JSH: Attacking the conclusion
James Harris
claiming it's wrong.
It's a cheat. It's a dodge. It's specious.
Claims of counterexamples against my work have by me repeatedly been
shown to be false.
In every case posters rely on a circular argument which basically
relies on the ring of algebraic integers not having the problem I've
proven it has.
Here's basically how it goes:
There are numbers that should properly be considered factors of 1.
However, in the ring of algebraic integers, these numbers are NOT
factors of 1.
Note: I did not say that they should be considered factors of 1 in
the ring of algebraic integers.
That's the problem. That is, the problem is that these numbers
provably should be considered to be factors of 1, and finding a ring
where they are is not even hard.
They are factors of 1 in a ring made up of numbers such that only 1
and -1 are integers units, which includes algebraic integers, and
other numbers besides them.
That's it. You have a ring where only 1 and -1 are integer units,
where that's the principal defining characteristic and you can show
that some irrational units in that ring, are not algebraic integers.
Simple.
But, when attacking my work, posters like Hall, Magidin, and Decker
assert that because these numbers are not units in the ring of
algebraic integers i.e. factors of 1 in that ring, then I must be
wrong.
But I've proven that I'm right using rather basic algebra.
That algebra shows that there is a ring of numbers where -1 and 1 are
the only integer units that is indeed LARGER than the ring of
algebraic integers.
That's it. That's the amazing conclusion that mathematicians have
been running from for so long.
It's actually kind of a neat fact.
James Harris
Virgil
jst...@msn.com (James Harris) wrote:
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>
> It's a cheat. It's a dodge. It's specious.
>
But going to the end of a paper and PROVING the conclusions are wrong is
an entirely different matter.
Which is what actually happened.
C. Bond
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>
> It's a cheat. It's a dodge. It's specious.
Which is why it hasn't happened. Your conclusion contradicts your own
premises. It is false.
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
Nope. Not ever. You simply reassert your original false position. Your
work has been repeatedly shown to be riddled with errors and ambiguities.
Worse, you are afflicted with terminal ignorance and have demonstrated
(repeatedly) that you must be continuously beaten with the "2x4 of
TRUTH" before enough sense seeps in for you to identify or acknowledge
your errors. Then you defend your corrected arguments as vigorously as
you did the previous erroneous ones.
> In every case posters rely on a circular argument which basically
> relies on the ring of algebraic integers not having the problem I've
> proven it has.
You have proven no such thing. Your argument is flawed.
> Here's basically how it goes:
>
> There are numbers that should properly be considered factors of 1.
Give an example and specify the ring. When you say, "There are number
that should be properly be considered factors of 1", do you mean that 1
is divisible by these numbers in some ring? What ring? Suppose we assume
the ring of rational numbers. Proceed.
> However, in the ring of algebraic integers, these numbers are NOT
> factors of 1.
So what?
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
Well, where then? In the ring of rational numbers?
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
What a laugh! All this means is that there are rings which do not support
every possible divisor of 1, i.e. reciprocals. Everyone knows that.
Divisibility is not an intrinsic requirement in the ring of algebraic
integers.
[snip usual nonsensical assertions about divisors, factors, rings, etc.
leading to a false conclusion]
> James "Often in error, but never in doubt!" Harris
--
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
W. Dale Hall
James Harris wrote:
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>
Anyone can claim anything. This is not a valid argument.
If you mean that a conclusion of any math paper can be proven
to be false, then you have made a specific claim that you
cannot support.
Your best response is simple: select a paper in mathematics
that people generally agree is valid, find any conclusion,
and prove that the conclusion is false.
Make a claim, support that claim. It's pretty simple,
isn't it?
> It's a cheat. It's a dodge. It's specious.
>
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
>
Really? You have shown that the polynomial products I've produced are
in error? You have shown that the fact that the products are what I
claim they are does not support my claim that the a's and 5 are never
coprime?
> In every case posters rely on a circular argument which basically
> relies on the ring of algebraic integers not having the problem I've
> proven it has.
>
You have proven no such thing. Let's set some of the details down
for the sake of definiteness:
The claim I am refuting is:
In the factorization
65 x^3 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)
one of the coefficients a_* is coprime to 5.
I show that each a_* has a non-unit factor in common with
5 in the ring of algebraic integers. I conclude that none
of the a_* can be coprime to 5.
My argument refuting this follows (I have omitted details of the
polynomials q,r,s. They will be provided upon request):
There are polynomials q(x),r(x),s(x) with integer coefficients
that satisfy the following conditions:
q(x)r(x) = P(x)A(x) + 5
r(x)s(x) = P(x)B(x) + x
where P(x) is the minimal polynomial of -a, namely
P(x) = x^3 - 12 x^2 + 65
and where A(x) and B(x) are polynomials with integer
coefficients. Thus, for -a = any root of P(x), we have
the following factorizations:
5 = q(-a)r(-a)
a = r(-a)s(-a)
These factorizations hold in *any* ring containing both the
ring of integers and the number a.
I have further shown that the number r(-a) has as the minimal
polynomial
MP_r = x^3 - 969 x^2 + 315 x + 5
and so (noting that this polynomial is irreducible over Q) is
an algebraic integer which is not a unit in the ring of
algebraic integers.
Thus, in the ring of algebraic integers, the numbers a and 5
have the common, non-unit factor r(-a).
None of the above algebra relies on anything other than simple
manipulation of polynomials (i.e., multiplication and addition), and
the elementary result that a unit of the ring of algebraic integers
must have a minimal polynomial with both the leading term and constant
term equal to (rational) integers.
If you have shown that any of those assumptions fails, then you need to
make that claim explicitly. I claim you haven't found any such result.
You have claimed that the argument is circular because the
condition that u and v be coprime in the ring of algebraic
integers entails that any common factor of u and v, in that
ring, be a unit (in the ring of algebraic integers), while
I have shown that r(-a) cannot be a unit in the ring of
algebraic integers.
There is no reason behind your assertion that this is a circular
argument. Simply saying "it assumes that the algebraic integers
do not have the problem [you have] proven it has" does nothing
to show circularity.
> Here's basically how it goes:
>
> There are numbers that should properly be considered factors of 1.
>
This is a non-mathematical statement. Make it mathematical, and you
will have something to say. That something may be correct or incorrect,
and I would imagine the latter, but so far you've just made an
esthetic or a moral assertion, not a mathematical one.
> However, in the ring of algebraic integers, these numbers are NOT
> factors of 1.
>
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
>
However, your claim is that the numbers 5 and a are coprime
*in the ring of algebraic integers*.
I have found a non-unit common factor *in that ring*, and for
the numbers to be coprime *in that ring* any such factor would
necessarily be a unit *in that ring*. This is a contradiction:
no number can be both a unit and a non-unit of the same ring.
Even you must admit that.
My argument holds in the ring of algebraic integers, and my result
holds in that ring.
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
>
No. As long as you are sticking with the esthetic "should be considered"
there is no "provably" to be had. It's not so hard to climb out of that
hole: just give a mathematical definition of the phrase
"should be considered to be a factor of 1",
and you'll have something (correct or not) to say. Until then, you're
not making sense.
> They are factors of 1 in a ring made up of numbers such that only 1
> and -1 are integers units, which includes algebraic integers, and
> other numbers besides them.
>
Didn't someone show that your desired ring had to have all integers
invertible (and thus to lead to the full field of algebraic numbers)?
> That's it. You have a ring where only 1 and -1 are integer units,
> where that's the principal defining characteristic and you can show
> that some irrational units in that ring, are not algebraic integers.
>
But you haven't demonstrated what that ring is. No one will argue that
there aren't plenty of rings between the ring of algebraic integers
and the ring of algebraic numbers. They are incredibly easy to find:
Take A, the ring of algebraic numbers, and {a1,a2,...,aN}
a set of N non-integral algebraic numbers. Then the ring
A[a1,a2,...,aN] = the set of all values P(a1,a2,...,aN)
where P(x1, x2,..., xN) is a polynomial in N variables
with coefficients in A.
What you haven't shown is that your notion (a) exists, and (b) has
the other properties you think it has. You can postulate all you like,
but until you prove some existence lemma, and demonstrate the
characteristics of this alleged ring, no one will pay it any mind.
> Simple.
>
> But, when attacking my work, posters like Hall, Magidin, and Decker
> assert that because these numbers are not units in the ring of
> algebraic integers i.e. factors of 1 in that ring, then I must be
> wrong.
>
You yourself have admitted that the numbers r(-a) are not units
in the ring of algebraic integers, haven't you?
I have thus found a non-unit common factor of a and 5, within that
ring, right?
It then follows, does it not, that a and 5 cannot be coprime in
the ring of algebraic integers?
You have called the following argument circular. Please
show precisely where the circularity lies, keeping in mind
that it does not specify the algebraic integers, yet applies
to that ring:
Let a,b,m,n,p be elements of R, a commutative ring
with identity, and suppose a and b have a common
factor within R, as follows:
a = mn
b = mp
Suppose also that a and b are coprime in R.
I will show that m must be a unit of R.
By definition, "a and b coprime in R" means
that every ideal of R containing both a and b
is R itself; by noting that the sum of the
principal ideals Ra + Rb is such an ideal
(i.e., it contains both a and b), we have
R = Ra + Rb
Specializing to the identity element 1 of R, we
find that there must be elements t,u of R with:
1 = ta + ub
Substituting:
1 = tmn + ump
= m(tn + up)
And we find that the common factor m is a unit.
