| Topic: |
Science > Physics |
| User: |
"James Harris" |
| Date: |
08 Nov 2003 02:53:38 PM |
| Object: |
Factorization dispute, again |
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
You know, it's like how in integers 1/2 doesn't exist. It's not an
integer, so it's not in the ring.
So you see, my argument is correct and simple, and mathematicians are
indeed running from a little gut check in their field. They're
pussies too scared to handle the truth.
But you should also understand, some people will be able to see that,
which is part of my plan. I can let mathematicians destroy themselves
proving they can't be trusted based on what they *see*, while they
forget what they can't see: the wearing down of the mathematician
mystique.
James Harris
http://mathforprofit.blogspot.com/
.
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| User: "flip" |
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| Title: [JSH] [Lunatic-Crank-Reposted Nonsense] Re: Factorization dispute, again |
08 Nov 2003 04:05:40 PM |
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"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0311081253.75c36043@posting.google.com...
So you see, my argument is correct and simple, and mathematicians are
indeed running from a little gut check in their field. They're
pussies too scared to handle the truth.
Mr. Harris,
still thinking that pissing out this flawed argument will ever make it true.
You are so sad and your "so-called" math work is even sadder!
Happy pissing!
.
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| User: "hantheman" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 07:02:10 AM |
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(James Harris) wrote in message news:<3c65f87.0311081253.75c36043@posting.google.com>...
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
Yeah, yeah. Weren't you supposed to leave this forum?
I guess the troll needs to be fed again.
.
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| User: "Mark Burlingame" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 10:16:35 PM |
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Here's a nugget for the "did you ever notice" bin:
Did you ever notice, how Mr. Harris gives refutations based on general
relations, whereas those who dispute his claims give exact, concrete
counter-examples?
MB
"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0311081253.75c36043@posting.google.com...
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
You know, it's like how in integers 1/2 doesn't exist. It's not an
integer, so it's not in the ring.
So you see, my argument is correct and simple, and mathematicians are
indeed running from a little gut check in their field. They're
pussies too scared to handle the truth.
But you should also understand, some people will be able to see that,
which is part of my plan. I can let mathematicians destroy themselves
proving they can't be trusted based on what they *see*, while they
forget what they can't see: the wearing down of the mathematician
mystique.
James Harris
http://mathforprofit.blogspot.com/
.
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| User: "Randy Poe" |
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| Title: Re: Factorization dispute, again |
11 Nov 2003 08:55:29 AM |
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"Mark Burlingame" <mbrennem@vt.edu> wrote in message news:<bopo17$2fkv$1@msunews.cl.msu.edu>...
Here's a nugget for the "did you ever notice" bin:
Did you ever notice, how Mr. Harris gives refutations based on general
relations, whereas those who dispute his claims give exact, concrete
counter-examples?
There are plenty of examples of the converse as well,
where somebody proves that something in general need not
be true, but James comes up with an example where it is
true and claims that proves the general result.
- Randy
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| User: "Mark Burlingame" |
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| Title: Re: Factorization dispute, again |
12 Nov 2003 08:06:24 AM |
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I guess my point is that this is part of his problem: he has a bad tendency
not to check calculations and to see if what he is deriving is actually
correct (and for a physicist this is odd).
Sure, the technique he uses is correct IF USED BY SOMEONE WHO IS COMPETENT
AND KNOWS HOW TO USE IT. But he isn't, and he doesn't. If I wanted to
disabuse someone of something in a manner which was clear and indisputable,
I would provide a simple direct counter-example. IF JSH tried to do this, he
would often realize that he is wrong (since he wouldn't be able to come up
with one, and his examples would show him that he ain't gettin what he
thinks he should be gettin), and prevent these back and forth posts ad
nauseum that end up with him realizing after weeks that he's made some
stupid assumption/error.
MB
"Mark Burlingame" <mbrennem@vt.edu> wrote in message
news:bopo17$2fkv$1@msunews.cl.msu.edu...
Here's a nugget for the "did you ever notice" bin:
Did you ever notice, how Mr. Harris gives refutations based on general
relations, whereas those who dispute his claims give exact, concrete
counter-examples?
MB
"James Harris" <jstevh@msn.com> wrote in message
news:3c65f87.0311081253.75c36043@posting.google.com...
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
You know, it's like how in integers 1/2 doesn't exist. It's not an
integer, so it's not in the ring.
