mm-1869
===
why did noone comment on HECKMAN 1st reply to me about HECKLING ME
right after the name by nature claim??
read it!   its a million to one shot right on cue.
   it is trivial that if a sequnce of functions in $H^{infty}$ on the
unit circle converges then it also converges as a sequence of functions
in $H_p$ on the unit circle for all $1leq p <infty$. are there
sufficient conditions on the sequence for the converse (i.e. a sequence
of functions converging in $H^p$ will also converge in $H^{infty}$) to
be true, especially for the case where $p=2$?
What's the point to all those dollar signs? You really think that
$p=2$ is somehow more clear than p=2?
Anyway, it's hard to imagine such a condition that's weak
enough to give an interesting result. The best I can do is
g = f.
If you have some specific condition in mind someone might
be able to tell you whether it's sufficient.
************************
David C. Ullrich
   posting-account=84XyagwAAAAn211ho6L11j1JK1UhtY3o
NOT
I don't think Euclid is assuming not q. The absurdity that he gets
is that a prime measure a unit, not that he finds a prime different
from A, B, C. Euclid is in a subsection of the proof where he is
trying to show that he has a new prime not equal to A, B, C.
[cut]
Let me simplify the subsection of the proof that we are now talking
about. I will change the theorem to a simpler version: namely, trying
to show that A does not measure A*B*C + 1.
I try to stay as close to Euclid's wording as possible.
Do you say the following proof is a proof
by contradiction?
I say the following is a proof by contrapositive.
Theorem.
   Let A, B, and C be some prime numbers.
   Let DE be the least number measured by A, B, and C.
   Let EF be the result of adding the unit DF to DE.
   I say that A does not measure EF.
Proof:
I say that A does not measure EF.
If possible, let it be so.
Now A measures DE. But it also measures EF.
Therefore A, being a number, measures the remainder, the unit
DF, which is absurd.
Therefore, A does not measure EF.
Q.E.D. 
The implications in the proof proceed as follows:
Now, taking the contrapositives, and noting
that A does not measure a unit is true, I can
conclude that A does not measure EF.
I say this is not a proof by contradiction.
Do you agree?
-- Bill Hale
   posting-account=dd5WWA0AAACAXyu9_CIdlUeToqGNlLRy
I will point out though that my claim was that the proof AS A WHOLE
is by contradiction, not contraposition; I wasn't talking about
individual sections. Just to recapitulate my claim, the proof is
of the form p, trying to prove q, assuming q is false, and deriving
a contradiction. I claim that any contradiction derived after 'q is
what subsection it is in.
Yes. I would say that it is a proof by contradiction.
p is 'Let A, B, and C be some prime numbers.
      Let DE be the least number measured by A, B, and C.
      Let EF be the result of adding the unit DF to DE.'
q is 'A does not measure EF'
Assume not-q. Then A measures DE. But it also measures EF. Therefore
A, being a number, measures the reaminder, the unit DF, which is
absurd.
Therefore, A does not measure EF. QED.
That is, p and not-q produces an absurdity. Isn't that what you called
contradiction? A contrapositive would occur if not q implied not p, not
that not q resulted in an absurdity.
I don't follow this last part. The proof as I understand Euclid to
intend it doesn't operate by taking a sequence of contrapostives as
present it. If you can remake the proof so that it involves these
contrapositives, that is not the same thing as discovering that his
proof actually operates by these contrapositives. Also, I'm not
sure I even accept the sequence of contrapositives. For example,
This is an interesting subtlety. To be honest, I was unfamiliar with
proof by contraposition. I always encountered proof by contradiction,
in the kind of general sense, I have a mathematical situation, and
given
that situation, p, want to prove q. So I assume not-q and derive a
contradiction. That and construction are mathematical proof as I know
it.
Maybe I'm simply forcing this into what I know, but it seems to be
it really is p (as I stated in both the overall case, the theorem as
a whole, and in the particular subsection you present above) and not-q
derive an absurdity, which is proof by contradiction as I know it.
java yet.
Ken
   posting-account=84XyagwAAAAn211ho6L11j1JK1UhtY3o
trying
I understood that. I'm concentrating on the subsection since it
leaves out a lot of extraneous distractions. I'm considering the
subsection to be the whole theorem for discussion at this
point. I see below that you say that the proof even in this case
is still a proof by contradiction. Thus, I think it is worthwhile
to pursue that special case, since I am saying that it is a
proof by contrapositive. In particular, I eliminated the
whole question of whether A,B, C are the only primes
since the modified theorem makes no such claim that
they are.
Thus, I wish to pursue the discussion only on the
modified theorem presented below.
called
not
Yes, this is a problem that needs to be addressed. I'm claiming that
Euclid presentation is just a stylized presentation of a proof
by contrapositive. I need to justify this more. But, I don't
think that we are at that point yet since you are still
saying that my explanation of his proof is just a proof
by contradiction (as you say below). I wish to get to
a point where we both agree that we have a proof
by contrapositive and not a proof by contradiction.
Then, we can discuss whether I just presented a different
proof than what Euclid presented.
DF is a particular unit. I'm not sure if that is what you are asking.
More precisely, you assume both p is true and not-q is true.
In particular, you cannot just take for granted that p is assumed true
for the proof. This is part of the stylized method of proving
the statement p implies q by contrapositive: assume not-q
is true, make some additional statements, finally conclude not-p
is true, qed. Technically, you need to say in the proof whether
or not you are assuming the hypothesis p: it is not automatically
taken for granted. Thus, my stylized method of proving
the statement p impies q directly is: Proof Assume p is true,
make some additional statements, finally conclude q is true, qed.
not-q
OK, let's continue on my modified theorem and see if I have a
proof by contrapositive rather than a proof by contradiction.
Before I do that, I would like to first point out some problems
that I am having with what I am saying.
First, as I have said before, I really prefer to have a theorem
in the form if p then q format, rather than in the form q
is true when I am talking about proof by contradiction.
We have the latter form: A does not measure EF. That is,
it is not stated in an implication format. I will correct
this problem below, although you might complain
that I am again rewording Euclid too much from
Second, the usual complaint against doing a proof by contradiction
is that you have no goal to seek except some contradiction among
numerous possible ones, whereas a direct proof of if p then q
has the goal of establishing q given p, and a contrapositive proof
has the goal of establishing not-p given not-q. In the proof of
the modified theorem I gave above, I worried about not having
a goal to achieve. I reworded the theorem so that a goal would
emerge. Again, you might complain that I am doing too much
rewriting. I would need to justify that I am only making cosmetic
changes. Briefly, I am just trying to make explicit what is going
on in the background for the proof by contrapositive so that
it is in the more standard style.
Here is my redoing of the theorem and its proof.
 I say the following is a proof by contrapositive.
 Theorem.
    Let A, B, and C be some numbers.
    Let DE be the least number measured by A, B, and C.
    Let EF be the result of adding the unit DF to DE.
    If A, B, and C are primes, then A does not measure EF.
 Proof:
 Suppose A  measures EF.
 Now A measures DE. But it also measures EF.
 Therefore A, being a number, measures the remainder, the unit
 DF.
Thus, A is not a prime (because it is a unit).
Hence, it is false that A, B, and C are primes.
 Q.E.D. 
Note that the contrapositive of the last statement in the theorem
is If A measures EF, then it is false that A, B, and C are primes.
Thus, I am proving the contrapositive. I assume the negative
of my conclusion (A measures EF). My goal is to prove the
negative of my hypothesis: A is not prime or B is not prime
or C is not prime. Note that I am not assuming A, B, and C
are primes in the proof even though they are given in
the hypothesis of the statement of the theorem.
Finally,  I say this is not a proof by contradiction.
Do you agree?
If so, then we can discuss whether I have modified
the proof of the original theorem beyond what Euclid
-- Bill Hale
   posting-account=dd5WWA0AAACAXyu9_CIdlUeToqGNlLRy
Hold all your thoughts! I can't respond just now but I'll try
to respond tom'w. You have raised further refinements that need
addressing I think is clear.
THanks.
Ken
Overflowing *sceptic* tank?  That brings up even more unpleasant
images than I'm sure you intended...
Yes with feasting magotts in the pool of water and notice to the a
of couple dozen month long guests, not to use much water.  The
exploitations incured upon the paying guests was why I left.
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
You were not writing about a septic tank.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
I was so.  But whoops, sceptic tanks are important coolants for a feverish
fervent society.
I don't really understand these sets. Can you clarify?
Also are these sets compact?
You need to use 'fixed width font' to see the diagrams.
Well it's your homework problem so it's up to you to
clarify the details.
Of course they are.
 to be released
http://stardrive.org
If you read the Arcos & Pereira paper, how do you understand the 
authors' position regarding the extraction of a localizable tensor 
vacuum stress-energy density
from the net gravitational-inertial stress-energy of the Einstein field 
in (1) their own teleparallel formalism; and (2) in the conventional 
curved-manifold formalism?
Jack, this question was specifically addressed to Art.
J: Yes, I know.
What A&P do, very nicely is to show the B field!
  Z: In the context of a *teleparallel* theory.
  J: They do not have my Vacuum ODLRO SBS idea, i.e. they do not write
  B = (Lp/2pi)dargVacuumODLRO
  argVacuumODLRO = Goldstone Phase of Higgs Field (post-inflation).
Z:  Right. Of course not.
J:  If B is closed non-exact 1-form that's A&P torsion flux without 
flux.
Z: So you are saying that your tetrad B can be modeled as a torsion 
field?
J: Yes, it's the potential for the A&P torsion field not the Shipov 
torsion field. B is like A in EM, F = dB is like the EM field tensor, 
but here it is the A&P tetrad substratum torsion field tensor.
As R. Kiehn points out the most general p-form has 3 possible 
independent components
form = exact + closed not-exact + non-closed
If B has a non-closed piece I call that flux with flux i.e.
F = dB
If B is only closed and not-exact
Stokes theorem says |B = 0 on a bounding cycle but need not be zero on a 
non-bounding cycle that is only a piece of the total boundary that 
consists of several cycles when the manifold of integration of dB is not 
simply-connected.
If, in addition, there is macro-quantum vacuum ODLRO (a new independent 
physical idea)
B = (Lp/2pi)dtheta in sense described in John Baez's book on Gauge 
Fields, Knots & Gravity, then the period integral of 1-form over the 
boundary fragment, which is nevertheless a closed loop (nonbounding 
1-cycle) is, from single-valuedness of the ODLRO order parameter
|B = NLp
N is integer winding number
This leads me to define another Ghost of a departed quantity (Bishop 
Berkeley)
NLp = (Area)dB'
Mean value theorem for integrals.
I put the ' to denote flux without flux. The Area is inside the 
non-bounding cycle and all the other inner non-bounding cycles isolating 
phase singularities where |ODLRO| = 0 (like vortex cores in Type II 
superconductor of real EM flux with flux) shrink to zero (I mean small). 
Actually HeII quantized circulations may be flux without flux in this 
sense.
Thus in cylindrical coordinates for |ODLRO| = 0 along z-axis
dB' = NLp/pir^2
Even though technically dB = 0.
NLp is like a finite integral of a zero integrand only because we have 
ignored the inner cycles that add up to -NLp of course. But as 
physicists we can do that since we only actually measure the outer 
non-bounding cycle in the case of quantized vortices in superfluid HeII.
So this is an archetype analogy for the complete nonlocality of gravity 
energy in ordinary vacuum /zpf = 0 i.e. gravity vacuum energy without 
gravity vacuum energy.
Integrating a local zero to get a finite total is not so crazy when 
the manifolds of integration are not simply-connected and the holes 
correspond to phase singularities, i.e. zeros in the ODLRO intensity.
The situation is like that of infinitesimals in calculus that are both 
zero and not zero in the heuristic sense. As physicists we don't care 
about what Feynman called rigor mortis. For example the Feynman path 
integrals are mathematically non-rigorous yet we get good results from 
them.
  If B is not closed that's torsion flux with flux
  i.e.
  F = dB  is the A&P torsion field (not same as Shipov's).
  I assume here G/H = S1 like Abelian Maxwell E&M, this is only a toy model
  dF = 0
  d*F = *J
  The local stress-energy tensor of this torsion field is
  ~ F/*F  i.e. a 4-form
  But NOTE if B is closed and non-exact, then strictly speaking LOCALLY
  F = dB = 0
Yet, from flux without flux the global integral is not zero even 
though the local density is!
  WHERE HAVE WE SEEN THAT BEFORE? :-)
Z: OK, I think I get it.
J: Yes, this is an interesting strategy. Penrose gives a good heuristic 
why total gravity ordinary vacuum energy really must be nonlocal so that 
Yilmaz's theory is asking the wrong question and so is Alex's. 
Valentini's new papers on the creative tension between locality & 
nonlocality add fuel to that fire. There are other arguments as well.
Flux without flux means
|B = ||dB = NLp,  N an integer
Strictly of course |B = 0 over a boundary, but here we take the 
incomplete integral over a non-bounding cycle and ignore the inner cycle 
that gives -NLp (Stoke's theorem) that shrinks to the topological defect 
Goldstone phase singularity where |Vacuum ODLRO| = 0 in the G/H order 
parameter space. Note, unlike micro-quantum theory, in macro-quantum 
theory the radius of S1 has independent physical meaning. Also see A. 
Valentini's latest papers on sub-quantum non-equilibrium signal 
nonlocality.  Let P be the non-equilibrium distribution, the non-local 
signal is an integral with kernel P - |psi|^2, in macro-quantum theory, 
I think this becomes ~ |ODLRO|^2, which of course is zero in nonlocal 
micro-quantum theory with presponse signal locality. That is, any 
system with ODLRO (in vacuum for virtual quanta E^2 =/= (pc)^2 + 
(mc^2)^2, in ground states  E^2 = (pc)^2 + (mc^2)^2 for real quanta) is 
sub-quantal non-equilibrium. Call this the Sarfatti Conjecture since 
our conscious minds are pumped open giant quantum robust ODLRO systems 
immune to warm wet environmental decoherence of the Zurek-Tegmark kind.
Z: OK, looks good. Very interesting idea.
J: Valentini has a formula in one of his new papers
Nonlocal signal ~ Integral of stuff [P(sub-quantum non-equil) - 
P(sub-quantal equil)]
I think I can argue as a sufficient, though maybe not necessary, 
condition for signal nonlocality in macro-quantum theory
|OLDRO|^2 = [P(sub-quantum non-equil) - P(sub-quantal equil)]
I get this from theory of reduced density matrices as given by L. 
Onsager & O. Penrose.
matrix in diagonal limit = Born probability density! (pair density 
matrix in case of fermions)
The macroscopic eigenvalue of first-order reduced density matrix is 
sub-quantal non-equilibrium in A. Valentini's sense and it is immune 
to warm wet decoherence. This is WHY Dick Bierman sees presponse in 
his data.
Z: I'm also curious about how you understand A&P's claim that that the 
teleparallel formalism is physically equivalent to the curved manifold 
formalism?
Again, this was addressed to Wagner.
In answer to R.Kiehn today
Clarification of notation
When I write
g(curved) =[(I + B)^T][(flat)]([I + B)]
That is just my lazy notation. I do not mean that as an equation in 
Cartan forms. All I mean by that is the well known
guv = eu^a(Minkowski)abev^b
where
eu^a = Iu^a + Bu^a
Iu^a is Kronecker Delta
B = dx^uBu^a(Pa(Goldstone Phase)/ih)
{Pa} is Lie Algebra of T4
  It means simply one version of EEP
  g(curved) = (I + B)(flat)(I + B)
Z: So you think that by physical equivalence A&P simply mean that they 
recover standard EEP? You think they mean equivalence as in 
equivalence principle? In other words -- pun intended?
J: Pun intended sure. What they mean is simply that starting from the 
zero curvature torsion tetrad substratum, one can use the formal EEP to 
derive standard vanilla 1916 GR curved torsionless model.
There is a kind of curvature/torsion duality here between the local 
gauge force and geometrodynamical pictures.
The torsion flat gauge force picture is the square root of the curved 
torsionless geometrodynamic picture.
So, the torsion picture is a covering theory of the geometrodynamics. It 
is closed to the Ashtekar picture!
That is, in the torsion tetrad substratum it is obvious how gravity is 
In 1905 special relativity
These are GLOBAL displacements in Minkowski space-time making up the 
RIGID group T4 (with x & y).
It is completely trivially obvious that Diff(4) GCT's
at a fixed PHYSICAL EVENT P = equivalence class of manifold points p ~ 
p', p =/= p'
is by definition the local gauging of T4 to Diff(4) with compensating 
gauge force potential Bu^a where
guv(curved) = eu^a(Minkowski)abev^b
Z: I think they mean something much broader -- more like matrix 
mechanics vs. wave mechanics.
J: Frankly I don't care what they mean. What I care about here is what I 
mean. However, I suspect that what I mean and they mean are not that far 
apart.
  This is exactly how curved space-time is Vacuum ODLRO emergent bottom 
  Back-of-the-envelope heuristics
  The quantized Goldstone vibrations in argVacuum ODLRO are not spin 2 
gravitons.
  OK.
  Are they like phonons?
More like phonons yes. They are not spin 2 at least not when G/H = S1 where
Vacuum ODLRO = PSI = Re^iS (Bohm Pilot Wave notation)
If Bu^a = X^a(PuS/ih) = X^aS,u
X^a is position operator conjugate to Pa in Minkowski space
i.e. X^a makes displacement in momentum Minkowksi space
Z = (X + Lp^2P/ih) & Z* =(X - Lp^2P/ih)
Give Wigner Phase Space Density Representation of Glauber & Squeezed 
macro-quantum coherent states
z = re^itheta
Use the formalism of quantum optics
D(z) = e^(zA* - z*A) Glauber coherent states
S(w) = e^(wA*A* - w*AA) Bogoliubov pairing transformation for squeezed 
states.
w vacuum ODLRO squeezing parameter ~ BCS Nambu/Jona-Lasino gap function, 
i.e. rest masses of lepto-quarks?
  No gravitons. No quantum foam. No quantum gravity.
Z:  Well, you do have a true QFT layer beneath your BEC 
macro-condensate, don't you? That describes the micro-behavior of 
off-mass-shell electrons and positrons?
Yes, it's their pairing that is the chaotic inflation phase 
transition. The pairing binding energy powers the hot phase of the Big 
Bang. It's all pretty obvious intuitively. It fits Lenny Susskind's 
landscape with WEP and also Max Tegmark's Level I & Level II (see his 
homepage for pretty pictures)
Who ordered that?
What about Duff's argument for quantum gravity?
What happens when you reach rock bottom?
Z: Yes, of course this is the Sakharov emergent-gravity thesis. Same in 
Hu's papers, for example.
J: Same in general philosophy not in detail. This stuff is in the air. 
Mine is the simplest of em all. Are all the rest over-complicated or is 
my model simpler than is possible like Puthoff's PV and Haisch's EM SED 
ZPE origin of inertia.
Is a theory of everything for everyone impossible? Do you have to be a 
fancy mathematician to grok it? Calabi-Yau and all that really needed? 
