view it as simply a generalised form of arithmetic (i.e.
> > you are always thinking of just its evaluation), you will get
> > different answers to quite basic questions like "Are these two
> > polynomials equal?" or "Is this polynomial a factor of that one?"
>
> Hmmm...I think you're trying to blow smoke.
I'm paraphrasing Sawyer: are you saying he's "blowing smoke"? (You
could at least have accused me of blowing smoke rings - that might
have been mildly funny.)
> Why don't you show just how well you understand Sawyer by explaining
> with
>
> P(x) = x + 1
>
> in the ring of integers?
>
> > So you need either: to explain which way you are thinking of your
> > polynomials, or to explain why Sawyer's carefully delineated
> > distinction doesn't exist.
>
> More than once I've talked about polynomials as a family of values.
OK: this sounds a little like the idea of a polynomial function: for
each integer, n say, there's another integer m, got by evaluating this
P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the
'bright pupils' tend to see introductory algebra.
> But at times I also speak of them as objects in their own right.
OK: this sounds a bit like the formal view of a polynomial. How Sawyer
suggests the 'bright pupils' tend to see introductory algebra.
(**Of course**, he's not saying that one view is right or wrong, or
weak or powerful, just that it turns out that different beginning
students grapple with this differently. It seems you do have some idea
of the distinction he's making.)
Well, forgive me for choosing a different example, but I'd like to
take:
Q(x) = 2x+1
When you consider this "in the ring of integers", does it have any
roots?
> > Um, actually, have I understood your latest claim correctly? You say
> > that the algebraic integers are incomplete, because you have x and y,
> > such that y is a factor of x, but x/y isn't an algebraic integer? So
> > in your new system of completing the ring of algebraic integers (is
> > this related to Objects, by the way? I'll called them CAIs for
> > completed algebraic integers for now) - is it the case that for any
> > CAIs, r and s, that r/s is also a CAI? So consider two general
> > polynomials P(x) and Q(x) "in" the ring of CAIs: since you say that
> > the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is
> > also a CAI?
>
> No.
Why ever not? Is there a false step in my outline proof? Can you show
me where it is? After all, we know proofs are irrefutable...
[snip: Arturo Magidin dealt with this]
> Damn. So you're just another jackass.
Aha!! I'll have my point back, please.
Brian Chandler
----------------
geo://Sano.Japan.Planet_3
Arturo Magidin
[.snip.]
>nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
>> Without loss of generality, suppose a1 is coprime
>> to 5 in the ring of algebraic integers. By definition
>> this means that there exist algebraic integers r and s
>> such that
>
>Notice the call to the definition.
Definitions are shorthand. In computer science terms, something you
claim to be familiar with, "coprime" is a macro; saying "by
definition" just means that you are expanding the macro. Saying "a1
and 5 are coprime" means "there exists r and s such that r*a1+s*5=1."
Just like saying "a is even" means, ->by definition of "even"<-, that
there exist an integer k such that a=2*k.
[.snip.]
>Which looks like a good example to show those who wondered how what
>I've shown highlights an error in *taught* mathematics.
>
>It's basically an abuse of Galois Theory.
It is an abuse of Galois Theory that someone who admits he does not
know anything about it will nonetheless insist he can spot when it is
being used correctly and when it is not.
[.snip.]
>Now I've explained more than once that I prove that x has y as a
>factor, but x/y is not an algebraic integer, which is a contradiction.
Which means you do not know what the word "factor" means.
That's all. You do not understand what "factor" means. What you write
above is exactly the same as saying that you have found and even
integer n such that n/2 is not an integer. It is an abuse of
terminlogy.
>So *obviously* there's a problem somewhere
Yes: the problem is that you are an ignoramus who refuses to learn
what the words he uses mean.
DEF. Let R be a commutative ring. Let x and y be elements of R. We say
"y is a factor of x (in R)" if and only if there exist z in R such
that y*z = x. If the ring R is understood from context, we can omit
"(in R)".
DEF. Let R be a commutative ring. An element x in R is called "a unity"
if and only if for all y in R, x*y=y*x = y. If so, we denote x by "1",
and we say that R is a "commutative ring with 1."
DEF. Let R be a commutative ring. We say that R is an INTEGRAL DOMAIN
if and only if it has a 1 for every x,y in R, if x*y=0, then x=0 or y=0.
DEF. Let R be an integral domain. The "FIELD OF FRACTIONS" of R is
constructed as follows:
(1) Let k be the collection of all formal symbols "x/y", with x,y
in R, y nonzero.
(2) Define an equivalence relation on k, by letting (x/y)~(z/w) if
and only if x*w=z*y in R.
(3) Let K be the set whose elements are the equivalence classes of
elements of k under ~. Denote the class of x/y by [x/y].
(4) Define an addition and a multiplication in K by:
[x/y] + [z/w] = [(x*w+z*y)/y*w]
[x/y] * [z/w] = [(x*z)/(y*w)].
THEOREM. The field of fractions of an integral domain is a field; the
0 is the class [0/1]. The one is [1/1]. R is a subring of K by the map
that identifies the element r in R with the element [r/1] of K.
THEOREM. Let R be an integral domain, and let x,y be elements of
R. Then y is a factor of x in R if and only if the element [x/y] of K
lies in the image of R. We express this by writing "x/y lies in R".
Therefore, what you have claimed is that there is an algebraic integer
which is not an algebraic integer. And that is nonsense on the same
level as saying that you can find an even number which is not a
multiple of 2.
Randy Poe
> > to 5 in the ring of algebraic integers. By definition
> > this means that there exist algebraic integers r and s
> > such that
>
Uh, yes. In verifying whether a1 is coprime to 5, we see
whether it meets the definition of "coprime to 5". Do
you really think that there's something wrong with that?
- Randy
Jim Ferry
neutral, disinterested, but mathematically compentent third party willing
to review one's work. Nice job.
James Harris wrote:
> Nora Baron wrote:
>>James Harris wrote:
>>
>>>Their names are quite familiar to me and to many of you who have kept
>>>up with the discussions that have raged about my mathematical work:
>>>
>>>Ullrich, Magidin, Wayne Brown, Christian Bau, etc.
>>>
>>>A group uniform in that they are unrelentingly negative, often take
>>>cheapshots, like Ullrich, a math professor at a state university, who
>>>loves the word "idiot" or likening me to a monkey with a typewriter.
Did he do that? That's not right. You're more like a monkey with a computer
hooked up to the internet. A monkey with only a typewriter is far more limited
in where he can fling his crap.
>>>and included the paper Advanced Polynomial Factorization, which is
>>>sweeping in its simplicity, while being the most important math paper
>>>of the year because it highlights an error in taught mathematics--a
>>>problem in "core".
What were some of the runners-up for "most important math paper of the
year"? Was it hard to judge, objectively, between those in the field of
"core mathematics" and those in, say, algebraic topology?
>>Without loss of generality, suppose a1 is coprime
>>to 5 in the ring of algebraic integers. By definition
>>this means that there exist algebraic integers r and s
>>such that
>
>
> Notice the call to the definition.
