|Can anyone describe the Stone-Cech compactification of
|the natural numbers topologically and algebraically,
|for example : is it still a semi-group. is it metrizable
|and what would be a definition of such a metric ?
it's still got all of the same finitary algebraic structure that the
natural numbers has, because the stone-czech compactification functor
from sets to topological spaces preserves finite cartesian products;
so in particular it's still a rig (= ring without negatives).
-- 
==== 
>Can anyone describe the Stone-Cech compactification of
>the natural numbers topologically and algebraically,
>for example : is it still a semi-group. is it metrizable
>and what would be a definition of such a metric ?
It's certainly not metrizable: Since it's compact, if
it were metrizable that would mean that the sequence
1, 2, 3, ... would have a convergent subsequence
n_1, n_2, .... And there are obviously no convergent
sequences of integers (if (n_j) is convergent
then for every bounded function f : N -> C the sequence
(f(n_j)) is convergent. Not likely.)
>
************************
David C. Ullrich
====
>
>|Can anyone describe the Stone-Cech compactification of
>|the natural numbers topologically and algebraically,
>|for example : is it still a semi-group. is it metrizable
>|and what would be a definition of such a metric ?
>
>
>it's still got all of the same finitary algebraic structure that the
>natural numbers has, because the stone-czech compactification functor
>from sets to topological spaces preserves finite cartesian products;
I take it that that means that in particular b(N) x b(N) is naturally
homeomorphic to b(NxN) ? I could be all wet, but offhand this
doesn't seem right:
If so then every bounded complex-valued function on NxN extends
continuously to b(N) x b(N). But let f(n,m) = 1 if n = m, 0 otherwise.
It seems to me that f does not extend to a function continuous on
NxN: If we had such an extension then f(n, y) = 0 for n in N and
y in b(N)N (?) and hence f(x,y) = 0 for x, y in b(N)N, so f vanishes
on the complement of NxN, and hence f is not continuous since
the diagonal of NxN is not closed.
???
>so in particular it's still a rig (= ring without negatives).
************************
David C. Ullrich
====
> Can anyone describe the Stone-Cech compactification of
> the natural numbers topologically and algebraically,
Let's write beta N for the Stone-Cech compactification.
Any such description lives way out in Axiom of Choice Land.
> for example : is it still a semi-group.
no: For the addition operation x+y: For fixed x you can extend y to all
of beta N, and for fixed y you can extend x to all of beta N, but not
both at once (while remaining jointly continuous).
There is the Bohr compactification of the group Z of integers, smaller
than beta Z, but still a group.
> is it metrizable
no: beta N is compact but not sequentially compact.  It is separable
but not second-countable.  It has power 2^c, which is much larger than
any separable metric space.
> and what would be a definition of such a metric ?
>
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
====
>
>
>|Can anyone describe the Stone-Cech compactification of
>|the natural numbers topologically and algebraically,
>|for example : is it still a semi-group. is it metrizable
>|and what would be a definition of such a metric ?
>
>
>it's still got all of the same finitary algebraic structure that the
>natural numbers has, because the stone-czech compactification functor
>from sets to topological spaces preserves finite cartesian products;
>
> I take it that that means that in particular b(N) x b(N) is naturally
> homeomorphic to b(NxN) ? I could be all wet, but offhand this
> doesn't seem right:
Not only does it not seem right, it isn't right.  As an example of how
excellent mathematicians can make mistakes, back in his first major paper 
in
1948, Ed Hewitt had a theorem which stated the b(X) x B(Y) is homeomorphic
to b(X x Y).  (This is the paper in which he introduced the concept of
realcompactness, and he didn't do much along that line after that
embarrassment.)  Long ago I read the paper to try to find his error, and he
used a preliminary lemma which had as another correlary that the 2-sphere 
is
homeomorphic to a torus.
lot about extending algebraeic structures to the Stone-Cech
compactification.  Here is a link to his home page, which include a list of
his publications.
http://members.aol.com/nhindman/
>
> If so then every bounded complex-valued function on NxN extends
> continuously to b(N) x b(N). But let f(n,m) = 1 if n = m, 0 otherwise.
> It seems to me that f does not extend to a function continuous on
> NxN: If we had such an extension then f(n, y) = 0 for n in N and
> y in b(N)N (?) and hence f(x,y) = 0 for x, y in b(N)N, so f vanishes
> on the complement of NxN, and hence f is not continuous since
> the diagonal of NxN is not closed.
>
> ???
>
>
>so in particular it's still a rig (= ring without negatives).
>
> ************************
>
> David C. Ullrich
>
====
>
>
>|Can anyone describe the Stone-Cech compactification of
>|the natural numbers topologically and algebraically,
>|for example : is it still a semi-group. is it metrizable
>|and what would be a definition of such a metric ?
>
>
>it's still got all of the same finitary algebraic structure that the
>natural numbers has, because the stone-czech compactification functor
>from sets to topological spaces preserves finite cartesian products;
>
> I take it that that means that in particular b(N) x b(N) is naturally
> homeomorphic to b(NxN) ? I could be all wet, but offhand this
> doesn't seem right:
>
> If so then every bounded complex-valued function on NxN extends
> continuously to b(N) x b(N). But let f(n,m) = 1 if n = m, 0 otherwise.
> It seems to me that f does not extend to a function continuous on
> NxN: If we had such an extension then f(n, y) = 0 for n in N and
> y in b(N)N (?) and hence f(x,y) = 0 for x, y in b(N)N, so f vanishes
> on the complement of NxN, and hence f is not continuous since
> the diagonal of NxN is not closed.
>
> ???
>
>
>so in particular it's still a rig (= ring without negatives).
>
> ************************
>
> David C. Ullrich
>
====
I seem to  be having problems with my server/reader.  Somehow, this didn't
seem to make it, but one without my additions did.
>
>
>
>|Can anyone describe the Stone-Cech compactification of
>|the natural numbers topologically and algebraically,
>|for example : is it still a semi-group. is it metrizable
>|and what would be a definition of such a metric ?
>
>
>it's still got all of the same finitary algebraic structure that 
the
>natural numbers has, because the stone-czech compactification functor
>from sets to topological spaces preserves finite cartesian products;
>
> I take it that that means that in particular b(N) x b(N) is naturally
> homeomorphic to b(NxN) ? I could be all wet, but offhand this
> doesn't seem right:
>
> Not only does it not seem right, it isn't right.  As an example of how
> excellent mathematicians can make mistakes, back in his first major paper
in
> 1948, Ed Hewitt had a theorem which stated the b(X) x B(Y) is 
homeomorphic
> to b(X x Y).  (This is the paper in which he introduced the concept of
> realcompactness, and he didn't do much along that line after that
> embarrassment.)  Long ago I read the paper to try to find his error, and
he
> used a preliminary lemma which had as another correlary that the 2-sphere
is
> homeomorphic to a torus.
>
> lot about extending algebraeic structures to the Stone-Cech
> compactification.  Here is a link to his home page, which include a list
of
> his publications.
>
> http://members.aol.com/nhindman/
>
>
>
>
> If so then every bounded complex-valued function on NxN extends
> continuously to b(N) x b(N). But let f(n,m) = 1 if n = m, 0 otherwise.
> It seems to me that f does not extend to a function continuous on
> NxN: If we had such an extension then f(n, y) = 0 for n in N and
> y in b(N)N (?) and hence f(x,y) = 0 for x, y in b(N)N, so f 
vanishes
> on the complement of NxN, and hence f is not continuous since
> the diagonal of NxN is not closed.
>
>
>
> ???
>
>
>so in particular it's still a rig (= ring without negatives).
>
> ************************
>
> David C. Ullrich
>
>
>
====
|
|>
|>|Can anyone describe the Stone-Cech compactification of|
|>|the natural numbers topologically and algebraically,
|>|for example : is it still a semi-group. is it metrizable
|>|and what would be a definition of such a metric ?
|>
|>
|>it's still got all of the same finitary algebraic structure that the
|>natural numbers has, because the stone-czech compactification functor
|>from sets to topological spaces preserves finite cartesian products;|
|
|I take it that that means that in particular b(N) x b(N) is naturally
|homeomorphic to b(NxN) ? I could be all wet, but offhand this
|doesn't seem right:
thanks for the correction.  (i figured someone would correct me if i
was wrong, but i guess i should have mentioned that i wasn't
completely sure about it anyway.)
actually i didn't even read the details of the corrections yet,
because i want to try figuring out my mistake first.  i know that i
occasionally get mixed up about the precise definition of stone-czech
compactification in the most general case, but this isn't the most
general case so that shouldn't be the problem here, i think.  i could
have sworn that there's _some_ functor floating around here somewhere
that preserves finite products, but maybe i got mixed up about which
one it is.
-- 
====
In a proof related to formal language theory i have a context-free grammar 
G = (V_N,V_T,S,F) 
where V_N is a set of non-terminals e.g {A, B, S}
i have to define the subsets of V_N as follows
U_1 = {X|X->lambda in F}
U_(i+1) = U_i {X|X->P in F for some P in U_i*} for i>=1
Now i try to realize. If V_N is {A,B,S} and i have a rule in F, for example
S->lambda, i obtain that U_1 = {S}, ok.
Let's proceed. I have to do the union of U_1 with a set where is present
another element of V_N, in particular the condition required (as you can
see) is that exists a rule X->P for some P in U_i*. But if here U_i* is
the set of all words over U_1 i have
U_i* = {S, SS, SSS, SSSS, .... S^t}; since U_1 = {S}
So, in this way, i can proceed with U_(i+1) only if in F there is a rule
like A->S or B->SS. Right?
====
I'd be grateful to anyone can give me a proof of this sentence:
The finite-dimension subspaces of an infinite-dimension space are
closed and without interior points.
====
> I'd be grateful to anyone can give me a proof of this sentence:
> The finite-dimension subspaces of an infinite-dimension space are
> closed and without interior points.
What sort of space is intended?  Topological vector space?
Without interior points:
Let V be an infinite-dimensional topological vector space.  Let W be a
finite-dimensional linear subspace.  Or, more generally, simply a
proper linear subspace.  Suppose W has an interior point w.  Then W-w
is a neighborhood of 0 (by continuity of addition), and since W is a
linear space, W-w = W.  Because scalar multiplication is continuous at
0, for any v in V there is a nonzero scalar t so that tv in W.  Again,
since W is linear, this means v in W.  This proves V = W, contradicting
the assumption that W is a _proper_ subspace.
Closed I leave to you.
====
> I'd be grateful to anyone can give me a proof of this sentence:
> The finite-dimension subspaces of an infinite-dimension space are
> closed and without interior points.
You've got to have a topology on the space for your sentence to make sense.
Since you didn't specify, I'll make it easy on myself and use a normed
space.
Suppose B is a normed space and S is a finite dimensional subspace with
basis
{s1, s2, ... , sn}.  Define a norm on C^n by
|| (a1, ... , an) || = || a1 s1 + ... + an sn ||, where this second norm is
the one in B.  The fact that all norms on C^n are equivalent shows that S
with the inherited B-norm is complete, therefore closed in B.
Suppose B is a normed space and S is a subspace and x is an element of S,
and a neighborhood U of x lies in S.  Then S also contains a neighborhood 
of
0, namely -x+U.  If y is any element of B, then there is a positive number 
r
such that ry belongs to -x+U; therefore ry belongs to S, therefore y 
belongs
to S, and it follows that S=B.
====
>and without interior points.
If P is an interior point of H then you can find a closed ball B(P,r) with
r>0 such that B(P,r) is in H, ie for any point X  such that N(P-X)<=r, X is
in H.
But if you take Q in S-H, N(P-(P-r*Q/N(Q))) = r, hence P-r*Q/N(Q) is in H
... contradiction.
-- 
Julien Santini,
CMI Technop.99le de Ch.89teau-Gombert, France
Home page: http://www.analgebra.com
====
> If P is an interior point of H then you can find a closed ball B(P,r) 
with
> r>0 such that B(P,r) is in H, ie for any point X  such that N(P-X)<=r, X
is
> in H.
> But if you take Q in S-H, N(P-(P-r*Q/N(Q))) = r, hence P-r*Q/N(Q) is in H
> ... contradiction.
For a normed vector space of course...
====
> Suppose B is a normed space and S is a subspace and x is an element of S,
> and a neighborhood U of x lies in S.  Then S also contains a neighborhood 
of
> 0, namely -x+U.  If y is any element of B, then there is a positive number 
r
> such that ry belongs to -x+U; therefore ry belongs to S, therefore y 
belongs
> to S, and it follows that S=B.
infinite-dimensional space is also normed.
But, if I'm not wrong, the hypothesis that the space is
infinite-dimensional is not relevant for the proof. In fact it should
be also true that:
A subspace of a normed infinite or finite-dimensional linear space is
closed and without interior points.
In particular when you prove that a subspace of a normed space is
without interior point ( except the case that subspace = space ) you
don't refer to its dimension. So the sentence should be always true.
====
Also, he didn't say what scalar field he is using.  A normed vector
space over the rationals, say ... ? Then a finite-dimensional subspace
need not be closed.
-- 
G. A. Edgar                               
http://www.math.ohio-state.edu/~edgar/
====
> Suppose B is a normed space and S is a subspace and x is an element of
S,
> and a neighborhood U of x lies in S.  Then S also contains a
neighborhood of
> 0, namely -x+U.  If y is any element of B, then there is a positive
number r
> such that ry belongs to -x+U; therefore ry belongs to S, therefore y
belongs
> to S, and it follows that S=B.
>
> infinite-dimensional space is also normed.
> But, if I'm not wrong, the hypothesis that the space is
> infinite-dimensional is not relevant for the proof. In fact it should
> be also true that:
> A subspace of a normed infinite or finite-dimensional linear space is
> closed and without interior points.
> In particular when you prove that a subspace of a normed space is
> without interior point ( except the case that subspace = space ) you
> don't refer to its dimension. So the sentence should be always true.
Right you are, halfway.  A *proper* subspace contains no interior points.
On the other hand, a linear subspace of a normed space does not have to be
closed.  Start with any normed linear space which is not complete; this
space is isomorphically isometric to a dense subspace of its completion.
More specifically, the space of polynomials is not closed in the space of
continuous functions (uniform norm) on [0, 1].
====
This is probably absolutely obvious, but I will post it anyway.
Let a() and b() be functions integrable in the domain.
For m = positive integer,
sum{j=1 to m}  integral{0 to j} a(y) (b(j+1-y) -b(j-y)) dy
= integral{0 to m} a(y) (b(m+1-y) -b(1-{y})) dy,
where {y} is fractional-part, = y - floor(y).
Leroy Quet
====
for linux. Since I don't have money, I can't buy linux versions of those 
softwares. I obtained the software called Octave, and it works very well 
, but it is a numerical software. Right now, I am trying to calculate 
divergence or curl, using some sort of a software since calculating 
those takes a long time. Could you give me suggestions on that?
====
This looks promising:
http://scilinux.sourceforge.net/mathpack.html
| for linux. Since I don't have money, I can't buy linux versions of those
| softwares. I obtained the software called Octave, and it works very well
| , but it is a numerical software. Right now, I am trying to calculate
| divergence or curl, using some sort of a software since calculating
| those takes a long time. Could you give me suggestions on that?
|
====
> for linux. Since I don't have money, I can't buy linux versions of those
> softwares. I obtained the software called Octave, and it works very well
> , but it is a numerical software. Right now, I am trying to calculate
> divergence or curl, using some sort of a software since calculating
> those takes a long time. Could you give me suggestions on that?
Mupad might be a good choice for you. Have a look at http://www.mupad.com. 
Also, visit the sci.math.symbolic usenet group.
Zdenek Hurak
====
How are questions normally posted to this group? I'll use LaTeX-ish
notation in this post. I'm not a mathematician, just an engineer with
a question.
I understand that for a vector r_i, that 
dr_i/dr_j = delta_ij.
I also think that for a tensor t_ij, that 
dt_ij/dt_kl = delta_ik delta_lj.
Is this correct?
What about when t_ij = t_ji? 
dt_ij/dt_kl = dt_ji/dt_kl = dt_ij/dt_lk, right?
So then does:
dt_ij/dt_kl = delta_ik delta_lj + delta_il delta_kj - delta_iklj?
Kevin Van Workum, PhD
X-Replace-Address: yes
reply-to: vanw a_t nist d_o_t gov
====
I am stuck on showing that this sytem of equations has unique nonzero
solution (well I think it does). 
A is n by n matrix with positive entries (maybe not necessary) and
largest magnitude eigenvalue > 1 . 
Show that there is exactly one nonzero x=(x1,x2,...,xn) with all entries
between 0 and 1 such that: 
(Ax)_1 = -log(1 - x1)  
(Ax)_2 = -log(1 - x2) 
.
.
. 
(Ax)_n = -log(1 - xn) 
where (Ax)_k is kth entry of vector Ax, 
(Ax)_k = A_{k1} x1 + A_{k2} x2 + ... + A_{kn} xn 
I can do it for n=1 but that's all. :-(  
Any ideas? 
====
I think I want A to have positive entries. Can a fixed point theorem be
applied if we bound it away from zero?
====
I'm in a distance education course, and am stuck on the following
question:
Find the equation of the line through (8,8) that is tangent to the
hyperbola y^2-3xy+2x^2=4.
We are currently doing a unit involving Newton's Method, Tangent line
approximation, and Tangent line approximation(increment form), so i am
assuming that i must use one of them to find the answer.
I've been trying to isolate variables and trying to do a system of
equations, etc., but can't get the answer no matter what i try.  If
anyone could just point me in the right direction with a first step
that would be great.  I don't need the answer, just a nudge in the
Scott Eliason
====
> I'm in a distance education course, and am stuck on the following
> question:
>
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
>
> We are currently doing a unit involving Newton's Method, Tangent line
> approximation, and Tangent line approximation(increment form), so i am
> assuming that i must use one of them to find the answer.
>
> I've been trying to isolate variables and trying to do a system of
> equations, etc., but can't get the answer no matter what i try.  If
> anyone could just point me in the right direction with a first step
> that would be great.  I don't need the answer, just a nudge in the
>
> Scott Eliason
Just guessing, I think this might well be a misprint.  The exact solution 
is
a mess.  I suggest that you work the same problem but using the point (3,7)
instead.  That point lies on the given hyperbola, so if you've done the
section Implicit Differentiation, you can use implicit differentiation 
to
get y' in terms of x and y, then you can plug in x=3, y=7 to get the slope
of the tangent line at that point.
====
> I'm in a distance education course, and am stuck on the following
> question:
> 
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
> 
> We are currently doing a unit involving Newton's Method, Tangent line
> approximation, and Tangent line approximation(increment form), so i am
> assuming that i must use one of them to find the answer.
> 
> I've been trying to isolate variables and trying to do a system of
> equations, etc., but can't get the answer no matter what i try.  If
> anyone could just point me in the right direction with a first step
> that would be great.  I don't need the answer, just a nudge in the
I don't think the intent of the problem is to  use any of the
three topics you've mentioned above.  It seems like more of a
review and synthesize the stuff you've already learned type
of problem.  
Imagine that you know the coordinates of the point on the 
hyperbola where the line is tangent.  Now express the slope
of that line in two different ways.
Bart
====
First find dy/dx by Implicit Differentiation.
2y(dy/dx)-3y-3x(dy/dx)+4x=0
(2y-3x)(dy/dx)=3y-4x
dy/dx=(3y-4x)/(2y-3x)
So m=-8/-8=1
Now we have to find the equation of a line with the slope of 1 and has the
point (8,8). The line is obviously y=x
Looking at the equation, I don't think (8,8) is on the curve. Perhaps a typo 
or
misprint?
David Moran
====
mensaje|n9ccb080a.0307141217.4ffbff2a@posting.google.com:
> I'm in a distance education course, and am stuck on the following
> question:
>
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
>
> We are currently doing a unit involving Newton's Method, Tangent line
> approximation, and Tangent line approximation(increment form), so i am
> assuming that i must use one of them to find the answer.
>
> I've been trying to isolate variables and trying to do a system of
> equations, etc., but can't get the answer no matter what i try.  If
> anyone could just point me in the right direction with a first step
> that would be great.  I don't need the answer, just a nudge in the
>
> Scott Eliason
I would do so:
A line trought (8, 8) with slope m, has the equation
y = m(x - 8) + 8
If that line is a tangent, it must have a unique intersection point with 
the
hyperbola. Substituing in the hyperbola equation,
(8 + m(x - 8))^2 - 3áx(8 + m(x - 8)) + 2x^2 = 4  ==>
(m^2 - 3ám + 2)x^2 + 8(1 - m)(2m - 3)x + 64(m - 1)^2 - 4 = 0
Solving for x,
x = 2(2(m - 1)(2m - 3) +/- Sqrt(5m^2 - 11m + 6))/(m^2 - 3m + 2)
The discriminant of that equation must be equal 0, in order to the solution
would be unique. Then,
5m^2 - 11m + 6 = 0  ==>
m1 = 6/5,   m2 = 1
Substituing m = 6/5, you get the tangent
y = (6x - 8)/5
with tangent point (3, 2).
Substituing m = 1, you get y = x for the 'tangent line', but without 
tangent
point, because x is indeterminated. If you substitute y = x in the 
hyperbola
equation, you get 0 = 4!.
What happen here?
It happen that the line y = x is a asymptote of the hyperbola, a tangent in
one of the its two infinite points.
-- 
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
====
> First find dy/dx by Implicit Differentiation.
> 2y(dy/dx)-3y-3x(dy/dx)+4x=0
> (2y-3x)(dy/dx)=3y-4x
> dy/dx=(3y-4x)/(2y-3x)
>
> So m=-8/-8=1
>
> Now we have to find the equation of a line with the slope of 1 and has 
the
> point (8,8). The line is obviously y=x
>
> Looking at the equation, I don't think (8,8) is on the curve. Perhaps a
typo or
> misprint?
>
> David Moran
The actual solution is not a mess at all, as some miscalculation led me to
believe.  If (x,y) is the point of tangency, then you have the equation
(3y-4x)/(2y-3x)=(y-8)/(x-8).  Solve that equation simultaneously with the
equation for the hyperbola, and you get a unique solution.  Both x and y
are small positive integers.
====
Why is this a hyperbola, why not an ellipse? Although that would not change
how this problem would be solved.
> I'm in a distance education course, and am stuck on the following
> question:
>
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
>
> We are currently doing a unit involving Newton's Method, Tangent line
> approximation, and Tangent line approximation(increment form), so i am
> assuming that i must use one of them to find the answer.
>
> I've been trying to isolate variables and trying to do a system of
> equations, etc., but can't get the answer no matter what i try.  If
> anyone could just point me in the right direction with a first step
> that would be great.  I don't need the answer, just a nudge in the
>
> Scott Eliason
====
> Why is this a hyperbola, why not an ellipse? Although that would not 
change
> how this problem would be solved.
IIRC, the shape of a curve satisfying the quadratic equation
  A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0, will be an ellipse, 
parabola or hyperbola as B^2 - 4*A*C is negative, zero, or positive, 
respectively, provided the equation defines any of these curves.
E.g., x^2 + y^2 + 1 = 0 does not define a real curve at all.
The expression B^2 - 4*A*C may be shown to be invariant under 
rotations of coordinates, which may always be rotated to form an 
equation with cross term (x*y term) equal to zero, and the above 
conditions for ellipse, prarabola and hyperbola clearly hold when 
B = 0.
> I'm in a distance education course, and am stuck on the following
> question:
>
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
====
>Why is this a hyperbola, why not an ellipse? Although that would not 
change
>how this problem would be solved.
>> I'm in a distance education course, and am stuck on the following
>> question:
>>
>> Find the equation of the line through (8,8) that is tangent to the
>> hyperbola y^2-3xy+2x^2=4.
Because the quadratic form y^2 - 3 x y + 2 x^2 is indefinite: the
corresponding matrix 
[   2   -3/2 ]
[ -3/2    1  ] 
has determinant -1/4 and therefore has one positive and one negative 
eigenvalue.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
====
> First find dy/dx by Implicit Differentiation.
> 2y(dy/dx)-3y-3x(dy/dx)+4x=0
> (2y-3x)(dy/dx)=3y-4x
> dy/dx=(3y-4x)/(2y-3x)
>
> So m=-8/-8=1
>
> Now we have to find the equation of a line with the slope of 1 and has 
the
> point (8,8). The line is obviously y=x
>
> Looking at the equation, I don't think (8,8) is on the curve. Perhaps a
>  typo or
> misprint?
>
> David Moran
> 
> The actual solution is not a mess at all, as some miscalculation led me 
to
> believe.  If (x,y) is the point of tangency, then you have the equation
> (3y-4x)/(2y-3x)=(y-8)/(x-8).  Solve that equation simultaneously with the
> equation for the hyperbola, and you get a unique solution.  Both x and y
> are small positive integers.
This is definitely an interesting question.  I tried to follow Mr.
Moran's work, but I encountered a problem when solving as a system of
equations.  If someone could help me out here that would be great. 
J.A. Sanders
====
> First find dy/dx by Implicit Differentiation.
> 2y(dy/dx)-3y-3x(dy/dx)+4x=0
> (2y-3x)(dy/dx)=3y-4x
> dy/dx=(3y-4x)/(2y-3x)
>
> So m=-8/-8=1
>
> Now we have to find the equation of a line with the slope of 1 and 
has
the
> point (8,8). The line is obviously y=x
>
> Looking at the equation, I don't think (8,8) is on the curve. Perhaps
a
>  typo or
> misprint?
>
> David Moran
>
> The actual solution is not a mess at all, as some miscalculation led me
to
> believe.  If (x,y) is the point of tangency, then you have the equation
> (3y-4x)/(2y-3x)=(y-8)/(x-8).  Solve that equation simultaneously with
the
> equation for the hyperbola, and you get a unique solution.  Both x and 
y
> are small positive integers.
>
> This is definitely an interesting question.  I tried to follow Mr.
> Moran's work, but I encountered a problem when solving as a system of
> equations.  If someone could help me out here that would be great.
>
> J.A. Sanders
Okay, to lay this thing to rest.  David Moran did at first what I did at
first, assumed (8,8) was on the curve, because this is the kind of problem
that you see in calculus books in the Implicit Differentiation chapter.
Starting with the equation
y^2-3xy+2x^2=4
we differentiate it implicitly to obtain
2y y' - 3x y' - 3y + 4x = 0
We solve this equation for y' and get
y' = (3y-4x)/(2y-3x)
If a line tangent to the hyperbola passes through
the point (x,y) lying on the hyperbola, then its slope
is given by (3y-4x)/(2y-3x).
We want such a line that also passes through (8,8).
Since (x,y) and (8,8) are points on the line, its slope
is also given by (y-8)/(x-8).
Thus we have the equation
(3y-4x)/(2y-3x)=(y-8)/(x-8).
Cross multiply and collect all terms on one side to obtain
8x - 4x^2 - 8y + 6xy - 2y^2 = 0   (*)
and now we wish to solve this simultaneously with
y^2- 3xy + 2x^2 = 4   (**)
Multiply (**) by 2 and add to (*).  This gives
8x - 8y = 8
y = x - 1
Substitute this in (**) and expand, and you get
x = 3, y = 2.
Use the trick of setting the constant term to 0 to find the
equations of the asymptotes.
y^2-3xy+2x^2=0 becomes
(y-x)(y-2x)=0
fact that the points (0,2) and (0,-2) are on the curve, you can
sketch a good graph of the hyperbola, and you will be convinced
by looking at the graph that (3,2) is the only solution.
I guess this isn't quite all, the original problem asked for
the equation of the tangent line.
====
Aha Now I see what you did, I stand corrected.
David Moran
====
The equation y^2 - 3*x*y + 2*x^2 = 4 is that of a hyperbola.
All the equations y^2 - 3*x*y + 2*x^2 = C, for real non-zero C, are 
hyperbolas with y^2 - 3*x*y + 2*x^2 = 0 giving their asymptotes, 
namely the asymptotes are the lines y = x and y = 2*x.
Since the point (8,8) is on one asymptote, but not on the other, 
there can be only one tangent through (8,8) to a finite point of the 
hyperbola and one tangent at infinity, namely the asymptote 
y = x.
A normal vector at (x,y) to y^2 - 3*x*y + 2*x^2 = 4 is given by the 
gradient [ -3*y+4*x, 2*y - 3*x], so a tangent vector at that point 
is [2*y - 3*x, 3*y - 4*x], and the slope of such a tangent line is
m = (3*y-4*x)/(2*y-3*x).
The slope of the line through (8,8) and (x,y) is m = (y-8)/(x-8).
Thus the point of tangency, (x,y) must satisfy both 
 y^2 - 3*x*y + 2*x^2 = 4 and (3*y-4*x)/(2*y-3*x) = (y-8)/(x-8).
and there is only one such real point, (3,2), and the slope of that 
tangent line is m = 6/5.
====
This problem works out pretty simple because (8,8) is on the asymptote y = 
x
of the hyperbola.
For the hyperbola we have
   y*2 - 3xy + x*2 - 4 = 0          (1) or
   (y - x)(y - 2x) = 4                  (1a)
Represent the tangent line through (8,8) by
   y - 8 = m(x - 8)                      (2).
    6x - 5y - 8 = 0
Dick Tjaden
> I'm in a distance education course, and am stuck on the following
> question:
>
> Find the equation of the line through (8,8) that is tangent to the
> hyperbola y^2-3xy+2x^2=4.
>
> We are currently doing a unit involving Newton's Method, Tangent line
> approximation, and Tangent line approximation(increment form), so i am
> assuming that i must use one of them to find the answer.
>
> I've been trying to isolate variables and trying to do a system of
> equations, etc., but can't get the answer no matter what i try.  If
> anyone could just point me in the right direction with a first step
> that would be great.  I don't need the answer, just a nudge in the
>
> Scott Eliason
====
><Quote>
>As of the early 1990s, most mathematicians believed that the
>Taniyama-Shimura conjecture was not accessible to proof. However, A.
>Wiles was not one of these. He attempted to establish the
>correspondence between the set of elliptic curves and the set of
>modular elliptic curves by showing that the number of each was the
>same. Wiles accomplished this by counting Galois representations and
>comparing them with the number of modular forms.
></Quote>
>  [...]
>My assessment is that Wiles commits the logical fallacy of Cum hoc
>ergo propter hoc.
> 
> Popular, secondhand sources inevitably oversimplify technical statements.
> Here they even cue you to the fact by putting counting in scare 
quotes.
> What you're doing is to take an informal statement in a secondary source
> literally, noticing that it is not perfectly accurate mathematically, and
> then concluding that the formal mathematics in the primary sources must
> be logically flawed.
Well in another thread someone posted a link to Wiles's paper, so I've
started looking over its introduction.
Here's an intriguing quote which I'm focusing on, though it may drag
me into greater details.
The key development in the proof is a new and surprising link between
two strong but distinct traditions in number theory,the relationship
between Galois representations and modular forms on the one hand and
the interpretation of special values of L -functions on the other.
p.2
Source: 
http://modular.fas.harvard.edu/21n/papers/Wiles,Modular_Elliptic_Curves_and_F
ermats_Last_Theorem.pdf
The special valus of L-functions are those 4 descriptors popping up
again.  My understanding is that mathematicians have reworked that
approach having found something shorter, but I'll focus on the
original.
 
