My math errors, healthy skepticism
Dirk Van de moortel
"James Harris" <jst...@msn.com> wrote in message news:3c65f87.03040...@posting.google.com...
> Ok, so the familiar story is that years ago I figured I'd try my hand
> at finding a short proof of Fermat's Last Theorem, using modern
> problem solving techniques.
Ladies And Gentlemen, The Euphemism Of The Millennium:
"modern problem solving techniques"
aka
"countless years of fruitless trial and error on Usenet"
Dirk Vdm
Arturo Magidin
James Harris <jst...@msn.com> wrote:
[.snip.]
>What you may not know is that these claims must rely on the claim of
>ONE mathematician who said that he proved that given a polynomial of
>degree n with integer coefficients for the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>the a's and b's must be algebraic integers.
This is an incorrect presentation of the result in question. The
result in question states that the a's and b's MAY BE CHOSEN to be
algebraic integers. Regardless of whether that result is true or
false, it is not the result you are claiming here.
[.snip.]
You claim this is incorrect. You have said several times that it may
be impossible to find algebraic integers (though you have never
produced an example that shows that it is impossible), and must
instead use what you have termed "objects."
Your current definition of "objects", as it appears in your webpage,
is:
"Objects are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative rings
for which they are members, where no member is a factor of an object
for which it is not a factor in all possible commutative rings with a
unit in which it and that object are members."
As written, there are no objects, because you are not specifying that
the ring structures must be compatible. I think that the definition
you are thinking of is the following:
"Let R be a ring contained in the complex numbers. Then R is an
OBJECT RING if and only if:
(a) If u is a unit in R, with inverse v, then ANY ring contained
in the complex numbers which contains u must also contain v, and ANY
ring contained in the complex numbers which contains v must also
contain v.
(b) If a and b are elements of R, and a divides b in R, then in ANY
ring S contained in the complex numbers, if a and b are in S, then
a divides b in S.
" A complex number r is an OBJECT if and only if there exists an
object ring R such that r is an element of R."
Is this what you had in mind? Or perhaps "complex numbers" replaced
with "algebraic numbers"?
(The reason your definition is no good as written, if interpreted
literally, is because nothing is an object. Say x is an object. We
define a ring consisting only of two elements, 0 and x, with addition
0+0 = 0 = x+x
0+x = x = x+0
and multiplication
0 = 0*x = x*0 = 0*0
x = x*x
Then x is a unit in R with itself as inverse. Now we define a new
ring, S, consisting of two elements, 1 and x, and addition and
multiplication given by:
x+x = 1+1 = 0
1+x = x+1 = 1
x = x*x = x*1 = 1*x
1 = 1*1.
Then x is not a unit in S. This is a "trivial" example in that we
have specified incompatible structures for R and s, which is why I
think you meant compatible structures in your definition; moreover,
you are always insisting that you are working inside the complex
numbers (maybe even the algebraic numbers), so it seems reasonable to
think that you were thinking only in terms of rings contained in the
complex numbers (or the algebraic numbers).
Note also that, read as written, you would be asking that both the
unit u and its inverse be in the ring; but if both u and v are in the
ring, with uv=1, then it is automatic that u is still a unit; so it
seems that you meant to write as I did above, that having one will
automatically imply having the other. Thus, 1/2 is not an "object"
because there are rings which contain its inverse (2), but not 1/2.)
There is also an important problem with your second clause in that it
uses "object" again to define "object", making it a circular
definition. I think what you meant to do is define "object ring", and
then you can define a complex number x as being an object if and only
if there is an object ring which contains x.
Are these accurate? And is your contention that given any polynomial
of degree n>0, with integer coefficients, a factorization into linear
terms may be found where the coefficients are all objects?
======================================================================
"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
Virgil
<3c65f87.03040...@posting.google.com>,
jst...@msn.com (James Harris) wrote:
[Sour grapes snipped]
>
>
> James Harris
a
>at finding a short proof of Fermat's Last Theorem, using modern
>errors on my part, including at least a couple of times where I
>proudly and loudly proclaimed that I'd found the proof only to find
>out later that I was wrong.
Don't forget the part about proudly and loudly proclaiming that the
people who point out your mistakes are liars.
--
"There are difficult logarithums in astrology."
-- Edmond Wollmann <io587.25266$Gj5.14...@typhoon.san.rr.com>
Xcott Craver
>
> What you may not know is that these claims must rely on the claim of
> ONE mathematician who said that he proved that given a polynomial of
> degree n with integer coefficients for the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
> the a's and b's must be algebraic integers.
That's really interesting. Can you provide a polynomial P(x)
for which this isn't the case?
-X
Don Taylor
> Ok, so the familiar story is that years ago I figured I'd try my hand
> at finding a short proof of Fermat's Last Theorem, using modern
> problem solving techniques. Parts of that story are quite a few
> errors on my part, including at least a couple of times where I
> proudly and loudly proclaimed that I'd found the proof only to find
> me actually ended up with me paying some guy over a hundred dollars I
> think it was, because I was confident enough that I made a bet that I
> bothered to pay up on.
I was the one who accepted the bet with Mr. Harris.
I think there should be a bit of explanation included with this. I
have mostly remained silent on this in the years since this was
done.
I spent three or four months, if I recall, dealing with this. THE
point of the bet was that Mr. Harris felt he was not getting a
careful point-by-point checking of his work, and I tended to agree
with him at the time. I agreed to the bet because I honestly felt
it would give him the evaluation that he was seeking and because I
thought it might possibly bring a conclusion to the work and postings
of Mr. Harris.
We agreed to a bet on the condition that a judge be found who would
do a fair and detailed evaluation of the work. I approached several
different people, explained the circumstances, and said that both
their mathematical skill and their ability to present their findings
in a way that would be accepted by Mr. Harris were essential.
Finally I did secure a judge, who I believe still continues to
remain nameless. A version of the proposed proof was specified and
a time limit of 30 days was set to complete the evaluation. This
individual spent a great deal of time studying the claimed proof
at the time, AND putting his evaluation in a form that would be
accepted by Mr. Harris.
I also received one outside contribution of a flaw in the claimed
proof. I reworded the description of that flaw and presented it.
I believe that at that time Mr. Harris stated that this flaw was
sufficient to consider the proof insufficient and the bet settled.
But the judge went on to provide a lengthy explanation, on the
grounds that if he felt it would be possible for him to find a way
to fill a gap then he would not consider this to be enough to declare
the proposed proof invalid. But the judge did find several substantial
gaps in the proof. If I recall correctly, there was some initial
questioning by Mr. Harris as to whether these were sufficient to
declare the proposed proof invalid.
Somewhere about this point in time the thirty day clock had expired
and I believe we were all accepting that there was enough evidence
to agree that the proposed proof, in its original form, was not
sufficient to be considered complete and valid. And then I made a
mistake. I agreed that we would extend the deadline. This started
the whole process again and in hindsight was probably a bad thing.
The judge again spent time going through the proof and spent even
more time creating an even longer evaluation. In this there was
no question that there were gaps in the proposed proof that were
felt could not be filled in any reasonable way, based on what Mr.
Harris had provided.
We agreed the bet was concluded, I made a lengthy posting to sci.math
announching the conclusions and Mr. Harris mailed me a check. The
judge was given somewhat less than half of the amount, but I can't
remember the precise figure. He deserved far more. I have a color
photocopy of the original and a frame to put it in but I never did
get that mounted and put on the wall here.
> Also part of that story has been a lot of harsh and personal criticism
> from people, some of whom have been provably mathematicians by
> checking up on the webpages of their universities, who over YEARS have
> made it their mission to see if they can't hurt my feelings some way
> or another. I've not reacted well to that, and now consider
> mathematicians in general to be of questionable morality based on my
> experiences with quite a few of them over the years.
I do not believe that I have EVER been harsh or critical of Mr.
Harris, with a single exception in one email during the bet with
Mr. Harris when my exasperation exceeded any common sense I had.
When Mr. Harris responded harshly to that I realized my error and
apologized to him for this. Based on the content of email between
Mr. Harris and myself during all this and on postings made by Mr.
Harris during and after this I felt that Mr. Harris believed that
he had been treated fairly and honestly. I hope he continues to
feel this, I would want nothing less than that.
I have said on many occasions that I do not ever wish to be rude
to Mr. Harris. I have said in email to Mr. Harris on a number of
occasions that I would give a substantial sum of money if I could
find a way to have the drive and motivation and dedication that he
has, and if at the same time I could have better interpersonal
skills than I presently have. I am completely sincere in this.
But I also do realize that his drive to pursue this may come at a
high personal cost to him.
I believe that most of what I have said here can be confirmed by
looking at the series of postings archived by google.
Thank you for your time and patience
(the email address is valid, I've been 'dont' on the net since BEFORE spam)
David C. Ullrich
>Ok, so the familiar story is that years ago I figured I'd try my hand
>at finding a short proof of Fermat's Last Theorem, using modern
>problem solving techniques.
Jesus. "Modern problem-solving techniques"? (Given that you
cross-posted this to alt-writing you should note that the hyphen
is required there, btw.)
What's an example of some of the modern problem-solving
techniques you used? Like not bothering to learn any of
the relevant math, then fiddling around with equations?
>Parts of that story are quite a few
>errors on my part, including at least a couple of times where I
>proudly and loudly proclaimed that I'd found the proof only to find
>out later that I was wrong. One of the more morale hurting cases for
>me actually ended up with me paying some guy over a hundred dollars I
>think it was, because I was confident enough that I made a bet that I
>bothered to pay up on.
>
>Also part of that story has been a lot of harsh and personal criticism
>from people, some of whom have been provably mathematicians by
>checking up on the webpages of their universities, who over YEARS have
>made it their mission to see if they can't hurt my feelings some way
>or another. I've not reacted well to that, and now consider
>mathematicians in general to be of questionable morality based on my
>experiences with quite a few of them over the years.
You consider mathematicians to be of questionable morality?
We're all very relieved to hear that - last I checked we were
all children of the Devil. This is a step up.
>You also now know that I DID find a nifty way to count prime numbers
>not in references, and you may have seen how mathematicians work to
>discount that work of mine, and explain why no one should bother
>putting it into math references, so that others, including future
>students like yourself, might see that way of counting primes, without
>maybe hearing about it on newsgroups from my posts!!!
>
>What you may also know as part of the story is my claim to actually
>having succeeded to finding a short proof of Fermat's Last Theorem,
>and that there are mathematicians who say otherwise.
"There are mathematicians who say otherwise" indeed. There are
no mathematicians who say other than otherwise.
>What you may not know is that these claims must rely on the claim of
>ONE mathematician who said that he proved that given a polynomial of
>degree n with integer coefficients for the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>the a's and b's must be algebraic integers.
For heaven's sake, that's not what the result says!
What James knows but readers in the plethora of
groups he posted this to is that he's at various times called
the result in question trivially true and false at other
times. There have been times that he's insisted it was
false, _simultaneously_ _using_ it in The Proof! Honest.
And times he's insisted it's true but that it was proved
hundreds of years ago.
>Now you can see another reason why I talk about that factorization a
>lot, as I'm a person who has found a short proof of Fermat's Last
>Theorem, who is being blocked from proper recognition of my find. And
>part of the story is the false claim of ONE mathematician of a proof.
>
>So you see, I know from my own experience that a mathematician can
>make a proof claim, obviously be wrong, and STILL be accepted by
>mathematicians even when his claim is false.
>
>Your math education is inadequate, and you can prove it to yourselves
>by talking about that mathematician's claim with ANY of your math
>professors.
First we need to get the claim right. The claim is not what you say.
The claim is that there _exists_ such a factorization where the
a's and b's are algebraic integers.
Now that we have the claim straight, why don't you explain to
us why it's false?
>And remember, that claim is THE basis for claims that my proof of
>Fermat's Last Theorem is false.
>
>ALL the people you see spending so much time and energy tracking my
>posting around Usenet to reply against me depend on that claim because
>without it, the objections raised to my proof fall away.
>
>In the end, the question is of your intelligence, and your ability to
>determine the truth.
And lemme guess, if we all fall down and say all hail the great
Harris, Finder of The Proof that will show we have the ability to
find the truth? Hint: You got that exactly backwards.
When you talk about the Truth this way it's disgusting.
Because you're the _only_ person I've _ever_ seen
_anywhere_ on usenet say out loud that he didn't
want to know about it if he was wrong.
For anyone who missed that, a quote from
824hn8$i61$1...@nntp9.atl.mindspring.net
, which you can find at
:
"Where I have p in the proof, use 3. That is, try it out with p=3.
But if it fails, don't tell me. I don't want to know."
An actual quote. You say out loud that you're not interested
in the truth, but lecture us about Truth. Gecgh.
>The test is of your mental abilities as
>students.
>
>What potential do you really have as mathematicians? Do you have the
>right stuff?
>
>
>James Harris
******************
David C. Ullrich
David C. Ullrich
wrote:
It's trivial to give counterexamples to what he states here. That's
because he simply mis-states the actual theorem.
A counterexample to the version he states would be
x^2 - 1 = (2x - 2)(x/2 + 1/2);
1/2 is not an algebraic integer.