>
>
>> r*a1 + s*5 = 1.
Oh yeah, look at that! When Nora uses the word "coprime" she takes it
to mean what its definition says it means. Whereas you use the word
more creatively, like an artist really. This powerful technique allows
you to prove, well, anything you want! But when will the stodgy
mathematical community come to accept your avant-garde vision?
>>Note that F12(5) = 5 and F12(1) = 1 because
>>the automorphism F12 leaves fixed the subfield
>>of rational numbers. Also F12(a1) = a2.
>>
>>Let r' = F12(r) and s' = F12(s). Note that
>>both r' and s' are algebraic integers. Thus eqn (1)
>>reduces to
>>
>> r'*a2 + s'*5 = 1.
>>
>>That is, a2 is also coprime to 5.
>>
>>Similarly one shows that a3 is coprime to 5.
>
>
> Which looks like a good example to show those who wondered how what
> I've shown highlights an error in *taught* mathematics.
>
> It's basically an abuse of Galois Theory.
Is any use of Galois Theory an abuse of it, or just those that debunk your
arguments?
Please don't let the fact that you don't know the first thing about Galois
Theory stop you from giving a ruling! We need to know ASAP whether to stop
stop teaching it.
> Finally, notice the attempt to disprove a proof with *another*
> argument, which is claimed to be a proof.
>
> It's like trying to fight a proof with a proof, but proofs can't fight
> each other.
Even proofs that use Galois theory? Okay, denote Nora's proof N, and let
N^* be [must stop fingers from typing . . . losing control . . .] the dual
proof [aarrghhh!]
> Now given my original post, which called for an objective review of
> the actual paper, which is available 24 hours a day, one might
> understand why I see Nora as a yet another jackass.
A jackass on the TV show "Jackass" recently attached a bungie cord to his
underwear, then jumped out of a tree in pursuit of the ultimate wedgie.
It was generally agreed that he found it.
To call you a jackass would be an insult to that guy, James.
C. Bond
[snip]
Putting aside, for a moment, the urge to comment about your histrionics
(which often reach epic proportions) and concentrating on the scientific
content of your work, it is clear that you are not yet qualified to enroll
in algebra 101. Your character flaws consistently foil any attempt on
your part to devise a rational argument suitable for presentation to
others, leaving you with the double stigma of being a lousy mathematician
and a worse human being.
If you really do have a degree in physics, which I doubt, I think you have
compelling grounds to demand your money back. There is no evidence of
scientific thought present in your flawed arguments or your passionate,
nay maniacal, defense of them.
On a positive note, you can probably still get a job as a crash dummy.
--
"Words mean exactly what I want them to mean!" -- the caterpillar in Alice
in Wonderland and, more recently, James Harris.
--
Democracy: The triumph of popularity over principle.
Nora Baron
>nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
>> > Their names are quite familiar to me and to many of you who have
kept
>> > up with the discussions that have raged about my mathematical
work:
>> >
>> > Ullrich, Magidin, Wayne Brown, Christian Bau, etc.
>> >
>> > A group uniform in that they are unrelentingly negative, often
take
>> > cheapshots, like Ullrich, a math professor at a state university,
who
>> > loves the word "idiot" or likening me to a monkey with a
typewriter.
>> >
>>
>> <snip>
>>
>> > > I have NOTHING to hide about my work. It's available 24 hours
a day,
>> > 7 days a week at
>> >
>> >
>> > and included the paper Advanced Polynomial Factorization, which
is
>> > sweeping in its simplicity, while being the most important math
paper
>> > of the year because it highlights an error in taught
mathematics--a
>> > problem in "core".
>> >
>>
>> Let's have a look at it. The main claim, it appears,
>> is that if P(x) is the polynomial
>>
>> P(x) = 65*x^3 - 12*x + 1
>>
>> and P(x) is factored in the form
>>
>> P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1)
>>
>> where a1, a2, and a3 are algebraic integers, then
>> at least one of a1, a2 or a3 is coprime to 5.
>>
>> Is this correct ?
>
>
>> Let x = 1/u. Then P(x) = 0 implies
>>
>> P(x) = 65/u^3 - 12/u + 1 = 0.
>>
>> Multiplying through by u^3 gives
>>
>> u^3 - 12*u^2 + 65 = 0.
>>
>> Call this polynomial Q(u). Let r1, r2, and r3 be
>> roots of Q. Note that a1, a2 and a3 happen to be
>> the negatives of r1, r2, and r3 - that is, a1 = -r1,
>> a2 = -r2, a3 = -r3 [the order does not matter].
>>
>> The polynomial Q(u) is irreducible. Let H be the
>> field of algebraic numbers; clearly r1, r2, and
>> r3 are in H. Let F12 be an automorphism of H such
>> that F12(r1) = r2, and F12 leaves fixed the subfield
>> of rational numbers. Such exists because of the
>> irreducibility of Q.
Reference for this:
http://www.math.niu.edu/~beachy/aaol/galois.html
especially Proposition 8.6.2 on that page. Or see
the textbook, Abstract Algebra, by John Beachy
and William Blair. An *excellent* book.
>> Note that F12(a1) = a2, since
>> a1 = -r1 and a2 = -r2.
>>
>> Without loss of generality, suppose a1 is coprime
>> to 5 in the ring of algebraic integers. By definition
>> this means that there exist algebraic integers r and s
>> such that
>
>Notice the call to the definition.
>
You have a problem with my using a
standard definition ???
>>
>> r*a1 + s*5 = 1.
>>
>> Now apply F12 to both sides:
>>
>> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1).
>>
>> Note that F12(5) = 5 and F12(1) = 1 because
>> the automorphism F12 leaves fixed the subfield
>> of rational numbers. Also F12(a1) = a2.
>>
>> Let r' = F12(r) and s' = F12(s). Note that
>> both r' and s' are algebraic integers. Thus eqn (1)
>> reduces to
>>
>> r'*a2 + s'*5 = 1.
>>
>> That is, a2 is also coprime to 5.
>>
>> Similarly one shows that a3 is coprime to 5.
>
>Which looks like a good example to show those who wondered how what
>I've shown highlights an error in *taught* mathematics.
>
>It's basically an abuse of Galois Theory.
>
What's the abuse?
>> Therefore if one of a1, a2, or a3 is coprime
>> to 5, they all are.
>>
>> But a1*a2*a3 = 65. Thus at least one of a1,
>> a2, and a3 is NOT coprime to 5. Thus a contradiction.
>>
>> The claim in the Advanced Polynomial Factorization
>> paper is therefore wrong.
>
><deleted>
>
>Now I've explained more than once that I prove that x has y as a
>factor, but x/y is not an algebraic integer, which is a
contradiction.
>
>So *obviously* there's a problem somewhere with what mathematicians
>are doing if they think they can prove something that's false.
>
I would say obviously there is a problem somewhere,
yes.
>Finally, notice the attempt to disprove a proof with *another*
>argument, which is claimed to be a proof.