Too soon to tell.
You return to the false globally flat massless conformal vacuum of 
spin1/2 & spin 1 QFT with no gravity and no inertia, i.e. all rest m = 0
Z: But is it kinematically false?
J: I do not know what that even means. The center of the cyclone where 
|ODLRO| = R = 0 thats FALSE VACUUM. So it's physical as real as anything 
is.
Z: Or just unstable if construed as a physical vacuum?
J: It's real and unstable. It's as real as riding a unicycle.
http://stardrive.org/cartoon/Saturn.html
Z: I still sense Einsteinian ambiguity and even confusion here.
J: The fault dear Z is in your mind and not beyond the stars.
Z:  What did Sakharov say about this emergence?
J: Not much. Like Bohr he mumbled some things about metric elasticity 
and if you read Adler Rev Mod Phys 1982 you will see more very 
complicated stuff. It's not clear which comes first however. It's 
Chicken & Egg. Been up so long it looks like down to them. Adler seems 
to get Einstein-Hilbert action from SBS & estimates of G & /, but it's 
far from obvious what he is doing. Also I don't see no tetrads there.
Z: What's wrong with a bi-metric approach at the macro-level? I can 
imagine that a bi-metric approach would fit your BEC model quite well if 
you were open to it.
J: Occam's razor. Parsimony. More with less. Excess Baggage. If it ain't 
broke don't fix it. Who ordered a second metric. That's an unimaginative 
act of desperation - not at all like Max Planck's.
  Z.
  E = pc for all real quanta
  This is what you get for |Vacuum ODLRO| = 0 in the core of a 
topological defect in G/H order parameter space.
  Note that the non-perturbative lepto-quark mass scale is
  i.e.
  1 Mev ~ 10^19Geve^-1/|x|
  e^-1/|x| ~ 10^-22
  Theta(x) = step function
V is the attractive interaction potential between the virtual electrons 
and positrons at the edge of the negative energy Dirac Sea with Fermi 
energy Ef = 0, but pf ~ h/L
rho is density of states of massless negative energy virtual electrons 
Fermi liquid at Ef per unit energy.
Admittedly Paul I suspect you are not the only one confused by this 
issue of how to separate intrinsic geometry from non-inertial local 
frame inertial g-forces especially since one cannot tell the difference 
without making local tidal geodesic deviation measurements.
[JS] What I am telling you is that you cannot use a GCT X when B =/= 0 
to make (LC)' = 0.
[Z] Then there is something wrong with your B. Of course you can always 
cancel net g-forces with a GCT.
[J] There is nothing wrong with B. In fact it is what you were looking 
for, but you looked in the wrong place - at the wrong level. Math is 
math. Diff(4) GCT non-dynamical Xu'^u connects non-geodesic LNIFu with 
non-geodesic LNIFu'. Indeed, you can always locally cancel the g 
inertial force of an LNIF by jumping to a coincident LIF, but you do 
that with a tetrad eu^a not with a GCT Xu'^u. The globally flat 
Minkowski space-time is degenerate i.e. eu^a = Kronecker delta Iu^a so 
that the u/a distinction vanishes and, indeed, in that case only, you 
can use GCT to get from an inertial frame to a non-inertial frame. Your 
intuition on Minkowski space-time does not carry over to curved 
space-time. Can I teach this Lazy Dog a new trick?
  in Newtonian limit.
  In the B = 0 degenerate case where a = u so to speak
  (Minkowski)uv = Iu^a(Minkowski)abIv^b
Now, take any Diff(4) GCT Xu'^u at all to get the PHONY 100% inertial 
force zero tidal warp pseudo-curved metric
  gu'v'(Minkowski) = Xu'^u(Minkowski)uvXv'^v
  B = 0
  Ruvwl = Ru'v'w'l' = 0 everywhere-when in Minkowski
These Diff(4) GCT Xu'^u describe infinite sets of pairs of coincident 
possible local observers u & u' in arbitrary relative motion at event P. 
Mathematically Xu'^u is the overlap transition function of two local 
coordinate charts x^u & x^u' in same neighborhood of event P. P is an 
equivalence ~ class of bare manifold points p i.e. P is the coset of ~ 
p's. The space of events P is a coset quotient space under gauge 
non-perturbative inflationary emergence of
B = (Lp/2pi)d(argVacuum ODLRO)
closed non-exact 1-form means period integrals of 1-form B on 
non-bounding 1-chain cycle need not vanish, but is NLp hence flux 
without flux? The boundary round a |Vacuum ODLRO| = 0 hole is two 
concentric opposite running cycles and we ignore the inner one to get 
flux without flux that is, in fact, measured by observers on the outer 
non-bounding 1-cycle. Goldstone phase = argVacuum ODLRO is not a unique 
0-form (it is a set of partially overlapping functions) because of 
|Vacuum ODLRO| = 0 topological defects in the G/H quotient order 
parameter space of the inflationary vacuum phase transition. The defects 
give non-trivial cohomology/homology/homotopies for all the p-forms, p = 
0,1,2,3,4 in 4D space-time. The macro-quantum order parameters out of 
which curved space-time emerges must be single-valued. It's not enough 
that |Vacuum ODLRO| match. The phases must match mod 2piN round a closed 
circuit in G/H that corresponds to some path in ordinary 3D-space. The 
number of circuits in each space need not be identical.
[JS] that Taylor series procedure
[Z]:  What Taylor series procedure?
[JS]: guv(P') = guv(P) + guv,w(P)(P'-P)^w + (1/2)guv,w,l(P)(P'-P)^w(P'-P)^l
Use the non-trivial tetrads e = (I + B)
Finding the local dynamical tetrad eu^a(P) requires the critical point
guv,w = (LC)uv,w - (LC)vu,w = 0 ? (from memory - error?)
This seems to require zero torsion to get orthodox EEP?
So what we really want is
gab,c = ea^ueb^vec^wguv,w = 0
P'is now in the tangent space
gab(P') ~ (Minkowski)ab + (1/2)gab,c,d(P)(P'-P)^c(P'-P)^d + ...
where second term on RHS is the actual tidal geodesic deviation 
(curvature) in the LIF.
Note also
guv(P) = (Iu^a + Bu^a(P))(Minkowski)ab(Iv^b + Bv^b(P))
and inversely
(Minkowski)ab = (Ia^u + Ba^u(P))guv(P)(Ib^v + Bb^v(P))
local all at same P
Therefore, the Taylor series can be reduced to expansions of the 
dynamical B field.
 be released
[snip 500 lines of trolled crap]
Jacko has the biggest dick in the world.  Good taste dictates that he
describe it rather than display it.
If you had a perpetual motion machine giant weapon that could boost
mankind to Tau Ceti and cure hemorrhoids forever, would you talk or
build?
Jacko's book is crap. It is nothing but crap. It is only crap. It is
purely crap. It is clarified, concentrated, distilled, condensed,
polished, packaged, and briskly shoved up your fundament while homing
in on your wallet crap.
It is the worst kind of pandering magnetic hernia truss multi-level
marketing to make you rich beyond your wildest dreams by tomorrow
without so much as a lick of personal inconvenience crap. It is Lotto
crap, Publisher's Clearinghouse crap, Home Shopping Network crap,
infomercial crap, Bre-X crap, hollow Earth crap, flat Earth crap,
Heaven's Gate crap, Chariots of the Gods crap, quartz crystal crap,
teleevangelical crap, thinner thighs in 30 days crap, other-ablement
crap, Virgin Mary on a taco crap, Elvis on a taco crap, War on Drugs
crap, War on Poverty crap, Homeland Severity crap, night basketball
crap, eat oat bran crap, breast enlargement cream crap, buy the
Brooklyn Bridge at only a tiny fraction of its true value crap,
1000-year old ginseng crap, 5000 yogic floaters to cleanse the world
crap, free energy crap, anti-gravity crap, hot young girls waiting for
your call crap, psychics hot line crap, never change your engine oil
again crap, interpret your dreams crap, sure-fire horse betting crap,
Superman squeezing a lump of coal into a blue white and perfect fully
faceted 290 carat diamond crap...
... Peace Dividend crap, Bosnian/Somali/Northern Ireland/Iraqi
pacification crap, peace in our time crap, Pacem in Terris crap,
Limits to Growth crap, Whip Inflation Now crap, I am not a crook crap,
nukular engineer crap, I didn't inhale crap, Health Care Crisis crap,
Department of Energy crap, marijuana will kill you crap, chastity
crap, tobacco isn't addictive crap, burning tobacco isn't carcinogenic
crap, children kidnapped over the Internet and sold into hebephile
pornography crap...
...MBA crap, human factors engineering crap, social engineering crap,
Project Head Start crap, thousand points of light crap, compassion
crap, nurturance crap, sensitivity counseling crap, Special Olympics
crap, learning disablement crap, Affirmative Action/Equal
Opportunity/diversity crap, multiculturalism crap, multilingualism
crap, Kwanzaa crap, Days of Rage crap, Nobel Prize in Literature crap,
automotive airbags crap, commuter lane crap, the E*L*E*C*T*R*I*C car
crap, the H*Y*D*R*O*G*E*N car crap, recycled post-consumer waste crap,
low volume flush toilet crap, fuel from biomass crap...
...Workers' Paradise crap, Democratic Republic of [you name it] crap,
New Soviet Man crap, Save the Earth crap, Furbish lousewort crap,
clothe animals to hide their genitalia crap, N-rays crap, Krebiozen
crap, amygdalin crap, smart drugs crap, coffee enemas to cure cancer
crap, cosmic consciousness crap, orgone energy crap, aura crap, astral
plane crap, astrology crap, ouija board crap...
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
 soonto  be released
Gee Al, when are you going to stop pussyfooting around and say what you 
really mean?
Bob Kolker
Bring me up to date on this one, Mesanator. Some years ago
there were a couple of unusually dry years in the central
plains. The Mississippi was so shallow that the paddle-
wheelers had to be parked. I heard that if Lake Michigan
dropped by another foot or two, Chicago would become
the stinkiest town in the Western Hemisphere, because
is a big raunchy mud-flat would appear along the
waterfront. Hear anything about that?
   posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK
are
Venus
contingency
The Mississippi is on the other side of the state, not my
problem either.
Why would Lake Michigan drop if the ice caps melt?
Would? Chicago has always been the stinkiest town in the
Western Hemisphere. The entire downtown area is elevated one
story above ground level for that very reason.
I think you are confused. Lake Michigan is deep. You must be
thinking of Cleveland where Lake Erie can disappear if the wind
is from the south.
The only thing I heard was that the comet Hale-Bopp was going
to cause an earthquake that would split open Lake Michigan and
let all the water come rushing out. I must have been out of town
when that happened.
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
Hi...
        This is, I'm sure a simple question... I just don't know how
to do it...
        I think I understand how to calculate, say the number of ways
of rearranging say AABBBCCCC... (9!/(2!3!4!)) right? (order _is_
relevant among different types)
But, say I wanted the number of ways to pick _3_ items from there?
How would I get that number, with 2A's, 3B's and 4C's?
(Or more basically, for a group consisting of any number of items,
with any number of those items possibly repeated, how do I find the
number of ways of picking n items, with order being significant.)
I can see that AABBBCCCC empirically can have 26 possible
ways of picking 3 items. I can see that's because it's 3^3-1, with
the -1 being the possibility of picking 3 A's... But, I'm
not sure how to genericize the calculation, which is what I'm
after. (for instance, it's less easy for me to see how to calculate
that to pick _4_ items from there, would have 71 possible ways,
even tho, I can enumerate those...)
Binesh Bannerjee
- -- 
One thing a computer can do, that most humans can't is be sealed up in
 a cardboard box and sit in a warehouse. -- Jack Handey
    PGP  Key: http://www.hex21.com/~binesh/binesh-public.asc
PGP Key fingerprint = 421D B4C2 2E96 B8EE 7190  A0CF B42F E71C 7FC3 AD96
    SSH2 Key: http://www.hex21.com/~binesh/binesh-ssh2.pub
OpenSSH  Key: http://www.hex21.com/~binesh/binesh-openssh.pub
BubbleBabble = 
xibeb-voges-havez-pabaf-debop-cylil-lelyc-viruv-bygeg-zotoh-dixex
 Fingerprint = 9d:7c:84:5d:80:e3:65:8d:ee:9e:a3:b9:56:0a:e9:ad
    SSH1 Key: http://www.hex21.com/~binesh/binesh-ssh1.pub
CipherKnight Seals:
        http://www.hex21.com/~binesh/binesh-seal.tar.bz2.cs256
        http://www.hex21.com/~binesh/binesh-seal.zip.cs256
        http://www.hex21.com/~binesh/binesh-certificate.gif.cs256
        Decrypt with CipherSaber2 N=256, Password=WelcomeJedi! (No 
quotes)
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.0 (GNU/Linux)
iD8DBQFCYH7UtC/nHH/DrZYRApEwAKDeIC/Nun9u/FyoAD34i1rjs/s1iACgtwwb
mxKxNJpxIttDWpR+Vz01UIA=
=6pej
-----END PGP SIGNATURE-----
X-RFC2646:  Original
[Binesh Bannerjee]
The question is simple enough, but I don't think there's an equally simple
answer (would be delighted to be shown wrong about that, though!).
Yup, that's right.
You actually have some irregular constraints on integer partitions here,
which is why I'm pretty sure a general solution is correspondingly messy.
You need to sum 3!/(a!b!c!) over all non-negative integers a, b, c
satisfying:
    a+b+c = 3
    a <= 2
    b <= 3
    c <= 4
it's tedious:
    a b c  6/(a!b!c!)
    - - -  ----------
    0 0 3           1
    0 1 2           3
    0 2 1           3
    0 3 0           1
    1 0 2           3
    1 1 1           6
    1 2 0           3
    2 0 1           3
    2 1 0         + 3
                  ---
                   26
Clever!
That's the rub, all right.
...
And another thing they can do that most humans can't is crunch easy but
I'll give a (Python 2.4) program for computing this stuff below.  Here's a
little interactive session then:
...     print nitems, solve([2, 3, 4], nitems)
1 3
2 9
3 26
4 71
5 180
6 410
7 805
8 1260
9 1260
`solve()` takes an optional boolean argument `verbose`, if you want to see
details.  For example, maybe it's curious that picking 8 items and 9 items
both came up with 1260 ways to do it.  So let's see:
[1, 3, 4] 280
[2, 2, 4] 420
[2, 3, 3] 560
1260
[2, 3, 4] 1260
1260
Looks right to me!
Or we can check the two examples you started with:
[0, 0, 3] 1
[0, 1, 2] 3
[0, 2, 1] 3
[0, 3, 0] 1
[1, 0, 2] 3
[1, 1, 1] 6
[1, 2, 0] 3
[2, 0, 1] 3
[2, 1, 0] 3
26
[0, 0, 4] 1
[0, 1, 3] 4
[0, 2, 2] 6
[0, 3, 1] 4
[1, 0, 3] 4
[1, 1, 2] 12
[1, 2, 1] 12
[1, 3, 0] 4
[2, 0, 2] 6
[2, 1, 1] 12
[2, 2, 0] 6
71
Apologies in advance, but if you don't know Python this will be hard to
follow -- the usual approach to generating combinatorial families in Python
involves generator functions, which allow producing one item at a time,
and that's an unusual feature (few programming languages have such a
facility).  Worse, I'm using recursive generators, just because it was the
easiest way to code it:
def generate_vectors(counts, n):
    assert 0 not in counts
        return
    nc = len(counts)
    # most_after[i] <- the most we can absorb after index i
    most_after = [0] * nc
    i = nc - 2
        most_after[i] = most_after[i+1] + counts[i+1]
        i -= 1
    # Generate all possibilities for distributing
    # remaining starting at index i.
    def gen_inner(i, remaining):
        if i == nc:
            assert remaining == 0
            yield []
            return
        # Can't put more at this index than the number
        # remaining, or the capacity of this index.
        max_todo = min(remaining, counts[i])
        # But have to do enough so that what's left fits
        # after this index.
        min_todo = max(0, remaining - most_after[i])
        for j in range(min_todo, max_todo + 1):
            for v in gen_inner(i+1, remaining - j):
                yield [j] + v
    for v in gen_inner(0, n):
        yield v
def fac(n):
    if n <= 1:
        return 1
    result = 2
    for i in xrange(3, n+1):
        result *= i
    return result
def solve(counts, n, verbose=False):
    result = 0
    nfac = fac(n)
    for v in generate_vectors(counts, n):
        prod = nfac
        for count in v:
            prod //= fac(count)
        if verbose:
            print v, prod
        result += prod
    return result
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Cross-posted to alt.sci.math.combinatorics ...
(I thought I had an answer, but I realized it doesn't work.)
     --- Christopher Heckman
   posting-account=tN6qcA0AAACUaa8g28ioHaW6IKJhmtDk
David Hawkins is credited with the following stochastic analog of the
sieve of Eratosthenes:
List the integers 2, 3, ..., N, where N is large. Circle the number 2
and cross out each subsequent number with probability 1/2. If m is the
first remaining number on the list not crossed off, circle m and cross
out every succeeding number with probability 1/m. Continue until all
numbers are either circled or crossed off.
One can establish that many facts about primes are exactly what one
would expect to be true for a typical sequence of circled integers at
the end of this process. For example, Hawkins showed [1] that the prime
number theorem holds in the sense of the weak law of large numbers. And
Wunderlich [2], embedding each of the finite probability spaces
implicit above into a single space, proved a strong law analog.
But my question concerns the paper of Hawkins [3] which introduced this
process and preceded the above results by 20 years. He proceeds roughly
as follows: let P_m be the probability that m is circled at the end of
this procedure, where m <= N. It is easy to check that this is
independent of the large integer N (as long as m <= N). Hawkins
proves that P_2=1 and
 P_{m+1} = P_m (1- P_m/m);
this is then used as a springboard to show the (true result) that P_m ~
1/log m, hence the expected number of integers remaining at the end of
the algorithm is ~ n/log n.
However, it seems this recurrence relation is incorrect (and a
different method is used in the more recent Hawkins and Wunderlich
papers, though neither paper points out the mistake in [3]). For
example, the recurrence relation predicts that P_5 = 215/576, but
actually P_5 = 107/288 (=214/576 !).
Unfortunately this incorrect recurrence relation has found its way into
[4] (and perhaps [5], but I should check before I make this slanderous
claim!). My question is this: the MathSciNet review of the first
edition of Harald Scheid's _Zahlentheorie_ says that it contains a
discussion of the random sieve. Does Scheid's _Zahlentheorie_ [6]
include this mistaken claim? Information about any edition would be
appreciated, but I would prefer hearing about the current (3rd)
edition.
If I do not get any responses then I can request the book by
interlibrary loan. But as I have several requests outstanding and this
seems like a simple matter to check, I am hoping it can be speedily
resolved by some kind soul who has the book handy.
Paul
[1] Hawkins, David Random sieves. II.  J. Number Theory  6  (1974),
192--200.
[2] Wunderlich, M. C. The prime number theorem for random sequences. J.
Number Theory 8 (1976), no. 4, 369--371.
[3] Hawkins, David. The random sieve.  Math. Mag.  31  1957/1958 1--3.
[4] Greger, Karl. Random sieving and the prime number theorem. Two-Year
College Math. Journal 5 (1974), no. 1, 41-46.
[5] Greger, Karl. Random sieves and the prime number theorem. (Swedish.
English summary) Nordisk. Mat. Tidskr. 21 (1973), 57--66, 127.
[6] Scheid, Harald. Zahlentheorie. (Number theory). 3. Aufl. (German)
Spektrum Lehrbuch. Heidelberg: Spektrum Akademischer Verlag. 497 S.
Quite a few people are working on automatic
speech-to-text conversion; new algos in that field
would have a visible impact. Maybe too we will see
some high-level algos for finding important needles in
haystacks. Here's a little about that:
http://www.darpa.mil/dso/thrust/math/tda.htm
I suppose we can hope for a breakthrough in the messy
area of molecular computations, and maybe the
simulation of fluids.
LH
X-RFC2646:  Original
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
  + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
That says it all.
Aristotle Polonium
http://pearl1.lanl.gov/periodic/elements/84.html
http://www.chemicalelements.com/elements/po.html
http://www.webspawner.com/users/radioactivethreat/
http://www.webelements.com/webelements/elements/text/Po/key.html
  + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
   posting-account=ZMkdMg0AAAD6SEeiN7BOQNA4Oo4dVYb_
what did that say exactly? care to elaborate?
        by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 