> -X
******************
David C. Ullrich
William Kunka
James wrote:
I was confident enough that I made a bet that I bothered to pay up on.
I'm eager to believe that you pay on 'most' lost bets. Bill
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7267c$2q41$1...@agate.berkeley.edu>...
>> [.newsgroups trimmed.]
>>
>> In article <4t499vk0te28rhere...@4ax.com>,
>> David C. Ullrich <ull...@math.okstate.edu> wrote:
>> >On 9 Apr 2003 09:16:44 -0700, jst...@msn.com (James Harris) wrote:
>>
>> [.snip.]
>>
>> >>What you may not know is that these claims must rely on the claim of
>> >>ONE mathematician who said that he proved that given a polynomial of
>> >>degree n with integer coefficients for the factorization
>> >>
>> >> P(x) = (a1 x + b1)...(an x + bn)
>> >>
>> >>the a's and b's must be algebraic integers.
>
>In case those of you in alt.writing or sci.skeptic are worried that
>it's all too technical for you, so far the discussion is not very
>technical at all as I'll explain by defining algebraic integers.
>
>They are simply roots to polynomials with integer coefficients where
>the lead coefficient is 1.
>
>That's it.
>
>For instance, the roots of x^2 + 2x + 1 are algebraic integers, and
>they are themselves -1, as x^2 + 2x + 1 = (x+1)(x+1), while the roots
>of 4x^2 + 4x + 1 are NOT algebraic integers because the lead
>coefficient is 4 and not 1.
>
>> >
>> >For heaven's sake, that's not what the result says!
>> >
>> >What James knows but readers in the plethora of
>> >groups he posted this to is that he's at various times called
>> >the result in question trivially true and false at other
>> >times. There have been times that he's insisted it was
>> >false, _simultaneously_ _using_ it in The Proof! Honest.
>> >And times he's insisted it's true but that it was proved
>> >hundreds of years ago.
>>
>> And the proofs that he was incorrect ->never<- depended on the truth
>> of the (correct) result. In fact, it was only his argument that
>> depended on the truth of the proposition that such a factorization
>> always exists.
>
>What some of you students may not know is that Ullrich is a math
>professor (apparently tenured) at Oklahoma State University (ok, I
>doubt that any of you are impressed by that school, but it's important
>information), while Magidin is apparently still affiliated with the
>math department at Berkeley, a far more impressive school, where he
>got his Ph.d in mathematics.
What does this have to do with anything?
And more to the point, why would you be bringing the question
of mathematical credentials into this? So that someone can point
out that you have an undergraduate degree in physics, hence
must understand the math better than those evil math professor
guys?
>So now that you're introduced to these two players and understand that
>they are REAL LIVE mathematicians, let's get down to what they're
>saying.
>
>In fact, I *did* at one time agree with Magidin, who is the person who
>has made claims of having proven that polynomials with integer
>coefficients could always be factored into algebraic integers in the
>way I've previously described. That is, I was convinced by him, with
>the help of others that given a polynomial P(x) of degree n, there
>must exist the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
Which by the way is a totally different statement from the one
that you made in the post at the top of the thread...
>I became convinced after making some ghastly mistakes with the
>factorization
>
> 5x^2 + 3x + 2 = (a1 x + b1)(a2 x + b2)
>
>where I sought to prove that the a's and b's couldn't be algebraic
>integers, when in fact they are.
>
>So the whole story is that I make mistakes! And that I can be
>convinced when real live mathematicians spend a lot of time and energy
>pushing some particular point, even when they're wrong.
>
>But at least now there's progress as now you know that Magidin is the
>mathematician who made the claim about the polynomial and algebraic
>integers.
>
>Oh, I later *proved* that Magidin's claim is false.
Uh, no you didn't. You're right when you say you made many
erroneous posts claiming to prove this, all of which were wrong.
You never gave a correct proof that the "claim" was false.
(Oh yeah? Exactly _where_ did this proof appear?)
>Moving on...
>
>> The proofs that he was incorrect proceded assuming that his assertion
>> that the factorization existed was correct; if the theorem in question
>> were false, then James' argument would fall by virtue of invoking a
>> false result.
>
>That doesn't make much sense,
No, it doesn't. That's because you were not making much sense
during the period when you were insisting the result was false,
and at the same time _using_ it in The Proof.
>and I don't feel like puzzling through
>it.
>
>The gist of what happens in the short FLT proof is that I get an
>expression which is not very simple, but it's not extraordinarily
>complicated, which can be looked at as if it were a non-monic
>polynomial i.e. has a lead coefficient other than 1 or -1 with integer
>coefficients.
>
>For instance, with p=3, I have
>
> (v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3
>
>which fits the bill. Notice that with respect to x, y and z you can
>look at it as if it were a polynomial, though technically it's not
>because v is the variable!!!
>
>NEAT!!! Yup, the short proof of FLT is fascinating in its
>peculiarities!!!
>
>You see, x, y and z are not variable because they are presumed
>counterexamples to FLT, so in fact they'd be some numbers, where x^3 +
>y^3 = z^3.
>
>The equations produced by my research are quite elegant and
>beautifully balanced.
>
>Too bad real live mathematicians like Ullrich and Magidin spend so
>much time attacking rather than appreciating mathematical elegance.
>
>> His current argument falls because (a) his definition of "object" is
>> nonsense as written, and he has declined to clarify what he really
>> means; and (b) he asserts that a factorization exists where the
>> corresponding coefficients are "objects". Pending a correct
>> definition, this claim cannot be justified. Once a correct definition
>> is given, this claim will depend on the actual properties of
>> "objects".
>
>Here you see Magidin trying to distract you from the real point
I thought the point underlying all this was that you've proved FLT.
The fact that the proof is unintelligible, because it uses an
incoherent definition of "object", is very relevant to that.
Here we see Harris trying to distract you from the main point...
>--the
>truth or falsehood of *his* assertion that given a polynomial P(x) of
>degree n with integer coefficients the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers exists.
He's _published_ a _proof_ of the result on usenet and the
web, and provided links to those proofs in this thread. You
haven't done anything to indicate that the proof is wrong,
and you haven't done anything to indicate that the result
itself is wrong.
>> (If my interpretation of what he means by "objects" is correct, then
>> the factorization with "object" coefficients does not ever exist, so
>> his argument would be wrong at an even earlier stage than the previous
>> incarnation).
>
>That's not relevant to this discussion, and is not the basis that most
>people have heard before for claims that my proof of FLT is false.
>
>But you see, Magidin jumps all over the map as I attempt to nail him
>down.
>
>He's a moving target.
>
>
>> [.rest deleted.]
>>
>> ======================================================================
>> "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
>
>Notice that Magidin takes pains to give his email address a second
>time though in posts where I've considered that he might be
>*requesting* emails or at least giving notice that he'd appreciate
>emails, he has replied that is not the case.
>
>My thinking has long been that he feels a need to boost his own posts
>by making certain that his affiliation with Berkeley is known.
And our feeling is that the things you deduce from the fact that
he includes this email address are totally wacky. He's simply
identifying himself, and offering a way that a reader can
contact him.
>And from what I've determined in the past from visiting the Berkeley
>website, his affiliation is genuine.
C. BOND
[snip]
>
> In case those of you in alt.writing or sci.skeptic are worried that
> it's all too technical for you, so far the discussion is not very
> technical at all as I'll explain by defining algebraic integers.
>
> They are simply roots to polynomials with integer coefficients where
> the lead coefficient is 1.
>
> That's it.
>
> For instance, the roots of x^2 + 2x + 1 are algebraic integers, and
> they are themselves -1, as x^2 + 2x + 1 = (x+1)(x+1), while the roots
> of 4x^2 + 4x + 1 are NOT algebraic integers because the lead
> coefficient is 4 and not 1.
>
I think the above statements and definitions are quite clear, but please
help a non-mathematician to fully grasp it.
The "roots of x^2 + 2x + 1 are algebraic integers" to use your example,
so -1 is therefore an algebraic integer. Also, if I understand you
correctly, the roots of 2x^2 + 3x + 1 are NOT algebraic integers because
the lead coefficient is not 1. But -1 is a root of this equation so it
is *not* an algebraic integer. How can -1 be an algebraic integer and
not an algebraic integer?
--
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
Zachary Turner
> >> mag...@math.berkeley.edu
> >
> >Notice that Magidin takes pains to give his email address a second
> >time though in posts where I've considered that he might be
> >*requesting* emails or at least giving notice that he'd appreciate
> >emails, he has replied that is not the case.
> >
> >My thinking has long been that he feels a need to boost his own posts
> >by making certain that his affiliation with Berkeley is known.
>
> And our feeling is that the things you deduce from the fact that
> he includes this email address are totally wacky. He's simply
> identifying himself, and offering a way that a reader can
> contact him.
>
> >And from what I've determined in the past from visiting the Berkeley
> >website, his affiliation is genuine.
What do you do if you feel threatened by someone's credentials? You attack
them, naturally and try to discredit them.
Xcott Craver
>
> The "roots of x^2 + 2x + 1 are algebraic integers" to use your
> example, so -1 is therefore an algebraic integer. Also, if I
> understand you correctly, the roots of 2x^2 + 3x + 1 are NOT algebraic
> integers because the lead coefficient is not 1.
Mr. Harris's language is not quite exact. The roots of 2x^2 +
etc don't *have* to be algebraic integers, but they can be.
As long as there is some monic polynomial somewhere of integer
coefficients, with x as a root, x is an algebraic integer.
-X
Xcott Craver
>
> That is, I was convinced by him, with the help of others that given a
>
> P(x) = (a1 x + b1)...(an x + bn)
>
James, to further put this in context: are you hoping this
statement is false because your latest proof of FLT (or whatever)
relies on it being false?
> Oh, I later *proved* that Magidin's claim is false.
>
> Moving on...
Meaning that you have found a counterexample?
-X
C. BOND
I think you must mean that Mr. Harris's statement was incorrect. There
was nothing inexact or ambiguous about the language in his post. He
stated quite clearly that the roots of his example equation were NOT
algebraic integers, because the leading coefficient was not one. I'm
beginning to see the problem with his *proofs".
William Hale
<cb...@ix.netcom.com> wrote:
> Xcott Craver wrote:
> >
> > C. BOND wrote:
> > >
> > > The "roots of x^2 + 2x + 1 are algebraic integers" to use your
> > > example, so -1 is therefore an algebraic integer. Also, if I
> > > understand you correctly, the roots of 2x^2 + 3x + 1 are NOT algebraic
> > > integers because the lead coefficient is not 1.
> >
> > Mr. Harris's language is not quite exact. The roots of 2x^2 +
> > etc don't *have* to be algebraic integers, but they can be.
> > As long as there is some monic polynomial somewhere of integer
> > coefficients, with x as a root, x is an algebraic integer.
> >
>
> I think you must mean that Mr. Harris's statement was incorrect.
I would prefer to say that his logic was incorrect. Namely, he
did not correctly form the negation of being an algebraic integer.
Mr. Harris correctly gives the definition for being algebraic
integers to be "roots to polynomials with integer coefficients where
the lead coefficient is 1".
He applies this definition correctly to show that -1 is an algebraic
integer by exhibiting the polynomial x^2 + 2x + 1, for which -1 is
a root.
He then says that "the roots of 4x^2 + 4x + 1 are NOT algebraic integers
because the lead coefficient is 4 and not 1". The "because" here is
not a sufficient reason to conclude that the roots of 4x^2 + 4x + 1
are not algebraic integers. He is not forming the negation correctly.
More formally, the definition of a number u being an algebraic
integer can be stated as follows: For some polynomial f with
integer coefficients where the lead coefficient is 1, f(u) = 0.
The negation of that statement, to get that the number u is not
an algebraic integer, is formed by negating f(u) = 0 and also
changing "for some" to "for all", yielding the negated statement
to be "For all polynomial f with integer coefficients where the
lead coefficient is 1, f(u) does not equal 0."
This is standard result in symbolic logic. To negate a statement
involving the quantifiers "for some" and "for all", you need
to change "for some" to "for all" and "for all" to "for some"
and then negate the proposition itself.
One probably first runs into this when taking calculus and limits.
The definition that lim (x->3) f(x) = 5 says that (For all e > 0)
(For some d > 0)( |x - 3| < d implies | f(x) - 5 | < e).
The negation of that is: (For some e > 0)(For all d > 0)
(it is false that |x - 3| < d implies | f(x) - 5 | < e).
There is a similar error when taking the negation of statements
involving "and" and "or". Besides negating the components statements
(which everyone realizes), you must also change "and" to "or" and
"or" to "and" (which some fail to do).
As an aside, to negate an implication like "p implies q", I first
change the implication to "not p, or q" and then negate that to
get "p and not q" as being the negation of "p implies q".
> There was nothing inexact or ambiguous about the language in his post.
> He stated quite clearly that the roots of his example equation were NOT
> algebraic integers, because the leading coefficient was not one. I'm
> beginning to see the problem with his *proofs".