>
You claimed something was true. I gave an argument
to show it wasn't.
Four possibilities:
1. I have an error in my argument. You however did
not point out any error.
2. You have an error in your argument. I did not
point it out in my post, but it has been pointed
out many times previously. Your argument depends
on the form that a factorization has in a degenerate,
singular case (when m = 0). You argue implicitly
that that same form of factorization must hold in the
nonsingular case in which you are actually interested.
It does not. This has been pointed out innumerable times.
Arturo and others have tried many times to get
you to understand it, but they have failed. I will
probably fail as well.
3. Both of us have errors in our arguments.
4. Neither of us has an error in our arguments. There
is a fundamental contradiction in mathematics. I
think we should just put this idea aside. It
is the least likely of the possibilities and it
does not obviously lead anywhere.
It's also possible that the fact I used about automorphisms,
which comes from Galois theory, is wrong. That too seems
unlikely since Galois theory has been around for over 160
years and looked at by many very smart people, and this
fact is encountered at the most basic level. I think that
too should be put aside.
>It's like trying to fight a proof with a proof, but proofs can't
fight
>each other.
>
We both present arguments. You say yours is a proof.
I say mine is. We are not both right. In the end
we have to try to convince other people and each other.
>Now given my original post, which called for an objective review of
>the actual paper, which is available 24 hours a day, one might
>understand why I see Nora as a yet another jackass.
>
Thanks for enlightening me on that. I did not give a
comprehensive review, but I addressed your main point
and I showed it was wrong. I was objective. I did not
address Lemma 1.1 because in my view it is trivial:
given the hypotheses, and noting that g is actually a
function of x, you then let the constant c = g(0) [you
did do that part], and then you define r(x) = g(x) - c.
End of proof.
But again, your central error is assuming that the form
of a factorization is the same in a singular case
as it is in the general case. It is always dangerous
to generalize based on degenerate cases or singularities.
You can probably guess why.
Nora B.
>
>
>James Harris
Thomas Wasell
>nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
>> > Their names are quite familiar to me and to many of you who have kept
>> > up with the discussions that have raged about my mathematical work:
>> >
>> > Ullrich, Magidin, Wayne Brown, Christian Bau, etc.
>> >
>> > A group uniform in that they are unrelentingly negative, often take
>> > cheapshots, like Ullrich, a math professor at a state university, who
>> > loves the word "idiot" or likening me to a monkey with a typewriter.
>> >
>
>
>It's nasty but maybe there's another route to resolution as I thought
>I'd come back and go after Nora's argument a little more directly.
You mean you actually thought that calling people "jackass" was a route to
resolution? Maybe you should give mathematics a rest and lend your services
to the diplomatic corps. There are tons of wars simply dying to get started
-- YOU could be one that gets them going!
[snip]
>> Now apply F12 to both sides:
>>
>> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1).
>>
>> Note that F12(5) = 5 and F12(1) = 1 because
>> the automorphism F12 leaves fixed the subfield
>> of rational numbers.
>
><deleted>
>
>Prove it.
[S]he just did! Is it the word "automorphism" you don't understand? Just
look it up, then! Or do you disbelieve the existence of such an
automorphism? What, exactly, is it that you want proven?
[snip]
>If Nora isn't being a jackass, then sorry for calling you a jackass
>Nora.
>
>But if your argument is right, you have shown Galois Theory to be
>false.
As has been pointed out before, there are at least four possibilities here:
1) Nora's argument is wrong;
2) Your argument (or alleged "proof") is wrong;
3) Both arguments are wrong;
4) Galois Theory is wrong.
Galois Theory has never been proven wrong; you have (almost) never been
proven un-wrong.
>So you see, it doesn't matter, so I don't mind.
>
>In the end, mathematicians have to face my proof, and not try to run
>away from it, or disprove it with another proof because you see,
>proofs don't duel.
So the proof that your proof is correct is the "fact" that it is a proof?
Therefore, any disproof of your proof is nessesarily a disproof of itself
and anything that has been used to derive the disproof, right? Is there ANY
way one can convince you that your alleged "proof" is actually wrong?
--
Thomas Wasell | A pencil with no point needs no eraser.
was...@bahnhof.se |
BraileTrail
>In article <3EEF5326...@btinternet.com> George Cox
><georg...@btinternet.com> writes:
> > > or likening me to a monkey with a typewriter.
> >
> > This I find harder to explain. Perhaps David once met a monkey with a
> > typewriter and found it to be an unpleasant experience?
>
>Eh, no. This was an allusion to the knowledge that a million monkeys
>at typewriters will at some time produce a work of Shakespeare. Even
>I (non-native English speaker) understood it.
Sorry Dirk, but you missed George's very droll British sense of humour.
I thought George's remark very funny, but maybe that just says something
about me!
Regards,
BraileTrail
--
David C. Ullrich
wrote:
>In article <3c65f87.03061...@posting.google.com>, James Harris
><jst...@msn.com> wrote:
>>nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
>>> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03061...@posting.google.com>...
[...]
>>
>>In the end, mathematicians have to face my proof, and not try to run
>>away from it, or disprove it with another proof because you see,
>>proofs don't duel.
>
>So the proof that your proof is correct is the "fact" that it is a proof?
That's correct. No matter _how_ many times he eventually has to
say "oh by the way, that was wrong, sorry", it's still true that the
next time he thinks he's proved something the fact that it's a proof
is what proves it must be correct, and that anyone who disputes
its correctness is a lying incompetent fool who should be fired
(recently abbreviated to "jackass".)
Truly remarkable.
>Therefore, any disproof of your proof is nessesarily a disproof of itself
>and anything that has been used to derive the disproof, right? Is there ANY
>way one can convince you that your alleged "proof" is actually wrong?
************************
David C. Ullrich
Brian Chandler
> On 19 Jun 2003, James Harris wrote:
> >nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
> >> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03061...@posting.google.com>...
<snippety-snoppety: Nora attempts to refute JSH's argument, in the
course of which...>
Nora:
> >> Without loss of generality, suppose a1 is coprime
> >> to 5 in the ring of algebraic integers. By definition
> >> this means that there exist algebraic integers r and s
> >> such that
JSH:
> >Notice the call to the definition.
> >
Nora:
> You have a problem with my using a
> standard definition ???
This is only a wild hypothesis, but here goes.
Suppose JSH thinks a proof is rather like a computer program. Then he
thinks of bits like for loops, assignments, ifs, and whatnot as "pure
logic". But notice his exact phrasing above - a "call to the
definition". Suppose he thought that a definition was rather like a
#define in C: can be used for unhygienic purposes, and is generally
deprecated. Wouldn't this explain a number of his wilder claims?
(Well I can't see how to account for the "abuse of Galois theory" bit,
but I think we have to go one step at a time.)
Brian Chandler
----------------
geo://Sano.Japan.Planet_3
http://imaginatorium.org/
JSH "Jackass" rating: 1 (so far)
Randy Poe
> That's false. It turns out that the *definition* of an algebraic
> integer as the ROOT of a monic polynomial with integer coefficients is
> what allows the ring to be incomplete.