1.9  primary) with ESMTP id j3G3X3h00922;
        Fri, 15 Apr 2005 23:33:03 -0400
        by support2.mathforum.org (8.12.10/8.12.10/The Math Forum, 
$Revision: 1.6  secondary) with SMTP id j3G3XAeK019797;
        Fri, 15 Apr 2005 23:33:37 -0400
        Sat, 16 Apr 2005 07:24:36 +0300
X-SpamTest-Info: Profile: Formal (167/041110)
X-SpamTest-Info: Profile: Detect Hard No RBL (4/030500)
 We tried contacting you awhile ago about your low interest morta(ge rate.
 You have qualified for the lowest rate in years...
 You could get over $380,000 for as little as $500 a month!
 Ba(d credit? Doesn't matter, low rates are fixed no matter what!
 
 To get a free, no obli,gation consultation click below:
 http://www.n0wwewillsave.com/sign.asp
 Lanny Hammond
 
 to be remov(ed:        http://www.n0wwewillsave.com/gone.asp
 this process takes one week, so please be patient. we do our 
 best to take your email/s off but you have to fill out a rem/ove
 or else you will continue to recieve email/s.
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
physics stuff.
Sukjah Roh
improving.
I'm no superstar but I know that a lot of students have this same
problem. I think part of the cause is that textbooks and treatises
must be rigorous, not relying on any appeal to intuition; but
any original work does require intuition, speculation, experiment.
In a curriculum, there may be no time for those.
And the professionals seem to be shy about saying that some
results are more important than others; that a new proof by a
schmoe is better than an earlier proof by a famous historic figure;
and so on. Many writers make no pedagogical remarks at all.
The mathematical field is rather self-consciously conservative, IMO.
Books like Polya's are embarrassingly few and far between.
But anyway we all learn by doing. Compare the work
of the masters, in the original, with the later polished
conventional textbook versions. You will see that while
the textbooks all look much the same, no two of the
masters are alike.
LH
I was stupid for the first question.
I temporary lost my mind...^^!
That is true. The only references I have handy are
Lang's _Algebra_ (the section on finitely generated
Abelian groups) and the algebra appendix in Volume 5
of Dieudonne's _Treatise on Analysis_. Why not just
make up your own examples of matrices?
Wish I could be more helpful, but this algebra will all
make perfect sense in the end. I'm surprised there have
been no other responses to this thread :(
LH
Easy one:
You are in the desert and you find a thread that has been
put there by a genie. Legend tells you that if you follow
it in one direction, it will lead you to the Minotaur, but
in the other direction it leads to the hidden treasure or
the captive maiden or whatever. The thread is marked
with 0's and 1's, periodically, in such a way that you
can tell the two directions apart. What is the smallest
possible period of that periodic function?
Not so easy:
Generalize to two dimensions.
Cool it, take it easy.  Don't let phantom fanstansies inflame your
passions for lust and riches.  As a matter of fact, you've got more
serious problems than that, like finding water.  The Minotaur is at both
ends, the one with rich busty maiden and the one without.  This you will
find out, as do all making deals with devils find out, at the end of the
long long hard to follow thread when you run out of water and find upon
return that the thin thread had small sections snipped off by desert
creatures and long sections caught up by the wind and tangled all about.
There you'll be, lost in Death Valley on a sunny summer day.
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
... Furthermore, one of the ends of the thread bends down into the sand
and goes into the inner shell of the Earth. (The Earth is holloy, btw.
Bet you didn't know that...) How did it get there? Well, there was this
tunnelling machine that a scientist built, and he took it for a ride
one day ...
... provided, that is, that you can tell which devil is lying and which
is telling by asking only one of them only one question ...
... and if you only went over one more dune, you'd find the telephone
wires, which would lead you back to the rocket base you took off from.
     --- Christopher Heckman
P.S. Quick quiz: Where did I get these three excerpts from?
It's simple. If you didn't get swindled, he lied about being the devil.
Jules Vern in Snorting off the Looking Glass.
just guessing
I'm assuming that you have no clue that 1 is right way and 0 = wrong way,
else, the problem is too easy
Right track??
Bob Pease
I probably didn't state it very well. Here's what it's supposed to be:
Z to {0,1} such that
i) f(x)=f(x+n) for all x in Z
ii) for any u in Z, the functions x mapsto f(x)
and x mapsto f(u-x) are distinct.
I tried to glam it up a little :)
Larry Hammick a .8ecrit :
The obvious solution n=6 ( ...01011001011001011001...) is the shortest. 
But what is your exact question in dimension 2?
denis feldmann
Like I said, easy. But I found it mildly surprising that 5 is not
sufficient.
If we allow three different marks, then of course 012012... works.
We want a doubly periodic function on ZxZ which has
no reflection or rotation symmetry. Variation: a
doubly periodic function on the set of complex
numbers of the form m+nt where tt+t+1=0 and
m,n are integers.
LH
   posting-account=wtMkxQwAAAClN0BUt6zcgY19YPKK-tfO
I tried some simple cases, your result seems correct.
I really appreciate your help. Nice weekend.:-)
   posting-account=wtMkxQwAAAClN0BUt6zcgY19YPKK-tfO
Sorry I made a mistake,
ln(n/(k-1))/ln(k/(k-1))=(ln(n)-ln(k-1))/(ln(k)-ln(k-1))<ln(n)/ln(k)=log(n,k)

is not correct. But still when n=6,k=4, we have
ceil(ln(n/(k-1))/ln(k/(k-1)))=3
Three colorings are not enough do distinguish all 4-subsets of a 6-set.
   posting-account=wtMkxQwAAAClN0BUt6zcgY19YPKK-tfO
Your analysis is fine, but I am afraid this lower bound is not tight.
Because
ln(n/(k-1))/ln(k/(k-1))=(ln(n)-ln(k-1))/(ln(k)-ln(k-1))<ln(n)/ln(k)=log(n,k)

seems a little bit too good.
Consider a simple case when n=6 and k=4, I don't think we can find two
colorings that will do the job.
Anyway your analysis is very helpful, I now know how to attach this
problem and where to find the knowledge I need.
I really appreciate your help. Nice weekend.:-)
   posting-account=wtMkxQwAAAClN0BUt6zcgY19YPKK-tfO
Great, thank you very much Mr. Heckman.
I will look into your proof carefully. Can I write to you later?
Fenghui
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Yes. My professional e-mail address is checkman@math.usa.edu.
(Actually, you need to reverse the orders of the letters in USA; this
is to throw off spambots.)
I can't shake the feeling that someone has answered this problem
before, though, and written a paper about it.
     --- Christopher Heckman
   posting-account=wtMkxQwAAAClN0BUt6zcgY19YPKK-tfO
That's what I was thinking too. This problem seems well-defined and
should have been studied.
Write to you later.
Fenghui
And it hasn't happened. All you have to show for your ten years on Usenet 
is
your reputation for incompetence, ignorance, unteachability, dishonesty,
immaturity, narcissism, and vindictiveness. Repulsive up.
X-RFC2646:  Original
[JSH]
Don't be so modest!  You've kept on explaining it too -- and explaining it, 

