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| Guest |
 Posted: Sun Feb 20, 2005 10:19 pm |
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The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
16z^4 + 8(y-x)z^2 + (y-x)^2 = 16z^2 y, so you have
16(1 - y)z^4 + 8(y-x)z^2 + (y-x)^2 = 0
and now let
2z* = sqrt(x) + sqrt(y), and you have
2z* - sqrt(x) = sqrt(y), squaring both sides gives
4z*^2 - 4z* sqrt(x) + x = y, and grouping to isolate the square root
gives
4z*^2 - y - x = 4z*^2 sqrt(x), and now squaring both sides gives
16z*^4 - 8(y+x)z*^2 + (y+x)^2 = 16z*^4 x, so you have
16(1-x)z*^4 - 8(y+x)z*^2 + (y+x)^2 = 0.
And partially solving both using the quadratic formula, I have
z^2 = (-(y-x)+/-sqrt((y-x)^2 - (1-y)(y-x))/(4-4y)
and
z*^2 = ((y+x)+/-sqrt((y+x)^2 - (1-x)(y+x))/(4-4x)
and you already know that if sqrt(x) is irrational, and sqrt(y) is
irrational, then z and z* are both irrational.
BUT 4z(z*) = x - y, and if x and y are both odd integers, where neither
equals 1, and chosen such that x-y has 4 as a factor, then z(z*) is an
integer.
However, as mathematicians typically interpret Galois Theory, both z
and z* are in some sense fractions where the 4 in the denominator
overrides any other algebraic integer factors of 4 in the numerator, so
how can they multiply together to give an integer?
That is, if the numerator has some algebraic integer factor of 4, then
that can cancel out, leaving the numerator coprime to 4, in BOTH cases,
but the two cases multiply TOGETHER to give an integer!
But, if conventional teaching is correct, they should produce a
fraction with a factor of 2 in the denominator, not an integer.
So what gives? Is mathematics broken? Nope.
The correct answer is easy: the current teaching on the subject is
wrong.
James Harris |
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| Justin |
| Posted: Sun Feb 20, 2005 10:19 pm |
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Guest
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jstevh@msn.com wrote:
: The discussion on sign conventions is more than an idle exercise, as
: the errors in thinking in what is typically taught explode outward in
: to ever bigger errors.
Yes, the fact that you can't understand that y=sqrt(x) and y^2=x are two
different equations with different solution sets has obviously exploded
outwards such that you can't solve a single sausage.
: Pay attention to your *emotions* when you get to the end.
Right...paying attention.
(snip gibberish)
: The correct answer is easy: the current teaching on the subject is
: wrong.
At this point I'm almost rolling on the floor with laughter. Is this the
emotion you were speaking of?
-justin |
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| Schweinkolben |
| Posted: Sun Feb 20, 2005 11:01 pm |
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Guest
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<jstevh@msn.com> wrote in message
news:1108955963.823156.255430@l41g2000cwc.googlegroups.com...
[quote:f1a6b73f1b]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
[/quote:f1a6b73f1b]
WRONG
(2*z) ^2 = x + y - 2*sqrt (x*y)
( 4*z^2 - x - y) = -2*sqrt (x*y)
(4*z^2 -x -y)^2 = 4 *(x*y)
Where did you get 16*z^2*y ? that is not 4 * (x*y)
The rest of it is wrong............. Try again. |
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| Manuel Petit |
| Posted: Sun Feb 20, 2005 11:40 pm |
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Guest
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Schweinkolben wrote:
[quote:facf807aa0]jstevh@msn.com> wrote in message
news:1108955963.823156.255430@l41g2000cwc.googlegroups.com...
The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
WRONG
(2*z) ^2 = x + y - 2*sqrt (x*y)
( 4*z^2 - x - y) = -2*sqrt (x*y)
(4*z^2 -x -y)^2 = 4 *(x*y)
Where did you get 16*z^2*y ? that is not 4 * (x*y)
The rest of it is wrong............. Try again.
