| Topic: |
Science > Physics |
| User: |
"James Harris" |
| Date: |
02 Aug 2003 07:43:38 AM |
| Object: |
Problem is a math limitation |
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
That may seem esoteric, but it's possible to make it more direct as
algebraic integers are the roots of monic polynomials with integer
coefficients, so if it were possible then someone, somewhere in the
world could give the polynomial which would define that value.
Its last coefficient would also have a factor that is 13.
If you simply divided off all non unit factors of 5 with the three
roots to get a degree 3 polynomial, you'd have a polynomial with a
last coefficient that is -13, that'd have a first coefficient that is
1.
That only leave 2 other coefficients.
Mathematicians are *incapable* of giving the remaining coefficients.
They are incapable of giving that polynomial.
Here's the mystery polynomial:
y^3 + ay^2 + by - 13
and mathematicians cannot give 'a' and 'b'.
I want that to soak in as it explains to you how there could be so
much arguing, as mathematicians have been dodging around a fascinating
limitation.
The problem has been in mathematics for over a hundred years and
apparently sat there because no one bothered to try something like
dividing off *some* factors of a root.
It's not surprising why.
Usually for practical concerns with a polynomial irreducible over
Q--that is one that doesn't factor over rationals--if it's in physics
you'd just get a numerical approximation.
Physicists might assume that mathematicians considered the problem as
they deal with "pure math" but unfortunately they did not, and instead
relied on an assumption:
Facing their inability using basic algebraic methods to distinguish
between the roots of irreducibles like y^3 + 12y - 65, mathematicians
simply declared that non unit factors of the constant term had to be
spread between all the roots.
But they can't prove it because they have no tools for checking.
So I came along and found a need to do something that mathematicians
simply don't bother to do, and that lead to me finding this intriguing
problem with algebraic integers. Now if you have algebraic integers
already, and use ring operations, yeah, you keep getting algebraic
integers, but if you *decompose* algebraic integers looking at factors
in common with a particular prime integer, then you run into this
problem.
People in physics wouldn't notice it because you'd just take a
numerical approximation for an irreducible, while mathematicians
usually focus on questions of reducibility and not decomposing the
roots.
And even if they did think about it, they tried a cop-out which was
simply to proclaim that being irreducible over Q spread factors of the
prime factors of the constant term throughout the roots, but now I've
pointed out that they are incapable of proving that a further
decomposition will give algebraic integers.
A simple analogy to what I'm discussing can be found with the ring of
evens, which is just the set of all even numbers. In that ring 2 and
6 are coprime, which may seem odd, but since 6/2 = 3 is NOT in the
ring as 3 is not even, 2 and 6 in the ring of evens are coprime to
each other. That is, in the ring of evens 2 is NOT a factor of 6. So
while you can decompose 6 in a higher ring--the ring of integers--in
the ring of evens, it's prime.
Ostensibly the ring of algebraic integers covered decompositions of
algebraic integers, but I can easily prove directly that it does not,
or leave you with the indirect proof of mathematicians being
*incapable* of demonstrating that the results of such a decomposition
are algebraic integers.
James Harris
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| User: "Bhuvan nospam" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 10:19:30 PM |
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"James Harris" <jstevh@msn.com> creamed on himself in message
news:3c65f87.0308020443.55bc2e66@posting.google.com...
Hey James:
Go ***** yourself and your 'mathematics'. Get a clue and see that nobody here
cares a ***** what your stupid theories are. You are a reject; get over
yourself and your stupid, useless 'theories'.
Did I say go ***** yourself yet??
~Bhuvan
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| User: "The Last Danish Pastry" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 01:12:40 PM |
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"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0308020443.55bc2e66@posting.google.com...
Turns out that with all the arguing a rather direct limitation of
your brain has come to light. It cannot comprehend mathematics.
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| User: "James Harris" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 04:11:54 PM |
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(James Harris) wrote in message news:<3c65f87.0308020443.55bc2e66@posting.google.com>...