and explaining it, and explaining it, in all the years since.  Although 
usually it comes off as more threat than explanation.
Correct me if I'm wrong, but aren't you the same JSH who's currently 
spinning increasingly goofy excuses for refusing to test his SF theorem 
against any real world example?
One can always hope. 
Please reply.....
   posting-account=U_k2tw0AAAAu2-GKN-e5XaBFENNMOWcM
hi sir bag o lng ko kabalo kng diin ko makadto.hehe..
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
sir. good day!
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
sir. good day!
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
sir. good day!
   posting-account=8mm2nA0AAABI6jvBKJh4IoeU9PqYaf2y
Hi sir. What's next?
   posting-account=-af3RQ0AAACoiDP4KdgjRigmRHMCfCum
How would a sleeping person know that he is really sleeping?
By waking up to the fact that he's sleeping. ;-)
On Sat, 16 Apr 2005 01:00:00 -0700, William Elliot
(For extra credit, do it in your sleep.)
-- 
Angus Rodgers
(angus_prune@ eats spam; reply to angusrod@)
Contains mild peril
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
As a kid I developed a way to escape nightmares by closing my eyes and
opening some none-dream eyes instead.  Then there was this dream where
I knew some dreadful monster was pursuing me, and I realized it was a
dream, did my trick, and woke up in bed.  Unfortunately, my bedroom
door opened next and the monster entered.
Make sure you pop your stacks completely.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
 Discussion, linux)
Just the other night, I had a scary dream causing me much anxiety
until (in the dream) I mentioned to my companions, Oh, don't worry.
It's just a dream.
I regularly know that I'm dreaming when I'm dreaming.  More to the
point, my character in the dream knows it too.
But maybe I'm funny.
-- 
Jesse F. Hughes
Quincy (age 3 1/2, looking at a picture): Are these people Canadians?
Me: Uh, no, they're Australian Aborigines.
Quincy: Do they fight Canadians?
   posting-account=uTJlLQ0AAADvD--djrc8ES7ChusAdXrv
hi sir.
   posting-account=9zctQA0AAAAGImNmKAGNwGeGeLzqvWlX
received
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
   posting-account=luV1mg0AAADLj6DMzV0PZ-xZbzzswihF
   posting-account=6-E0EQ0AAACRMqICATg0tYJKQ9nm9qDx
   posting-account=6-E0EQ0AAACRMqICATg0tYJKQ9nm9qDx
I finally got it!
   posting-account=8mm2nA0AAABI6jvBKJh4IoeU9PqYaf2y
   posting-account=W66UjQwAAACT9lLzOGlKHfSsy-Cao9MN
hello!
   posting-account=9zctQA0AAAAGImNmKAGNwGeGeLzqvWlX
at last, nakita ko
   posting-account=z4I1OgwAAABkd0NAZizQpJ91OfZ2BSHx
Message sending...
   posting-account=Qvqs9Q0AAACJtCOlDHetBZkDxNj2i_Y4
Message sending failed...
   posting-account=RnPpTg0AAAC-R0_dR8NfKJKlrKJMfGe3
hi sir! message received...
   posting-account=oP2MXQsAAACN8tJy3-OiG7Jka6E1eEu0
Dei Gloriam.In English it means for the greater glory of God. It is
not a slang.
   posting-account=Qvqs9Q0AAACJtCOlDHetBZkDxNj2i_Y4
hi sir!!!
   posting-account=k7jjEA0AAAAjs2dUhBQxfnQDcpe3ik-Q
hi sir - ryanpogi :-)
   posting-account=k7jjEA0AAAAjs2dUhBQxfnQDcpe3ik-Q
hi sir! - ryan
   posting-account=k7jjEA0AAAAjs2dUhBQxfnQDcpe3ik-Q
Help for those who are math wizard pls help me answer this fact
The answer is yes.
   posting-account=uTJlLQ0AAADvD--djrc8ES7ChusAdXrv
what do you mean by yes? countable or unccountable?
Both, it depends on which infinity you're talking about. ;)
haroldpaykwa@yahoo.com a .8ecrit :
yes.
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Yes, it's countable or uncountable.
     --- Christopher Heckman
What is the greatest paradox mathematics has made?
   posting-account=z4I1OgwAAABkd0NAZizQpJ91OfZ2BSHx
We cannot tell yet...more paradoxes will come out...perhaps much
greater than the previous ones...
   posting-account=U-QWKw0AAABwXH5x6Cl-rchzsTO_TAyk
Math is the slave and queen of all sciences.
   posting-account=k7jjEA0AAAAjs2dUhBQxfnQDcpe3ik-Q
ambot nell...basta ang nabal-an ko gwapo ko hehehe
   posting-account=k7jjEA0AAAAjs2dUhBQxfnQDcpe3ik-Q
follow the ff steps...
1. find start a new topic link.
2. fill up the subject and the topic u wish to post...
easy guys!!! c u!!!
Go back to your chat room with the text messaging shorthand
(and the triple bangs). In a technical newsgroup posters
usually adhere to a higher language standard.
--
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
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Who let the effin' newbie in?
     --- A. C. Fenderson
Hi everyone! ayawan ko search!!! help!
   posting-account=uTJlLQ0AAADvD--djrc8ES7ChusAdXrv
Mga B's where are U?
 and Atom Totality Re: If gravity is replaced by a Cosmic Coulomb force
(snipped)
In the Atom Totality theory another one of these Nested mathematics
arises. The Universe is one big atom and the stuff inside this big atom
are simply more atoms. So the whole is the same as its pieces. There
cannot be a mathematical formula or equation to capture this whole is a
part and parts are the whole.
Many of us believe in a Unification of the Forces of Physics. So in the
unification all the forces become versions of one force. I say it is the
Coulomb force. So that StrongNuclear and WeakNuclear are Coulomb. And
gravity is a Coulomb force. Here again we have a example of parts making
a whole and the whole is its part. Not as profound as the Atom Totality
whole and parts but nonetheless an example.
Finally I give a third example of this Nested mathematical form which
cannot be made into equations. We cannot write an equation for pi since
it is transcendental. So can we say that an Atom Totality is
mathematically transcendental since you cannot write an equation where
parts combine to be the whole yet the whole is one of its parts? No, of
course not.
Similarly, can we write a equation for the Unification of Forces as one
Coulomb force. Of course not.
Now onto the exciting challenge of the Maxwell theory or Maxwell
Equations. They were considered old and Classical Physics with the
intonation or insinuation or depreciation or low tone mocking that the
Maxwell Equations were on the verge of extinction just as Newtonian
Mechanics was extincted by Quantum Mechanics, or to be more fair that
the Newtonian Mechanics was made an algorithm of first approximation of
physics.
But, now, consider what happens when we call the force of gravity a
fiction force and replace it with a Cosmic Coulomb force. So that the 94
protons of the Atom Totality bend space and the mass of the observable
universe is the bits and pieces of the last 6 electrons (the galaxies
are dots of the electron dot cloud). So we have no need of gravity for a
Cosmic Coulomb force is the force of gravity.
All of a sudden we have to add another term to the Maxwell Equations of
a Cosmic Coulomb force. And thereby elevating the Maxwell theory into
another one of those Nested math forms that math equations cannot
capture the physics. To have transcendentalized the Maxwell Equations.
To have elevated the Maxwell theory so that it is no longer Classical
Physics but the very heart of physics. And that no program of Feynman or
Landau and Lifschitz are able to derive from other axioms.
So that when you add a term of a Cosmic Coulomb to the existing Maxwell
Equations is like trying to form a equation of the Universe where 1
atom whole has many atoms inside itself to compose the whole. So we
transcendentalize the Maxwell theory by adding a term of a Cosmic
Coulomb.
So I see three of these patterns of a Nested mathematics. I see it in
the Atom Totality theory where the whole is 1 atom which is composed of
many atoms inside itself. And with the Unification of Forces of Physics
in that all the forces are really just a version of the Coulomb force.
And finally with the Maxwell theory of injecting a Cosmic Coulomb force
when one of its laws is already the Coulomb force creates another Nested
mathematics.
So logically this sounds correct. Because if the Atom Totality is true,
then we should expect to see this Nested pattern elsewhere. We should
see it in a Unification of Forces and finally we should see it in
Physics most praised set of axioms which are the Maxwell Equations.
Truth in physics ends up with a pattern and that pattern (whole is one
of its parts and its parts combine to make the whole) should not be a
one of a kind pattern but be repeated elsewhere. In physics the theory
of the cosmos pattern repeats itself in the Unification of Forces and
repeats the pattern in the axioms of physics which are the Maxwell
Equations.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
 Forces and Atom Totality Re: If gravity is replaced by a Cosmic Coulomb 
 force
I should make myself more clear here. Pi is transcendental meaning it is
not algebraic. Equations have pi in them but pi is not algebraic. And
Lagrangian and Hamiltonian are algebraics. So by injecting a Cosmic Coulomb
force that replaces gravity into the Maxwell Equations makes the Maxwell
Equations as transcendental or nonalgebraic.
We have another example of this feature in the Uncertainty Principle which
duality as given by the Uncertainty Principle is nonalgebraic and gives
rise to Quantum Tunnelling.
You are not going to derive the Uncertainty Principle from some other set
of axioms using Lagrangian. Likewise, when this Cosmic Coulomb term is
added to the Maxwell Equations will put the Maxwell Equations as
nonalgebraic just as pi is, and just as the Uncertainty Principle is.
A Cosmic Coulomb force replacing gravity into the Maxwell Equations will
elevate those Equations on a par to Quantum Mechanics and the Maxwell
Equations will no longer be thought of as a axiom set of Classical Physics.
The Maxwell Equations will no longer be Classical Physics but rather the
very heart of physics and on par with everything in Quantum Mechanics.
Now some may complain that the Maxwell Equations already has the Coulomb
Law and so why should there be an overarching Coulomb term added to the
Maxwell Equations. And my most simple answer is that the Maxwell Equations
of old exist in a experimental condition where gravity does play a role in
what those equations are. The old Maxwell equations ignores gravity and
sees gravity as some entity outside of EM theory. But when we see gravity
as a Coulomb force itself then Faradays law and Ampere's law have missing
the term of a overarching Cosmic Coulomb that allows for a Faraday's law or
an Ampere's law to even exist. In the past we ignored gravity because we
thought it was distinct from EM and that its force strength was too weak to
influence the laws of EM. But we cannot ignore gravity as a Cosmic Coulomb
because that is EM itself. EM creates gravity and so the old Maxwell
Equations are incomplete. There would not be a Faradays Law or Ampere Law
if a Cosmic Coulomb did not exist to create an environment to experiment
and find out what those laws are in the first place.
If gravity were distinct from EM then the old Maxwell Equations stand as
not needing correction. But when gravity is a force that is one of the
Maxwell Equations then the old Maxwell Equations have to be added a new
term of an overarching Coulomb force that creates Faraday's and Ampere's
laws.
Feynman and Landau and Lifschitz in the 20th century could derive the old
Maxwell Equations from diverse other principles such as Principle of Least
Action and Fermat's Principle of Least Time and using the algebraics of
Lagrangian or Hamiltonians. But the revised Maxwell Equations where gravity
term of a Cosmic Coulomb is added onto the old Maxwell Equations, that they
are no longer algebraic. There is no longer a derivation of the Maxwell
Equations using Lagrangians, just as there is no derivation of Quantum
Tunneling or Uncertainty Principle.
The new Maxwell Equations with a Cosmic Coulomb term becomes the heart of
physics and is NON derivable from other axioms. The new Maxwell Equations
become the root axiom set.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
 Forces  and Atom Totality Re: If gravity is replaced by a Cosmic Coulomb
 force
No, the dots making up a galaxy arm- the *stars*-
are the dots of the electron cloud.
They extend from the 'Black Hole' or proton
out to the edge in a spiral shape.
Each star is made from atoms. Each of these atoms in its
turn contains a Black Hole, or proton,
at its center and is surrounded by 'dots' that
make up electron clouds.
These dots are smaller stars yet, made out of smaller
atoms, each of which has
a 'Black Hole' at its center...........
Nested.
John
http://users.accesscomm.ca/john/
GALAXY MODEL For The ATOM
 ofForces  and Atom Totality Re: If gravity is replaced by a Cosmic 
 Coulombforce
In the 1990s when I told a friend of mine that the Atom Totality theory
makes God into a pantheism where the Cosmos itself is God. Where Science
is God, and God is Science. And that the closest thing to God is the
nucleus of the Atom Totality. And my friend replied that people are
nutty and that they will find a place to inject their Jesus or inject
their Holy Ghost or inject their miracles and supernatural and
mysticism.
So I guess, from John above that not only does the Atom Totality theory
put the religious and supernatural folk on a run for cover or a run for
injection but it puts the String theorist on a run for cover and run for
injection. And it puts the string theorists as nutty or even nuttier
than the religious folk.
Way back in the time of Democritus, those early Atomists were often
burned at the stake for being atheists. Democritus Atom theory was
viewed as too atheistic. And some people like Hypatia were burned for
knowing science is greater than religion.
No, John, I don't think string people could ever make a stand of
injection into the Atom Totality theory but will disappear and vanish
almost overnight when the Atom Totality theory catches momentum.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
   posting-account=z4I1OgwAAABkd0NAZizQpJ91OfZ2BSHx
Do you have any proof for 1 = -1?
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Yeah. 1 = abs(-1) = | -1 | = det(-1) = -1.
Newbie troll ...
     --- Christopher Heckman
   posting-account=z4I1OgwAAABkd0NAZizQpJ91OfZ2BSHx
Do you have any proof for 1 = -1?
   posting-account=DaXtvAwAAACPATqzmVZ4JJbgwUGjL51g
Hi:
This is true in any field, say, of characteristic 2, for example: in
Z_2:= Z/2Z.
Tonio
Differentiate undefined and indeterminate.
Is it true that zero (0) is now considered as a counting number?
Yes, but don't count on it. ;-)
   posting-account=uTJlLQ0AAADvD--djrc8ES7ChusAdXrv
hi mam lea. . ..
   posting-account=uTJlLQ0AAADvD--djrc8ES7ChusAdXrv
sorry indi ko kabalo
   posting-account=RnPpTg0AAAC-R0_dR8NfKJKlrKJMfGe3
i wanna know...please help...
Let me remark on the mistakes in the published theses of my 
students, Guckenheimer and Nitecki mentioned above, Thom's 
mistake referred to in Problem 10 and Poincare's mistake 
in the Poincare conjecture of Problem 2. Mistakes happen 
frequently in published mathematics; I certainly have made 
my share. Especially in the early development of a subject, 
one is likely to err. Here oftentimes not only are the main 
concepts confused, but even the definitions are ambiguous. 
Thus there is the need to heed well the establishment of 
good foundations for a new subject. These considerations 
are a motivating factor in my efforts to help create a more 
solid foundations for numerical analysis (as in BCSS or Smale,
1990b).
Let me point out a story about Poincare as told by Diacu-Holmes 
(1996) or Barrow-Green (1996). When Poincare discovered a 
mistake in his theory of celestial mechanics, he had the
destroyed; but at the same time he discovered a phenomenon 
in dynamics which is now called chaos. About 60 years later,
oblivious of these events, I foolishly hypothesized (see Smale, 
1998) that chaos did not exist in dynamics!
 Steve Smale, Mathematical Problems for the Next Century
-- 
Angus Rodgers
(angus_prune@ eats spam; reply to angusrod@)
Contains mild peril
        ETAtAhUAs3nLFVDD61r+C2f0JkOEFjmqoxMCFGf7g10gU0T4TqUU5bH5SqTMwcEe 
I read in one of Martin Gardner's books that Legendre pulled one off.
The claim, which got published, was that the equation a^3+b^3 = 6c^3 had
no solutions in integers.  In fact a = 17, b = 37, c = 21 does solve the
equation.  Presumably Legendre had proven no solution for sufficiently
large integers but failed to thoroughly check the smaller ones.
Today peer reviewers are better; Wiles' originally flawed proof of FLT
was corrected BEFORE going into print.  But they are not perfect.
Though it's not mathematics, the possibilities that still exist today
are demonstrated by a chemistry/metallurgy boner in 2001 (IIRC).  Water
vapor was found to have reduced copper oxide to the metal!  Neither the
author nor the reviewers figured out why until after it was published in
Scripta Materialia.  (The water vapor had reacted with silicon in the
apparatus the authors were using, and hydrogen thus formed was the
reductant.)
--OL
There was a famous one by Lebesgue, when he said
(J. de Mathematiques (6) 1 (1905) pp. 191-192)
that for an infinite decreasing sequence of subsets of R^2, the
intersection of their projections is equal to the projection
of their intersection.
A few unsound proofs of Poincare's conjecture
have reached print, including one by Poincare' himself,
and another by Whitehead.
I don't remember his name, but some unfortunate guy
did a lot of  what appeared to be good work
on what we now call the Feit-Thompson theorem,
only to find, after his paper appeared, that his several
lemmas (all of which were okay) could not
simultaneously apply.
I heard that Serge Lang, in one his books, sets out to
prove a certain theorem by induction, proves it for n=1,
and then moves on to the next theorem! Can anyone
confirm this tale, or is it a fable?
LH
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Giving James Harris a Usenet account.
Seriously? People trying to add 1 + 1/2 + 1/3 + 1/4 + ..., thinking the
sum is a real number.
     --- Christopher Heckman
Proginoskes a .8ecrit :
On the othe hand, 1+2+3+4+5+...= -1/12 has some merits :-)
   posting-account=DaXtvAwAAACPATqzmVZ4JJbgwUGjL51g
Hi:
It's hard to know, but Lame's proof of Fermat's Last Theorem back in
1847 is one of the most well known mistakes. It also helped to push
forward many subjects that right now are common in mathematics, so it
was a very nice mistake.
Tonio
Ps. Of course, Yang Mills' answer is just an idiocy: so far non serious
mathematician has even claimed pi is an infinite number (what's that
in this context, anyway?), and the set of prime numbers (in the ring of
integers) is infinite, as can be easily proved by any regular high
school student.
   posting-account=ZMkdMg0AAAD6SEeiN7BOQNA4Oo4dVYb_
that pi is an infinite number and that there are an infinite number of
primes.  also mentionable is that random data cant be compressed in a
lossless way. all of these assumptions will be proven false in the
future.
 Discussion, linux)
[..snip utter incoherence..]
Herc, you've reverted to your own private means of communication.  If
there was a point to all that, it alludes me.
Unless the point is this: It is so remarkably unlikely that someone
named Gamble would *ever* over the course of many years on the
internet reply to someone who has said nutty things regarding coin
flips that the latter person must be God.
The Ullrich thing is just utterly silly.  His middle initial is C.
Gosh.  Nowhere does he mention wealth, but instead that someone might
be convinced he's God.  If only one person thinks so, I guess that
doesn't make David rich.
I think I'll drop out of this thread.  Really, you are too obviously
in need of serious psychiatric help.  I don't feel particularly good
about these conversations with a delusional man.  
Seriously.
-- 
Jesse F. Hughes
Your approach so far has not depended on the key result of Dedekind's
that is challenged by my work, so you can proceed without concern.
                      -- James S. Harris reassures a concerned critic.
X-RFC2646:  Original
Now see what you,ve done Herc. I told you to stop sucking cane toads!
                                    Marty from Oz
   posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L
You guys are unbelievable.  Open your eyes this is year 1 UNI stats
problem.  It takes 20 mintues to detect a correlation.
This is Gambles 1st post in one of my threads.
Ah, the reverse pyramid scam.  Excellent use of it,
by the way.
--
 -john
***********************
IT EITHER :    REFLECTS HIS ALIAS  (just when he posts to me)
OR  IT DOESNT
GET IT YET?
Fair enough Davids post is cryptic since its really a double post with
Gamble before his name ties in so I'll skip Davids example (SEE I'LL BE
RICH ~ his 1st post to me) from now on.
Its very obvious very simple proof of macro quantum entanglement.
There's 20 CIA agents with me right now because of it, my permanent
24/7 buddies.  There's 10,000 people in Townsville Australia WHO KNOW
exactly what I'm saying, they all say... the internet still hasn't
caught on, 1000 mathematicians can't see a simple rhyming correlation
with every second reply to me.
Here's another..... use your brain on the puzzle, not to worm out of
it.  I've heard literally 1000 skeptics explain they're way out of even
investigating the data, and NOBODY look at the data.
how can I put this, I'm in bare feet....
A street (usenet) covered in broken glass has numerous houses
(newsgroups) and I run up to a house and start asking for help,
numerous expletives and I try the next house, numerous excuses and I
try the next, so I try the next, a big party full of people, and I'm
standing at the door shuffling around on broken glass, waving, waving,
day after day waving.
Who replied to me?
Ghost in the machine
William Elliot
Barb Knox
George Green
Gib Bogle
JUST WORK IT OUT.   NO MATHS HERE, NO THEORY, NO EXPERMENTAL PROTOCOL,
JUST A SIMPLE QUESTION
Herc
 Discussion, linux)
[one last response since Herc has made a touch more sense.]
Why didn't you post this in the beginning, you twit?  You would have
seemed somewhat more coherent (but not sane).
Certainly I didn't remember this post.
-- 
So why are mathematicians NOT what most people suppose?  Why are they
not these brilliant and wonderful people who act in favor of humanity
instead of against it? -- James S. Harris, on public confusion about
                                       mathematicians and superheroes.
In sci.logic, HeadRush
on Wed, 13 Apr 2005 18:43:11 +1000
Not necessarily.  I for one would suspect some sound
backwash from the nasal and/or mouth shaping areas to
filter down back into the general area of the larynx.
Not that that helps Herc any, though admittedly the only
knowledge I have regarding any sort of throat mike comes
from a work of fiction: A Life For The Stars, _Cities in
Flight_, James Blish.  Of course this was a device worn by
the user, presumably willingly, as opposed to some sort
of wallmounted infralaser scanner unit (which probably
wouldn't work all that well if the user is moving about,
as the scanner would have to deal with both the larynx and
the back of the user's neck -- or even the user's head,
if he tilts it in a certain way while, say, looking down
at the floor and/or reading a book).
(James Blish hypothesized that it would substantially modify
the consonants.)
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
   posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L
nope.  the nerves to the throat/mouth, the muscles they activate, the
surrounding air are all PERFECT CONDUCTORS for every thought at the
seat of our minds.  there is neglibible loss of sound quality.  you
know NASA has put mikes on throats and detected 20 words already, there
is effectively NO loss of signal from our minds to the surrounding air.
to the government we are ALL exactly like Fred Savage in The Wonder
Years (that was the 1st truman show about me and Nicole Zeller in the
6th grade).  Every show we see Fred and listen to every thought he
ponders.  THAT IS REAL.
IN FACT, its so sensitive it hears your SUBCONSCIOUS myriad of high
speed thoughts.  The governmnet can choke your own thoughts by
deciphering whole sentences, whole stories you're just thinking of in a
split second and READ OUT YOUR THOUGHTS TO YOU 1 SECOND BEFORE you
think it yourself.
IN FACT, they can tell you when you're about to fart, and they do, all
day.
You haven't heard it, you all think you're god's gift to mankind and
your own thoughts are sacred.  YOU ARE WRONG.  Stop believing in your
ancient history.  Give me ONE REASON why there would be ANY losses from
thought to sound.  Your speech is just LOUD thought, nothing more and
when you 'think to yourself' you have switched off the 'mind to voice'
switch but in fact its just a leaky transistor.
Herc
X-RFC2646:  Original
Isn't this just the maximum leaf spanning tree problem?
-- 
David Eppstein
Computer Science Dept., Univ. of California, Irvine
http://www.ics.uci.edu/~eppstein/
 Maxwell derivation
Feynman uses the Lagrangian also. So is the derivation of Feynman the same 
as
the derivation by Landau & Lifschitz so that basically one copied from the
other.
Second question, a tough one. Suppose we say that the force of gravity is a
fiction and replace it with a Cosmic Coulomb force where astro bodies move 
in
a bent space caused by the protons of a Atom Totality nucleus. Galaxies are
bits and pieces of the electrons where dots of the electron dot cloud are
galaxies. Question: can we nest a new term of a Cosmic Coulomb, just as
Maxwell added a new term to Ampere's Law, only a new term in each of the
Maxwell Equations of a Cosmic Coulomb. If we can do that, then those 
Maxwell
Equations would be non-derivable by any Lagrangian.
So if we say that the force of gravity was all along a part of the Maxwell
theory and incorporate the Cosmic Coulomb into the Maxwell Equations would 
it
not make the Maxwell Equations non-derivable.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
X-RFC2646:  Original
HI all,
I am reading a book about telegraph signal: derived from a Poisson process 
N(t): if the number in interval (0, t) is even, then X(t)=1, if the number 
is odd, then X(t)=-1...
It says that
P(X(t)=1)=exp(-lamda*t)*cosh(lamda*t)
P(X(t)=-1)=exp(-lamda*t)*sinh(lamda*t)
E(X(t))=exp(-2*lamda*t)
I agreed all the above ...
But here is my problem: For E(X(t1)X(t2)), I computed it as:
Assuming t1<t2,
E(X(t1)X(t2))=E(X^2(t1))+E(X(t1))*E(X(t2)-X(t1))
=1+exp(-2*lamda*t1)*exp(-2*lamda*(t2-t1))
=1+exp(-2*lamda*t2)
=1+exp(-2*lamda*max(t1, t2))
But the books says it is
E(X(t1)X(t2))=exp(-2*lamda*|t2-t1|),
Why?
Majorem Dei Gloriam For the Greater Glory of God. Kung wala ka sang
has anyone a idea how to simplify the following product in maple notation:
product((-exp(-2*I*Pi*k*n/m))^(I),k=1..m-1);
Christia Ruether
What's your definition of z^i, for a complex number z?
Jose Carlos Santos
   posting-account=JOpQKQ0AAACoSlSj3VatnSa2ToFlP3qq
Hi Everyone,
This is my first time posting to the group. I am a senior math major at
Reed College in Portland, OR in the United States. For my senior
thesis, I am attempting to write an extremely basic and easy
introduction to Lie groups via matrix groups. My goal is to create a
text that is simple enough and self-contained enough so that a talented
sophomore or junior math/physics major can understand it without any
real trouble. I believe that the majority of the books on Lie theory
are well beyond the reach of most undergraduate students. My hope is
that I can introduce people to the important basic ideas while assuming
knowledge of only basic linear algebra and calculusanalysis. My thesis
is influenced to a great degree by the excellent Lie theory books by
Wulf Rossmann and Brian Hall, but I have gone to much greater lengths
than these to insure that readers with less of a math background can
fully grasp almost everything.
I am now done with a draft of the first 3 chapters (about 90 pages),
and I am looking for people who would be willing to spend some time
reading what I have and making suggestions as to what could be made
clearer or easier, or a more elementary proof of some result I use. If
you are willing (or just curious as to what it looks like), please
email me (jeffrey.emanuel@gmail.com) or post a brief response, and I
will email you a ~1.5 MB pdf file. If you read it, you may very well
find the style long-winded and redundant, but try to imagine reading it
as an unsophisticated sophomore would.
The 3 chapters are mostly complete, although there is one missing
proof. The proof is the so-called Jordan decomposition, which says that
any matrix X can be written in the form X=S+N where SN=NS, N is
nilpotent, and S is semi-simple (i.e., diagonalizable). I have already
spent quite a while proving Schur's theorem, which states that any
matrix X is unitarily equivalent to a triangular matrix (X=PTP^-1, P
unitary, T triangular). This seems to get me part of the way there, and
I believe there is a way to get SN=NS and uniqueness without too much
additional work. This way has the benefit of leveraging previous
understanding. I am trying to find the easiest and most basic proof for
this. If I can't find another one, I will prove that S and N are
actually polynomials in X, which implies it immediately; however, this
proof would then be the most technical thing in the thesis, so I am
unwilling to do this. Any other suggestions would be most welcome.
The thesis is due on April 29th, so the sooner you can respond, the
more time I can spend implementing your suggestions. Anyone who gives
me significant help will of course be thanked in the acknowledgements.
Jeffrey Emanuel
 Lie groups?
I hope that, besides Rossmann's and Hall's books, you've also studied
Monthly 90, 600-624 (1983) and 91, 247 (1984)), as well as the books
Matrix groups by Morton L. Curtis and Matrix Groups: An Introduction
to Lie Group Theory by Andrew Baker.
Jose Carlos Santos
   posting-account=JOpQKQ0AAACoSlSj3VatnSa2ToFlP3qq
Mr. Santos,
I have actually been corresponging with Roger Howe on the thesis. Also,
I've avoided the Baker book after reading a pretty negative review of
it. I've never seen the Curtis book before, though, so I'll look for it.
On 16 Apr 2005 12:06:27 -0700, Jeffrey  Emanuel
I've know nothing about this area, but I've only seen negative
opinions from Open University students who were taking a course
(now discontinued) based on Curtis' book.
One of the students remarks:
Matrix Groups: An Introduction to Lie Group Theory, Andrew Baker.
This is work in progress, but the draft book can be downloaded from
http://www.maths.gla.ac.uk/~ajb/course-notes.html
The pdf file is about 1Mb in size and may be viewed and printed using
Acrobat or Ghostview. In his introduction the author acknowledges a debt to
Curtis' book, and the book covers much the same ground as our course. I've
printed out and skimmed through the chapter on exponentials and one-
parameter subgroups, and it's much clearer and more thorough than Curtis.
Another says:
Very Basic Lie Theory, Roger Howe, 24pp, American Mathematical Monthly,
November 1983 is mentioned in nearly all the books and papers on Lie theory
I have looked at, including our set book and course notes.
Now I have looked at it I can see why. It is very clear and well written,
and fills in a lot of the gaps in Curtis. It is particularly good on
one-parameter subgroups.
Another:
It looks as if we just have to accept that Curtis does only prove a couple
of special cases but uses the general result several times.
And finally:
Set book: Matrix Groups, ML Curtis, 210pp (typed, roughly equivalent to
100pp properly set):
Ultimately this is a disappointment. It is barely more than a set of
lecture notes, typed with hand annotations, and even the corrected
version has more typos and mathematical errors than I've ever seen in a
text book. It would be fine as an accompaniment to an ordinary university
lecture course, but doesn't really work as a self-study text. I think the
subject was poorly served with text books at our level when the course was
written, but there now seem to be several better modern books, and a
particularly good one is due to be published shortly.
I don't know for sure what book he was referring to, but elsewhere the
same student says:
Lectures on Lie Groups and Lie Algebras, Roger Carter, Graeme Segal and
Ian Macdonald, London Mathematical Society student texts, CUP, 
£18
This book consists of three separate but related sections - Lie Algebras
and root systems, Lie Groups, and Linear Algebraic Groups - which the
authors have developed from their lecture courses given at the LMS
conference on Lie theory in 1993. All write very clearly, and start at our
level, but they do take their subjects quite a lot further. It's cleared up
quite a few things I didn't understand in Curtis.
(That book was published in 1995, so it can hardly be the one he meant.)
-- 
Angus Rodgers
(angus_prune@ eats spam; reply to angusrod@)
Contains mild peril
Pls help me summarize the history of firiday the 13th. .
Don't do it.
   http://users.pandora.be/vdmoortel/dirk/Stuff/Superstition.html
Dirk Vdm
Helo. . .
Can i ask 10 pythatgorean tripples..
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
   posting-account=SAlCkQwAAADOQfb8GuNFkWcQtC01OYCg
In fact there are infinitely many Pythagorean triples - see (11) in
http://mathworld.wolfram.com/PythagoreanTriple.html
Yes, you can.
Jose Carlos Santos
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
Ask, and you (shall or shall not) receive:
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
3,4,5
Matt
X-RFC2646:  Original
I have a problem and I hope someone can help me to continue my work : )
 I have to represent an extension field F(3^6m) as follow:
F(3^6m)=F(3^m)[x] / (x^6+ x + 2)
How can I do it with NTL?
ZZ_p::init(to_ZZ(3));
ZZ_pX Polyn;
BuildIrred(Polyn,to_ZZ(7));
ZZ_pE::init(Polyn);                      //  I'm in F(3^7)
ZZ_pEX Polyn2;
 SetCoeff(Polyn2,0,2);
 SetCoeff(Polyn2,1,1);
 SetCoeff(Polyn2,6,1);
// Now I would use this Monic and irriducible poly as my generator. how?
I tried with:
ZZ_pE::init(Polyn2);
but it's wrong :/
Hope to receive an answer
Ty very much
Claudio
gswork@mailcity.com says...
Well, that may well be a good approach to some issues/problems.  On the 
other hand, a unified concept of information/data may be better for 
others.  I tend to lean towards the latter but maybe that's just a 
function of the things I have been thinking about.  
Yes, that seems to encapsulate one difference - although maybe not the 
essential difference, as the above is largely a matter of encryption.
 