[/quote:facf807aa0]
Actually he got that part right, and you are wrong:
2z = sqrt(x) - sqrt(y)
4z^2 = x + y - 2 sqrt(xy)
4z^2 + y = x + y + y - 2sqrt(xy)
4z^2 + y - x = 2y - 2sqrt(xy)
4z^2 + y - x = 2 sqrt(y) (sqrt(y) - sqrt(x))
4z^2 + y - x = 4 sqrt(y) z
(4z^2 + y - x)^2 = 16 yz^2
manuel,
[quote:facf807aa0]
[/quote:facf807aa0] |
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| Guest |
| Posted: Mon Feb 21, 2005 1:21 am |
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jstevh@msn.com wrote:
[quote:5a2be07c2a]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
16z^4 + 8(y-x)z^2 + (y-x)^2 = 16z^2 y, so you have
16(1 - y)z^4 + 8(y-x)z^2 + (y-x)^2 = 0
[/quote:5a2be07c2a]
No, that doesn't follow. It would follow if the z^2 on the right hand
side were a z^4. |
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| anon |
| Posted: Mon Feb 21, 2005 2:54 am |
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Guest
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Let's replace taking square root by multiplication by two.
We start from the equation
x = -2y
If y=1 then we only have one solution, that is x=-2
Squaring both sides we get
x^2 = 4y^2
and if y=1 we get the solutions x=2 and x=-2
Freaky! There is clearly something wrong with the
way multiplication by 2 is taught. |
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| W. Dale Hall |
| Posted: Mon Feb 21, 2005 5:34 am |
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Guest
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jstevh@msn.com wrote:
[quote:7395776679]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
[/quote:7395776679]
Even if you knew what you were talking about, none of what you say
makes any sense. For instance, you clearly have no idea of "what is
typically taught". You are demonstrably ignorant of mathematics, its
methods, how it is taught, and how it is done. The article here is
a case in point.
[quote:7395776679]Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
[/quote:7395776679]
This, not surprisingly, is an unsupported assertion. Put up or shut up.
[quote:7395776679]
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
[/quote:7395776679]
4 z^2 = x + y - 2 sqrt(x) sqrt(y).
4 z^2 + y - x = 2y - 2 sqrt(x) sqrt(y)
= 2 sqrt(y)(sqrt(y) - sqrt(x))
= - 4 sqrt(y)z
So, yes, it's true that
(4z^2 + y - x)^2 = 16z^2 y,
at least for positive x and y.
[quote:7395776679]16z^4 + 8(y-x)z^2 + (y-x)^2 = 16z^2 y, so you have
16(1 - y)z^4 + 8(y-x)z^2 + (y-x)^2 = 0
and now let
2z* = sqrt(x) + sqrt(y), and you have
2z* - sqrt(x) = sqrt(y), squaring both sides gives
4z*^2 - 4z* sqrt(x) + x = y, and grouping to isolate the square root
gives
4z*^2 - y - x = 4z*^2 sqrt(x), and now squaring both sides gives
16z*^4 - 8(y+x)z*^2 + (y+x)^2 = 16z*^4 x, so you have
16(1-x)z*^4 - 8(y+x)z*^2 + (y+x)^2 = 0.
And partially solving both using the quadratic formula, I have
z^2 = (-(y-x)+/-sqrt((y-x)^2 - (1-y)(y-x))/(4-4y)
and
z*^2 = ((y+x)+/-sqrt((y+x)^2 - (1-x)(y+x))/(4-4x)
and you already know that if sqrt(x) is irrational, and sqrt(y) is
irrational, then z and z* are both irrational.
[/quote:7395776679]
Why should sqrt(x) and sqrt(y) irrational force sqrt(x) - sqrt(y) to
be irrational? For instance, if x = y = 2, z = 0. I suppose if x and
y are distinct integers that aren't squares, then both sqrt(x)-sqrt(y)
and sqrt(x) + sqrt(y) would be irrational. Perhaps that's what you
meant here.