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
That may seem esoteric, but it's possible to make it more direct as
algebraic integers are the roots of monic polynomials with integer
coefficients, so if it were possible then someone, somewhere in the
world could give the polynomial which would define that value.
Its last coefficient would also have a factor that is 13.
If you simply divided off all non unit factors of 5 with the three
roots to get a degree 3 polynomial, you'd have a polynomial with a
last coefficient that is -13, that'd have a first coefficient that is
1.
That only leave 2 other coefficients.
Mathematicians are *incapable* of giving the remaining coefficients.
They are incapable of giving that polynomial.
Here's the mystery polynomial:
y^3 + ay^2 + by - 13
and mathematicians cannot give 'a' and 'b'.
<deleted>
Um, yeah they can.
Scratch that, forget about it, looks like I'm still searching for a
way to beat these people. Damn it. A proof should be enough. But
while they keep lying I have to keep looking for alternate proofs. Oh
well, that's just the way it is.
But I will get them. And when I do, I will make certain that people
remember that they sat all that time lying and had to know they were
lying.
After all I have PROOFS!!! Since when did a person with proofs have
to spend so much time trying to find proofs to prove their proofs?
James Harris
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| User: "W. Dale Hall" |
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| Title: Re: Problem is a math limitation |
04 Aug 2003 08:11:05 PM |
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I prefer not to make mistakes.
When I make mistakes, I prefer not to post them.
When I make mistakes, and I post them anyhow, I feel guilty until I've
done this.
W. Dale Hall wrote:
... a buncha stuff ...
For the present case,
y^3 + 12 y - 65,
when one takes the extension of Q by adjoining a single root, the
resulting field has class number 3. Therefore, when attempting to
find a factor in common between the root (I'll call it p) and 5,
the ideal <r,5> is not principal. It was that feature that made the
preceding problem turn out so simply, but in this case I wasn't so
lucky. So, rather than take the ideal <p,5>, I had to take its cube:
<p,5>^3, which *is* principal, and generated by the element
g = 4 p^2 + 12 p + 85.
In other words, it isn't the case that
p = q r
5 = r s
for some q,r,s in the integers of the extension Q(p), but rather
p^3 = q r
125 = r s
and in this case, that "r" is the above number g (depending on which
root p one takes, one obtains a different g).
The numbers q and s are as follows:
q = -4 p^2 + 8 p + 13
s = -4 p^2 + 20 p - 23
In order to find the appropriate common factors for p and 5, I need to
take a cube root. In terms of the field Q(p), I need to pass to its
cubic extension Q(p,g^(1/3)). As the field Q(p) is already of degree
3 over Q, the new field is of degree 9 over Q. So already I'm in luck,
since I don't need to take off my shoes to do the counting.
I was able to find the minimal polynomials of q and s. Since I don't
really care about the remaining factor(s) of 5 (plural, since each
choice of p gives a separate choice for r=g, and thus for s), I'll only
provide the minimal polynomial of q:
Irred(q) : x^3 + 135 x^2 + 12315 x + 2197.
Notice the constant 2197: it's 13^3.
The above formula for Irred(q) is incorrect. That is Irred(-q). While
no one's counting, I figure I'll post this correction anyway:
Irred(q) : x^3 - 135 x^2 + 12315 x - 2197.
OK, folks, now get back to what it is you were already doing before
I barged in to interrupt the show.
... the rest deleted ...
Dale
Dale, again.
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| User: "Edward Keyes" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 10:10:06 AM |
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In article <3c65f87.0308020443.55bc2e66@posting.google.com>,
jstevh@msn.com (James Harris) wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
This is probably a silly thing to point out, but if you divide off
*any* factor from a root _that has that factor_, as is stated above,
the result will indeed be an algebraic integer. This is from the
definition of what a factor is.
The problem, of course, is when you try to divide off things when the
factor _isn't_ present, because you think it _should_ be present for
some un-mathematical reason.