Hence my belief that in a many-worlds 'multiverse' all information must 
have identical status.
- Gerry Quinn
Hi all,
I'm trying to create a computer algorithm to solve a 15 puzzle which
should be able to find the solution with the least amount of moves
given a certain amount of memory and time.
(An example of a 15 puzzle)
01 02 03 04
05 06 07 08
09 10 11 12
13 14 15
http://en.wikipedia.org/wiki/15_puzzle
I know this is NP problem, But i'd like to know if there is a method
to estimate (on average) how many moves it takes to solve a shuffled
puzzle. Or if there is data available from people who have solved the
(near) best path for some shuffled puzzles. Basically i am looking for
a number to compare my results with to see if my algorithm is any
good.
Rick
I just did some web searching and found this:
1999: The parallel search bench zram and its applications
by A. Br.9fngger, A. Marzetta, K. Fukuda, and J. Nievergelt
Available here:
http://www.ifor.math.ethz.ch/publications/oldpublications
They proved that the worst cases are 80 moves from solved.
Jaap
to your posting sooner.  I only just noticed that someone had answered.
I don't know what incidence relations are.
The 3DLDF language is modelled on Donald Knuth's METAFONT language.
In 3DLDF, polygons are represented as `points' connected by
connectors.
The `points are represented as homogeneous coordinates while
the connectors are  represented as `strings'.
In 3DLDF, `polygons' are a kind of `path'.  For example,
this is a portion from the output produced by the command
`show unit_pentagon;' that shows the points (without the
w-coordinate) and the connectors:
(0.000000, 0.000000, 1.000000) -- (-0.951057, 0.000000, 0.309017) --
(-0.587785, 0.000000, -0.809017) -- (0.587785, 0.000000, -0.809017) --
(0.951057, 0.000000, 0.309017) -- cycle;
A polyhedron is represented as a set of polygons, so that
the vertices are duplicated.  Aside from the duplication
itself, the problem is that the
coordinates of the points representing the same vertex
aren't the same, because I use the trigonometric functions
for rotating the polygons into place, and the precision used
for representing floating point numbers isn't high enough to
ensure that the resulting coordinates are the same, given
different rotations.
On the other hand, I've since discovered that it may be an
advantage to have multiple polygons for use in certain
operations on the polyhedra.  For example, I'm currently
working on surface hiding (or occlusion) and I'm interested
in the figures that result from bisecting a polyhedron using
a plane.
question, I've been given some references to books on
polyhedra.  I think the ones by Coxeter will be particularly
helpful.  I'm currently working on occlusion of plane
figures, so it will be awhile before I get around to
changing the way polyhedra are generated.
Laurence Finston
Daryl S. Kabatoff, Certified Usenet Kook (TM) of the 33rd Order, with
Tinfoil Stars and Banana Tree Clusters, has returned to Usenet after an all
too-short absence.
Snipping the numerology crap, those who aren't familiar with this loser 
need
to read the paragraph below to get an idea what you're dealing with if you
ever let this wackjob get hold of your personal e-mail address.
FULL CANVAS JACKET - Noteworthy Unhinged Lunatic Rants
http://www.ratbags.com/ranters/kabatoff010304.htm
Usenet has the tendency to provide a public forum for those who would
normally be scribbling in a closet. For example, take Daryl Shawn
Kabatoff. For the last few years, he's methodically gathered statistics
from various sources, ranging from local newspaper obituary pages to the
food court of the Saskatoon Midtown Plaza mall. With all the raw data he's
collected, he's attempting to prove daily that our full names are in
mathematical harmony with our birthdays.
His rants normally focus on a single individual he's met or read about,
starting with calculations related to their birthdate and full names,
blending in whatever other personal information about their family members,
descends into a mix of numerology, religious zealotry, and personal 
torment.
I've never seen anything like it.
With all the prime numbers, Fibonacci sequences and biblical references,
it's like reading the notebooks of Maximillian Cohen and John Nash 
combined.
Unsurprisingly, several posts unfold to reveal a history of painful mental
illness. If you have some time, take a look. I've detailed his posting
history and a several sample posts below.
http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml
Busker's mom wants police to pick up son
Man ejected from Fringe 'very ill'
Sarah Gibb
Special to The StarPhoenix
The mother of a busker ejected from the Fringe Festival for posting 
personal
information of Fringe visitors online has appealed to the police and 
doctors
to have her son picked up because he is very ill.
Daryl Shawn Kabatoff has recently posted angry Internet messages, saying he
wants local children to die and has revealed that he's trying to obtain
guns, the StarPhoenix has learned. He is now under police investigation,
said Insp. Jeff Bent, spokesperson for Saskatoon Police Service. Kabatoff's
posts, which are full of references to religion, sex and death, contain the
names, birthdays, descriptions and workplaces of dozens of Saskatoon
residents and their families, including in some instances their deceased
children. He obtained some of the information from children's obituaries
published in the StarPhoenix. The busker says in some posts that he hopes
the other children in the families die too.
Kabatoff, who sat on a Mexican blanket at the intersection of 11th Street
and Broadway during the Fringe, recently offered on a bulletin board to
trade one of his trademark numerological name analyses for a rifle or
shotgun. In another post, he says he is acquiring a firearms licence. In
response to the contents of Kabatoff's posts, obtained by the StarPhoenix,
Insp. Bent said: (We) are going to assign some investigators to this
immediately . . . I appreciate your call because after the fact people say,
'Why wasn't something done?' 
The balding, pony-tailed 46-year-old man, who stood for mayor in 1982 and
1985, is a well-known figure in Saskatoon's shopping malls and coffee 
shops.
Kabatoff believes people's names are given to them by God and are 
harmonized
with their dates of birth. He says he can reveal this relationship using
mathematics and for years has asked passers-by to let him analyse their
names. An online search has revealed Kabatoff has posted around 1,000 
highly
disturbing messages on Usenet sites since 2001. Usenet is a system of
worldwide bulletin boards, also called newsgroups, where people post
messages and correspond about shared interests. Kabatoff's messages contain
the personal details of scores of Saskatoon residents, often teenage girls
he refers to as nubile sweeties after meeting them in the food court at
Midtown Plaza, the Mall at Lawson Heights, the Broadway Roastery and
Starbucks.
Kabatoff is notorious in the Usenet community. One site calls him one of
Usenet's most elusive and creepy psychopaths. Another user asks: At 
this
stage, what can be done under law to stop him getting a gun from any
source? Several Usenet members implore someone to alert Saskatoon police 
to
S. Kabatoff using a Shaw e-mail address and Saskatoon box number, states
that he is looking for: A) a rifle for deer hunting this fall, B) a 
shotgun
for the early deer-hunting season, C) a .22-caliber (sic) rifle for
improving my shooting skills during the summer. The post, entitled 
Wanting
firearm and sent to the calgary.forsale newsgroup, continues: I'll 
offer
you a trade . . . I will show you gorgeous harmony between your name and
birthday and numbers in the Bible . . . In exchange for your unwanted
firearm I'll provide you with a short essay that you will treasure far
more. The next day, June 16, the same post appeared on the
talk.politics.guns newsgroup, this time called Trade for Gun.
Usenet post from him states: I am in the process of applying for and
obtaining a license (sic) to acquire a rifle . . . for hunting animals and
not people. If somebody . . . trades a gun for my work before I obtain the
license (sic), that gun will be received by a friend who is legally allowed
to possess such . . . 
Attempts to reach Kabatoff for comment were unsuccessful. [NO , MS.
GIBB. HOW DO YOU COMMUNICATE WITH SOMEBODY WHO'S IN ORBIT SOMEWHERE BETWEEN
URANUS AND PLUTO??? - SdSD]
In a telephone interview Monday, Kabatoff's mother, who asked not to be
named, sobbed as she begged doctors and the police for help.He is not 
being
treated at all. The doctors say they can't force him . . . When he is well,
he is a likeable person (who) is so nice. But when he's ill, he's very ill.
He needs . . . the police or somebody to take him in for treatment. He 
did
go to a hospital a number of times. He just absolutely refuses the
treatment. Shall I pack up and leave or put my head in the sand or what,
because I just can't take any of this anymore. I wish he would just be
picked up. She added that she does not believe her son is dangerous.
Kabatoff's long and angry posts contain names, birthdays, physical
descriptions of people, where they work and the names and birthdays of 
their
family members. The details are freely given to Kabatoff by his 
unsuspecting
customers when he offers to analyse their names. His most recent post, 
dated
Aug. 8, names a 24-year-old woman he met on the first day of this year's
Fringe Festival. He refers to her and her friends as cheap, filthy
assholes. Kabatoff also trawls through obituaries, mainly in the
StarPhoenix, looking for details of teenagers who have recently died. He
posts the information on bulletin boards along with his analysis of the
deceased's name, sometimes adding that he is glad the young person has died
and that he hopes the family loses their other children too. He claims to
attend the funerals of these teenagers so he can meet nubile sweeties 
and
eat free food.
One of his posts reads: (S)ince nubile sweeties are found in greatest
abundance at the funerals of high school students, then it is the funerals
of high school students that make the very, very best funerals, especially
if there is food . . . I stuff my face (and my pockets) with all the good
food and look at all the pretty nubile sweeties and have the time of my 
life
. . .
Five Corners on Broadway Avenue. (Name withheld) stopped to provide stats .
. . He said he had no money for me . . . You people have absolutely zero
respect for me . . . I in turn have zero respect for you and so don't
require your permission to post these stats on the Usenet. Furthermore I am
on my knees and begging God to . . . terminate the lives of these children.
If I find any of these family members in the obituaries I will cheer with
utter glee . . . and I will post your stats again.
Kabatoff appears to keep an elaborate cross-referencing system on his 
laptop
computer of the families whose stats he obtains, so he can compare the
names with published obituaries. A recent post based on a StarPhoenix
obituary begins: Oh dear. (Name withheld) hung herself in her basement
bedroom at her parent's home on (date withheld). Her brother (name 
withheld)
provided stats on May 9th 2002, he hangs around Broadway Avenue, the street
in Saskatoon with the poster poles shaped like penises. I first met him
about 10 years ago. The post ends with Kabatoff calling the family and 
the
rest of the community, the God-damned  of the earth.
Some of Kabatoff's most disturbing posts are based on the obituaries of
children. Several of the families he has targeted lost their children at
birth. His posts mock the parents and their grief. A message dated July 6,
crying out to God against your family and am requesting that He . . .
terminates the lives of your other children . . . If I see any other (name
withheld) in the obituaries I will cheer with utter glee and post your 
stats
again. A post to the parents of a 15-year-old boy killed in an accident
reads: Well now I  on you, I rejoice in your loss of (name withheld)
and pray for the loss of (your daughter).
To the parents of another young Saskatoon resident who died, Kabatoff
could have supported me overseas for months.
Kabatoff, who claims to have worked in the 1980s as a social worker in
Whitecourt, Alta., and says he has a degree in anthropology, was reported 
to
police during this year's Fringe Festival by a woman who gave him her
personal details before being warned by passers-by that Kabatoff would post
them. Nancy (not her real name) told the StarPhoenix: This man is preying
on people. The public needs to be warned. After the StarPhoenix ran a 
story
Saturday about Nancy's experience, Kabatoff's Fringe busking licence was
withdrawn. Readers wanting to check whether Kabatoff has posted their names
you your name, then click on groups.
Kabatoff himself moderates a newsgroup at alt.dear.whitehouse. His mother
says she has lived with this turmoil for years. She asks for understanding.
Heaven help you and your children if they ever have mental health
problems, she said.
Remember when you ran away
And I got on my knees
And begged you not to leave
Because I'd go berserk
Well you left me anyhow
And then the days got worse and worse
And now you see I've gone
Completely out of my mind
And they're coming to take me away ha-haaa
They're coming to take me away ho ho hee hee ha haaa
To the funny farm
Where life is beautiful all the time
And I'll be happy to see those nice young men
In their clean white coats
And they're coming to take me away ha haaa
You thought it was a joke
And so you laughed
You laughed when I said
That losing you would make me flip my lid
Right? You know you laughed
I heard you laugh. You laughed
You laughed and laughed and then you left
But now you know I'm utterly mad
And they're coming to take me away ha haaa
They're coming to take me away ho ho hee hee ha haaa
To the happy home with trees and flowers and chirping birds
And basket weavers who sit and smile and twiddle their thumbs and toes
And they're coming to take me away ha haaa
I cooked your food
I cleaned your house
And this is how you pay me back
For all my kind unselfish, loving deeds
Ha! Well you just wait
They'll find you yet and when they do
They'll put you in the A.S.P.C.A.
You mangy mutt
And they're coming to take me away ha haaa
They're coming to take me away ha haaa ho ho hee hee
To the funny farm where life is beautiful all the time
And I'll be happy to see those nice young men
In their clean white coats
And they're coming to take me away
To the happy home with trees and flowers and chirping birds
And basket weavers who sit and smile and twiddle their thumbs and toes
And they're coming to take me away ha
haaa!http://www.twin-music.com/azlyrics/n_file/napoleon.html
   posting-account=SAlCkQwAAADOQfb8GuNFkWcQtC01OYCg
(or
and 3)
The most simple algorithm I can think about is this (I think it should
work):
First we find out which elements are equal. This is the same as to find
2-connected components of a digraph - there are some known algorithms
for this.
Then we identify all vertices which are in the same component to one
point. We can use the usal topological sort on the modified graph.
Martin
X-RFC2646:  Original
There are infinite vector spaces. But is it corret to say that there exist 
only 2 type of v.s. (i.e. those which are direct sum of Q copies and those 
which are direct sum of Z_p copies) ?
Yes. If V is a vector space over some field k, let k' be the smallest
subfield of k. Then k' is isomorphic either to Q or to some Z_p. In the
first case, V is a Q-vector space and in the second case V is a
Z_p-vector space. In each case, V can be written as a direct sum of
subspaces isomorphic to k'. Notice that I'm using the axiom of choice
here.
Jose Carlos Santos
   posting-account=n-_z4Q0AAABVC_ZJcXR_vWidvNMIMsX0
I am so darn grateful that my God is so darn different than yours. Live
and let live.
If hes different, how do you know what you believe is right?
B
   posting-account=n-_z4Q0AAABVC_ZJcXR_vWidvNMIMsX0
the eternal zzzzzzzz.....
   posting-account=n-_z4Q0AAABVC_ZJcXR_vWidvNMIMsX0
the eternal zzzzzzzz.....
X-CompuServe-Customer: Yes
X-Coriate: interspeed.co.nz
X-Ecrate: tanandtanlawyers.com
X-Pose: George Cox
X-Punge: Micro$oft
X-Sanguinate: The MVS Guy
X-Terminate: SPA(GIS)
X-Tinguish: Mark Griffith
X-Treme: C&C,DWS
04/15/2005
You can't. The best that you can do is to find examples of how
Mathematics changed the world in the past and ask the students why
they believe it won't happen again.
BTW, what is the practical importance of Art? Baseball? Music?
Theatre?
-- 
Unsolicited bulk E-mail subject to legal action.  I reserve the
right to publicly post or ridicule any abusive E-mail.  Reply to
domain Patriot dot net user shmuel+news to contact me.  Do not
reply to spamtrap@library.lspace.org
Commercial art, professional baseball, popular music and movie theaters
are very important money makers.  Other forms of art, music and theater
can go begging.  Money is now the measure of all things human and divine
as dictated by corporate social engineering of the last twenty years.
Studies showing inclusion of art and music classes in schools improves
students' test scores aren't well liked by educational theorists.
assume a1*a2*...*an=1
does there exist b1,b2,...,bn  such that:
a1(b1/b2)=1
a2(b2/b3)=1
    .
    .
    .
an(bn/b1)=1
if exist, how to find these bi (i=1~n) ?
--
   posting-account=lnE5XAwAAACI22BecqT-BfkBhz0ffSRp
Not only do such bi exist, there are an infinite number of sets.
proof of if there is one solution, there are infinitely many:
assume b1,b2,...,bn is a solution. Since the constraining equations
(non-zero) value, x, then you are simply multiplying the numerator and
denominator of a fraction by the same value, leaving it unchanged.
Thus, x*b1,x*b2,...,x*bn is another solution.
Proof that a solution exists:
a1 (b1/b2)=1, so b1=b2/a1
similarly, bk=b(k+1)/ak, k<n and bn=b1/an
Chose b1 arbitary.
Then, b(n-1)=bn/a(n-1)=b1/(an*a(n-1))
b(n-2)=b(n-1)/a(n-2)=b1/(an*a(n-1)*a(n-2))
.
.
.
b2=b3/a2=b1/(an*a(n-1)*...*a3*a2
b1=b2/b1=b1/(an*a(n-1)*...*a3*a2*a1)=b1
Thus, any b1 you choose (except, of course, b1=0) will work, and that
will force the values of all the other bi. The fact that b1=b1 in the
last equation shows that the equations are all consistant.
example, with n=4:
let a1=1/4, a2 = 1/3, a3=2, a4=6
choose b1 = x
b1=b2/a1, so b2=b1*a1=x/4
b2=b3/a2, so b3= b2*a2=x/12
b3=b4/a3, so b4=b3*a3=x/6
b1/b2=x/(x/4)=4=1/a1
b2/b3=(x/4)/(x/12)=3=1/a2
b3/b4=(x/12)/(a/6)=1/2=1/a3
b4/b1=(x/6)/x=1/6=1/a4
So, everything checks out.
   posting-account=9wVPIwwAAAAonf5Dj39AQaTL2sJYvErF
The title of the paper is A Mechanical Proof of the Unsolvability of
the Halting Problem by Robert S. Boyer and J. Strother Moore.  The
URL is http://www.cs.utexas.edu/users/boyer/unsolv.pdf
It begins, We describe a proof by a computer program of the
unsolvability of the halting problem.
The word proof is defined in dictionaries of Mathematics as
listed below.  In general it is a deductive, rigorous, compelling
demonstration (mathematical argument) that some statement is true.  In
the context of the above, a mechanical proof of the unsolvability of
the Halting Problem would be:
A deductive, rigorous, compelling demonstration (mathematical argument)
that the Halting Problem is unsolvable, created by a computer program.
I ask the question:  (a) What is the compelling argument that Boyer and
Moore have produced that demonstrates that this theorem is true?  (b)
Where in this paper or elsewhere is it presented?
Moreover, such a compelling argument would have to have been created by
a computer program - yet nothing in this paper is output produced by
any computer program.
My answers: (a) They have not presented any compelling argument that
demonstrates that the theorem is true.  (b) There is no location
because no such argument has been presented.
C-B
Definitions of Proof
1. a deductive demonstration of a mathematical statement
2. a set of arguments used to deduce a mathematical theorem from a set
of axioms
3. a chain of reasoning using rules of inference, ultimately based on a
set of axioms, that leads to a conclusion.
4. a sequence of statements that show a particular theorem to be true
5. a rigorous mathematical argument which unequivocally demonstrates
the truth of a given proposition
6. a valid argument should be compelling in the sense that it is
capable of convincing someone about the truth of the conclusion
7. a rigorous compelling argument . . . a demonstration that given
certain axioms, some statement of interest is necessarily true
References
1. McGraw-Hill Dictionary of Mathematics
2. NTC Dictionary of Mathematics
3. Penguin Dictionary of Mathematics
4. Barron's Dictionary of Mathematical Terms
5. http://www.mathworld.com/
6. http://www.wikipedia.org/
7. http://encyclopedia.thefreedictionary.com
8. ALSO: http://www.wordiq.com/definition/True
9. ALSO: http://www.planetmath.org
10. ALSO: http://www.mathacademy.com
Suppose such a map exists. Then it can't be surjective on the first
homology group.  So it lifts to a cover. Etc.
Yes, that's what I persisted in not seeing.
To fill in the etc. part for the OP, the lifting to a cover
X' then provides a map of positive degree to a surface of genus
        For a d-fold cover of a surface of genus g, the Euler
        characteristic is
        d*(2-2g) =  2 - 2(d(g-1)+1)
        so the surface has genus d(g-1)+1, or (g=2): d+1.
        Degree is multiplicative wrt composition of maps, so
        if the total degree is 2, one must have a degree 1
        map to a surface of genus 3.
Since the rank of H_1 is 2g, we see that the cover X' has first
Betti number (here, it's 6) greater than that of X (which is 4).
However, for a map of positive degree to exist, the Betti numbers
of the domain must be no less than those of the codomain. This
gives the contradiction.
-- 
Dale.
   posting-account=9wVPIwwAAAAonf5Dj39AQaTL2sJYvErF
 . . .
If we give an example of a Y that X produces, then we have proven that
X is capable of producing a Y.  If we don't give an example, only
stupid, na.95ve and overly-emotional people don't realize that we have
operator and Modus Tollens.)
C-B
Repetition is not going to make this any less ridiculous.
************************
David C. Ullrich
   posting-account=jEbKGQ0AAADZF1UpkDsHa5gkWBqABnUE
that
Stupid prattler.
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
The factoring problem looks tractable from an approch in the set of
rationals using a transform that I call the surrogate factoring
theorem.
How it works is very simple despite all the arguing that's been taking
place lately:
Given f_1, f_2, g_1 and g_2 rationals, where
f_1 f_2 = M^2 T, and g_1 g_2 = j^2 T
where M and j are non-zero integers, and
T = M^2 - j^2
the surrogate factoring theorem supplies a link between f_1, f_2 and
g_1, g_2, so you have something like
{f_1, f_2} : SF : {g_1 , g_2}
which means that given f_1 and f_2, you have values for g_1 and g_2,
linked by the theorem.
It's a one-to-one linking between infinite sets, and it has relevance
to factoring because even though trivially in rationals every number
but 0 is a factor of every other number, you can focus on the numerator
of rationals as that is an integer, and check to see if it's a trivial
or non-trivial factor.
I've used M for the number you wish to factor, but the mathematics
allows you to go the other way as well, so you can use M or j to check.
That's extraordinary because I have a way to make factoring meaningful
in the set of rationals, as there is no indication that the link is
picky, and in fact it cannot be.
That's because you also have
{g_1, g_2} : SF : {f_1 , f_2}
as the order is not important.  In terms of using the SF theorem,
switching the order changes the sign of T.
So, since the full set of factors of j^2 T and M^2 T are linked, you
have both trivial and non-trivial factors.
Some posters have argued rather obsessively that you tend to get
trivial factors, but how?
Given that there are an equal number of trivial and non-trivial
factors--an infinite number of each--how could you logically tend to
get only trivial factors?
My assessment of the situation is that because the political and social
implications of the mathematics is rather huge, posters are not
bothering with logic and mathematics, but simply taking a social
position.
Some people might still not quite understand what I mean by trivial and
non-trivial factors from the set of rationals, so I'll explain more.
Noting that in the set of rationals every member but 0 is a factor of
every other member, it seems natural to call ALL factors there trivial
factors.
But the reality is that you can look at a fraction, like
7(3)/2
which is a factor of 15, and get that 3 is a non-trivial factor of 15
by focusing on the numerator.
That hardly matters though if you don't have a transform like the SF
theorem because how would you pick rationals to get non-trivial
factors?
It seems reasonable to suppose that the extra is unnecessary, and that
it's better to just work in integers.
But that's wrong.
You can instead go to the set of rationals, with a transform, and with
factors linked, it is meaningful to consider factors that can be
fractions, as you can still separate them into trivial and non-trivial
factors.
So despite its simplicity, with mathematicians focused on integers,
they would never find it, unless they did like me and stepped into
rationals, which I did from the outset with the quadratic
yx^2 + Ax - M = 0
which is what I started with, back in December of last year, and the
derivation of the transform now begins with
yz^2 - Az + j^2 = 0
and also uses
yx^2 - Ax + M^2 = 0
which is the value of basic research.  I started with a generic
quadratic that looked one way, and the research forced slightly
different forms.
I started with a non-monic quadratic which necessarily has solutions
that tend to not be integers, and went from there, while someone
trained to believe that you can't do meaningful factoring in rationals
might never have tried such a basic approach.
That's the power of training and education to block knowledge.
But learning modern problem solving techniques is a way for training
and education to open up areas that were blocked.
You need to learn modern problem solving theory.
That's a must for any serious researcher as otherwise you have these
blindspots that can keep even rather basic results from being
known--just from what you've been taught.
James Harris
 Discussion, linux)
Really, James, these counterarguments are getting a bit pathetic.
If you think that the method gives non-trivial factors reasonably
often, then there's an easy way to prove this claim.  Show us how the
method works on a few numbers.
Instead, you've argued that either the method works or there's a
reason it doesn't.  Since no one has given a reason it doesn't work,
we may conclude that it works.
It's not the kind of airtight argument that sparks a revolution.
-- 
But he himself was not to blame for his vices. They grew out of a 
personal
defect in his mother. She did her best in the way of flogging him while an
infant... but, poor woman! she had the misfortune to be left-handed, and a
child flogged left-handedly had better be left unflogged. -- E.A. Poe
Then why don't you actually solve it?
So your method gives an infinite amount of rational numbers. Let's call
r_1, r_2, r_3, ... to the rational numbers that you get and let b_1,
b_2, b_3, ... be their numerators. For each numerator, you check whether
or not it is a trivial factor, right?!
Just one question, if you don't mind: what makes you think that there's
some advantage in checking with the natural numbers b_1, b_2, b_3, ...
instead of 1, 2, 3, ... ?
Which mathematical problem have you actually solved?
Jose Carlos Santos
...to whom?...
The arguing has been primarily about *whether* it works. The best way to
resolve that is to actually solve a specific factoring problem with it,
showing *how* it works.
--
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
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
You mean, that given a number n, you can then choose a number p, and n
readily factors into p*(n/p), and you can then just check whether n/p
is an integer?  Something like that, only lots more complicated?
Obviously there is only a finite number of nontrivial factors, those
that lead to an integral factorization.
That sounds like you are not bothering with logic and mathematics, but
simply taking a social position.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
X-RFC2646:  Original
[David Kastrup, to JSH]
Yup.
...
[JSH]
Alas, would that it were so.  Given n, a composite integer to be factored,
every non-zero rational i/j is a factor of n (in the rationals, of course),
and the set of rationals i/j such that 1 < gcd(i, n) < n is in fact
countably infinite.  That's the set of non-trivial factors in 
JSH-speak.
The set of i/j for which gcd(i, n) is 1 or n is also countably infinite, 
and
is the set of trivial factors in JSH-speak.  James has talked himself 
into
believing (or pretends to -- who really knows?) that because those sets 
have
the same cardinality, there's a 50% chance of succeeding on each try.
Of course that's the same as this argument:  the set of integers i such 
that
1 < gcd(i, n) < n is countably infinite, and the set of integers for which
that doesn't hold is also countably infinite.  Therefore we have a 50%
chance of finding a non-trivial integer factor by picking a gcd candidate 
at
random directly.
Why he might think only one of those arguments is correct is baffling.  But
given that's the corner he painted himself into, it's not surprising that 
he
thinks the overly complicated version is the correct one (he does a lot of
that -- perhaps layering complications gives more room for wishful hope).
I certainly don't object to the notion that James can whip up something
that actually works from time to time. After all, his prime counting
algorithm does get correct results. But it's difficult to determine
exactly what he is getting at about 99% of the time because of his
sloppy language and militant hostility to defining his terms. Frankly, I
even doubt that he would acknowledge your formal characterization of
'non-trivial rational factor'  above -- although it seems to fit his
usage.
However, if you have the time and inclination to post a worked example
using his SF method with an explanation of how you interpret  the
ambiguous points, it would serve as a touchstone for others to find out
1) whether the method does work, 2) whether the method can be made to
work by suitable modifications, 3) whether the method can be established
by rigorous proof, 4) whether it compares well with existing methods.
James will certainly do none of these.
Care to try? It might blow away some of the fog.
--
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
X-RFC2646:  Original
...
[JSH]
...
[Proginoskes]
Does this refer to the oblique hints in earlier posts that you may have 
found a way to get something useful out of an SF variation, but didn't want 

