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| JSH... |
 Posted: Thu Jul 02, 2009 12:16 pm |
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Guest
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On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
But not even God can choose for you here if you go with mathematical
logic. Of course, you can like so many human beings before you simply
choose to believe in what you've been taught, even though it is very,
very wrong.
The mystical nature of the disagreement with me is remarkable but I
think it's natural for human beings to turn to spirits or mystical
forces in the face of something difficult to understand. That's what
your ancestors did.
But then you're no longer really mathematicians.
James Harris |
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| Bacle... |
| Posted: Thu Jul 02, 2009 1:24 pm |
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Guest
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[quote:17c8219b74]
Notice the ring of algebraic integers does not fail
gracefully and its
error is not fixable as your alternative is to
introduce some mystical
force to choose out of infinity, like, maybe God?
[/quote:17c8219b74]
[quote:17c8219b74]
(Posters have been remarkably vague in past arguments
about how a
particular factor of 7 arises, often simply repeating
what is taught
in math classes about ways to find the factors at any
particular x,
but remember there are an infinity of choices for
EVERY x, so the
choice problem gives you infinity at every point. So
who chooses?
God?)
But not even God can choose for you here if you go
with mathematical
logic. Of course, you can like so many human beings
before you simply
choose to believe in what you've been taught, even
though it is very,
very wrong.
The mystical nature of the disagreement with me is
remarkable but I
think it's natural for human beings to turn to
spirits or mystical
forces in the face of something difficult to
understand. That's what
your ancestors did.
And not those of _your_ inferior species, right?[/quote:17c8219b74]
[quote:17c8219b74]But then you're no longer really mathematicians.
Get that corn cob of your ass , loser. You will[/quote:17c8219b74]
never be a mathematician, notwithstanding the fact
that you have ripped off material from other
posters.
[quote:17c8219b74]James Harris
[/quote:17c8219b74]
Asshole. |
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| KMF... |
| Posted: Thu Jul 02, 2009 2:09 pm |
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Guest
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[quote:51b61ce474]
EVERY x, so the
choice problem gives you infinity at every point. So
who chooses?
God?)
[/quote:51b61ce474]
So you accuse others of having mystical beliefs and
you bring up god?. As a hypocrit, you are not very
subtle.
[quote:51b61ce474]
But not even God can choose for you here if you go
with mathematical
logic. Of course, you can like so many human beings
before you simply
choose to believe in what you've been taught, even
though it is very,
very wrong.
The mystical nature of the disagreement with me is
remarkable but I
think it's natural for human beings to turn to
spirits or mystical
forces in the face of something difficult to
understand. That's what
your ancestors did.
Asshole: weren't you the one constantly bring up[/quote:51b61ce474]
prophecies of doom, of cannibalism, etc.?.
Do explain the rational process by which you
came to predict them.
Liar. Hypocrit. Self-deluded. Coward.
[quote:51b61ce474]But then you're no longer really mathematicians.
You never were nor will be one. You don't even have[/quote:51b61ce474]
the physics degree you claim you have. You failed
back then, and keep failing now.
Looking forward to 15 more years of failures
and whining.
[quote:51b61ce474]James Harris
[/quote:51b61ce474]
Loser. |
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| Fishcake... |
| Posted: Thu Jul 02, 2009 7:08 pm |
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I got it!
I finally understand your problem.
You have no idea what "algebraic integers" are.
Now I understand why you are so confused. |
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| fishfry... |
| Posted: Thu Jul 02, 2009 9:06 pm |
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Guest
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In article
<c6471c0c-e30a-4d77-9fb3-4bd4270dffe1 at (no spam) x6g2000prc.googlegroups.com>,
JSH <jstevh at (no spam) gmail.com> wrote:
[quote:7acdf4baef]On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
[/quote:7acdf4baef]
Unfortunately, infinity has been called to Washington to help calculate
the national debt, and isn't available for comment. |
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| Mackly... |
| Posted: Thu Jul 02, 2009 10:51 pm |
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Guest
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"JSH" <jstevh at (no spam) gmail.com> wrote in message
news:c6471c0c-e30a-4d77-9fb3-4bd4270dffe1 at (no spam) x6g2000prc.googlegroups.com...
[quote:6fb9731e32]On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
[/quote:6fb9731e32]
trivial
[quote:6fb9731e32]
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
[/quote:6fb9731e32]
[quote:6fb9731e32]7(x^2 + 3x + 2) = (x + 1)(7x + 14)
[/quote:6fb9731e32]
7(x^2 + 3x + 2) = 7*(x + 1)(x + 2)
OK now what ?