So you see, my argument is correct and simple, and mathematicians are
indeed running from a little gut check in their field. They're
pussies too scared to handle the truth.
But you should also understand, some people will be able to see that,
which is part of my plan. I can let mathematicians destroy themselves
proving they can't be trusted based on what they *see*, while they
forget what they can't see: the wearing down of the mathematician
mystique.
James Harris
http://mathforprofit.blogspot.com/
.
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| User: "Dik T. Winter" |
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| Title: Re: Factorization dispute, again |
08 Nov 2003 06:39:10 PM |
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In article <3c65f87.0311081253.75c36043@posting.google.com> (James Harris) writes:
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so
it can be divided by 2. Your reasoning about "constant term" would
lead to the result P(x)/2 = (x + 1)(x/2 + 1), because now the
"constant terms" match. My position is that there are functions w1(x)
and w2(x), defined as follows:
w1(x) = gcd(x + 1, 2)
w2(x) = gcd(x + 2, 2)
such that
P(x)/2 = [(x + 1)/w1(x)] * [(x + 2)/w2(x)]
is a valid factorisation. According to your reasoning above (paraphrase):
"...as you may think it doesn't matter if w1(x) has some factor of 2,
despite *seeing* (1/w1(x)) but you see, as 2 is coprime to 1 in the
ring of integers, if w1(x) isn't coprime to 2, (1/w1(x)) does not exist
in the ring."
So, although both factors in the factorisation are integer, according to
you it is not a valid factorisation, so the ring of integers is flawed.
So you see, my argument is correct and simple, and mathematicians are
indeed running from a little gut check in their field. They're
pussies too scared to handle the truth.
So, it is your thinking that the ring of integers is flawed?
With your polynomial the situation is similar. I have given pretty
*explicit* definitions of the functions w1(x) to w3(x) such that
(5 a1(x) + 7)/w1(x) ,
(5 a2(x) + 7)/w2(x) and
(5 b3(x) + 22)/w3(x)
are all algebraic integer for all integer x.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
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| User: "James Harris" |
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| Title: Re: Factorization dispute, again |
09 Nov 2003 07:39:04 AM |
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"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
In article <3c65f87.0311081253.75c36043@posting.google.com> (James Harris) writes:
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so
it can be divided by 2. <deleted>
Why? What I've done is do a basic simplification going from
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
to (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22,
which is done simply enough by multiplying everything out and letting
the terms with x as a factor cancel each other out.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You're a crank Dik Winter, who has been refuted quite simply but you
keep talking as if convincing *others* changes mathematical truth.
It does not.
James Harris
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| User: "William Hughes" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 12:51:57 AM |
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(James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You keep saying that 22/w_3(x) need not be in the ring.
No-one is disputing this. The problem is that you seem to think
that the fact that 22/w_3(x) need not be in the ring is of great
importance. Indeed, you seem to assume that it is obvious why
22/w_3(x) must be in the ring. It appears that you are reasoning thus:
The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x))
cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)
The constant term must be in the ring, so 22/w_3(x)
must be in the ring.
The problem is that the constant term of (b_3(x)/w_3(x) + 22/w_3(x))
is 22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in the
ring for values of x other than 0 does not matter.
- "William Hughes"
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| User: "James Harris" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 08:34:23 AM |
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(William Hughes) wrote in message news:<4d5e4663.0311092251.58373fe2@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You keep saying that 22/w_3(x) need not be in the ring.
No-one is disputing this. The problem is that you seem to think
that the fact that 22/w_3(x) need not be in the ring is of great
importance. Indeed, you seem to assume that it is obvious why
22/w_3(x) must be in the ring. It appears that you are reasoning thus:
The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x))
cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)
The constant term must be in the ring, so 22/w_3(x)
must be in the ring.
You're lying William Hughes, as my exact reasoning was posted. I just
multiply out the factorization, and simplify.
Now readers should note that posters like William Hughes are cranks,
so of course they have to ignore the actual facts, but consider what I
posted before:
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
The problem is that the constant term of (b_3(x)/w_3(x) + 22/w_3(x))
is 22/w_3(0)= 22/1 = 22. Whether or not 22/w_3(x) is in the
ring for values of x other than 0 does not matter.
- "William Hughes"
Denial of basic algebra is such a sad thing to display to the world as
William Hughes demonstrates a rather odd irrationality, to all those
readers who go through my posts.