to say how until you had more time to investigate?
before a week passed.  If it's not too much bother (and I'll sure understand 

if it is), I'd like to hear what you had in mind, and how that turned out. 
Besides just sating my curiousity, the educational value of these threads 
for non-participants would (I think) benefit from an example of negative 
results obtained & accepted dispassionately.
... [much good advice and explanation snipped] ...
LOL!  It's amazing to what levels we eventually sink when trying to have a 
discussion with James.  You can rest assured that the _point_ of the above 
...
The current version of that post goes on to explain what open means:
    However, there are people who are primarly from the newsgroup
    sci.math who work diligently to confuse others about my work,
    and a few have come over to make posts, and I've deleted their
    posts and banned them.
    I will continue to do so, as these people have proven to me that
    rational discourse is what they're fighting against.
    ...
    Some posters from the sci.math newsgroup will simply not be given
    a chance here as I've had enough time to see what they're about,
    so if they come here and post, they will simply be banned and
    their posts deleted off as a matter of policy.
According to the above, he isn't the only poster -- he's just the only 
poster not all of whose posts he deleted.  All James All The Time is the 
one sure path to Truth, I guess.
...
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
Yes, it does. I'm planning on writing up a thread where I take JSH's
ideas and formalize them mathematically, then show what it is about SF
which may save the idea itself.
It's not so much that I got negative results; it's that if I were to
plow ahead, it would become incredibly messy. Again, by formalizing
JSH's ideas, maybe someone can understand what I've meant when I said
I found the KEY, and then they can possibly come up with new ideas
and continue formally.
Actually, that last part should say then that 1 is a factor of N.
Yes, and if he would take one math class that expresses the ideas of
proof (i.e., a logic class, or even check out a book about mathematical
proof), then he could build a bridge between what he's trying to do,
and the rest of the world.
His biggest fault is that he simply doesn't understand the methods or
terminology. If he could be humble for a moment and admit to himself
that he's in over his head, that would help a lot. He could read a
book, try to prove some of the exercises in the book, and post his
solutions on Usenet to see if he's getting it. Sure, some people may be
out to get him, but most would probably be glad to see him trying to
understand what needs to be done and help him along.
Someone said in one of these posts that JSH uses words like rational
factors because they sound good ... Maybe the analogy would be a
five-year old, who can write letters of the alphabet, but can't spell,
can recognize the output, but there's no rhyme or reason.
... which goes to show that JSH can't discuss his Surrogate Factoring
idea rationally. By that logic, he should ban himself from his group.
Of course, he doesn't understand the irony of this ...
True. I jumped the gun on that one. (James, that's two mistakes I've
made and admitted to, in the past couple of days! See, it's easy to
do!)
It's like the Catholic church banning books that criticize Christianity
/ Catholicism, because they might confuse the devout followers.
Actually, they were originally called newgroups, as in new
misinformation. 8-)
     --- Christopher Heckman
X-RFC2646:  Original
[JSH]
That's not the problem.  The problem is that your theorem doesn't appear 
to have more _content_ than, say, this theorem:  given rational r, 2*r 
is 
also rational.  You do a lot more algebra in yours, but the relationship you 