[quote:6fb9731e32]
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
[/quote:6fb9731e32]
so what? your just multiplying some number by another
[quote:6fb9731e32]
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
[/quote:6fb9731e32]
and........
[quote:6fb9731e32]
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
[/quote:6fb9731e32]
yes, you like the 7..........
[quote:6fb9731e32]
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
[/quote:6fb9731e32]
could you make that sentance shorter?
[quote:6fb9731e32]
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
[/quote:6fb9731e32]
one 7 goes in, one 7 comes out
no contridiction!
[quote:6fb9731e32]
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
[/quote:6fb9731e32]
7 is no good. Use 19, its much better.
[quote:6fb9731e32]
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
[/quote:6fb9731e32]
it is the negitive root you keep missing. Negitive square root.
[quote:6fb9731e32]
But not even God can choose for you here if you go with mathematical
logic. Of course, you can like so many human beings before you simply
choose to believe in what you've been taught, even though it is very,
very wrong.
[/quote:6fb9731e32]
You may call me god, several others do.
[quote:6fb9731e32]
The mystical nature of the disagreement with me is remarkable but I
think it's natural for human beings to turn to spirits or mystical
forces in the face of something difficult to understand. That's what
your ancestors did.
[/quote:6fb9731e32]
your ancestors were all trolls.
[quote:6fb9731e32]
But then you're no longer really mathematicians.
James Harris
[/quote:6fb9731e32]
and you're a mathematiticn now |
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| Mark Murray... |
| Posted: Fri Jul 03, 2009 2:26 am |
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JSH wrote:
[quote:820501473a]a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
[/quote:820501473a]
Do you know what an Algebraic Integer _is_?
14 = 2 x 7 (two primes)
But, dividing 14 by an arbitrary algebraic integer gives
14/(5 - sqrt(3) i) = 5/2 + 1/2 sqrt(3) i
So
14 = (5 - sqrt(3) i)(5/2 + 1/2 sqrt(3) i) (the product of two
arbitrary algebraics
having no factors in
common with 2 or 7)
Any contradiction there?
M |
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| Bacle... |
| Posted: Fri Jul 03, 2009 9:24 am |
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And I have to fully agree with KMF above:
You question others' motives. How about
your motives?. Fame, recognition. |
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| marcus_b... |
| Posted: Sat Jul 04, 2009 8:19 am |
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Guest
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On Jul 2, 5:16 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote:88690071d7]On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
[/quote:88690071d7]
Here is a proof, for x = 1, that a_1(x) and a_2(x) are both
not divisible by 7.
Both a_1(1) and a_2(1) are roots of a^2 - 6a + 35.
We can let a_1(1) = 3 + sqrt(-26), a_2(1) = 3 - sqrt(-26)
Let u = -307 - 30 * sqrt(-26) and v = -307 + 30 * sqrt(-26)
Then the following can be readily checked by ordinary
arithmetic:
(1) a_1(1)^6 / u = -109 + 12 * sqrt(-26)
(2) a_2(1)^6 / v = -109 - 12 * sqrt(-26)
(3) u * v = 7^6
(4) a_1(1)^6 * a_2(1)^6 / (u * v) = 15625 = 5^6.
(5) u and v are both algebraic integers
(6) Neither u nor v are units in the ring of
algebraic integers.
Further, note that because of (4) above and because
5 and 7 are coprime, you can conclude that a_1(1)^6 / u
and a_2(1)^6 / v have no factors in common with 7.
Conclusions:
a_1(1)^6 has a factor in common with 7, namely u.
a_2(1)^6 has a factor in common with 7, namely v.
Neither a_1(1) nor a_2(1) are divisible by 7, since
neither u nor v are units and a_1(1)*a_2(1)/(u*v) = 5^6
is coprime to 7.
The key thing here is, if you have a proof contradicting
this, you are contradicting not some old theorems in algebraic
number theory or Galois theory. You are contradicting plain
old arithmetic. If there is a problem, it goes beyond just a
problem with the ring of algebraic numbers. If you are right,
you have arrived at an inconsistency in all of mathematics - you
have essentially proved that 1 + 1 = 3.
Either that, or your "proof" is wrong. Which do you think
is more likely?
Marcus.
[quote:88690071d7]But not even God can choose for you here if you go with mathematical
logic. Of course, you can like so many human beings before you simply
choose to believe in what you've been taught, even though it is very,
very wrong.