Thus, if a,b are coprime elements of R, then
any common factor is a unit of R.
Equivalently (via the contrapositive): if any
two elements of R have a common factor that
is not a unit then they cannot be coprime.
> But I've proven that I'm right using rather basic algebra.
>
You have tricked yourself into believing you've proven something,
by virtue of having imprecise definitions and overly vague arguments.
If you were correct, then your result would have been correct. Your
result is flawed, and that in itself indicates your method is flawed.
> That algebra shows that there is a ring of numbers where -1 and 1 are
> the only integer units that is indeed LARGER than the ring of
> algebraic integers.
>
You say there is a LARGER ring. There are many LARGER rings. The
existence of a LARGER ring is absolutely irrelevant for the validity
of a proof within the smaller ring of algebraic integers.
Yet you're saying that a and 5 are coprime in this alleged LARGER ring?
I don't address any LARGER ring. My result holds in the ring of
algebraic integers; yours doesn't. Simply saying that the ring
of algebraic integers has some problem, therefore no proof in that
context is valid, is no argument at all, unless you can prove that the
set of algebraic integers fails to be a commutative ring or fails
to be integrally closed (neither of which is likely, given the number
of times the relevant proofs have been revisited).
> That's it. That's the amazing conclusion that mathematicians have
> been running from for so long.
>
> It's actually kind of a neat fact.
>
>
> James Harris
Whatever.
Dale.
W. Dale Hall
W. Dale Hall wrote:
>
... a buncha stuff...
requiring at least the following two editorial corrections:
> None of the above algebra relies on anything other than simple
> manipulation of polynomials (i.e., multiplication and addition), and
> the elementary result that a unit of the ring of algebraic integers
> must have a minimal polynomial with both the leading term and constant
(1) coefficient---^^^^^
> term equal to *units, in the ring of* (rational) integers.
(2) Insert this ^^^^^^^^^^^^^^^^^^^^^
Sorry for the poor proofreading.
Dale.
... the rest deleted ...
Will Twentyman
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
>
> It's a cheat. It's a dodge. It's specious.
Claim: any number of the form 2n+1 where n is a positive integer is prime.
Proof: [a lovely "proof" by induction that I won't bore anyone with here]
Counterexample: n=4 yields 2n+1 = 9, which is a composite.
The proof is irrelevant. The claim is wrong, so the proof is wrong.
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
You mean when the roots were computed, and the non-unit factors they had
in common with 5 were computed, there were arithmatic mistakes? Why
didn't you compute the *real* roots and common factors then?
> Here's basically how it goes:
>
> There are numbers that should properly be considered factors of 1.
Where? As it stands, this statement is meaningless. Perhaps a better
statement would be: There are numbers that you want to be considered
factors of 1 in a particular ring that you have not yet constructed. If
you can clearly define which factors you want to include, then you can
begin constructing the ring and see whether it is a ring that is already
known, such as the algebraic numbers.
> However, in the ring of algebraic integers, these numbers are NOT
> factors of 1.
Ok.
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
Ok.
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
>
> They are factors of 1 in a ring made up of numbers such that only 1
> and -1 are integers units, which includes algebraic integers, and
> other numbers besides them.
This is not a proper definition. Is pi in the ring? Is it a unit in
the ring? You definition does not give sufficient information to answer
these questions. How does this definition show that the numbers you
want to be units are? Which numbers are those?
> That's it. You have a ring where only 1 and -1 are integer units,
> where that's the principal defining characteristic and you can show
> that some irrational units in that ring, are not algebraic integers.
>
> Simple.
Is pi in the ring?
> But, when attacking my work, posters like Hall, Magidin, and Decker
> assert that because these numbers are not units in the ring of
> algebraic integers i.e. factors of 1 in that ring, then I must be
> wrong.
You have completely misunderstood their comments. They have said that
you have not defined your ring, and they have said your claims about the
roots of a particular polynomial are incorrect.
> But I've proven that I'm right using rather basic algebra.
>
> That algebra shows that there is a ring of numbers where -1 and 1 are
> the only integer units that is indeed LARGER than the ring of
> algebraic integers.
>
> That's it. That's the amazing conclusion that mathematicians have
> been running from for so long.
>
> It's actually kind of a neat fact.
--
Will Twentyman
email: wtwentyman at copper dot net
Brian Quincy Hutchings
He'll read something & use its "conclusions" as a generic reference
for all "Liberals," which are the same as Democrats,
which are now as bad as Commies used to be, but
it's still sophistry -- that he calls, "my principles."
jst...@msn.com (James Harris) wrote in message news:<3c65f87.04061...@posting.google.com>...
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
--Give Earth a Trickier Dick Cheeny -- out of office, after gigayears!
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Michael Hochster
: ANYONE can attack any math paper by just going to the conclusion and
: claiming it's wrong.
: It's a cheat. It's a dodge. It's specious.
: Claims of counterexamples against my work have by me repeatedly been
: shown to be false.
: In every case posters rely on a circular argument which basically
: relies on the ring of algebraic integers not having the problem I've
: proven it has.
: Here's basically how it goes:
: There are numbers that should properly be considered factors of 1.
As far as I know, and I readily admit that I have not followed
even a fraction of the discussion, you have never made clear what
you mean by "should properly be."
Dik T. Winter
> ANYONE can attack any math paper by just going to the conclusion and
> claiming it's wrong.
There is a difference between claiming a conclusion is wrong and just
plain proving a conclusion is wrong.
> Claims of counterexamples against my work have by me repeatedly been
> shown to be false.
Eh?
> In every case posters rely on a circular argument which basically
> relies on the ring of algebraic integers not having the problem I've
> proven it has.
What is the problem in the ring of algebraic integers? You say that
factors of numbers can not be split along other numbers. It has been
shown that that can be done.
> There are numbers that should properly be considered factors of 1.
What do you mean with "should properly be considered"?
> Note: I did not say that they should be considered factors of 1 in
> the ring of algebraic integers.
>
> That's the problem. That is, the problem is that these numbers
> provably should be considered to be factors of 1, and finding a ring
> where they are is not even hard.
But it is hard.
> They are factors of 1 in a ring made up of numbers such that only 1
> and -1 are integers units, which includes algebraic integers, and
> other numbers besides them.
This is not enough of a definition of the ring to be even workable.
What do you mean with "integer units"?
> That's it. You have a ring where only 1 and -1 are integer units,
> where that's the principal defining characteristic and you can show
> that some irrational units in that ring, are not algebraic integers.
Well, first give a definition of "integer units" and after that we
can continue.
> That algebra shows that there is a ring of numbers where -1 and 1 are
> the only integer units that is indeed LARGER than the ring of
> algebraic integers.
Again, what do you mean with "integer units"?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
James Harris
I repeatedly explain but there's always someone to keep coming at me
claiming they can't understand, but here goes again. I'm going to
explain yet again to the sci.math newsgroup, and let's see if you
people don't try to come back later and claim I haven't.
You can define a ring by a membership requirement that no number in
that ring other than -1 or 1 is a unit integer.
Some seem dedicated to simply claiming that ring is coincident with
the ring of algebraic integers, which I've proven to be false.
In that larger ring there are units that are NOT algebraic integers.
I've proven that the ring is not coincident with the ring of algebraic
integers with a rather basic argument which can be seen in my paper
Advanced Polynomial Factorization.
Rather than face the argument, certain posters like W. Dale Hall, Rick
Decker, and Arturo Magidin have speciously attacked the conclusion
basically relying on the fact that there are certain numbers provably
not in the ring of algebraic integers that are units in the more
inclusive ring.
They simply, and repeatedly, prove that certain numbers are not units
IN THE RING OF ALGEBRAIC INTEGERS and then claim that proves I'm
wrong.
That they are deliberately trying to attack algebra itself can be seen
by the fact that they stay away from the arguments in the paper,
choosing instead to attack the conclusion or the messenger, whether
it's me or the Southwest Journal of Pure and Applied Mathematics, as
seen by their email campaign against my paper.
Notice they did NOT PAUSE despite the paper having passed peer review.
Notice how clearly the mathematical establishment, rules, procedures
and logic mean little to them as they're dedicated in their mission.
They will not stop as you can see from what they're doing now.
Their position requires that mathematics itself be inconsistent, while
my position only requires that you accept algebra.
They are anti-mathematicians, evil incarnate, dedicated to undermining
intellectual development in this area. If you never thought such
people could actually exist, outside of myths or legends, welcome to
the real world.
James Harris
C. Bond
> Michael Hochster <mic...@rgmiller.Stanford.EDU> wrote in message news:<cat7eb$jam$1...@news.Stanford.EDU>...
> > James Harris <jst...@msn.com> wrote:
> > : ANYONE can attack any math paper by just going to the conclusion and
> > : claiming it's wrong.
> >
> > : It's a cheat. It's a dodge. It's specious.
> >
> > : Claims of counterexamples against my work have by me repeatedly been
> > : shown to be false.
> >
> > : In every case posters rely on a circular argument which basically
> > : relies on the ring of algebraic integers not having the problem I've
> > : proven it has.
> >
> > : Here's basically how it goes:
> >
> > : There are numbers that should properly be considered factors of 1.
> >
> > As far as I know, and I readily admit that I have not followed
> > even a fraction of the discussion, you have never made clear what
> > you mean by "should properly be."