>
> Now any rational person looking over that definition for an algebraic
> integer would be at a loss to see how the definition of "factor"
> applies.
No, a rational person would say this:
"Let c factor into a*b in the ring of algebraic integers.
Since algebraic integers are roots of monic polynomials with
integer coefficients, I conclude that a and b are roots of
monic polynomials with integer coefficients."
What "loss" did you think a rational person would be at? There
is NO OTHER MEANING of factor!
- Randy
William Hale
jst...@msn.com (James Harris) wrote:
> nora...@hotmail.com (Nora Baron) wrote in message
news:<36024859.03061...@posting.google.com>...
> > >> r*a1 + s*5 = 1.
> > >>
> > >> Now apply F12 to both sides:
> > >>
> > >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1).
> > >>
> > >> Note that F12(5) = 5 and F12(1) = 1 because
> > >> the automorphism F12 leaves fixed the subfield
> > >> of rational numbers. Also F12(a1) = a2.
>
Which statement(s) do you need more justification for:
a) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1)
b) F12(5) = 5
c) F12(1) = 1
d) F12(a1) = a2
-- Bill Hale
Arturo Magidin
In article <3c65f87.03062...@posting.google.com>,
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bct5bc$bkp$1...@agate.berkeley.edu>...
>> In article <3c65f87.03061...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [.snip.]
>>
>> >nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
>>
>> >> Without loss of generality, suppose a1 is coprime
>> >> to 5 in the ring of algebraic integers. By definition
>> >> this means that there exist algebraic integers r and s
>> >> such that
>> >
>> >Notice the call to the definition.
>>
>> Definitions are shorthand. In computer science terms, something you
>> claim to be familiar with, "coprime" is a macro; saying "by
>> definition" just means that you are expanding the macro. Saying "a1
>> and 5 are coprime" means "there exists r and s such that r*a1+s*5=1."
>
>Other readers will remember that I replied *again* noting that my
>issue with the call to the definition here didn't matter.
Will you ->EVER<- learn about distribution in usenet? That message did
not make it to my server until after I had posted this. Your innuendos
notwithstanding.
[.snip.]
>> Which means you do not know what the word "factor" means.
>>
>> That's all. You do not understand what "factor" means. What you write
>> above is exactly the same as saying that you have found and even
>> integer n such that n/2 is not an integer. It is an abuse of
>> terminlogy.
>
>That's false. It turns out that the *definition* of an algebraic
>integer as the ROOT of a monic polynomial with integer coefficients is
>what allows the ring to be incomplete.
Please provide a definition of "incomplete".
Because, if we use the words "definition", "algebraic integer", and
"factor" in their usual sense, then what you have said is nonsense.
>Now any rational person looking over that definition for an algebraic
>integer would be at a loss to see how the definition of "factor"
>applies.
DEFINITION. Let c be a complex number. Then c is an "algebraic
integer" if and only if there exists a monic polynomial f(x) with
integer coefficients such that c is a root of f(x).
DEFINITION. Let R be any commutative ring. Let x and y be elements of
R. Then "y is a factor of x in R" if and only if there exists z in R
such that y*z = x.
THEOREM. The collection of all algebraic integers form an integral
domain, contained in the field of complex numbers.
Which of the above do you disagree with?
Remember: definitions cannot be "right" or "wrong"; they are only
shorthand.
Now, you claim that:
(a) You have algebraic integer y and x;
(b) that "y is a factor of x" in the ring of algebraic integers;
(c) that there does not exist a z in the ring of algebraic integers
such that y*z = x.
Therefore, you are speaking nonsense.
[.snip.]
>> >So *obviously* there's a problem somewhere
>>
>> Yes: the problem is that you are an ignoramus who refuses to learn
>> what the words he uses mean.
>
>Well, you're a jackass.
I see you do not contest my statement above. For the record, your
appreciation of whether or not I am, as you claim, "a jackass" is
immaterial.
>Ok, so let's say readers wish to believe that I've been refuted.
Why do you not comment on anything of the above? Because you do not
understand it. Despite being mathematics.
>Why did Magidin go through the trouble?
Read the .sig.
>And what can I say here? That's he's full of it? That he's boldly
>lying yet again?
You could say that. That seems to be your only defense when faced with
undeniable evidence of your errors.
>But when you look over what he presented and think to yourself, "wow,
>that looks like what I figure math stuff should look like", what am I
>to do?
Read it, try to understand it, and if there is something you do not
understand, ask. Nobody is asking you to take it on faith. We only ask
that you take the time to LEARN the stuff you try to pontificate aboue.
>Even if I take him apart piece by piece if you *believe* it doesn't
>matter.
Because you never do "take apart piece by piece", except in your
imagination. Your complaints are always either wrong, nonsense, or
based on ignorance and misunderstanding.
Brian Quincy Hutchings
seriously,
"bright pupil" is sort of an oxymoron,
considering that Jimmy's eyes are always dilated.
imagin...@despammed.com (Brian Chandler) wrote in message news:<f2c35871.03061...@posting.google.com>...
> I'm paraphrasing Sawyer: are you saying he's "blowing smoke"? (You
> could at least have accused me of blowing smoke rings - that might
> OK: this sounds a little like the idea of a polynomial function: for
> each integer, n say, there's another integer m, got by evaluating this
> P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the
> 'bright pupils' tend to see introductory algebra.
> Q(x) = 2x+1
>
> When you consider this "in the ring of integers", does it have any
> roots?
> [snip: Arturo Magidin dealt with this]
>
> > Damn. So you're just another jackass.
>
> Aha!! I'll have my point back, please.
--les ducs d'Enron!
http://quincy4board.homestead.com/
Funny.html (schoolboard stuffin')
Alan Morgan
[snip]
>As I've stated more than once and as anyone can verify who looks over
>my paper, I prove that a number I'll call x has a number I'll call y
>as a factor,
When you state that x has y as a factor, what do you mean? Please be
precise. I'm not looking for examples, I'm looking for a definition.
I want a way to take two items, x and y, and determine if y is a factor
of x.
Alan
--
Defendit numerus
Brian Quincy Hutchings
that always halts, no matter what the input!
David C. Ullrich <ull...@math.okstate.edu> wrote in message news:<e6q5fv4v39bcat2un...@4ax.com>...
> That's correct. No matter _how_ many times he eventually has to
> say "oh by the way, that was wrong, sorry", it's still true that the
> next time he thinks he's proved something the fact that it's a proof
> is what proves it must be correct, and that anyone who disputes
> its correctness is a lying incompetent fool who should be fired
> (recently abbreviated to "jackass".)
> >Therefore, any disproof of your proof is nessesarily a disproof of itself
> >and anything that has been used to derive the disproof, right? Is there ANY
> >way one can convince you that your alleged "proof" is actually wrong?
--les ducs d'Enron!
Nora Baron
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
> > On 19 Jun 2003, James Harris wrote:
>
>
> > >
> > >Notice the call to the definition.