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| User: "C. Bond" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 08:58:45 AM |
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James Harris wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
If it is your claim that any of the three roots have 'non unit factors in
common with 5' then the burden is on you to produce and post them. Which
root has them? What are they? The roots of this polynomial are known and
have been posted. Now post the 'non unit factors in common with 5' which
you claim they include.
[snip psychotic techno-babble]
This post is, as usual, a red herring. Your so-called 'proof' of FLT has
been conclusively refuted and it cannot be redeemed by some off-the-wall
side trip. Get a grip.
--
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
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| User: "Gib Bogle" |
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| Title: Re: Problem is a math limitation |
02 Aug 2003 05:13:42 PM |
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C. Bond wrote:
James Harris wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
If it is your claim that any of the three roots have 'non unit factors in
common with 5' then the burden is on you to produce and post them. Which
root has them? What are they? The roots of this polynomial are known and
have been posted. Now post the 'non unit factors in common with 5' which
you claim they include.
JSH has _never_ (in my observation) responded to a request like this,
just as he _never_ responds to disproofs of his assertions. My guess is
that he doesn't respond because he is not mathematically equipped to do so.
Gib
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| User: "Norm Dresner" |
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| Title: I need a URL WAS Re: Problem is a math limitation |
02 Aug 2003 08:15:39 PM |
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"Gib Bogle" <bogle@too.much.spam.ihug.co.nz> wrote in message
news:bghd58$hvi$1@lust.ihug.co.nz...
C. Bond wrote:
James Harris wrote:
Blah , Blah, Blah
y^3 + 12y - 65
etc
The roots of this polynomial are known and
have been posted.
I apologize for not having read every last post, but I did miss this.
Can some kind soul please give me a URL for the post that includes these
roots?
Thanks
Norm
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| User: "C. Bond" |
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| Title: Re: I need a URL WAS Re: Problem is a math limitation |
03 Aug 2003 10:26:38 AM |
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Norm Dresner wrote:
"Gib Bogle" <bogle@too.much.spam.ihug.co.nz> wrote in message
news:bghd58$hvi$1@lust.ihug.co.nz...
C. Bond wrote:
James Harris wrote:
Blah , Blah, Blah
y^3 + 12y - 65
etc
The roots of this polynomial are known and
have been posted.
I apologize for not having read every last post, but I did miss this.
Can some kind soul please give me a URL for the post that includes these
roots?
Thanks
Norm
There were a number of posts on this subject which a google search will turn
up. However, be advised that James is very sloppy in his posts and the signs
and exponents of the polynomials he writes change all the time. At the time
I posted the roots, the polynomial was: x^3 + 12x^2 - 65. Note that the
degree of the term with coefficient = 12 is different from James' post
above, and that his post above represents another one of his careless
errors. Anyway, here is an excerpt from one post giving the roots of x^3 +
12x^2 = 65:
The exact values of a1, a2 and a3 for the polynomial problem you cite are
given below.
Let w = (1/3)ArcTan[Sqrt[12415]/63], a number whose value is approx.
0.352065133...
Then,
a1 = -4-8Cos[w] (approx. -11.50930064..)
a2 = -4+4Cos[w]+4Sqrt[3]Sin[w] (approx. 2.143751171..)
a3 = -4+4Cos[w]-4Sqrt[3]Sin[w] (approx. -2.634450526..)
not necessarily in any preferred order.
HTH.
--
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
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| User: "C. Bond" |
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| Title: Re: Problem is a math limitation |
03 Aug 2003 09:17:42 AM |
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James Harris wrote:
"C. Bond" <cbond@ix.netcom.com> wrote in message news:<3F2BC395.E32F3C7E@ix.netcom.com>...
James Harris wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
If it is your claim that any of the three roots have 'non unit factors in
common with 5' then the burden is on you to produce and post them. Which
root has them? What are they? The roots of this polynomial are known and
have been posted. Now post the 'non unit factors in common with 5' which
you claim they include.
The posters statements are irrational.
On the contrary. It is you who are irrational. (See below)
Given that the last
coefficient is -65, the three roots, which I'll call r_1, r_2, and r_3
multiply to give 65.