>
> I repeatedly explain but there's always someone to keep coming at me
> claiming they can't understand, but here goes again. I'm going to
> explain yet again to the sci.math newsgroup, and let's see if you
> people don't try to come back later and claim I haven't.
>
> You can define a ring by a membership requirement that no number in
> that ring other than -1 or 1 is a unit integer.
>
> Some seem dedicated to simply claiming that ring is coincident with
> the ring of algebraic integers, which I've proven to be false.
Hold it! Who claimed this? The ring of integers satisfies your requirement. It isn't necessary to take the
larger ring of algebraic integers to accomodate your membership requirement.
Virgil
wrote:
> In article <3c65f87.04061...@posting.google.com> jst...@msn.com
> (James Harris) writes:
> > ANYONE can attack any math paper by just going to the conclusion and
> > claiming it's wrong.
>
> There is a difference between claiming a conclusion is wrong and just
> plain proving a conclusion is wrong.
>
> > Claims of counterexamples against my work have by me repeatedly been
> > shown to be false.
>
> Eh?
>
> > In every case posters rely on a circular argument which basically
> > relies on the ring of algebraic integers not having the problem I've
> > proven it has.
>
> What is the problem in the ring of algebraic integers? You say that
> factors of numbers can not be split along other numbers. It has been
> shown that that can be done.
>
> > There are numbers that should properly be considered factors of 1.
>
> What do you mean with "should properly be considered"?
>
> > Note: I did not say that they should be considered factors of 1 in
> > the ring of algebraic integers.
> >
> > That's the problem. That is, the problem is that these numbers
> > provably should be considered to be factors of 1, and finding a ring
> > where they are is not even hard.
>
> But it is hard.
Actually, in any field containing all of the numbers in question they
will all but zero be factors of 1, but then every non-zero numbers is a
factor of every other, and there are no non-units at all other than
zero, and such fields are not at all what James' alleges that his
mythical ring is like.
David Kastrup
> I repeatedly explain but there's always someone to keep coming at me
> claiming they can't understand, but here goes again. I'm going to
> explain yet again to the sci.math newsgroup, and let's see if you
> people don't try to come back later and claim I haven't.
>
> You can define a ring by a membership requirement that no number in
> that ring other than -1 or 1 is a unit integer.
No, you can't. For one thing it does not make sense to talk about
"unit integer" in the definition of a ring, since either we are
talking about integers, in which case this already defines the ring,
or not, in which case talking about "unit integers" does not make
sense. So you probably mean "unit" instead of "unit integer".
Anyway, with that definition, Z\4, the numbers modulo 4, meet the
requirement, as just 1 and -1 are units. The integers themselves meet
the requirement, too. Z\4 and Z are different rings, so you can't
define both that way.
> Some seem dedicated to simply claiming that ring is coincident with
> the ring of algebraic integers, which I've proven to be false.
You can't talk about "that ring" before it is properly defined, and
your definition above holds for a whole lot of bunch.
> In that larger ring there are units that are NOT algebraic integers.
You have not defined a larger ring yet.
> I've proven that the ring is not coincident with the ring of
> algebraic integers with a rather basic argument which can be seen in
> my paper Advanced Polynomial Factorization.
You can't prove anything of that complexity about "that ring" before
giving a definition that unambiguously defines it. Certainly no such
thing results from merely your "definition" property.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
David Kastrup
> I repeatedly explain but there's always someone to keep coming at me
> claiming they can't understand, but here goes again. I'm going to
> explain yet again to the sci.math newsgroup, and let's see if you
> people don't try to come back later and claim I haven't.
>
> You can define a ring by a membership requirement that no number in
> that ring other than -1 or 1 is a unit integer.
No, you can't. For one thing it does not make sense to talk about
"unit integer" in the definition of a ring, since either we are
talking about integers, in which case this already defines the ring,
or not, in which case talking about "unit integers" does not make
sense. So you probably mean "unit" instead of "unit integer".
Anyway, with that definition, Z\4, the numbers modulo 4, meet the
requirement, as just 1 and -1 are units. The integers themselves meet
the requirement, too. Z\4 and Z are different rings, so you can't
define both that way. Actually, Z\2, Z\3, Z\6 all fit that
requirement as well.
> Some seem dedicated to simply claiming that ring is coincident with
> the ring of algebraic integers, which I've proven to be false.
You can't talk about "that ring" before it is properly defined, and
your definition above holds for a whole lot of bunch.
> In that larger ring there are units that are NOT algebraic integers.
You have not defined a larger ring yet.
> I've proven that the ring is not coincident with the ring of
> algebraic integers with a rather basic argument which can be seen in
> my paper Advanced Polynomial Factorization.
You can't prove anything of that complexity about "that ring" before
Bill Dubuque
>
> You can define a ring by a membership requirement that no number
> in that ring other than -1 or 1 is a unit integer. Some seem
> dedicated to simply claiming that ring is coincident with
> the ring of algebraic integers, which I've proven to be false.
That's by no means a precise definition. It seems you are interested
in considering a ring J of algebraic numbers such that J /\ Q = Z.
In which case see my 2003-08-12 post [1] where I already mentioned
rings of this sort. As I explain there, if you drop the requirement
that the ring be closed under conjugation then J may indeed contain
numbers that aren't algebraic integers while still preserving the
condition that J /\ Q = Z. However, as I mention there, such rings
will be of little value for your intended purposes.
> I've proven that the ring is not coincident with the ring of algebraic
> integers with a rather basic argument which can be seen in my paper
> Advanced Polynomial Factorization.
In said paper [2] the word "ring" occurs only 3 times, and each time
it occurs as part of the larger phrase "ring of algebraic integers".
Since it mentions no other rings, surely it contains no mention of
your elusive "object ring" J, so your above claim is clearly false.
-Bill Dubuque
[1] http://google.com/groups?selm=y8zllty36hr.fsf%40nestle.ai.mit.edu
[2] http://google.com/groups?selm=y8zisf8b0ho.fsf%40nestle.csail.mit.edu
Christopher J. Henrich
Harris <jst...@msn.com> wrote:
> You can define a ring by a membership requirement that no number in
> that ring other than -1 or 1 is a unit integer.
>
This doesn't seem to me to work. The way that one might get a ring
with some special property is this:
1. Specify which numbers belong to the structure. Examples:
1A: "All numbers of the form a+bi where a and b are rational integers"
(and i is the square root of -1).
1B. "All numbers z such that z is a root of a polynomial whose leading
coefficient is 1."
2. You may have to prove that the set so selected actually is a ring.
Basically this means to show that the sum and product of two elements
of the set also belng to the set. If this seems obvious, you can skip
it (but be prepared to fill in the details).
3. Now, you may prove that the special property you want is satisfied.
The point is that specification of what numbers are in the ring is
prior to proving properties of the ring itself.
--
Chris Henrich
James Harris
> jst...@msn.com (James Harris) writes:
>
> > I repeatedly explain but there's always someone to keep coming at me
> > claiming they can't understand, but here goes again. I'm going to
> > explain yet again to the sci.math newsgroup, and let's see if you
> > people don't try to come back later and claim I haven't.
> >
> > You can define a ring by a membership requirement that no number in
> > that ring other than -1 or 1 is a unit integer.
>
> No, you can't. For one thing it does not make sense to talk about
> "unit integer" in the definition of a ring, since either we are
> talking about integers, in which case this already defines the ring,
> or not, in which case talking about "unit integers" does not make
> sense. So you probably mean "unit" instead of "unit integer".
I'll refute your statement quickly:
Consider 3, it is an integer, correct?
Now consider 3 in any ring. It is STILL an integer, correct?
If it is a unit in that ring, then it is an integer unit, correct?
Knowing that 3 is an integer, and that 3 is in a ring, does NOT mean
that ring is the ring of integers.
Why you would assert otherwise is a mystery unless you're just trying
like so many of you do to hide the truth.
> Anyway, with that definition, Z\4, the numbers modulo 4, meet the
> requirement, as just 1 and -1 are units. The integers themselves meet
> the requirement, too. Z\4 and Z are different rings, so you can't
> define both that way. Actually, Z\2, Z\3, Z\6 all fit that
> requirement as well.
That's a finite sized ring. Besides, it doesn't change my point.
You're just throwing up distractions.
>
> > Some seem dedicated to simply claiming that ring is coincident with
> > the ring of algebraic integers, which I've proven to be false.
>
> You can't talk about "that ring" before it is properly defined, and
> your definition above holds for a whole lot of bunch.
>
Now your tactics are clear. You try to question how an integer can
still be an integer *and* a unit, and now claim that there's a
definition problem.
However, consider the requirement for an infinite sized ring that -1
and 1 are the only integers in that ring that are units in that ring.
What's wrong with that requirement?
> > In that larger ring there are units that are NOT algebraic integers.
>
> You have not defined a larger ring yet.
>
For people who don't understand how sci.math posters operate, they
will play silly games with me, like David Kastrup here, where they
make statements that are almost childishly obtuse.
Then they act stubborn.
LOTS of them will post in concert, making strange and irrational
statements, with bold claims, and they will do this building up
threads to hundreds of posts.
If I ignore the irrational posters, they just start posting that they
refuted my claims and that *I* am the irrational one.
Since there are a lot of them, they can just hide the actual
discussions by sheer volume, and keep up the charade.