[quote:7395776679]BUT 4z(z*) = x - y, and if x and y are both odd integers, where neither
equals 1, and chosen such that x-y has 4 as a factor, then z(z*) is an
integer.
[/quote:7395776679]
Big deal.
[quote:7395776679]However, as mathematicians typically interpret Galois Theory, both z
and z* are in some sense fractions where the 4 in the denominator
overrides any other algebraic integer factors of 4 in the numerator, so
how can they multiply together to give an integer?
[/quote:7395776679]
I don't see a any argument here. Perhaps you could provide it.
You might note that 2z and 2z* are both algebraic integers, and
if anything, one would have
z = (sqrt(x) - sqrt(y))/2
z* = (sqrt(x) + sqrt(y))/2.
The numerators are algebraic integers, and you can plainly see that
the denominators are 2, not 4.
It's easy enough to multiply here:
zz* = (x-y)/4.
Note that I did't really need two separate quadrics to deduce this.
[quote:7395776679]That is, if the numerator has some algebraic integer factor of 4, then
that can cancel out, leaving the numerator coprime to 4, in BOTH cases,
but the two cases multiply TOGETHER to give an integer!
[/quote:7395776679]
Here's what (a miniscule amount of) Galois theory has to say: the
minimal polynomial of 2z is the same as the minimal polynomial of 2z*.
A little arithmetic shows that it is this (using w for the variable):
(w+sqrt(x)+sqrt(y))(w+sqrt(x)-sqrt(y))(w-sqrt(x)+sqrt(y))(w-sqrt(x)-sqrt(y))
Simplifying, you should get this:
P(w) = w^4 - 2(x+y) w^2 + (x-y)^2.
If x is congruent to y modulo 2, i.e., x-y = 2k for some rational
integer, no root of this polynomial is coprime to 2. To see this,
let u be a root of P(w). Then
u^4 = 2u^2 (x+y) - (x-y)^2
= 2(u^2 (x+y) + 2 k^2)
so u^4 is a then multiple of 2 in the ring of integers of the splitting
field of P(w). Unless 2 is invertible in that ring, u^4 cannot be
coprime to 2.
What I'm aiming for is to show that u itself cannot be coprime to 2.
Here's a tiny lemma:
If p and q are coprime in R, a commutative ring with unity,
then p^2 and q are also coprime in R.
Proof:
Let ap + bq = 1, where a and b are in R.
Then we have
(ap+bq) ap + bq = 1
since we have merely multiplied ap by 1,
but the expanding the product yields
a^2 p^2 + bq ap + bq = 1
or
a^2 p^2 + (ap + 1)b q = 1
This is of the form
A p^2 + Bq = 1
with A and B in R.
Thus,
p^2 is coprime to q in R.
Corollary: If p and q are coprime in R, then p^4 and q are
also coprime in R.
Thus, if u, a root of P(w), were coprime to 2, then w^4 would
also be coprime to 2.
As a result, since u^4 is not coprime to 2 (being a multiple of 2),
we see that u cannot be coprime to 2 in that ring.
[quote:7395776679]But, if conventional teaching is correct, they should produce a
fraction with a factor of 2 in the denominator, not an integer.
[/quote:7395776679]
So how come I find that the roots of the above quadric (which is
the minimal polynomial for 2z and for 2z*) are not coprime to 2 in
the very case you're citing? Further, since neither 2z nor 2z* is
coprime to 2 when x-y is divisible by 2, why is it so surprising
that the product zz* be an ordinary integer when x-y is divisible
by 4?
[quote:7395776679]So what gives? Is mathematics broken? Nope.
The correct answer is easy: the current teaching on the subject is
wrong.
[/quote:7395776679]
Get a clue. You don't know what the subject is, you've never been
exposed to the "current teaching on the subject", and still you
imagine you're qualified to make such a claim.