> > >
> >
> >
> > You have a problem with my using a
> > standard definition ???
>
>
> It was a mistake. I later posted in a second reply noting that fact.
>
> However, I notice that the second reply didn't get much attention.
>
>
> > >>
> > >> r*a1 + s*5 = 1.
> > >>
> > >> Now apply F12 to both sides:
> > >>
> > >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1).
> > >>
> > >> Note that F12(5) = 5 and F12(1) = 1 because
> > >> the automorphism F12 leaves fixed the subfield
> > >> of rational numbers. Also F12(a1) = a2.
>
>
Keep reading - I gave a reference to a website and a book
that explain this very clearly.
Oh, I see you deleted this reference. It was:
http://www.math.niu.edu/~beachy/aaol/galois.html
See esp. section 8.6.2.
> Also as noted repeatedly, proofs don't duel.
>
> If I'm wrong then there's an error in my proof.
>
> Trying to attack one proof by claiming it contradicts with another is
> useless.
>
> You can cast doubt that way, but to prove a "proof" false, you have to
> deal with the actual argument within it.
>
Actually, I don't agree with this. True, it is good to
identify where a person has made an error when they claim
to have proved something. But if you provide a proof that
what they are claiming is wrong, you have headed off not
only their current proposed argument, but also all future
arguments in support of that same claim. In any case, I
did both. I did say explicitly where your error occurs
and why it is wrong - see below.
>
> > >>
> > >> Let r' = F12(r) and s' = F12(s). Note that
> > >> both r' and s' are algebraic integers. Thus eqn (1)
> > >> reduces to
> > >>
> > >> r'*a2 + s'*5 = 1.
> > >>
> > >> That is, a2 is also coprime to 5.
> > >>
> > >> Similarly one shows that a3 is coprime to 5.
> > >
> > >Which looks like a good example to show those who wondered how what
> > >I've shown highlights an error in *taught* mathematics.
> > >
> > >It's basically an abuse of Galois Theory.
> > >
> >
> > What's the abuse?
>
> My belief has been that posters like yourself have been abusing Galois
> Theory as I don't believe it's wrong. I think you're cheating, which
> is why you wish to go to some other argument rather than deal with the
> one I've presented.
>
> So I see it as an abuse.
>
I still don't see the abuse. If I'm cheating, please
tell me where.
> Alternatively, Galois Theory IS wrong.
>
> Or you may *believe* I'm wrong, but if I'm wrong there would be a way
> to show it using my work.
>
Yes - see below.
> That is, if mathematicians are experts, which by definition they are,
> and I'm not a mathematician, but claim to have a proof, it hardly
> makes sense for mathematicians to use an alternative argument, as I'll
> point out the potential for abuse.
>
> It'd seem more rational for mathematicians to save their energy and
> attack my work itself.
>
I think both approaches should be taken - it's one thing
to point to a specific place in your argument and say,
"You have not justified this", to which you might answer
"It's obvious and anyone can see it and it doesn't need to
be justified." That can get frustrating. It's of value
also to say, "Here is an argument that shows your
argument, *whatever it is*, must be wrong. I chose to
take the latter course initially.
> > >> Therefore if one of a1, a2, or a3 is coprime
> > >> to 5, they all are.
> > >>
> > >> But a1*a2*a3 = 65. Thus at least one of a1,
> > >> a2, and a3 is NOT coprime to 5. Thus a contradiction.
> > >>
> > >> The claim in the Advanced Polynomial Factorization
> > >> paper is therefore wrong.
> > >
> > ><deleted>
> > >
> > >Now I've explained more than once that I prove that x has y as a
> > >factor, but x/y is not an algebraic integer, which is a
> contradiction.
> > >
> > >So *obviously* there's a problem somewhere with what mathematicians
> > >are doing if they think they can prove something that's false.
> > >
> >
> >
> > I would say obviously there is a problem somewhere,
> > yes.
>
> Then it hardly makes sense for you to spend time and effort not
> addressing my central claim which is that I have a proof.
>
> After all, if you found an error in that proof, you could make a VERY
> short post, and it'd be over instantly--no more discussion needed.
>
> CERTAINLY if I refused to acknowledge the truth, then I might continue
> arguing,
Yes - I think that is what will happen - anyway, keep
reading -
> but at least other posters would know there was a discovered
> error which they could view themselves to make their own decision.
>
> Now then are you a mathematician?
>
Yep.
> If so, then you are a math expert by definition. Then if there is an
> error in my paper you should be able to point it out, and quit wasting
> so much time trying to fight my proof with claims of other proofs of
> your own.
>
Keep reading.
> > >Finally, notice the attempt to disprove a proof with *another*
> > >argument, which is claimed to be a proof.
> > >
> >
> > You claimed something was true. I gave an argument
> > to show it wasn't.
>
> You fought a conclusion of one argument, which I claim is a proof,
> with another argument which you claim is a proof.
>
> I challenge you to find an error in my argument; you challenge me to
> find an error in yours.
>
I didn't challenge you. I just pointed out that you
hadn't. Anyway, see below.
> It's a waste of time.
>
I have nothing to lose in this. You are the one who
wants to be recognized for great discoveries. You cannot
afford to ignore proofs that you are wrong.
> I'm not a mathematician. My argument is simpler.
>
> So now there are dueling claims. You claim you gave an argument--I
> presume you'd claim it is correct--proving me wrong, I say you're
> wrong.
>
> Isn't anyone else bothered by the implication that mathematics is a
> mess?
>
> What's with all the debate?
>
> Are all mathematicians such losers that they can't handle a claim from
> an admitted non-mathematician, about a paper that's on display 24
> hours a day?
>
> ARE YOU ALL LOSERS??!!!
>
> Yup, I'm frustrated.
>
> > Four possibilities:
> >
> > 1. I have an error in my argument. You however did
> > not point out any error.
>
> I'm not a mathematician. Why in the hell should I necessarily be able
> to find an error in your argument?
>
It's a simple, short argument. You know enough math to
evaluate it. You are perfectly capable of finding an
error in it if there is one.
Might be worth noting that none of the other posters
here have ventured to say whether my argument is right or
wrong. It could easily be wrong, eh?
> Dueling claims continue.
>
I don't get this emphasis on "duelling claims". Why are
you stressing that so much? It's standard procedure if
one guy says he has a proof, and another guy says he is
wrong, the second guy either finds a counterexample (which
often involves a proof of its own) or a counter-proof.
> > 2. You have an error in your argument. I did not
> > point it out in my post, but it has been pointed
> > out many times previously. Your argument depends
> > on the form that a factorization has in a degenerate,
> > singular case (when m = 0). You argue implicitly
> > that that same form of factorization must hold in the
> > nonsingular case in which you are actually interested.
> > It does not. This has been pointed out innumerable times.
> > Arturo and others have tried many times to get
> > you to understand it, but they have failed. I will
> > probably fail as well.
>
> If that's true then you can reference actual statements in the paper.
