mm-81
===
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 de?ition of such a metric ?it's still got all of the
same ?itary algebraic structure that thenatural numbers
has, because the stone-czech compacti?ation functorfrom sets
to topological spaces preserves ?ite cartesian products;so in
particular it's still a rig (= ring without negatives).--

===
>Can anyone describe the Stone-Cech compacti?ation
of>the natural numbers topologically and algebraically,>for
example : is it still a semi-group. is it metrizable>and what
would be a de?ition of such a metric ?It's certainly not
metrizable: Since it's compact, i? were metrizable that
would mean that the sequence1, 2, 3, ... would have a
convergent subsequencen_1, n_2, .... And there are obviously
no convergentsequences of integers (if (n_j) is
convergentthen 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 compacti?ation of>|the
natural numbers topologically and algebraically,>|for example
: is it still a semi-group. is it metrizable>|and what would
be a de?ition of such a metric ?>it's still got all of the
same ?itary algebraic structure that the>natural numbers
has, because the stone-czech compacti?ation functor>from
sets to topological spaces preserves ?ite cartesian
products;I take it that that means that in particular b(N) x
b(N) is naturallyhomeomorphic to b(NxN) ? I could be all wet,
but offhand thisdoesn't seem right:If so then every bounded
complex-valued function on NxN extendscontinuously 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 onNxN: If we
had such an extension then f(n, y) = 0 for n in N andy in
b(N)N (?) and hence f(x,y) = 0 for x, y in b(N)N, so f
vanisheson the complement of NxN, and hence f is not
continuous sincethe 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 compacti?ation of> the natural
numbers topologically and algebraically,Let's write beta N
for the Stone-Cech compacti?ation.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 ?ed x you
can extend y to allof beta N, and for ?ed y you can extend
x to all of beta N, but notboth at once (while remaining
jointly continuous).There is the Bohr compacti?ation of the
group Z of integers, smallerthan beta Z, but still a
group.> is it metrizableno: beta N is compact but not
sequentially compact. It is separablebut not
second-countable. It has power 2^c, which is much larger
thanany separable metric space.> and what would be a
de?ition of such a metric ?>-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
>|Can anyone
describe the Stone-Cech compacti?ation of>|the natural
numbers topologically and algebraically,>|for example : is
it still a semi-group. is it metrizable>|and what would be
a de?ition of such a metric ?>it's still got all of
the same ?itary algebraic structure that the>natural
numbers has, because the stone-czech compacti?ation functor>
>from sets to topological spaces preserves ?ite 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 howexcellent
mathematicians can make mistakes, back in his ?st major paper
in1948, Ed Hewitt had a theorem which stated the b(X) x B(Y)
is homeomorphicto b(X x Y). (This is the paper in which he
introduced the concept ofrealcompactness, and he didn't do
much along that line after thatembarrassment.) Long ago I
read the paper to try to ?d his error, and heused a
preliminary lemma which had as another correlary that the
2-sphere ishomeomorphic to a torus.lot about extending
algebraeic structures to the Stone-Cechcompacti?ation. Here
is a link to his home page, which include a list ofhis
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
compacti?ation of>|the natural numbers topologically and
algebraically,>|for example : is it still a semi-group. is
it metrizable>|and what would be a de?ition of such a
metric ?>it's still got all of the same ?itary
algebraic structure that the>natural numbers has, because
the stone-czech compacti?ation functor>from sets to
topological spaces preserves ?ite 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'tseem to make it, but one
without my additions did.>>|Can anyone describe
the Stone-Cech compacti?ation of>|the natural numbers
topologically and algebraically,>|for example : is it
still a semi-group. is it metrizable>|and what would be a
de?ition of such a metric ?>it's still got all
of the same ?itary algebraic structure that the>
>natural numbers has, because the stone-czech compacti?ation
functor>from sets to topological spaces preserves ?ite
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 ?st major paperin> 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 ?d his error, andhe> used a preliminary lemma which
had as another correlary that the 2-sphereis> homeomorphic to
a torus.>> lot about extending algebraeic structures to the
Stone-Cech> compacti?ation. Here is a link to his home page,
which include a listof> 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
compacti?ation of||>|the natural numbers topologically and
algebraically,|>|for example : is it still a semi-group. is
it metrizable|>|and what would be a de?ition of such a
metric ?|>|>|>it's still got all of the same ?itary
algebraic structure that the|>natural numbers has, because
the stone-czech compacti?ation functor|>from sets to
topological spaces preserves ?ite 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
?ured someone would correct me if iwas wrong, but i guess i
should have mentioned that i wasn'tcompletely sure about it
anyway.)actually i didn't even read the details of the
corrections yet,because i want to try ?uring out my mistake
?st. i know that ioccasionally get mixed up about the
precise de?ition of stone-czechcompacti?ation in the most
general case, but this isn't the mostgeneral case so that
shouldn't be the problem here, i think. i couldhave sworn
that there's _some_ functor ?g around here somewherethat
preserves ?ite products, but maybe i got mixed up about
whichone 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 de?e
the subsets of V_N as followsU_1 = {X|X->lambda in
F}U_(i+1) = U_i {X|X->P in F for some P in U_i*} for
i>=1Now i try to realize. If V_N is {A,B,S} and i have a rule
in F, for exampleS->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 presentanother element of V_N, in particular the
condition required (as you cansee) is that exists a rule X->P
for some P in U_i*. But if here U_i* isthe set of all words
over U_1 i haveU_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 rulelike A->S or B->SS. Right?
===
I'd be grateful
to anyone can give me a proof of this sentence:The
?ite-dimension subspaces of an in?ite-dimension space
areclosed and without interior points.
===
> I'd be grateful
to anyone can give me a proof of this sentence:> The
?ite-dimension subspaces of an in?ite-dimension space are>
closed and without interior points.What sort of space is
intended? Topological vector space?Without interior
points:Let V be an in?ite-dimensional topological vector
space. Let W be a?ite-dimensional linear subspace. Or, more
generally, simply aproper linear subspace. Suppose W has an
interior point w. Then W-wis a neighborhood of 0 (by
continuity of addition), and since W is alinear space, W-w =
W. Because scalar multiplication is continuous at0, 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,
contradictingthe 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
?ite-dimension subspaces of an in?ite-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
normedspace.Suppose B is a normed space and S is a ?ite
dimensional subspace withbasis{s1, s2, ... , sn}. De?e a
norm on C^n by|| (a1, ... , an) || = || a1 s1 + ... + an sn
||, where this second norm isthe one in B. The fact that all
norms on C^n are equivalent shows that Swith 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 of0, namely -x+U. If y is any element of B,
then there is a positive number rsuch that ry belongs to
-x+U; therefore ry belongs to S, therefore y belongsto S,
and it follows that S=B.
===
>and without interior points.If P
is an interior point of H then you can ?d a closed ball
B(P,r) withr>0 such that B(P,r) is in H, ie for any point X
such that N(P-X)<=r, X isin 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, FranceHome page:
http://www.analgebra.com
===
> If P is an interior point of H
then you can ?d 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, Xis>
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.in?ite-dimensional space is also normed.But, if I'm
not wrong, the hypothesis that the space isin?ite-dimensional
is not relevant for the proof. In fact it shouldbe also true
that:A subspace of a normed in?ite or ?ite-dimensional
linear space isclosed and without interior points.In
particular when you prove that a subspace of a normed space
iswithout interior point ( except the case that subspace =
space ) youdon't refer to its dimension. So the sentence
should be always true.
===
Also, he didn't say what scalar
?ld he is using. A normed vectorspace over the rationals,
say ... ? Then a ?ite-dimensional subspaceneed 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 ofS,>
and a neighborhood U of x lies in S. Then S also contains
aneighborhood of> 0, namely -x+U. If y is any element of B,
then there is a positivenumber r> such that ry belongs to
-x+U; therefore ry belongs to S, therefore ybelongs> to S,
and it follows that S=B.>> in?ite-dimensional space is also
normed.> But, if I'm not wrong, the hypothesis that the space
is> in?ite-dimensional is not relevant for the proof. In fact
it should> be also true that:> A subspace of a normed in?ite
or ?ite-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 beclosed. Start with any normed linear space
which is not complete; thisspace is isomorphically isometric
to a dense subspace of its completion.More speci?ally, the
space of polynomials is not closed in the space ofcontinuous
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 - ?).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-ishnotation in this post. I'm not a mathematician, just
an engineer witha 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,
PhDX-Replace-Address: yesreply-to: vanw a_t nist d_o_t
gov
===
I am stuck on showing that this sytem of equations has
unique nonzerosolution (well I think it does). A is n by n
matrix with positive entries (maybe not necessary) andlargest
magnitude eigenvalue > 1 . Show that there is exactly one
nonzero x=(x1,x2,...,xn) with all entriesbetween 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 ?ed point theorem beapplied if we
bound it away from zero?