Do you still wonder why no one believes your pronouncements?
[quote:7395776679]
James Harris
[/quote:7395776679]
Dale. |
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| Guest |
| Posted: Mon Feb 21, 2005 6:39 am |
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Oh, yeah, I had some minor errors in what I gave last night, that don't
remove the conclusion, which is far reaching.
Corrections below...
jstevh@msn.com wrote:
[quote:01ec4f54d1]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
16z^4 + 8(y-x)z^2 + (y-x)^2 = 16z^2 y, so you have
[/quote:01ec4f54d1]
16z^4 - 8(y+x)z^2 + (y-x)^2 = 0
[quote:01ec4f54d1]and now let
2z* = sqrt(x) + sqrt(y), and you have
2z* - sqrt(x) = sqrt(y), squaring both sides gives
4z*^2 - 4z* sqrt(x) + x = y, and grouping to isolate the square root
gives
[/quote:01ec4f54d1]
4z*^2 - y + x = 4z* sqrt(x), and now squaring both sides gives
16z*^4 - 8(y-x)z*^2 + (y+x)^2 = 16z*^2 x, so you have
16z*^4 - 8(y+x)z*^2 + (y+x)^2 = 0.
[quote:01ec4f54d1]And partially solving both using the quadratic formula, I have
[/quote:01ec4f54d1]
z^2 = ((y+x)+/-sqrt((y+x)^2 - (y-x)^2)/4
[quote:01ec4f54d1]and
[/quote:01ec4f54d1]
z*^2 = (-(y+x)+/-sqrt((y+x)^2 - (y+x)^2)/4
if I didn't make any more mistakes.
[quote:01ec4f54d1]and you already know that if sqrt(x) is irrational, and sqrt(y) is
irrational, then z and z* are both irrational.
BUT 4z(z*) = x - y, and if x and y are both odd integers, where
neither
equals 1, and chosen such that x-y has 4 as a factor, then z(z*) is
an
integer.
However, as mathematicians typically interpret Galois Theory, both z
and z* are in some sense fractions where the 4 in the denominator
overrides any other algebraic integer factors of 4 in the numerator,
so
how can they multiply together to give an integer?
That is, if the numerator has some algebraic integer factor of 4,
then
that can cancel out, leaving the numerator coprime to 4, in BOTH
cases,
but the two cases multiply TOGETHER to give an integer!
But, if conventional teaching is correct, they should produce a
fraction with a factor of 2 in the denominator, not an integer.
So what gives? Is mathematics broken? Nope.
The correct answer is easy: the current teaching on the subject is
wrong.
James Harris[/quote:01ec4f54d1] |
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| David Kastrup |
| Posted: Mon Feb 21, 2005 7:19 am |
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Guest
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jstevh@msn.com writes:
[quote:1b92127bcc]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
[/quote:1b92127bcc]
Wiles' work explodes because (-1)^2 = (+1)^2 ? Wow. Could you point
out exactly where Wiles' reasoning goes astray because of that?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum |
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| fishfry |
| Posted: Mon Feb 21, 2005 11:42 am |
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Guest
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In article <1108955963.823156.255430@l41g2000cwc.googlegroups.com>,
jstevh@msn.com wrote:
[quote:a2dbbe8be6]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
[/quote:a2dbbe8be6]
Could you please show exactly how WIles's proof is flawed? |
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| Will Twentyman |
| Posted: Mon Feb 21, 2005 2:47 pm |
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Guest
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jstevh@msn.com wrote:
[quote:0e9a8f6174]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
then you get
(4z^2 + y - x)^2 = 16z^2 y , which is
16z^4 + 8(y-x)z^2 + (y-x)^2 = 16z^2 y, so you have
16(1 - y)z^4 + 8(y-x)z^2 + (y-x)^2 = 0
and now let
2z* = sqrt(x) + sqrt(y), and you have
2z* - sqrt(x) = sqrt(y), squaring both sides gives
4z*^2 - 4z* sqrt(x) + x = y, and grouping to isolate the square root
gives
4z*^2 - y - x = 4z*^2 sqrt(x), and now squaring both sides gives
16z*^4 - 8(y+x)z*^2 + (y+x)^2 = 16z*^4 x, so you have
16(1-x)z*^4 - 8(y+x)z*^2 + (y+x)^2 = 0.