After all, I'm just talking about multiplying out and simplifying, and
I'll post again because these cranks have a bad habit of creative
deletion what I said previously:
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring
of algebraic integers if w_3(x) isn't coprime to 22.
And for those of you who wondered, yes, when faced with a *very* basic
refutation of their positions posters who argue with me tend to just
ignore the truth. Later they come back with the same position.
They're irrational and are the real cranks as you can't reason with
them.
They believe false things but want desperately to believe and be
believed.
James Harris
http://mathforprofit.blogspot.com/
.
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| User: "William Hughes" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 01:06:39 PM |
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(James Harris) wrote in message news:<3c65f87.0311100634.4def506@posting.google.com>...
wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0311092251.58373fe2@posting.google.com>...
(James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You keep saying that 22/w_3(x) need not be in the ring.
No-one is disputing this. The problem is that you seem to think
that the fact that 22/w_3(x) need not be in the ring is of great
importance. Indeed, you seem to assume that it is obvious why
22/w_3(x) must be in the ring. It appears that you are reasoning thus:
The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x))
cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)
The constant term must be in the ring, so 22/w_3(x)
must be in the ring.
You're lying William Hughes, as my exact reasoning was posted. I just
multiply out the factorization, and simplify.
Your reasoning that 22/w_3(x) need not be in the ring has been posted
many, many times. No-one disagrees with your conclusion.
Your reasoning that 22/w_3(x) must be in the ring has never been posted.
The above is my guess as to what your reasoning is.
Quiz time:
What is the constant term of the following three functions?
U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7)
V(x) = sqrt(x^2 + x)/7 + 7/7
W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7)
(hint: The constant term of b(x) is b(0) )
- "William Hughes"
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| User: "James Harris" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 04:49:34 PM |
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(William Hughes) wrote in message news:<4d5e4663.0311101106.55f4a56a@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311100634.4def506@posting.google.com>...
(William Hughes) wrote in message news:<4d5e4663.0311092251.58373fe2@posting.google.com>...
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You keep saying that 22/w_3(x) need not be in the ring.
No-one is disputing this. The problem is that you seem to think
that the fact that 22/w_3(x) need not be in the ring is of great
importance. Indeed, you seem to assume that it is obvious why
22/w_3(x) must be in the ring. It appears that you are reasoning thus:
The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x))
cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)
The constant term must be in the ring, so 22/w_3(x)
must be in the ring.
You're lying William Hughes, as my exact reasoning was posted. I just
multiply out the factorization, and simplify.
Your reasoning that 22/w_3(x) need not be in the ring has been posted
many, many times. No-one disagrees with your conclusion.
Actually, it IS in the ring, which is my point William Hughes. And
you deleted out the factorization and simplification which shows that
fact!!!
Extraordinary behavior which is indeed crank.
I give you the information, explain it clearly, and you try to just
ignore it.
Your reasoning that 22/w_3(x) must be in the ring has never been posted.
The above is my guess as to what your reasoning is.
The assertions are over the ring of algebraic integers, so operations
are to be in that ring. I show that you're pushed out of that ring in
a surprising way.
Quiz time:
What is the constant term of the following three functions?
Irrelevant to the issue of 22/w_3(x) having to be in the ring of
algebraic integers.
U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7)
V(x) = sqrt(x^2 + x)/7 + 7/7
W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7)
(hint: The constant term of b(x) is b(0) )
- "William Hughes"
What I did was multiply out the factorization and simplify to show
that it's impossible for w_3(x) to not be coprime to 7 as then
22/w_3(x) is NOT in the ring of algebraic integers.
It's simple, direct, and basic, and it refutes attempts at claiming
that w_3(x) can share non-unit factors with 7 in the ring of algebraic
integers.
James Harris
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| User: "William Hughes" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 09:33:39 PM |
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(James Harris) wrote in message news:<3c65f87.0311101449.33a9689a@posting.google.com>...
wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0311101106.55f4a56a@posting.google.com>...
(James Harris) wrote in message news:<3c65f87.0311100634.4def506@posting.google.com>...
wpihughes@hotmail.com (William Hughes) wrote in message news:<4d5e4663.0311092251.58373fe2@posting.google.com>...
(James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You keep saying that 22/w_3(x) need not be in the ring.