> It's illegitimate to fault Wiles's argument on the basis of secondary
> sources.  If you think there is something wrong with Wiles's argument,
> tell us specifically which claims in his paper, or in his joint paper
> with Richard Taylor, are wrong.  I assume you *have*, of course, read
> and understood both papers?  That you are not simply relying on secondary
> sources because the primary sources are too advanced for you?
Oh the primary source is *way* too advanced for me in detail.
However, it's intriguing to see if the logical error pops out as
easily as I expect it should, or if it's buried behind a lot of
technical jargon.
I'm considering that question now with the source.
James Harris
====
James Harris, it appears that your current strategy for saving face (read: 
posting crap) to the internet, having
completely failed to provide a valid proof of FLT, is to attack the proof 
provided by Wiles. This is standard
fare for cranks -- if you can't create, try to destroy. But please spare us 
the disassociated hand-waving
arguments. Stick to a specific point instead of continually leaping 
triumphantly to false conclusions. You have
been so thoroughly discredited that the only hope of redeeming yourself is 
to actually do something right.
--
There are two things you must never attempt to prove: the unprovable -- and 
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
clearly, for the greater glory of his church-school sponsors, so
as not to tip-off the many, competent to beat him
to the punch (such as Ribet).  of course,
it could also be flawed; or,
simply inelegant.
how would you characterize the sum-total
of your now 8-year mission, monsieur Harris,
minus all of the vituperative garbarge?
   of course,
such an approach may be feasible for teaching math, although
it's hard to imagine the student-body that'd tolerate that sort
of harangue.  (of course,
in real life, one probably would be forced (or happy)
to modify one's approach, if the students were at-all hominid .-)
 
> Ah, well, this quotation helps explain where you got your impression of
> what Wiles was claiming.  You are perfectly correct in analyzing this
> sentence by pointing out that on the one hand we have elliptic curves,
> and on the other hand we have modular forms, and there is no 
superset
> of objects of which elliptic curves and modular forms are both members.
> So just because you get this L-series thingy from an elliptic curve,
> and you can get the same L-series thingies from modular forms, how can
> this possibly imply anything like all elliptic curves are modular?
> 
> Good question.
 
> As of the early 1990s, most mathematicians believed that the
> Taniyama-Shimura conjecture was not accessible to proof. However, A.
> Wiles was not one of these. He attempted to establish the
> correspondence between the set of elliptic curves and the set of
> modular elliptic curves by showing that the number of each was the
> same. Wiles accomplished this by counting Galois representations and
> comparing them with the number of modular forms.
 
> My assessment is that Wiles commits the logical fallacy of Cum hoc
> ergo propter hoc.
> 
> (Source http://users.tru.eastlink.ca/~brsears/reafault.htm )
 
> Why would Wiles deceive his colleagues?  Why haven't more people
> thought that relevant?  Why be surprised that a man obsessed and
> isolated for several years living a deception might delude himself
> into believing a logically flawed approach might work?
--A church-school McCrusade (Blair's ideals?):
Harry-the-Mad-Potter want's US to kill Iraqis?...
For a 1000-year anglo-american hegemony?
HEY, JIMMY; LET'S US and SU FIGHT -then-PM of England & Zbiggy
 http://www.tarpley.net/bush25.htm (Thyroid Storm ch.)
  http://www.rwgrayprojects.com/synergetics/plates/plates.html
   http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX
====
> James Harris, it appears that your current strategy for saving face (read: 
posting crap) to the internet, having
> completely failed to provide a valid proof of FLT, is to attack the proof 
provided by Wiles. This is standard
> fare for cranks -- if you can't create, try to destroy. But please spare 
us the disassociated hand-waving
> arguments. Stick to a specific point instead of continually leaping 
triumphantly to false conclusions. You have
> been so thoroughly discredited that the only hope of redeeming yourself is 
to actually do something right.
> 
Human beings have a flaw: they can be thoroughly convinced of false
things.
The reason the primary newsgroup for this thread is sci.logic is that
logicians are tasked with being careful in a way that even
mathematicians aren't.
Mathematicians can get away with leaps and assertions because of the
relevance of mathematics to the real world, where physics has been
powered by mathematical models.
But logic is cold, hard, and less amenable to social pressure.
I've outlined a specific logical flaw in Wiles's approach, which is
that it depends on the logical fallacy of trying to find conditions
through association.
That is the logical fallacy is, Cum hoc ergo propter hoc.
So does Wiles get around the lack of a logical basis by using a finite
subset to find a restriction on an infinite set?
The technique is called lifting and is something like infinite
descent made somewhat famous by Fermat himself.
It is an intriguing question, so I'll back off somewhat while I
consider whether or not he somehow gets around the logical fallacy to
find a way to break it.
Now mathematicians apparently are quite certain that Wiles succeeded
and I applaud their energy.  However, I must also rely on my
understanding that human beings have a certain flaw: an ability to
believe almost anything.
Logic, however, is perfect.
James Harris
====
> Human beings have a flaw: they can be thoroughly convinced of false
> things.
AS is James Steven Harris mistakenly convinced of his own genius.
====
> Now let's suppose that slowly it sinks in that his path is logically
> flawed.
Are we to understand that, after years of promoting your error-ridden 
proofs of FLT as irrefutable,
you now are proclaiming that Wiles' proof is wrong, and that you are the 
only one who has succeeded?
--
What a maroon! -- Bugs Bunny
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> Human beings have a flaw: they can be thoroughly convinced of false
> things.
I just love the way you state the most commonplace trivialities as if
they were some profound new insight.
> So does Wiles get around the lack of a logical basis by using a finite
> subset to find a restriction on an infinite set?
> The technique is called lifting and is something like infinite
> descent made somewhat famous by Fermat himself.
> It is an intriguing question, so I'll back off somewhat while I
> consider whether or not he somehow gets around the logical fallacy to
> find a way to break it.
You are not qualified to determine whether or not Wiles was successful.
Only a limited number of people on the face of the earth are qualified
(and willing to take the necessary time and effort) to do so.  Until you
are able to read (and understand!) his proof for yourself, you'll just
have to take their word for it.
Or not.  Your opinion has no importance to anyone but yourself, so
believe whatever you want.
-- 
Wayne Brown                | When your tail's in a crack, you improvise
fwbrown@bellsouth.net      |  if you're good enough.  Otherwise you give
                           |  your pelt to the trapper.
e^(i*pi) = -1  -- Euler  |           -- John Myers Myers, 
Silverlock
====
[snip]
>Wiles's supposed accomplishment rests
>*solely* on the assertion of a relatively small group of people that
>his work is correct.
James Harris's supposed accomplishments rest *solely* on the assertion
of one person that his work is correct.
-- 
Yup, you guessed it.  If worse comes to worse, I *will* turn to the
 Army to help me with mathematicians.
  -- James Harris <3c65f87.0304191552.511ad5b4@posting.google.com>
====
> 
> [snip]
> 
>Wiles's supposed accomplishment rests
>*solely* on the assertion of a relatively small group of people that
>his work is correct.
> 
> James Harris's supposed accomplishments rest *solely* on the assertion
> of one person that his work is correct.
What can you do?  Time after time I've faced false assertions, and
time after time people have been unreasonable in the face of
rationality.  Sure I set out a few years back to find some answers to
some math problems, and made a LOT of mistakes along the way.  Yup,
I've made a lot of mistakes.
But right now I'd like some cogent answers to what follows:
I've presented a problem with the logical foundation of Wiles's work
as it relies on association to prove a condition.  More useful
discussion has worked things down to the assertion that Wiles used a
finite set, and lifting to prove the equality of two infinite sets,
where the equality supposedly forces the condition.
There is, however, no reason for the condition given, and no claim of
a reason, where the condition, from my understanding, is that every
elliptic curve is a modular elliptic curve.
The question raised before Wiles's work being whether or not an
elliptic curve could not be modular, where mathematicians had related
various elliptic curves that they tested to something called modular
forms, and found that every one they tested was modular.  Then the
mathematicians Taniyama and Shimura conjectured that every elliptic
curve was modular, which my understanding means, they are associated
with a modular form, where that association can be described by 4
descriptors.
My understanding is that Wiles's work depends on association.
The logical fallacy I've put forward as a challenge against his work
is,
Cum hoc ergo propter hoc.
James Harris
====
[snip]
> My understanding is that Wiles's work depends on association.
>
> The logical fallacy I've put forward as a challenge against his work
> is,
>
> Cum hoc ergo propter hoc.
>
> James Harris
Your 'understanding' is a misunderstanding. What you have put forward is,
Argumentum ad ignorantum.
You have no standing to challenge anyone about anything. Get over it, James 
Harris. You
have been thoroughly debunked. Remove your faulty and error-ridden attempts 
at proving FLT
from public view. You are polluting the internet with your posts.
--
The only thing more pitiful than beating a dead horse is trying to ride 
one.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> there is nothing to differentiate your complaints about Wiles
> from complaints about photons, DNA, the Jurassic  era,
> evolution, relativity, and the rest.
>
> [...] Wiles's supposed accomplishment rests *solely*
> on the assertion of a relatively small group of people that
> his work is correct.
There is no difference in that respect between Wiles and photons,
DNA, etc.   Where differences exist, they tend to favor Wiles' proof
over the other situations.  Experimental evidence, for example, can be
checked more easily, unambiguously and objectively in Wiles' situation
than the others.
> Yet to take one of your examples--photons--and consider that the
> existence of photons has been theorized for some time, but was proven
> by experiment.
It was not *proven* by experiment: another respect in which Wiles'
work is qualitatively more reliable than photons, DNA, and the rest.
The physics experiments were consistent with certain theoretical models,
but of course, you have not even come close to verifying the immense
chain of experimental and theoretical reasoning leading to the current 
models
with photons.  Instead, you rely on textbooks, fourth-hand (if that)
accounts, and the assertions of the Science Establishment.
And the 64 dollar question is why you happily parrot the party line
on matters of photons, DNA, relativity, evolution, the existence of
the Iraq War and Sikkim and Napoleon --- but intone high skepticism
concerning Wiles.
> Since that time from lasers to spectral analysis the theory has fit
> with reality.
Lasers only hurt your cause, as to check that a laser (resp. spectrometry)
experiment actually corroborates photons you would have to check 
matters
of chemistry, crystallography (geology!), engineering, manufacture, and so
on all the way down.  The only way out of this is to accept various 
assertions
on faith from the Evil Scientific Establishment, and the question arises why 
you
are such a sheep and conformist when it comes to non-Wiles but raise 
hightened
standards concerning Wiles (who of courses passes all the scientific 
standards
for photons, etc, and then some).
====
> 
> there is nothing to differentiate your complaints about Wiles
> from complaints about photons, DNA, the Jurassic  era,
> evolution, relativity, and the rest.
>
> [...] Wiles's supposed accomplishment rests *solely*
> on the assertion of a relatively small group of people that
> his work is correct.
> 
> There is no difference in that respect between Wiles and photons,
> DNA, etc.   Where differences exist, they tend to favor Wiles' proof
> over the other situations.  Experimental evidence, for example, can be
> checked more easily, unambiguously and objectively in Wiles' situation
> than the others.
> 
> 
> Yet to take one of your examples--photons--and consider that the
> existence of photons has been theorized for some time, but was proven
> by experiment.
> 
> It was not *proven* by experiment: another respect in which Wiles'
> work is qualitatively more reliable than photons, DNA, and the rest.
> The physics experiments were consistent with certain theoretical models,
> but of course, you have not even come close to verifying the immense
> chain of experimental and theoretical reasoning leading to the current 
models
> with photons.  Instead, you rely on textbooks, fourth-hand (if that)
> accounts, and the assertions of the Science Establishment.
My degree is in physics.  I did physics experiments in school.
 