That is,
r_1 r_2 r_3 = 65
and what I've pointed out is that mathematicians failed to ever check
and see what would happen if you divided off 5, to get
r_1 r_2 r_3/5 = 13.
That is, mathematicians are incapable of explicitly showing 'a' and
'b' where
y^3 + ay^2 + by - 13
where the three roots are those that would result from dividing out
factors in common with 5 from the roots of y^3 + 12y - 65.
You have claimed (again) that the roots of y^3 + 12y - 65 have "factors in common with 5",
presumably within the ring of algebraic integers, but you have not proven that. In fact, none
of these roots have "factors in common with 5" within the ring of algebraic integers.
It's a fascinatingly simply problem to demonstrate, which
mathematicians missed for over a hundred years.
You haven't demonstrated anything. You merely made a claim (something like the claim that
there are unicorns in the zoo).
[snip psychotic techno-babble]
Attacking the messenger is not a good idea.
You aren't being attacked -- you are being exposed.
Physicists may be very forgiving to mathematicians, but clearly now
mathematicians are in a position where they need to give facts, not
personal attacks.
The facts are there, on the mathematician's side. The unsupported claims are also there, on
your side.
This post is, as usual, a red herring. Your so-called 'proof' of FLT has
been conclusively refuted and it cannot be redeemed by some off-the-wall
side trip. Get a grip.
If so, give 'a' and 'b' for the polynomial I've described.
Give me the an algebraic integer which divides the roots of your polynomial and the constant
5. Otherwise, shut up.
There's no reason for debate, histrionics or personal attacks.
There's no debate. You made a claim. It's your responsibility to support it.
What I want to remind though is that I've *proven* my case using
rather basic algebraic methods, which posters have challenged, when
pushing their challenge logically shows the problem I've described.
They got away with pointing people in the wrong direction.
You have not proven your case. Your "proof" has been completely refuted.
I suggest those of you who wish to know the truth read my paper
Advanced Polynomial Factorization.
James Harris
They'd be better off reading Grimm's Fairy Tales.
--
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
.
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| User: "Nora Baron" |
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| Title: Re: Problem is a math limitation |
03 Aug 2003 12:37:07 PM |
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"C. Bond" <cbond@ix.netcom.com> wrote in message news:<3F2D1986.A0ED939A@ix.netcom.com>...
James Harris wrote:
"C. Bond" <cbond@ix.netcom.com> wrote in message news:<3F2BC395.E32F3C7E@ix.netcom.com>...
James Harris wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
If it is your claim that any of the three roots have 'non unit factors in
common with 5' then the burden is on you to produce and post them. Which
root has them? What are they? The roots of this polynomial are known and
have been posted. Now post the 'non unit factors in common with 5' which
you claim they include.
The posters statements are irrational.
On the contrary. It is you who are irrational. (See below)
Given that the last
coefficient is -65, the three roots, which I'll call r_1, r_2, and r_3
multiply to give 65.
That is,
r_1 r_2 r_3 = 65
and what I've pointed out is that mathematicians failed to ever check
and see what would happen if you divided off 5, to get
r_1 r_2 r_3/5 = 13.
That is, mathematicians are incapable of explicitly showing 'a' and
'b' where
y^3 + ay^2 + by - 13
where the three roots are those that would result from dividing out
factors in common with 5 from the roots of y^3 + 12y - 65.
You have claimed (again) that the roots of y^3 + 12y - 65 have "factors in common with 5",
presumably within the ring of algebraic integers, but you have not proven that. In fact, none
of these roots have "factors in common with 5" within the ring of algebraic integers.
Just to show that not all of us mathematicians are united in
one big conspiracy to deny the Truth of Harris's argument, I am
going to have to disagree with you on this. Let r1, r2, and r3
be the roots. The fact that
r1*r2*r3 = 65
implies that at least one of the roots DOES have an algebraic-
integer factor in common with 5. Put another way, not all of
r1, r2, and r3 can be coprime to 5 in the ring of algebraic
integers. Put another way (though this requires some deeper
math), if r1 is a root which is not coprime to 5, there exist
nonunit algebraic integers s, t, and v such that
r1 = s * v and
5 = t * v.