The mystical nature of the disagreement with me is remarkable but I
think it's natural for human beings to turn to spirits or mystical
forces in the face of something difficult to understand. That's what
your ancestors did.
But then you're no longer really mathematicians.
James Harris[/quote:88690071d7] |
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| JSH... |
| Posted: Sat Jul 04, 2009 3:15 pm |
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On Jul 4, 11:19 am, marcus_b <marcus_bruck... at (no spam) yahoo.com> wrote:
[quote:34338de19e]On Jul 2, 5:16 pm, JSH <jst... at (no spam) gmail.com> wrote:
On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
Here is a proof, for x = 1, that a_1(x) and a_2(x) are both
not divisible by 7.
Both a_1(1) and a_2(1) are roots of a^2 - 6a + 35.
We can let a_1(1) = 3 + sqrt(-26), a_2(1) = 3 - sqrt(-26)
Let u = -307 - 30 * sqrt(-26) and v = -307 + 30 * sqrt(-26)
Then the following can be readily checked by ordinary
arithmetic:
(1) a_1(1)^6 / u = -109 + 12 * sqrt(-26)
(2) a_2(1)^6 / v = -109 - 12 * sqrt(-26)
(3) u * v = 7^6
(4) a_1(1)^6 * a_2(1)^6 / (u * v) = 15625 = 5^6.
(5) u and v are both algebraic integers
(6) Neither u nor v are units in the ring of
algebraic integers.
Further, note that because of (4) above and because
5 and 7 are coprime, you can conclude that a_1(1)^6 / u
and a_2(1)^6 / v have no factors in common with 7.
Conclusions:
a_1(1)^6 has a factor in common with 7, namely u.
a_2(1)^6 has a factor in common with 7, namely v.
Neither a_1(1) nor a_2(1) are divisible by 7, since
neither u nor v are units and a_1(1)*a_2(1)/(u*v) = 5^6
is coprime to 7.
The key thing here is, if you have a proof contradicting
this, you are contradicting not some old theorems in algebraic
number theory or Galois theory. You are contradicting plain
old arithmetic. If there is a problem, it goes beyond just a
[/quote:34338de19e]
Nope.
The same techniques you just used would work to "prove" that neither
of the roots of
x^2 + 3x + 2 = 0
has 2 as a factor--if you don't resolve the square root.
Trivial.
The algebra is easy but the psychology is complex as the error allows
people like you to do pretend mathematics!
Your very sense of self may be wrapped up in the bogus math that you
do, so it's an uphill slog to get you past your denial, into the world
of doing valid mathematics.
James Harris |
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| Fishcake... |
| Posted: Sat Jul 04, 2009 6:03 pm |
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Guest
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[quote:96a7ed6eb2]On Jul 4, 11:19Â am, marcus_b
marcus_bruck... at (no spam) yahoo.com> wrote:
On Jul 2, 5:16 pm, JSH <jst... at (no spam) gmail.com> wrote:
On my math blog Penny Hassett had a comment that
got me to thinking
and after pondering at her suggestion what others
believe to be true
in their disagreements with me about a certain
serious mathematical
issue, I concluded that infinity could help to
clear the air.
So I have simple examples yet again to start
(believe me it gets
really hard later so please pay close attention
to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic
integers. Â And notice that
reflects CHOICE. Â My choice for a simple example
as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact
there are an
infinity of possible variations all equally
valid.
Key here also to notice is that there are an
infinity of choices FOR
EVERY x, and remember the ring is the ring of
algebraic integers.
So far easy and there should be no disagreement
with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the
ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) +
2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0,
both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and
you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) +
14)
which is just one more human choice out of
infinity.
And there are an INFINITY of possibilities for
EVERY x. Â Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. Â I did. Â I chose.
But that's where the fights start and the arguing
and the dead math
journal comes into the picture as what follows
from mathematical logic
and the rigidity of infinity is a conclusion that
is devastating
emotionally to thousands of mathematicians around
the world which is
that there is a core error: the ring of algebraic
integers gives a
false result.
Because now you have that one of the a's has 7 as
a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers
NEITHER of the roots
can have 7 as a factor and there is shown the
direct contradiction.
Notice the ring of algebraic integers does not
fail gracefully and its
error is not fixable as your alternative is to
introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past
arguments about how a
particular factor of 7 arises, often simply
repeating what is taught
in math classes about ways to find the factors at
any particular x,
but remember there are an infinity of choices for
EVERY x, so the
choice problem gives you infinity at every point.