No-one is disputing this. The problem is that you seem to think
that the fact that 22/w_3(x) need not be in the ring is of great
importance. Indeed, you seem to assume that it is obvious why
22/w_3(x) must be in the ring. It appears that you are reasoning thus:
The constant term of (b_3(x) + 22) is 22
The constant term of (b_3(x)/w_3(x) + 22/w_3(x))
cannot be b_3(x)/w_3(x) so it must be 22/w_3(x)
The constant term must be in the ring, so 22/w_3(x)
must be in the ring.
You're lying William Hughes, as my exact reasoning was posted. I just
multiply out the factorization, and simplify.
Your reasoning that 22/w_3(x) need not be in the ring has been posted
many, many times. No-one disagrees with your conclusion.
Actually, it IS in the ring, which is my point William Hughes. And
you deleted out the factorization and simplification which shows that
fact!!!
You seem to be claiming that
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
shows that (22/2_3(x)) is in the ring of algebraic integers.
Actually, it doesn't show anything.
Either by using the identity
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
and the fact that (w_1(x))(w_2(x))(w_3(x)) = 49
or by direct use of the fact that (w_1(x))(w_2(x))(w_3(x)) = 49
we obtain
(7)(7)(22) / ((w_1(x))(w_2(x))(w_3(x))) = 22
in the ring of algebraic integers. We can rewrite this
in the complex numbers as
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
but this no more shows that (22/w_3(x)) is in the ring
of algebraic integers than
(6)(35)(99)/((15)(18)(7)) =(6/15)(35/18)(99/7) = 11
shows that (99/7) is an integer.
You didn't attempt my quiz. Please try, It isn't
hard.
What is the constant term of the following three functions?
U(x) = sqrt(x^2 + x)/sqrt(x+7) + 7/sqrt(x+7)
V(x) = sqrt(x^2 + x)/7 + 7/7
W(x) = sqrt(x^2 +x)/(x^2 +2x +7) + 7/(x^2 +2x +7)
(hint: The constant term of b(x) is b(0) )
- "William Hughes"
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| User: "Dik T. Winter" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 05:59:28 PM |
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In article <3c65f87.0311101449.33a9689a@posting.google.com> (James Harris) writes:
....
The assertions are over the ring of algebraic integers, so operations
are to be in that ring. I show that you're pushed out of that ring in
a surprising way.
But so are you! Because a1(x) is *not* divisible in the algebraic integers.
You say that is a flaw of he algebraic integers. The question is, which
pushing out is the better push.
With:
h1(x) = gcd(5 a1(x) + 7, 49)
h2(x) = gcd(5 a2(x) + 7, 49)
h3(x) = gcd(5 b3(x) + 22, 49)
k(x) = 49/( h1(x).h2(x).h3(x) )
l1(x) = gcd(k(x), h1(x))
l2(x) = gcd(k(x)/l2(x), h2(x))
l3(x) = k(x)/( l1(x).l2(x) ) { Perhaps l3(x) = 1 for all x, just be save.}
w1(x) = h1(x)/l1(x)
w2(x) = h2(x)/l2(x)
w3(x) = h3(x)/l3(x)
where all functions defined above are functions from the algebraic
integers to the algebraic integers.
P(x)/49 = [(5 a1(x) + 7)/w1(x)] [(5 a2(x) + 7)/w2(x)] [(5 b3(x) + 22)/w3(x)]
all three factors are algebraic integers for *all* x. But when we set:
w1(x) = w2(x) = 7 and w3(x) = 1
in the above factorisation, we find that *not* all three factors are
algebraic integers.
That 22/w3(x) is *not* necessarily an algebraic integer is irrelevant.
You do not need to calculate that value. But *even if you wish to
calculate that intermediate result*, it is irrelevant.
In general the case is that when you say about a function P(x) that it
is a function from the ring of algebraic integers to the algebraic
integers, it is not necessary that all intermediate results must also
be in that ring, as long as the final result is in that ring. And
as functions, all three factors with the w's as I defined them *are*
functions from the algebraic integers to the algebraic integers. So
P(x) is factored in three functions that are from the algebraic
integers to the algebraic integers. No flaw of the algebraic integers
in view.
....
What I did was multiply out the factorization and simplify to show
that it's impossible for w_3(x) to not be coprime to 7 as then
22/w_3(x) is NOT in the ring of algebraic integers.