In fact, as been shown by several of us in several different
ways, NONE of r1, r2, or r3 is coprime to 5 in the
ring of algebraic integers.
Side note: Harris is wrong that no one can write down the
polynomial that he describes. It has been done by W. Dale Hall
in another thread. Contrary to what Harris claims, the constant
term is 13^3, not 13 itself.
It's a fascinatingly simply problem to demonstrate, which
mathematicians missed for over a hundred years.
You haven't demonstrated anything. You merely made a claim (something like the claim that
there are unicorns in the zoo).
[snip psychotic techno-babble]
Attacking the messenger is not a good idea.
You aren't being attacked -- you are being exposed.
Physicists may be very forgiving to mathematicians, but clearly now
mathematicians are in a position where they need to give facts, not
personal attacks.
The facts are there, on the mathematician's side. The unsupported claims are also there, on
your side.
This post is, as usual, a red herring. Your so-called 'proof' of FLT has
been conclusively refuted and it cannot be redeemed by some off-the-wall
side trip. Get a grip.
If so, give 'a' and 'b' for the polynomial I've described.
Give me the an algebraic integer which divides the roots of your polynomial and the constant
5. Otherwise, shut up.
You have mixed this up a bit. What Harris claims (incorrectly)
is that sqrt(5) divides two of the roots, and the third root is
coprime to 5 in the algebraic integers. David Ullrich has given
a nice simple proof that this is wrong. However other proofs have
been given that EACH of the roots has an algebraic integer factor
in common with 5. I am not sure that anyone has specified exactly
what those factors are. They are roots of monic polynomials, possibly
of degree higher than 3, and in general it may not be possible to
write down the roots themselves. However, the monic polynomials
of which they are roots can be calculated.
Nora B.
There's no reason for debate, histrionics or personal attacks.
There's no debate. You made a claim. It's your responsibility to support it.
What I want to remind though is that I've *proven* my case using
rather basic algebraic methods, which posters have challenged, when
pushing their challenge logically shows the problem I've described.
They got away with pointing people in the wrong direction.
You have not proven your case. Your "proof" has been completely refuted.
I suggest those of you who wish to know the truth read my paper
Advanced Polynomial Factorization.
James Harris
They'd be better off reading Grimm's Fairy Tales.
--
There are two things you must never attempt to prove: the unprovable -- and the obvious.
--
Democracy: The triumph of popularity over principle.
.
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| User: "C. Bond" |
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| Title: Re: Problem is a math limitation |
03 Aug 2003 04:47:41 PM |
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Nora Baron wrote:
"C. Bond" <cbond@ix.netcom.com> wrote in message news:<3F2D1986.A0ED939A@ix.netcom.com>...
James Harris wrote:
"C. Bond" <cbond@ix.netcom.com> wrote in message news:<3F2BC395.E32F3C7E@ix.netcom.com>...
James Harris wrote:
Turns out that with all the arguing a rather direct limitation of
modern mathematics was being missed, which is that given the roots of
a polynomial like
y^3 + 12y - 65
mathematicians are incapable of proving that if you divide off, any
non unit factors in common with 5 from a root that has them that the
result is an algebraic integer.
If it is your claim that any of the three roots have 'non unit factors in
common with 5' then the burden is on you to produce and post them. Which
root has them? What are they? The roots of this polynomial are known and
have been posted. Now post the 'non unit factors in common with 5' which
you claim they include.
The posters statements are irrational.
On the contrary. It is you who are irrational. (See below)
Given that the last
coefficient is -65, the three roots, which I'll call r_1, r_2, and r_3
multiply to give 65.
That is,
r_1 r_2 r_3 = 65
and what I've pointed out is that mathematicians failed to ever check
and see what would happen if you divided off 5, to get
r_1 r_2 r_3/5 = 13.