derive in the end appears to be no more interesting for factoring than that 

2*r is rational whenever r is rational.
No, I was advising you to _test_ your claims.  You've repeatedly made a 
mathematical claim that the chance of finding a non-trivial factor via your 

theorem is about 50%.  That claim is false.  For whatever reason, you've 
been unable to understand the many explanations of this people have given 
you (my _guess_ is that you're consumed by wishful thinking here again, but 

it doesn't really matter why -- false is false).
The simple, obvious, and honorable approach for you then is to _try_ it. 
Then you'll see, with your own eyes, that it isn't true -- or you can hope 
that you'll find evidence that it is true.  Either way you would make actual 

progress.
The aim of that isn't to be practical, it's to help you to see that one 
of 
your pure math claims is false.  Most others here already know (or would 
bet a great deal on it, and reasonably so) that it's false, which is why 
nobody else here is going to do this part of your work for you this time 
around.
My concern remains that _your_ key mathematical claim (indeed, the only 
claim of any conceivable real interest here) is wrong.  Your key claim in my 

eyes is the one about the probability of finding a non-trivial factor; an 
algebraic relation between rational factors is of no interest at all to 
me 
(neither on pure math nor practical grounds) unless an interesting 
(on 
either pure math or practical grounds) probability proof follows 
from 
it. 
X-RFC2646:  Original
[JSH]
[Tim Peters]
[JSH]
It is an easy point, so try a bit harder indeed.  I'd be happy (so far as 
this little example goes) had you, for example, written this instead:
    For every trivial rational factor, you can either multiply it by
    a prime factor of M or divide a prime factor of M off, and you get a
    non-trivial factor -- or, if you don't, divide M by the result of
    that, and then you'll get a non-trivial factor.
original statement.  If you had written this instead, then it would be true 

that p^2 q^2 wouldn't work as a counterexample.  It remains a counterexample 

to what you did write.
Now look at Christopher's original example: a p^2 q^2 / b, where a and b are 

relatively prime to both p and q.  You can divide that by q to get (a p^2 
q)/b.  That's useless.  We can use your new strategy and divide M by that 
too, but the result is still a trivial factor:  b/(a p).
I see that in a later message, you made up yet another rule:
    However those are not useless factors, since you can jsut divide
    off powers of pq and get a non-trivial factor.
That wasn't mentioned in your original statement either -- but still isn't 
enough to cover all cases.
Is it hard to repair this?  Of course not.  Will you repair it?  If I were 
you I wouldn't bother -- as I said the first time, it's a minor point (its 
only use was to support the idea that there's a bijection between trivial 
and non-trivial factors, but that's dead to easy to show without this). 
X-RFC2646:  Original
If w is a (k-1)-form on a k-dimensional compact manifold, how can I prove 
that the integral over M of dw is 0.  Give an counterexample if M is not 
compact?
James 
Using the divergence theorem? (Or Stokes' theorem, or whatever
the differential-forms version is called...) What is the boundary
of M?
************************
David C. Ullrich
X-RFC2646:  Original
(a) Show that Q (rationals) is neither open nor closed in R.
(b) Find the interior, the exterior, the boundary, and the closure of Q.
For (a) is seems clear that is is not open, since if you take any open ball 

B(p, r) there will be some non rational numbers in that ball.  How to prove 

it though?
For (b) the interior is the union of all open sets contained in Q.  But, 
from (a) whenever we make an open ball, we get some non-rational numbers in 

that ball--so we will have no open sets.  So the interior is the empty set ? 

Similar argument to show the exterior is the empty set.  Thus the boundary 
is Q.  The closure of Q (intersection of closed sets in Q) is the empty 
set?
yes
yes
no
no
For each large enough natural number n, p + sqrt(2)/n belongs to B(p,r).
Yes, but i wouldn't say that we will have no open sets. I'd say that
we will have no open subsets of R contained in Q.
Correct.
Wrong. The closure of Q is the intersection of all closed subsets of R
which contain Q. Therefore, it is equal to...
Jose Carlos Santos
No, that's not correct. Sorry about that. :-(
Jose Carlos Santos
 of Primes proof
--- quoting the webpage
http://www.utm.edu/research/primes/notes/proofs/infinite/euclids.html
---
Euclid's Proof of the Infinitude of Primes (c. 300
                      BC)
                      Euclid may have been the first to give a proof
that
                      there are infinitely many primes. Even after 2000
                      years it stands as an excellent model of
reasoning.
                      Below we follow Ribenboim's statement of Euclid's
                      proof [Ribenboim95, p. 3], see the page There are
                      Infinitely Many Primes for several other proofs.
                      Theorem.
                      There are infinitely many primes. Proof.
                      Suppose that p1=2 < p2 = 3 < ... < pr are all of
the
                      primes. Let P = p1p2...pr+1 and let p be a prime
                      dividing P; then p can not be any of p1, p2, ...,
pr,
                      otherwise p would divide the difference
P-p1p2...pr=1,
                      which is impossible. So this prime p is still
another
                      prime, and p1, p2, ..., pr would not be all of the
                      primes.
 It is a common mistake to think that this proof says
                      the product p1p2...pr+1 is prime.  The proof
actually
                      only using the fact that there is a prime dividing
                      this product (see primorial primes).
                      The proof above is actually quite a bit different
from
as
                      abstract objects, but the ancient Greeks
understood
                      them as lengths of line segments (multiples of
some
                      unit line segment).  Where we talk of
divisibility,
                      measuring (dividing) length b if some integer
numbers
                      of segments of length a makes a total length equal
to
                      b.
                      The ancient Greeks also did not have our modern
notion
                      of infinity.  School children now easily
understand
                      lines as infinite, but the ancients were again
more
                      concrete.  For example, they viewed lines as
something
                      that could be extended indefinitely (not something
                      infinite that we view just part of).  For this
reason
                      Euclid could not have written there are
infinitely
more
                      than any assigned multitude of prime numbers.
                      style which would be unacceptable today--giving an
                      example rather than handling the general case.  It
was
                      clear he understood the general case, he just did
not
                      have the notation to express it.  His proof of
this
                      theorem is one of those cases.
                      Below is a proof closer to that which Euclid
                      but still using our modern concepts of numbers and
                      proof.  See David Joyce's pages for an English
                      translation of Euclid's actual proof.
                      Theorem.
                      There are more primes than found in any finite
list of
                      primes. Proof.
                      Call the primes in our finite list p1, p2, ...,
pr.
                      Let P be any common multiple of these primes plus
one
                      (for example, P = p1p2...pr+1).  Now P is either
prime
                      or it is not.  If it is prime, then P is a prime
that
                      was not in our list.  If P is not prime, then it
is
                      divisible by some prime, call it p.  Notice p can
not
                      be any of p1, p2, ..., pr, otherwise p would
divide 1,
                      which is impossible.  So this prime p is some
prime
                      that was not in our original list.  Either way,
the
                      original list was incomplete.
                      You can see that it is a very sort step from this
                      proof, to the version given above.  It is also
worth
                      noting that what is found in this proof is another
                      prime, one not in the given initial set.  It may
even
                      be smaller than some of those in the initial set.
For
                      example, if we begin with the set:
                      {2, 3, 7, 43, 13, 139, 3263443},
                      then P = 547.607.1033.32051.  The new prime found
                      would be 547, 607, 1033 or 32051, all of which are
                      smaller than the last prime in the original set.
                      Another prime page by Chris K. Caldwell
                      (caldwell@utm.edu)
--- end quoting the webpage at
www.utm.edu/research/primes/notes/proofs/infinite/euclids.html ---
Mistake (1) of Chris Caldwell is that he forgot to give me credit for
having pointed out in the decade of the 1990s that Euclid's proof method
was that of the Direct method and not the Reductio Ad Absurdum. Give
credit where credit is due and I think UTM.edu believes in that credo.
Academic journals pride themselves in giving source credit to ideas and
their original authors.  Below I give one of my first posts to the
Internet regarding Euclid's Infinitude of Primes Proof for which by 1994
I had started to number and to point out the mistakes of authors of
Euclid's Infinitude of Primes proof.
Mistake (2) of Chris Caldwell is that he never mentions set and set
cardinality and that in the mind of Euclid he was increasing set
cardinality as a direct proof method to generalize to the infinite. Any
finite set can be increased generalizes, by logic, to the infinite. The
Ancient Greek Mathematicians did have in their minds, even though they
never said it explicitly in their writings that they were using what we
call in our modern day as set theory. The very word Elements alludes
to set theory. So Chris is correct in his above by his talk that the
minds of the ancient Greek mathematicians had different concepts and
different notions is true. Those ancient Greek mathematicians never had
the Indirect Method or Reductio Ad Absurdum because that form of logic
was unknown to Pythagorus, unknown to Euclid and unknown to Archimedes.
They solved problems in the Direct method of geometrical construction
and generalized to the infinite.
So somewhere in the history of mathematics has the Reductio Ad Absurbum
been falsely ascribed to the Ancient Greeks when they never had this
method in their mind. I believe it was around the time of Gauss that the
ancient Greeks were falsely ascribed to have had and to have used
consciously the Reductio Ad Absurdum. I doubt that even Aristotle, the
author of Aristotleian Logic, or one can say the first creator of the
subject of Logic even had or used Reductio Ad Absurdum (not sure of that
claim).
So the mistake (2) of Chris Caldwell's Euclid Infinitude of Primes
attempt is that he neglects to ever mention set theory and set
cardinality increase that Euclid applied to make his proof.
Mistake (3) is that Ribenboim which Chris uses as reference was unable
to give a valid proof of Euclid's Infinitude of Primes and where
Ribenboim even believes Euclid's IP was Indirect method as apparently
Chris Caldwell also must believe Euclid gave an indirect proof because
the above version starts out with Suppose the primes are finite.
Mistake (4) is when Chris Caldwell says  It is a common mistake to
think that this proof says the product p1p2...pr+1 is prime.  In the
Direct proof method of set cardinality increase this new number of P!+1
is not necessarily prime, that is true, but in the Indirect Method which
Ribenboim uses and which Chris uses above since he copied directly from
Ribenboim then in the Indirect Method or Reductio ad Absurdum then P!+1
is necessarily and automatically prime because the supposition restricts
the Space of the primes to a finite set and so in that restricted space
P!+1 is necessarily prime.
But I want to leave on a good note with Chris Caldwell's attempt of
Euclid's IP above. In that Chris is the first author who talks in some
length about the misinterpretation of our modern day mathematics and the
history of mathematics and to what the Ancient Greek mathematicians
really and actually knew. Chris raises the questions as to whether
Euclid really knew about Reductio Ad Absurdum.
I believe the Ancient Greeks never had the Reductio Ad Absurdum,
although there maybe faint glimmers of the notion of this proof method
that none of the Ancient Greeks had it in their conscious mind. And that
their method was predominantly that of geometrical construct and a
direct method onto those geometrical constructions.
So Chris is good in pointing out the wide gap or gulf of methods and
concepts that prevailed during the Ancient Greeks and our modern day
concepts.
--- quoting one of my old posts of 1993 & 1994 ---
Ludwig.Plutonium@dartmouth.edu (Ludwig Plutonium)
GIVE INVALID PROOFS
         A proof that the counting numbers, or positive integers are
infinite is a direct proof method. The proof is this-- the counting
numbers never end because to the largest counting number a new counting
number is constructed by adding 1. Stated in terms of set theory it is
this-- to the cardinality of any finite set of positive integers a
larger cardinality is obtained by adding 1 to the largest member of the
set, producing a new positive integer not contained in the original
finite set.  Therefore a new set having a larger cardinality is always
produced.
         What follows is a correction of most proofs of the infinitude
of primes.  Most proofs are invalid since they use both the direct and
the indirect proof method in the same argument.
         Most attempts at proving the infinitude of primes neglect the
Fundamental Theorem of Arithmetic, also called the Unique Prime
Factorization Theorem (UPFAT), which is essential in the indirect proof
method, and it is also used in the algorithm for the direct proof
method.  The Fundamental Theorem of Arithmetic is critical in all proof
attempts of the infinitude of primes by the indirect method.
         Euclid intended to give a direct proof of the infinitude of
primes inferring from his terminology of the theorem-- Prime numbers are
more than any assigned multitude of prime numbers.  This is my rendition
of a correct Euclid's proof. My proof rendition is analogous to the
direct proof method for the infinitude of positive integers above.
         PROOF:  Given any finite set of primes p1,p2,p3,p4, . .pn,
construct a new number by taking p1xp2xp3xp4x . . xpn and adding 1.
This new number is not divisible by p1,p2,p3,p4, . .,pn  for each leaves
a remainder of 1.  Either this new number is itself a prime, or it has a
prime factor not equal to any p1,p2,p3,p4, . .,pn via the unique prime
factorization theorem. Thus to any finite set of primes another prime
number is constructible which increases the cardinality of the original
finite set.   Thus to the cardinality of any finite set of prime
numbers, a larger cardinality is constructible. Thus the set of primes
is infinite. Q.E.D.
         The Indirect proof method for the infinitude of primes follows.
PROOF:
  1) Suppose the set of all primes ends.
  2) Then there is a last and largest prime number which exists, call it
pL.
  3) Thus 2,3,5, . . ,pL are all the primes that exist. (Take note that
a valid indirect proof must use existence.)
  4) Now take the product of all the primes that exist and subtract 1,
so (2x3x5x . . xpL) -1=r , and take the product of all the primes that
exist and add 1, so (2x3x5x . . xpL) +1=s giving two new numbers r and s
which both are larger than pL.
  5) Consider r and s. What kinds of numbers are they? Any specific
positive integer can either be composite or prime. Can r be composite?
  (Notice no second supposition is made. Instead the reasoning just
  flows.)
  6)  Since r was built-up from every prime number that exists and
singularly taking each prime that exists, i.e., P1,P2,..PL and
dividing into r.
 7)  Each prime that exists leaves a remainder.
 8) Hence r  is not decomposible by the Unique Prime Factorization
Theorem into primes that exist.
 9) Hence by the definition of what a prime number is, nondecomposible,
r (and likewise s) is prime.
 10) The one and only one supposition that the set of all primes ends is
contradicted
 11) Therefore primes are infinite.  Q.E.D.
         Hardy in his book A Mathematician's Apology  gives a wrong and
invalid proof of the infinitude of primes by using both the direct and
the indirect method simultaneously in one proof.
         Stillwell's book Math and Its History  on page 29 gives a
version of the direct proof but it lacks the concluding part of the
argument that given any finite subset of primes then a larger
cardinality set is constructible and hence the set of all primes is
infinite. The line stating p  p1,p2, . . .,pn is not needed, since in
the direct proof method, the construction of another prime number which
is not in the original finite subset is what is wanted.  For example,
the set {2,17} which has cardinality of 2, produces the prime numbers 5
and 7, for (2x17)+1 =35 which factors into 5 and 7, both of which are
smaller than 17, yet yielding a new set with cardinality of 4.  The
inclusion of this line hints that the author was flipping to the
indirect proof method.  The fact that p is greater than any prime in the
original list is crucial to the indirect proof method and not the direct
proof method.
         Courant and Robbins book What is Math  on page 22 give an
invalid proof rendition. The authors made an error in saying that
Euclid's proof method was an indirect method; it was not. Euclid's proof
method was a direct proof method.
         Dunham's book Journey through Genius  on pages 73-74 has
Euclid's proof as the way Archimedes gave it. It was a direct proof
concluded with: Hence any finite set of primes can always be
supplemented by yet another.  Dunham, although he does not mention
cardinality of sets, makes no mistake through his interpretation of the
proof, and I believe it is because he stayed as close to the originator
himself, Archimedes correcting Euclid.
         Eves book AN INTRODUCTION TO THE HISTORY OF MATH, sixth
edition, on pages 148-149 starts-out by employing the indirect proof
method, then he switches to the direct method by looking for a prime
factor, never mentioning the fundamental theorem of arithmetic, and at
the end switches back to an indirect proof method.
         Davis & Hersh's book THE MATH EXPERIENCE   on pages 210-212
starts-out by employing the indirect proof method, then they switch to
the direct method by looking for a prime factor, never mentioning the
fundamental theorem of arithmetic, and at the end switch back to an
indirect proof method.
         Kline's book MATH THOUGHT FROM ANCIENT TO MODERN TIMES  on page
80 gives a flip-flop attempt of a proof, some of it is the direct method
and the other the indirect.  It is difficult to unscramble this mixed-up
attempt.
   I have asked whether Gauss gave a version of the proof of the
Infinitude of Primes. I have the hunch that Gauss's proof was invalid.
--- end quoting one of my old posts of 1993 & 1994 ---
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
They need not be listed and they cannot be listed. They are not.
A finite ordinal marks a position, a finite cardinal marks an amount.
Ordinal: The 3rd man. Cardinal 3 man in a boat. Ordinals can be
interpreted as ordered sets in set theory, e.g. by v. Neuman: n =
{0,1,2, ..., n-1}. Cantor uses the expression Ordnungstyp
(order-type). Cardinal n = I{0,1,2,...,n-1}I.
In finity both n are equal. As long as only finite ordinal numbers are
concerned, there is no possibility to obtain an infinite cardinal
number.
Genuine mathematics did exist before 1873 and it will survive set
theory.
Hippasos, the inventor of he irrational, is said to have been buried
in absence. So it may be a good finish for irrational numbers to be
beloved in absence.
Unless you learn how to quote my name or initials, this was my last
answer to you.
No one can, but most do not bother to take some for all. Below you
state just the oposite of your present claim.
You see your self-contraditive speaking. It leads to dirty
mathematics.
There is none. Perhaps you wanted to express by P: For all n: Each n
is included in at least one of these sums. But you said (correctly)
all n are included. The difference to your second example is the
ordering of N. IFF it makes sense to talk about all n at all, then, by
the ordering and the definition of from 1 to n we have: 1 is included
in all sums. 2 is included in all but the first sum. 3 is included in
all but two sums. And so on ... . All are included in one sum.
Otherwise you would not get all n.
I do not pretend that it makes sense to talk about all n. But if you
do, then each of them is a natural number which must be included in
this process.
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
Do you have eyes in your head?
Ross
Do you have eyes in your head?
You're either a finitist, in all things, or not.
Ross
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
Do you have eyes in your head?
You're either a finitist, in all things, or not.
Ross
in
I think *for mathematical reasons*, that such sequences do exist. Each
series can be subdivided into partial sums. Cantor's series for
irrational numbers do that, for instance: 3.1; 3.14; 3.141; ... . The
limit is never in the bijection with the indices. It is most easily to
be seen by considering the sequence (1/n) the limit of which has not a
natural number as an index. The reason for the flourishing of set
theory is that definitions of limits and their realisability in
enumerated sequences are frequently intermingled. In short: Dirty
mathematics.
no surprise.
The limit of (1/n) is not in the sequence. Hence there is no index
at all.
There is no limit definition in the axioms of set theory.
hardly.
F. N.
-- 
xyz
Dik T. Winter - WM
 
The large number of contributions is welcome, but forces me to put
some together.
  