===
I'm in a distance education
course, and am stuck on the followingquestion:Find the
equation of the line through (8,8) that is tangent to
thehyperbola y^2-3xy+2x^2=4.We are currently doing a unit
involving Newton's Method, Tangent lineapproximation, and
Tangent line approximation(increment form), so i amassuming
that i must use one of them to ?d the answer.I've been
trying to isolate variables and trying to do a system
ofequations, etc., but can't get the answer no matter what i
try. Ifanyone could just point me in the right direction with
a ?st stepthat would be great. I don't need the answer, just
a nudge in theScott 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 ?d 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 ?st step> that would be great. I don't need the
answer, just a nudge in the>> Scott EliasonJust guessing, I
think this might well be a misprint. The exact solution isa
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 thesection Implicit Differentiation, you
can use implicit differentiation toget y' in terms of x and
y, then you can plug in x=3, y=7 to get the slopeof 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 ?d 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 ?st step> that would be great. I don't need the
answer, just a nudge in theI don't think the intent of the
problem is to use any of thethree topics you've mentioned
above. It seems like more of areview and synthesize the
stuff you've already learned typeof problem. Imagine that
you know the coordinates of the point on the hyperbola where
the line is tangent. Now express the slopeof that line in two
different ways.Bart
===
First ?d dy/dx by Implicit
Differentiation.2y(dy/dx)-3y-3x(dy/dx)+4x=0(2y-3x)(dy/dx)=3y-
4xdy/dx=(3y-4x)/(2y-3x)So m=-8/-8=1Now we have to ?d the
equation of a line with the slope of 1 and has thepoint
(8,8). The line is obviously y=xLooking at the equation, I
don't think (8,8) is on the curve. Perhaps a typo
ormisprint?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 ?d 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 ?st
step> that would be great. I don't need the answer, just a
nudge in the>> Scott EliasonI would do so:A line trought (8,
8) with slope m, has the equationy = m(x - 8) + 8If that line
is a tangent, it must have a unique intersection point with
thehyperbola. Substituing in the hyperbola equation,(8 + m(x
- 8))^2 - 3x(8 + m(x - 8)) + 2x^2 = 4 ==>(m^2 - 3m + 2)x^2 +
8(1 - m)(2m - 3)x + 64(m - 1)^2 - 4 = 0Solving 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 solutionwould be unique. Then,5m^2 - 11m + 6 = 0 ==>m1
= 6/5, m2 = 1Substituing m = 6/5, you get the tangenty = (6x
- 8)/5with tangent point (3, 2).Substituing m = 1, you get y
= x for the ?tangent line', but without tangentpoint,
because x is indeterminated. If you substitute y = x in the
hyperbolaequation, you get 0 = 4!.What happen here?It happen
that the line y = x is a asymptote of the hyperbola, a tangent
inone of the its two in?ite points.-- Ignacio Larrosa
Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
> First
?d 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 ?d 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 atypo or>
misprint?>> David MoranThe actual solution is not a mess at
all, as some miscalculation led me tobelieve. 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 theequation for the hyperbola, and you
get a unique solution. Both x and yare small positive
integers.
===
Why is this a hyperbola, why not an ellipse?
Although that would not changehow 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 ?d 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 ?st
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 de?es any of these curves.E.g., x^2 + y^2 + 1 = 0
does not de?e 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 inde?ite: thecorresponding 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.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
> First ?d 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 ?d 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 de?itely an interesting question. I tried
to follow Mr.Moran's work, but I encountered a problem when
solving as a system ofequations. If someone could help me out
here that would be great. J.A. Sanders
===
> First ?d
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 ?d the equation of a line with the slope of 1 and
hasthe> point (8,8). The line is obviously y=x>>
Looking at the equation, I don't think (8,8) is on the curve.
Perhapsa> typo or> misprint?>> David Moran>>
> The actual solution is not a mess at all, as some
miscalculation led meto> 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 withthe> equation for the hyperbola, and you
get a unique solution. Both x and y> are small positive
integers.>> This is de?itely 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.
SandersOkay, to lay this thing to rest. David Moran did at
?st what I did at?st, assumed (8,8) was on the curve,
because this is the kind of problemthat you see in calculus
books in the Implicit Differentiation chapter.Starting with
the equationy^2-3xy+2x^2=4we differentiate it implicitly to
obtain2y y' - 3x y' - 3y + 4x = 0We solve this \
equation for
y' and gety' = (3y-4x)/(2y-3x)If a line tangent to \
the
hyperbola passes throughthe point (x,y) lying on the
hyperbola, then its slopeis 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 slopeis 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 obtain8x - 4x^2 - 8y + 6xy -
2y^2 = 0 (*)and now we wish to solve this simultaneously
withy^2- 3xy + 2x^2 = 4 (**)Multiply (**) by 2 and add to
(*). This gives8x - 8y = 8y = x - 1Substitute this in (**)
and expand, and you getx = 3, y = 2.Use the trick of setting
the constant term to 0 to ?d theequations of the
asymptotes.y^2-3xy+2x^2=0 becomes(y-x)(y-2x)=0fact that the
points (0,2) and (0,-2) are on the curve, you cansketch a
good graph of the hyperbola, and you will be convincedby
looking at the graph that (3,2) is the only solution.I guess
this isn't quite all, the original problem asked forthe
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 ?ite point of the
hyperbola and one tangent at in?ity, 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 ism = (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 = xof 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 = 0Dick
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 ?d 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 ?st
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 ?Well in another thread someone posted a link
to Wiles's paper, so I'vestarted looking over its
introduction.Here's an intriguing quote which I'm \
focusing
on, though it may dragme into greater details.The key
development in the proof is a new and surprising link
betweentwo strong but distinct traditions in number
theory,the relationshipbetween Galois representations and
modular forms on the one hand andthe interpretation of
special values of L -functions on the other.p.2Source:
http://modular.fas.harvard.edu/21n/papers/Wiles,Modular_
Elliptic_Curves_and_Fermats_Last_Theorem.pdfThe special
valus of L-functions are those 4 descriptors popping upagain.