>
> --
> 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
-- Bill Hale
Zachary Turner
"Zachary Turner" <__NOzturner...@hotmail.com> wrote in message
news:3e9594f1$1$25087$4c41...@reader1.ash.ops.us.uu.net...
Note that I was not saying 'you' as in "Well, what does David Ullrich do
when he feels threatened by someone's credientials? etc etc" It was the
indefinite "you", as in "what does one do when one feels etc etc"
Randy Poe
> James Harris wrote:
> [snip]
> >
> > In case those of you in alt.writing or sci.skeptic are worried that
> > it's all too technical for you, so far the discussion is not very
> > technical at all as I'll explain by defining algebraic integers.
> >
> > They are simply roots to polynomials with integer coefficients where
> > the lead coefficient is 1.
> >
> > That's it.
> >
> > For instance, the roots of x^2 + 2x + 1 are algebraic integers, and
> > they are themselves -1, as x^2 + 2x + 1 = (x+1)(x+1), while the roots
> > of 4x^2 + 4x + 1 are NOT algebraic integers because the lead
> > coefficient is 4 and not 1.
> >
> James,
>
> I think the above statements and definitions are quite clear, but please
> help a non-mathematician to fully grasp it.
>
> The "roots of x^2 + 2x + 1 are algebraic integers" to use your example,
> so -1 is therefore an algebraic integer. Also, if I understand you
> correctly, the roots of 2x^2 + 3x + 1 are NOT algebraic integers because
> the lead coefficient is not 1. But -1 is a root of this equation so it
> is *not* an algebraic integer. How can -1 be an algebraic integer and
> not an algebraic integer?
Because James still doesn't quite understand the concepts.
See Arturo's correct explanation. The roots of a primitive IRREDUCIBLE
polynomial are not algebraic integers. Any algebraic integer is
trivially the root of many polynomials with non-unity leading
coefficient. Proof: Let a be an algebraic integer, and let P(x) be
a polynomial with integer coefficients for which a is a root.
Consider the polynomial P(x)*(3x+2).
Your polynomial factors as 2x^2 + 3x + 1 = (2x+1)(x+1), so it's
reducible and proves nothing about the roots.
On the other hand (x+1) is primitive and irreducible, and -1 is
a root, so it is an algebraic integer.
2x+1 is primitive, irreducible, and has leading coefficient 2,
so its root, -1/2, is not an algebraic integer.
- Randy
Roy
Incidentally, all integers are algebraic integers, since any integer 'r'
is a root of
x^2 - (r+1)x + r
Roy
David C. Ullrich
Um, there's a much easier way to show that the integer r is an
algebraic integer...
>Roy
******************
David C. Ullrich
Arturo Magidin
[.My apologies to sci.skeptic and alt.writing; note followups.]
In article <585ab5d8.03041...@posting.google.com>,
Randy Poe <rpo...@yahoo.com> wrote:
>"C. BOND" <cb...@ix.netcom.com> wrote in message news:<3E94C8D4...@ix.netcom.com>...
[.snip.]
>See Arturo's correct explanation.
I only posted it to sci.math and alt.math.undergrad, where this thread
belongs.
>The roots of a primitive IRREDUCIBLE
>polynomial are not algebraic integers.
Careful, Randy. primitive, irreducible, polynomials with integer
coefficients and leading coefficient different from 1 and -1.
John R Ramsden
> hand at finding a short proof of Fermat's Last Theorem, using
> modern problem solving techniques.
>
James, I don't know if you heard, but Saddam Hussein has just fired
his information minister Mohammed Saeed al-Sahhaf for not lying hard
enough. (Shame really - I was looking forward to his victory speech,
flanked by US marines ".. and let that be a lesson to the infidels!".)
But anyway there's a new guy now, called Ali al-Jameshish Harrishi,
and boy can he tell some whoppers. So I'd be careful from now on,
because as you must know all too well, people on sci.math can be
very cruel sometimes, and if you start fibbing and exaggerating
again it won't be long before people cotton on to that name...
Cheers
---------------------------------------------------------------------------
John R Ramsden (j...@adslate.com)
---------------------------------------------------------------------------
"I want you to look at the man on either side of you. In six months'
time only one of you three will be left. But if you are the lucky man,
I promise you this - you will be two ranks higher."
Air Marshall Sir Arthur "Bomber" Harris (1942).
Andrzej Kolowski
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7267c$2q41$1...@agate.berkeley.edu>...
> > [.newsgroups trimmed.]
> >
> > In article <4t499vk0te28rhere...@4ax.com>,
> > David C. Ullrich <ull...@math.okstate.edu> wrote:
> >
> > >>What you may not know is that these claims must rely on the claim of
> > >>ONE mathematician who said that he proved that given a polynomial of
> > >>degree n with integer coefficients for the factorization
> > >>
> > >> P(x) = (a1 x + b1)...(an x + bn)
> > >>
> > >>the a's and b's must be algebraic integers.
>
> it's all too technical for you, so far the discussion is not very
> technical at all as I'll explain by defining algebraic integers.
>
> They are simply roots to polynomials with integer coefficients where
> the lead coefficient is 1.
>
> That's it.
>
> For instance, the roots of x^2 + 2x + 1 are algebraic integers, and
> they are themselves -1, as x^2 + 2x + 1 = (x+1)(x+1), while the roots
> coefficient is 4 and not 1.
>
> > >
> > >For heaven's sake, that's not what the result says!
> > >
> > >What James knows but readers in the plethora of
> > >groups he posted this to is that he's at various times called
> > >the result in question trivially true and false at other
> > >times. There have been times that he's insisted it was
> > >false, _simultaneously_ _using_ it in The Proof! Honest.
> > >And times he's insisted it's true but that it was proved
> > >hundreds of years ago.
> >
> > of the (correct) result. In fact, it was only his argument that
> > depended on the truth of the proposition that such a factorization
> > always exists.
>
> What some of you students may not know is that Ullrich is a math
> professor (apparently tenured) at Oklahoma State University (ok, I
> doubt that any of you are impressed by that school, but it's important
> information), while Magidin is apparently still affiliated with the
> math department at Berkeley, a far more impressive school, where he
> got his Ph.d in mathematics.
>
I think the affiliation now extends only to having an e-mail account.
> So now that you're introduced to these two players and understand that
> they are REAL LIVE mathematicians, let's get down to what they're
> saying.
>
> In fact, I *did* at one time agree with Magidin, who is the person who
> has made claims of having proven that polynomials with integer
> coefficients could always be factored into algebraic integers in the
> way I've previously described. That is, I was convinced by him, with
> the help of others that given a polynomial P(x) of degree n, there
> must exist the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
> where the a's and b's are algebraic integers.
>
> I became convinced after making some ghastly mistakes with the
> factorization
>
> 5x^2 + 3x + 2 = (a1 x + b1)(a2 x + b2)
>
> where I sought to prove that the a's and b's couldn't be algebraic
> integers, when in fact they are.
>
Or at least they could be.
> So the whole story is that I make mistakes! And that I can be
> convinced when real live mathematicians spend a lot of time and energy
> pushing some particular point, even when they're wrong.
>
Interesting statement.
> But at least now there's progress as now you know that Magidin is the
> mathematician who made the claim about the polynomial and algebraic
> integers.
>
> Oh, I later *proved* that Magidin's claim is false.
>
Reference???
I have never seen any such proof from you. Someone else has claimed
that Magidin's theorem is an old result. I have worked out a proof
also. All proofs I have seen depend ultimately on an old and deep
result of Dedekind. You at various times have claimed to have a
proof or claimed that it was obvious and trivial. Your argument
for FLT actually requires Magidin's theorem. If you have a disproof,
your own argument is in trouble.
> Moving on...
>
> > The proofs that he was incorrect proceded assuming that his assertion
> > that the factorization existed was correct; if the theorem in question
> > were false, then James' argument would fall by virtue of invoking a
> > false result.
>
> That doesn't make much sense, and I don't feel like puzzling through
> it.
>
> The gist of what happens in the short FLT proof is that I get an
> expression which is not very simple, but it's not extraordinarily
> complicated, which can be looked at as if it were a non-monic
> polynomial i.e. has a lead coefficient other than 1 or -1 with integer
> coefficients.
>
> For instance, with p=3, I have
>
> (v^3+1)z^6 - 3v x^2 y^2 z^2 - 2 x^3 y^3
>
> which fits the bill. Notice that with respect to x, y and z you can
> look at it as if it were a polynomial, though technically it's not
> because v is the variable!!!
>
Oh yes. The famous old switch-the-polynomial-variable-but-claim-the
constant-term-is-still-the-same argument.
> NEAT!!! Yup, the short proof of FLT is fascinating in its
> peculiarities!!!
>
> You see, x, y and z are not variable because they are presumed
> counterexamples to FLT, so in fact they'd be some numbers, where x^3 +
> y^3 = z^3.
>
> The equations produced by my research are quite elegant and
> beautifully balanced.
>
All equations by definition are balanced. Things that are not
balanced are cleverly called "inequalities". What does
"beautifully balanced" mean?
> Too bad real live mathematicians like Ullrich and Magidin spend so
> much time attacking rather than appreciating mathematical elegance.
>
> > His current argument falls because (a) his definition of "object" is
> > nonsense as written, and he has declined to clarify what he really
> > means; and (b) he asserts that a factorization exists where the
> > corresponding coefficients are "objects". Pending a correct
> > definition, this claim cannot be justified. Once a correct definition
> > is given, this claim will depend on the actual properties of
> > "objects".
>
> Here you see Magidin trying to distract you from the real point--the
> truth or falsehood of *his* assertion that given a polynomial P(x) of
> degree n with integer coefficients the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
> where the a's and b's are algebraic integers exists.
>
I quote your definition of "object", from your web site:
"Objects are members of commutative rings where any
unit and its multiplicative inverse are units in all
possible commutative rings for which they are members,
where no member is a factor of an object for which it
is not a factor in all possible commutative rings with
a unit in which it and that object are members."
I defy you to make sense of this "definition":
1. Note that it is circular. It is all one long sentence.
The word "object" is thus defined in terms of "object".
2. Identify specifically at least one number which is an
object but not an integer. And prove it.
3. Prove or disprove: all algebraic integers are objects.
4. Prove or disprove: all objects are algebraic integers.
5. Prove or disprove: sqrt(2) is an object.
>
> > (If my interpretation of what he means by "objects" is correct, then
> > the factorization with "object" coefficients does not ever exist, so
> > his argument would be wrong at an even earlier stage than the previous
> > incarnation).
>
> That's not relevant to this discussion, and is not the basis that most
> people have heard before for claims that my proof of FLT is false.
>
The "proof" on your web page says on the third line:
"All that follows is in an *object* ring."
Therefore any problems, inconsistencies, or gaps in your definition
of objects are legitimate reasons to dismiss your "proof".
It is thus very definitely relevant.
> But you see, Magidin jumps all over the map as I attempt to nail him
> down.
>
> He's a moving target.
>
[non-math deleted].
Andrzej
C. BOND
>
> "C. BOND" <cb...@ix.netcom.com> wrote in message news:<3E94C8D4...@ix.netcom.com>...
> > [snip]
> > >
> > > In case those of you in alt.writing or sci.skeptic are worried that
> > > it's all too technical for you, so far the discussion is not very
> > > technical at all as I'll explain by defining algebraic integers.
> > >
> > > They are simply roots to polynomials with integer coefficients where
> > > the lead coefficient is 1.
> > >
> > > That's it.
> > >
> > > For instance, the roots of x^2 + 2x + 1 are algebraic integers, and
> > > they are themselves -1, as x^2 + 2x + 1 = (x+1)(x+1), while the roots
> > > of 4x^2 + 4x + 1 are NOT algebraic integers because the lead
> > > coefficient is 4 and not 1.
> > >
> > James,
> >
> > I think the above statements and definitions are quite clear, but please
> > help a non-mathematician to fully grasp it.
> >
> > The "roots of x^2 + 2x + 1 are algebraic integers" to use your example,
> > so -1 is therefore an algebraic integer. Also, if I understand you
> > correctly, the roots of 2x^2 + 3x + 1 are NOT algebraic integers because
> > the lead coefficient is not 1. But -1 is a root of this equation so it
> > is *not* an algebraic integer. How can -1 be an algebraic integer and
> > not an algebraic integer?
>
> monic polynomial with integer coefficients.
>
> With 4x^2 + 4x + 1 you do not have such a polynomial. But with 2x^2 +
> 3x + 1, you have 2x^2 + 3x + 1 = (2x + 1)(x+1), where you'll notice
> that x+1 is a monic polynomial with integer coefficients.
>
> What I did was give the definition and allow you to use your mental
> faculties to do the rest.
And what I did was provide a counter-example to your statement, which
was (and I quote): "..the roots of 4x^2 + 4x + 1 are NOT algebraic
integers because the lead coefficient is 4 and not 1."
> It's not complicated.
I, too, think it is not complicated. However, your prose needs work if
you expect to appeal to rational minds.
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b742rv$abg$1...@agate.berkeley.edu>...
>> [.newsgroups trimmed again; this has nothing to with sci.skeptic or
>> alt.writing.]
>
>Actually it does as they're included as an audience in a rather simple
>and direct test of honesty among mathematicians.