> And the 64 dollar question is why you happily parrot the party line
> on matters of photons, DNA, relativity, evolution, the existence of
> the Iraq War and Sikkim and Napoleon --- but intone high skepticism
> concerning Wiles.
Wiles's work would mean a workaround to the logical fallacy called,
Cum hoc ergo propter hoc.
Ultimately, if Wiles's work is correct then it does not have any
logical flaws, but checking it potentially involves going through each
step in his work, which is a formidable task.  If he did find a proof,
then I think it interesting on logical grounds that there is a
workaround i.e. that Cum hoc ergo propter hoc is not actually a
logically fallacious approach.
Now as for physics results, like many people trained in physics, I
keep a skeptical eye on theory, and depend on things I've personally
checked, or that are very unlikely to be wrong that have been checked
by others.  Physicists can be hard-liners to the extent that they
don't believe physics they haven't personally checked.  I'm not.  Like
how I believe that nuclear weapons work.  But still realize that the
absolute truth may be something other than what I've learned.
In mathematics, absolute truth *can* be determined, just like a
logical argument can be checked against certain rules for internal
consistency.
 
> Since that time from lasers to spectral analysis the theory has fit
> with reality.
> 
> Lasers only hurt your cause, as to check that a laser (resp. 
spectrometry)
> experiment actually corroborates photons you would have to check 
matters
> of chemistry, crystallography (geology!), engineering, manufacture, and 
so
> on all the way down.  The only way out of this is to accept various 
assertions
> on faith from the Evil Scientific Establishment, and the question arises 
why you
> are such a sheep and conformist when it comes to non-Wiles but raise 
hightened
> standards concerning Wiles (who of courses passes all the scientific 
standards
> for photons, etc, and then some).
As a person with a science degree, I guess you'd consider me a part of
the Evil Scientific Establishment.
It's actually more fun attacking them than just sitting around
believing in them.  Because you learn a lot in the attack, and your
guarantee from math and logic is that the proof doesn't care.
To a math proof, you do not exist as a relevant entity.
James Harris
====
> Wiles's work would mean a workaround to the logical fallacy called,
> Cum hoc ergo propter hoc.
No, it wouldn't.
[snip]
> ... I feel like I can
> speak confidently on the subject.
You speak confidently whether you know what you're talking about or not. 
Confidence is not your
problem, honesty and credibility are.
> To a math proof, you do not exist as a relevant entity.
You do not exist as a relevant entity.
--
There are two things you must never attempt to prove: the unprovable -- and 
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> 
> there is nothing to differentiate your complaints about Wiles
> from complaints about photons, DNA, the Jurassic  era,
> evolution, relativity, and the rest.
>
> [...] Wiles's supposed accomplishment rests *solely*
> on the assertion of a relatively small group of people that
> his work is correct.
> 
> There is no difference in that respect between Wiles and photons,
> DNA, etc.   Where differences exist, they tend to favor Wiles' proof
> over the other situations.  Experimental evidence, for example, can be
> checked more easily, unambiguously and objectively in Wiles' situation
> than the others.
> 
> 
> Yet to take one of your examples--photons--and consider that the
> existence of photons has been theorized for some time, but was proven
> by experiment.
> 
> It was not *proven* by experiment: another respect in which Wiles'
> work is qualitatively more reliable than photons, DNA, and the rest.
> The physics experiments were consistent with certain theoretical 
models,
> but of course, you have not even come close to verifying the immense
> chain of experimental and theoretical reasoning leading to the current 
models
> with photons.  Instead, you rely on textbooks, fourth-hand (if 
that)
> accounts, and the assertions of the Science Establishment.
> 
> My degree is in physics.  I did physics experiments in school.
>  
> And the 64 dollar question is why you happily parrot the party line
> on matters of photons, DNA, relativity, evolution, the existence of
> the Iraq War and Sikkim and Napoleon --- but intone high skepticism
> concerning Wiles.
> 
> Wiles's work would mean a workaround to the logical fallacy called,
> Cum hoc ergo propter hoc.
> 
> Ultimately, if Wiles's work is correct then it does not have any
> logical flaws, but checking it potentially involves going through each
> step in his work, which is a formidable task.  If he did find a proof,
> then I think it interesting on logical grounds that there is a
> workaround i.e. that Cum hoc ergo propter hoc is not actually a
> logically fallacious approach.
> 
> Now as for physics results, like many people trained in physics, I
> keep a skeptical eye on theory, and depend on things I've personally
> checked, or that are very unlikely to be wrong that have been checked
> by others.  Physicists can be hard-liners to the extent that they
> don't believe physics they haven't personally checked.  I'm not.  Like
> how I believe that nuclear weapons work.  But still realize that the
> absolute truth may be something other than what I've learned.
> 
> In mathematics, absolute truth *can* be determined, just like a
> logical argument can be checked against certain rules for internal
> consistency.
>
How? How can absolute truth be determined? About a month ago you
essentially:
1) proof of an absolute kind, presumably stated in the symbolism of
formal logic, and
2) proof that merely convinces other mathematicians, presumably stated
in some meta-language (like English).
Furthermore, your position is that proof of the second kind is what
most mathematicians produce, and is not good enough. You go on to say
that you produce proofs of the 1st kind.
Question: How does one determine that a proof or mathematical
argument is absolutely and irrefutably correct?
The validity must be checked by 1) God, 2) a machine, or 3) a human
being, as a proof cannot check itself.
I think (1) is out, for the time being anyway.
What about (2)? Well, we could encode some axioms and rules of
inference, but it occurs to me that a few problems could arise. First,
the algorithm may take an unreasonable amount of time to reach a
decision. Second, hardware failure, electrical surges, sunspot
activity, running the program under Microsoft Windows, etc. could
cause erroneous results. Third, and perhaps most importantly, a human
being (or beings) must write the software. Therefore, any errors
caused by people could conceivably appear here. That leaves us with
option (3). As we all know, people make mistakes. They make mistakes
writing proofs. The publisher/editors of a journal may make a mistake
mistake by erroneously believing the proof.
Who has the final and ultimate authority to say that a given argument
is valid or not? Surely, not one person. There is so much mathematics,
no one person can know it all.
So, a proof then must be judged by the readers. If there is a
disagreement, then the sides may argue their cases until one side
prevails and convinces the other, at least within a given mathematical
system.
Therefore, in this sense, all proofs are of the second type. We must
strive to convince other mathematicians. That is all there is --
simply because there is no other means of asserting the validity of a
mathematical argument. It really is an appeal to the gallery.
We must also consider that mathematics may be inconsistent. According
to Kurt Godel, this is a possibility (at least for mathematical
systems strong enough to support integer arithmetic.)
So much for proofs being irrefutable, absolute, perfect, eternal, etc.
ad nauseum.
> Since that time from lasers to spectral analysis the theory has fit
> with reality.
> 
> Lasers only hurt your cause, as to check that a laser (resp. 
spectrometry)
> experiment actually corroborates photons you would have to check 
matters
> of chemistry, crystallography (geology!), engineering, manufacture, and 
so
> on all the way down.  The only way out of this is to accept various 
assertions
> on faith from the Evil Scientific Establishment, and the question arises 
why you
> are such a sheep and conformist when it comes to non-Wiles but raise 
hightened
> standards concerning Wiles (who of courses passes all the scientific 
standards
> for photons, etc, and then some).
> 
> As a person with a science degree, I guess you'd consider me a part of
> the Evil Scientific Establishment.
> 
> experiences in the military, where I actually had the honor of giving
> a lecture on the physics of lasers to the medical personnel at Madigan
> Army Medical Center, including the surgeons, other doctors and nurses,
> for their medical continuing education credits, I feel like I can
> speak confidently on the subject.
> 
> My position on Wiles is about logic.  Emotional response is not
> necessary as I assure you that if Wiles found a proof then there is no
> need for concern.  If he did not, why fight for a false belief?
> 
> Math proofs are indestructible, incorruptible, and irrefutable.
> 
> It's actually more fun attacking them than just sitting around
> believing in them.  Because you learn a lot in the attack, and your
> guarantee from math and logic is that the proof doesn't care.
> 
> To a math proof, you do not exist as a relevant entity.
> 
> 
> James Harris
====
> there is nothing to differentiate your complaints about Wiles
> from complaints about photons, DNA, the Jurassic  era,
> evolution, relativity, and the rest.
> Yet to take one of your examples--photons--and consider that the
> existence of photons has been theorized for some time, but was
> proven by experiment.
>
> It was not *proven* by experiment: another respect in which Wiles'
> work is qualitatively more reliable than photons, DNA, and the rest.
> The physics experiments were consistent with certain theoretical 
models,
> but of course, you have not even come close to verifying the immense
> chain of experimental and theoretical reasoning leading to the current 
models
> with photons.  Instead, you rely on textbooks, fourth-hand (if 
that)
> accounts, and the assertions of the Science Establishment.
>
> My degree is in physics.  I did physics experiments in school.
Ask the school for a refund.
First, the existence of photons is not a precisely formulated 
statement
as in the case of Wiles' proof, let alone one provable by experiment.  
There
are of course theoretical models (not necessarily well-defined or known
to be logically consistent, by the way) within which one can single out
certain objects as photons.
Second, your student experiments in optics could not possibly replicate
the mountain of theoretical and experimental steps involved in building
up any of the theoretical model(s) involving photons.  Instead, you
accepted on trust assertions by textbook authors, professors and similar
purveyors of the Social Truth that you like to castigate, amounting to
a certification-by-authority that the apparatus you were doing the 
experiments
with actually corresponded to the theory in the manner claimed.
You did not produce the relevant gases, crystals, apparatus, electricity, 
..
involved in laser experiments, nor did you do the experimentation
needed to corroborate the values of relevant physical and chemical
parameters listed in the CRC handbook, and so on all the way down.
What actually happened is that a long and social process
of knowledge-accumulation occurred and you took the results on trust.
In particular, if your experiments gave a wrong result, the conclusion
would be that you made a mistake, not that photons' existence is in doubt;
a pure assertion of authority by the Scientific Establishment concerning 
its
Social Truth, which you accept without any objection in all the non-FLT
situations.
Note that your repeatedly discredited objections in this thread about
Wiles' logic are irrelevant, as you also object to Ribet's proof without
giving any particular reason to doubt it.   The matter is simply one of an
obvious double standard produced for the occasion, where social 
certification
by a small network of experts counts as OK for photons, DNA,
evolution, relativity, the Jurassic era (or the existence of Napoleon and
George W Bush), etc --- but somehow the information that experts
have certified Ribet's and Wiles' work is cast as suspicious.
====
> 
> My degree is in physics.  I did physics experiments in school.
Yet you seem never to have encountered the SR thought
experiment called the superluminal scissors or had any
idea what I was talking about in sci.physics when I
explained how a 5 m/sec water jet can be used to create
an illusion of arbitrarily fast, even superluminal
motion.
           - Randy
====
> 
> there is nothing to differentiate your complaints about Wiles
> from complaints about photons, DNA, the Jurassic  era,
> evolution, relativity, and the rest.
>  
> Yet to take one of your examples--photons--and consider that the
> existence of photons has been theorized for some time, but was
> proven by experiment.
>
> It was not *proven* by experiment: another respect in which Wiles'
> work is qualitatively more reliable than photons, DNA, and the rest.
> The physics experiments were consistent with certain theoretical 
models,
> but of course, you have not even come close to verifying the immense
> chain of experimental and theoretical reasoning leading to the current 
models
> with photons.  Instead, you rely on textbooks, fourth-hand (if 
that)
> accounts, and the assertions of the Science Establishment.
>
> My degree is in physics.  I did physics experiments in school.
> 
> Ask the school for a refund.
I had a full tuition scholarship.
 