My understanding is that mathematicians have reworked
thatapproach having found something shorter, but I'll focus
on theoriginal. > 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 speci?ally
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 aseasily as I expect it should, or
if it's buried behind a lot oftechnical 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, havingcompletely failed
to provide a valid proof of FLT, is to attack the proof
provided by Wiles. This is standardfare for cranks -- if you
can't create, try to destroy. But please spare us the
disassociated hand-wavingarguments. Stick to a speci? point
instead of continually leaping triumphantly to false
conclusions. You havebeen 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, soas not
to tip-off the many, competent to beat himto the punch (such
as Ribet). of course,it could also be ? or,simply
inelegant.how would you characterize the sum-totalof your now
8-year mission, monsieur Harris,minus all of the vituperative
garbarge? of course,such an approach may be feasible for
teaching math, althoughit's hard to imagine the student-body
that'd tolerate that sortof 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 ?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/?es/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
speci? 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 ?hey can be
thoroughly convinced of falsethings.The reason the primary
newsgroup for this thread is sci.logic is thatlogicians are
tasked with being careful in a way that evenmathematicians
aren't.Mathematicians can get away with leaps and assertions
because of therelevance of mathematics to the real world,
where physics has beenpowered by mathematical models.But
logic is cold, hard, and less amenable to social
pressure.I've outlined a speci? logical ? Wiles's
approach, which isthat it depends on the logical fallacy of
trying to ?d conditionsthrough 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 ?itesubset
to ?d a restriction on an in?ite set?The technique is called
lifting and is something like in?itedescent made somewhat
famous by Fermat himself.It is an intriguing question, so I'll
back off somewhat while Iconsider whether or not he somehow
gets around the logical fallacy to?d a way to break it.Now
mathematicians apparently are quite certain that Wiles
succeededand I applaud their energy. However, I must also
rely on myunderstanding that human beings have a certain ? ab
ility tobelieve almost anything.Logic, however, is
perfect.James Harris
===
> Human beings have a ?hey 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>
?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
?hey can be thoroughly convinced of false> things.I just
love the way you state the most commonplace trivialities as
ifthey were some profound new insight.> So does Wiles get
around the lack of a logical basis by using a ?ite> subset
to ?d a restriction on an in?ite set?> The technique is
called lifting and is something like in?ite> 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> ?d a
way to break it.You are not quali?d to determine whether or
not Wiles was successful.Only a limited number of people on
the face of the earth are quali?d(and willing to take the
necessary time and effort) to do so. Until youare able to
read (and understand!) his proof for yourself, you'll
justhave to take their word for it.Or not. Your opinion has
no importance to anyone but yourself, sobelieve whatever you
want.-- Wayne Brown | When your tail's in a crack, you
improvisefwbrown@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 assertionof 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, andtime after time people have been
unreasonable in the face ofrationality. Sure I set out a few
years back to ?d some answers tosome 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 workas it relies on association to prove a
condition. More usefuldiscussion has worked things down to
the assertion that Wiles used a?ite set, and lifting to
prove the equality of two in?ite sets,where the equality
supposedly forces the condition.There is, however, no reason
for the condition given, and no claim ofa reason, where the
condition, from my understanding, is that everyelliptic curve
is a modular elliptic curve.The question raised before Wiles's
work being whether or not anelliptic curve could not be
modular, where mathematicians had relatedvarious elliptic
curves that they tested to something called modularforms, and
found that every one they tested was modular. Then
themathematicians Taniyama and Shimura conjectured that every
ellipticcurve was modular, which my understanding means, they
are associatedwith a modular form, where that association can
be described by 4descriptors.My understanding is that Wiles's
work depends on association.The logical fallacy I've put
forward as a challenge against his workis,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 HarrisYour ?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. Youhave been thoroughly
debunked. Remove your faulty and error-ridden attempts at
proving FLTfrom 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' proofover the other situations. Experimental
evidence, for example, can bechecked more easily,
unambiguously and objectively in Wiles' situationthan 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 immensechain of experimental and
theoretical reasoning leading to the current modelswith
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 lineon matters of photons, DNA, relativity,
evolution, the existence ofthe Iraq War and Sikkim and
Napoleon --- but intone high skepticismconcerning Wiles.>
Since that time from lasers to spectral analysis the theory
has ?> with reality.Lasers only hurt your cause, as to check
that a laser (resp. spectrometry)experiment actually
corroborates photons you would have to check mattersof
chemistry, crystallography (geology!), engineering,
manufacture, and soon all the way down. The only way out of
this is to accept various assertionson faith from the Evil
Scienti? Establishment, and the question arises why youare
such a sheep and conformist when it comes to non-Wiles but
raise hightenedstandards concerning Wiles (who of courses
passes all the scienti? standardsfor 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 anylogical ?but checking it potentially
involves going through eachstep in his work, which is a
formidable task. If he did ?d a proof,then I think it
interesting on logical grounds that there is aworkaround i.e.
that Cum hoc ergo propter hoc is not actually alogically
fallacious approach.Now as for physics results, like many
people trained in physics, Ikeep a skeptical eye on theory,
and depend on things I've personallychecked, or that are very
unlikely to be wrong that have been checkedby others.
Physicists can be hard-liners to the extent that theydon't
believe physics they haven't personally checked. I'm \
not.
Likehow I believe that nuclear weapons work. But still
realize that theabsolute truth may be something other than
what I've learned.In mathematics, absolute truth *can* be
determined, just like alogical argument can be checked
against certain rules for internalconsistency. > Since that
time from lasers to spectral analysis the theory has ?>
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 Scienti? 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 scienti? standards> for
photons, etc, and then some).As a person with a science
degree, I guess you'd consider me a part ofthe Evil
Scienti? Establishment.It's actually more fun attacking
them than just sitting aroundbelieving in them. Because you
learn a lot in the attack, and yourguarantee 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 con?ently on the subject.You speak con?ently
whether you know what you're talking about or not. Con?ence
is not yourproblem, 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 ?but checking it potentially involves going
through each> step in his work, which is a formidable task. If
he did ?d 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
youessentially:1) proof of an absolute kind, presumably
stated in the symbolism offormal logic, and2) proof that
merely convinces other mathematicians, presumably statedin
some meta-language (like English).Furthermore, your position
is that proof of the second kind is whatmost mathematicians
produce, and is not good enough. You go on to saythat you
produce proofs of the 1st kind.Question: How does one
determine that a proof or mathematicalargument is
absolutely and irrefutably correct?The validity must be
checked by 1) God, 2) a machine, or 3) a humanbeing, 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 o?ference, but it occurs to me that a few
problems could arise. First,the algorithm may take an
unreasonable amount of time to reach adecision. Second,
hardware failure, electrical surges, sunspotactivity, running
the program under Microsoft Windows, etc. couldcause erroneous
results. Third, and perhaps most importantly, a humanbeing (or
beings) must write the software. Therefore, any errorscaused
by people could conceivably appear here. That leaves us
withoption (3). As we all know, people make mistakes. They
make mistakeswriting proofs. The publisher/editors of a
journal may make a mistakemistake by erroneously believing
the proof.Who has the ?al and ultimate authority to say
that a given argumentis 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
adisagreement, then the sides may argue their cases until one
sideprevails and convinces the other, at least within a given
mathematicalsystem.Therefore, in this sense, all proofs are
of the second type. We muststrive to convince other
mathematicians. That is all there is --simply because there
is no other means of asserting the validity of amathematical
argument. It really is an appeal to the gallery.We must
also consider that mathematics may be inconsistent.
Accordingto Kurt Godel, this is a possibility (at least for
mathematicalsystems 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 ?> 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 Scienti? 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 scienti? 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
Scienti? 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 con?ently 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 ?ht 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 statementas in the case of Wiles'
proof, let alone one provable by experiment. Thereare of
course theoretical models (not necessarily well-de?ed or
knownto be logically consistent, by the way) within which one
can single outcertain objects as photons.Second, your
student experiments in optics could not possibly replicatethe
mountain of theoretical and experimental steps involved in
buildingup any of the theoretical model(s) involving photons.
Instead, youaccepted on trust assertions by textbook authors,
professors and similarpurveyors of the Social Truth that you
like to castigate, amounting toa certi?ation-by-authority
that the apparatus you were doing the experimentswith
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 experimentationneeded to corroborate the values of
relevant physical and chemicalparameters listed in the CRC
handbook, and so on all the way down.What actually happened
is that a long and social processof knowledge-accumulation
occurred and you took the results on trust.In particular, if
your experiments gave a wrong result, the conclusionwould
be that you made a mistake, not that photons' existence is in
doubt;a pure assertion of authority by the Scienti?