>
>> In article <3c65f87.03041...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [.snip.]
>>
>> >In fact, I *did* at one time agree with Magidin, who is the person who
>> >has made claims of having proven that polynomials with integer
>> >coefficients could always be factored into algebraic integers in the
>> >way I've previously described. That is, I was convinced by him, with
>> >the help of others that given a polynomial P(x) of degree n, there
>> >must exist the factorization
>> >
>> > P(x) = (a1 x + b1)...(an x + bn)
>> >
>> >where the a's and b's are algebraic integers.
>> >
>> >I became convinced after making some ghastly mistakes with the
>> >factorization
>> >
>> > 5x^2 + 3x + 2 = (a1 x + b1)(a2 x + b2)
>> >
>> >where I sought to prove that the a's and b's couldn't be algebraic
>> >integers, when in fact they are.
>> >
>> >So the whole story is that I make mistakes! And that I can be
>> >convinced when real live mathematicians spend a lot of time and energy
>> >pushing some particular point, even when they're wrong.
>> >
>> >But at least now there's progress as now you know that Magidin is the
>> >mathematician who made the claim about the polynomial and algebraic
>> >integers.
>> >
>> >Oh, I later *proved* that Magidin's claim is false.
>>
>> Really? I must have missed the counterexample. All you did was assert
>> it was wrong.
>
>So you still claim that your assertion of proof is correct?
What's the _error_ in the proof?
>For those who have forgotten Magidin claims to have *proven* that
>given a polynomial P(x) of degree n, there exists the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
>
>It might sound really complicated but it's not. It's like if you have
>
> 23x^34 + 45x^3 + 11x^2 + 137 = (a1 x + b1)...(a34 x + b34)
>
>Magidin claims that he has proven that you can find a's and b's that
>themselves MUST be roots of polynomials with a lead coefficient of 1,
>which would make them algebraic integers.
>
>> Perhaps you can post the counterexample and the proof that it is a
>> counterexample?
>
>I'm not interested in your efforts to sidetrack things.
Classic Harris. You say someone's wrong about something,
and then when they ask you to back up your statement they're
trying to sidetrack things. Remarkable.
[...]
******************
David C. Ullrich
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b77k54$1dur$1...@agate.berkeley.edu>...
>> In article <3c65f87.03040...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> My continuing apologies to the denizens of sci.skeptic and
>> alt.writing. Newsgroups trimmed. Note follow-ups.
>
>You can run but you can't hide Magidin.
>
>The discussion is STILL one that people on alt.writing and sci.skeptic
>can follow.
>
>And I need people from groups with healthy skepticism as I have seen
>up close how easily you manipulate math groups, even when I catch you
>directly in mathematical falsehoods, as they seem to only care that
>you're a mathematician, and I'm not, not if you actually tell them the
>truth.
And that<guffaw> is why you need to crosspost this stuff to
alt.writing and sci.skeptic? You've noticed that in _those_
groups people don't have<giggle> those sorts of biases,
and hence they<huh?> take your work seriously?
Got it.
******************
David C. Ullrich
Xcott Craver
>
> You can run but you can't hide Magidin.
Who's running? We're all waiting, quite patiently in fact,
for you to name a single polynomial that can not be factored
the way Dr. Magidin describes.
You said you proved him wrong. Where's that proof? Do you have
a counterexample? What is it? Why stall for so long?
-X
Jim Burns
>
[...]
> You can run but you can't hide Magidin.
>
> can follow.
>
> And I need people from groups with healthy skepticism as I have seen
> up close how easily you manipulate math groups, even when I catch you
> directly in mathematical falsehoods, as they seem to only care that
> you're a mathematician, and I'm not, not if you actually tell them the
> truth.
Mr. Harris --
You inadvertently posted the following to sci.math only.
HTH,
Jim Burns
James Harris wrote in "Web complete" 12 Apr 2003 07:46:33 -0700
<3c65f87.03041...@posting.google.com>:
:
: Mathematicians made claims now they will be held to them.
:
: Accountability is not just a word.
:
: Putin things together at this point is required.
:
: Secondaries take permanent lead. Tertiaries on stand-by.
:
: It's been fun, but it's time for me to retire.
:
: I trust your judgement. Be safe, be sound, and remember that with
: absolute power comes absolute responsibility.
:
: James Harris
Jim Burns
Brian Quincy Hutchings
in several months, and maybe I should,
just to make sure.
I must have missed, lo these many years
from when I first was briefly amuzed
by this ne'er-ending story, the point
at which monsieu Magadin typed it out in plain ASCII,
"James, you punk-ass accountant/programmer/bathroom-attendant,
you don't know **** about Galois theory!"
for myself,
I'd have to defend le pauvre petit
from his would-be enemies beyond the grave
(and they hardly ever visit him
in the cemetary .-)
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03041...@posting.google.com>...
> You can run but you can't hide Magidin.
>
> The discussion is STILL one that people on alt.writing and sci.skeptic
> can follow.
>
> And I need people from groups with healthy skepticism as I have seen
> up close how easily you manipulate math groups, even when I catch you
> directly in mathematical falsehoods, as they seem to only care that
> you're a mathematician, and I'm not, not if you actually tell them the
> truth.
> For other readers don't worry about Galois Theory as the main point is
> that Magidin is trying to back off from the basis that he and others
> used to attack my approach to proving Fermat's Last Theorem.
> I repeat, this is not a movie, you've been pulled into a real life
> drama where mathematicians don't have a script and have to question
> their ability to successfully lie to the world, versus what it might
> mean to actually tell the truth, especially the whole truth.
[NB: an "we told you, so," in my sig
of about 2 years' standing (tlansrated
into French, recently, by me,
however badly.]
http://www.eia.doe.gov/emeu/cabs/canada.html
--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...
La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:
(FOSSILISATION [McCainanites?] (TM/sic))/
BORE/GUSH/NADIR "@" http://www.tarpley.net/aobook.htm.
Http://www.tarpley.net/bushb.htm (content partiale, below):
17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81
23 -- Le FIN d'HISTOIRE
24 -- L'ORDEUR du MONDE NOUVEAU
25 -- THYROID STORK !?!
Wayne Brown
> James Harris wrote:
>>
> [...]
>> You can run but you can't hide Magidin.
>>
>> The discussion is STILL one that people on alt.writing and sci.skeptic
>> can follow.
>>
>> And I need people from groups with healthy skepticism as I have seen
>> up close how easily you manipulate math groups, even when I catch you
>> directly in mathematical falsehoods, as they seem to only care that
>> you're a mathematician, and I'm not, not if you actually tell them the
>> truth.
> Mr. Harris --
> You inadvertently posted the following to sci.math only.
> HTH,
> Jim Burns
Here's another article Mr. Harris neglected to crosspost to alt.writing
and sci.skeptic. I'm reproducing it here, with expletives deleted,
for his convenience:
>>From: jst...@msn.com (James Harris)
>>Newsgroups: sci.math,sci.physics
>>Subject: JSH: Ok, I'm a loser
>>Date: 22 Dec 2002 20:34:06 -0800
>>
>>It's finally settled in that I'm just some pathetic loser. If I
>>weren't so pathetic I'd just go away gracefully, but I'll send one
>>more post, or who am I kidding, my patheticness is so great that I'll
>>probably post yet again.
>>
>>I'm disgusting. I'm just a pile of ****. I should just die like so
>>many of you have said. I hate myself. I despise this life. I'm
>>nothing but a sick joke to be made fun of by those of you who have
>>real educations. People who actually know something, when I know
>>nothing. I'm just nothing.
>>
>>If I hadn't been such a disgusting human being I'd have come to this
>>realization years ago instead of wasting your time.
>>
>>My life is nothing. I know nothing. I'm worth nothing. I'm just
>>****.
>>
>>Please forgive me. All your attacks were justified.
>>
>>
>>James Harris
I refuse to use the sort of language Mr. Harris uses, even in a quote.
But anyone who wishes to see the uncensored text of his article may do
so at:
http://groups.google.com/groups?selm=3c65f87.0212222034.d5959fd%40posting.google.com
--
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"
Arturo Magidin
Note follow-ups. My continuing apologies to sci.skeptic and alt.writing.
In article <3c65f87.03041...@posting.google.com>,
James Harris <jst...@msn.com> wrote:
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b742rv$abg$1...@agate.berkeley.edu>...
>> [.newsgroups trimmed again; this has nothing to with sci.skeptic or
>> alt.writing.]
>
>Actually it does as they're included as an audience in a rather simple
>and direct test of honesty among mathematicians.
>
>> In article <3c65f87.03041...@posting.google.com>,
Leading question. I made no "assertion of proof." I provided a proof,
both in the newsgroup, and in a manuscript. Here it is again:
http://www.math.umt.edu/~magidin/preprints/gauss.ps (postscript)
http://www.math.umt.edu/~magidin/preprints/gauss.pdf (PDF)
When people have pointed out your arguments are wrong, you always
demand that they point out exactly which line in the argument is
wrong. I formally invite you to adhere to your own standards and point
out the exact line of the proofs which is incorrect.
>For those who have forgotten Magidin claims to have *proven* that
>given a polynomial P(x) of degree n, there exists the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers.
>
>It might sound really complicated but it's not. It's like if you have
>
> 23x^34 + 45x^3 + 11x^2 + 137 = (a1 x + b1)...(a34 x + b34)
>
>Magidin claims that he has proven that you can find a's and b's that
>themselves MUST be roots of polynomials with a lead coefficient of 1,
>which would make them algebraic integers.
Actually, I have proven that the a's and b's ->exist<-. The argument
provides an algorithm of sorts, but which can be very hard to actually
follow. Thus, I would say that "claims [...] you can find" is a
misrepresentation of both the theorem and its proof. That sounds like
I have offered an algorithm to find them; I've done no such
thing. I've provided a proof of existence.
>> Perhaps you can post the counterexample and the proof that it is a
>> counterexample?
>
>I'm not interested in your efforts to sidetrack things.
Let's see: you claim I'm wrong. I ask you to show your work. You claim
this is sidetracking things.
Is this an accurate summary of my attempt at "sidetracking"?
>> Failing that, perhaps instead of pontificating you can answer the
>> specific questions I asked elsewhere in this thread about your
>> definition of "object"?
>
><deleted>
>
>The point here is your claim,
I thought the point was your supposed proof. This result's relevance
to your argument is only that it is necessary for what I have called
elsewhere your second branch of the current argument. If the result is
false, then your second branch is dead at the outset. If the result is
right... then nothing. No proof that your second branch was incorrect
depended on this result.
[.rest deleted.]
Arturo Magidin
>> James Harris <jst...@msn.com> wrote:
>>
>> My continuing apologies to the denizens of sci.skeptic and
>> alt.writing. Newsgroups trimmed. Note follow-ups.
>
>
>The discussion is STILL one that people on alt.writing and sci.skeptic
>can follow.
>
>And I need people from groups with healthy skepticism as I have seen
>up close how easily you manipulate math groups, even when I catch you
>directly in mathematical falsehoods, as they seem to only care that
>you're a mathematician, and I'm not, not if you actually tell them the
>truth.
Fair enough. I find it surprising, then, than you removed all the
content from my post before posting your vitriolic reply. I shall
attempt to rememdy this problem now. A link to the original post can
be found on:
http://groups.google.com/groups?selm=b77k54%241dur%241%40agate.berkeley.edu
I include here the text of the reply removed by James. My apologies to
sci.skeptic and alt.writing.
--- BEGIN INSERT ---
The current attempt seems to have three definite stages. I am linking
all three together, and separate from any prior attempts, because they
are characterized by the initial phase of the argument.
STAGE 1. Use congruences to derive, from a counterexample x^p+y^p=z^p,
a polynomial expression in x,y,z, and an integer v, which is
defined in a specific way depending on x,y,z. This stage has
remained mostly unchanged.
STAGE 2. After obtaining the polynomial, factor it in a specific
way. There have been three distinct branches for the
argument here; I will address them separately below.
STAGE 3. Analyze the coefficients of that factorization.
Stage 3 of course is meant to culminate in a contradiction, thereby
proving that no counterexample can exist.
This family of arguments seems to have begun sometime in 2001. They
derive almost directly from a previous attempt, that was ongoing at
the beginning of 2001 and which I will not address. See for example:
[1] Mon 05 Feb 2001 "FLT Proof: Simple enough for all",
author jst...@my-deja.com
http://groups.google.com/groups?selm=95nb7m%24no0%241%40nnrp1.deja.com
(Note: for some reason, I do not know how to cut and paste URLs, so
I'm copying them by hand; I'm including date, title, and author so
that if I have miscopied the URL, you can try to find it yourself.)
Phase I is already clearly present by May 2001, for example:
[2] 28 May 2001 "Re: Details on my FLT proof", author jst...@msn.com
http://groups.google.com/groups?selm=3c65f87.0105281758.4247c3ca%40posting.google.com
Since the most analyzed case is p=3, I will restrict my comments to
that case. I do note, however, that James most of the time presents
his arguments in full generality for arbitrary odd prime p.