> > I've proven that the ring is not coincident with the ring of
> > algebraic integers with a rather basic argument which can be seen in
> > my paper Advanced Polynomial Factorization.
>
> You can't prove anything of that complexity about "that ring" before
> giving a definition that unambiguously defines it. Certainly no such
> thing results from merely your "definition" property.
One of their favorite tactics, since I've talked about using the
requirement that -1 and 1 are the only integers that are units, is to
just claim that I haven't defined *anything* at all, or claim the
requirement is ambiguous.
These tactics have been used by quite a few people, from people like
Rick Decker , Arturo Magidin, and Dik Winter to posters like David
Kastrup, C. Bond, and Nora Baron.
It's actually kind of fun in a way. It's like they keep saying all
this crazy stuff and getting away with it, which lets me know just how
unimportant to most of you rational discussion actually is.
Posting on sci.math is NEVER about anything of import for most of you.
To most of you, it's just a social venue for what is basically
unimportant chatter.
James Harris
Richard Tobin
Christopher J. Henrich <chen...@monmouth.com> wrote:
>> You can define a ring by a membership requirement that no number in
>> that ring other than -1 or 1 is a unit integer.
>This doesn't seem to me to work.
It doesn't, but there's nothing wrong with that sort of definition in
general. For example, "the ring with two elements" is a perfectly
good definition of a ring. To show that it's a good definition, you
have to prove (a) that there is such a ring and (b) that there is only
one.
You would also have to say what you mean by "unit integer".
-- Richard
Michael Hochster
:> : There are numbers that should properly be considered factors of 1.
:>
:> As far as I know, and I readily admit that I have not followed
:> even a fraction of the discussion, you have never made clear what
:> you mean by "should properly be."
: You can define a ring by a membership requirement that no number in
: that ring other than -1 or 1 is a unit integer.
I don't understand this (what's a unit integer?), but let's suppose for
the sake of argument that you have clearly defined some ring R of which
the algebraic integers are a proper subset.
: In that larger ring there are units that are NOT algebraic integers.
Ok. That is also true if the larger ring is all complex numbers.
What's the point?
James Harris
> :>
> :> : There are numbers that should properly be considered factors of 1.
> :>
> :> As far as I know, and I readily admit that I have not followed
> :> even a fraction of the discussion, you have never made clear what
> :> you mean by "should properly be."
>
>
> : You can define a ring by a membership requirement that no number in
> : that ring other than -1 or 1 is a unit integer.
>
> I don't understand this (what's a unit integer?), but let's suppose for
> the sake of argument that you have clearly defined some ring R of which
> the algebraic integers are a proper subset.
Since I've seen several posters show confusion here, I've switched to
saying integers that are units.
For instance, if 1/2 is in the ring then 2 is a unit, as 2(1/2) = 1.
In such a ring, there are then more integers that are units than -1
and 1, as 2 is an integer, and then also a unit.
>
> : In that larger ring there are units that are NOT algebraic integers.
>
> Ok. That is also true if the larger ring is all complex numbers.
> What's the point?
In the ring of complex numbers ALL integers except 0 are units!
The rule that -1 and 1 are the only units that are integers defines a
very specific ring which includes the ring of algebraic integers;
however, it is provably BIGGER than the ring of algebraic integers.
ALL claims of error with my work depend on a denial of that fact.
The math is actually kind of neat. In fact, I'm lucky that
mathematicians missed this neat thing for over a hundred years. It's
rather cool.
Oh, yeah, that's why my paper passed formal peer review. It's
correct.
The people objecting to it are wrong. The math doesn't lie.
James Harris
Jesse F. Hughes
> You would also have to say what you mean by "unit integer".
I'm puzzled by the fact that so many people can't guess what James
means here.
He is only interested in rings R such that Z c R. A "unit integer" of
R is a unit u of R such that u is in Z.
Yeah, he should define the term, but it's easy enough to guess in this
case.
--
"Even if [...] a communistic regime should come [to China], the old
tradition [...] will break Communism and change it beyond recognition,
rather than Communism [...] break the old tradition. It must be so."
-- Lin Yutang on "Socialism with Chinese characteristics" in 1935
James Harris
> ric...@cogsci.ed.ac.uk (Richard Tobin) writes:
>
> > You would also have to say what you mean by "unit integer".
>
> I'm puzzled by the fact that so many people can't guess what James
> means here.
>
> He is only interested in rings R such that Z c R. A "unit integer" of
> R is a unit u of R such that u is in Z.
>
> Yeah, he should define the term, but it's easy enough to guess in this
> case.
Thanks! The requirement that the only integers that are units are -1
and 1 creates a special grouping of numbers. Notice that the
requirement holds for integers and the ring of algebraic integers, but
not for complex numbers.
That *single* requirement can be used to understand why Hall, Magidin,
and Decker have repeatedly failed to produce a mathematically correct
rebuttal to my paper.
They have been using the assumption that all rings where -1 and 1 are
the only integers in the ring that are units are subsets of the ring
of algebraic integers.
My work proves that assumption to be false. But while people let them
get away with their false assumption, they get away with pushing the
lie that my work is incorrect.
Now since I've tried to explain this before, I have reason to believe
that Magidin, and his cronies have been deliberately trying to obscure
the truth.
My hope is that their email assault on my paper--especially
understanding what will happen to Ioannis Argyros's career thanks to
them--might finally turn off some of you from supporting these people.
You're not really hurting me by fighting correct mathematics. After
all, think about it. I have these mathematical proofs. I have a
greater reward already than any human being or groups of human beings
can give me.
But I do find it fascinating that people who spend energy and time
trying to learn mathematics, or at least talk about it, would use so
much energy and time to fight it.
Mathematical truth is not personal. Math doesn't play favorites. It
doesn't pick sides. But I think some of you HATE that about math, and
hate it, for letting me make the discoveries I have.
You've personalized mathematics into some human being that you can
despise for letting me be correct, when I found my results because I
learned not to personalize mathematics as a human being, but
appreciate it as logical without regard to human needs or emotions.
James Harris
David Kastrup
> The rule that -1 and 1 are the only units that are integers defines
> a very specific ring which includes the ring of algebraic integers;
Rubbish and hogwash. That "definition" is already satisfied by
taking ordinary integers. So it definitely does _not_ define a "very
specific ring which includes the ring of algebraic integers".
> however, it is provably BIGGER than the ring of algebraic integers.
Ordinary integers are a counterexample, so you are provably having no
clue what you are talking about.
C. Bond
[snip]
> The rule that -1 and 1 are the only units that are integers defines a
> very specific ring which includes the ring of algebraic integers;
> however, it is provably BIGGER than the ring of algebraic integers.
>
> ALL claims of error with my work depend on a denial of that fact.
Wrong! The errors that have been identified have nothing to do with the existence (or non-existence) of such
a ring. They have to do with your claims that there are number which *should be* algebraic integers but which
are not, and your conclusion that the ring of algebraic integers is flawed -- which it is not.
This tasteless and false revisionist history of your previously hotly defended claims and false conclusions
reveals a deep seated dishonesty. If you want to back down from your previous claims and assert a 'new' set,
you ought to begin by clearing the record you previously established.
Is it now your fundamental claim that it is possible to construct a ring which includes the algebraic
integers, but which has additional members, and whose only units which are integers are +1 and -1? Do you no
longer claim that there are numbers which *should be* algebraic integers but which are not? Do you no longer
claim that ring of algebraic integers is somehow "incomplete" or otherwise flawed?
> James "If you can't defeat 'em, dazzle 'em with footwork!" Harris
Andrzej Kolowski
> claiming it's wrong.
>
On one level, Harris is right.
The conclusion of 'APF' does not follow from the main result
claimed in APF.
Why?
Because the main result is about polynomials of the form
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +3m) x^3 - 3(-1 + mf^2) ux^2 + u^3 f),
where f is a non unit, nonzero algebraic integer coprime to 3 and x,
and u is a non unit non zero algebraic integer coprime to f.
Harris's main result about such polynomials is applied to
65 x^3 - 12 x + 1.
But this polynomial does not have the same form as P(m). In
particular, it has constant term 1 rather than constant term
u^3 f^3.
Therefore it is *could* be that this 'conclusion' to APF is wrong,
but the 'main result' is right.
Everyone, perhaps including Harris, seems to have overlooked this.
His main result does not apply to the polynomial for which Dale
Hall, among others, provides irrefutable counterarguments.
Has Harris even read his own paper?
Here is what Harris does. In the expression above for P(m),
he lets f = sqrt(5), m = 1, and u = 1. He writes:
[*] (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3(-1 + mf^2) xu^2 + u3 = 65x^3 -12x + 1.
But go back and look at the expression for P(m) above. It is
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 +3m) x^3 - 3(-1 + mf^2) ux^2 + u^3 f)
= (m^3 f^6 - 3m^2 f^4 + 3m f^2)x^3 - 3(-1 + mf^2) xu^2 f^2 + u3 f^3.
It does not agree with the left side of [*].
Harris was just plain careless in multiplying through by f^2.
Harris has re-submitted 'APF' to another journal. Does this
version include this careless error and the erroneous conclusion?
OK, let's look at things on another level.
Harris trots out a mantra which says "Proofs cannot duel."
What he wants is not a counterexample to what he claims. He
thinks he has a valid proof. Dale Hall of course thinks he has
a valid counterproof. Harris has not found an error in what Dale
has done. He says Hall is using 'circular reasoning', but he
does not point out where. He says Hall's argument is a 'cheat',
that it is 'specious'.