> First, the existence of photons is not a precisely formulated 
statement
> as in the case of Wiles' proof, let alone one provable by experiment.  
There
> are of course theoretical models (not necessarily well-defined or known
> to be logically consistent, by the way) within which one can single out
> certain objects as photons.
> 
> Second, your student experiments in optics could not possibly replicate
> the mountain of theoretical and experimental steps involved in building
> up any of the theoretical model(s) involving photons.  Instead, you
> accepted on trust assertions by textbook authors, professors and similar
> purveyors of the Social Truth that you like to castigate, <deleted>
Oh please, you've been beaten.  That's what's annoying about Usenet as
some loser will state a case, get their ass kicked, but STILL keep
coming back as if nothing happened.
My *degree* is in physics.  I went to school on a full-tuition
scholarship, and you stepped into my field with your assertions, got
your ass kicked but refuse to back down.
Now *emotion* is not necessary when it comes to Wiles's work.  If he
found a proof I can assure you that it is irrefutable.  That's why
it'd be a proof.  All this emotion just annoys me, as part of the fun
of science and mathematics is attacking things that are supposedly
proven.
It's GREAT fun not just accepting what people tell you.  But that's
what really annoys me about mathematicians as time after time I get
yahoo's replying back in defense of mathematics, using tactics.
But you see, not a single REAL mathematician in the world gets excited
about an attack on a proof.  No mathematician worth the title would
get even a little concerned, nor would they lose sleep, or find
themselves emotional about some person--any person--any time--any
place--who decides to go after a math proof.
That's because a math proof is indestructible, incorruptible, and
irrefutable.
It just doesn't care if you attack it, and no real mathematician would
care either.
Now I've discovered math proofs, which is why I'm not concerned about
people refuting them because they are proofs.  And in fact people who
call themselves mathematicians can't touch them, so they come up with
extraneous stuff, or make claims of finding their own proofs to refute
my proofs, but you see, proofs don't duel.
And you know what?  I think that feature of mathematics terrifies some
people who call themselves mathematicians.  Mathematics does NOT care
what you call yourself.  It DOES NOT care that you have a mortgage. 
It DOES NOT CARE that you really, really, really want people to like
you and think you're a great mathematician.
Now I've made a specific claim against Wiles's work.  If he found a
proof the claim can be answered, but even if it is answerable then it
must be true that he has found a way around what is considered to be a
logically fallacious approach.  Logicians should thank him in that
case for correcting them.
My challenge is a logical one.  Wiles's work fails and is not a proof
as it is an argument by Cum hoc ergo propter hoc.
James Harris
====
Is there an easy way to come up with all the schedules where n teams
play each other once?
I tried several avenues.  I don't have the time or expertise to tackle
this and thought this is the newsgroup with the expertise that may know.
I hope you are not offended.
I think it can be reduced to symmetric matrices that are 2nd roots of
unity with zero diagonal, but I may be wrong.
thanks
====
> Is there an easy way to come up with all the schedules where n teams
> play each other once?
There are tables of the solution available somewhere or other. At
least that's how my local YMCA schedules its basketball leagues.
-- 
Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/Bukharin.html
                            To solve Linear Programs:  .../LPSolver.html
r           c                                 A game:  .../Keynes.html
 v         s a           Whether strength of body or of mind, or wisdom, or
  i       m   p          virtue, are found in proportion to the power or 
wealth
   e     a     e         of a man is a question fit perhaps to be discussed 
by
    n   e       .        slaves in the hearing of their masters, but highly
     @ r         c m     unbecoming to reasonable and free men in search of
      d           o      the truth.    -- Rousseau
====
> Is there an easy way to come up with all the schedules where n teams
> play each other once?
> 
> I tried several avenues.  I don't have the time or expertise to tackle
> this and thought this is the newsgroup with the expertise that may know.
> I hope you are not offended.
> 
> I think it can be reduced to symmetric matrices that are 2nd roots of
> unity with zero diagonal, but I may be wrong.
> 
> thanks
Have a look at:
  http://mathforum.org/library/drmath/view/54715.html
====
>Is there an easy way to come up with all the schedules where n teams
>play each other once?
>
On the surface, this seems to translate geometrically as the typical 
diagonals
in the polygon of n sides type of counting.  
Make a regular polygon; the number of verices equals the number of sides.  
how
many diagonals are present, including the sides?   Now make another regular
polygon using one more vertex, and equivalently, one more side.  Count the
diagonals and include the sides.  Keep doing this until you have a pattern 
for
which to give a form.  
G    C
====
>> Is there an easy way to come up with all the schedules where n teams
>> play each other once?
>> 
>> I tried several avenues.  I don't have the time or expertise to tackle
>> this and thought this is the newsgroup with the expertise that may know.
>> I hope you are not offended.
>> 
>> I think it can be reduced to symmetric matrices that are 2nd roots of
>> unity with zero diagonal, but I may be wrong.
>> 
>> thanks
>
>Have a look at:
>
>  http://mathforum.org/library/drmath/view/54715.html
Now, I hope I can get my computer to do this.
====
>>Is there an easy way to come up with all the schedules where n teams
>>play each other once?
>>
>
>On the surface, this seems to translate geometrically as the typical 
diagonals
>in the polygon of n sides type of counting.  
>
>Make a regular polygon; the number of verices equals the number of sides.  
how
>many diagonals are present, including the sides?   Now make another 
regular
>polygon using one more vertex, and equivalently, one more side.  Count the
>diagonals and include the sides.  Keep doing this until you have a pattern 
for
>which to give a form.  
>
>G    C
====
>Is there an easy way to come up with all the schedules where n teams
>play each other once?
>
>I tried several avenues.  I don't have the time or expertise to tackle
>this and thought this is the newsgroup with the expertise that may know.
>I hope you are not offended.
>
>I think it can be reduced to symmetric matrices that are 2nd roots of
>unity with zero diagonal, but I may be wrong.
Please check my post at
If n is odd, add a dummy team as a bye to make the number even.
Rob Johnson
take out the trash before replying
====
Good day,
I am processing batches of data stored in arrays, and I have seen
examples of vector mathematics (rather than loops) being used to do
the processing.
For instance, for performing a linear regression on a set of data X
and Y (to find the best-fit slope 'k') I am used to seeing the
mean-squared-error (MSE) being expressed as a summation of the
squared-errors of each data pair, but I have also seen the MSE
expressed as in dot-product form which is equivalent.  In the end, the
result is k = (x'x)^(-1) * x'y.
To get this optimized result, a partial derivative with resprct to k
is taken on the MSE function.  I can follow along, but I am unsure
about some of the procedures, especially with regard to getting the
_order_ of the vectors expressed properly to ensure dimensional
agreement (e.g. how to expand (kx-y)*(kx-y)' , how to apply chain
rules when taking derivatives), but I don't know what the forms or
tecniques are called.
This is not the vector algebra or vector calculus that I have been
exposed to.  Or is it?  I'd appreciate knowing so that I could have a
more fruitful search on the topic.
-Jagan
====
> Good day,
> 
> I am processing batches of data stored in arrays, and I have seen
> examples of vector mathematics (rather than loops) being used to do
> the processing.
> 
> For instance, for performing a linear regression on a set of data X
> and Y (to find the best-fit slope 'k') I am used to seeing the
> mean-squared-error (MSE) being expressed as a summation of the
> squared-errors of each data pair, but I have also seen the MSE
> expressed as in dot-product form which is equivalent.  In the end, the
> result is k = (x'x)^(-1) * x'y.
> 
> To get this optimized result, a partial derivative with resprct to k
> is taken on the MSE function.  I can follow along, but I am unsure
> about some of the procedures, especially with regard to getting the
> _order_ of the vectors expressed properly to ensure dimensional
> agreement (e.g. how to expand (kx-y)*(kx-y)' , how to apply chain
> rules when taking derivatives), but I don't know what the forms or
> tecniques are called.
> 
> This is not the vector algebra or vector calculus that I have 
been
> exposed to.  Or is it?  I'd appreciate knowing so that I could have a
> more fruitful search on the topic.
You see a lot of manipulations of this sort (e.g.,
taking the derivative of a matrix expression) in
optimization theory. You might for instance check into
the theory of quadratic programming, which is the theory
of optimizing a general multivariate quadratic function
under linear equality and inequality constraints.
Least-squares is one particularly easy quadratic problem:
an unconstrained minimization with a positive definite
coefficient matrix.
When I was first exposed to this kind of manipulation,
I found it helpful to work out expressions in terms
of individual components and summations. Going back
and forth between those kinds of things and their matrix
equivalents is really helpful to get facility with the
algebra.
Here are a couple of particularly useful identities
for you for free (prime ' means transpose).
1) x'Qx = sum (i) q_ii * x_i^2
      + sum(i!=j) (q_ij + q_ji) * x_i * x_j
If Q is not symmetric, there is always a different matrix
P which gives the same function but is symmetric:
P = 1/2(Q + Q'). Component wise, p_ij = (q_ij+q_ji)/2
for j!=i, and p_ii = q_ii, and
  x'Px = sum(i) p_ii * x_i^2 + sum(i!=j) 2*p_ij*x_i*x_j
This tells you the correspondence between a general multivariate
quadratic and its symmetric coefficient matrix.
2) If b is a vector, b'x is a scalar. Thus it is
its own transpose, i.e. b'x = x'b.
3) The gradient grad(f) where f is a scalar is a vector
whose i-th component is df/dx_i.
  grad(b'x) = b
Proof: d(b'x)/dx_i = d/dx_i sum(k) b_k * x_k = b_i.
The i-th component of grad(b'x) is b_i.
4) grad(x'Px) = 2Px
Proof: Left to reader.
         - Randy
====
> [snip]
> To get this optimized result, a partial derivative with resprct to k
> is taken on the MSE function.  I can follow along, but I am unsure
> about some of the procedures, especially with regard to getting the
> _order_ of the vectors expressed properly to ensure dimensional
> agreement (e.g. how to expand (kx-y)*(kx-y)' , how to apply chain
> rules when taking derivatives), but I don't know what the forms or
> tecniques are called.
This is vector calculus, but using a trick that you may not have seen.
Notation: the derivative of a function F:R^n->R^m at x is DF(x).
DF(x) is a linear map DF(x):R^n->R^m.  Given a vector h in R^n,
DF(x)(h) is a vector in R^m.
If F is a linear function, then DF(x)(h) = F(h) .  This helps you
compute even when DF(x) itself is hard to write down using vector (or
matrix) notation.  This is the trick: rather than work with DF(x),
work with DF(x)(h).
For example, consider the inner product F(x,y) = x'y .  (Vectors in
R^n are column vectors; prime ' denotes transpose.)  Then the partial
derivatives with respect to the first argument (x) and the second
argument (y) can be expressed by
  D_1F(x,y)(h) = h'y
  D_2F(x,y)(h) = x'h .
The chain rule (using . to represent composition) is
  D(F.G)(x) = DF(G(x)) . DG(x)
so
  D(F.G)(x)(h) = DF(G(x))(DG(x)(h)) .
The product rule for scalar functions of vectors is
  D(F*G)(x)(h) = DF(x)(h)*G(x) + F(x)*DG(x)(h) .
The product rule for inner products <,> of vectors is
  D<F,G>(x)(h) = <DF(x)(h),G(x)> + <F(x),DG(x)(h)> .
(You should prove these, and derive similar rules for other things
such as the cross product in R^3.)
An excellent example of this, which illustrates both the concept and
how difficult it can be to write down DF(x) rather than DF(x)(h), is
the formula for the derivative of the matrix inversion function.  Let
F(X) = X^{-1}, where X is an invertible matrix.  Then let
  G(X) := X F(X) - I = 0 ;
then, computing the derivative and applying the product rule,
  DG(X)(H) = H F(X) + X DF(X)(H) = 0
so
  X DF(X)(H) = - H F(X) .
Left-multiplying by F(X)=X^{-1},
  DF(X)(H) = - F(X) H F(X) = - X^{-1} H X^{-1} .
(Recall that matrix multiplication does not commute, so this is not
the same as -X^{-2}H ; in particular, the scalar formula DF(X)=-X^{-2}
does not hold for matrices.)
Kevin.
====
testing
====
> testing
F-
====
|-|erc says...
>YES I showed this in Cantor's disproof thread, by definition
>computable reals are computable, which means they can
>be given a TM number.
But not every TM number represents a computable real. To get
a bijection between the natural numbers and computable reals,
you have to set up a correspondence such that each natural
number corresponds to exactly one computable real, and every
computable real corresponds to exactly one natural number.
A computable version of Cantor's proof shows that such a
correspondence cannot be computable.
More specifically, let C = the set of all natural numbers m such
that m represents a TM number for a computable real number. Let
f be a one-to-one mapping from C onto N (the set of all natural
numbers). Then f cannot be a computable function. So there is no
computable bijection between the computable real numbers
and the naturals.
   
--
Daryl McCullough
Ithaca, NY
====
> I had thought that I was reading a message from a human being;
> but now I think I'm hallucinating.
Liar.
====
> If you're going to mock the very idea of a world
> of computation, which you seem to be doing, then
> there's little point for me to respond to you.
> 
> Good. Then why did you?
Courtesy. He did ask a question.
====
> On the contrary, Cantor's proof goes through perfectly
> well if you only consider computable mathematical
> objects: There does not exist a computable bijection
> between the natural numbers and the computable reals.
That's true, and relevant: you do not thereby 
conclude that the computable reals are
uncountable.
But would you really call that Cantor's proof?
 <874r1dem8d.fsf@phiwumbda.localnet> <vhsihi3vtgo5f6@corp.supernews.com>
 <8765ltbnw8.fsf@phiwumbda.localnet>
 <25bac3c0.0307231415.1043d316@posting.google.com>
 <87znj4l6p9.fsf@phiwumbda.localnet>
 <25bac3c0.0307241131.795f970d@posting.google.com>
 <87n0f38zl9.fsf@phiwumbda.localnet>
 <25bac3c0.0307250950.43084790@posting.google.com>
====
>
>> Infinitary objects exist in the sense that we
>> can construct finitary approximations to the objects.
>> 
>> Is this or is this not a rejection that R exists? 
>
> It really does depend on what you mean by exists.
>
>
>> Is R constructed by
>> finitary approximations in your view?  
>
> The notion of R as it is defined within the axiom
> systems inspired by Cantor's theory cannot be
> constructed by finitary approximations. In fact,
> that's what the diagonalization argument proves.
>
> Nevertheless, we can construct something very
> close to our intuitive notions of R by finitary
> approximations. 
But, the definition of R precedes Cantor.  Take Dedekind's definition,
say, or Cauchy's, and it is easy to derive that the usual canonical
decimal representation works.  Cantor did *not* offer a new or
different definition of R.  Nor does the definition of R come from
Cantor's work.  Rather, Cantor proved an important theorem about the
existing definition of R.
You need to check your history.
> I would claim that Cantor's theory leads to
> counter-intuitive conclusions about R, though
> I realized that those who have spent years
> studying Cantor's theory may have changed their
> intuitions.
Yes and no.  Cantor's theory doesn't lead to counter-intuitive
conclusions, but that's because the theory of R sufficient for
Cantor's result is not due to Cantor.  In other words, I agree that
the prevailing theory of R in Cantor's time to the present leads to at
least one counter-intuitive (i.e., surprising) result, namely that 
|N| < |R|, but you seem to be showing historical confusion in calling
it Cantor's theory[1].  
One ought to have a better reason for tossing out the completeness of
R than that he didn't expect |N| < |R|.
Footnotes: 
[1]  I'm not a historian, and could be persuaded that I am mistaken.
For this, one must provide either historical references or
well-respected secondary sources.
-- 
Come on people!!!  The US just blew up a lot of people in Iraq, don't
you realize that a person with my exposure might just end up dead, by
mysterious circumstances? 
  --James Harris, on the dangers of proving Fermat's last theorem
====
david_lawrence_petry@yahoo.com (David Petry) says...
>
>
>> On the contrary, Cantor's proof goes through perfectly
>> well if you only consider computable mathematical
>> objects: There does not exist a computable bijection
>> between the natural numbers and the computable reals.
>
>That's true, and relevant: you do not thereby 
>conclude that the computable reals are
>uncountable.
Yes, I would. X is uncountable means the same thing as
there does not exist a bijection between X and N.
Of course, whether a set is uncountable or not depends on
the set of bijections you are willing to consider. If you
only consider computable bijections, then the computable
reals are uncountable. If you allow definable bijections
(definable in ZF, for instance, or in type theory) then
the computable reals become countable. 
>But would you really call that Cantor's proof?
It is structurally the same proof, except that instead of
assume that f(n) is an enumeration of all reals
you say assume that f(n) is a computable enumeration
of all the computable reals.
The big difference between computable objects and noncomputable
objects is not Cantor's theorem, which is the same in both cases.
It is the theorem that Any subset of a countable set is countable.
That theorem is true classically, but it isn't true constructively.
To see that it isn't true constructively, let NH = the set of
Turing machine indices n such that {n}(n) never halts. There is no
constructive bijection between NH and N. So NH is uncountable by
constructive functions. But NH is a subset of N.
--
Daryl McCullough
Ithaca, NY
====
>> So far nobody has taken up my challenge problem of determining which
>  dyadic
>> rationals belong to the Cantor set.
>>
> Is there a decision predicate for p/q which is faster than caclulating 
the
> ternary expansion?
> 
> Phil
Yes. Here's one that is faster by O(1) (so not much faster, I admit).
Choose Q. Store all rationals p/q, where p/q is in the Cantor set, and
q<Q, in a big hashtable. Then the algorithm goes:
given p/q:
inCantor(p,q) {
  if q<Q return inHashTable(p,q);
  return inCantorViaTernaryExpansion(p,q);
}
Not very useful in general, I admit, but technically, it is 'faster'.
On the other hand, it would be VERY useful if the rationals of
interest are bounded in their denominator.
====
Starblade Darksquall
> Proof:
>
> 1) Ax(BxC) != (AxB)xC
> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = -A(C*B)+B(C*A)
> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
> when Bx(AxC) = 0, then clearly cross products are noncommunative.
> 2) AxA = 0
> This follows from the definition of the cross product. The cross
> product of parallel vectors is zero.
> 3) (A+B)xC = AxC+BxC
> Since the vector cross product IS distributive over addition. This is
> easy enough to verify.
> 4) Ax(BxC)+Bx(CxA)+Cx(AxB) = 0
> Changing it to (B(A*C)-C(A*B))+(C(B*A)-A(B*C))+(A(C*B)-B(C*A)) makes
> it obvious that this identity is true, since that the dot product is
> communative.
> 5) AxB=-BxA
> This is as easy to verify as 2.
>
> So clearly, we have known about a type of lie algebra for longer than
> we let on. In fact, we use lie algebra type mathematics in dealing
> with even classical physics all the time, particularly in
> electromagnetics.
Er, sort of. The old vector product AxB is the same as *(A^B) where * is 
the
Hodge dual and ^ is the (antisymmetric) exterior product. (I'm not sure if
Hodge dual is standard jargon, but it's easy to see what the term is
intended to mean.)
LH
====
> 
> Proof:
> 
> 1) Ax(BxC) != (AxB)xC
> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = -A(C*B)+B(C*A)
> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
> when Bx(AxC) = 0, then clearly cross products are noncommunative.
> 
> Hmmm. That's not one of the Lie algebra axioms!
Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms. If [A,B] = AxB
then you get the above result.
(...Starblade Riven Darksquall...)
====
>> 
>> Proof:
>> 
>> 1) Ax(BxC) != (AxB)xC
>> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = -A(C*B)+B(C*A)
>> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
>> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
>> when Bx(AxC) = 0, then clearly cross products are noncommunative.
>> 
>> Hmmm. That's not one of the Lie algebra axioms!
> 
> Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms. 
No it isn't. There are Lie algebras L for which [x[yz]] = [[xy]z]
for all x, y, z in L.
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
 The League of Gentlemen
====
> Starblade Darksquall
> Proof:
>
> 1) Ax(BxC) != (AxB)xC
> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = -A(C*B)+B(C*A)
> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
> when Bx(AxC) = 0, then clearly cross products are noncommunative.
> 2) AxA = 0
> This follows from the definition of the cross product. The cross
> product of parallel vectors is zero.
> 3) (A+B)xC = AxC+BxC
> Since the vector cross product IS distributive over addition. This is
> easy enough to verify.
> 4) Ax(BxC)+Bx(CxA)+Cx(AxB) = 0
> Changing it to (B(A*C)-C(A*B))+(C(B*A)-A(B*C))+(A(C*B)-B(C*A)) makes
> it obvious that this identity is true, since that the dot product is
> communative.
> 5) AxB=-BxA
> This is as easy to verify as 2.
>
> So clearly, we have known about a type of lie algebra for longer than
> we let on. In fact, we use lie algebra type mathematics in dealing
> with even classical physics all the time, particularly in
> electromagnetics.
> Er, sort of. The old vector product AxB is the same as *(A^B) where * is 
the
> Hodge dual and ^ is the (antisymmetric) exterior product. (I'm not sure 
if
> Hodge dual is standard jargon, but it's easy to see what the term is
> intended to mean.)
> LH
Are you talking about tensors, or something else entirely?
(...Starblade Riven Darksquall...)
====
> 
>> 
>> Proof:
>> 
>> 1) Ax(BxC) != (AxB)xC
>> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = 
-A(C*B)+B(C*A)
>> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
>> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
>> when Bx(AxC) = 0, then clearly cross products are noncommunative.
>> 
>> Hmmm. That's not one of the Lie algebra axioms!
> 
> Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms. 
> 
> No it isn't. There are Lie algebras L for which [x[yz]] = [[xy]z]
> for all x, y, z in L.
That's not what I heard.
http://mathworld.wolfram.com/LieAlgebra.html
Take a look there.
And even if they're wrong, and I'm wrong, about that, then that still
doesn't make the results of my proof (that the cross product is a form
of lie algebra) any less valid.
(...Starblade Riven Darksquall...)
====
>> 
>>> 
>>> Proof:
>>> 
>>> 1) Ax(BxC) != (AxB)xC
>>> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) =
>>> -A(C*B)+B(C*A) = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B)
>>> since A(B*C) != C(A*B) except for special cases, namely, when
>>> A(B*C)-C(A*B) = 0, IE, when Bx(AxC) = 0, then clearly cross 
products
>>> are noncommunative.
>>> 
>>> Hmmm. That's not one of the Lie algebra axioms!
>> 
>> Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms.
>> 
>> No it isn't. There are Lie algebras L for which [x[yz]] = [[xy]z]
>> for all x, y, z in L.
> 
> That's not what I heard.
> 
> http://mathworld.wolfram.com/LieAlgebra.html
> 
> Take a look there.
Their defintion of nonassociative algebra is rather dubious.
Did you find any examples of associative Lie algebras (your exercise).
 
> And even if they're wrong, and I'm wrong, about that,
Yup, and yup!
> then that still
> doesn't make the results of my proof (that the cross product is a form
> of lie algebra) any less valid.
And doesn't make this example any less of an exercise in chapter 1
of any Lie algebra texttbook :-)
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
 The League of Gentlemen
====
> 
>> 
>>> 
>>> Proof:
>>> 
>>> 1) Ax(BxC) != (AxB)xC
>>> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) =
>>> -A(C*B)+B(C*A) = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B)
>>> since A(B*C) != C(A*B) except for special cases, namely, when
>>> A(B*C)-C(A*B) = 0, IE, when Bx(AxC) = 0, then clearly cross 
products
>>> are noncommunative.
>>> 
>>> Hmmm. That's not one of the Lie algebra axioms!
>> 
>> Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms.
>> 
>> No it isn't. There are Lie algebras L for which [x[yz]] = [[xy]z]
>> for all x, y, z in L.
> 
> That's not what I heard.
> 
> http://mathworld.wolfram.com/LieAlgebra.html
> 
> Take a look there.
> 
> Their defintion of nonassociative algebra is rather dubious.
I didn't get it. Their definition reads 'a nonassociative algebra is
an algebra that ain't associative'. Isn't that true?
> 
> Did you find any examples of associative Lie algebras (your exercise).
I think there are plenty of associative Lie algebras. Just the
generators must commute that's all I guess.
>  
> And even if they're wrong, and I'm wrong, about that,
> 
> Yup, and yup!
> 
> then that still
> doesn't make the results of my proof (that the cross product is a form
> of lie algebra) any less valid.
> 
> And doesn't make this example any less of an exercise in chapter 1
> of any Lie algebra texttbook :-)
I guess Cross products are Lie Algebras [or is it Cross Algebra?]
They satisfy:
1) Linearity in both factors
2) [A,B]=-[B,A] 
3) Jacobi Identity
====
>> 
>> Their defintion of nonassociative algebra is rather dubious.
> 
> I didn't get it. Their definition reads 'a nonassociative algebra is
> an algebra that ain't associative'. Isn't that true?
There we are then. So must Lie algebras be nonassociative?
 