Establishment concerning itsSocial Truth, which you accept
without any objection in all the non-FLTsituations.Note that
your repeatedly discredited objections in this thread
aboutWiles' logic are irrelevant, as you also object to
Ribet's proof withoutgiving any particular reason to doubt
it. The matter is simply one of anobvious double standard
produced for the occasion, where social certi?ationby a
small network of experts counts as OK for photons,
DNA,evolution, relativity, the Jurassic era (or the existence
of Napoleon andGeorge W Bush), etc --- but somehow the
information that expertshave certi?d 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 thoughtexperiment called the superluminal
scissors or had anyidea what I was talking about in
sci.physics when Iexplained how a 5 m/sec water jet can be
used to createan illusion of arbitrarily fast, even
superluminalmotion. - 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-de?ed 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
assome loser will state a case, get their ass kicked, but
STILL keepcoming back as if nothing happened.My *degree* is
in physics. I went to school on a full-tuitionscholarship,
and you stepped into my ?ld with your assertions, gotyour
ass kicked but refuse to back down.Now *emotion* is not
necessary when it comes to Wiles's work. If hefound a proof I
can assure you that it is irrefutable. That's whyit'd \
be a
proof. All this emotion just annoys me, as part of the funof
science and mathematics is attacking things that are
supposedlyproven.It's GREAT fun not just accepting what
people tell you. But that'swhat really annoys me about
mathematicians as time after time I getyahoo's replying back
in defense of mathematics, using tactics.But you see, not a
single REAL mathematician in the world gets excitedabout an
attack on a proof. No mathematician worth the title wouldget
even a little concerned, nor would they lose sleep, or
?dthemselves emotional about some person--any person--any
time--anyplace--who decides to go after a math proof.That's
because a math proof is indestructible, incorruptible,
andirrefutable.It just doesn't care if you attack it, and no
real mathematician wouldcare either.Now I've discovered math
proofs, which is why I'm not concerned aboutpeople refuting
them because they are proofs. And in fact people whocall
themselves mathematicians can't touch them, so they come up
withextraneous stuff, or make claims of ?ding their own
proofs to refutemy proofs, but you see, proofs don't duel.And
you know what? I think that feature of mathematics terri?s
somepeople who call themselves mathematicians. Mathematics
does NOT carewhat you call yourself. It DOES NOT care that
you have a mortgage. It DOES NOT CARE that you really,
really, really want people to likeyou and think you're a
great mathematician.Now I've made a speci? claim against
Wiles's work. If he found aproof the claim can be answered,
but even if it is answerable then itmust be true that he has
found a way around what is considered to be alogically
fallacious approach. Logicians should thank him in thatcase
for correcting them.My challenge is a logical one. Wiles's
work fails and is not a proofas 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 teamsplay each other once?I
tried several avenues. I don't have the time or expertise to
tacklethis 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
ofunity 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. Atleast 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.htmlr 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 ? 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.> thanksHave 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 diagonalsin the
polygon of n sides type of counting. Make a regular polygon;
the number of verices equals the number of sides. howmany
diagonals are present, including the sides? Now make another
regularpolygon using one more vertex, and equivalently, one
more side. Count thediagonals and include the sides. Keep
doing this until you have a pattern forwhich 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.htmlNow, 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 atIf n is odd, add a dummy team as a bye to
make the number even.Rob Johnsontake out the trash before
replying
===
Good day,I am processing batches of data stored
in arrays, and I have seenexamples of vector mathematics
(rather than loops) being used to dothe processing.For
instance, for performing a linear regression on a set of data
Xand Y (to ?d the best-? slope ?k') I am used to seeing
themean-squared-error (MSE) being expressed as a summation of
thesquared-errors of each data pair, but I have also seen the
MSEexpressed as in dot-product form which is equivalent. In
the end, theresult is k = (x'x)^(-1) * x'y.To get \
this
optimized result, a partial derivative with resprct to kis
taken on the MSE function. I can follow along, but I am
unsureabout some of the procedures, especially with regard to
getting the_order_ of the vectors expressed properly to ensure
dimensionalagreement (e.g. how to expand (kx-y)*(kx-y)' , how
to apply chainrules when taking derivatives), but I don't
know what the forms ortecniques are called.This is not the
vector algebra or vector calculus that I have beenexposed
to. Or is it? I'd appreciate knowing so that I could have
amore 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
?d the best-? 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)
inoptimization theory. You might for instance check intothe
theory of quadratic programming, which is the theoryof
optimizing a general multivariate quadratic functionunder
linear equality and inequality constraints.Least-squares is
one particularly easy quadratic problem:an unconstrained
minimization with a positive de?itecoef?ient matrix.When I
was ?st exposed to this kind of manipulation,I found it
helpful to work out expressions in termsof individual
components and summations. Going backand forth between those
kinds of things and their matrixequivalents is really helpful
to get facility with thealgebra.Here are a couple of
particularly useful identitiesfor 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_jIf Q is not symmetric, there is always a
different matrixP which gives the same function but is
symmetric:P = 1/2(Q + Q'). Component wise, p_ij =
(q_ij+q_ji)/2for 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_jThis tells you the
correspondence between a general multivariatequadratic and
its symmetric coef?ient matrix.2) If b is a vector, b'x is a
scalar. Thus it isits own transpose, i.e. b'x = x'b.3) \
The
gradient grad(f) where f is a scalar is a vectorwhose i-th
component is df/dx_i. grad(b'x) = bProof: 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) = 2PxProof: 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 youcompute even when DF(x) itself is hard to write down
using vector (ormatrix) 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 inR^n are column
vectors; prime ? denotes transpose.) Then the
partialderivatives with respect to the ?st argument (x) and
the secondargument (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 thingssuch as
the cross product in R^3.)An excellent example of this, which
illustrates both the concept andhow dif?ult it can be to
write down DF(x) rather than DF(x)(h), isthe formula for the
derivative of the matrix inversion function. LetF(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) = 0so 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 notthe same as
-X^{-2}H ; in particular, the scalar formula
DF(X)=-X^{-2}does not hold for
matrices.)Kevin.
===
testing===> testingF-===|-|erc
says...>YES I showed this in Cantor's disproof thread, by
de?ition>computable reals are computable, which means they
can>be given a TM number.But not every TM number represents a
computable real. To geta bijection between the natural numbers
and computable reals,you have to set up a correspondence such
that each naturalnumber corresponds to exactly one computable
real, and everycomputable real corresponds to exactly one
natural number.A computable version of Cantor's proof shows
that such acorrespondence cannot be computable.More
speci?ally, let C = the set of all natural numbers m
suchthat m represents a TM number for a computable real
number. Letf be a one-to-one mapping from C onto N (the set
of all naturalnumbers). Then f cannot be a computable
function. So there is nocomputable bijection between the
computable real numbersand the naturals. --Daryl
McCulloughIthaca, 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 areuncountable.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>
===
>
In?itary objects exist in the sense that we>> can
construct ?itary 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>> ?itary approximations in your view? >> The notion of R
as it is de?ed within the axiom> systems inspired by
Cantor's theory cannot be> constructed by ?itary
approximations. In fact,> that's what the diagonalization
argument proves.>> Nevertheless, we can construct something
very> close to our intuitive notions of R by ?itary>
approximations. But, the de?ition of R precedes Cantor. Take
Dedekind's de?ition,say, or Cauchy's, and it is easy \
to
derive that the usual canonicaldecimal representation works.
Cantor did *not* offer a new ordifferent de?ition of R. Nor
does the de?ition of R come fromCantor's work. Rather,
Cantor proved an important theorem about theexisting
de?ition 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-intuitiveconclusions, but that's because the theory
of R suf?ient forCantor's result is not due to Cantor. In
other words, I agree thatthe prevailing theory of R in
Cantor's time to the present leads to atleast one
counter-intuitive (i.e., surprising) result, namely that |N|
< |R|, but you seem to be showing historical confusion in
callingit Cantor's theory[1]. One ought to have a better
reason for tossing out the completeness ofR 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 orwell-respected
secondary sources.-- Come on people!!! The US just blew up a
lot of people in Iraq, don'tyou realize that a person with my
exposure might just end up dead, bymysterious 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 asthere does
not exist a bijection between X and N.Of course, whether a
set is uncountable or not depends onthe set of bijections you
are willing to consider. If youonly consider computable
bijections, then the computablereals are uncountable. If you
allow de?able bijections(de?able in ZF, for instance, or in
type theory) thenthe computable reals become countable. >But
would you really call that Cantor's proof?It is
structurally the same proof, except that instead ofassume
that f(n) is an enumeration of all realsyou say assume that
f(n) is a computable enumerationof all the computable
reals.The big difference between computable objects and
noncomputableobjects 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 ofTuring machine indices n
such that {n}(n) never halts. There is noconstructive
bijection between NH and N. So NH is uncountable
byconstructive functions. But NH is a subset of N.--Daryl
McCulloughIthaca, 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?> PhilYes. 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, andq<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 o?terest 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 de?ition 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 theHodge dual and ^ is the (antisymmetric)
exterior product. (I'm not sure ifHodge dual is standard
jargon, but it's easy to see what the term isintended 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] = AxBthen 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 de?ition 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.)> LHAre 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.htmlTake
a look there.And even if they're wrong, and I'm wrong, \
about
that, then that stilldoesn't make the results of my proof
(that the cross product is a formof 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 de?tion of nonassociative algebra is rather
dubious.Did you ?d 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 1of 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 de?tion of nonassociative algebra is
rather dubious.I didn't get it. Their de?ition reads ?a
nonassociative algebra isan algebra that ain't \
associative'.
Isn't that true?> Did you ?d any examples of associative
Lie algebras (your exercise).I think there are plenty of
associative Lie algebras. Just thegenerators 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 factors2) [A,B]=-[B,A] 3) Jacobi Identity
===
>
Their de?tion of nonassociative algebra is rather
dubious.> I didn't get it. Their de?ition 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 ?d 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 ?d 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 liealgebras.--
===
>
Their de?tion of nonassociative algebra is rather
dubious.> I didn't get it. Their de?ition 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 ?d 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 lackof 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 Liealgebra' or something. Somehow got it
messed up.