>
You should be able to tell EXACTLY what I am talking
about from my reference to a degenerate case (where your
polynomial, which is usually of degree 3, becomes a linear
function, i.e., a polynomial of degree 1) and my reference
to m = 0. What I have given would be sufficient for any
normal person to understand what is wrong. Therefore I
refuse to give you more help than that at this point.
You have the choice of (1) actually doing some work to
figure out what I am talking about, or (2) resigning
yourself to the the fact that I have specified exactly
where your error is and that, separately, I have a
counterargument which proves you are wrong. I have led
the horse to water. If the horse wishes to die of thirst,
there is not much I can do about it.
Well, theoretically, there is one alternative. You
could give a detailed proof of why properties of a
factorization in a degenerate case imply that those same
properties hold in nondegenerate cases. That is, you can
fill in the main gap in your argument. Or try to.
And, theoretically, there is another alternative. You
could find an explicit error in MY argument. Do that and
I will go away.
Finally, if after studying it for a while you still
don't get what I am saying, let me know.
> What I see is an unsupported statement, and worse, you have an appeal
> to authority by mentioning Arturo Magidin.
>
By no means! I mentioned Arturo just to jog your
memory. 'Authority' here is not an issue. My statements
stand strictly on their own. For all I know, Arturo
thinks my argument is B.S..
> It seems to me that you are unsure of your own statement, and possibly
> wish to convince others without presenting actual evidence.
>
Not at all. I thought what I said was more than sufficient.
Apparently not. See above.
> > 3. Both of us have errors in our arguments.
> >
> > 4. Neither of us has an error in our arguments. There
> > is a fundamental contradiction in mathematics. I
> > think we should just put this idea aside. It
> > is the least likely of the possibilities and it
> > does not obviously lead anywhere.
>
> There is no fundamental contradiction in mathematics.
>
> I have presented the most logical possibility before which is that
> mathematicians are lying.
>
Doesn't seem all that likely that ALL of us would lie.
In any case, *I'm* definitely not lying.
> > It's also possible that the fact I used about automorphisms,
> > which comes from Galois theory, is wrong. That too seems
> > unlikely since Galois theory has been around for over 160
> > years and looked at by many very smart people, and this
> > fact is encountered at the most basic level. I think that
> > too should be put aside.
>
> I don't care how long it's "been around", if it's false it's false.
>
> If it's true it's true.
>
> And that condition is independent of time.
>
> However, I should point out that Galois Theory is a field theory.
>
All I used here was a little introductory theorem about
field automorphisms. A key feature is that automorphisms
exist which transform one root of an irreducible polynomial
into another root, they will leave the set of algebraic
integers invariant. That is not really a "field"
property, but it's relevant and useful here.
> My work is not over fields.
>
> Still either you screwed up that F12 bit above, or Galois Theory is
> wrong.
>
Or your argument is wrong. How have you excluded that
possibility?
> I'm interested in other replies that address whether you screwed up
> Nora.
>
You are? Well, people are perfectly free to make such
replies.
> > >It's like trying to fight a proof with a proof, but proofs can't
> fight
> > >each other.
> > >
> >
> > We both present arguments. You say yours is a proof.
> > I say mine is. We are not both right. In the end
> > we have to try to convince other people and each other.
>
> <deleted>
>
> That's stupid.
>
Is it? If so, why do you spend so much time trying
to convince everyone here that your arguments are correct?
> A proof begins with a truth and proceeds by logical steps to a
> conclusion which then must be true.
>
A correct proof does that. An incorrect proof either
leaves out a step or has an erroneous step. An argument
is not a proof just because its author says it is a proof.
> So all that's necessary is to start at the beginning and trace out
> each step.
>
> That IS possible with my paper, and I've done it which is why I get to
> have fun calling people like you out, and also calling you a jackass.
>
> You'll have a harder time tracing out your argument, but it might help
> you out to try, if you have the ability Nora.
>
See above.
> Yup, are you smart enough Nora?
>
Nah. I'm just a jackass. Why do you think anyone would
be convinced by what a jackass would say?
Nora B.
> Screw convincing people. People are often stupid enough to be
> convinced of just about anything.
>
>
> James Harris
Randy Poe
>rpo...@yahoo.com (Randy Poe) wrote in message news:<585ab5d8.03062...@posting.google.com>...
>> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03062...@posting.google.com>...
>> > That's false. It turns out that the *definition* of an algebraic
>> > integer as the ROOT of a monic polynomial with integer coefficients is
>> > what allows the ring to be incomplete.
>> >
>> > Now any rational person looking over that definition for an algebraic
>> > integer would be at a loss to see how the definition of "factor"
>> > applies.
>>
>> No, a rational person would say this:
>>
>> "Let c factor into a*b in the ring of algebraic integers.
>> Since algebraic integers are roots of monic polynomials with
>> integer coefficients, I conclude that a and b are roots of
>> monic polynomials with integer coefficients."
>
>Sure, *if* it so factors.
If it doesn't, neither a nor b is a factor of c.
>
>You're running in a circle.
>
>As I've stated more than once and as anyone can verify who looks over
>my paper, I prove that a number I'll call x has a number I'll call y
>as a factor,
The expression "y is a factor in the algebraic integers" means "x/y is
an algebraic integer".
> where they both are algebraic integers, but it turns out
>that x/y is NOT an algebraic integer.
Then y is not a factor of x.
Once again, the term "y is a factor of x in the algebraic integers"
means "x/y is an algebraic integer". That's what it means to every but
you.
What does it mean to you? In particular, what does it mean for y NOT
to be a factor of x in the algebraic integers? Just tell me that. If I
have two arbitrary algebraic integers, what test to I apply to tell if
y is a factor of x?
I'm going to hammer on that question for awhile to see if you answer
it.
>Here's a test of your ability to think rationally Randy Poe, can you
>see how nothing you have said prevents that possibility?
No I can't say that, since "x/y is an algebraic integer" and "y is a
factor of x in the algebraic integers" are the same statement.
>If you say that x factors into ab in the ring of algebraic integers,
>then of course x/a or x/b is an algebraic integer.
Yep.
>But I didn't say that x factors into y times some algebraic integer.
Then y is not a factor of x in the algebraic integers. See how that
works?
>I said that x *provably* has y as a factor in the ring of algebraic
>integers, but that leads to a contradiction as x/y is not in the ring,
>so the ring isn't complete. That is, x/y is *supposed* to be in the
>ring, but provably is not.
OK, I think I see what you're saying. You think you have a proof that
x/y is an algebraic integer. And another proof that it isn't.
Contradiction.
So what is your line of argument that x/y is an algebraic integer?
- Randy
David C. Ullrich
>rpo...@yahoo.com (Randy Poe) wrote in message news:<585ab5d8.03062...@posting.google.com>...
>> jst...@msn.com (James Harris) wrote in message news:<3c65f87.03062...@posting.google.com>...
>> > That's false. It turns out that the *definition* of an algebraic
>> > integer as the ROOT of a monic polynomial with integer coefficients is
>> > what allows the ring to be incomplete.
>> >
>> > Now any rational person looking over that definition for an algebraic
>> > integer would be at a loss to see how the definition of "factor"
>> > applies.