For the case of p=3, phase I can be summarized as follows:
Suppose x^3+y^3=z^3 is a Fermat counterexample. Let f be a prime,
(which may be assumed to be greater than 3), which is a common factor
of x and z-y; write x=f^j*u, where u is coprime to f. For each
positive integer n, let
v=-1+mf^{2j}, where m is a positive integer, chosen so that
x^2+y^2 + vz^2 = 0 (mod f^{n+2j}).
Then from x^3+y^3=z^3, it follows that:
(v^3+1)z^3 - 3v(z^2x^2y^2) -2x^3y^3 = 0 (mod f^{n+2j}).
At this point, I believe that everyone more or less agrees that this
derivation can be made. At times, there were some arguments on whether
the specific steps were correct or not as presented (i.e., whether the
justification given was correct), but I believe we all agree that one
may indeed obtain that expression. This is the end of Phase I, upon
obtaining the polynomial
f(w) = (v^3+1)w^3 - 3v*w - 2.
At this point we enter Phase II.
The first branch of Phase II was the first presented. By considering
the polynomial f(w) given above, we factor it as:
f(w) = (sqrt(v+1)w + b1)(sqrt(v+1)w + b2)( (v^2-v+1)w + b3)
with b1, b2, b3 complex numbers. This occurs, for example, in
[3] 30 Sep 2001 "FLT Proof Exposition for p=3, Part 2 revised", from
jst...@msn.com
http://groups.google.com/groups?selm=3c65f87.0109301551.79fg7a58%posting.google.com
James derived a number of congruences from this. Among others, I
challenged some of the justifications and some of the conclusions he
provided. Those conclusions were in some instances correct, even when
the justification was lacking. The approach was formalized by Rick
Decker, and his exposition of the conclusions can be found at
http://big-oh.cs.hamilton.edu/~rdecker/FLT/
Much of the discussion regarding this branch focused on two items: the
justification for the steps James was taking, and whether or not a
contradiction was truly reached at the end.
However, James has abandoned this branch, so I will not discuss it
further.
The second branch of the argument occurs when James instead chooses to
factor
f(w) = (v^3+1)w^3 - 3v*w - 2 = (a1*w+b1)(a2*w+b2)(a3*w+b3),
where a1, a2, a3, b1, b2, b3 are chosen to be algebraic integers. The
question of whether this could be done in general had been asked by
Dave Pritchard, in part due to some side-discussions occurring in
the threads on James's argument. The question occurs on:
[4] 11 Jul 2001, "Two Algebra Questions" from dav...@hotmail.com
(Dave Pritchard),
http://groups.google.com/groups?selm=c7de0075.0107112016.6f7b7437%40posting.google.com
After first stating that I thought it was not always possible, and
giving an incorrect example, I posted an argument showing it is always
possible.
[5] 15 Jul 2001, "Re: Two Algebra Questions (answer?)",
mag...@math.berkeley.edu (Arturo Magidin)
http://groups.google.com/groups?selm=9iqqkh%241v9v%241%40agate.berkeley.edu
I and David McKinnon have since written this up, submitted it to The
Monthly, received a referee report, and heavily rewritten it; the
current version is available at
http://www.math.umt.edu/~magidin/preprints/gauss.ps (postscript)
http://www.math.umt.edu/~magidin/preprints/gauss.pdf (PDF)
Let me write it explicitly for reference:
THEOREM 1. Let f(x) be a polynomial with integer coefficients. Then
there exists a factorization of f(x) into (not necessarily monic)
linear terms, each with algebraic integer coefficients.
The rewriting has mostly concerned adding an historical survey and
loosening the discussion; the argument is still, essentially, the one
in [5].
It is unclear to me when James abandoned the
a1=a2=sqrt(v+1), a3=v^2-v+1 factorization. He seems to be trying to
->deduce<- that this is "the" factorization that must be obtained in
[6] 18 Aug 2001 "How my current FLT argument works" from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.010818609.4b113efb%40posting.google.edu
but he once again simply sets a1=sqrt(v+1)=a2 on
[7] 4 Oct 2001 "FLT Proof Exposition for p=3: Part 2 revised
(Repost)", from jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0110041541.54e279d7%40posting.google.com
He is still using this factorization in November; the issues that were
being argued over on branch one are the subject of
[8] 14 Dec 2001 "Elementary FLT Proff: Credibility of objectors" from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0112141022.28a92bd%40posting.google.com
so it seems reasonable to conclude that on December 2001, James was
still pursuing branch 1 of the argument. I have found further posts in
January and February of 2002. However, on
[9] 12 Apr 2002 "JSH: FLT Proof status and why I'm angry." from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0204121655.385acae5%40posting.google.com
James complaints that he was not made aware of algebraic integers
before. So I'm guessing that sometime around March of 2002, James
switched over to branch 2.
In branch 2, the main disagreement lies on the divisibility properties
of a1, a2, a3. James asserts that exactly two of them must be
divisible (in the ring of all algebraic integers) by f, and the third
must be coprime to f.
Bengt and Rupert offered an explicit calculation for the polynomial
f(w) = 65w^3 - 12w - 2
(so v = 4, f=5, m=1, j=1), and obtained that all of a1/5, a2/5, and
a3/5 were roots of a nonmonic, primitive, irreducible polynomials with
integer coefficients. For example,
[10] 7 Jun 2002 "Re: Simple solutions in complex mathematical systems"
from rupertm...@yahoo.com (Rupert)
http://groups.google.com/groups?selm=d6af759.0206071708.f999b2a%40posting.google.com
Using the following result, they concluded that James assertion was
incorrect:
THEOREM 2. Let f(x) be an irreducible polynomial with integer
coefficients. Assume that f(x) is primitive. Then the roots of f(x)
are algebraic integers if and only if the leading coefficient of f(x)
is either 1 or -1.
James, after agreeing that the calculations performed by Bengt and
Rupert were correct, said that the problem was in Theorem 2, which was
not a theorem at all, and was instead false. He presented several
attempts at providing a counterexampe.
[11] 9 Jun 2002 " "Impossible" polynomial revisited ", from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0206090957.60bcc0ec%40posting.google.com
[12] 10 Jun 2002 "Irreducible non monic polynomial with algebraic
integer factor" from jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0206100644.23f03421%40posting.google.com
He still claimed Theorem 2 was false on
[13] 20 Dec 2002 "Disproof of assertion about non monic roots" from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0212201002.8b52805%40posting.google.com
He had been offered many proofs of Theorem 2 since June 2. One copy of
the proof he finally accepted was in:
[14] 21 Dec 2002 "Re: JSH: Assessing the truth / O(N) complexity" from
mag...@math.berkeley.edu (Arturo Magidin)
http://groups.google.com/groups?selm=au0bgc%24muo%241%40agate.berkeley.edu
And his agreement that it was indeed correct is in:
[15] 20 Dec 2002 "JSH: FLT withdrawal while I think" from
jst...@msn.com (James Harris)
http://groups.google.com/groups?selm=3c65f87.0212202140.bb5ff17%40posting.google.com
In addition to the explicit counterexamples that had been calculated
by Bengt and Rupert, I had offered an argument using Galois Theory
that stated that none of a1, a2, a3 could be divisible by f when the
resulting polynomial f(w) is irreducible over Q. For example,
[16] 16 Oct 2002, "Re: Current discussion on short FLT Proof", from
mag...@math.berkeley.edu (Arturo Magidin)
http://groups.google.com/groups?selm=aojvfr%24jus%241%40agate.berkeley.edu
At this point (Dec. 2002), James said the counterexample offered by
Rupert and Bengt was inapplicable; and abandoned branch 2 to
concentrate on branch 3.
In branch 3, the factorization is asserted to be
f(w) = (a1*w+b1)(a2*w+b2)(a3*w+b3)
where a1, a2, a3, b1, b2, b3 are "objects".
Although I have requested for a clarification of the definition of
"objects" several times, I have not received a response.
Now, James is claiming that the truth or falsity of Theorem 1 is at
issue, and is in fact something which is relied upon in arguments
against his proposed argument. While he was implicitly using the
argument, James was also saying that it was false. During the last few
months of 2002, James first said it was trivial, and attempted to
present a proof; then he said it was false and dropped the issue.
However, the only argument that relies on the truth of Theorem 1 is
James attempt in branch 2. If Theorem 1 is false, then all of branch 2
must be either abandoned, or the existence of the factorization with
a1, a2, a3, b1, b2, b3 algebraic integers must be proven for either
the specific polynomial, or a family of polynomials that includes
his. The rebuttals by me, Bengt, and Rupert, did not depend on whether
such a factorization existed; from assuming that a factorization
exists, we provided arguments to show that none of the a1, a2, a3
would be divisible by f. However, the rebuttals of Bengt and Rupert
DID depend on Theorem 2, which was also challenged by James.
Where do things stand now? Branch 1 and 2 have apparently been
abandoned. James is now saying that Theorem 1 is false, and that the
objections raised to branch 2 depend on Theorem 1 being
right. However, branch 2 itself is void if Theorem 1 is incorrect. As
such, I suspect that his memory is playing tricks and he is confusing
Theorem 1 with Theorem 2.
Branch 3 is a big unknown. The existence of a factorization into
linear terms with "object" coefficients depends on the definition of
"object", and properties to be derived (once again, James attempts to
establish that two of a1, a2, a3 are divisible by f in the "object
ring" and the third is coprime) depend on that definition. James has
already withdrawn at least one definition (which required units to
have complex norm equal to 1), but has not expanded on the definition
that currently appears on his page, despite repeated requests.
In any case, the validity of Branch 3 does not depend on Theorem 1
either.
As such, I do not understand why James asserts that the validity of
Theorem 1 is somehow the key to the criticisms to his argument.
======================================================================
"If Yeats was right when he wrote 'The best lack all conviction, while
the worst are full of passionate intensity,' then investigations by
Boris Fischoff and others on overconfidence suggest that most of us
aren't very admirable. [...] We're so often cocksure of our decisions,
actions, and beliefs because we fail to look for counterexamples, pay
no attention to alternative views and their consequences, distort our
memories and the evidence, and are seduced by our own explanatory
schemes."
-- John Allen Paulos, _Once Upon a Number: The Hidden Mathematical
Logic of Stories_
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
I did indeed introduce some typing errors when copying the URLs. Here
are the corrections:
>[3] 30 Sep 2001 "FLT Proof Exposition for p=3, Part 2 revised", from
>jst...@msn.com
>http://groups.google.com/groups?selm=3c65f87.0109301551.79fg7a58%posting.google.com
Should be:
http://groups.google.com/groups?selm=3c65f87.0109301551.79fd7a58%40posting.google.com
>[6] 18 Aug 2001 "How my current FLT argument works" from
>jst...@msn.com (James Harris)
>http://groups.google.com/groups?selm=3c65f87.010818609.4b113efb%40posting.google.edu
Should be:
http://groups.google.com/groups?selm=3c65f87.0108181609.4b113efb%40posting.google.com
>[15] 20 Dec 2002 "JSH: FLT withdrawal while I think" from
>jst...@msn.com (James Harris)
>http://groups.google.com/groups?selm=3c65f87.0212202140.bb5ff17%40posting.google.com
Should be:
http://groups.google.com/groups?selm=3c65f87.0212202140.bb6ff17%40posting.google.com
My apologies for the mistakes.
Arturo Magidin
Andrzej Kolowski <akol...@hotmail.com> wrote:
[.snip.]
>> while Magidin is apparently still affiliated with the
>> math department at Berkeley, a far more impressive school, where he
>> got his Ph.d in mathematics.
>
> I think the affiliation now extends only to having an e-mail account.
And being an alum. I am charged an annual fee for keeping the account
open, but it is very stable, and offers me trn, which my other e-mail
acount does not, so I've kept it and use it for newsreading and
mailing lists.
>> Oh, I later *proved* that Magidin's claim is false.
>>
>
> Reference???
>
> I have never seen any such proof from you. Someone else has claimed
> that Magidin's theorem is an old result.
It was probably known, if not explicitly then implicitly from similar
results on more general domains (gcd domains, or domains which satisfy
the Riesz Interpolation Property).
> I have worked out a proof
> also. All proofs I have seen depend ultimately on an old and deep
> result of Dedekind. You at various times have claimed to have a
> proof or claimed that it was obvious and trivial. Your argument
> for FLT actually requires Magidin's theorem. If you have a disproof,
> your own argument is in trouble.
Only his second main attempt, which he has now abandoned in favor of
working with "objects." However, the difficulties with the meaning of
"object" of course remain.
[.snip.]
Arturo Magidin
>> Frankly, I do not believe this belongs in alt.writing or sci.skeptic;
>> as such, I have trimmed newsgroups. If you wish to reply to those
>> newsgroups, then it would seem to me that the honesty and integrity
>> you value so much would demand that you retain my full comments in
>> your reply, so as to give those people a truly "complete picture" of
>> the discussion. Otherwise, some might wonder what it is you have to
>> hide, for example, by removing in its entirety the substance of my
>> post before claiming I am lying...
>>
>> It's up to you, of course.
>
><deleted>
>
>The key points are quite simple, and I've had more experience than I'd
>like with your tendency to use a lot of verbiage and details to
>obscure the truth.