 So who chooses?
God?)
 Here is a proof, for x = 1, that a_1(x) and
a_2(x) are both
not divisible by 7.
 Both a_1(1) and a_2(1) are roots of a^2 - 6a +
35.
 We can let a_1(1) = 3 + sqrt(-26), a_2(1) = 3 -
sqrt(-26)
 Let u = -307 - 30 * sqrt(-26) and v = -307 + 30 *
sqrt(-26)
 Then the following can be readily checked by
ordinary
arithmetic:
  (1) a_1(1)^6 / u = -109 + 12 * sqrt(-26)
  (2) a_2(1)^6 / v = -109 - 12 * sqrt(-26)
  (3) u * v = 7^6
  (4) a_1(1)^6 * a_2(1)^6 / (u * v) = 15625 = 5^6.
  (5) u and v are both algebraic integers
  (6) Neither u nor v are units in the ring of
    algebraic integers.
 Further, note that because of (4) above and
because
5 and 7 are coprime, you can conclude that a_1(1)^6
/ u
and a_2(1)^6 / v have no factors in common with 7.
 Conclusions:
 a_1(1)^6 has a factor in common with 7, namely u.
 a_2(1)^6 has a factor in common with 7, namely v.
 Neither a_1(1) nor a_2(1) are divisible by 7,
since
neither u nor v are units and a_1(1)*a_2(1)/(u*v) =
5^6
is coprime to 7.
 The key thing here is, if you have a proof
contradicting
this, you are contradicting not some old theorems
in algebraic
number theory or Galois theory. Â You are
contradicting plain
old arithmetic. Â If there is a problem, it goes
beyond just a
Nope.
The same techniques you just used would work to
"prove" that neither
of the roots of
x^2 + 3x + 2 = 0
has 2 as a factor--if you don't resolve the square
root.
[/quote:96a7ed6eb2]
Resolve the square root? I have no idea what you are saying.
I think the point was that the roots were not integers, such as
x^2 + 2x + 2 = 0
[quote:96a7ed6eb2]
Trivial.
The algebra is easy but the psychology is complex as
the error allows
people like you to do pretend mathematics!
Your very sense of self may be wrapped up in the
bogus math that you
do, so it's an uphill slog to get you past your
denial, into the world
of doing valid mathematics.
James Harris
[/quote:96a7ed6eb2]
James. You have demonstrated a number of times that you have no idea what algebraic integers are. If you've been doing this for 15 years and failed, perhaps you should try to find another hobby. |
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| W. Dale Hall... |
| Posted: Sat Jul 04, 2009 11:51 pm |
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Guest
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JSH wrote:
[quote:92542a5700]On Jul 4, 11:19 am, marcus_b <marcus_bruck... at (no spam) yahoo.com> wrote:
On Jul 2, 5:16 pm, JSH <jst... at (no spam) gmail.com> wrote:
On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
Here is a proof, for x = 1, that a_1(x) and a_2(x) are both
not divisible by 7.
Both a_1(1) and a_2(1) are roots of a^2 - 6a + 35.
We can let a_1(1) = 3 + sqrt(-26), a_2(1) = 3 - sqrt(-26)
Let u = -307 - 30 * sqrt(-26) and v = -307 + 30 * sqrt(-26)
Then the following can be readily checked by ordinary
arithmetic:
(1) a_1(1)^6 / u = -109 + 12 * sqrt(-26)
(2) a_2(1)^6 / v = -109 - 12 * sqrt(-26)
(3) u * v = 7^6
(4) a_1(1)^6 * a_2(1)^6 / (u * v) = 15625 = 5^6.
(5) u and v are both algebraic integers
(6) Neither u nor v are units in the ring of
algebraic integers.
Further, note that because of (4) above and because
5 and 7 are coprime, you can conclude that a_1(1)^6 / u
and a_2(1)^6 / v have no factors in common with 7.
Conclusions:
a_1(1)^6 has a factor in common with 7, namely u.
a_2(1)^6 has a factor in common with 7, namely v.
Neither a_1(1) nor a_2(1) are divisible by 7, since
neither u nor v are units and a_1(1)*a_2(1)/(u*v) = 5^6
is coprime to 7.
The key thing here is, if you have a proof contradicting
this, you are contradicting not some old theorems in algebraic
number theory or Galois theory. You are contradicting plain
old arithmetic. If there is a problem, it goes beyond just a
Nope.
The same techniques you just used would work to "prove" that neither
of the roots of
x^2 + 3x + 2 = 0
has 2 as a factor--if you don't resolve the square root.