>>
>> No, a rational person would say this:
>>
>> "Let c factor into a*b in the ring of algebraic integers.
>> Since algebraic integers are roots of monic polynomials with
>> integer coefficients, I conclude that a and b are roots of
>> monic polynomials with integer coefficients."
>
>Sure, *if* it so factors.
>
>You're running in a circle.
>
>As I've stated more than once and as anyone can verify who looks over
>my paper, I prove that a number I'll call x has a number I'll call y
>as a factor, where they both are algebraic integers, but it turns out
>that x/y is NOT an algebraic integer.
Yes, you've stated this more than once. As has been stated in
reply many times, this is sheer nonsense, unless you clarify
that you mean "factor" in some sense other than "factor
in the algebraic integers", and explain what you _do_ mean.
[...]
>
>First I need to have rational people who can handle a logical
>argument.
>
>Not people whose brains can't handle the facts.
It appears that there's not a single mathematician on the
planet who can handle a rational argument. Very curious.
(Hmm, maybe there's another explanation...)
>James Harris
************************
David C. Ullrich
Wayne Brown
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<bct5bc$bkp$1...@agate.berkeley.edu>...
>> James Harris <jst...@msn.com> wrote:
>> >So *obviously* there's a problem somewhere
>>
>> Yes: the problem is that you are an ignoramus who refuses to learn
>> what the words he uses mean.
> Well, you're a jackass.
The problem is that you have *no* evidence to support what you're saying
about Arturo Magidin, but there are *thousands* of your articles over
a period of *years* that prove what he's saying about you.
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"
Wayne Brown
> Newsgroups trimmed.
> In article <3c65f87.03062...@posting.google.com>,
> James Harris <jst...@msn.com> wrote:
>>Other readers will remember that I replied *again* noting that my
>>issue with the call to the definition here didn't matter.
> Will you ->EVER<- learn about distribution in usenet? That message did
> not make it to my server until after I had posted this. Your innuendos
> notwithstanding.
James is completely ignorant of USENET. That's the trouble with Google
Groups; it gives idiots access to USENET without requiring them to have
even the minimum knowledge about it that even the most clueless users
used to have.
>>Why did Magidin go through the trouble?
> Read the .sig.
Oops, that requires USENET (or at least email) knowledge. You'd better
explain to him what a .sig is. (It won't do any good, of course, as he
never learns from explanations or definitions.)
W. Dale Hall
James Harris wrote:
> nora...@hotmail.com (Nora Baron) wrote in message news:<36024859.03061...@posting.google.com>...
... stuff deleted ...
>
> As requested in my second reply, you need to give more detail here.
>
> Also as noted repeatedly, proofs don't duel.
>
> If I'm wrong then there's an error in my proof.
>
> Trying to attack one proof by claiming it contradicts with another is
> useless.
>
> You can cast doubt that way, but to prove a "proof" false, you have to
> deal with the actual argument within it.
>
>
>
>>>>Let r' = F12(r) and s' = F12(s). Note that
>>>>both r' and s' are algebraic integers. Thus eqn (1)
>>>>reduces to
>>>>
>>>> r'*a2 + s'*5 = 1.
>>>>
>>>>That is, a2 is also coprime to 5.
>>>>
>>>>Similarly one shows that a3 is coprime to 5.
>>>
>>>Which looks like a good example to show those who wondered how what
>>>I've shown highlights an error in *taught* mathematics.
>>>
>>>It's basically an abuse of Galois Theory.
>>>
>>
>> What's the abuse?
>
>
> My belief has been that posters like yourself have been abusing Galois
> Theory as I don't believe it's wrong. I think you're cheating, which
> is why you wish to go to some other argument rather than deal with the
> one I've presented.
>
> So I see it as an abuse.
>
> Alternatively, Galois Theory IS wrong.
>
> Or you may *believe* I'm wrong, but if I'm wrong there would be a way
> to show it using my work.
>
> That is, if mathematicians are experts, which by definition they are,
> and I'm not a mathematician, but claim to have a proof, it hardly
> makes sense for mathematicians to use an alternative argument, as I'll
> point out the potential for abuse.
>
> It'd seem more rational for mathematicians to save their energy and
> attack my work itself.
>
>
>>>>Therefore if one of a1, a2, or a3 is coprime
>>>>to 5, they all are.
>>>>
>>>>But a1*a2*a3 = 65. Thus at least one of a1,
>>>>a2, and a3 is NOT coprime to 5. Thus a contradiction.
>>>>
>>>>The claim in the Advanced Polynomial Factorization
>>>>paper is therefore wrong.
>>>
>>><deleted>
>>>
>>>Now I've explained more than once that I prove that x has y as a
>>>factor, but x/y is not an algebraic integer, which is a
>>
>> contradiction.
>>
>>>So *obviously* there's a problem somewhere with what mathematicians
>>>are doing if they think they can prove something that's false.
>>>
>>
>>
>>I would say obviously there is a problem somewhere,
>>yes.
>
>
> Then it hardly makes sense for you to spend time and effort not
> addressing my central claim which is that I have a proof.
>
> After all, if you found an error in that proof, you could make a VERY
> short post, and it'd be over instantly--no more discussion needed.
>
> CERTAINLY if I refused to acknowledge the truth, then I might continue
> arguing, but at least other posters would know there was a discovered
> error which they could view themselves to make their own decision.
How about this:
Let z be any root of the polynomial:
p(x) = x^3 - 12 x^2 + 65,
and define these polynomials:
q(x) = 8 x^2 - 76 x -185
r(x) = 8 x^2 - 4 x - 45
s(x) = 4 x^2 - 37 x - 104
Then the following are trivial, if tedious, to show:
5 = q(z)*r(z).
z = r(z)*s(z).
Finally, r(z) has minimal polynomial
MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5
This shows that
1. 5 and z [remember z is *any* root of p(x)] have a mutual
factor, r(z).
2. r(z) is an algebraic integer.
3. The remaining factor of 5 is another algebraic integer q(z).
4. The remaining factor of z is another algebraic integer s(z).
5. r(z) is NOT a unit.
Besides this, the fact that the SAME polynomials q,r,s suffice to
produce the common factor between z and 5 is in accord with what one
expects from Galois theory; the common minimal polynomial for the
(varying) factor of 5 is similarly in accord with what one would
expect from Galois theory.
Since the a's in the factorization of
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
are precisely equal to (-1) times the roots of
p(x) = x^3 - 12 x^2 + 65 = (x - r1)(x - r2)(x - r3),
this result about the roots of p(x) directly shows that JSH is wrong.
>
> Now then are you a mathematician?
>
> If so, then you are a math expert by definition. Then if there is an
> error in my paper you should be able to point it out, and quit wasting
> so much time trying to fight my proof with claims of other proofs of
> your own.
>
The above shows that the so-called "Primary argument" in your paper is
in error. I may spend the time and effort to isolate the first error in
that section, but I will make no promises at this time.