===
> |> Did you ?d 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 ?d 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 bythe 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
de?ition 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 crossproduct is a lie algebra. let A, B in
R^3, then the hodge dual (thestar * operator) mapsA^B to A
cross B, and vice versa. as such, it is a linear
spaceisomorphism between R^3 and linear space of bivectors.
regardingelectromagnetism, it can certainly be formulated
using the exterioralgebra, 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 ?d the remainder of 37^100 when divided by 29.I have
played around with 37 = 8 (mod 29) by raising it to powers,
but Ihave yet to discover the right path. I
tried37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have
also thought aboutresidue classes, but I just can't seem to
make any connections. Couldanybody give me an
idea?TIALurch
===
>I am working on a problem from Dummit &
Foote's book, Abstract Algebra.>The problem is to ?d 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 tosuccessive powers, modulo
29?What is 8^{28} (mod 29), according to Fermat's Little
Theorem? What is8^{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
Magidinmagidin@math.berkeley.edu
===
>I am working on a
problem from Dummit & Foote's book, Abstract Algebra.>The
problem is to ?d 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.caDepartment 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 ?d 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 ?d 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 ?d 37^100 mod 29 quite quickly.
===
> I am
working on a problem from Dummit & Foote's book, Abstract
Algebra.> The problem is to ?d 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>> LurchBeing not familiar with this stuff,
I did some trial and errorwith 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 = 23Does anybody know whether this is a
(special caseof 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 mistakewith the exponents. Sorry.Lurch> I am
working on a problem from Dummit & Foote's book, Abstract
Algebra.> The problem is to ?d 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 ?d 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 ?st
line of yours: For every prime p, x^p == x (mod p) for all
integers xand I got stuck. Help!Any elementary on-line
textbook (pdf or ps) you couldrecommend?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 ?st
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 VdmMaybe
Course 311 - Abstract Algebra, Part I: Topics in Number
Theory:http://www.maths.tcd.ie/~dwilkins/Courses/311/
311NumTh.pdfby D.R. Wilkinsor follow the links given in
Annotated Web Links forKenneth H. Rosen's Elementary Number
Theory
Book:http://www.aw-bc.com/rosen/resources.htmle.g.Fermat's
Little Theoremhttp://www.cut-the-knot.org/blue/Fermat.shtmlor
search the homepages of number
theorists:http://www.numbertheory.org/ntw/list.htmlMany of
them have lecture notes online.Hugo Pfoertner
===
[...]> Any
elementary on-line textbook (pdf or ps) you could> recommend?>
> Dirk VdmOnline number theory lecture
notes:http://www.numbertheory.org/ntw/lecture_notes.htmlHugo
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. WilkinsJust what I needed.Apparently this thing For
every prime p, x^p == x (mod p) for all integers xis
Fermat's theorem. Not so trivial. no wonder I didn't \
?da
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.shtmlYep, had found
this one with google:
http://www.cut-the-knot.org/blue/Modulo.shtmlThis was exactly
what I was looking for.Had this stuff is almost 30 years ago -
Good refresherfor the basics.>> or search the homepages of
number theorists:>
http://www.numbertheory.org/ntw/list.html>> Many of them have
lecture notes online.>> Hugo PfoertnerDirk Vdm
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===
> So I have been trying to prove that ?st
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 modp, so whenever x is
not zero (mod p)x^(p-1) = 1 (mod p) (Lagrange's theorem)
whencex^p = x (mod p)which also holds for x=0 (mod
p).
===
without the aid of a calculator/computer, or Fermat's
little theorem. Inthe book, all he discusses is modular
arithmetic. FLT is not mentioned 'tilpage 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 remainderright? So, if, for example, I take 3^2 mod 2
the theorem says I should get3, but isn't it 1? I mean 9 mod
2 should be 1, right? Futhermore, if Itake 37^29 = 37 mod 29
and I try to raise this congruence by powers until Iget my
100, then my remainder gets larger and larger. So, if my
remainderis supposed to be 23, I doubt that 37^x is going to
be my answer. What theheck 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 ?d 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
29Dirk Vdm
===
fairly soon, but I was starting to get a little
frustrated. On the otherhand, they say close only counts in
horseshoes and handgrenades; so, whoknows? 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 refreshinglook this
afternoon at http://www.cut-the-knot.org/blue/Modulo.shtmland
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 numbertheory ;-)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 29Or 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 de?ition 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 congruenceSince 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 ?d 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
?ure 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 thislet's search for a small power of 2 that equals
+-1 (mod 29).We search for numbers equal +-1 (mod 29) that
factor intopowers 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 =
2^5so 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 thatboth answers are correct, modulo
2.If you want to skip Fermat's Little Theorem, then study how
the powersof your number cycle moudlo 29. For example, if I
wanted to ?ure outwhat the last digit of 3^{200}, what I
would do is consider the powersof 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 is0, 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. Thatmeans that37^{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 ?ure out how much37^{16} is.And
second, every time you get a number larger than 29, YOU
REDUCEMODULO 29, to get a smaller number and work with that;
usually, yourbest 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 Magidinmagidin@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-signedarithmetic.What emerged from my researches
is that not only is data limited bythe data-set from whence it
came, but the DIMENSIONS of data arelimited to the dimensions
from whence they came.Thus there is a fundamental problem
with Euler's EXP(iX) in that itbegins with a number (UNARY)
that has FREEDOM TO MOVE in the iXdimension (MONAL), and
converts to a PAIR of numbers (BINARY) withFREEDOM TO MOVE in
two dimensions (BINAL). These are Cos(X) andiSin(X).When you
try to go back to the unary state, you have not PEGGED
orSPECIFIED or LIMITED your remaining freedom - you have left
it tochance. Thus the answer you arrive at is just one of
many.Here is an example of a data set:12If I ask you to
continue the set, you might
say1,2,3,4,51,2,4,8,161,2,6,24,120These are a LINEan
EXPONENTIAL (base 2)a FACTORIAL.A Chebyshev approximation
might ?d Y=0+XTaking data from that poynomial might
give11.11.21.31.41.51.61.71.81.92WHAT AN ABUNDANCE OF
DATA!Convert this to a TENTH-ORDER polynomial approximation,
you getZero + 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 dittois no better than the ?st-order.At
position 1.5, we would have got1.5 for a straight
line1.414213562 for exponential 21.329340388 for a fractional
factorial.Given only two pieces of data, the Chebyshev could
not distinguishthese - and took the SIMPLEST. It delivered
the straight line.Interpolation, giving the in-betweens,
does not add to the data. Itis an exercise in REDUNDANCY,
and the eleven pieces of data stillcontain only two starting
facts within them.The METAPHYSICS of my reasoning with
complex numbers similarlystates that the the TWO dimensions
of the complex plane still onlycontain 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
pagehttp://wehner.org/euler and for thinking about it.I
describe the facts about the world as I see them. I do not
makethings up. However, such situations as the PROBLEM OF
MULTIPLESOLUTIONS require an abstract reasoning approach.That
problem will not go away - and those who slavishly apply
complexmathematics without using common sense will come
unstuck.Charles Douglas Wehner
===
I just thought about
something and hoped someone could help me explainwhy I'm
wrong in thinking it or why I could be right...I thought
really simple, Take a sphere and begin walking on it (itnever
ends). When I want to cut the knot I just do something
radicaland go right through the sphere or something. So are
we thinking thatsmall? This is somewhat metaphor I had on
in?ite numbers...Anyone who can help me ?d 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 in?ite numbers...>>Anyone who can
help me ?d that red line again?So in 1 dimension you have
in?ite numbers because of 1 dimensionalthinking.. In another
you have hypernumbers that could resemble thein?ite
===
>
So in 1 dimension you have in?ite numbers because of 1
dimensional> thinking.. In another you have hypernumbers that
could resemble the> in?iteWell, that wouldn't really work,
atleast if you're talking about thereal number system. Cantor
proved that |R^n (that's R to the n, theset of real, ordered
n-tuples, so the dimension is n) has the samecardinality as
|R. That is, there is a function which maps everypoint in
|R^n to a point in |R uniquely. So |R and |R^n have the
samenumber of points.You may want to read about the Riemann
sphere, however. Imagineplacing a sphere on the origin of the
complex plane (or just |R^2, thestandard real plane). Imagine
a line connecting the top of the sphereto a point on the
plane. For any point on the plane, this line willintersect
the sphere at a unique point. Now, what happens as
thedistance from the origin of the plane gets very far? The
line tendsto the tangent line at the top of the sphere! So it
only intersectsthe sphere at one point. In complex analysis,
we call this the pointat in?ity, and we call the plane
with this point added theaugmented plane.Check
outhttp://tinyurl.com/haxyThe ?st picture is of the Riemann
Sphere, the rest are of functionsmapped onto the Riemann
sphere.Alex
===
>> So in 1 dimension you have in?ite
numbers because of 1 dimensional>> thinking.. In another you
have hypernumbers that could resemble the>> in?ite>>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 in?ity, and we
call the plane with this point added the>augmented
plane.>>Check out>http://tinyurl.com/haxy>>The ?st picture
is of the Riemann Sphere, the rest are of functions>mapped
onto the Riemann sphere.>>AlexI hope I'll get new inspiration
from it :D
===
>Alex>Check out>http://tinyurl.com/haxy>>The
?st 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
inthe other. I have a box function on my old Panasonic
RK-P400C plotter that doesthat. 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 feeCivil 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 ?ace. 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 ?nd 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 ?d the shape of the universe, we
must ?d 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 ?d 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 ?ace. 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
asranks: n rank(H_n(T^3)) 0 1 1 3 2 3 3 1So, the Euler
characteristic is 1 - 3 + 3 - 1 = 0.This could have been seen
several ways: ?st, the Euler characteristicof any closed,
orientable manifold of odd dimension is zero (by
PoincareDuality). Second, the Euler characteristic of any
product X x Y, where Xhas zero Euler characteristic, is
zero.Regarding handles, for dimensions greater than 2,
handles come in alldimensions from 0 to n (for an
n-dimensional space). The 3-torus T^3has (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
cantake 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 ?nd that's not what \
we've
observed!Flat metrics exist on tori of all dimensions. After
all, the n-torus isthe 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 ?d the shape of the universe, we
must ?d 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 thetangent
bundle. If I recall correctly, that only requires a zero
eulercharacteristic (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 ofthe Lorentz group
with the group of Euclidean translations, so if youget a
Lorentz structure on R^4, compatible with the translation
group,you should automatically get the requisite Poincare'
structure on anyquotient by a discrete subgroup of the
translation group [such as howone obtains the torus]. I'm no
expert on how these particular groups arehandled in the
physics world, so I would gladly defer to a
knowledgeableperson 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 ?d out how
to make one of those things I>>described.Perhaps you should
use the term topology instead. Many references areavailable
for the topology of manifolds, including those with
Lorentzand/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 de?e 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 ?st
honest operators that I can come up with. The three sign
system obviously invokes rotational phenomena. Whereasthe
reals use a negation to toggle back and forth in magnitude
fromone extremity to another we now have more. The minus
operator (-) willtoggle in one direction. The (+) operator
will toggle in the otherdirection, and the star operator goes
all the way around to preservethe original sign. These
rotations are countable and can formulate aninteger 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 ?st.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 itaround 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 theneed 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 threadand using
the standard arithmetical laws (they do feel right for
themoment) ( + 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 minusdirection. 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 mathematicssystem.