Yes, but there is *no* 22/w3(x) in the factorisation above. It
is (5 b3(x) + 22)/w3(x). Or do you claim that it is invalid "in
the integers" to say that (x + 5)/2 is an integer for odd integer x,
because (x + 5)/2 = (x/2 + 5/2) and 5/2 is not an integer?
A pretty strange restriction I would say.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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| User: "Arturo Magidin" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 06:11:23 PM |
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In article <Ho5un4.2H0@cwi.nl>, Dik T. Winter <Dik.Winter@cwi.nl> wrote:
[.snip.]
Or do you claim that it is invalid "in
the integers" to say that (x + 5)/2 is an integer for odd integer x,
because (x + 5)/2 = (x/2 + 5/2) and 5/2 is not an integer?
A pretty strange restriction I would say.
Perhaps a better example would be to take (x^2+3x+2)/2, which is
always an integer for integer value of x; you can certainly write
(x^2+3x+2)/2 = (x^2/2) + (3x/2) + 1
even though it is not always true that x^2/2 and 3x/2 are integers.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
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| User: "Dik T. Winter" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 05:27:05 PM |
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In article <3c65f87.0311100634.4def506@posting.google.com> (James Harris) writes:
....
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring
of algebraic integers if w_3(x) isn't coprime to 22.
Well, you know, the w's as I have defined them for you get
w1(x)w2(x)w3(x) = 49,
so I do not understand how it is possible that
(7/w1(x))(7/w2(x))(22/w3(x)) = 22.
(Note that in most cases this involves a short excursion to the
algebraic numbers...)
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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| User: "Dik T. Winter" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 06:36:51 PM |
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In article <Ho5t55.GJ1@cwi.nl> "Dik T. Winter" <Dik.Winter@cwi.nl> writes:
In article <3c65f87.0311100634.4def506@posting.google.com> (James Harris) writes:
...
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
You see, the cranks can't get arouund 22/w_3(x) NOT being in the ring
of algebraic integers if w_3(x) isn't coprime to 22.
Well, you know, the w's as I have defined them for you get
w1(x)w2(x)w3(x) = 49,
so I do not understand how it is possible that
Of course I intended "impossible" here.
(7/w1(x))(7/w2(x))(22/w3(x)) = 22.
(Note that in most cases this involves a short excursion to the
algebraic numbers...)
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
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| User: "Arturo Magidin" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 06:07:52 PM |
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In article <Ho5t55.GJ1@cwi.nl>, Dik T. Winter <Dik.Winter@cwi.nl> wrote:
[.snip.]
Well, you know, the w's as I have defined them for you get
w1(x)w2(x)w3(x) = 49,
so I do not understand how it is possible that
(7/w1(x))(7/w2(x))(22/w3(x)) = 22.
You mean, "how is it possible that [...] does not hold"?
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
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| User: "Dik T. Winter" |
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| Title: Re: Factorization dispute, again |
09 Nov 2003 09:01:18 PM |
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In article <3c65f87.0311090539.74f25eeb@posting.google.com> (James Harris) writes:
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote in message news:<Ho275A.4Bt@cwi.nl>...
In article <3c65f87.0311081253.75c36043@posting.google.com> (James Harris) writes:
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
But 22/w_3(x) *need* not exist in the ring. It is (5 b3(x)+22)/w3(x)
that must exist in the ring (and which does exist in the ring. I have
no idea where you got the idea that 22/w3(x) must be in the ring.
Consider P(x) = (x + 1)(x + 2), in the integers. It is always even, so
it can be divided by 2. <deleted>
Why?
Why? Because your reasoning would lead to the conclusion that the ring
of integers is flawed because it does not contain numbers that should
be in that ring. But you are afraid to answer the questions I posed in
the part you deleted.
Why? What I've done is do a basic simplification going from
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
to (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22,
which is done simply enough by multiplying everything out and letting
the terms with x as a factor cancel each other out.
Now you wish to avoid that Dik Winter because your assertions can't be
true in the ring of algebraic integers as if w_3(x) isn't coprime to
7, there's a contradiction, as 22/w_3(x) can't be in the ring then.
You *must* be stupid. As I wrote above, it is not necessary that 22/w3(x)
is in the algebraic integers. If you think so you should answer the
questions in the part you deleted. Because if 22/w3(x) must be in
the algebraic integers, for *exactly the same reasons* 1/2 must be
in the integers (see P(x) = (x + 1)(x + 2), which is divisible by 2
in the integers).