But he doesn't find an error. Saying it is a 'cheat' or 'specious'
or a 'circular argument' is not equivalent to finding an error.
He himself has a 'proof'. He does not find an error in it either.
The problem is, Hall has not pointed out an error in Harris's
main proof either. Hall only showed that Harris's application of his
main result to 65 x^3 - 12 x + 1 is wrong. As noted above, this
could very well be because Harris was careless in applying his
main conclusion: not because the main conclusion itself was wrong.
So Harris says, "Proofs cannot duel." Hall's proof is a proof.
Harris thinks his proof is a proof. They don't 'duel' because
they are about different things. Both could be right.
Harris is not going to be satisfied until someone convinces him
that there is an error in his 'proof' of his main result. Showing
that a misapplication of his main result is wrong does not do that.
So where is Harris's error? IS there an error in his main
result ?
There is.
Contrary to what Harris says, the error has been pointed out before.
DOZENS of times. He notes that P(m) factorizes in a certain way
when m = 0. He then jumps to the conclusion that the same form
of factorization must be true when m <> = 0.
He gives no proof of this. m = 0 is a degenerate case. In that
case a certain associated polynomial is reducible. In general, when
m > 0, it is irreducible. Reducible polynomials factor differently
than irreducible ones. The central result is this: If Q(y) is
a monic polynomial with integer coefficients, and the constant term
is divisible by a prime q, then every root of Q(y) has a nonunit
algebraic integer factor in common with q. Put another way, no
root of Q(y) is coprime to q in the algebraic integers.
[Harris denies this. He says that, e.g., if Q(y) = y^3 + 12*y^2 - 65,
then one of the three roots of Q(y) is coprime to 5.]
This fact shows that Harris's main result is wrong.
Here is the precise excerpt from APF (the version that appeared
briefly in the electronic journal) where Harris makes his mistake:
" ... given g1 = a1 x + uf where with m = 0, g1 gives a factor
of f IT MUST HAVE THAT SAME FACTOR IN GENERAL, proving that
two of the a's have a factor that is f."
I added caps to highlight Harris's erroneous leap of logic.
Thus we have:
1. The main result of APF is wrong.
2. The exact spot where Harris's reasoning goes bad
has been identified.
Moreover, Rick Decker and Keith Ramsay provided examples back
in January-February which showed that Harris's reasoning is
incorrect. Harris started 20 or more threads in sci.math, frantically
trying to squirm out of Decker's argument. He failed.
Decker's example showed that the central argument of APF was wrong.
That's why Harris was so worried about it. If he couldn't refute
Decker, he knew that APF should be withdrawn.
He could not refute Decker or Ramsay, no more than he was able to
refute Hall.
So he went quiet for a long time.
But somehow now his solipsism has prevailed. APF "passed peer review",
although it is clear now that there was never any peer review,
and that the editor of the electronic journal to which he submitted
it accidentally put it in the wrong electronic pile.
But the damage was done. For a day or so, APF was published.
People here wrote the editor saying it was obviously, clearly wrong.
The editor hastily withdrew it and gave no explanation: a cowardly
act, no matter what the motivation.
But this stupid mistake gave Harris all he needed. "It passed
peer review." Therefore someone must have thought it was right.
The editor must have been lying when he said it was a 'technical mistake.'
He was beaten into withdrawing the paper and then lying by sci.math
"bullies". This gave Harris the excuse to ignore all the counter-arguments:
Decker's, Ramsay's, Hall's, Magidin's, whoever. Never mind that
he could not refute any of these. Some anonymous reviewer, he thought,
gave it the stamp of approval. That outweighs everything: Galois
theory, basic algebraic number theory, even obvious arithmetic.
Andrzej
Andrzej Kolowski
> claiming it's wrong.
>
Why?
u^3 f^3.
There is.
Thus we have:
Andrzej
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<87ekocu...@phiwumbda.org>...
>> ric...@cogsci.ed.ac.uk (Richard Tobin) writes:
>>
>> > You would also have to say what you mean by "unit integer".
>>
>> I'm puzzled by the fact that so many people can't guess what James
>> means here.
>>
>> He is only interested in rings R such that Z c R. A "unit integer" of
>> R is a unit u of R such that u is in Z.
>>
>> Yeah, he should define the term, but it's easy enough to guess in this
>> case.
>
> Thanks! The requirement that the only integers that are units are -1
> and 1 creates a special grouping of numbers. Notice that the
> requirement holds for integers and the ring of algebraic integers, but
> not for complex numbers.
I did not intend to support any of your claims regarding this property
of yours. In particular, I see no evidence that any of your opponents
have assumed that there is no ring AlgInt c R c C such that R has no
"unit integers" besides -1 or 1.
On the contrary, Arturo has repeatedly pointed out that (1) this is
obvious and trivial and (2) you haven't proved it yourself.
> My hope is that their email assault on my paper--especially
> understanding what will happen to Ioannis Argyros's career thanks to
> them--might finally turn off some of you from supporting these
> people.
Yeah, well, wake me up when you have word about Argyros's career.
> You've personalized mathematics into some human being that you can
> despise for letting me be correct, when I found my results because I
> learned not to personalize mathematics as a human being, but
> appreciate it as logical without regard to human needs or emotions.
*Who* has personalized mathematics?
I seem to remember *someone* personalizing mathematics, even calling
it "Mother Mathematics". Google shows five hits for
"mother mathematics" author:james author:harris
Some of those is for jst...@mindspring.com. That's you, too, right?
I know that there have been some forgeries of your posts, so I'm not
sure.
Anyway, I am not supporting your arguments here, aside from pointing
out that the meaning of "unit integers" is not too hard to guess from
context. To be sure, explicit definition is always better than having
to guess from context.
--
"[R]eality has a fascinating ability to check us when we get a little too
big for our britches... Make no mistake. There isn't a mathematician alive
today that I can't now touch, and not a mathematical career on the planet
that I can't now affect." --James Harris, render of worlds
Jesse F. Hughes
> jst...@msn.com (James Harris) writes:
>
>> The rule that -1 and 1 are the only units that are integers defines
>> a very specific ring which includes the ring of algebraic integers;
>
> Rubbish and hogwash. That "definition" is already satisfied by
> taking ordinary integers. So it definitely does _not_ define a "very
> specific ring which includes the ring of algebraic integers".
Although he hasn't said so in so many words, it's likely that he's
interested in a maximal subring of C that (1) contains the algebraic
integers and (2) satisfies his unit conditions.
If I'm not mistaken, Arturo has shown that there are many such maximal
subrings, so James's "definition" doesn't suffice. Also, Bill Dubuque
showed that if the ring is mumble mumble mumble (closed under
conjugates?) then mumble mumble mumble happens (where the second group
of mumbles is something bad). Arturo recently mentioned this result
(Bill too, I think), but I don't recall the details. Maybe it was
that the only ring matching that description was the algebraic
integers themselves.
--
Jesse Hughes
"Surround sound is going to be increasingly important in future
offices."
-- Microsoft marketing manager displays his keen insight
Jesse F. Hughes
> But somehow now his solipsism has prevailed. APF "passed peer review",
> although it is clear now that there was never any peer review,
> and that the editor of the electronic journal to which he submitted
> it accidentally put it in the wrong electronic pile.
I don't see how that's clear at all.
What makes you say so?
Did James receive any referee reports? If so, perhaps he can post
them. If not, I still wouldn't know how your story is clearly
correct.
--
Jesse F. Hughes
"[I]t's the damndest thing. There's something wrong with every last
one of you, and I *never* thought that was a possibility. But now I
feel it's the only reasonable conclusion." --JSH sees some sorta light
Quinn Tyler Jackson
> them. If not, I still wouldn't know how your story is clearly
> correct.
*After* publication (report to that point: "correct mathematics is correct
mathematics") he received a "referee's report" that was a modified[1]
version of a post made here *after* publication -- that had somehow been
sitting on the editor's pile from months before.
<ahem -- for the sake of avoiding suits>
To the best of my knowledge.
</ahem>
This is what I refer to as a bend in the editorial time-space continuum.
--
Quinn
[1] Modified such that Hall's reference to the paper being cricized had
already been published had magically vanished from the "report".
Richard Henry
"Quinn Tyler Jackson" <qui...@shaw.ca> wrote in message
news:rp0Bc.811358$oR5.521877@pd7tw3no...
> > Did James receive any referee reports? If so, perhaps he can post
> > them. If not, I still wouldn't know how your story is clearly
> > correct.
>
> *After* publication (report to that point: "correct mathematics is correct
> mathematics") he received a "referee's report" that was a modified[1]
> version of a post made here *after* publication -- that had somehow been
> sitting on the editor's pile from months before.
>
> <ahem -- for the sake of avoiding suits>
> To the best of my knowledge.
> </ahem>
>
> This is what I refer to as a bend in the editorial time-space continuum.
I am assuming that your <knowledge> of this was because it is what James
told you.
Brian Quincy Hutchings
jst...@msn.com (James Harris) wrote in message news:<3c65f87.04061...@posting.google.com>...
> You've personalized mathematics into some human being that you can
> despise for letting me be correct, when I found my results because I
> learned not to personalize mathematics as a human being, but
> appreciate it as logical without regard to human needs or emotions.
--Give Earth a Trickier Dick Cheeny -- out of office, after gigayears!