And partially solving both using the quadratic formula, I have
z^2 = (-(y-x)+/-sqrt((y-x)^2 - (1-y)(y-x))/(4-4y)
and
z*^2 = ((y+x)+/-sqrt((y+x)^2 - (1-x)(y+x))/(4-4x)
and you already know that if sqrt(x) is irrational, and sqrt(y) is
irrational, then z and z* are both irrational.
[/quote:0e9a8f6174]
How do we know that? Considering there are plenty of pairs of
irrationals that add or subtract to give a rational number, and that
quotients of irrationals can be rational, it is not obvious to me that x
and y couldn't be chosen from among the reals to have irrational square
roots but leave z and/or z* as rationals.
For example: in your original equation for z, if y = 3 and
x=(5+sqrt(3))^2, then x and y have irrational square roots, yet z = 5/2.
[quote:0e9a8f6174]BUT 4z(z*) = x - y, and if x and y are both odd integers, where neither
equals 1, and chosen such that x-y has 4 as a factor, then z(z*) is an
integer.
[/quote:0e9a8f6174]
Which is certainly possible.
[quote:0e9a8f6174]
However, as mathematicians typically interpret Galois Theory, both z
and z* are in some sense fractions where the 4 in the denominator
overrides any other algebraic integer factors of 4 in the numerator, so
how can they multiply together to give an integer?
[/quote:0e9a8f6174]
I'm not sure what you mean by that, but if x=5 and y=1, then your
formulas give that z=(sqrt(5) - 1)/2 and z* = (sqrt(5) + 1)/2, which
multiply together to give 1. Perhaps if you spelled out this
interpretation of Galois Theory it would help.
[quote:0e9a8f6174]That is, if the numerator has some algebraic integer factor of 4, then
that can cancel out, leaving the numerator coprime to 4, in BOTH cases,
but the two cases multiply TOGETHER to give an integer!
[/quote:0e9a8f6174]
Let's be more careful here: apparently you are suggesting that both need
to be reducible in the quotient ring over the algebraic integers. That
is not necessarily the case. Then again, I'm not sure what you are saying.
[quote:0e9a8f6174]But, if conventional teaching is correct, they should produce a
fraction with a factor of 2 in the denominator, not an integer.
So what gives? Is mathematics broken? Nope.
The correct answer is easy: the current teaching on the subject is
wrong.
[/quote:0e9a8f6174]
It's hard to say with the above discourse. BTW, what was the point of
all that algebra at the beginning? It looked like you were just
obfuscating the simple expressions you started with.
--
Will Twentyman
email: wtwentyman at copper dot net |
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| Guest |
| Posted: Mon Feb 21, 2005 6:23 pm |
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jstevh@msn.com wrote:
[quote:d8a570f034]The discussion on sign conventions is more than an idle exercise, as
the errors in thinking in what is typically taught explode outward in
to ever bigger errors.
Ultimately the errors are huge enough to sink supposed "proofs" like
Wiles's work on Taniyama-Shimura, and oddly enough, it's easy to show
the explosion by moving forward with an variation on an example in my
sign conventions thread.
Pay attention to your *emotions* when you get to the end.
Now let
2z = sqrt(x) - sqrt(y)
deleted
[/quote:d8a570f034]
Crap. The approach doesn't work. I was in some weird mood last night,
and still didn't catch it this morning.
A while back I actually went through a rather dismal period when I
spent a lot of time manipulating basic polynomials trying to break
through sign ambiguity to prove a problem with the ring of algebraic
integers.