You're a crank Dik Winter, who has been refuted quite simply but you
keep talking as if convincing *others* changes mathematical truth.
You dodge all my questions, because you are afraid to answer them. Now,
who is the crank?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.
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| User: "Brian Quincy Hutchings" |
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| Title: Re: Factorization dispute, again |
10 Nov 2003 05:13:46 PM |
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who are these "*others*" that you refer to?... perhaps,
you have an audience that is not known to the rest of us,
the desingated Peanut Gallery of would-be critics & helpmeets.
or is it just the entire, 'virtual" crowd of all
of the folks who are in the googolplex,
that *could* lurk on your bifurcating threads -- or
maybe they "should?"
mea culpa, dood.
jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0311090539.74f25eeb@posting.google.com>...
keep talking as if convincing *others* changes mathematical truth.
--ils dcues d'Enron!
http://larouchepub.com/
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| User: "Alan Morgan" |
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| Title: Re: Factorization dispute, again |
09 Nov 2003 01:53:11 PM |
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In article <3c65f87.0311081253.75c36043@posting.google.com>,
James Harris <jstevh@msn.com> wrote:
Notice,
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where you see that the constant terms match as now you have 7(7)(22) =
1078, which is the constant term of the polynomial
49(300125 x^3 - 18375 x^2 - 360 x + 22).
Various people have debated me about what happens when you divide off
49, where for some odd reason, some of them seem to believe that you
can have
w_1(x), w_2(x), and w_3(x) such that w_1(x) w_2(x) w_3(x) = 49, and
(5 a_1(x) + 7)/w_1(x) (5 a_2(x) + 7)/w_2(x) (5 b_3(x) + 22)/w_3(x) =
300125 x^3 - 18375 x^2 - 360 x + 22
where the w's vary as x varies, which is a rather naive notion.
Why is it naive?
w1(x) = x+7
w2(x) = x^2+7
w3(x) = 22/((x+7)(x^2+6))
That's because you can multiply *everything* out, and simplify to get
(7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22
As can I (understand that I'm not claiming that these particular
functions will work for this problem, I'm just pointing that that
it isn't naive to assume that they don't have to be constant).
which should be simple enough for all of you.
Now those of you who usually work in the field of complex numbers may
think that it's not a big deal, as you may think it doesn't matter if
w_3(x) has some factor factor of 7, despite *seeing* (22/w_3(x)) but
you see, as 22 is coprime to 7 in the ring of algebraic integers, if
w_3(x) isn't coprime to 7, (22/w_3(x)) does not exist in the ring.
Doesn't matter. What matters is whether or not (5 b_3(x) + 22)/w_3(x)
exists in the ring.
Let's try a simpler example: f(x) = x(x+1) over the integers. Clearly
f(x) is even. So, can we assume that either x/2 is an integer or
(x+1)/2 is an integer? The answer is "yes", but you can't assume it
is always the same one. IOW, there exist a1 and a2 such that x/a1(x)
and (x+1)/a2(x) are both integers and a1(x)a2(x)=2. But you can't
assume that just because a1(0) = 2 and a2(0) = 1 that a1(x)=2 and
a2(x)=1.
And note that 1/2 doesn't exist in the ring, but that doesn't mean
that a2(x) can't equal 2 every now and then.
Alan
--
Defendit numerus
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| User: "Uncle Al" |
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| Title: Re: Factorization dispute, again |
08 Nov 2003 05:35:18 PM |
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James Harris wrote:
Notice,
[snip]
http://www.crank.net/harris.html
It's not every braying jackass that gets a whole page at crank.net
Stupid is as stupid does. There is no fixing stupid because it is not
broken. Harris would be much happier as a priest proclaiming god
amdist heretics and buggering altar boys to His greater glory.
Hey Harris, your village called: Its idiot is missing.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
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| User: "Gregory L. Hansen" |
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| Title: Re: Factorization dispute, again |
09 Nov 2003 07:24:57 AM |
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In article <3FAD7DB6.A06BA798@hate.spam.net>,
Uncle Al <UncleAl0@hate.spam.net> wrote:
There is no fixing stupid because it is not
broken.
What an intruguing comment! I'm not sure I get it, but it had me giggling
while I was tryin to figure it out.
--
"Let us learn to dream, gentlemen, then perhaps we shall find the
truth... But let us beware of publishing our dreams before they have been
put to the proof by the waking understanding." -- Friedrich August Kekulé
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