Arturo Magidin
Jesse F. Hughes <je...@phiwumbda.org> wrote:
>I did not intend to support any of your claims regarding this property
>of yours. In particular, I see no evidence that any of your opponents
>have assumed that there is no ring AlgInt c R c C such that R has no
>"unit integers" besides -1 or 1.
>
>On the contrary, Arturo has repeatedly pointed out that (1) this is
>obvious and trivial and (2) you haven't proved it yourself.
There seems to be some confusion. Dale has repeatedly asked where this
alleged assumption has been used in his argument; he also once asked
for explicit examples. I gave him some, and pointed out to some
comments of Bill Dubuque that may be relevant.
--
======================================================================
"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
Jesse F. Hughes
>> Did James receive any referee reports? If so, perhaps he can post
>> them. If not, I still wouldn't know how your story is clearly
>> correct.
>
> *After* publication (report to that point: "correct mathematics is correct
> mathematics") he received a "referee's report" that was a modified[1]
> version of a post made here *after* publication -- that had somehow been
> sitting on the editor's pile from months before.
But is that the only report he ever saw?
It's very unusual for an article to be published without the author
having previously received reports.
--
Jesse F. Hughes
"It's easy folks. Just talk about my approach to your favorite
mathematician. If they can't be interested in it, they've
demonstrated a lack of mathematical skill." -- James Harris
Jesse F. Hughes
And did Andrzej have better knowledge of the facts when he wrote the
following?
,----
| But somehow now his solipsism has prevailed. APF "passed peer review",
| although it is clear now that there was never any peer review,
| and that the editor of the electronic journal to which he submitted
| it accidentally put it in the wrong electronic pile.
`----
That seems to be fairly blatant second-guessing.
Arturo Magidin
>
>> jst...@msn.com (James Harris) writes:
>>
>>> The rule that -1 and 1 are the only units that are integers defines
>>> a very specific ring which includes the ring of algebraic integers;
>>
>> Rubbish and hogwash. That "definition" is already satisfied by
>> taking ordinary integers. So it definitely does _not_ define a "very
>> specific ring which includes the ring of algebraic integers".
>
>Although he hasn't said so in so many words, it's likely that he's
>interested in a maximal subring of C that (1) contains the algebraic
>integers and (2) satisfies his unit conditions.
>
>If I'm not mistaken, Arturo has shown that there are many such maximal
>subrings, so James's "definition" doesn't suffice. Also, Bill Dubuque
>showed that if the ring is mumble mumble mumble (closed under
>conjugates?) then mumble mumble mumble happens (where the second group
>of mumbles is something bad).
If you ask that the ring contain the algebraic integers and be closed
under Galois conjugates (that is, if it contains the root of an
irreducible polynomial with coefficients in Q, then it contains all
roots of that polynomial), then Bill Dubuque pointed out that the only
such subring of the algebraic numbers which intersects Q at Z is the
algebraic integers.
This because if you pick such any algebrac number not an algebraic
integer which is in the ring, consider its minimal monic polynomial
over Q. At least one of the coefficients must be a non-integer
rational, and it can be expressed as a symmetric function of the
roots. Thus, the ring would contain at least one non-integer rational,
contradicting the condition that R intersect Q be equal to Z.
Arturo Magidin
Andrzej Kolowski <akol...@hotmail.com> wrote:
[.snip.]
>He gives no proof of this. m = 0 is a degenerate case. In that
>case a certain associated polynomial is reducible. In general, when
>m > 0, it is irreducible. Reducible polynomials factor differently
>than irreducible ones. The central result is this: If Q(y) is
>a monic polynomial with integer coefficients, and the constant term
^
irreducible over Q,
>is divisible by a prime q, then every root of Q(y) has a nonunit
>algebraic integer factor in common with q. Put another way, no
>root of Q(y) is coprime to q in the algebraic integers.
In fact, you may replace q by any positive rational power of any
integer other than 1, -1, and 0 (i.e., any algebraic integer which is
not a unit, not zero, and all of whose Galois conjugates are
associates of each other).
Jesse F. Hughes
> In article <87acyzv...@phiwumbda.org>,
> Jesse F. Hughes <je...@phiwumbda.org> wrote:
>
>>I did not intend to support any of your claims regarding this property
>>of yours. In particular, I see no evidence that any of your opponents
>>have assumed that there is no ring AlgInt c R c C such that R has no
>>"unit integers" besides -1 or 1.
>>
>>On the contrary, Arturo has repeatedly pointed out that (1) this is
>>obvious and trivial and (2) you haven't proved it yourself.
>
> There seems to be some confusion. Dale has repeatedly asked where this
> alleged assumption has been used in his argument; he also once asked
> for explicit examples. I gave him some, and pointed out to some
> comments of Bill Dubuque that may be relevant.
Sorry for the confusion. Especially the obviously specious claim that
you've commented on this "repeatedly". Given your current policy,
that just ain't plausible.
--
Jesse Hughes
"Basically there are two angry groups. I am a harsh force of
one. Against me is a society of mathematicians. So far it's been a
draw." -- JSH gives another display of keen insight.
Jesse F. Hughes
> In article <87y8mjt...@phiwumbda.org>,
> Jesse F. Hughes <je...@phiwumbda.org> wrote:
>>David Kastrup <d...@gnu.org> writes:
>>
>>> jst...@msn.com (James Harris) writes:
>>>
>>>> The rule that -1 and 1 are the only units that are integers defines
>>>> a very specific ring which includes the ring of algebraic integers;
>>>
>>> Rubbish and hogwash. That "definition" is already satisfied by
>>> taking ordinary integers. So it definitely does _not_ define a "very
>>> specific ring which includes the ring of algebraic integers".
>>
>>Although he hasn't said so in so many words, it's likely that he's
>>interested in a maximal subring of C that (1) contains the algebraic
>>integers and (2) satisfies his unit conditions.
>>
>>If I'm not mistaken, Arturo has shown that there are many such maximal
>>subrings, so James's "definition" doesn't suffice. Also, Bill Dubuque
>>showed that if the ring is mumble mumble mumble (closed under
>>conjugates?) then mumble mumble mumble happens (where the second group
>>of mumbles is something bad).
>
> If you ask that the ring contain the algebraic integers and be closed
> under Galois conjugates (that is, if it contains the root of an
> irreducible polynomial with coefficients in Q, then it contains all
> roots of that polynomial), then Bill Dubuque pointed out that the only
> such subring of the algebraic numbers which intersects Q at Z is the
> algebraic integers.
It's easier to state in the mumble mumble mumble version.
Thanks.
--
Jesse Hughes
"So far as this negative attitude toward life is concerned, Buddhism
is merely Taoism a little touched in its wits."
-- Lin Yutang, /My Country and My People/
Rick Decker
Jesse F. Hughes wrote:
> akol...@hotmail.com (Andrzej Kolowski) writes:
>
>
>>But somehow now his solipsism has prevailed. APF "passed peer review",
>>although it is clear now that there was never any peer review,
>>and that the editor of the electronic journal to which he submitted
>>it accidentally put it in the wrong electronic pile.
>>
>
> I don't see how that's clear at all.
>
> What makes you say so?
>
> Did James receive any referee reports? If so, perhaps he can post
> them. If not, I still wouldn't know how your story is clearly
> correct.
>
Perhaps, but there is compelling evidence that no one looked at
the paper that appeared in SWJPAM, or at the very least that
anyone who did had no influence on the version that appeared.
The paper that appeared in SWJPAM is a farrago of errors--not
only errors of logic, but also errors of formatting that would
be blatantly obvious even to a reader with no knowledge of
mathematics. There are literally dozens of wrong fonts, misplaced
punctuation, and TeX-related goofs, as have been pointed out
here. Andrzej's claim that there never was any peer review
has, I think, at least a possibility of being true in light
of the errors in the paper that appeared.
Regards,
Rick
James Harris
> jst...@msn.com (James Harris) writes:
>
> > je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<87ekocu...@phiwumbda.org>...
> >> ric...@cogsci.ed.ac.uk (Richard Tobin) writes:
> >>
> >> > You would also have to say what you mean by "unit integer".
> >>
> >> I'm puzzled by the fact that so many people can't guess what James
> >> means here.
> >>
> >> He is only interested in rings R such that Z c R. A "unit integer" of
> >> R is a unit u of R such that u is in Z.
> >>
> >> Yeah, he should define the term, but it's easy enough to guess in this
> >> case.
> >
> > Thanks! The requirement that the only integers that are units are -1
> > and 1 creates a special grouping of numbers. Notice that the
> > requirement holds for integers and the ring of algebraic integers, but
> > not for complex numbers.
>
> I did not intend to support any of your claims regarding this property
> of yours. In particular, I see no evidence that any of your opponents
> have assumed that there is no ring AlgInt c R c C such that R has no
> "unit integers" besides -1 or 1.
I didn't say you did. I thanked you for the clarifying definition.
I think there's something really wrong with a lot of you.
>
> On the contrary, Arturo has repeatedly pointed out that (1) this is
> obvious and trivial and (2) you haven't proved it yourself.
I'm pointing out a false assumption that your "Arturo" is using in his
claims of rebuttals to my work. That false assumption of his is one
I've told him of *repeatedly*, and in fact, I can *easily* shoot down
his, Hall's, and Decker's claims by pointing out that they're using
that hidden assumption that rings defined by -1 and 1 are always
subsets of the ring of algebraic integes.