For some reason, every once in a while, despite knowing it doesn't
work, I find myself trying the same approaches, over and over again,
only to realize that they can't work, like I learned before.
And yet, even then, I find myself just doodling for a while, redoing
the calculations as if it's...comforting.
Weird, but long story short, the approach I took here doesn't work to
show the problem.
James Harris |
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| W. Dale Hall |
| Posted: Mon Feb 21, 2005 7:49 pm |
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jstevh@msn.com wrote:
... stuff deleted ...
[quote:e9ebd0251c]
Crap. The approach doesn't work. I was in some weird mood last night,
and still didn't catch it this morning.
A while back I actually went through a rather dismal period when I
spent a lot of time manipulating basic polynomials trying to break
through sign ambiguity to prove a problem with the ring of algebraic
integers.
For some reason, every once in a while, despite knowing it doesn't
work, I find myself trying the same approaches, over and over again,
only to realize that they can't work, like I learned before.
And yet, even then, I find myself just doodling for a while, redoing
the calculations as if it's...comforting.
Weird, but long story short, the approach I took here doesn't work to
show the problem.
[/quote:e9ebd0251c]
You know, if you actually spelled out a *specific* claim about the
algebraic integers that mathematicians maintain to be true, but that
you can show is false, you might stand a chance of making more than
illusory gains. Instead, you're stumbling over needlessly complicated
polynomial equations to show stuff that could be shown not to be so,
using much simpler methods.
You might note that to date, you have demonstrated *zero* actual
flaws of the ring of algebraic integers. Your rants about numbers
being "properly units" in that ring, without so much as a hint of
a definition, or your more recent blather about numbers being divisible
by some algebraic integer while not being algebraic integers themselves,
are indications that you're flailing, just hoping one of your wild
swings will connect.
Given your occasional tendency to cite current popular films to drive
home a point, perhaps you should watch "Million Dollar Baby". In that
movie, there is a pathetic character who regularly announces that he
is challenging the (whatever weight class) champion. When the guys in
the gym get him into the ring, he is dispatched easily, if a bit
cruelly; Morgan Freeman's character comes to this guy's aid, and
easily takes care of the bully, but it is clear that the kid is not
what he thinks himself to be. Instead, he is living a dream that
is just not consistent with any likely reality, and the folks who
run the gym just take pity on him.
You might have a little in common with that kid; you talk big, claim
to challenge the whole world of mathematics. But "girly, tough ain't
enough", to quote Clint Eastwood's character.
You wouldn't need to wave your hands so wildly if you understood what
you were talking about. As long as you imagine that your precious
insights and trick formulas will contradict proofs that have stood
for 100 or more years, you will remain at this pre-adolescent state.
You should be aware, by now, that just as the algebra of polynomials
unifies and generalizes ordinary arithmetic, so does the study of
rings and polynomial algebras unify and generalize the study of
polynomials. Any peculiar polynomial you come up with will still be
subject to the constraints that apply from these "higher" algebraic
methods. All indications that your low-level manipulations give will
simply never be able to contradict extant proofs in algebraic number
theory, or in Galois theory for that matter.
[quote:e9ebd0251c]
James Harris
[/quote:e9ebd0251c]
Dale. |
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| Jesse F. Hughes |
| Posted: Tue Feb 22, 2005 1:34 am |
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"W. Dale Hall" <mailtowd-hall@pacbell.net> writes:
[quote:4619badea1]You know, if you actually spelled out a *specific* claim about the
algebraic integers that mathematicians maintain to be true, but that
you can show is false, you might stand a chance of making more than
illusory gains.
[/quote:4619badea1]
That's easy.
Mathematicians believe that the algebraic integers are not quirky.
James has shown that, on the contrary, they are quirky.
Non-quirkiness has been the key step in important proofs for over a
hundred years, of course.
Try to keep up.