===
>
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 ?st honest operators that
I can come up with.It's not clear what T is, nor in this
summary did you include anydescription 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 anoperator 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 0And so
the system is multiplicatively identicalto the union of the
three rays through the originin the complex plane through the
cube roots of unity.Perhaps some day we'll get a notion of
addition forthese. 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 de?ing T in terms of R.If you
could de?e R in the same terms that you propose then I
wouldhappily modify it to encompass T. But since such a
de?ition of Rwould read: ({-,+} x { x in R | x > 0}) /
{0}you would de?e 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
andso 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 possiblesigns (-,+,*).Would an equivalent to the number
line suf?e?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
linesintersecting and so is the natural symbol to use for the
next sign inthe 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 con?ith 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 inthread).> Also think
about 0 = +0 = -0 = *0 and do you have convention, +r = r.I
think that an unsigned zero is ?e and that it is equivalent
to thesigned zeros. I see this as an exception and would like
to see signpreserved symbollically on all non-zero numbers.
Therefore +r does notequal 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 takeseveral
forms.To clear up your controversy lets de?e 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 minusoperator 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 wayof 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 sothat there is less wading through
redundant stuff.I do believe you understand very well the
problem that I am workingon.-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 toY.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
datagoing 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 in?ity it would return.The COMPLEX
mathematics states that it shifts SIDEWAYS in complexspace,
and so leaves the real world. Multiplication by -i1 would
bringit back.NEITHER is a total answer to the problem of
vanishing data. Perhapsthere is the possibility of a new
Three-signed arithmetic emergingto tackle such
problems.However, at a ?st perusal I cannot quite see where
this is heading.It would be useful to have real-world
problems solved or PARTLY-solvedby the Three-signed
arithmetic. Such worked examples would helpfocus 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 beexplained by the
hump in the Eucalculus curve.That curve is itself no more
than the reciprocal of Euler's Gammafunction - so it is an
extension of mainstream maths.In this present case, we seem
to be leaving the customary de?itionof + 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
thefundaments to be found - with no discussion. This would
help in anystudy 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
datagoing 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 in?ity it would return.The COMPLEX
mathematics states that it shifts SIDEWAYS in complexspace,
and so leaves the real world. Multiplication by -i1 would
bringit back.NEITHER is a total answer to the problem of
vanishing data. Perhapsthere is the possibility of a new
Three-signed arithmetic emergingto tackle such
problems.However, at a ?st perusal I cannot quite see where
this is heading.It would be useful to have real-world
problems solved or PARTLY-solvedby the Three-signed
arithmetic. Such worked examples would helpfocus 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 beexplained by the
hump in the Eucalculus curve.That curve is itself no more
than the reciprocal of Euler's Gammafunction - so it is an
extension of mainstream maths.In this present case, we seem
to be leaving the customary de?itionof + 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
thefundaments to be found - with no discussion. This would
help in anystudy 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
datagoing 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 in?ity it would return.The COMPLEX
mathematics states that it shifts SIDEWAYS in complexspace,
and so leaves the real world. Multiplication by -i1 would
bringit back.NEITHER is a total answer to the problem of
vanishing data. Perhapsthere is the possibility of a new
Three-signed arithmetic emergingto tackle such
problems.However, at a ?st perusal I cannot quite see where
this is heading.It would be useful to have real-world
problems solved or PARTLY-solvedby the Three-signed
arithmetic. Such worked examples would helpfocus 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 beexplained by the
hump in the Eucalculus curve.That curve is itself no more
than the reciprocal of Euler's Gammafunction - so it is an
extension of mainstream maths.In this present case, we seem
to be leaving the customary de?itionof + 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
thefundaments to be found - with no discussion. This would
help in anystudy 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 in?ity 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 ?st perusal I cannot
quite see where this is heading.This will hopefully
eventually replace space-time of classicalphysics.In a
physical system the concept of differentiating out a
constantdoesn't seem to be a problem. If you look at the
velocity of a car itdoesn't matter where the car started
travelling, if the resultingvelocity is the same for multiple
cases in different positions, thenthe velocities should match
in the math. In this way the constantcould be seen as
ethically positive and legititmates comparing thevelocity of
cars at differing positions.I think more interesting is that
there does seem to exist a physicalfundamental from which the
calculus operates, in this example it is aspatial position,
which has no further derivative according toclassical
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 someonewill come across this and ?d an
application. I've discovered thatthe Y-space covers the plane
and can be graphed on a piece of paper. Aunit circle takes on
foreign proportions and yet is still the unitcircle. I guess
I can't demonstrate things like unit circles without
awebsite.> 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 de?ition> 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 justi?d in this example until the
fundamentals arein place.This example accepts that a
magnitude can be cast into T-space and canbe brought back out
with the use of signed unit values in T.Challenges arise when
summation is incorporated into this. I've takena stab at that
type of product for Y space, which you will ?d inthis 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 guessI just wanted
somebody to shoot this down and it hasn't happened yet.>
Charles Douglas Wehnersome 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.>
-TimI had a few thoughts on the subject. Have you heard of
the QUATERNIONSof 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 ?d thatsomebody got there ?st.
The effect on my mind is not SHAME, butquite the REVERSE. I
?d that the great men of science thought alongthe same lines
as I do - so on those occasions when I am second, Ifeel
FLATTERED.I have ALSO found things that nobody ever found
before me. That isbecause ones con?ence 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 whathappens.Quaternions appear to contain
lots of real numbers and so cannot beequivalent.Have you ever
tried graphing in a plane like this? + plus pole + + + + + + .
. . . . . . p1 = - 7 + 4 + . + . . + . . 0 - - - - - - - - - -
minus pole . * origin . p3 . * . . * . * . * . * . . . . . . .
. . p2 = - 9 * 6 * * * * star poleThis is Y space as a plane.