Now to you my ability to shatter the bogus claims of "Arturo" may be a
source of personal pain, but I can assure you that doesn't change the
mathematics.
He is just wrong, no matter how much that pains you or him.
> > My hope is that their email assault on my paper--especially
> > understanding what will happen to Ioannis Argyros's career thanks to
> > them--might finally turn off some of you from supporting these
> > people.
>
> Yeah, well, wake me up when you have word about Argyros's career.
>
I sent a math paper to the Southwest Journal of Pure Mathematics which
had it peer reviewed. From what I was told by Ioannis Argyros, the
chief editor of that journal, it had favorable reviews and was to be
published.
So far, everything is by the book then, and it's just another math
paper to be published.
Then someone posted the information on sci.math and Magidin, Hall, and
Decker among others conspired to mount an email assault against the
paper, and they carried out their plans.
Argyros yanked my paper *immediately* which means there's no chance he
had time to carefully review their claims of error, and then lied to
me in an email about why, where he's caught by the fact that he
included text posted on Usenet, and he emailed me after yanking the
paper.
Now when it's revealed that he yanked a correct paper and did so after
being lead astray by *Usenet* posters, what career do you think he
will have left?
James Harris
Rick Decker
James Harris wrote:
<snip>
> I'm pointing out a false assumption that [...] "Arturo" is using in his
> claims of rebuttals to my work. That false assumption of his is one
> I've told him of *repeatedly*, and in fact, I can *easily* shoot down
> his, Hall's, and Decker's claims by pointing out that they're using
> that hidden assumption that rings defined by -1 and 1 are always
> subsets of the ring of algebraic integes.
>
We have no less than four posts within the past week explicating
Dale's argument and asking where this "hidden assumption" is used.
You haven't responded. Just in case you missed it, I'll repost a
walkthrough of Dale's counterexample:
1. You assume in your paper that you have a factorization
65 x^2 - 12 x + 1 = (a_1 x + 1)(a_2 x + 1)(a_2 x + 1)
where a_1, a_2, a_3 are algebraic integers.
2. From this, the values a_1, a_2, a_3 can be seen to be roots of
y^3 + 12 y^2 - 65
(Note that this differs slightly from Dale's polynomial above.)
3. From each of the a values, Dale constructs three new numbers,
q = 8 a^2 - 76 a - 185
r = 8 a^2 - 4 a - 45
s = 4 a^2 - 37 a - 104
(So, for instance, one of the a values is approximately 2.1435
which gives us the corresponding r value of about 0.3403.)
4. Now Dale forms the products qr and sr by straightforward
multiplication and shows that qr = 5 and sr = -a.
(Note 1: Dale has sr = a, because his polynomial in (2)
differs from mine, as I mentioned.)
(Note 2: These multiplications make heavy use of the fact
that the a values satisfy a^3 + 12 a^2 - 65 = 0, so we
can simplify a^3 by replacing it with -12a^2+65 and likewise
can write a^4 as 144a^2+65a-780).
5. Thus, for each of the a values, the corresponding r value
is a common factor of 5 and a.
6. We observe that for each a (an algebraic integer by (2))
the corresponding q, r, and s are algebraic integers.
7. We then verify (again by simple multiplication) that
each of the three r values satisfy
r^3 - 969 r^2 + 315 r + 5 = 0
and note that this is an irreducible monic polynomial with constant
term unequal to 1 or -1. Thus, each of the r values is an
algebraic integer non-unit.
8. The conclusion is that in the ring of algebraic integers
each of the a values has a non-unit factor in common with 5.
In other words, none of the three a values are coprime to
5 in the ring of algebraic integers.
Pray tell me where this "assumption" of a different ring is used.
Regards,
Rick
Andrzej Kolowski
It's not guessing. Feel free to ask the editor.
Andrzej
Quinn Tyler Jackson
> > To the best of my knowledge.
> > </ahem>
> >
> > This is what I refer to as a bend in the editorial time-space continuum.
>
> I am assuming that your <knowledge> of this was because it is what James
> told you.
It's what he's already told everyone here -- if you'd read between the lines
of those messages he's already posted on the topic.
<ahem -- to the best of my knowledge inserted here>
Of course, Solution to FLT may also mean "Faster than Light Travel" and
someone might have such a fast box that emails sent after publication
suddenly become emails sent before, minus a few key sentences that refer to
publication already being a fait accompli. Wiles should win a Nobel in
Physics if such is the case, eh what?
</ahem>
--
Quinn
#include <standard_disclaimers.here>
James Harris is Nuts
>
(snip)
>
>I am assuming that your <knowledge> of this was because it is what James
>told you.
>
Don't forget, Quinn Tyler Jackson is a friend and
booster of James Harris. Perhaps they discussed
it with their "closed list" of "several
mathematicians" (yeah, right).
Watch the evasion from both of them!
James Harris is Nuts(TM)
Quinn Tyler Jackson
> booster of James Harris. Perhaps they discussed
> it with their "closed list" of "several
> mathematicians" (yeah, right).
>
> Watch the evasion from both of them!
Dude, dude, dude.
Our lists, unlike those in Usenet (at least the unmoderated ones such as
this) are governed by civility guidelines to which the posters agree to be
bound:
http://www.iomas.com/gina/ultrahiq/mega-society/MegaBoard/MegaBoardGL.htm
Anyone can find those guidelines with a little help from what Einstein has
already posted here.
There are no secrets -- but stop insisting that we do your legwork for you.
That's not evasion -- that's life such as it is. Unless someone is a member
of Ultranet, they don't get onto the list in question. It's not for us to
point out the names of those members of the list who do not themselves point
out their membership or we ask specifically for permission. This is called
CONFIDENTIALITY, not evasion. Why *should* we ask permission to post such
names? A quoi bon?
You are obsessed with finding out the names of those mathematicians on the
list who read James' paper on the algo that he found the error in that they
admitted to not noting the error. BIG DEAL. It happens all the time, dude --
it's not a quantum flux of credibility that human beings miss mistakes.
I tell you what, dude. Somewhere behind your pseudonym *is* a real person.
Not one I particularly think has much to like about him -- but a person just
the same. So I will stop calling you a computer program. That was silly of
me. I'll stop insisting you post under a human name in order to be treated
like one. I'll stop referring to you as Eliza. Kind of silly of me to have
started that in the first place -- I know.
I won't even expect you to retract your statements about me. What's done is
done.
But stop, please, obsessing about this. Frankly, you are starting to
frighten me. You really are. I *wish* I were being melodramatic about
this -- that I was exaggerating my feelings in this regard -- but you are
starting to make me feel creepy -- like you're invading my personal space. A
past incident with an anonymous critic lead to that person somehow finding
my unlisted personal number and phoning me at home. I don't want to project
that other poster's behavior onto you -- but frankly, your obsession with
this stuff is really going to far for my comfort levels.
So, since I really do know that there is a human being behind that pseudo,
sitting on some computer at home -- please, exercise some restraint. I don't
care what you think, but show some moderation. Show me that I have nothing
to worry about from your behavior, that you're really just treating this as
a personal crusade, and it isn't something beyond that.
Like I say -- I wish I were being melodramatic, but you've carried this a
bit far, dude.
--
Quinn
Jesse F. Hughes
> je...@phiwumbda.org (Jesse F. Hughes) wrote in message news:<87acyzv...@phiwumbda.org>...
>> jst...@msn.com (James Harris) writes:
>>
>> >
>> > Thanks! The requirement that the only integers that are units are -1
>> > and 1 creates a special grouping of numbers. Notice that the
>> > requirement holds for integers and the ring of algebraic integers, but
>> > not for complex numbers.
>>
>> I did not intend to support any of your claims regarding this property
>> of yours. In particular, I see no evidence that any of your opponents
>> have assumed that there is no ring AlgInt c R c C such that R has no
>> "unit integers" besides -1 or 1.
>
> I didn't say you did. I thanked you for the clarifying definition.
Just making sure that we understood each other, that's all. The rest
of your post made me suspect that you thought I was a convert.
>> On the contrary, Arturo has repeatedly pointed out that (1) this is
>> obvious and trivial and (2) you haven't proved it yourself.
>
> I'm pointing out a false assumption that your "Arturo" is using in his
> claims of rebuttals to my work.
Actually, my "Arturo" ("my" Arturo?) pointed out I was
mischaracterizing the situation. He has not repeatedly pointed out
(1) and (2), although I'm pretty sure *someone* has.
[...]
>> > My hope is that their email assault on my paper--especially
>> > understanding what will happen to Ioannis Argyros's career thanks to
>> > them--might finally turn off some of you from supporting these
>> > people.
>>
>> Yeah, well, wake me up when you have word about Argyros's career.
>>
>
> I sent a math paper to the Southwest Journal of Pure Mathematics which
> had it peer reviewed. From what I was told by Ioannis Argyros, the
> chief editor of that journal, it had favorable reviews and was to be
> published.
>
> So far, everything is by the book then, and it's just another math
> paper to be published.
There is some confusion on this point. When you were told it would be
published, did you receive copies of the referee reports? I've never
had anything published prior to receiving (and reacting to) referee
reports.
If you received no reports, then either this journal has a very odd
publication procedure, or everything is not "by the book".
>
> Then someone posted the information on sci.math and Magidin, Hall, and
> Decker among others conspired to mount an email assault against the
> paper, and they carried out their plans.