--
"Looking at their behavior I see them endangering not only their own
futures, but that of their families, and now, considering my latest
result, the future of people all over the world." -- James S. Harris,
on the shortsightedness of his mathematical critics |
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| Arturo Magidin |
| Posted: Tue Feb 22, 2005 11:09 am |
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In article <1109090136.282500.251120@z14g2000cwz.googlegroups.com>,
Nora Baron <norabaron@hotmail.com> wrote:
[.snip.]
[quote:ca42383483]My theory is that Harris indulges in 'magical thinking'.
[/quote:ca42383483]
I've commented on this before. E.g.,
I commented some months ago that you seem to think that
terminology and definitions are like magic spells and
incantantions: you do not need to understand them, you just need
to say the words in the right order with the right inflection, and
all your problems will disappear like magic.
http://groups-beta.google.com/group/alt.math.undergrad/msg/194bf9d074af2c50
(July 2 2003)
(My guess: James thinks that definitions are magic spells: if you
can come up with the correct definition, it will protect you
against harm. The first part of the definition, about units, is
his attempt at "I won't have to worry about f being a unit,
because I know it is not a unit in the integers; that will show
the bastards who argued against me before"; the second part of his
definition is his attempt at "I won't have to worry about
divisibility, because I'm ensuring that 'a divides b' will be true
everywhere if it is true anywhere; that willshow the bastards who
kept pestering me about what 'factor' meant".
[)]
http://groups-beta.google.com/group/alt.math.undergrad/msg/73b717180979d7ac
(April 15 2003)
The problem, James, is that you are dealing with definitions as if
they were magic spells. You think that if you can just come up
with the right combination of words, everything will get fixed
magically. But nowhere on your page do you ever use explicitly
any properties of "objects" or "object rings", and the only place
where you use them implicitly (in the factorization) is a big huge
gap. It is based on your incorrect conclusion that such a
factorization need not exist for algebraic integers. You don't
know what you want, you don't know why you want it, so it is small
wonder you cannot give a correct definition.
http://groups-beta.google.com/group/sci.math/msg/8ac3752d915e3253
(January 15 2003)
[quote:ca42383483]It
resembles the cargo-cult thinking in south Pacific islands, where
people would construct little models that looked like airplanes
in the hope that the airplanes would return again and bestow
various goodies as they had done during the war. Harris constructs
what looks to him like mathematics. He has faith that whatever
he writes down is going to work because he has this special 'gift'
that I think goes back to his junior high days when he was
identified as 'gifted' and given special treatment, and then
again when he got a scholarship to Vanderbilt - he concluded that
because of his gift, he didn't need to work hard like other
people - his ideas were just automatically golden. Perhaps it's
an example of the downside of labelling people as 'gifted'.
[/quote:ca42383483]
There also seems to be a bit of the same kind of approach as Alexander
Abian used to take: he'll come up with the grand ideas, everyone else
will do the grunt work and work out the details for him (while
preserving ultimate glory and rewards for the one who came up with the
grand idea, of course). "Refereee reports" are expected to point out
specific errors, and to explain how to fix any such and make the
argument right. Any gaps are to be filled by others, and simply
pointing them out is not valid.
Finally, there is an utter incomprehension of how mathematics is
taught and learned once one gets beyond lower division calculus and
related courses. I've commented on this before as well, though I can't
seem to find the right phrases to find one of those posts in
google. The continual comments about "teaching math" and "errors
presented" and so on make it painfully obvious that he believes math
instruction occurs in the same general format as in calculus: certain
rules and theorems are stated but not proven, and students are asked
to "take them for granted" (for example, the Intermediate Value
Theorem is seldom proven in lower division courses in the U.S., or the
Extreme Value Theorem; they are merely asserted and students are asked
to take them on faith and use them). The notion that in an upper
division course the normal process is to state theorems and give
proofs in detail, and students are asked, nay, required to make sure
every single line is clear and valid, seems completely foreign to
him. It could be part of that same labeling you mentioned (having
taken "advanced" courses before finishing high school, for example).
--
======================================================================
"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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