It is simpler than both cartesian 2Dand complex values in
that it has just the three way branch instead ofa four way
branch. As you can see parallelograms at angle 2pi/3resolve
the entire plane symmetrically. A reduced Y value always
hasat most a pair of magnitudes. Every position in the plane
can beresolved.The dimensionality of Y is quite a
conundrum. Becauseinformationally there are two magnitudes
informationally it is twodimensional, 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 athree-signed
system we should require an equal magnitude for each poleto
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 willbe convinced. The
trouble is that we all think in context of realnumbers.Much
of this thinking does not extend when more signs are used in
theconstruction.I started this all in the context of the
question Why R X R X R Xt?.This is the spacetime of
classical physics which even string theoristsare not
destroying. I consider Y X Y to be a competitor to R X R X R
Xt.But also if you add yet another sign ( four signs ) then
the extensionwould 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 RX R, especially if the
cancellation effect is looked upon asaccumulation, which then
provides a basis for time. I think it is wiseto continue
opening up the three-signed can of worms before moving onto
the four-signed can. It is my hope that dynamics will be
found in Ywhich 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 ?d that> somebody got there ?st. The effect on my
mind is not SHAME, but> quite the REVERSE. I ?d 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 con?ence 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 ?st question of data going missing and being
recovered camefrom this statement of mine:> The
TRANSFINITE mathematics states that it becomes vanishingly
small,> but if it could be scaled up by a special in?ity it
would return.> Here, we can draw a NUMBER LINE
-3,-2,-1,0,1,2,3and put the factorials above the positive
part:1,1,2,60,1,2,3To ?d 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 ?d the factorial of 1
from 2!, we divide by 2.We can also go back up.To ?d 0! from
1! we divide by 1, and can go back.To ?d (-1)! - note the
brackets - we divide by 0.ANYTHING divided by zero is said to
be INFINITE.Zero times in?ite is said to be INDETERMINATE.So
we cannot go back up!!!!!However, at the console we de?ed
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-in?ity by it.So we can go back up.(-2)! is -(console
in?ity)(-3)! is +(1/2)(console-in?ity)&c.And so, by strict
de?ition of the nature of the zeroes andof the in?ities, we
do not have to lose information byde?ing it as
indeterminate.This is one of my discoveries in the as-yet
unpublishedfantasy maths - where fantasy means anything
containingeasy-to-use numbers (FUN NUMBERS) and the FOLLIES
zero andin?ity.Another name is the TRANSFINITE MATHS.However,
as I made my discoveries by one route it was pointedout to me
by an eminent mathematician that Georg Cantor hadfound
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, whengiven
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.0.999999999999999999999999999999999Think of it being
applied as the cosine passed through zero atthe point Pi/2But
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 tore-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 cosinebeing part of a PAIR. The cosine is
REAL, and is the SIGNAL -but there is a hidden sine
accompanying it. This we call thePHASE.The PHASE is hidden
because it exists only in the IMAGINARYworld. However, it
de?es the SLOPE of the original cosine.Thus, if the cosine
began at Y=1, and the units of the X-axisare RADIANS, the
slope will be a downward slope of 1 as thecosine goes through
zero. By means of its phase, a cosine can be allowed to pass
throughzero 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-tOmega 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
de?es the slope.Such complications are necessary because the
simple mathematics will not always de?e what we want.In the
trans?ite case, we have no need for theimaginary numbers. In
the complex maths we have noneed for the trans?ite.At least
in theory.Perhaps there are problems that can only be
solvedby BOTH.Then we have the mad maths of Boole. 1+1=1At
?st 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 andno Internet for me to write this on.The
Quaternions are sometimes used in elementary-I cannot ACCEPT
a new mathematical system withoutseeing its bene?s.I cannot
REJECT a new mathematical system forfear of rejecting
something good.If I had time to delve and delve, I might
knowwhether the THREE-SIGNED ARITHMETIC is usefulor not.But
it is not mine, and I am busy. I will, however,if there is an
adequate summary, look back now andthen 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 fewexhaustive examples from
scratch? My guess is that it is probably isomorphicto
something we're already familiar with, in which case there
will be lotsof useful things you could draw on that already
are known to understand howT 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 de?ed as a
magnitude with twosigns( -, + ), T is a magnitude with three
signs( -, +, * ).Examples of elements in T are: - 1.23 +
2.134 * 6.54Now 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 ?d 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 newspace Y because Sum( t1, t2 ) is not
generally in T. So even thoughthe subject in this thread is T
space it is only a starting step toget 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 ofthree-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 productsfollow 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 ?ld where
a new psychology of mathematicsis needed.The graphics are
elaborate, convoluted, impenetrable. Given thegraphics, we
cannot tell how they are made. Given the method, we
aresurprised 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 ?d
the ORDER within the CHAOS. Charles Douglas Wehner
===
I'm
looking at timesheets scheduling and optimisation task: a
shopmanager would like to specify a number of employees
working in anygiven time interval, and employees would like
to stay within theirpreferable time constraints.I think such
optimisation task should be quite common, but I justdon't
know the category of algorithms it falls to, or at least
where Ican 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 integerlinear programming. The general
?ld in which this is done is operations research, and you
might try the sci.op-research newsgroup.Robert Israel
israel@math.ubc.caDepartment 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 ?ld is Operations Research.It may also be of
interest that some complex examples of timetablingproblems
are solved with Evolutionary or Genetic Algorithms, and
therehas 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 asyour 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 ?ld 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
currenttitle in the late January part of Category theory vs.
Set theory questions
http://mathforum.org/discuss/sci.math/a/t/311474there:[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
havemy own additional comments to add. In [3] I introduced a
hierarchy of fragments of Z = Zermelo'stheory (separation, no
replacement), namely the hierarchy havingsigma-m separation as
a fragment of full separation, (as m varies inthe hierarchy),
this hierarchy below a hierachy under ZF havingcorresponding
sigma-m fragments of replacement. I noted in [3] a Lowenhiem
Skolem argument to get the replacementpart of the hierachy
had strictly increasing increasing strength atevery 4th
level. In [3] I noted I did not know how consistencystrength
increased in the separation part of the hierarchy. I think we
can argue consistency strength goes up in the separtionpart,
ie sigma-m+4 separation |- Con(sigma-m separtion) for atail
of m's anyway. Namely, arguing in the sigma-m+1 separation
theory, suppose there isa proof of a contradiction in the
sigma-m Z fragment. Then we canmake this a proof of
contradiction from a ?ite fragment of theseaxioms over pure
?st order logic. But then by Gentzen cutelimination for pure
?st order logic, we can also ?d a proof whereall formulas
are substitution instances of subformulas of premises, iethe
?ite fragment of the theory. So we could ?d a proof of a
contradiction where all formulas usedin 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 atail of m's) to cover
the complexity of all the Z axioms other thanseparation, then
sigma-m separtion itself is I think sigma-m+2 iethe
parameters can be free variables, one quanti?r to say
theseparating set exists, and another quanti?r inside to say
all membersare in that set according to the condition, then
the considtion itselfwhich is sigma-m. So we have a proof of
contradiction using only sigma-m+2 formulas.Now in our
working theory we de?e sigma-m+2 truth for the universe.This
is a sigma-m+2 de?ition. Then we run an induction argument
onthe proof, using that truth de?ition, that all formulas in
the proofare true on all interpretations of variables. I think
one morequanti?r to handle all interpretations and then one
more to codethe induction. I am sliding over details here so
maybe m+4 as a claimed above andseem to be getting with this
sketch isn't exactly right, but I thinkthis is enough to show
some ?ite step up is enough, ie what I saidwas 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 ZFmodel, and make it into a topos, how to recover from the
topos(working model theoretically above the topos) an
isomorphic copy ofthe original ZF model: namely suitable
equivalence classes ofwell-founded trees representing the
epsilon relation on transitiveclosures. That construction in
turn could make sense in any arbitraryelementary topos,
producing not necessarily a full ZF model but amodel of a ZF
fragment, as I discussed in [2]. I suggested one
reformulation of full ZF into topos theory would beto
axiomatize in topos language that the result of that
constructionsatis?s 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
andonly bounded separation, and the other usual axioms of ZF.