>
>Here are the main points which should be accessible to all readers:
>
>1. I'm not a mathematician but have been using modern problem solving
>techniques in personal math research, and I claim to have made great
>math discoveries.
>
>2. Mathematicians have behaved heinously in replying to me, and have
>shown to me that they have problems with morality, especially honesty.
>
>3. Their problems can be demonstrated with a simple false
>proposition, which is that given a polynomial P(x) with integer
>coefficients of degree n, the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
Don't forget:
3.3 You claimed this result is related to Galois Theory. It is not.
3.5 You claimed this result was key in the posts in which people argue
your arguments was wrong. It is not.
3.7 I indicated that you are probably confusing it with another
result, which I quoted but you deleted, which ->does<- rely on
Galois Theory.
Is it possible you are confused about which theorem you claim is
wrong, why you claim it is wrong, and how it is related to your
arguments? And that all this week, when you have been calling me all
sorts of names, you were actually... incorrect?
Or is it possible that your supposed concern for truth does not even
extend to reading my reply in full? Nah, couldn't be that. Not with
the integrity and honesty you have always shown.
[.rest deleted.]
======================================================================
"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
David C. Ullrich
You really think that nobody's noticed that you have not
given even the slightest _hint_ why this is false?
I mean some months ago you posted a lot of counterexamples
showing it was false - every one was wrong, and you eventually
admitted each one was wrong. In the present spate of posts
you have not given _any_ evidence it's false, valid or not.
>4. Magidin revealed himself as the mathematician I was talking about
>who'd made a false claim of proof, and he continually attempts to
>distract by bringing up side topics,
Yeah, like stating the result is true and pointing out you've given
no evidence it's false is a "side topic" when you're claiming it's
false (and then jumping to conclusions about how mathematicians
are all evil.)
> as I've seen him do in the past,
>and by trying to make longer and longer posts, which would be
>difficult for people to read all the way through without getting
>confused.
Readers who haven't been watching James for years should be
advised that this is standard: He asks a question, or says something
false. People answer the question, or explain what's erroneous
about what he said. He doesn't follow the explanation, so he gets
a more detailed explanation. This pattern repeats a few times,
until the explanation of something that would be very simple
to explain if he knew what he was talking about has become
extremely long and complicated. And at that point he declares
it too long and confusing, like that somehow showed it was wrong.
>5. Math undergrads can ask their own math professors about Magidin's
>claim to see what those professors say.
Why do you keep saying this? Do you know of a professor somewhere
who's going to say it's false? Hint: Like I said the other day, 99.9%
of the math professors in the world will say they don't know if you
ask (the reason for this being that they don't know.) Some of them
will say they don't know whether it's true but they doubt it - those
ones will be wrong. It _may_ be that there's a professor somewhere
who will say that it's false - I'd like to know _who_ you think is
going to say that.
>6. Skeptical readers can do their own web searches to see what's on
>the web about factoring polynomial to see if Magidin's claims are
>supported *anywhere* by anything they can come across.
What an idiotic suggestion. The result seems to be new (it's been
suggested that it's at least implicit in older work, but if I recall
correctly nobody's found a place in the literature where it's
given explicitly.) If it's a new result it's not _going_ to be
"supported" by web pages (except of course that it _is_
supported by the _proof_ on the web pages that Magadin's
given links to many times...)
>Or they can
>look at his replies with a skeptical eye and knowledge of human
>nature. After all, why am I getting such response from a math Ph.d
>educated at Berkeley?
Because you're falsely claiming that his result is false, and he
evidently has nothing better to do than argue with idiots on
the internet (which trait of course he shares with a lot of us,
for example me.)
>7. Clearly I am acting from my certainty that Magidin has made a
>false claim of proof, and I'm showing people who might be curious that
>even with relatively simple propositions a mathematician when caught
>telling a falsehood can rely on other mathematicians to NOT reveal the
>truth, as I show that mathematicians represent a gang, which follows
>rules that might surprise others.
Guffaw. This would all be much clearer if you'd give even a _hint_
as to _why_ you think the result is false.
It's clear why you feel this needs to go to sci.skeptic. The point
to the "skeptic" is that people will believe whatever you say,
whether you give any evidence or not.
Come to think of it, maybe part of the problem is that you
really _don't_ understand what the word "skeptic" is
supposed to mean here. Maybe you think it refers to
skepticism regarding things said by the Establishment,
so that anything that contradicts the Standard Story
is going to be recieved as clearly true. That's not
what the "skeptic" means - it means one doesn't believe
_anything_, whether it's what everybody thinks or
what nobody thinks, without being given a good reason
to believe it.
>James Harris
******************
David C. Ullrich
Xcott Craver
>
> The key points are quite simple, and I've had more experience than I'd
> like with your tendency to use a lot of verbiage and details to
> obscure the truth.
If it's so damn simple, why don't you just show us that counter-
example?
Talk about verbiage! You could have ended this discussion a long
time ago by posting this one single polynomial (assuming you have
it, and of course I suspect that you don't.)
All this blather about corrupt academic establishments seems like
a very long stall tactic.
-X
Brian Quincy Hutchings
of course, not knowing Galois-theory is not a big deal,
til you might want to use it, as you might, since
it deals with the symmetries of equaitons; remember that
Fermat had no recourse to the young republican fire-brand,
either (as in, me, three).
but is that any reason to dyss a nice mathematician
like Magidin, because he seems to know it, well-enough?...
what role would he play in your mathology (sik),
aside of Generic Bad Guy?
most mathematicians have their specialties, and
may be, you could develop one, other than Big Jerk ...
or, the Trickster?
mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7elmk$dhc$1...@agate.berkeley.edu>...
> 3.7 I indicated that you are probably confusing it with another
> result, which I quoted but you deleted, which ->does<- rely on
> Galois Theory.
[NB: an "we told you, so," in my sig
Rupert
> Now then, do you claim to have proven that given a polynomial P(x)
> with integer coefficients of degree n, the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
> must exists where the a's and b's are algebraic integers?
>
>
I posted the link to the proof once already, quarter-wit...
http://www.math.berkeley.edu/~magidin/preprints/gauss.ps
http://www.math.berkeley.edu/~magidin/preprints/gauss.pdf
> James Harris
Arturo Magidin
>> In article <3c65f87.03041...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7cect$2q7j$1...@agate.berkeley.edu>...
[.non math deleted.]
>> >Here are the main points which should be accessible to all readers:
>> >
>> >1. I'm not a mathematician but have been using modern problem solving
>> >techniques in personal math research, and I claim to have made great
>> >math discoveries.
>> >
>> >2. Mathematicians have behaved heinously in replying to me, and have
>> >shown to me that they have problems with morality, especially honesty.
>> >
>> >3. Their problems can be demonstrated with a simple false
>> >proposition, which is that given a polynomial P(x) with integer
>> >coefficients of degree n, the factorization
>> >
>> > P(x) = (a1 x + b1)...(an x + bn)
>> >
>> >must exist where the a's and b's are algebraic integers.
>>
>> Don't forget:
>>
>> 3.3 You claimed this result is related to Galois Theory. It is not.
>
>Hmmm...is there hope? I don't think you believe in the truth Magidin.
>
>I think you are still trying to run.
>
>That claim is what so many have been hanging onto for some time now in
>disputing my proof of Fermat's Last Theorem.
What claim?
>But what matters for the purpose of this discussion is whether or not
>you're withdrawing that claim.
>
>Are you now admitting that you have NOT proven that given a polynomial
>P(x) with integer coefficients of degree n, the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>must exists where the a's and b's are algebraic integers?
I really have to ask: what exactly was it I said that led you to the
conclusion that I was "withdrawing" the claim? Was it that I said it
has nothing to do with Galois Theory?
No, for the record, I have given a proof of this result. Neither you
nor anyone else has produced a counterexample (together with a proof
that it is a counterexample), nor found any errors on the proof.
Nonetheless, the result has nothing to do with Galois Theory.
[.more nonmath deleted.]
>> 3.5 You claimed this result was key in the posts in which people argue
>> your arguments was wrong. It is not.
>
>That discussion is for another thread.
No. It was from this thread.
> Quit running Magidin. I'm
>getting bored with chasing you down.
I cannot help what you feel bored with. I on my part feel bored at
having to point out the truth to you so often. I do not think you are
lying, I simply think you are utterly confused, but have too much
invested in your mental image to risk checking it.
>> 3.7 I indicated that you are probably confusing it with another
>> result, which I quoted but you deleted, which ->does<- rely on
>> Galois Theory.
>
>Sorry, I'm not interested in red herrings. Why don't you just try and
>stick with the relevant facts?
That ->is<- a relevant fact. Your exposition of how the theorem above
relates to arguments against your proposed proof is simply wrong. You
are mistaken when you claim the result on factorization in the ring of
all algebraic integers has been used to challenge your argument. It
has not. Ever.
>> Is it possible you are confused about which theorem you claim is
>> wrong, why you claim it is wrong, and how it is related to your
>> arguments? And that all this week, when you have been calling me all
>> sorts of names, you were actually... incorrect?
>
>What is at issue here now is whether or not a
>mathematician--you--could succeed in claiming to have a proof when
>that claim was false.
So you say.
>The point being that a mathematician, even with a relatively simple
>claim, can be wrong, and there's no safeguard to others who might
>believe that person besides tracking them down, as I'm doing with you.
Easy enough to solve: give a counterexample and prove it is a
counterexample. No need for all this grandstanding.
[.snip.]
>You can try to run Magidin. But you can't hide. Your attempts at
>distraction don't interest me.
I wouldn't dream of trying to distract you from making a fool of
yourself. Please go ahead.
>Now then, do you claim to have proven that given a polynomial P(x)
>with integer coefficients of degree n, the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>must exists where the a's and b's are algebraic integers?
I not only "claim to have proven", I have given a complete proof,
several times. Feel free to point out an error any time. Or to
continue your distractions.
William Kunka
You can try to run Magidin. But you can't hide. Your attempts at
distraction don't interest me.
Apparently the details of your medication schedule don't interest
you,either. Bill
Wayne Brown
You understand James and his tactics quite well.
Arturo Magidin
[.snip.]
>The claim that given a polynomial P(x) of degree n with integer
>coefficients that the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers must exist.
[.snip.]
>The point for everyone is that there's this simple enough assertion,
>which I say is false, and Magidin claims to have proven it.
Well, let's be more precise. You CLAIM it is false, but provide
nothing except your say so to back up your claim.
I say I have proven it, and provide an explicit link to a full proof
of the result for anybody to go and read, check, verify, and point out
any mistakes they find.
http://www.math.umt.edu/~magidin/preprints/gauss.ps (postscript)
http://www.math.umt.edu/~magidin/preprints/gauss.pdf (PDF)
You are being dishonest when you misrepresent the situation as being
merely something I "claim to have proven."
The argument provided may be wrong: you are welcome to point out
EXACTLY WHERE THE ERROR IS. That's why there's a link to the full
argument.
Now, where is YOUR argument, so that others may judge its quality
objectively?
David C. Ullrich
>mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7h4lp$16co$1...@agate.berkeley.edu>...
>> In article <3c65f87.03041...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7elmk$dhc$1...@agate.berkeley.edu>...
>> >> In article <3c65f87.03041...@posting.google.com>,
>> >> James Harris <jst...@msn.com> wrote:
>> >> >mag...@math.berkeley.edu (Arturo Magidin) wrote in message news:<b7cect$2q7j$1...@agate.berkeley.edu>...
>>
[...]
>> >
>> >That claim is what so many have been hanging onto for some time now in
>> >disputing my proof of Fermat's Last Theorem.
>>
>> What claim?
>
>The claim that given a polynomial P(x) of degree n with integer
>coefficients that the factorization
>
> P(x) = (a1 x + b1)...(an x + bn)
>
>where the a's and b's are algebraic integers must exist.
You claim this claim is false, but you give _no_ proof
that it's false. Not even a hint.
[...]
>
>For the readers of alt.math.undergrad, alt.writing, and sci.skeptic,
>note that I'm verifying that a mathematician did make this claim and
>in fact you can see now that he is continuing to make it.
>
>The math undergrads have the luxury of actually talking to their math
>teachers and asking them about Magidin's claim of proof.
Idiot. Magidin's proof has not been published anywhere - those
math teachers could not possibly say whether it's correct since
they haven't seen it.
[...]
>
>Magidin claims to have proven that a polynomial can always be
>expressed as a factorization in linear terms with the coefficients
>being algebraic integers.
>
>His claim is false so he can't have a proof.
Except for the proof that he's posted on sci.math and on
various web sites, you mean.
[...]
>
>Ah, but that's what make mathematics so grand as it doesn't depend on
>what I or anyone else *says* but on what we can prove!!!
Which is precisely why your current tirade is so ridiculous: Anyone,
even someone with no knowledge of the math at all, can see that
you're not even _trying_ to give a proof that his "claim" is false,
you're just repeating it over and over.
>> >The point being that a mathematician, even with a relatively simple
>> >claim, can be wrong, and there's no safeguard to others who might
>> >believe that person besides tracking them down, as I'm doing with you.