>> Did you find any examples of associative Lie algebras (your exercise).
> 
> I think there are plenty of associative Lie algebras. Just the
> generators must commute that's all I guess.
Then your answer is no :-)
>>  
> I guess Cross products are Lie Algebras [or is it Cross Algebra?]
No. 
<pedantry>
The cross product is the operation for one particular Lie algebra,
not a Lie algenra itself.
</pedantry>
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
 The League of Gentlemen
====
|> Did you find any examples of associative Lie algebras (your exercise).
|
|I think there are plenty of associative Lie algebras. Just the
|generators must commute that's all I guess.
actually there's plenty of examples of associative but non-abelian lie
algebras.
-- 
====
> 
>> 
>> Their defintion of nonassociative algebra is rather dubious.
> 
> I didn't get it. Their definition reads 'a nonassociative algebra is
> an algebra that ain't associative'. Isn't that true?
> 
> There we are then. So must Lie algebras be nonassociative?
>  
>> Did you find any examples of associative Lie algebras (your exercise).
> 
> I think there are plenty of associative Lie algebras. Just the
> generators must commute that's all I guess.
> 
> Then your answer is no :-)
> 
:)
Yeah, I didn't get it. Ok, I'll be looking for them. 
But Lie algebras have this Jacobi Identity. They sure give us the lack
of associativity ... don't they?
>>  
> I guess Cross products are Lie Algebras [or is it Cross Algebra?]
> 
> No. 
> 
> <pedantry>
> 
> The cross product is the operation for one particular Lie algebra,
> not a Lie algenra itself.
> 
> </pedantry>
Yes. Indeed. I intended to say 'I guess cross product is a Lie
algebra' or something. Somehow got it messed up.
====
> 
> |> Did you find any examples of associative Lie algebras (your exercise).
> |
> |I think there are plenty of associative Lie algebras. Just the
> |generators must commute that's all I guess.
> 
> 
> actually there's plenty of examples of associative but non-abelian lie
> algebras.
Are there any that are not nilpotent of class 2?
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
 The League of Gentlemen
====
|
|> 
|> |> Did you find any examples of associative Lie algebras (your 
exercise).
|> |
|> |I think there are plenty of associative Lie algebras. Just the
|> |generators must commute that's all I guess.
|> 
|> 
|> actually there's plenty of examples of associative but non-abelian lie
|> algebras.
|
|Are there any that are not nilpotent of class 2?
i think a one-line proof rules that out, using the leibniz identity
[[a,b],c] = [[a,c],b]+[a,[b,c]]
.
-- 
====
> 
> Proof:
> 
> 1) Ax(BxC) != (AxB)xC
> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = 
-A(C*B)+B(C*A)
> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
> when Bx(AxC) = 0, then clearly cross products are noncommunative.
> 
> Hmmm. That's not one of the Lie algebra axioms!
> 
> Yes it is. [A,[B,C]] != [[A,B],C] is one of the axioms. If [A,B] = AxB
> then you get the above result.
Idiot! Let A=B=C and then both sides equal 0 in _any_ Lie algebra by
the antisymmetry axiom!
> 
> (...Starblade Riven Darksquall...)
---- David
====
> Starblade Darksquall
> Proof:
>
> 1) Ax(BxC) != (AxB)xC
> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = 
-A(C*B)+B(C*A)
> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) !=
> C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,
> when Bx(AxC) = 0, then clearly cross products are noncommunative.
> 2) AxA = 0
> This follows from the definition of the cross product. The cross
> product of parallel vectors is zero.
> 3) (A+B)xC = AxC+BxC
> Since the vector cross product IS distributive over addition. This is
> easy enough to verify.
> 4) Ax(BxC)+Bx(CxA)+Cx(AxB) = 0
> Changing it to (B(A*C)-C(A*B))+(C(B*A)-A(B*C))+(A(C*B)-B(C*A)) makes
> it obvious that this identity is true, since that the dot product is
> communative.
> 5) AxB=-BxA
> This is as easy to verify as 2.
>
> So clearly, we have known about a type of lie algebra for longer than
> we let on. In fact, we use lie algebra type mathematics in dealing
> with even classical physics all the time, particularly in
> electromagnetics.
> Er, sort of. The old vector product AxB is the same as *(A^B) where * is 
the
> Hodge dual and ^ is the (antisymmetric) exterior product. (I'm not sure 
if
> Hodge dual is standard jargon, but it's easy to see what the term 
is
> intended to mean.)
> LH
> 
> Are you talking about tensors, or something else entirely?
> 
> (...Starblade Riven Darksquall...)
it is a well known fact that the linear space R^3 with the cross
product is a lie algebra. let A, B in R^3, then the hodge dual (the
star * operator) maps
A^B to A cross B, and vice versa. as such, it is a linear space
isomorphism between R^3 and linear space of bivectors. regarding
electromagnetism, it can certainly be formulated using the exterior
algebra, or, in more compact form, the clifford algebra on R^3.
                                            M.T.
====
I am working on a problem from Dummit & Foote's book, Abstract Algebra.
The problem is to find the remainder of 37^100 when divided by 29.
I have played around with 37 = 8 (mod 29) by raising it to powers, but I
have yet to discover the right path.  I tried
37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought about
residue classes, but I just can't seem to make any connections.  Could
anybody give me an idea?
TIA
Lurch
====
>I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
>The problem is to find the remainder of 37^100 when divided by 29.
>
>I have played around with 37 = 8 (mod 29) by raising it to powers, but I
>have yet to discover the right path.  I tried
>
>37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought about
>residue classes, but I just can't seem to make any connections.  Could
>anybody give me an idea?
As you note, 37 = 8 (mod 29). What happens to 8 as you raise it to
successive powers, modulo 29?
What is 8^{28} (mod 29), according to Fermat's Little Theorem? What is
8^{29} (mod 29)? 
If a^b = 1 (mod c), then how much is a^{k*b} (mod c)?
======================================================================
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
====
>I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
>The problem is to find the remainder of 37^100 when divided by 29.
>I have played around with 37 = 8 (mod 29) by raising it to powers, but I
>have yet to discover the right path.  I tried
>37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  
Umm...  Please review the laws of exponents.
>I have also thought about
>residue classes, but I just can't seem to make any connections.  Could
>anybody give me an idea?
obtained by four squarings.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
====
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of 37^100 when divided by 29.
> 
> I have played around with 37 = 8 (mod 29) by raising it to powers, but I
> have yet to discover the right path.  I tried
> 
> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought 
about
> residue classes, but I just can't seem to make any connections.  Could
> anybody give me an idea?
Start with 37^1 (mod 29) = 8 (mod 29). 
Then 37^2 (mod 29) = 37*8 (mod 29) = 296 (mod 29) = 6 (mod 29). 
Then 37^3 (mod 29) = 37*6 (mod 29) = ...
and so on. Repeat until you have 37^100 mod 29. 
There is a shortcut to this: If you go on with the calculation, you will 
eventually get the same result again. For example, you might find that 
37^2 (mod 29) = 37^10 (mod 29). (I am not saying these two are the same, 
but there are only 29 possible results, so at some point you must get 
the same result that you got before). If you found that 37^2 (mod 29) = 
37^10 (mod 29), for example, then you would know that 37^18, 37^26, 
37^34 and so on are the same again, so you could find 37^100 mod 29 
quite quickly.
====
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of 37^100 when divided by 29.
>
> I have played around with 37 = 8 (mod 29) by raising it to powers, but I
> have yet to discover the right path.  I tried
>
> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought 
about
> residue classes, but I just can't seem to make any connections.  Could
> anybody give me an idea?
>
> TIA
>
> Lurch
Being not familiar with this stuff, I did some trial and error
with a high precision calculator and this is what I found:
Everything mod 29:
37^100
   = 8^100
   = 2^300
   = 2^(300-28)       ???
   = 2^(300-2*28)       ???
   = 2^(300-10*28)     ???
   = 2^(20)
   = 1048576
   = 23
Does anybody know whether this is a (special case
of some) theorem:
    2^k = 2^(k-p+1)    (mod p)
?
Dirk Vdm
====
couldn't see the connection :(  And, yes Robert I do see my bonehead 
mistake
with the exponents.  Sorry.
Lurch
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of 37^100 when divided by 29.
>
> I have played around with 37 = 8 (mod 29) by raising it to powers, but I
> have yet to discover the right path.  I tried
>
> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought 
about
> residue classes, but I just can't seem to make any connections.  Could
> anybody give me an idea?
>
> TIA
>
> Lurch
>
>
====
 Dirk Van de moortel 
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of 37^100 when divided by 29.
>
> I have played around with 37 = 8 (mod 29) by raising it to powers, but 
I
> have yet to discover the right path.  I tried
>
> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought 
about
> residue classes, but I just can't seem to make any connections.  Could
> anybody give me an idea?
>
> TIA
>
> Lurch
> 
> Being not familiar with this stuff, I did some trial and error
> with a high precision calculator and this is what I found:
> 
> Everything mod 29:
> 37^100
>    = 8^100
>    = 2^300
>    = 2^(300-28)       ???
>    = 2^(300-2*28)       ???
>    = 2^(300-10*28)     ???
>    = 2^(20)
>    = 1048576
>    = 23
> 
> Does anybody know whether this is a (special case
> of some) theorem:
>     2^k = 2^(k-p+1)    (mod p)
> ?
> 
> Dirk Vdm
> 
>
Yes. 
For every prime p,  x^p == x (mod p) for all integers x,
so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).
Then it follows for all primes p and positive integers x,y and k,
that x^(y + k*(p-1)) == x^y (mod p)
Thus, if p is prime and u == v (mod p-1)
one gets x^u == x^v (mod p) for all integers x.
====
>  Dirk Van de moortel
>
[snip]
> Does anybody know whether this is a (special case
> of some) theorem:
>     2^k = 2^(k-p+1)    (mod p)
> ?
>
> Dirk Vdm
>
>
>
> Yes.
>
> For every prime p,  x^p == x (mod p) for all integers x,
> so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).
>
> Then it follows for all primes p and positive integers x,y and k,
> that x^(y + k*(p-1)) == x^y (mod p)
>
>
> Thus, if p is prime and u == v (mod p-1)
> one gets x^u == x^v (mod p) for all integers x.
virtually *nothing* of it.
Well, maybe I remember this:
   x = y (mod p)  <==>
           There is a whole number k such that:  x-y = k*p.
and that's where it ends.
So I have been trying to prove that first line of yours:
  For every prime p,  x^p == x (mod p) for all integers x
and I got stuck. Help!
Any elementary on-line textbook (pdf or ps) you could
recommend?
Dirk Vdm
====
> 
>  Dirk Van de moortel
>
> 
> [snip]
> 
> Does anybody know whether this is a (special case
> of some) theorem:
>     2^k = 2^(k-p+1)    (mod p)
> ?
>
> Dirk Vdm
>
>
>
> Yes.
>
> For every prime p,  x^p == x (mod p) for all integers x,
> so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).
>
> Then it follows for all primes p and positive integers x,y and k,
> that x^(y + k*(p-1)) == x^y (mod p)
>
>
> Thus, if p is prime and u == v (mod p-1)
> one gets x^u == x^v (mod p) for all integers x.
> 
> virtually *nothing* of it.
> Well, maybe I remember this:
>    x = y (mod p)  <==>
>            There is a whole number k such that:  x-y = k*p.
> and that's where it ends.
> 
> So I have been trying to prove that first line of yours:
>   For every prime p,  x^p == x (mod p) for all integers x
> and I got stuck. Help!
> 
> Any elementary on-line textbook (pdf or ps) you could
> recommend?
> 
> Dirk Vdm
Maybe Course 311 - Abstract Algebra, Part I: Topics in Number Theory:
http://www.maths.tcd.ie/~dwilkins/Courses/311/311NumTh.pdf
by D.R. Wilkins
or follow the links given in Annotated Web Links for
Kenneth H. Rosen's Elementary Number Theory Book:
http://www.aw-bc.com/rosen/resources.html
e.g.
Fermat's Little Theorem
http://www.cut-the-knot.org/blue/Fermat.shtml
or search the homepages of number theorists:
http://www.numbertheory.org/ntw/list.html
Many of them have lecture notes online.
Hugo Pfoertner
====
[...]
> 
> Any elementary on-line textbook (pdf or ps) you could
> recommend?
> 
> Dirk Vdm
Online number theory lecture notes:
http://www.numbertheory.org/ntw/lecture_notes.html
Hugo Pfoertner
====
[snip]
> Maybe Course 311 - Abstract Algebra, Part I: Topics in Number Theory:
> http://www.maths.tcd.ie/~dwilkins/Courses/311/311NumTh.pdf
> by D.R. Wilkins
Just what I needed.
Apparently this thing
     For every prime p,  x^p == x (mod p) for all integers x
is Fermat's theorem. Not so trivial. no wonder I didn't find
a simple proof :-)
> or follow the links given in Annotated Web Links for
> Kenneth H. Rosen's Elementary Number Theory Book:
> http://www.aw-bc.com/rosen/resources.html
>
> e.g.
> Fermat's Little Theorem
> http://www.cut-the-knot.org/blue/Fermat.shtml
Yep, had found this one with google:
    http://www.cut-the-knot.org/blue/Modulo.shtml
This was exactly what I was looking for.
Had this stuff is almost 30 years ago - Good refresher
for the basics.
>
> or search the homepages of number theorists:
> http://www.numbertheory.org/ntw/list.html
>
> Many of them have lecture notes online.
>
> Hugo Pfoertner
Dirk Vdm
 <7h7Sa.20204$F92.2218@afrodite.telenet-ops.be>
Face: 
iVBORw0KGgoAAAANSUhEUgAAADAAAAAwBAMAAAClLOS0AAAAJ1BMVEX+/NhQ4Kr6/ft4q2fn
 