[/quote:92542a5700]
This is not true.
[quote:92542a5700]Trivial.
[/quote:92542a5700]
Then provide the proof, if you think you're up to it.
[quote:92542a5700]The algebra is easy but the psychology is complex as the error allows
people like you to do pretend mathematics!
[/quote:92542a5700]
As the King of Pretend Mathematics, you should know?
[quote:92542a5700]Your very sense of self may be wrapped up in the bogus math that you
do, so it's an uphill slog to get you past your denial, into the world
of doing valid mathematics.
[/quote:92542a5700]
Here's a challenge.
You've said that one could "prove" , using the techniques marcus_b has
used, that neither root of the polynomial
x^2 + 3x + 2
is divisible by 2, presumably in the ring of algebraic integers.
You've even said it's trivial.
So, prove it.
I say you can't do it. I know that I can't, I know that none of the
mathematicians that I know of can prove it.
But you claim it can be done, and that it is trivial. So, put your
alleged mathematics where your mouth is, and provide the proof.
Can't be so hard, if it's trivial. Maybe you'll get lucky!
[quote:92542a5700]
James Harris
[/quote:92542a5700]
Dale |
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| JSH... |
| Posted: Sun Jul 05, 2009 6:07 am |
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Guest
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On Jul 4, 10:51 pm, "W. Dale Hall"
<wdunderscorehallatpacbelldotnet at (no spam) last> wrote:
[quote:995a3dde67]JSH wrote:
On Jul 4, 11:19 am, marcus_b <marcus_bruck... at (no spam) yahoo.com> wrote:
On Jul 2, 5:16 pm, JSH <jst... at (no spam) gmail.com> wrote:
On my math blog Penny Hassett had a comment that got me to thinking
and after pondering at her suggestion what others believe to be true
in their disagreements with me about a certain serious mathematical
issue, I concluded that infinity could help to clear the air.
So I have simple examples yet again to start (believe me it gets
really hard later so please pay close attention to the easy part):
7(x^2 + 3x + 2) = (7x + 7)(x + 2)
and let the ring be the ring of algebraic integers. And notice that
reflects CHOICE. My choice for a simple example as the 7 can be
factored an INFINITY of ways!!!
After all,
7(x^2 + 3x + 2) = (x + 1)(7x + 14)
is ALSO just as valid mathematically, and in fact there are an
infinity of possible variations all equally valid.
Key here also to notice is that there are an infinity of choices FOR
EVERY x, and remember the ring is the ring of algebraic integers.
So far easy and there should be no disagreement with the above!
Now to the hard mathematics.
Now we'll TRY to assume the ring is still the ring of algebraic
integers and now a harder example:
7(175x^2 - 15x + 2) = (5(7)b_1(x) + 7)(5b_2(x) + 2)
where
7b_1(x) = a_1(x) and b_2(x) = a_2(x) + 1
and the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0.
Notice the b's are chosen such that when x=0, both the b's equal 0,
which is to say, they are normalized.
Now just like before you have human choice, and you can imagine moving
that 7 around easily enough:
7(175x^2 - 15x + 2) = (5b_1(x) + 1)(5(7)b_2(x) + 14)
which is just one more human choice out of infinity.
And there are an INFINITY of possibilities for EVERY x. Every x has
an infinite number of choices.
So how does the mathematics choose?
It doesn't. I did. I chose.
But that's where the fights start and the arguing and the dead math
journal comes into the picture as what follows from mathematical logic
and the rigidity of infinity is a conclusion that is devastating
emotionally to thousands of mathematicians around the world which is
that there is a core error: the ring of algebraic integers gives a
false result.
Because now you have that one of the a's has 7 as a factor by the
choice argument.
But the a's are roots of
a^2 - (7x-1)a + (49x^2 - 14x) = 0
and at x=1, that gives
a^2 - 6a + 35 = 0
and provably in the ring of algebraic integers NEITHER of the roots
can have 7 as a factor and there is shown the direct contradiction.
Notice the ring of algebraic integers does not fail gracefully and its
error is not fixable as your alternative is to introduce some mystical
force to choose out of infinity, like, maybe God?
(Posters have been remarkably vague in past arguments about how a
particular factor of 7 arises, often simply repeating what is taught
in math classes about ways to find the factors at any particular x,
but remember there are an infinity of choices for EVERY x, so the
choice problem gives you infinity at every point. So who chooses?
God?)
Here is a proof, for x = 1, that a_1(x) and a_2(x) are both
not divisible by 7.