>
>>>Finally, notice the attempt to disprove a proof with *another*
>>>argument, which is claimed to be a proof.
>>>
>>
>> You claimed something was true. I gave an argument
>> to show it wasn't.
>
>
> You fought a conclusion of one argument, which I claim is a proof,
> with another argument which you claim is a proof.
>
> I challenge you to find an error in my argument; you challenge me to
> find an error in yours.
>
> It's a waste of time.
>
> I'm not a mathematician. My argument is simpler.
>
I have shown *directly* that your conclusion is in error. I may look
closer to find the initial error. It is not my responsibility to wipe
your rumpus every time you make poopy, though, so I may just leave you
all stinky.
> So now there are dueling claims. You claim you gave an argument--I
> presume you'd claim it is correct--proving me wrong, I say you're
> wrong.
>
But I've shown *directly* that your conclusion is incorrect. Is that any
sort of thing you should pay attention to?
> Isn't anyone else bothered by the implication that mathematics is a
> mess?
>
Doesn't it bother you to be just plain wrong?
AGAIN????
> What's with all the debate?
>
> Are all mathematicians such losers that they can't handle a claim from
> an admitted non-mathematician, about a paper that's on display 24
> hours a day?
>
Are you such a loser that you need to harangue, taunt, cast aspersions,
and make a nuisance of yourself? All for reasons that wouldn't be
acceptable if you were in fact correct, yet you are apparently NEVER
correct.
> ARE YOU ALL LOSERS??!!!
>
> Yup, I'm frustrated.
>
>
>> Four possibilities:
>>
>> 1. I have an error in my argument. You however did
>> not point out any error.
>
>
> I'm not a mathematician. Why in the hell should I necessarily be able
> to find an error in your argument?
>
> Dueling claims continue.
>
I have shown you to be wrong. Irrefutably. Do the arithmetic.
The above numbers are r(z) are common factors of 5 and z (where z
ranges over the roots of the polynomial p(x)), so they are also
common factors between 5 and your ai's . Each ai has a factor in
common with 5. Theses factor are (1) integral over Q, (2) not units.
>
>> 2. You have an error in your argument. I did not
>> point it out in my post, but it has been pointed
>> out many times previously. Your argument depends
>> on the form that a factorization has in a degenerate,
>> singular case (when m = 0). You argue implicitly
>> that that same form of factorization must hold in the
>> nonsingular case in which you are actually interested.
>> It does not. This has been pointed out innumerable times.
>> Arturo and others have tried many times to get
>> you to understand it, but they have failed. I will
>> probably fail as well.
>
>
> If that's true then you can reference actual statements in the paper.
>
The actual statement is this:
Therefore, with the factorization
65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)
one fo the a'2 is coprime to 5.
> What I see is an unsupported statement, and worse, you have an appeal
> to authority by mentioning Arturo Magidin.
>
What I see is a pattern of obsession with the evil Arturo Magidin, and
an inability of JSH to distinguish between poo and poonola.
> It seems to me that you are unsure of your own statement, and possibly
> wish to convince others without presenting actual evidence.
>
It seems to me that you should go back and find out where you are wrong,
and perhaps learn to verify a proof *before* calling people names.
>
>> 3. Both of us have errors in our arguments.
>>
>> 4. Neither of us has an error in our arguments. There
>> is a fundamental contradiction in mathematics. I
>> think we should just put this idea aside. It
>> is the least likely of the possibilities and it
>> does not obviously lead anywhere.
>
>
> There is no fundamental contradiction in mathematics.
>
> I have presented the most logical possibility before which is that
> mathematicians are lying.
>
Oops, in the presence of evidence, the most logical possibility is the
one that's in agreement with the evidence. To date, all the evidence
we have is your track record, as compared to the track record of the
contributors of sci.math in demolishing your false claims.
These track records suggest something quite different from what you're
suggesting as being most likely. Could it be that you're blinded by
your obsessive need to show up those fargin iceholes?
>
>> It's also possible that the fact I used about automorphisms,
>>which comes from Galois theory, is wrong. That too seems
>>unlikely since Galois theory has been around for over 160
>>years and looked at by many very smart people, and this
>>fact is encountered at the most basic level. I think that
>>too should be put aside.
>
>
> I don't care how long it's "been around", if it's false it's false.
>
> If it's true it's true.
>
Galois theory has been tested FAR more than
(1) Relativity,
(2) Quantum mechanics,
(3) Object math [snrk]
> And that condition is independent of time.
>
> However, I should point out that Galois Theory is a field theory.
>
Again, you betray abysmal ignorance. You must imagine that results
derived from field considerations cannot be used to make statements
about subrings of that field. That's pretty much an indictment of
the level at which you operate, and that level will place a strict
limit on how far you can proceed, except by sheer accident.
> My work is not over fields.
>
> Still either you screwed up that F12 bit above, or Galois Theory is
> wrong.
>
Or, as I have just shown *directly*, you are just plain wrong.
Couldn't get your brain wrapped around that concept, now, could you?
> I'm interested in other replies that address whether you screwed up
> Nora.
>
Nora is correct. JSH is incorrect.
>
>>>It's like trying to fight a proof with a proof, but proofs can't
>>
>> fight
>>
>>>each other.
>>>
>>
>> We both present arguments. You say yours is a proof.
>>I say mine is. We are not both right. In the end
>>we have to try to convince other people and each other.
>
>
> <deleted>
>
> That's stupid.
>
> A proof begins with a truth and proceeds by logical steps to a
> conclusion which then must be true.
>
Oh, but, er, didn't I just show you how wrong you were?
Your conclusion IS WRONG!
> So all that's necessary is to start at the beginning and trace out
> each step.
>
Too bad you screwed up, isn't it?
> That IS possible with my paper, and I've done it which is why I get to
> have fun calling people like you out, and also calling you a jackass.
>
But your arguments lead you to incorrect conclusions. Why on earth would
you continue with such poor results? Doesn't it embarass you to make a
fool of yourself with such amazing regularity?
> You'll have a harder time tracing out your argument, but it might help
> you out to try, if you have the ability Nora.
>
No, standard mathematics is well-enough constructed, and comes with
sufficient infrastructure (by means of an array of lemmas & theorems,
standard examples and techniques) that allow one not only to produce
proofs but also to isolate errors.
> Yup, are you smart enough Nora?
>
It's all about how *smart* someone is, isn't it? The world
isn't always just about how *smart* people are, but sometimes
it's about producing correct results. No matter how *smart*
you think you are, as long as you continue to be wrong while
insisting you're right, as long as you continue to insult those
who *are* right, you're going to be chasing your own tail.
How long do you intend to keep this stupid game going? Lots of us
use this as a form of perverse entertainment; my guess is that you
are also using it in that way, believing somehow that you're fooling
someone. If you truly believe you're correct, you'll do something to
make that manifest, such as getting yourself some education in the
fields where you pretend to have ability.
> Screw convincing people. People are often stupid enough to be
> convinced of just about anything.
>
It's clear you can be convinced of anything you scribble down.
>
> James Harris
Dale