+6z//v8SJRjZ6Y5p3HJD5vX++/76+olRdUI5lXl0AAACP0lEQVR4nFWUv2vbUBDHjVvI0sUUAoYM
 
4hwjEFlSgyLwm5yhSyaBF2/G4DagodTQDtFmKG/wolHQIaiYwqObhg7qEPPAcvD9Ub17kuynGzR8
 
P+/dj3d36kxPNlNkW+iAi19Adc5gGgIoBaSPsOtZAOgwqOwaA7nWkxboEngWBynlRreAtwUQC8m2
 
t8CMQ8f9KyQdlxagwN34NU2uDlJo3waZ87BLyS5FWfitGPdGT39GRRsMNMtJgm0QutX5BKUBKuN6
 
6eOKWg9qADXQ/Ub3DRgCDAlcC6OnVDfuKkBXKFNxa3Sq4ceuV7liXxD3Gl3M+2kF6H1g9NroG5yn
 
NQAPbmo/Wm4+uY81yE4Fky40uM0NgMHOJPSeXnUxscCgKThYIzXwBAaj5iEkfqBCTzFuor7R/WBP
 
s2GDEntpIsbF/oVaDt7veZpU4JhjD8dFkWOHS6UbFfhzLFbLu4Lsr55kcHb1Sxd5pBkUuQYLqIsy
 
X1CT2VZl9wzAG+CCBsyQf6VnVf68+kw1HA2JXywA8E5L6d9VZNSky53NeF45ZTLRAOV4SjnfDnKN
 
Tya1sytuoSOYjFvAU7QwcCEC6l4LDKfTLVX8lvdl8VTkjzaY0Z03yPvSBgrIm/MxklKUqxbgSaQe
====
> So I have been trying to prove that first line of yours:
>   For every prime p,  x^p == x (mod p) for all integers x
> and I got stuck. Help!
The set {1,2,...,p-1} is a group of order p-1 under multiplication mod
p, so whenever x is not zero (mod p)
x^(p-1) = 1 (mod p) 
(Lagrange's theorem) whence
x^p = x (mod p)
which also holds for x=0 (mod p).
====
without the aid of a calculator/computer, or Fermat's little theorem.  In
the book, all he discusses is modular arithmetic.  FLT is not mentioned 
'til
page 97.  I am on page 10.
Even with FLT, I can't seem to get it.  If a^p = a mod p, a is the 
remainder
right?  So, if, for example, I take 3^2 mod 2 the theorem says I should get
3, but isn't it 1?  I mean 9 mod 2 should be 1, right?  Futhermore, if I
take 37^29 = 37 mod 29 and I try to raise this congruence by powers until I
get my 100, then my remainder gets larger and larger.  So, if my remainder
is supposed to be 23, I doubt that 37^x is going to be my answer.  What the
heck am I missing here?  (Besides a brain)
Lurch
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of 37^100 when divided by 29.
>
> I have played around with 37 = 8 (mod 29) by raising it to powers, but I
> have yet to discover the right path.  I tried
>
> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29).  I have also thought 
about
> residue classes, but I just can't seem to make any connections.  Could
> anybody give me an idea?
>
> TIA
>
> Lurch
>
>
====
> without the aid of a calculator/computer, or Fermat's little theorem.  In
> the book, all he discusses is modular arithmetic.  FLT is not mentioned 
'til
> page 97.  I am on page 10.
hm, maybe something like this?
x = 37^100 (mod 29)
    = 8^100 (mod 29)          casted out 29
    = 64^50 (mod 29)
    = 6^50 (mod 29)            casted out 2*29 = 58
    = 36^25 (mod 29)
    = 7^25 (mod 29)            casted out 29
    = 7*7^24 (mod 29)
    = 7*49^12 (mod 29)
    = 7*20^12 (mod 29)       casted out 29
    = 7*400^6 (mod 29)
    = 7*23^6 (mod 29)
    = 7*529^3 (mod 29)
    = 7*7^3 (mod 29)            casted out 29
    = 49*49 (mod 29)
    = 20*20 (mod 29)
    = 400 (mod 29)
    = 23 (mod 29)                 casted out 29
Dirk Vdm
====
fairly soon, but I was starting to get a little frustrated.  On the other
hand, they say close only counts in horseshoes and handgrenades; so, who
knows?  I think that your method will do the trick.
Lurch
>
out
> without the aid of a calculator/computer, or Fermat's little theorem.
In
> the book, all he discusses is modular arithmetic.  FLT is not mentioned
'til
> page 97.  I am on page 10.
>
> hm, maybe something like this?
> x = 37^100 (mod 29)
>     = 8^100 (mod 29)          casted out 29
>     = 64^50 (mod 29)
>     = 6^50 (mod 29)            casted out 2*29 = 58
>     = 36^25 (mod 29)
>     = 7^25 (mod 29)            casted out 29
>     = 7*7^24 (mod 29)
>     = 7*49^12 (mod 29)
>     = 7*20^12 (mod 29)       casted out 29
>     = 7*400^6 (mod 29)
>     = 7*23^6 (mod 29)
>     = 7*529^3 (mod 29)
>     = 7*7^3 (mod 29)            casted out 29
>     = 49*49 (mod 29)
>     = 20*20 (mod 29)
>     = 400 (mod 29)
>     = 23 (mod 29)                 casted out 29
>
> Dirk Vdm
>
>
====
> fairly soon, but I was starting to get a little frustrated.  On the other
> hand, they say close only counts in horseshoes and handgrenades; so, who
> knows?  I think that your method will do the trick.
I wouldn't have found it if I hadn't taken a nice refreshing
look this afternoon at
    http://www.cut-the-knot.org/blue/Modulo.shtml
and specially at the solved elementary problems in
    http://www.cut-the-knot.org/blue/chinese.shtml
        (the 'casting out' business)
Here's a little proof I just found for this 'casting out p':
         x = a (mod p)
    ==>  x = kp + a
    ==>  x = kp+np + a-np
    ==>  x = (k+n)p + a-np
    ==>  x = a-np  (mod p)
and a more general case:
         x = a^m (mod p)
    ==>  x = kp + a^m
    ==>  x = kp + Poly[a,n,p,m]*p + (a-np)^m
    ==>  x = (k+Poly[a,n,p,m])*p + (a-np)^m
    ==>  x = (a-np)^m  (mod p)
hm, I think I have been wrong never having liked number
theory ;-)
Dirk Vdm
====
>> without the aid of a calculator/computer, or Fermat's little theorem.  
In
>> the book, all he discusses is modular arithmetic.  FLT is not mentioned 
'til
>> page 97.  I am on page 10.
>
>hm, maybe something like this?
>x = 37^100 (mod 29)
>    = 8^100 (mod 29)          casted out 29
>    = 64^50 (mod 29)
>    = 6^50 (mod 29)            casted out 2*29 = 58
>    = 36^25 (mod 29)
>    = 7^25 (mod 29)            casted out 29
>    = 7*7^24 (mod 29)
>    = 7*49^12 (mod 29)
>    = 7*20^12 (mod 29)       casted out 29
>    = 7*400^6 (mod 29)
>    = 7*23^6 (mod 29)
>    = 7*529^3 (mod 29)
>    = 7*7^3 (mod 29)            casted out 29
>    = 49*49 (mod 29)
>    = 20*20 (mod 29)
>    = 400 (mod 29)
>    = 23 (mod 29)                 casted out 29
Or simpler:
    37^100 (mod 29)
  = 8^100 (mod 29)
  = 2^300 (mod 29)
  = 2^20 (mod 29)    casted out 28*10 by F
  = 32^4 (mod 29)
  = 3^4 (mod 29)
  = 3*27 (mod 29)
  = 3*-2 (mod 29)
  = -6 (mod 29)
  = 23 (mod 29).
Each of these steps is quite easy to verify in one's head.
 -- Erick
====
> Even with FLT, I can't seem to get it.  If a^p = a mod p, a is the 
remainder
> right?  So, if, for example, I take 3^2 mod 2 the theorem says I should 
get
> 3, but isn't it 1?  I mean 9 mod 2 should be 1, right?  Futhermore, if I
> take 37^29 = 37 mod 29 and I try to raise this congruence by powers until 
I
> get my 100, then my remainder gets larger and larger.  So, if my 
remainder
> is supposed to be 23, I doubt that 37^x is going to be my answer.  What 
the
> heck am I missing here?  (Besides a brain)
The remainder may always be taken as a non-negative integer less 
than the modulus. 
For modulus 2, that means zero or 1 (even or odd).
Note that, according to one definition of congruences, 
 9 == 1 (mod 2)  just means (9 - 1) is divisible by 2, which it is.
In the following = is ordinary equality and == is congruence
Since 37 == 8 (mod 29), you know that 37^29 == 8^29 == 8 (mod 29).
Further, x^(29-1) = x^28 == 1 (mod 29), for GCD(x,29) = 1.
Consider that 100 = 3*28 + 16 == 16 (mod 28),
so that now  37^100 == 37^ 16 == 8^16 (mod 29).
Also 8^2 = 64 == 6 (mod 29)
and  6^2 = 36 == 7 (mod 29) 
and  7^2 = 49 == 20 (mod 29)
and 20^2 = 400 == 23 (mod 29 
so 8^16 = (((8^2)^2)^2)^2 == 23 (mod 29).
All done by hand without need for electronic aids.
====
>
> I am working on a problem from Dummit & Foote's book, Abstract 
Algebra.
> The problem is to find the remainder of  37^100  when divided by  29.
>
> I have played around with  37 = 8 (mod 29)  by raising it to powers, but
> I have yet to discover the right path. [...] I'd like to figure this out
> without the aid of a calculator/computer, or Fermat's little theorem.  
> In the book, all he discusses is modular arithmetic.  FLT isn't mentioned
> 'til page 97.  I am on page 10. [...]
Mod 29:  37^100 = 8^100 = 2^300. To easily compute this
let's search for a small power of 2 that equals +-1 (mod 29).
We search for numbers equal +-1 (mod 29) that factor into
powers of 2 times small known powers of 2, such as 3 = 2^5.
Mod 29:  1 = 6*5 = 6(-24) = -2^4 3^2 = -2^14  via  3 = 2^5
so   2^300 = 2^(14*21+6) = (-1)^21 2^6 = -6 = 23.
The search succeeds quickly however you do it, e.g.
        -1 = 4*7  = 4(6^2) =  2^4 3^2
         1 = 8*11 = 8(-18) = -2^4 3^2
         1 = 9*13 = 9(-16) = -2^4 3^2
        -1 = 12*12 ...
-Bill Dubuque
====
>without the aid of a calculator/computer, or Fermat's little theorem.  In
>the book, all he discusses is modular arithmetic.  FLT is not mentioned 
'til
>page 97.  I am on page 10.
>
>Even with FLT, I can't seem to get it.  If a^p = a mod p, a is the 
remainder
>right?  So, if, for example, I take 3^2 mod 2 the theorem says I should 
get
>3, but isn't it 1?  
Yes, and yes. 3=1 (mod 2), after all, so it's hardly surprising that
both answers are correct, modulo 2.
If you want to skip Fermat's Little Theorem, then study how the powers
of your number cycle moudlo 29. For example, if I wanted to figure out
what the last digit of 3^{200}, what I would do is consider the powers
of 3 modulo 10:
3 = 3 (mod 10)
3^2 = 9 (mod 10)
3^3 = 7 (mod 10)
3^4 = 1 (mod 10)
3^5 = 3 (mod 10)
At this point, it should become obvious that 3^a = 3^{a+4} (mod 10)
for all a, so I just need to take the residue of 200 modulo 4; this is
0, so 3^{200} = 3^0 = 1 (mod 10).
>I mean 9 mod 2 should be 1, right?  Futhermore, if I
>take 37^29 = 37 mod 29 and I try to raise this congruence by powers until 
I
>get my 100, then my remainder gets larger and larger.
First, you would not do that; what you would do is note that 37^{28} =
1 (mod 29), so that 37^{k*28} = 1 (mod 29) for all integers k. That
means that
37^{100} = 37^{84+16} = 37^{84}*37^{16} = 37^{3*28}*37^{16} =
1*37^{16}=37^{16} (mod 29), so you just need to figure out how much
37^{16} is.
And second, every time you get a number larger than 29, YOU REDUCE
MODULO 29, to get a smaller number and work with that; usually, your
best bet would be a number between -13 and 15, to keep them small.
======================================================================
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
====
> This is all very interesting, including your link. 
> I followed much of it but must confess my ability to read others work
> is limited. It's mostly well established stuff and seems very solid.
> You are very well read. 
> I appreciate the simplicity with which you approach these things. 
> I have a puzzling coordinate system that might be of interest to you.
> Please search for three-signed arithmetic in sci.math. 
> 
I am always keen on novelties - I shall look up your three-signed
arithmetic.
What emerged from my researches is that not only is data limited by
the data-set from whence it came, but the DIMENSIONS of data are
limited to the dimensions from whence they came.
Thus there is a fundamental problem with Euler's EXP(iX) in that it
begins with a number (UNARY) that has FREEDOM TO MOVE in the iX
dimension (MONAL), and converts to a PAIR of numbers (BINARY) with
FREEDOM TO MOVE in two dimensions (BINAL). These are Cos(X) and
iSin(X).
When you try to go back to the unary state, you have not PEGGED or
SPECIFIED or LIMITED your remaining freedom - you have left it to
chance. Thus the answer you arrive at is just one of many.
Here is an example of a data set:
1
2
If I ask you to continue the set, you might say
1,2,3,4,5
1,2,4,8,16
1,2,6,24,120
These are 
a LINE
an EXPONENTIAL (base 2)
a FACTORIAL.
A Chebyshev approximation might find Y=0+X
Taking data from that poynomial might give
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
WHAT AN ABUNDANCE OF DATA!
Convert this to a TENTH-ORDER polynomial approximation, you get
Zero 
+ 1(X to the 1) 
+ Zero(X to the 2)
+ Zero(X to the 3)
+ Zero(X to the 4)
+ Zero(X to the 5)
+ Zero(X to the 6)
+ Zero(X to the 7)
+ Zero(X to the 8)
+ Zero(X to the 9)
+ Zero(X to the 10)
So this TENTH-ORDER CHEBYSHEV APPROXIMATION to a FIRST-ORDER ditto
is no better than the first-order.
At position 1.5, we would have got
1.5         for a straight line
1.414213562 for exponential 2
1.329340388 for a fractional factorial.
Given only two pieces of data, the Chebyshev could not distinguish
these - and took the SIMPLEST. It delivered the straight line.
Interpolation, giving the in-betweens, does not add to the data. It
is an exercise in REDUNDANCY, and the eleven pieces of data still
contain only two starting facts within them.
The METAPHYSICS of my reasoning with complex numbers similarly
states that the the TWO dimensions of the complex plane still only
contain the ONE dimension of the real world that we started from.
Something's got to give!
I am grateful to you for taking the time to view my page
http://wehner.org/euler and for thinking about it.
I describe the facts about the world as I see them. I do not make
things up. However, such situations as the PROBLEM OF MULTIPLE
SOLUTIONS require an abstract reasoning approach.
That problem will not go away - and those who slavishly apply complex
mathematics without using common sense will come unstuck.
Charles Douglas Wehner
====
I just thought about something and hoped someone could help me explain
why I'm wrong in thinking it or why I could be right...
I thought really simple, Take a sphere and begin walking on it (it
never ends). When I want to cut the knot I just do something radical
and go right through the sphere or something. So are we thinking that
small? This is somewhat metaphor I had on infinite numbers...
Anyone who can help me find that red line again?
====
>
>I just thought about something and hoped someone could help me explain
>why I'm wrong in thinking it or why I could be right...
>
>
>I thought really simple, Take a sphere and begin walking on it (it
>never ends). When I want to cut the knot I just do something radical
>and go right through the sphere or something. So are we thinking that
>small? This is somewhat metaphor I had on infinite numbers...
>
>Anyone who can help me find that red line again?
So in 1 dimension you have infinite numbers because of 1 dimensional
thinking.. In another you have hypernumbers that could resemble the
infinite
====
> 
> So in 1 dimension you have infinite numbers because of 1 dimensional
> thinking.. In another you have hypernumbers that could resemble the
> infinite
Well, that wouldn't really work, atleast if you're talking about the
real number system.  Cantor proved that |R^n (that's R to the n, the
set of real, ordered n-tuples, so the dimension is n) has the same
cardinality as |R.  That is, there is a function which maps every
point in |R^n to a point in |R uniquely.  So |R and |R^n have the same
number of points.
You may want to read about the Riemann sphere, however. Imagine
placing a sphere on the origin of the complex plane (or just |R^2, the
standard real plane).  Imagine a line connecting the top of the sphere
to a point on the plane.  For any point on the plane, this line will
intersect the sphere at a unique point.  Now, what happens as the
distance from the origin of the plane gets very far?  The line tends
to the tangent line at the top of the sphere!  So it only intersects
the sphere at one point.  In complex analysis, we call this the point
at infinity, and we call the plane with this point added the
augmented plane.
Check out
http://tinyurl.com/haxy
The first picture is of the Riemann Sphere, the rest are of functions
mapped onto the Riemann sphere.
Alex
====
>
>> 
>> So in 1 dimension you have infinite numbers because of 1 dimensional
>> thinking.. In another you have hypernumbers that could resemble the
>> infinite
>
>Well, that wouldn't really work, atleast if you're talking about the
>real number system.  Cantor proved that |R^n (that's R to the n, the
>set of real, ordered n-tuples, so the dimension is n) has the same
>cardinality as |R.  That is, there is a function which maps every
>point in |R^n to a point in |R uniquely.  So |R and |R^n have the same
>number of points.
>
>You may want to read about the Riemann sphere, however. Imagine
>placing a sphere on the origin of the complex plane (or just |R^2, the
>standard real plane).  Imagine a line connecting the top of the sphere
>to a point on the plane.  For any point on the plane, this line will
>intersect the sphere at a unique point.  Now, what happens as the
>distance from the origin of the plane gets very far?  The line tends
>to the tangent line at the top of the sphere!  So it only intersects
>the sphere at one point.  In complex analysis, we call this the point
>at infinity, and we call the plane with this point added the
>augmented plane.
>
>Check out
>http://tinyurl.com/haxy
>
>The first picture is of the Riemann Sphere, the rest are of functions
>mapped onto the Riemann sphere.
>
>Alex
I hope I'll get new inspiration from it :D
====
>Alex
>Check out
>http://tinyurl.com/haxy
>
>The first picture is of the Riemann Sphere, the rest are of functions
>mapped onto the Riemann sphere.
>
Wow. I never knew you could put a graphic in one table entry and a paragraph 
in
the other. I have a box function on my old Panasonic RK-P400C plotter that 
does
that. Very cool. Makes for easy reading. Maybe a little too much 
whitespace.
Yours,
Doug Goncz, Replikon Research, Seven Corners, VA 
Fair use and Usenet distribution without restriction or fee
Civil and criminal penalties for circumvention of any embedded encryption
====
> A three dimensional torus has no boundaries. But how many handles does
> it have? It should have more than one. The euler characteristic is
> X(M)=2-2h-b. Plugging in h = 2 and b = 0 you get X(M) = -2, which is
> NOT flat space. You get another nonzero result (-4) when you plug on h
> = 3.
> 
> I got this from rotating a torus in 4th dimensional space, or
> stretching it out to a torus-cylynder and then curling it around
> itself. I imagine it would be like creating something like a CD and
> then making another torus like thing out of that, except that you're
> only taking a projection of the torus onto a plane, which is the CD
> like thing.
> 
> Therefore, Hawking's hypothesis that the universe is a torus is wrong!
> Otherwise space wouldn't be flat, and that's not what we've observed!
> 
> We'd need a n-th dimensional entity with 1 handle. A torus has more
> than one handle, so it cannot be the shape of the universe!
> Furthermore, if the universe were a torus, it would NOT be
> rotationally invariant.
> 
> If we want to find the shape of the universe, we must find a 3d
> topography (or possibly 4d) with one handle and which is rotationally,
> translationally, and lorentz invariant, meaning Poincaire invariant. I
> imagine that such a thing exists in multidimensional geometry.
> 
> BTW, just a question... how do you describe a handle in terms of
> topography? I know a boundary is described as a point where the lines
> cross into nothingness, but what would a handle 'look' like in
> topography? Maybe we can find out how to make one of those things I
> described.
> 
> (...Starblade Riven Darksquall...)
Excuse me while I add another newsgroup. :P
(...Starblade Riven Darksquall...)
====
> 
>>A three dimensional torus has no boundaries. But how many handles does
>>it have? It should have more than one. The euler characteristic is
>>X(M)=2-2h-b. Plugging in h = 2 and b = 0 you get X(M) = -2, which is
>>NOT flat space. You get another nonzero result (-4) when you plug on h
>>= 3.
>>
No, the Euler characteristic is the alternating sum of the (integral)
Betti numbers; the homology of T^3 is all free abelian, with these as
ranks:
        n       rank(H_n(T^3))
        0            1
        1            3
        2            3
        3            1
So, the Euler characteristic is 1 - 3 + 3 - 1 = 0.
This could have been seen several ways: first, the Euler characteristic
of any closed, orientable manifold of odd dimension is zero (by Poincare
Duality). Second, the Euler characteristic of any product X x Y, where X
has zero Euler characteristic, is zero.
Regarding handles,  for dimensions greater than 2, handles come in all
dimensions from 0 to n (for an n-dimensional space).  The 3-torus T^3
has (at least) one 0-handle, 3 1-handles, 3 2-handles, and 1 3-handle.
More handles could exist, as long as the appropriate cancellation can
take place when passing to homology.
>>I got this from rotating a torus in 4th dimensional space, or
>>stretching it out to a torus-cylynder and then curling it around
>>itself. I imagine it would be like creating something like a CD and
>>then making another torus like thing out of that, except that you're
>>only taking a projection of the torus onto a plane, which is the CD
>>like thing.
>>
>>Therefore, Hawking's hypothesis that the universe is a torus is wrong!
>>Otherwise space wouldn't be flat, and that's not what we've observed!
Flat metrics exist on tori of all dimensions. After all, the n-torus is
the quotient of R^n under a discrete group of rigid translations.
>>
>>We'd need a n-th dimensional entity with 1 handle. A torus has more
>>than one handle, so it cannot be the shape of the universe!
>>Furthermore, if the universe were a torus, it would NOT be
>>rotationally invariant.
>>
>>If we want to find the shape of the universe, we must find a 3d
>>topography (or possibly 4d) with one handle and which is rotationally,
>>translationally, and lorentz invariant, meaning Poincaire invariant. I
>>imagine that such a thing exists in multidimensional geometry.
I imagine you're meaning that there is a Lorentz structure on the
tangent bundle. If I recall correctly, that only requires a zero euler
characteristic (so the tangent bundle splits into a sum of a line bundle
& a complementary 3-dimensional bundle, if you're thinking of a space-
time manifold).  I think the Poincare' group is a semidirect product of
the Lorentz group with the group of Euclidean translations, so if you
get a Lorentz structure on R^4, compatible with the translation group,
you should automatically get the requisite Poincare' structure on any
quotient by a discrete subgroup of the translation group [such as how
one obtains the torus]. I'm no expert on how these particular groups are
handled in the physics world, so I would gladly defer to a knowledgeable
person on this matter.
>>
>>BTW, just a question... how do you describe a handle in terms of
>>topography? I know a boundary is described as a point where the lines
>>cross into nothingness, but what would a handle 'look' like in
>>topography? Maybe we can find out how to make one of those things I
>>described.
Perhaps you should use the term topology instead. Many references are
available for the topology of manifolds, including those with Lorentz
and/or Poincare' structures.
>>
>>(...Starblade Riven Darksquall...)
> 
> 
> Excuse me while I add another newsgroup. :P
> 
> (...Starblade Riven Darksquall...)
Dale.
====
> I have played with this concept and would like to further explore it.
> Or could you dispel its validity?
> I will only define its construction.
> There are three branches from an origin.
> Each branch has its own sign.
> I label these signs { -, +, * }.
> Some numbers in this realm are { -1.234, +2.345, *3.456 }.
> I call this space T.
T = ({'-', '+', '*'} x { r in R | r > 0 }) / {0}
> Please do not think of these values in terms of vectors.
set theory product AxB = { (a,b) | a in A, b in B }, / union
> Operators must be created.
>
Do it.
====
Rotational Operators
---------------------
I propose the following sign functions in three-signed T space. 
These rotational operators change the sign of a number. 
In the table that follows s represents the sign of an element in T.
         s  |  - s  |  + s  |  * s  
      ~~~~~~|~~~~~~~|~~~~~~~|~~~~~~~
         -  |   +   |   *   |   -
            |       |       | 
         +  |   *   |   -   |   +
            |       |       |
         *  |   -   |   +   |   *
  