>>
>> Easy enough to solve: give a counterexample and prove it is a
>> counterexample. No need for all this grandstanding.
>>
>> [.snip.]
>
>That's not necessary as I have a mathematical proof.
Then why aren't you telling us how the proof works?
>However that's
>not a topic for this thread, but for the other threads I created, so
>no whining in reply here please.
>
>> >You can try to run Magidin. But you can't hide. Your attempts at
>> >distraction don't interest me.
>>
>> I wouldn't dream of trying to distract you from making a fool of
>> yourself. Please go ahead.
>
>Funny, not.
Wrong again. You _are_ making a fool of yourself (and yes,
it's hilarious.)
>> >Now then, do you claim to have proven that given a polynomial P(x)
>> >with integer coefficients of degree n, the factorization
>> >
>> > P(x) = (a1 x + b1)...(an x + bn)
>> >
>> >must exists where the a's and b's are algebraic integers?
>>
>> I not only "claim to have proven", I have given a complete proof,
>> several times. Feel free to point out an error any time. Or to
>> continue your distractions.
>
>That concludes my interest with Magidin in this thread.
>
>The point for everyone is that there's this simple enough assertion,
>which I say is false, and Magidin claims to have proven it.
And he's _shown_ you the proof many times - you've given no
hint what step in the proof is wrong, and no hint as to a
counterexample.
>Now then, he's the mathematician as he has a Ph.d from Berkeley, a
>respected institution of higher learning.
>
>Supposedly mathematicians offer protection from false claims of proof
>by one of their own, but I'm not worried about any other
>mathematicians popping into this thread to tell you that Magidin is
>wrong.
>
>That's because I don't expect much from mathematicians.
Arturo Magidin
>> In article <3c65f87.03041...@posting.google.com>,
>> James Harris <jst...@msn.com> wrote:
>>
>> [.snip.]
>>
>>
>> >The claim that given a polynomial P(x) of degree n with integer
>> >coefficients that the factorization
>> >
>> > P(x) = (a1 x + b1)...(an x + bn)
>> >
>> >where the a's and b's are algebraic integers must exist.
>>
>> [.snip.]
>>
>> >The point for everyone is that there's this simple enough assertion,
>> >which I say is false, and Magidin claims to have proven it.
>>
>> Well, let's be more precise. You CLAIM it is false, but provide
>> nothing except your say so to back up your claim.
>
>that you behaved oddly at best.
You have never provided a counterexample. You hav enever pointed out
any error in my proof. Your "explanation" amounts to saying that if I
were right you would be wrong, and since you know you are right,
therefore I must be wrong. Of course, whether or not you are right is
precisely the point at issue, so it cannot be used to support your
conclusion.
>Now I've posted proof on alt.math.undergrad and sci.math so it's
>available for the math students and other readers who may now wonder
>how you could not understand it.
Be sure to let me know whether anyone agrees with you on this.
>> I say I have proven it, and provide an explicit link to a full proof
>> of the result for anybody to go and read, check, verify, and point out
>> any mistakes they find.
>>
>> http://www.math.umt.edu/~magidin/preprints/gauss.ps (postscript)
>> http://www.math.umt.edu/~magidin/preprints/gauss.pdf (PDF)
>>
>> You are being dishonest when you misrepresent the situation as being
>> merely something I "claim to have proven."
>
>Well anyone who reads my exposition, which relies on rather basic
>algebra, must now wonder what is the matter with you.
They have before. Everyone keeps asking me why I bother with you, when
it is so clear to everyone who is right.
>> The argument provided may be wrong: you are welcome to point out
>> EXACTLY WHERE THE ERROR IS. That's why there's a link to the full
>> argument.
>
>I strongly suggest that the students look at Magidin's work, and
>consider very carefully what it means for a mathematician educated at
>Berkeley to so *easily* get away with false claims when what's at
>stake is the WORLD'S knowledge of a *short* proof of Fermat's Last
>Theorem.
Will you be taking your ->own<- suggestion? Have YOU ->EVER<- looked
at the work which you now question?
Or is it just like with the theorem on roots of nonmonic primitive
irreducible polynomials? Remember: you spent almost a year saying I
was an idiot and a liar for saying the result was true, when you were
positive it was false. What was is that happened in the end? Do you
remember?
>> Now, where is YOUR argument, so that others may judge its quality
>> objectively?
>
>I've posted it on alt.math.undergrad and sci.math and the title of the
>thread is "Key demonstration for my math discussions", and again I
>want students to consider that I've explained to Magidin before.
That post does not contain a refutation of the result in question, nor
does it contain a counterexample to the result in question. It
contains several assertions and several misguided comments, analysing
a family of polynomials at a singular point and trying to deduce
generic conclusions from it, for example.
In the past, you have demanded that people contesting your work not
rely on independent arguments establishing your conclusions to be
wrong, but instead that they should point out exactly where your
argument contains an error. Once again, I invite you to stop being a
hypocrite and adhere to your own standards. You claim my proof of that
result is wrong. What's more, you claim I know it is wrong, and you
claim that everyone who has read it is engaged in a massive conspiracy
where they know it is wrong and yet keep it quiet in order to opress
you. I do not require you to give evidence of the entire claim; I
simply wonder whether you are capable of pointing out a specific line
in my argument which is incorrect, or provide an explicit clear
counterexample to the result, together with a complete proof that it
is a counterexample.
======================================================================
"Destiny is a funny thing. Once I thought I was destined to become
Emperor of Greenland, sole monarch over its 52,000 inhabitants. Then
I thought I was destined to build a Polynesian longship in my garage.
I was wrong then, but I've got it now."
-- The Tick ("The Tick", by Ben Edlund)
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Randy Poe
> I strongly suggest that the students look at Magidin's work,
Why do you strongly suggest students do, but you won't yourself?
Have you ever read that paper? What page does the first error
in the proof occur?
> and
> consider very carefully what it means for a mathematician educated at
> Berkeley to so *easily* get away with false claims when what's at
> stake is the WORLD'S knowledge of a *short* proof of Fermat's Last
> Theorem.
>
> What you have now is *reality* and not fantasy about the world of
> mathematics.
>
> > Now, where is YOUR argument, so that others may judge its quality
> > objectively?
> >
>
> I've posted it on alt.math.undergrad and sci.math and the title of the
> thread is "Key demonstration for my math discussions", and again I
> want students to consider that I've explained to Magidin before.
Here's what I found there:
> It might be easier to see with P(v), which from before is
>
> P(v) = (v^3+1)z^6 - 3v x^2 y^2 z^2 - 2x^3 y^3.
>
> So now let's play that angle so I have
>
> P(m) = (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
>
> where the a's and b's are somewhat mysterious numbers though some
> people have been arguing that they've proven they are algebraic
> integers.
First problem: Nobody's been claiming this. This is not the
Magidin-McKinnon Theorem. M-M states only that there is a
factorization in algebraic integers, not that any factorization
you write down is guaranteed to be in algebraic integers.
You've been told that.
Second problem: Nothing below that point even *pretends* to
be a counterargument. Nothing is stated about a factorization
being or not being in algebraic integers. There is that
old stuff about "factors of f^j" and "factors of sqrt(5)"
which as usual again fails to mention what ring they are factors
in. But we know it's not the ring of algebraic integers.
So what paragraph(s) did you think were the proof that M-M
is false? Pretend I'm a complete idiot and identify the
particular paragraph and tell me why it says something
different about the existence of factorizations into
linear terms with algebraic integer coefficients. Because
I just can't find it. All I see is the one misstatement
above.
- Randy
David C. Ullrich
[...]
>
>I am one of the more important discoverers in mathematical history,
>but future students will have the luxury of knowing that, and may be
>puzzled by your behavior now.
>
>If you make the wrong decision they may believe that there was
>something wrong with the students of today, as if maybe the society of
>this time was especially corrupt.
I've explained this before. You don't seem to get it. But it's a true
fact whether you get it or not: Someone who knows nothing about
the math who read those last two paragraphs would decide you
must be a raving lunatic.
Honest.
>In your defense, I say that many of you may simply be true believers
>having to deal with a challenge to your religion--your math religion.
>
>For some of you, what you're facing may be harder than if you were a
>Christian having to question Christianity.
>
>Your FAITH is being tested.
>
>> Now, where is YOUR argument, so that others may judge its quality
>> objectively?
>>
>> ======================================================================
>> "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
>
>I've posted it on alt.math.undergrad and sci.math and the title of the
>thread is "Key demonstration for my math discussions", and again I
>want students to consider that I've explained to Magidin before.
>
>Now then, you can see why to me it's like arguing with someone
>claiming that the sun won't come up tomorrow and having people believe
>them!!!
>
>I can only assume that people *wish* to believe mathematicians, just
>as people wish to believe in other people in so many areas.
>
>But mathematics is about truth. The truth is before you.
>
>Can you handle it?
Brian Quincy Hutchings
it's not constituted by seven years of work
on newsgroups ??
anyways,
it's nice to have this sort of life-and-death (to all mathematicians,
thank *o*) heart-to-heart,
to take one's mind off of the war (etc.)
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03041...@posting.google.com>...
> > http://www.math.umt.edu/~magidin/preprints/gauss.ps (postscript)
> > http://www.math.umt.edu/~magidin/preprints/gauss.pdf (PDF)
> Well anyone who reads my exposition, which relies on rather basic
> algebra, must now wonder what is the matter with you.
> I strongly suggest that the students look at Magidin's work, and
> consider very carefully what it means for a mathematician educated at
> Berkeley to so *easily* get away with false claims when what's at
> stake is the WORLD'S knowledge of a *short* proof of Fermat's Last
> Theorem.
> For some of you, what you're facing may be harder than if you were a
> Christian having to question Christianity.
--les ducs d'Enron!
http://www.tarpley.net
Nat Silver
> Luckily for me the mathematics is simple enough ...
Unluckily for you, doing mathematics can be extremely
difficult, as is the case with constructing a proof of Fermat's
Last Theorem, which relatively few mathematicians can even
understand. Only very talented people, devoted to
mathematics and driven by singular purpose to work for
many years, can ever hope to achieve the kind of success
that you have deluded yourself into believing is your own.
Dr. Flonkenstein
in an ultimate attempt to be funny
and witty at once, that summum of the evolution
of human race, jst...@msn.com (James Harris) wrote:
>Qnc...@netscape.net (Brian Quincy Hutchings) wrote in message news:<bde404c9.03041...@posting.google.com>...
>> so, how short is the proof (assuming that
>> it's not constituted by seven years of work
>> on newsgroups ??
>> anyways,
>> it's nice to have this sort of life-and-death (to all mathematicians,
>> thank *o*) heart-to-heart,
>> to take one's mind off of the war (etc.)
>
>Well I don't know how serious you are but I'll take this opportunity
>to point out some things and I will answer your question.
>
>First, I'd like to emphasize to readers that I've given all the
>evidence needed for mathematicians to determine that I have indeed
>found a short proof of Fermat's Last Theorem.
>
At the same time you proved that pigs can fly too, WOW!!!!
[AUK] added!
--
Dr Flonkenstein
Alcatroll Labs Inc. Flame, troll and
(TINAL) antispaem bots
development.
mhm 27x12 MEOW MEOW ARMY TINMMA
ICQ 146303664 TNPJRL # -1
http://www.insurgent.org/~alcatroll
Highpriest of the Romath Religion (TINRR)
Professor Kookology at BUNGMUNCH Univ. (TINBU)
Flonk leader #19
Wereldsman's hate lits #7
Political Compass :Economic Left/Right: -3.25,
Authoritarian/Libertarian: -3.13
================================================================
rapist k00k "Bob" <bobx...@hotmail.com> points out that rape
is no crime, in article <3DC86E94...@hotmail.com>
"Sex is a normal, pleasurable body function. Rape is
normal pleasurable body function, even sometimes including
orgasm, with a disagreement over who is to decide."
================================================================
Lynn K. Circle explains English grammar in
Messg-Id <3e9c7ae4$0$1967$a726...@news.hal-pc.org>:
"At least he learned proper English grammar in school, as
well as being too mature attempt anything approaching
your pathetic grade-school level attempts at humor."
dre
<ull...@math.okstate.edu> wrote:
>On 15 Apr 2003 16:01:23 -0700, jst...@msn.com (James Harris) wrote:
>
>
>Idiot. Magidin's proof has not been published anywhere...
This isn't exactly true - Magadin published it and it's available on a
web page on the internet. The link has been posted here about a
bajillion times.
David C. Ullrich
Let's recall the context:
>>>The math undergrads have the luxury of actually talking to their math
>>>teachers and asking them about Magidin's claim of proof.
>>
>>Idiot. Magidin's proof has not been published anywhere - those
>>math teachers could not possibly say whether it's correct since
>>they haven't seen it.
The professors that James is exhorting the students of the world
to ask about the result have _not_ seen it.