Both a_1(1) and a_2(1) are roots of a^2 - 6a + 35.
We can let a_1(1) = 3 + sqrt(-26), a_2(1) = 3 - sqrt(-26)
Let u = -307 - 30 * sqrt(-26) and v = -307 + 30 * sqrt(-26)
Then the following can be readily checked by ordinary
arithmetic:
(1) a_1(1)^6 / u = -109 + 12 * sqrt(-26)
(2) a_2(1)^6 / v = -109 - 12 * sqrt(-26)
(3) u * v = 7^6
(4) a_1(1)^6 * a_2(1)^6 / (u * v) = 15625 = 5^6.
(5) u and v are both algebraic integers
(6) Neither u nor v are units in the ring of
algebraic integers.
Further, note that because of (4) above and because
5 and 7 are coprime, you can conclude that a_1(1)^6 / u
and a_2(1)^6 / v have no factors in common with 7.
Conclusions:
a_1(1)^6 has a factor in common with 7, namely u.
a_2(1)^6 has a factor in common with 7, namely v.
Neither a_1(1) nor a_2(1) are divisible by 7, since
neither u nor v are units and a_1(1)*a_2(1)/(u*v) = 5^6
is coprime to 7.
The key thing here is, if you have a proof contradicting
this, you are contradicting not some old theorems in algebraic
number theory or Galois theory. You are contradicting plain
old arithmetic. If there is a problem, it goes beyond just a
Nope.
The same techniques you just used would work to "prove" that neither
of the roots of
x^2 + 3x + 2 = 0
has 2 as a factor--if you don't resolve the square root.
This is not true.
Trivial.
Then provide the proof, if you think you're up to it.
[/quote:995a3dde67]
Assume for the sake of argument that x^2 + 3x + 2 = 0 is irreducible
over Z.
Now assume that only one root has 2 as a factor, and let x = 2y, then
4y^2 + 6y + 2 = 0, and dividing 2 off gives:
2y^2 + 3y + 1 = 0
which is also assumed to be irreducible over Z, then it cannot be true
that 2 is a factor, as then the polynomial is non-monic with integer
coefficients assumed to be irreducible over Z.
Note that *reducibility* is the one criteria so as I noted above, if
you do not resolve the square root, then the same type argument works
to show that 2 is not a factor of x^2 + 3x + 2 = 0.
So the argument is actuality is about reducibility.
I have to note for readers who don't know that the poster W. Dale Hall
is one of the posters who conspired online against my paper and is in
fact THE poster who the chief editor quoted in his email to me
claiming a reviewer had claimed a mistake with my paper.
He didn't realize that I'd seen the material on sci.math so knew he
was lying. Contacts with another editor at the now defunct journal
indicate that the journal DID use two reviewers as it states but that
the chief editor became convinced by the emails from the sci.math'ers
that his reviewers had made a mistake!!!
Posters on sci.math have repeatedly claimed without providing evidence
that the journal simply lied about using two reviewers but that is
unsubtantiated opinion from people with a vested interest in lying--
obviously sci.math'ers do not want to be known as journal killers.
So if you didn't know, here is one of the sci.math conspirators still
arguing against the result on the newsgroup.
He is one of the key people in the killing of the SJWPAM journal.
Another person known to have sent an email to the editors based on his
posting on sci.math was Arturo Magidin. So he is one of the other
people who was key in killing the journal.
If you didn't know they were two of the key conspirators, now you do.
James Harris |
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| mjc... |
| Posted: Sun Jul 05, 2009 9:28 am |
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Guest
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JSH starts:
Assume for the sake of argument that x^2 + 3x + 2 = 0 is irreducible
over Z.
(((end quote)))
It seems to me that x^2 + 3x + 2 = (x+1)(x+2).
So the assumption is false.
What have I missed?
Martin Cohen |
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| Musatov... |
| Posted: Sun Jul 05, 2009 12:30 pm |
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Guest
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On Jul 5, 12:28 pm, mjc <mjco... at (no spam) acm.org> wrote:
[quote:e7a9fea2cb]JSH starts:
Assume for the sake of argument that x^2 + 3x + 2 = 0 is irreducible
over Z.
(((end quote)))
It seems to me that x^2 + 3x + 2 = (x+1)(x+2).
So the assumption is false.
What have I missed?
Martin Cohen
[/quote:e7a9fea2cb]
x-{x-1-[x-2-(x-3-x+5)]} x's cancel out;
--
Martin Musatov |
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