For example:
                   - (-1.234) = +1.234.
                   + (-1.234) = *1.234.
                   * (-1.234) = -1.234.
These are the first honest operators that I can come up with. 
The three sign system obviously invokes rotational phenomena. Whereas
the reals use a negation to toggle back and forth in magnitude from
one extremity to another we now have more. The minus operator (-) will
toggle in one direction. The (+) operator will toggle in the other
direction, and the star operator goes all the way around to preserve
the original sign. These rotations are countable and can formulate an
integer counting system. The utility of this is not known to me.
Sorry I go so slowly.
-Tim
====
> 
> What is {-0.0, +2.5, *5.3} x {-0.0, +1.3, *2.0}?
I'd like to just do an arithmetical product first.
The x in your notation I will treat like the question:
   What is 7.0 x 5.0?
Where the answer would be 35.0 for real numbered values.
I feel reasonably sure that the answer is ( + 8.64 * 7.35 ).
I still am not entirely comfortable with this but will defend it:
Notation is a bit of an issue and I am sorry that I am changing it
around a little bit. I don't see the need for commas in an element.
Assuming we are doing an arithmetical product I also don't see the
need for the x.
I restate your question: What is ( + 2.5 * 5.3 )( + 1.3 * 2.0 )?.
The elements in parenthesis are elements in Y.
Each element in Y is in effect two elements in T summed.
Using the math of rotational operators that I just put on this thread
and using the standard arithmetical laws (they do feel right for the
moment)
   ( + 2.5 * 5.3 )( + 1.3 * 2.0 ) 
 = + + (2.5)(1.3) + * (2.5)(2.0) * + (5.3)(1.3) * * (5.3)(2.0)
 = - 3.25 + 5.0 + 6.89 * 10.6
 = - 3.25 + 11.89 * 10.6
 = + 8.64 * 7.35 .
To extend this example in general some more notation might be helpful:
 minus( y1 ) = the magnitude of the component of y1 in the minus
direction.
   e.g. minus( * 3.4 + 2.0 ) = 0, minus( - 1.2 * 2.3 ) = 1.2.
Similarly for plus() and star().
Now the arithmetical product y3 of two values y1 and y2 in Y is:
y3 = (  - minus(y1)star(y2) - star(y1)minus(y2) -plus(y1)plus(y2)
        + minus(y1)minus(y2) + star(y1)plus(y2) + plus(y1)star(y2)
        * minus(y1)plus(y2) * plus(y1)minus(y2) * star(y1)star(y2)
      ).
y3 will reduce to a clean value in Y.
Do you follow?
If you see any problems I hope you will let me know.
It's really not clear to me just how much of traditional mathematics
system.
====
> Rotational Operators
> ---------------------
> I propose the following sign functions in three-signed T space.
> These rotational operators change the sign of a number.
> In the table that follows s represents the sign of an element in T.
>
>          s  |  - s  |  + s  |  * s
>       ~~~~~~|~~~~~~~|~~~~~~~|~~~~~~~
>          -  |   +   |   *   |   -
>          +  |   *   |   -   |   +
>          *  |   -   |   +   |   *
Same as addition of integers modulus 3.
        +       2       1       0
        2       1       0       2
        1       0       2       1
        0       2       1       0
> For example:
>                    - (-1.234) = +1.234.
>                    + (-1.234) = *1.234.
>                    * (-1.234) = -1.234.
>
> These are the first honest operators that I can come up with.
It's not clear what T is, nor in this summary did you include any
description of T.  I proposed ({-,+,*} x { x in R | x > 0}) / {0}
So would you give a construction of T, a mathematical expression,
in contrast to a verbal description prone to vagueness.
> The three sign system obviously invokes rotational phenomena. Whereas
> the reals use a negation to toggle back and forth in magnitude from
> one extremity to another we now have more. The minus operator (-) will
> toggle in one direction. The (+) operator will toggle in the other
> direction, and the star operator goes all the way around to preserve
> the original sign. These rotations are countable and can formulate an
> integer counting system. The utility of this is not known to me.
>
Huh?  Why not just extend -r = -1*r and discuss -1, +1, *1 and is there an
operator a*b and if so, make that something else as * is already taken.
Also think about 0 = +0 = -0 = *0 and do you have convention, +r = r.
How is * parallel to -, which is the inverse of + ?
====
> 
>> Rotational Operators
>> ---------------------
>> I propose the following sign functions in three-signed T space.
>> These rotational operators change the sign of a number.
>> In the table that follows s represents the sign of an element in T.
>>
>>          s  |  - s  |  + s  |  * s
>>       ~~~~~~|~~~~~~~|~~~~~~~|~~~~~~~
>>          -  |   +   |   *   |   -
>>          +  |   *   |   -   |   +
>>          *  |   -   |   +   |   *
> 
> Same as addition of integers modulus 3.
> +     2       1       0
> 2     1       0       2
> 1     0       2       1
> 0     2       1       0
And so the system is multiplicatively identical
to the union of the three rays through the origin
in the complex plane through the cube roots of unity.
Perhaps some day we'll get a notion of addition for
these. That would be fun :-)
-- 
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
 The League of Gentlemen
====
> 
> It's not clear what T is, nor in this summary did you include any
> description of T.  I proposed ({-,+,*} x { x in R | x > 0}) / {0}
I would say that this is valid. 
But it is not fundamental.
I am averse to defining T in terms of R.
If you could define R in the same terms that you propose then I would
happily modify it to encompass T. But since such a definition of R
would read:
  ({-,+} x { x in R | x > 0}) / {0}
you would define R in terms of R and that would be a paradox.
Is there a clean mathematical basis for magnitude?
Or does magnitude always come from a distance function?
My gut feeling is that magnitude is a simplistic and clean concept and
so should be at the basis, not at a higher level.
> So would you give a construction of T, a mathematical expression,
> in contrast to a verbal description prone to vagueness.
T is one magnitude(an unsigned number) with one of three possible
signs (-,+,*).
Would an equivalent to the number line suffice?
Just draw a branch and label each extremity -, +, *.
Take a unit length and mark each branch from the vertex. 
Plop a point anywhere on the lines and you can measure it.
The star symbol has been chosen because it has three lines
intersecting and so is the natural symbol to use for the next sign in
the progression -,+,?.
> Huh?  Why not just extend -r = -1*r and discuss -1, +1, *1 and is there 
an
> operator a*b and if so, make that something else as * is already taken.
This is confusing me. Yes (-1)(*r) = -r, where r is a magnitude.
I understand that calling signs rotational operators is distasteful.
There is no conflict with using magnitude one with them.
Factors and multiplication work consistently:
  -r = (r)( -1 ) = (r)(-1)(*1).
 if (a)(b) = (r) then (-a)(-b) = +r, (*a)(-b) = -r.
Oddly, addition is a larger quagmire (see reply to Chapman, next in
thread).
> Also think about 0 = +0 = -0 = *0 and do you have convention, +r = r.
I think that an unsigned zero is fine and that it is equivalent to the
signed zeros. I see this as an exception and would like to see sign
preserved symbollically on all non-zero numbers. Therefore +r does not
equal r.
Unsigned number symbols should always be magnitudes.
> How is * parallel to -, which is the inverse of + ?
I'm not sure what you mean here. 
I don't believe there is any parallel.
I see what you are getting at.
The sign operations yield rotational phenomena.
This gets close to the integer counting system, which could take
several forms.
To clear up your controversy lets define one form:
  - rotates the sign + 1/3.
  + rotates the sign + 2/3.
  * rotates the sign + 3/3.
Now, although the plus operator appears to be the inverse of the minus
operator the integer counting system has dispelled that notion.
Example:
         (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
I don't know that the integer system has any utility. It's just a way
of looking at the sign.
I'm sorry that this isn't more convincing.
The star operator in T appears unique as the plus operator in R does.
for t1 in T: *t1 = t1.
just as for r1 in R: +r1 = r1.
This is not just a convention.
When you view these postings do you view the entire thread?
Perhaps I am using the Usenet system differently than you.
I view these postings from google groups through a web browser.
This allows a view of an entire thread. Therefore I snip old info so
that there is less wading through redundant stuff.
I do believe you understand very well the problem that I am working
on.
-Tim
====
 
> And so the system is multiplicatively identical
> to the union of the three rays through the origin
> in the complex plane through the cube roots of unity.
> 
> Perhaps some day we'll get a notion of addition for
> these. That would be fun :-)
Strict summation in T is uncomfortable. This is why I proposed Y,
which is a sum of two Ts.
It may be that sum( t1, t2 ) is not always reducible.
On the real number line the choice is simpler and is always reducible.
If I choose to make sum( t1, t2 ) reducible then I break the link to
Y.
The crux of the matter lies in cancellation and superposition.
What is summation supposed to be?
Should sum( -2.3, *2.3) be zero?
Or should sum( -2.3, +2.3, *2.3 ) be zero?
I suggest the latter and so T space slides away in favor of Y space.
In this way T is just a stepping stone to Y.
====
> 
> 
>   - rotates the sign + 1/3.
>   + rotates the sign + 2/3.
>   * rotates the sign + 3/3.
> 
> Now, although the plus operator appears to be the inverse of the minus
> operator the integer counting system has dispelled that notion.
> Example:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
There is a fundamental problem in the established mathematics of data
going missing. One example is the constant during a differentiation.
Where does it go?
The TRANSFINITE mathematics states that it becomes vanishingly small,
but if it could be scaled up by a special infinity it would return.
The COMPLEX mathematics states that it shifts SIDEWAYS in complex
space, and so leaves the real world. Multiplication by -i1 would bring
it back.
NEITHER is a total answer to the problem of vanishing data. Perhaps
there is the possibility of a new Three-signed arithmetic emerging
to tackle such problems.
However, at a first perusal I cannot quite see where this is heading.
It would be useful to have real-world problems solved or PARTLY-solved
by the Three-signed arithmetic. Such worked examples would help
focus the mind, to see where we are heading.
On a mathematical page on my own website, I introduced the Eucalculus
- and promptly showed an electronic circuit whose gain can only be
explained by the hump in the Eucalculus curve.
That curve is itself no more than the reciprocal of Euler's Gamma
function - so it is an extension of mainstream maths.
In this present case, we seem to be leaving the customary definition
of + and - behind. Then we get:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
(1)(2)(1)(2) (multiplication) based on the old or the new maths?
This is not a criticism, just a question.
It would be useful to know where there is a complete synopsis of the
fundaments to be found  - with no discussion. This would help in any
study of this system.
Charles Douglas Wehner
====
> 
> 
>   - rotates the sign + 1/3.
>   + rotates the sign + 2/3.
>   * rotates the sign + 3/3.
> 
> Now, although the plus operator appears to be the inverse of the minus
> operator the integer counting system has dispelled that notion.
> Example:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
There is a fundamental problem in the established mathematics of data
going missing. One example is the constant during a differentiation.
Where does it go?
The TRANSFINITE mathematics states that it becomes vanishingly small,
but if it could be scaled up by a special infinity it would return.
The COMPLEX mathematics states that it shifts SIDEWAYS in complex
space, and so leaves the real world. Multiplication by -i1 would bring
it back.
NEITHER is a total answer to the problem of vanishing data. Perhaps
there is the possibility of a new Three-signed arithmetic emerging
to tackle such problems.
However, at a first perusal I cannot quite see where this is heading.
It would be useful to have real-world problems solved or PARTLY-solved
by the Three-signed arithmetic. Such worked examples would help
focus the mind, to see where we are heading.
On a mathematical page on my own website, I introduced the Eucalculus
- and promptly showed an electronic circuit whose gain can only be
explained by the hump in the Eucalculus curve.
That curve is itself no more than the reciprocal of Euler's Gamma
function - so it is an extension of mainstream maths.
In this present case, we seem to be leaving the customary definition
of + and - behind. Then we get:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
(1)(2)(1)(2) (multiplication) based on the old or the new maths?
This is not a criticism, just a question.
It would be useful to know where there is a complete synopsis of the
fundaments to be found  - with no discussion. This would help in any
study of this system.
Charles Douglas Wehner
====
> 
> 
>   - rotates the sign + 1/3.
>   + rotates the sign + 2/3.
>   * rotates the sign + 3/3.
> 
> Now, although the plus operator appears to be the inverse of the minus
> operator the integer counting system has dispelled that notion.
> Example:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
There is a fundamental problem in the established mathematics of data
going missing. One example is the constant during a differentiation.
Where does it go?
The TRANSFINITE mathematics states that it becomes vanishingly small,
but if it could be scaled up by a special infinity it would return.
The COMPLEX mathematics states that it shifts SIDEWAYS in complex
space, and so leaves the real world. Multiplication by -i1 would bring
it back.
NEITHER is a total answer to the problem of vanishing data. Perhaps
there is the possibility of a new Three-signed arithmetic emerging
to tackle such problems.
However, at a first perusal I cannot quite see where this is heading.
It would be useful to have real-world problems solved or PARTLY-solved
by the Three-signed arithmetic. Such worked examples would help
focus the mind, to see where we are heading.
On a mathematical page on my own website, I introduced the Eucalculus
- and promptly showed an electronic circuit whose gain can only be
explained by the hump in the Eucalculus curve.
That curve is itself no more than the reciprocal of Euler's Gamma
function - so it is an extension of mainstream maths.
In this present case, we seem to be leaving the customary definition
of + and - behind. Then we get:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
(1)(2)(1)(2) (multiplication) based on the old or the new maths?
This is not a criticism, just a question.
It would be useful to know where there is a complete synopsis of the
fundaments to be found  - with no discussion. This would help in any
study of this system.
Charles Douglas Wehner
====
> There is a fundamental problem in the established mathematics of data
> going missing. One example is the constant during a differentiation.
> 
> Where does it go?
> 
> The TRANSFINITE mathematics states that it becomes vanishingly small,
> but if it could be scaled up by a special infinity it would return.
> 
> The COMPLEX mathematics states that it shifts SIDEWAYS in complex
> space, and so leaves the real world. Multiplication by -i1 would bring
> it back.
> 
> NEITHER is a total answer to the problem of vanishing data. Perhaps
> there is the possibility of a new Three-signed arithmetic emerging
> to tackle such problems.
> 
> However, at a first perusal I cannot quite see where this is heading.
This will hopefully eventually replace space-time of classical
physics.
In a physical system the concept of differentiating out a constant
doesn't seem to be a problem. If you look at the velocity of a car it
doesn't matter where the car started travelling, if the resulting
velocity is the same for multiple cases in different positions, then
the velocities should match in the math. In this way the constant
could be seen as ethically positive and legititmates comparing the
velocity of cars at differing positions.
I think more interesting is that there does seem to exist a physical
fundamental from which the calculus operates, in this example it is a
spatial position, which has no further derivative according to
classical physics.
> It would be useful to have real-world problems solved or PARTLY-solved
> by the Three-signed arithmetic. Such worked examples would help
> focus the mind, to see where we are heading.
I agree completely. I am working on it. I'm also hoping that someone
will come across this and find an application. I've discovered that
the Y-space covers the plane and can be graphed on a piece of paper. A
unit circle takes on foreign proportions and yet is still the unit
circle. I guess I can't demonstrate things like unit circles without a
website.
> 
> On a mathematical page on my own website, I introduced the Eucalculus
> - and promptly showed an electronic circuit whose gain can only be
> explained by the hump in the Eucalculus curve.
> 
> That curve is itself no more than the reciprocal of Euler's Gamma
> function - so it is an extension of mainstream maths.
> 
> In this present case, we seem to be leaving the customary definition
> of + and - behind. Then we get:
>          (-1)(+2)(-1)(*2) = -4. (total sign is 7/3)
> 
> (1)(2)(1)(2) (multiplication) based on the old or the new maths?
I suggest that the new math does still match the old in this way.
Isolating magnitude and sign should not be a problem.
I guess I am not justified in this example until the fundamentals are
in place.
This example accepts that a magnitude can be cast into T-space and can
be brought back out with the use of signed unit values in T.
Challenges arise when summation is incorporated into this. I've taken
a stab at that type of product for Y space, which you will find in
this thread up higher.
> 
> This is not a criticism, just a question.
> 
> It would be useful to know where there is a complete synopsis of the
> fundaments to be found  - with no discussion. This would help in any
> study of this system.
Yes. This thread is not an effective presentation any longer. I guess
I just wanted somebody to shoot this down and it hasn't happened yet.
> 
> Charles Douglas Wehner
some more impressive results.
-Tim
====
> 
> This thread is not an effective presentation any longer. I guess
> I just wanted somebody to shoot this down and it hasn't happened yet.
> 
> 
> some more impressive results.
> 
> -Tim
I had a few thoughts on the subject. Have you heard of the QUATERNIONS
of William Rowan Hamilton?
If your +, -, and * are none-other than the i, j, and k of Hamilton,
then you have REDISCOVERED the Quaternions.
That's OK. I have often discovered something only to find that
somebody got there first. The effect on my mind is not SHAME, but
quite the REVERSE. I find that the great men of science thought along
the same lines as I do - so on those occasions when I am second, I
feel FLATTERED.
I have ALSO found things that nobody ever found before me. That is
because ones confidence grows with every discovery - new or not.
Do investigate the Quaternions:
           http://mathworld.wolfram.com/Quaternion.html 
Charles Douglas Wehner
====
I think this construction is more primitive than the quaternions.
I'm merely adding another sign to the real numbers and seeing what
happens.
Quaternions appear to contain lots of real numbers and so cannot be
equivalent.
Have you ever tried graphing in a plane like this?
 +  plus pole
  +
   +
    +
     +
      +
       + . . . . . . . p1 = - 7 + 4
        +             .
         +             . 
        . +             .
       .   0 - - - - - - - - - - minus pole
      .   * origin          .
  p3 .   *                 .
      . *                 .
       *                 .
      *                 .
     * . . . . . . . . . p2 = - 9 * 6
    *
   *
  *
 * star pole
This is Y space as a plane. It is simpler than both cartesian 2D
and complex values in that it has just the three way branch instead of
a four way branch. As you can see parallelograms at angle 2pi/3
resolve the entire plane symmetrically. A reduced Y value always has
at most a pair of magnitudes. Every position in the plane can be
resolved.
The dimensionality of Y is quite a conundrum. Because
informationally there are two magnitudes informationally it is two
dimensional, yet the value is just one three-signed element in Y.
The question why a pair? should be in the mind at this point.
The answer is that in order to obtain a zero by cancellation in a
three-signed system we should require an equal magnitude for each pole
to provide that cancellation.
For a magnitude x
   
   In R : - x + x = 0.
    
   In Y : - x + x * x = 0.
   
   In Y : y1 = - x + x is not zero!
   We need y1 * x to get to zero.
If you believe that summation is superposition then I think you will
be convinced. The trouble is that we all think in context of real
numbers.
Much of this thinking does not extend when more signs are used in the
construction.
I started this all in the context of the question Why R X R X R X
t?.
This is the spacetime of classical physics which even string theorists
are not destroying. I consider Y X Y to be a competitor to R X R X R X
t.
But also if you add yet another sign ( four signs ) then the extension
would yield at most three magnitudes by the same law of cancellation.
And so four-signed arithmetic may also be a valid competitor to R X R
X R, especially if the cancellation effect is looked upon as
accumulation, which then provides a basis for time. I think it is wise
to continue opening up the three-signed can of worms before moving on
to the four-signed can. It is my hope that dynamics will be found in Y
which will surprise us.
> 
> I had a few thoughts on the subject. Have you heard of the QUATERNIONS
> of William Rowan Hamilton?
> 
> If your +, -, and * are none-other than the i, j, and k of Hamilton,
> then you have REDISCOVERED the Quaternions.
> 
> That's OK. I have often discovered something only to find that
> somebody got there first. The effect on my mind is not SHAME, but
> quite the REVERSE. I find that the great men of science thought along
> the same lines as I do - so on those occasions when I am second, I
> feel FLATTERED.
> 
> I have ALSO found things that nobody ever found before me. That is
> because ones confidence grows with every discovery - new or not.
> 
> Do investigate the Quaternions:
>            http://mathworld.wolfram.com/Quaternion.html 
>
> Charles Douglas Wehner
====
MISSING.
erased it by mistake.
just waiting until I had time for a fuller answer.
I have to answer it here.
The first question of data going missing and being recovered came
from this statement of mine:
> 
> The TRANSFINITE mathematics states that it becomes vanishingly small,
> but if it could be scaled up by a special infinity it would return.
> 
Here, we can draw a NUMBER LINE    -3,-2,-1,0,1,2,3
and put the factorials above the positive part:
1,1,2,6
0,1,2,3
To find the factorial of 2 from that of 3 - which is 6 -
we DIVIDE by 3. It gives 2.
But we can always go back up again by multiplying by 3.
To find the factorial of 1 from 2!, we divide by 2.
We can also go back up.
To find 0! from 1! we divide by 1, and can go back.
To find (-1)! - note the brackets - we divide by 0.
ANYTHING divided by zero is said to be INFINITE.
Zero times infinite is said to be INDETERMINATE.
So we cannot go back up!!!!!
However, at the console we defined zero by typing it in.
We DO NOT KNOW WHAT IT IS, but we know WHERE IT CAME FROM.
We know therefore that (-1)! is 1 divided by CONSOLE-ZERO.
We call this CONSOLE-INFINITY.
As we still have console-zero on our number-line, 
we can still multiply console-infinity by it.
So we can go back up.
(-2)! is -(console infinity)
(-3)! is +(1/2)(console-infinity)
&c.
And so, by strict definition of the nature of the zeroes and
of the infinities, we do not have to lose information by
defining it as indeterminate.
This is one of my discoveries in the as-yet unpublished
fantasy maths - where fantasy means anything containing
easy-to-use numbers (FUN NUMBERS) and the FOLLIES zero and
infinity.
Another name is the TRANSFINITE MATHS.
However, as I made my discoveries by one route it was pointed
out to me by an eminent mathematician that Georg Cantor had
found something very similar a hundred years ago by another route. 
The FOLLIES are almost the same as the ALEPHS of Cantor.
I am not ashamed. We sought in Nature, and we BOTH found.
I also stated:
> The COMPLEX mathematics states that it shifts SIDEWAYS in complex
> space, and so leaves the real world. Multiplication by -i1 would bring
> it back.
> 
Here we could consider a cosine. It starts at 1 and descends.
However, I want you to imagine some micro-polynomial that, when
given the number 1 computes the cosine a trace further - that is,
0.999999999975625 or whatever.
Think of a pico-polynomial that is even smaller in its steps.
0.999999999999999999999999999999999
Think of it being applied as the cosine passed through zero at
the point Pi/2
But 2Cos(X) goes through zero just as 1Cos(X) and 3Cos(X) do.
And we are not allowed special zeroes.
As the waveform creeps through zero, how does it know how to
re-emerge in such a way as to create a symmetrical cosine,
instead of Cos(X) above the zero and 2Cos(X) below?
In electronics engineering, we like to think of such a cosine
being part of a PAIR. The cosine is REAL, and is the SIGNAL -
but there is a hidden sine accompanying it. This we call the
PHASE.
The PHASE is hidden because it exists only in the IMAGINARY
world. However, it defines the SLOPE of the original cosine.
Thus, if the cosine began at Y=1, and the units of the X-axis
are RADIANS, the slope will be a downward slope of 1 as the
cosine goes through zero. 
By means of its phase, a cosine can be allowed to pass through
zero without getting lost.
So - with RESTRICTIONS - we can use complex maths.
We use the Euler equation Exp(iX)=Cos(X)+iSin(X)
However, i means CURRENT in electronics - so we use j.
X is too vague, so we use omega-t
Omega is the angular frequency. t is the time.
So we write Exp(j.omega.t)=Cos(omega.t)+jSin(omega.t)
Here, the jSin(omega.t) is actually the INVERTED slope.
Yet it still defines the slope.
Such complications are necessary because the simple 
mathematics will not always define what we want.
In the transfinite case, we have no need for the
imaginary numbers. In the complex maths we have no
need for the transfinite.
At least in theory.
Perhaps there are problems that can only be solved
by BOTH.
Then we have the mad maths of Boole. 1+1=1
At first sight, it is ridiculous.
Then somebody explains that 1 is TRUE.
He translates TRUE AND TRUE IS TRUE.
Now he seems REALLY to have gone bananas.
Yet without Boole, there would be no computer and
no Internet for me to write this on.
The Quaternions are sometimes used in elementary-
I cannot ACCEPT a new mathematical system without
seeing its benefits.
I cannot REJECT a new mathematical system for
fear of rejecting something good.
If I had time to delve and delve, I might know
whether the THREE-SIGNED ARITHMETIC is useful
or not.
But it is not mine, and I am busy. I will, however,
if there is an adequate summary, look back now and
then to see if something important is emerging.
I reserve judgement.
Charles Douglas Wehner.
====
I'm still largely confused by T and its operations. Could you give us a few
exhaustive examples from scratch? My guess is that it is probably 
isomorphic
to something we're already familiar with, in which case there will be lots
of useful things you could draw on that already are known to understand how
T works.
====
> I'm still largely confused by T and its operations. Could you give us a 
few
> exhaustive examples from scratch? My guess is that it is probably 
isomorphic
> to something we're already familiar with, in which case there will be 
lots
> of useful things you could draw on that already are known to understand 
how
> T works.
Whereas the real numbers (R)can be defined as a magnitude with two
signs( -, + ), T is a magnitude with three signs( -, +, * ).
Examples of elements in T are:
   - 1.23
   + 2.134
   * 6.54
Now consider summation in T.
The sum of the example numbers reads: 
   ( - 1.23 + 2.134 * 6.54 ).
It is a mistake to say that - 1.23 * 6.54 = * 5.31.
This style of cancellation will not work.
The order of operations would be too strict.
If we use the law:
   Sum( t1, t2, t3 ) = Sum( Sum( t1, t2 ), t3 ) = Sum( Sum( t3, t1 ),
t2 )
we will find the problem.
Consider the following using the fraudulent style of cancellation: 
   ( - 1.23 + 2.134 ) * 6.54 = ( * 6.54 - 1.23 ) + 2.134
   ( + 0.904 * 6.54 ) = ( * 5.31 + 2.134 )
   ( * 5.636 ) = ( * 3.176 ). 
This is nonsense.
The problem occurred at the concept of cancellation.
Should * 1 - 1 = 0?
Should - 1 + 1 = 0 ?
Or should  - 1 + 1 * 1 = 0?
The latter is the correct selelection.
For a magnitude x:
   In R  
      - x + x = 0.
   In T 
     