******************
David C. Ullrich
Brian Quincy Hutchings
you backwater would-be mathematicians, if
you don't want to be included in James' class-action suit.
like he will remind you,
as you get off o'the boat:
There are sufficient lawyers in the USA,
for me to get SSI, if not neccesarily enough
for me to prove Fermat's "last" theorem. that is certianly enough,
for me to put an immediate halt to the completion
of my undergrad education.
on the other hand,
I forever get to be a *student* of James, so,
this really is the best of all possible universes!
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03041...@posting.google.com>...
> I want to point out that non-English speaking mathematicians clearly
> have a defense as many of them may know nothing of my work.
Rupert
> So I'm the one person who is using mathematics continually to refute
> other people like yourself!!!
>
Actually, Arturo Magidin is using mathematics to refute your claim to
have shown his theorem to be false.
> So I'm beating mathematicians at their own game, and in response I get
> words, words and more words. What losers you are.
>
Arturo Magidin has a proof of his result, and so far you have given us
nothing but words.
Draw what conclusion you will about what a loser you are.
<snip>
Christopher Henrich
Harris <jst...@msn.com> wrote:
>
> I offer to readers the example of
>
> 2z^6 - 3x^2 y^2 z^2 - x^3 y^3 =
>
> (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
>
> where I made a simple change to an expression I've been using.
>
I get the following factors:
z^2 - xy
sqrt(2)z^2 + b1 xy
sqrt(2)z^2 + b2 xy
where b1 = (-1 + sqrt(3))/sqrt(2)
and b2 = (1 + sqrt(3))/sqrt(2).
It may not be obvious that b1 and b2 are algebraic integers, but it is
nevertheless true. Indeed,
b1^2 = 2 - sqrt(3)
b2^2 = 2 + sqrt(3).
From this we find that b1 and b2 are two roots of the polynomial
b^4 - 4 * b^2 + 1.
In short, we have a factorization of the kind that Arturo Magidin
claims always exists. It seems that Magidin was not wrong about this
case.
It may be useful, though boring, to retate what the argument was all
about. Arturo Magidin asserts that if P(X) is any polynomial with
integer coefficients, then it may be decomposed into linear factors
whose coefficients are algebraic integers. James Harris asserts that
this is not true.
James Harris can prove his case by displaying a polynomial P(X) which
cannot be factored thus. The example he has offered is not a
counterexample to Arturo Magidin's assertion.
--
Chris Henrich
The Last Danish Pastry
news:3c65f87.03042...@posting.google.com...
> [snip]
Dear shit-for-brains James,
See
http://tinylink.com/?QxAZiXEGVl
Please tell us more about how you are going to get the army to attack
mathematicians.
--
Clive Tooth
http://www.clivetooth.dk
Rupert
> rupertm...@yahoo.com (Rupert) wrote in message news:<d6af759.03041...@posting.google.com>...
> > jst...@msn.com (James Harris) wrote in message news:<3c65f87.03041...@posting.google.com>...
> > <snip>
> > > So I'm the one person who is using mathematics continually to refute
> > > other people like yourself!!!
> > >
> >
> > Actually, Arturo Magidin is using mathematics to refute your claim to
> > have shown his theorem to be false.
>
>
No, it's not false that Arturo Magidin has presented a mathematical
proof of his claim.
<snip>
> I've given the mathematics that proves my case. And now I give a
> related example so that people can see how it works a little more
> directly.
>
It's true that you have now given something resembling a mathematical
argument. However, as discussed elsewhere, the factorization you give
is simply wrong.
<snip>
W. Dale Hall
James Harris wrote:
> Christopher Henrich <chen...@monmouth.com> wrote in message news:<200420031339512870%chen...@monmouth.com>...
>
>>In article <3c65f87.03042...@posting.google.com>, James
>>Harris <jst...@msn.com> wrote:
>>
>>
>>>I offer to readers the example of
>>>
>>> 2z^6 - 3x^2 y^2 z^2 - x^3 y^3 =
>>>
>>> (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
>>>
>>>where I made a simple change to an expression I've been using.
>>
>
> That should be
>
> 2z^6 - 3x^2 y^2 z^2 + x^3 y^3
>
>
>>I get the following factors:
>>
>>z^2 - xy
>>
>>sqrt(2)z^2 + b1 xy
>>
>>sqrt(2)z^2 + b2 xy
>>
>>where b1 = (-1 + sqrt(3))/sqrt(2)
>
>
> Should be
>
> b1 = (1 - sqrt(3))/sqrt(2).
>
>
>>and b2 = (1 + sqrt(3))/sqrt(2).
>
>
> Should be
>
> b2 = -(1+sqrt(3))/sqrt(2).
>
>
>>It may not be obvious that b1 and b2 are algebraic integers, but it is
>>nevertheless true. Indeed,
>>
>>b1^2 = 2 - sqrt(3)
>>
>>b2^2 = 2 + sqrt(3).
>>
>>From this we find that b1 and b2 are two roots of the polynomial
>>
>>b^4 - 4 * b^2 + 1.
>>
>>In short, we have a factorization of the kind that Arturo Magidin
>>claims always exists. It seems that Magidin was not wrong about this
>>case.
>
>
> No, and that's why I used it because then I could show a case where
> you do get a factorization with algebraic integers to demonstrate that
> EXACTLY TWO have f^j as factors, which is the case.
>
Of course, you realize that your polynomial fails to be irreducible. The
factorization exists with algebraic integer coefficients, as Magidin
has proven (and as you are futilely attempting to refute), and the fact
that you have two of the a's with a factor not shared by the third (not
related to Magidin's result) is due to the reducibility of the
polynomial. That much is guaranteed by Galois theory, not at all related
to the result of Magidin & McKinnon.
Who's trying to mislead people by such an example? This shows your
utter lack of concern about the truth, right?
> Now you can change from f=sqrt(2) to f=sqrt(5), and quickly get a
> polynomial which refutes Magidin's claims.
>
What, are you back on this polynomial?
65 x^3 - 12 x - 2
Haven't you any shame at all? You have admitted, not only
that this polynomial factors as Magidin says, but also that
NONE of the coefficients of x in the linear factors (in which
the coefficients are constrained to be algebraic integers)
is divisible by sqrt(5), in the ring of algebraic integers.
You have admitted as much.
Now you claim otherwise.
Why behave this way?
> It's easy, trivial, and irrefutable.
>
> Note that v=-1+mf^{2j}, and m=1, j=1, for both cases.
>
> That's the only change.
>
>
>>It may be useful, though boring, to retate what the argument was all
>>about. Arturo Magidin asserts that if P(X) is any polynomial with
>>integer coefficients, then it may be decomposed into linear factors
>>whose coefficients are algebraic integers. James Harris asserts that
>>this is not true.
>
You have yet to produce a counterexample. For instance, the above cubic
65 x^3 - 12 x - 2
factors as follows (I've copied directly from the article
From: Bengt Månsson (ben...@telia.com)
Subject: Re: JSH: FLT Proof status and why I'm angry.
Newsgroups: sci.math
Date: 2002-04-20 12:07:43 PST
in which Månsson used the variable W)
65W^3 - 12W - 2 =
= ((k^2 - k·l + l^2)·W - 2^(1/3))
·((e1·k^2 - k·l + e2·l^2)·W - 2^(1/3))
·((e2·k^2 - k·l + e1·l^2)·W - 2^(1/3))
where
k = ((sqrt(65) + 1)/2)^(1/3)
l = ((sqrt(65) - 1)/2)^(1/3)
e1 = (-1 + sqrt(-3))/2
e2 = (-1 - sqrt(-3))/2
The coefficients in each of the linear factors are easily determined to
be algebraic integers.
On the unrelated claim that exactly two of the a's (coefficients of W in
the above factorization) are divisible by sqrt(5), Månsson also showed
that the values ai/sqrt(5) are all roots of this polynomial:
125·x^18 - 30819·x^12 + 2326623·x^6 - 4826809
which is readily seen to be a cubic in X = x^6:
125·X^3 - 30819·X^12 + 2326623·X - 4826809
Further, it is not difficult (once you recognize the values 125 = 5^3
and 4826809 = 13^6) to see that this polynomial has no rational roots,
so (being a cubic) must be irreducible over the rationals. Thus, its
roots cannot be algebraic integers.
>
> It's not *generally* true as it's trivially true that an infinite
> number of case can be factored.
>
> For instance,
>
> x^3 + 3x^2 + 3x + 1
>
> is such a case.
>
> That doesn't prove Magidin is right, now does it?
>
>
>>James Harris can prove his case by displaying a polynomial P(X) which
>>cannot be factored thus. The example he has offered is not a
>>counterexample to Arturo Magidin's assertion.
>
>
> However, by shifting to f=sqrt(5) versus f=sqrt(2), you can quickly
> arrive at a polynomial that *does* refute his claim as I've shown.
>
But you *HAVEN'T* shown that at all.
> It's trivial at this point.
>
Trivial that you haven't done as you claim!
> Or as they say, academic.
>
It might be, if you ever could construct a counterexample.
You're batting precisely 0 at this point, and it might be informative
if you would only work your way through the paper you seem to be trying
to trash. It's not terribly technical; the mathematics is standard, and
specialized terminology is kept pretty much to a minimum.
>
> James Harris
Dale
Dik T. Winter
> > > I offer to readers the example of
> > > 2z^6 - 3x^2 y^2 z^2 - x^3 y^3 =
> > > (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
> That should be
> 2z^6 - 3x^2 y^2 z^2 + x^3 y^3
Both your original polynomial and your changed polynomial are reducible
over the integers. (In the first z^2 + xy is a factor, in the second
z^2 - xy is one.)
The claim has *always* been that if the polynomial was *irreducible*
your claim would not hold.
To give the claim in full:
Given a polynomial P(x) with integer coefficient, irreducible over the
integers. There exist (Magidin-McKinnon) a factorisation with
algebraic integers ai and bi such that P(x) = prod (ai.x + bi).
Given such a factorisation, we find from Galois theory that all of the
ai contain algebraic integer factors of the leading coefficient of the
original polynomial. And stronger, *all* of the ai contain algebraic
integer factors of *all* of the (integer) prime factors of the leading
coefficient.
Note the keyword: *irreducible over the integers*. The claim does
obviously not hold when the polynomial is reducible over the integers,
so spouting reducible polynomials as counter-examples does not hold
water.
Now you might show that your polynomials are always reducible over the
integers, but I think that is a tough question.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Arturo Magidin
>In article <3c65f87.03042...@posting.google.com> jst...@msn.com (James Harris) writes:
> > > > I offer to readers the example of
> > > > 2z^6 - 3x^2 y^2 z^2 - x^3 y^3 =
> > > > (a1 z^2 + b1 xy)(a2 z^2 + b2 xy)(a3 z^2 + b3 xy)
>...
> > That should be
> > 2z^6 - 3x^2 y^2 z^2 + x^3 y^3
>
>Both your original polynomial and your changed polynomial are reducible
>over the integers. (In the first z^2 + xy is a factor, in the second
>z^2 - xy is one.)
>
>The claim has *always* been that if the polynomial was *irreducible*
>your claim would not hold.
>
>To give the claim in full:
> Given a polynomial P(x) with integer coefficient, irreducible over the
> integers.
And primitive. Otherwise, you can throw in a new divisor to any linear
term without bothering the others.
[...]
dre
>Let's recall the context:
>
>>>>The math undergrads have the luxury of actually talking to their math
>>>>teachers and asking them about Magidin's claim of proof.
>>>
>>>Idiot. Magidin's proof has not been published anywhere - those
>>>math teachers could not possibly say whether it's correct since
>>>they haven't seen it.
>
>The professors that James is exhorting the students of the world
>to ask about the result have _not_ seen it.
>
available at the web site.
Professors may have seen it and don't care. They may have seen it and
discussed it with the students, and the students are satisfied with
the proof. Or try this one - students may not care about this little
math feud except in the <real> context of the entertainment value of
watching people go at each other over this like pit bulls for however
many hundreds of messages (my personal favorite).
Look, you don't have to resort to using false claims to beat James
down or prove that he's an idiot in this one. Just use what you have
to argue with that's real, and I'll happily go back to being a
spectator.
David C. Ullrich
You have no conception of how things work here. It's hard enough
for a professor to keep up with the results in his field that are
published in journals. Do you really think that most professors
are aware of all the mathematics that's been posted on every
web site in the world?
In fact the vast majority of math professors in the world are
totally unaware of the result, because it has not been published.
After it actually appears in the Monthly it will _still_ be true that
the vast majority of math professors in the world will be totally
unaware of the result, because they're not going to read
that paper. But at _that_ time there _will_ be a substantial
minority of people who _are_ aware of the result.
******************
David C. Ullrich
dre
Oh, I've got a good concept of what's going on here. In spite of all
this info you gave me about the lives of math professors in this world
and their math habits that I don't give a good shit about, the real
point is still that:
Magadin's proof is available to people who are following this, math
students (pfff), professors (heh) and all.
You are <probably> right about the fact that bunches of math
professors <probably> haven't seen it. But not because it isn't
available. Why don't you look for another reason why tons of math
professors haven't come out of the woodwork and flocked to alt.math to
post? Or even the original point from James, which you glossed over,
that math students haven't posted their profesors reactions from
asking them about it? The reason should be plain as day, and it don't
got nothing to do with math.