Or maybeZFC, parallel discussions with or without AC. Let that
theory be S. Given any model M of S, (this including the case
of M modelingextensions of T to stronger fragments of ZFC),
we can form thecorresponding topos T(M). Inside this topos,
we can do theconstruction 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 toposfrom the S theory: T(S(T(M))). Then
this ToSoT(M) is isomorphic to T(M), by a de?ableisomorphism
pushing everything through the de?itions. The statement that
this de?ition is an isomorphism is a statementin topos
language in the language of T(M). So if we have some set
theory S' extending base set theory S, andwe want a
corresponding topos style theory to correspond to S', ie
wewant a theory to correspond to all T(M) ?s for M an S'
model, itwould be reasonable that such a topos style theory
should include thestatement that ToS(my underlying universe)
is isomorphic as statedabove to my underlying universe, since
every T(M) is like this. So suppose S' is some set theory
extending S, and suppose T' isa candidate extension of
elementary topos theory, which is supposed tocorrespond to
S', and in particular has that last property I justsuggested
is reasonable to expect of S'. Given C' a model of \
S' (I
would want to call this T' since it is atopos, but I have
been using T to name the construction betweentheories, so I
use base C to suggest category), so given C' a modelof \
S',
we can form the set theory model S(C'), then form the
toposToS(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
de?ableisomorphism 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') satis?s allof S', ie since \
each of
these are isomorphic to the correspondingS(C') and the
assumption on T' as being a topos theoretic codingof \
S' is
all S(C') model S' for C' modeling \
T'. Any model M' of S' can
induce a corresponding topos T(M'), andthis is topos having
S(T(M')) modeling S', ie such S(T(M')) \
areisomorphic to M'
and M' is assumed to model S'. If T', being \
a topos theory
reformulation of S', is supposed to besatis?d 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 modelof
T'. Two paragraphs ago, we had over T' models \
C' the 4 level
tower reachingup to SoToS(C') has isomorphism \
SoToS(C') to
S(C'), and by thelast paragrach, any S' model \
M' can be
arranged to be isomorphic tothe S(C') level of that tower, ie
take C' = T(M') obtaining an Smodel, and then the 2nd \
level of
the tower there SoT(M') isisomorphic 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)
isisomorphic by that de?ition 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
concludethat S' the underlying set theory proves something
about the zigzigtowers contructed over it. Now the clincher.
In the topos style theories we can make acorresponding notion
of sigma-m and pi-m, namely complexity onalternating blocks of
quanti?rs over objects and arrows. Suppose, T' being a topos
theoretic reformulation of set theory,proves that S(my
universe) satisi?s 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 zigzagsback to
isomorphism, ie SoT(my universe) isomorphic my universe.So we
put this single sentence as an new axiom into
thereaxiomatization, ok to add as axiom since it is a theorem
of thetarget theory. Then we de?e in the S' language the T(my
universe) construction,and we axiomatize that this constructed
category satis?s T'. So, by the assumed property of \
T', that
it proves S(my universe)satsi?s S', in S' we get ,
interpreting that last proof overthe T' axioms coded into our
new reaxiomatization of S',we get in our S' \
re-axiomatiztion
that SoT(my universe) satis?s S'. But S' had my \
universe
isomorphic to SoT(my universe). So our theory proves my
universe satsi?s S'. So we really have reaxiomatized \
S' this
way. Suppose the T' theory was all axiomatized at level
sigma-m, ie forthe compelixity notion of quantifying over
objects and arrows. Then, when we copy that axiomatization
into S' style, the objects andarrows get de?ed as sets. So
this copying from T' axioms into S'language for our
reaxiomatization carries sigma-m axioms (topostheory) to
sigma-m axioms (set theory) in the reaxiomatization ofS'. We
also need to have the sentence in the S' reaxiomatization
sayingthe tower over S' zigzaged to isomorphism, that is one
sentence havingsome speci? ?ite bound, call it b. Then our
reaxiomatization of S' has complexity of axioms at
mostsigma-max(m,b). Now suppose our starting S' was up to
level at least sigma-n inthe separation hierachy from [3].
(This includes all levels of thereplacement hierachy from
[2], which starts at the top of theseparation hierarchy.) So
the sigma-n S' theories has been reaxiomatized as a theory
withaxioms of complexity at most sigma-max(m,b). But if n >=
max(m,b) + 2 or whatever from above, S' can provethe
consistency of the reaxiomatization, along the lines of
theargument I sketched above over smaller separation. (The
previousargument with +4 : 2 was from the separation axiom,
and 2 to processit, now use 2 for processing). But the
reaxiomatization recovers the original S', so S' \
would
proveits own consistency. We conclude max(m,b) > n - 2. What
this says, is beyond the b base cases, you need almost as
manyalternating quanti?rs in a topos style theory recoding
of set theoryas the original set theory had. Zermelo's theory
Z is level sigma-omega, ie it includes axioms ofeach level
sigma-m. So a topos theory version of Zermelo would need
unboundedalternations of quanti?r. Usual elementary toposes
are Caretsian closed categories withsubobject classi?r, and
then the other stuff as [2]-[3] to jazzthem up more toward
set theory: AC, natural numbers object. All these things are
phrased in terms of an existential quanti?rfor the new
object (if the theory adds a symbol then by thecompleteness
theorem there should be a corresponding theory in thebase
language with corresponding existential quanti?r, at least
for?itely axiomatized theories which I think the axioms of
topos theorycan be arramged to be). Anyway, one outer
existential quanti?r is at most complexity upsigma-m
hierarchy by 1. Then in those base symbols, we have to assert
various functors haveadjunctions. This can be expressed by all
?ite diagrams of acertain form have an appropriate universal
arrow. Universal quanti?rson the vertices of the diagram.
Then we have to say there is a uniquearrow to ?l in the
diagram to make something happen. Saying uniquearrow is a
couple more quanti?rs, a conjunct of an existential andoff
to the side two nested universals expressing uniqueness. The
property inside the universal property: sometimes it is
justthat the diagram commutes. That requires no quanti?rs.
Forde?ing subobject classi?r there is a de?ition that
there is aunique arrow to complete a square making the result
a pull-back. Sothose outer quanti?rs as above handle unique
existence, and thepull-back property is itself another
universal property, which willitself will be unique existence
over a diagram commuting. So unraveelling it all, you only get
a ?ite nesting aquanti?rs. Well not surprsing since it is to
be a sentence. Thepoint here is that the intuive de?itions
given in mathematicalEnglish do not unravel into a schematum
over growing sigma-mcomplexity, everything in the intuitive
de?ition just expands out toone or two nestings of
quanti?r. So the point is, cartesian closed category with
subobjectclassi?r, and whatever ?titely more stuuf you add:
naturalnumbers object: yeah some objects and some adjunctions
and someunique existences, no way are we going to build up to
in?itesigma-m complexity, as we need to get back to Zermelo
theory. Think of the ordinary de?tions in category theory.
It is alwaysthat a handful of things have adjunctions,
basically just iteratingover and over again that diagrams
have unique arrows completing themto a previously de?ed
property. Well, from [2]-[3] we already had that usual
elementary toposes donot recapture Zermelo or ZF. But this is
more striking, that alltheories in the style of category
theory don't. In [2] I suggested a try at a topos style
version of ZF. Namelyto axiomatize that S(my universe) |= ZF.
So that theory does getsigma-omega, because the separation and
replacement axioms saying thatbecome unbounded sigma-m accross
the schematum, this sigma-m in thesense of topos theory as I
de?ed above. As I say though, this is a different style that
usually done incategory theory. So the reference I found. The
Kock and Reyes paper in the Handbookof Mathematical Logic, ed
Jon Barwise. Their section 7 is above thissort 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 tobe
something like my S(C) construction to make a set theory
modelfrom a topos. They say there that Osius de?es a theory
ZO, which is a settheory corresponding exactly in this sense
to ordinary elementarytoposes. They don't de?e ZO, they
refer to Osius. But they say ZOis intermediate between
Zermelo-Thiele and ZF. They don't de?eZermelo-Thiele. I
couldn't ?d that on the web, but I did ?d
mathematicianThiele, who did set theory and is from around
the time of Zermelo. When I ?st read that, I assumed
Zermelo-Thiele and ZO werede?ed in terms of limited
complexity separation etc, as my ownchecking seems to show
comes up here. But with Thiele being old time, I wonder if
Zermelo-Thiele is justZermelo theory? But back then, Zermelo
originally phrased separationas using a de?ite property to
separate out. I recall it was maybe1918 Skolem and 1922
Fraenkel sometime in there that this was redoneas a schamatum
in ?st order logic. And the original Zermelo was Ithink 1908.
So with everything so vague as those original versions,
theywouldn't have had anything like sigma-n levels. But if
Zermel-Thiele = Zermelo's Z, and ZO is intermediatebetweeen
Zermelo-Thiele and ZF, and ZO corresponds to usual
toposes,how come we have ZO including all the sigma-omega
complexity of Zcorresponding to toposes which supposedly by
the above have boundedcomplexity?-- David Libert
ah170@FreeNet.Carleton.CA