      - x + x * x = 0.
Some values are not reducible. Thich leads me to describing a new
space Y because Sum( t1, t2 ) is not generally in T. So even though
the subject in this thread is T space it is only a starting step to
get to Y space, which is general three-signed arithmetic.
Y is a general sum of T which is always reducible to at most a pair of
three-signed magnitudes. Now Sum( y1, y2 ) is always in Y.
This leads to ( - 1.23 + 2.134 * 6.54 ) = ( + 0.904 * 5.31 ).
   where ( - 1.23 + 1.23 * 1.23 ) has been cancelled out.
This is all that I have time for right now. Arithmetical products
follow much more easily.
====
> This is nonsense.
Yes, now you're talking sense.
Max de Macs
====
I think I can appreciate your criticisms of the meaning to the math.
But the graphics are inspiring.
> So I look forward to a simple T-space solution to chaos.
> 
> Charles Douglas Wehner
====
> I think I can appreciate your criticisms of the meaning to the math.
> But the graphics are inspiring.
> 
NOT a criticism.
> So I look forward to a simple T-space solution to chaos.
CHAOS theory is indeed a field where a new psychology of mathematics
is needed.
The graphics are elaborate, convoluted, impenetrable. Given the
graphics, we cannot tell how they are made. Given the method, we are
surprised that something so simple produces such elaboration.
Nature is full of this - elaborate patterns created by simple means.
The skill that is sought is to find the ORDER within the CHAOS.
 
Charles Douglas Wehner
====
I'm looking at timesheets scheduling and optimisation task: a shop
manager would like to specify a number of employees working in any
given time interval, and employees would like to stay within their
preferable time constraints.
I think such optimisation task should be quite common, but I just
don't know the category of algorithms it falls to, or at least where I
can pick up bits of information on this subject.
Any advice is greatly appreciated,
and thanks in advance,
Vlad
====
>I'm looking at timesheets scheduling and optimisation task: a shop
>manager would like to specify a number of employees working in any
>given time interval, and employees would like to stay within their
>preferable time constraints.
>I think such optimisation task should be quite common, but I just
>don't know the category of algorithms it falls to, or at least where I
>can pick up bits of information on this subject.
Yes, this is quite common.  The usual formulations involve integer
linear programming.  The general field in which this is done is 
operations research, and you might try the sci.op-research newsgroup.
Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2
====
> I'm looking at timesheets scheduling and optimisation task: a shop
> manager would like to specify a number of employees working in any
> given time interval, and employees would like to stay within their
> preferable time constraints.
> I think such optimisation task should be quite common, but I just
> don't know the category of algorithms it falls to, or at least where I
> can pick up bits of information on this subject.
Dr. Israel mentioned that the general field is Operations Research.
It may also be of interest that some complex examples of timetabling
problems are solved with Evolutionary or Genetic Algorithms, and there
has been discussion within the past year of this in comp.ai.genetic.
xanthian.
-- 
====
There are several plug-ins for Excel which will deal with this as long as
your problem is not too large.
>
>
>I'm looking at timesheets scheduling and optimisation task: a shop
>manager would like to specify a number of employees working in any
>given time interval, and employees would like to stay within their
>preferable time constraints.
>
>I think such optimisation task should be quite common, but I just
>don't know the category of algorithms it falls to, or at least where I
>can pick up bits of information on this subject.
>
> Yes, this is quite common.  The usual formulations involve integer
> linear programming.  The general field in which this is done is
> operations research, and you might try the sci.op-research newsgroup.
>
> Robert Israel                                israel@math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
====
  Jesse Hughes and I had a discussion along the lines of the current
title in the late January part of
      Category theory vs. Set theory questions
      http://mathforum.org/discuss/sci.math/a/t/311474
there:
[2]   David Libert   Jan 25, '01
      Re: Category theory vs. Set theory questions
      http://mathforum.org/discuss/sci.math/a/m/311474/311527
[3]   David Libert   Jan 26, '01
      Re: Category theory vs. Set theory questions
      http://mathforum.org/discuss/sci.math/a/m/311474/311535
  I have noticed a literature reference relating to this, and I have
my own additional comments to add.
  In [3] I introduced a hierarchy of fragments of  Z  =  Zermelo's
theory (separation, no replacement),  namely the hierarchy having
sigma-m separation as a fragment of full separation, (as m varies in
the hierarchy),  this hierarchy below a hierachy under  ZF  having
corresponding sigma-m fragments of replacement.
  I noted in [3] a Lowenhiem Skolem argument to get the replacement
part of the hierachy had strictly increasing increasing strength at
every 4th level.  In [3] I noted I did not know how consistency
strength increased in the separation part of the hierarchy.
  I think we can argue consistency strength goes up in the separtion
part, ie  sigma-m+4  separation  |-  Con(sigma-m separtion)  for a
tail of m's anyway.
  Namely, arguing in the sigma-m+1 separation theory, suppose there is
a proof of a contradiction in the sigma-m Z fragment.  Then we can
make this a proof of contradiction from a finite fragment of these
axioms over pure first order logic.  But then by Gentzen cut
elimination for pure first order logic, we can also find a proof where
all formulas are substitution instances of subformulas of premises, ie
the finite fragment of the theory.
  So we could find a proof of a contradiction where all formulas used
in the proof are no more complexity than the premises axioms.
Allowing a few m's at the start to get started  (ie I claimed for a
tail of m's)  to cover the complexity of all the Z axioms other than
separation, then sigma-m separtion itself is  I think  sigma-m+2  ie
the parameters can be free variables, one quantifier to say the
separating set exists, and another quantifier inside to say all members
are in that set according to the condition, then the considtion itself
which is sigma-m.
  So we have a proof of contradiction using only sigma-m+2 formulas.
Now in our working theory we define sigma-m+2 truth for the universe.
This is a sigma-m+2 definition.  Then we run an induction argument on
the proof, using that truth definition, that all formulas in the proof
are true on all interpretations of variables.  I think one more
quantifier to handle all interpretations  and then one more to code
the induction.
  I am sliding over details here so maybe  m+4 as a claimed above and
seem to be getting with this sketch isn't exactly right, but I think
this is enough to show some finite step up is enough, ie what I said
was 4, if 4 doesn't really work some other small integer should.
  Regarding another point, in [2] I noted how if you start with a ZF
model, and make it into a topos, how to recover from the topos
(working model theoretically above the topos) an isomorphic copy of
the original ZF model:  namely suitable equivalence classes of
well-founded trees representing the epsilon relation on transitive
closures.
  That construction in turn could make sense in any arbitrary
elementary topos, producing not necessarily a full ZF model but a
model of a ZF fragment, as I discussed in [2].
  I suggested one reformulation of full ZF into topos theory would be
to axiomatize in topos language that the result of that construction
satisfies ZF, or  ZFC or whatever.
  More comments along these lines.  Let use consider the weak theory,
as from the base of [3]'s hierarchy, ZF style, but no replacement and
only bounded separation, and the other usual axioms of ZF.  Or maybe
ZFC, parallel discussions with or without AC.  Let that theory be S.
  Given any model M of S, (this including the case of M modeling
extensions of T to stronger fragments of ZFC), we can form the
corresponding topos  T(M).  Inside this topos, we can do the
construction above of a S model again:  S(T(M)).  In fact,  S(T(M))
will be isomorphic to M, but that is not my main point now.
  Inside  S(T(M)),  we can once more construct the corresponding topos
from the S theory:   T(S(T(M))).
  Then this  ToSoT(M)  is isomorphic to  T(M), by a definable
isomorphism pushing everything through the definitions.
  The statement that this definition is an isomorphism is a statement
in topos language in the language of  T(M).
  So if we have some set theory  S' extending base set theory S,  and
we want a corresponding topos style theory to correspond to S', ie we
want a theory to correspond to all  T(M) 's  for  M an S' model,  it
would be reasonable that such a topos style theory  should include the
statement that  ToS(my underlying universe)  is isomorphic as stated
above  to my underlying universe,   since every  T(M)  is like this.
  So suppose  S' is some set theory extending S,  and suppose  T'  is
a candidate extension of elementary topos theory, which is supposed to
correspond to S', and in particular has that last property I just
suggested is reasonable to expect of S'.
  Given  C' a model of S'  (I would want to call this T' since it is a
topos, but I have been using T to name the construction between
theories, so I use base C to suggest category),  so given C' a model
of S',  we can form the set theory model  S(C'),  then form the topos
ToS(C'),  and then inside that topos we can go back to set theory:
SoToS(C').  Since S' provided  ToS(C')  is isomorphic to  C',
we get  SoToS(C')  is isomorphic to  S(C'),  by a definable
isomorphism in  C'.
  So if  S'  is at some level of the hierarchies as discussed above,
the T' theory is proving as a schematum that  SoToS(C')  satisfies all
of S',  ie since each of these are isomorphic to the corresponding
S(C')  and  the assumption on  T'  as being a topos theoretic coding
of S'  is  all   S(C') model  S'  for    C' modeling  T'.
  Any model  M' of S'  can induce a corresponding topos  T(M'), and
this is topos having  S(T(M'))  modeling  S',  ie  such  S(T(M'))  are
isomorphic to M' and M' is assumed to model S'.
  If  T', being a topos theory reformulation of S', is supposed to be
satisfied by any topos C' with  S(C') modeling  S',  since  C' = T(M')
meets this condition, by the last paragraph,  T(M')  should be a model
of T'.
  Two paragraphs ago, we had over T' models C' the 4 level tower reaching
up to  SoToS(C')  has isomorphism  SoToS(C')  to  S(C'),  and by the
last paragrach,  any S' model M'  can be arranged to be isomorphic to
the S(C') level of that tower,  ie take  C' = T(M') obtaining an S
model, and then the 2nd level of the tower there  SoT(M')  is
isomorphic to M.
  So we have for any M' an S' model,  SoT(M) isomorphic to M.
  So by the completeness theorem,  S' proves that SoT(my universe) is
isomorphic by that definition to  my universe.
  So from the fact that  T' is a topos style reformulation of S'
having proeprties such a theory can be expected to have, we conclude
that  S'  the underlying set theory proves something about the zigzig
towers contructed over it.
  Now the clincher.  In the topos style theories we can make a
corresponding notion of sigma-m and pi-m,  namely complexity on
alternating blocks of quantifiers over objects and arrows.
  Suppose, T' being a topos theoretic reformulation of set theory,
proves that  S(my universe) satisifies  S'  (as a schematum).
  Then here is a way to reaxiomatize S', in the set theoretic S'
language.  Above we had that S' proves that the tower over it zigzags
back to isomorphism,  ie  SoT(my universe) isomorphic  my universe.
So we put this single sentence as an new axiom into the
reaxiomatization, ok to add as axiom since it is a theorem of the
target theory.
  Then we define in the S' language the  T(my universe) construction,
and we axiomatize that this constructed category satisfies  T'.
  So, by the assumed property of T',  that it proves  S(my universe)
satsifies  S',  in  S'  we get  , interpreting that last proof over
the T' axioms coded into our new reaxiomatization of S',
we get in our S' re-axiomatiztion  that  SoT(my universe) satisfies S'.
  But S' had my universe  isomorphic to  SoT(my universe).
  So our theory proves  my universe satsifies S'.
  So we really have reaxiomatized S' this way.
  Suppose the T' theory was all axiomatized at level sigma-m,  ie for
the compelixity notion of quantifying over objects and arrows.
  Then, when we copy that axiomatization into S' style, the objects and
arrows get defined as sets.  So this copying from T' axioms into S'
language for our reaxiomatization carries  sigma-m axioms (topos
theory)  to  sigma-m axioms (set theory)  in the reaxiomatization of
S'.
  We also need to have the sentence in the S' reaxiomatization saying
the tower over S' zigzaged to isomorphism, that is one sentence having
some specific finite bound, call it b.
  Then our reaxiomatization of S' has complexity of axioms at most
sigma-max(m,b).
  Now suppose our starting  S' was up to level  at least sigma-n in
the separation hierachy from [3].  (This includes all levels of the
replacement hierachy from [2], which starts at the top of the
separation hierarchy.)
  So the sigma-n S' theories has been reaxiomatized as a theory with
axioms of complexity at most  sigma-max(m,b).
  But if  n >=  max(m,b) + 2   or whatever from above,  S' can prove
the consistency of the reaxiomatization, along the lines of the
argument I sketched above over smaller separation.  (The previous
argument with +4 :  2 was from the separation axiom, and 2 to process
it, now use 2 for processing).
  But the reaxiomatization recovers the original S', so S' would prove
its own consistency.
  We conclude  max(m,b) > n - 2.
  What this says, is beyond the b base cases, you need almost as many
alternating quantifiers in a topos style theory recoding of set theory
as the original set theory had.
  Zermelo's theory Z  is level  sigma-omega, ie it includes axioms of
each level sigma-m.
  So a topos theory version of Zermelo would need unbounded
alternations of quantifier.
  Usual elementary toposes  are Caretsian closed categories with
subobject classifier, and then the other stuff as  [2]-[3]  to jazz
them up more toward set theory:  AC, natural numbers object.
  All these things are phrased in terms of an existential quantifier
for the new object  (if the theory adds a symbol then by the
completeness theorem there should be a corresponding theory in the
base language with corresponding existential quantifier, at least for
finitely axiomatized theories which I think the axioms of topos theory
can be arramged to be).
  Anyway, one outer existential quantifier is at most complexity up
sigma-m hierarchy by 1.
  Then in those base symbols, we have to assert various functors have
adjunctions.  This can be expressed by all finite diagrams of a
certain form have an appropriate universal arrow.  Universal quantifiers
on the vertices of the diagram.  Then we have to say there is a unique
arrow to fill in the diagram to make something happen.  Saying unique
arrow is a couple more quantifiers,  a conjunct of an existential and
off to the side two nested universals expressing uniqueness.
  The property inside the universal property:  sometimes it is just
that the diagram commutes.  That requires no quantifiers.  For
defining subobject classifier there is a definition that there is a
unique arrow to complete a square making the result a pull-back.  So
those outer quantifiers as above handle unique existence, and the
pull-back property is itself another universal property, which will
itself will be unique existence over a diagram commuting.
  So unraveelling it all, you only get a finite nesting a
quantifiers.  Well not surprsing since it is to be a sentence.  The
point here is that the intuive definitions given in mathematical
English do not unravel into a schematum over growing sigma-m
complexity, everything in the intuitive definition just expands out to
one or two nestings of quantifier.
  So the point is,  cartesian closed category with subobject
classifier, and whatever fintitely more stuuf you add:  natural
numbers object:  yeah some objects and some adjunctions and some
unique existences,  no way are we going to build up to infinite
sigma-m complexity, as we need to get back to Zermelo theory.
  Think of the ordinary defintions in category theory.  It is always
that a handful of things have adjunctions, basically just iterating
over and over again that diagrams have unique arrows completing them
to a previously defined property.
  Well, from [2]-[3]  we already had that usual elementary toposes do
not recapture Zermelo or ZF.  But this is more striking, that all
theories in the style of category theory don't.
  In [2] I suggested a try at a topos style version of ZF.  Namely
to axiomatize that  S(my universe) |= ZF.  So that theory does get
sigma-omega, because the separation and replacement axioms saying that
become unbounded  sigma-m accross the schematum, this sigma-m in the
sense of topos theory as I defined above.
  As I say though, this is a different style that usually done in
category theory.
  So the reference I found.  The Kock and Reyes paper in the Handbook
of Mathematical Logic, ed Jon Barwise.  Their section 7 is above this
sort of thing.  They mention papers from the early 1970's by Cole,
Mitchell and Osius.  They give a sketch for the Osius one, it seems to
be something like my S(C)  construction  to make a set theory model
from a topos.
  They say there that Osius defines a theory  ZO, which is a set
theory corresponding exactly in this sense to ordinary elementary
toposes.  They don't define ZO, they refer to Osius.  But they say ZO
is intermediate between  Zermelo-Thiele  and  ZF.  They don't define
Zermelo-Thiele.
  I couldn't find that on the web, but I did find mathematician
Thiele, who did set theory and is from around the time of Zermelo.
  When I first read that, I assumed Zermelo-Thiele and ZO  were
defined in terms of limited complexity separation etc, as my own
checking seems to show comes up here.
  But with Thiele being old time, I wonder if Zermelo-Thiele is just
Zermelo theory?  But back then, Zermelo originally phrased separation
as using a definite property to separate out.  I recall it was maybe
1918 Skolem and  1922 Fraenkel  sometime in there that this was redone
as a schamatum in first order logic.  And the original Zermelo was I
think 1908.
  So with everything so vague as those original versions, they
wouldn't have had anything like sigma-n levels.
  But if Zermel-Thiele = Zermelo's Z,  and  ZO is intermediate
betweeen Zermelo-Thiele and ZF, and  ZO corresponds to usual toposes,
how come we have  ZO including all the  sigma-omega complexity of Z
corresponding to toposes which supposedly by the above have bounded
complexity?
-- 
David Libert          ah170@FreeNet.Carleton.CA