That is, mathematicians are incapable of explicitly showing 'a' and
'b' where
y^3 + ay^2 + by - 13
where the three roots are those that would result from dividing out
factors in common with 5 from the roots of y^3 + 12y - 65.
You have claimed (again) that the roots of y^3 + 12y - 65 have "factors in common with 5",
presumably within the ring of algebraic integers, but you have not proven that. In fact, none
of these roots have "factors in common with 5" within the ring of algebraic integers.
Just to show that not all of us mathematicians are united in
one big conspiracy to deny the Truth of Harris's argument, I am
going to have to disagree with you on this. Let r1, r2, and r3
be the roots. The fact that
r1*r2*r3 = 65
implies that at least one of the roots DOES have an algebraic-
integer factor in common with 5. Put another way, not all of
r1, r2, and r3 can be coprime to 5 in the ring of algebraic
integers. Put another way (though this requires some deeper
math), if r1 is a root which is not coprime to 5, there exist
nonunit algebraic integers s, t, and v such that
r1 = s * v and
5 = t * v.
In fact, as been shown by several of us in several different
ways, NONE of r1, r2, or r3 is coprime to 5 in the
ring of algebraic integers.
Side note: Harris is wrong that no one can write down the
polynomial that he describes. It has been done by W. Dale Hall
in another thread. Contrary to what Harris claims, the constant
term is 13^3, not 13 itself.
It's a fascinatingly simply problem to demonstrate, which
mathematicians missed for over a hundred years.
You haven't demonstrated anything. You merely made a claim (something like the claim that
there are unicorns in the zoo).
[snip psychotic techno-babble]
Attacking the messenger is not a good idea.
You aren't being attacked -- you are being exposed.
Physicists may be very forgiving to mathematicians, but clearly now
mathematicians are in a position where they need to give facts, not
personal attacks.
The facts are there, on the mathematician's side. The unsupported claims are also there, on
your side.
This post is, as usual, a red herring. Your so-called 'proof' of FLT has
been conclusively refuted and it cannot be redeemed by some off-the-wall
side trip. Get a grip.
If so, give 'a' and 'b' for the polynomial I've described.
Give me the an algebraic integer which divides the roots of your polynomial and the constant
5. Otherwise, shut up.
You have mixed this up a bit. What Harris claims (incorrectly)
is that sqrt(5) divides two of the roots, and the third root is
coprime to 5 in the algebraic integers.
Boy, I *am* mixed up!. What I saw was (pasted from above):
[JSH]
Given that the last
coefficient is -65, the three roots, which I'll call r_1, r_2, and r_3
multiply to give 65.
That is,
r_1 r_2 r_3 = 65
and what I've pointed out is that mathematicians failed to ever check
and see what would happen if you divided off 5, to get
r_1 r_2 r_3/5 = 13.
That is, mathematicians are incapable of explicitly showing 'a' and
'b' where
y^3 + ay^2 + by - 13
where the three roots are those that would result from dividing out
factors in common with 5 from the roots of y^3 + 12y - 65.
David Ullrich has given
a nice simple proof that this is wrong. However other proofs have
been given that EACH of the roots has an algebraic integer factor
in common with 5. I am not sure that anyone has specified exactly
what those factors are. They are roots of monic polynomials, possibly
of degree higher than 3, and in general it may not be possible to
write down the roots themselves. However, the monic polynomials
of which they are roots can be calculated.
Nora B.
Thanks for the 'heads up'. Maybe posting those polynomials you mentioned would be worth the
effort.--
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
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| User: "W. Dale Hall" |
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| Title: Re: Problem is a math limitation (clarification) |
03 Aug 2003 10:53:50 AM |
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W. Dale Hall wrote:
... a buncha stuff deleted ...
I need to make a bit of a clarification, due to my sloppy terminology.
I was rolling along writing this stuff:
For the present case,
y^3 + 12 y - 65,
... rambling discourse omitted ...
I was able to find the minimal polynomials of q and s. Since I don't
really care about the remaining factor(s) of 5 (plural, since each
choice of p gives a separate choice for r=g, and thus for s), I'll only
provide the minimal polynomial of q:
Irred(q) : x^3 + 135 x^2 + 12315 x + 2197.