>
> Argyros yanked my paper *immediately* which means there's no chance he
> had time to carefully review their claims of error, and then lied to
> me in an email about why, where he's caught by the fact that he
> included text posted on Usenet, and he emailed me after yanking the
> paper.
>
> Now when it's revealed that he yanked a correct paper and did so after
> being lead astray by *Usenet* posters, what career do you think he
> will have left?
Well, I suppose we have to wait until the consensus is that the paper
is correct, huh?
I think I'll just wait and see what happens. Your hammer must be very
powerful, judging by the length of the swing.
--
"...you are around so that I have something else to do when I'm not
figuring something important out. I was especially intrigued on this
iteration by cursing, which I think I'll continue at some later date
as it's so amusing." --- James S. Harris
Jesse F. Hughes
>> That seems to be fairly blatant second-guessing.
>
> It's not guessing. Feel free to ask the editor.
Oh? So you've asked him yourself?
Well, that's different then. While it doesn't mean that your account
is the obviously correct account, it does mean it isn't mere
speculation. You should've said so.
--
Jesse F. Hughes
"I have written many words to sci.math, some of them are not even
meaningless." --Ross Finlayson
David C. Ullrich
Hughes) wrote:
>"Quinn Tyler Jackson" <qui...@shaw.ca> writes:
>
>>> Did James receive any referee reports? If so, perhaps he can post
>>> them. If not, I still wouldn't know how your story is clearly
>>> correct.
>>
>> *After* publication (report to that point: "correct mathematics is correct
>> mathematics") he received a "referee's report" that was a modified[1]
>> version of a post made here *after* publication -- that had somehow been
>> sitting on the editor's pile from months before.
>
>But is that the only report he ever saw?
>
>It's very unusual for an article to be published without the author
>having previously received reports.
I once sumbitted a short article containing a slightly amusing
but inappropriately silly phrase (something about how "although
these distributions are not infinitely divisible they are in
fact pretty darn divisible".) I intended to change that in the
version that finally appeared. But they went ahead and published
it without ever sending me a referee's report - I never had a
chance to make the change.
************************
David C. Ullrich
Wayne Brown
>
> Now to you my ability to shatter the bogus claims of "Arturo" may be a
> source of personal pain, but I can assure you that doesn't change the
> mathematics.
>
> He is just wrong, no matter how much that pains you or him.
I'm sure you're not causing any "pain" to Prof. Magidin or anyone else
here. You do inspire considerable *disgust*, though, especially when you
make these cowardly attacks on someone you know will not respond to you.
I'm not surprised you had Google delete those articles in which you
admitted just how sick, twisted and narcissistic you really are. In fact,
I was expecting it, which is why I (and others) made our own copies.
You can be sure you'll *never* be able to live them down, because there
always will be some of us ready to quote them again. You're despicable,
Harris, and you'll never escape from that being public knowledge.
It was *so* pleasant in sci.math while you were away. Please return to
your megalomaniac mutual admiration society at your earliest convenience,
and leave sci.math unpolluted with your drivel.
--
Wayne Brown (HPCC #1104) | "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"
James Harris
> akol...@hotmail.com (Andrzej Kolowski) writes:
>
> >> That seems to be fairly blatant second-guessing.
> >
> > It's not guessing. Feel free to ask the editor.
>
> Oh? So you've asked him yourself?
>
> Well, that's different then. While it doesn't mean that your account
> is the obviously correct account, it does mean it isn't mere
> speculation. You should've said so.
Well now there's a hint that someone talked to the editor Ioannis
Argyros, who may be trying to float yet another story.
The trouble is in the details of his emails to me, including the fact
that I'd emailed him several times *before* I received word the paper
passed peer review.
There's enough dialogue back and forth between me and editors,
including the chief editor Argyros, at SWJPAM to dismiss any claim
that my paper was accidentally accepted, outright.
Given that Argyros clearly lied in one of his emails to me, I wouldn't
be surprised if he turned to someone else now to try and cover his
ass.
But the main thing is that my paper IS correct. No one has yet shown
an error in it, and Argyros didn't bother to even try. He just sent
me some of Hall's crap that Hall posted the day before and claimed
that was a report from a reviewer.
His word now doesn't mean much. But I find it interesting if anyone
on this list is supposedly speaking for him.
That is relevant information.
James Harris
Will Twentyman
Perhaps you should look at the ring of 2x2 matrices, where
1 = [1 0] , -1 = [-1 0] , 0 = [0 0]
[0 1] [ 0 -1] [0 0]
1, 0, -1 are not always used to represent integers.
--
Will Twentyman
email: wtwentyman at copper dot net
Arturo Magidin
David C. Ullrich <ull...@math.okstate.edu> wrote:
>On Sat, 19 Jun 2004 22:31:48 +0200, je...@phiwumbda.org (Jesse F.
>Hughes) wrote:
[.snip.]
>>It's very unusual for an article to be published without the author
>>having previously received reports.
>
>I once sumbitted a short article containing a slightly amusing
>but inappropriately silly phrase (something about how "although
>these distributions are not infinitely divisible they are in
>fact pretty darn divisible".) I intended to change that in the
>version that finally appeared. But they went ahead and published
>it without ever sending me a referee's report - I never had a
>chance to make the change.
Just to add a few more data points, I have had a couple of papers
published where I never received a referee report, and one in which I
was simply told of one suggestion by the referee. In one where I
never received a referee report I was simply adviced by the editor
that the referee had recommended publication and that the paper was
accepted (this was at a very good journal, and the paper was
substantial at over 70 pages as submitted, and over 60 as published).
Richard Henry
"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
news:cb4r7v$rep$1...@agate.berkeley.edu...
>
> Just to add a few more data points, I have had a couple of papers
> published where I never received a referee report, and one in which I
> was simply told of one suggestion by the referee. In one where I
> never received a referee report I was simply adviced by the editor
> that the referee had recommended publication and that the paper was
> accepted (this was at a very good journal, and the paper was
> substantial at over 70 pages as submitted, and over 60 as published).
Did they use a smaller font?
Arturo Magidin
<rph...@home.com> wrote:
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
>news:cb4r7v$rep$1...@agate.berkeley.edu...
>>
>> substantial at over 70 pages as submitted, and over 60 as
>> published).
>
>Did they use a smaller font?
Different formatting and different margins. Lines were slightly
longer, each page was slightly longer (smaller bottom margin) and
subsection titles were in-line rather than displayed; little things
like that add up considerably given the length at issue.
Jesse F. Hughes
> In article <vskad09bgubnj4rp2...@4ax.com>,
> David C. Ullrich <ull...@math.okstate.edu> wrote:
>>On Sat, 19 Jun 2004 22:31:48 +0200, je...@phiwumbda.org (Jesse F.
>>Hughes) wrote:
>
> [.snip.]
>
>>>It's very unusual for an article to be published without the author
>>>having previously received reports.
>>
>>I once sumbitted a short article containing a slightly amusing
>>but inappropriately silly phrase (something about how "although
>>these distributions are not infinitely divisible they are in
>>fact pretty darn divisible".) I intended to change that in the
>>version that finally appeared. But they went ahead and published
>>it without ever sending me a referee's report - I never had a
>>chance to make the change.
>
> Just to add a few more data points, I have had a couple of papers
> published where I never received a referee report, and one in which I
> was simply told of one suggestion by the referee. In one where I
> never received a referee report I was simply adviced by the editor
> that the referee had recommended publication and that the paper was
> accepted (this was at a very good journal, and the paper was
> substantial at over 70 pages as submitted, and over 60 as published).
Obviously, both you and David have more and different experience than
I. Thanks for the corrections.
--
Jesse F. Hughes
"I think the burden is on those people who think he didn't have
weapons of mass destruction to tell the world where they are."
-- White House spokesman Ari Fleischer
W. Dale Hall
James Harris wrote:
... stuff deleted ...
> But the main thing is that my paper IS correct. No one has yet shown
> an error in it, and Argyros didn't bother to even try. He just sent
> me some of Hall's crap that Hall posted the day before and claimed
> that was a report from a reviewer.
>
Crap?
Crap?
CRAP?????
It's arithmetic, son. Would that you knew as much.
... stuff deleted ...
> That is relevant information.
>
>
> James Harris
Dale.
James Harris is Nuts
>Frankly, you are starting to
>frighten me. You really are. I *wish* I were being melodramatic about
>this -- that I was exaggerating my feelings in this regard -- but you are
>starting to make me feel creepy -- like you're invading my personal space
(snip)
>but frankly, your obsession with
>this stuff is really going to far for my comfort levels.
>
(snip)
>please, exercise some restraint. I don't
>care what you think, but show some moderation. Show me that I have nothing
>to worry about from your behavior, that you're really just treating this as
>a personal crusade, and it isn't something beyond that.
>
>Like I say -- I wish I were being melodramatic, but you've carried this a
>bit far, dude.
>
>--
>Quinn
>
GROW UP, YOU BABY!
Why don't you read the archives of your buddy James Harris
and his YEARS of SHIT postings, insults, threats, distortions of
reality, etc. etc. etc.?
Then you'll see the lost life and mind of James Harris,
poster boy for cranks!
I dare you - read the archives, and then come back and
tell this newsgroup what you think of your friend James Harris.
Or don't YOU have the balls to face this newsgroup about
your friend?
Until then ...
James Harris is Nuts(TM)