Notice the constant 2197: it's 13^3.
and came to this part, where I attempted to explain why the
minimal polynomial of g, which [if you've been following] is
gcd(p^3, 5^4), where p is a root of
x^3 + 12 x - 65
and 5 is, well "five". I think I found that p and 5 don't have
any nontrivial common divisors in the integers of Q(p), but p^3
and 5^3 generate a principal ideal <g>.
I'm trying to explain why it is that, although the precise value
of g is a function of p, its minimal polynomial is *independent*
of which root p you choose:
Why should that be the case? Here's my understanding:
Each choice of a root p has the ideal <p, 5>^3 principal, generated
by the element
g = 4 p^2 + 12 p + 85.
of the field Q(p). While that element *does* change with p, and while
the element q = -4 p^2 + 8 p + 13 *does* change with p, the minimal
polynomial of q *does not* change with p. Because it's characteristic
of the field Q(p), if that polynomial were to change with p, then we
would be able to distinguish Q(p1) from Q(p2), with p1, p2 distinct
roots of the original polynomial. However, the fields Q(p1) and Q(p2)
are canonically isomorphic in a way that fixes the elements of Q.
The clumsy terminology here was in my saying that the minimal polynomial
of g was "characteristic" of Q(p). Not *the* characteristic of Q(p).
See, clumsy terminology.
What I meant to be saying was that the field Q(p), or rather its ring of
integers O_p somehow *determined* that minimal polynomial, and while p
certainly plays a part in that calculation, it does so in a way that is
carried over from one O_p1 to another O_p2 [p1, p2 roots of the original
polynomial], under the action of the Galois group of its splitting
field: that is, you can do the arithmetic showing g1 divides p1^3
and 5^3, all in one extension Q(p1), and apply Gal( x^3+12x-65 ) to
get the arithmetic showing g2 divides p2^3 and 5^3, over in Q(p2).
The actual minimal polynomial of the quotient
p/gcd(p,5)
would be found from the above cubic Irred(q), by replacing x with x^3,
since q generates <p, 5>^3:
Irred(p/gcd(p,5)) = x^9 + 135 x^6 + 12315 x^3 + 2197.
We get 13^3, since the product of the values p_i/gcd(p_i,5) for i=1:3
should yield 13; since we're led to a cube root above, we obtain 3
factors of 13 (one for each of the 3 cube roots).
The rest of the discussion is deleted, since I've already taken enough
room.
Dale
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| User: "W. Dale Hall" |
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| Title: Re: Problem is a math limitation (clarification) |
03 Aug 2003 11:14:03 AM |
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Damned sloppy proofreading (mine, of course).
Perhaps *this* is correct:
W. Dale Hall wrote:
... you've seen all this already ...
The clumsy terminology here was in my saying that the minimal polynomial
of g was "characteristic" of Q(p). Not *the* characteristic of Q(p).
See, clumsy terminology.
It should have been the minimal polynomial of q = p^3/g. Sorry about
that.
... the rest of an already too-long ramble omitted ...
Dale
Dale
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| User: "C. Bond" |
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| Title: Re: Problem is a math limitation (clarification) |
03 Aug 2003 11:58:35 AM |
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W. Dale Hall wrote:
Damned sloppy proofreading (mine, of course).
Perhaps *this* is correct:
W. Dale Hall wrote:
... you've seen all this already ...
The clumsy terminology here was in my saying that the minimal polynomial
of g was "characteristic" of Q(p). Not *the* characteristic of Q(p).
See, clumsy terminology.
It should have been the minimal polynomial of q = p^3/g. Sorry about
that.
... the rest of an already too-long ramble omitted ...
Dale
Dale
I think James 'slipped you a mickey' when he posted the polynomial y^3 + 12y -
65. It should be, y^3 + 12y^2 - 65. Recall that it was derived from
65x^3-12x-1, and in previous, long past, posts the term with the coefficient
12 was y^2.
--
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
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