DTW:
Which is a limit (although not explicitly stated).  And position oo is
The limit of the sequence (1/n) is 0, but it is never reached by any
n. The limit of Cantor's antidiagonal is also never reached by any
line number n. But Cantors proof considers only the enumerated lines.
Therefore the limit of the antidiagonal number is not relevant for
this proof.
DTW:
You are wrong, as has already been written before.  Cantor uses
*implicitly*
the theorem that any sequence of decimal digits defines a real number,
and
The limit of the sequence (1/n) is 0, but it is never reached by any
n. The limit of Cantor's antidiagonal is also never reached by any
line number n. But Cantors proof considers only the enumerated lines.
Therefore the limit of the antidiagonal number is not relevant for
this proof.
DTW:
Where in that limit process do you find (for instance) 1/9?  And where
does it occur in the list?  Note that a limit is not necessarly an
element of the sequence.
WM:
Exactly, therefore your above statement about Cantor's implicit use of
he limit is wrong. But if not, my sequence can use it too.
DTW:
Yes, numbers from the list come close, very close...
 WM:
 What is the difference?
DTW:
That the limit is not in the list.  No difference.  Cantors number
(which is a limit) is not in the list.
WM:
Not in my and not in his!
DTW:
Wrong.  All real numbers the non-zero decimals of which have only
finitely
many natural indices can be reflected.  In *all* real numbers *all*
decimals are indexed by natural indices.  Also 1/9 and pi.  Or can you
give me the index of a decimal of 1/9 that is not indexed by a natural
number?
WM:
 1/9 is a limit of a sequence of partial sums, i.e. of a sequence. It
is the same as with 1/n. The limit 0 is not indexed by any natural
number. So the limit 1/9 is not indexed by any natural number. Would
you distinguish between limts of sequences, such which are taken and
such which are not?
Further, why should we not reflect a decimal indexed by any natural
number? We could exactly determine its place as for index 2 in 0.07
DTW:
Nope.  Limit here means growing without bound, and you do *not* reach
infinity.
WM:
OK. 1/9 is a limit of a sequence of partial sums, i.e. of a sequence.
It is the same as with 1/n. The limit 0 is not indexed by any natural
number. So the limit 1/9 is not indexed by any natural number. Would
you distinguish between limits of sequences, such which are taken and
such which are not? If not,1/9 is not antidiagonal. But every smaller
series SUM 10^-k is in some row.
Virgil - WM
V:
If those are intended to be binary representations, then that list
omits
every real in (0,1) whose binary representation requires infinitely
many
non-zero digits.
WM:
There are numbers with as many digits as has Cantors diagonal, i.e.
all natural numbers appear as indices.
V:
While there is a flaw in assuming that the diagonal process MUST work 
with binary representations, it is fixable in binary , and does not
hold
for any base greater than 3. Using bases of 3 or more, the 
diagonalization process can be made, and has been made, correct.
WM:
You should only learn by my remark, that you are not the only
mathematician who is unable to determine some genuine errors in set
theory.
V:
How is 0.333... different from SUM_{1<= n < oo} 3/10^n  ?
WM:
If it were not, why then should the definition of 0.333... include n
from 1 to oo?
V:
Reaching for but never acheiving.
WM:
Right. The sequence (1/k) does never achieve its limit 0 for any k e
N. If we want to have 0, we cannot but take 1/oo = 0. It is the same
as with the infinity in the indices of 0.111... 0 1/9. Finite indices
will never achieve it. Therefore, it is not on the diagoal of the
Cantor list, which is enumerated by natural numbers only.
V:
On the right hand side(RHS), we are dealing with a nested sequence of 
set in which the last is always the union of all the sets up to that 
point.
WM:
No. The last element, n, does not belong to any of the foregoing sets.
V:
The union of any falily of sets is defined to be that set containing 
tose elements and only those which are members of  at least of the
sets
in the family. Thus the union of all right hand side sets is well 
defined.
WM:
No. It is not different from the LHS. But also the LHS is well
defined. And the equation holds for every natural number n, which is
well defined.
V:
There is no corresponding definition the left hand side, as the 
definition of the limit of a sequence says that no such limit exists
for
the sequence s_n = n.
WM:
I am talking about existing natural numers. Not about limits of
sequnces.
It does involve oo. Because we have for any finite natural number n:
   n
SUM 1/2^k  =  1  -  1/2^n < 1
k = 1
independent of the question how many terms contribute to the sum. The
criterion for obtaining the Sum = 1 is not the infinitude of terms to
add but that in the limit 1/2^n = 0 which is impossible for any
natural number.
Bu you can see this fact more clearly by the sequence 1/k which
requires k = oo to assume its limit 0.
        
        
NOTHING in your following reply defines what you mean by 
Sum[k in N] 1/2^k.    Defining this expression is *your*
obligation before continuing to discuss it further. 
We might discuss this point further *after* you 
DEFINE WHAT YOU MEAN BY  SUM[k in N] 1/2^k.
Your reply does NOT provide a definition ...
We're still waiting for your definition of  SUM[k in N] 1/2^k.
--r.e.s.
Virgil - WM
  
The large number of contributions is welcome, but forces me to put
some together.
WM:
 Not in that sum which stretches over all n.
V:
That sum is exactly like that excessively fecund woman who is every 
man's mother, it does not exist.
Folgetext zu diesem Beitrag schreiben
WM:
Your example is completely missing the point (as usual), i.e. the SUM
in a linearly ordered set.
Form the (converging) SUM a_k over k = 1 to n.
Do this for every finite n, without leaving out any.
This implies that one of the sums includes all n e N.
If you disagree, tell me which n is left out.
V:
Nobody is saying that the number yoou create is not a real.
What we are saying is that any number with infinitely many 1s in it 
created from a list in which each number has only finitely many 1s in
it
is not in the list. Which  shows that your list is not a surjection
from
N to R.
WM
The digits of diagonal and bottom of each finite triangle
1
11
111
give the value diagonal / bottom = 1. Should this become false in the
infinite. Further, do you see that every 1 of the diagonal stems from
a line?
There is no need for the limit os a convergent sequence of partial
sums
to appear in the sequence itself. So the question is irrelevant.
WM:
It is exactly the same as with my number 0.111... which does neither
appear in a line nor on the diagonal.
V:
If a given list is already known not to be surjective, no more is
needed.
It is only for lists not initially known not to be surjective that 
anything more is needed.
WM:
Cantor's assertion does not refer to a particular list but to all
lists.
Try to think better.
V:
But each and every sum_to_n omits infinitely many terms of the 
infinite sequence. Since every one omits infinitely many terms, they
all omit infinitely many terms.
WM:
V:
Not I.
WM:
Which one can be left out?
 
V:
Since every number in the list you propose has only finitely many 2s
as
place values but has more than any finite number of 4s, then a number 
with no 4s and having more than any finite number of 2s is not in the 
list.
If there are no lines supplying ALL digits of the diagonal (each line
one digit) then the diagonal cannot exist.
V:
So the Cantor costruction works for that list, as it does for every
list.
WM:
By construction, it does not.
V:
What do you mean by the other way around?
WM:
First you take an epsilon, then you choose n_0.
V:
And for each n the infinitely many terms beyond n are all omitted.
WM:
Which is omitted in every case?
V:
In every finite summation (of n terms), infinitely many terms MUST be 
omitted.
WM:
Which n is omitted n every case?
V:
And in every finite summation (of n terms), infinitely many terms MUST
be omitted.
WM:
Which n is omitted n every case?
V:
You are arguing that since every man has a mother that every man must 
have the same mother.
WM:
Your example is missing the point, i.e. the SUM in a linearly ordered
set.
Form the (converging) SUM a_k over k = 1 to n.
Do this for every finite n, without leaving out any.
This implies that one of the sums includes all n e N.
If you disagree, tell me which n is left out.
V:
If you must have usefulness, of what use is a baby?
WM:
Natural numbers enumerate. If you are unable to say which of two comes
first, you cannot enumerate by them. They are no numbers.
V:
For example, if one knows that the nth term of a series is always in
[0,
1/n^2] then one knows that the sequence of partial sums has a limit, 
though one may be totally unable to ennumerate it by finding its
exact
value.
WM:
But the enumeration of the digits is impossible. 0.111... = 1/9 has,
therefore, not only enumerated digits. But Cantor's lines are
enumerated.
Matt Gutting - WM 
The large number of contributions is welcome, but forces me to put
some together.
MG:
Here's an analogy: It is true that all countries are included in at
least
one continent (just as all n are included in at least one sum), but
it
is false that at least one continent includes all countries (as it
is false
that at least one sum includes all n.
WM:
Yes, but your example is missing the point, i.e. the SUM in a linearly
ordered set.
Form the (converging) SUM a_k over k = 1 to n.
Do this for every finite n, without leaving out any.
This implies that one of the sums includes all n e N.
If you disagree, tell me which n is left out.
MG:
But there is no line consisting solely of ones; the antidiagonal
consists solely
of ones.
WM:
Where do you know that from? All numbers which have a finite distance
from the decimal point are ones. How will it be in the infinite. In
particular, there is always a line necessary for the diagonal digit to
exist.
The digits of diagonal and bottom of each finite triangle
1
11
111
give the value diagonal / bottom = 1. Should this become false in the
infinite. Further, do you see that every 1 of the diagonal stems from
a line?
MG:
Why must all the digits of a number be known in order for the number
to
be useful? Very few digits of pi were known until quite recently, but
that did not stop pi from being quite useful, even in ancient times.
WM:
Not pi was used but some rational approximations.
MG:
How ought one decide what makes a number (or a system of mathematics)
useful?
WM:
Not the usefulness is the mathematical criterion of a number, but
whether you know its relation with 1 (and by that with other numbers).
MG:
The antidiagonal has more ones than any given line.
WM:
But its ones stem all from lines.
MG:
In that case, your definition of natural number is, to say the
least,
*significantly* different from the standard mathematical definition.
It appears,
then, that you and I are not talking about the same thing when we say
number. For a mathematician, a number (or any other object)
deserves its
name if and only if it fulfills the criteria set for objects to be
called
by that name. (And, as I've posted elsewhere, I fail to see why it
should
be necessary that the nth digit of a number be knowable in order for
the number
to be acceptable as a number.)
WM:
Natural numbers enumerate. If you are unable to say which one of two
comes first, you cannot enumerate by them. So they are not numbers.
MG:
Virgil was using the definition of limit; you are using the Cauchy
criterion.
It comes down to the same thing, though. In each case, you pick an
epsilon
*first*. Then you check that there's an n satisfying the condition
*for that
epsilon*. 
WM:
have to show then
I S(n) - S I < epsilon. That is the condition for series.
MG:
He's not been understanding this, Franziska. And I think what you've
put your finger on is precisely the point: He's unable to think of the
sequence as a fixed entity, but has to consider it as if it were
constructed
one number at a time.
WM:
Every series can be taken apart in partial sums. And only all of them
are in the diagonal. The limit would require a row enumerated by oo.
Remember the sequence (1/n) and its limit 0.
MG:
Your assertion was that there was at least one sum that includes all
n.
The falsification of this is not that there is that n, which is not
included in any of these sums, but that there is no sum that includes
all n. But this is trivial to see. The nth sum includes n numbers;
therefore
any given sum does not include the next term in the series. 
WM:
Therefore it is useless to define some process for all natural
numbers. There is always the next one not included.
MG:
That is,
for any given sum one can find a number not included in the sum. This
disproves your assertion.
WM
No, it disproves Cauchy's requirement, to show something for all n.
And it disproves Cantor's statement, that something can be done for
all n. It is simply nonsense.
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
I don't see how. There's no requirement that some *one* thing be done
that includes all n. There's only a requirement that something be done
that includes n = 1, something else be done that includes n = 2, ...
Yes, but your example is missing the point, i.e. the SUM in a linearly 
ordered set.
Form the (converging) SUM a_k over k = 1 to n.
Do this for every finite n, without leaving out any.
This implies that one of the sums includes all n e N.
If you disagree, tell me which n is left out.
NO, it is Meucken who misses the point
, i.e. the SUM in a linearly 
This, if it is a converging sequence of partial sums, defines a limiting 
value. That limiting value, if it exists, is the only meaning of the 
infinite sum. 
If the definition were really as Meucken imagines it to be then there 
would be no issue of convergence, and SUM_{n in N} 2^n would have as 
much a real number value as does SUM_n in N} 1/2^n.
Of which partial sum? Since the infinite sum can only exist if the 
partial sums, taken as a sequence, converge, the only possible meaning 
of the infinite sum is the limit, if it exists, of the partial sums.
And since it is required to be the limit of a sequence, that is all that 
it needs to be.
   posting-account=alQKkAwAAAA82xoCXcVIo1q-o-rv2IW-
sums
which
value of n
s1 != s2
s1 != s2
different things.
Hmm, well, you are missing the point fairly spectacularly.
Q1: Do you, or do you not agree that everyone has a mother?
Answer: Yes or No
(I'll assume 'Yes'!)
Q2: OK, what is her name?
 i.e. the SUM in a linearly ordered set.
(How many of these finite n are there?)
No it doesn't. For any n, I can show you its mother, sorry, I mean a
sum including it. I can't show you the single (finite) sum that
includes all of them, because it doesn't exist.
Oh, ok, try a different approach. How about 10^(100^283227442)? Could
that be the answer to your interminably repeated, and interminably
stupid question?
Brian Chandler
http://imaginatorium.org
Sequences cause no problems, because all n are finite. Series cause
problems, because N is said to be infinite. Please repeat your proof
for the infinite Sum 1/2^k = 1, and you will see where you fail.
Every series is no more than the sequence of its partial sums,  so if 
sequences cause no problems, series cannot either.
Please repeat your proof
Wrong! The series Sum 1/2^k is, in fact, identical to the sequence of 
partial sums, and it has a value if and only if that sequence has a 
limit and then the value and the limit cooicide.
s_n = SUM_{1<+k<=n} 1/2^k = 1 - 1/2^n
Let me reassure you: The set N _is_ complete. It contains every natural
number n and also its successor n + 1, but no more.
Since you unconvincibly deny what some axioms say, you should take into
consideration that your refusal may be the cause of the purported
inconsistency of set theory.
F. N.
-- 
xyz
in
Where have you seen it? And did you carefully look at it. No one
missing?
To be honest: The natural numbers are incomplete by definition.
Remember Hilbert's hotel. A complete set of natural numbers would lack
the most important property of natural numbers. It would be the same
as actually finished infinity.
 It contains every natural
I do not deny what they say, but what, according to your
interpretation, they should say. They do not say, for instance, that
any Cantor list (sequence) must be complete and that any infinite
rectangle must have a diagonal.
If
1
11
111
...
has a diagonal, then
1
2
11
22
111
222
...
has not and vice versa.
You know my answer.
By _what_ definition? Please post the definition you have in mind.
The set of natural numbers _is_ _not_ a natural number.
To what definition do you refer when writing actually finished
infinity?
 
What does complete in this context mean? That every n e N is mapped
to a number in the sequence? In this sense the sequence (list) is
complete, since it is _defined_ to map every n e N to a number.
What is the name of the logic you are applying here?
F. N.
-- 
xyz
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
One can't deny axioms.  They are not true or false.  You are free to
replace any of them with anything different, or even nothing at all.
If you do, you simply don't get to talk about the system built on the
unmodified axioms.  And that is M.9fckenheim's fallacy: you can't say
give me the natural numbers, but without the successor or induction
axioms.
Inconsistency with intuition is nothing that axioms are bothered
about.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
That's Cantor's and your fallacy. You can't have something complete
which, by definition, is incomplete. That is typical for the actually
finished infinity. An axiom saying so is wrong - not because it is in
contradiction with another one, but because it is contadicting itself.
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
You are babbling.  What definition would that be?
You are babbling.
Which axiom would that be?
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
IMHO axioms are condidered true. You probably mean that choosing a
certain set of propositions to be axioms is neither true nor false.
True.
Indeed.
Positively not.
F. N.
-- 
xyz
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
No, that is not the case.  What you are calling axioms, the
non-logical axioms, are actually theorems, of tautology, and the
excluded middle.
Here's what I'm worried about:  Andy.
Oh wait, I _am_ Andy.
Goedel says so!
Ross
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
Not as much true as given, not as much considered as
prescribed.  They define the systems you are talking about.  Within
a given system, they are trivially true, following from the axioms
of the system.  Outside of a given system, they are arbitrary.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
I mean that internal truth.
F. N.
-- 
xyz
Well, that certainly explains a lot of the frothing about axioms
on sci.math: clearly some people don't take their presriptions.
Lee Rudolph
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
That's because they're proscribed from doing so.
Matt
  But most mathematicians don't talk about systems at all.
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
Oh...
Your axioms are stupid!  Who are you, to constrain the multiverse?
Am I bad if I think you're stupid for trying to say those things?  I'm
tested, say I have a 145 IQ.  The SAT says 160.  My words appear to say
185...
Try to not say those things.
Ross Andrew Finlayson
--
Did you or did you not?  You're god-damned right I did! - Sugar Ray
Yes I am!
 
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi
$t^
Well, there are quite a few offers around Test your IQ, for just
$....  In my opinion, paying constitutes failure.
Not too hard.
-- 
David Kastrup, Kriemhildstr. 15, 44793 Bochum
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
No!
I want you to think why it is imperative that you _don't_ apply axioms
to your theory.
Consider Goedel.  Goedel says:  if you apply axioms to your theory, (in
this case of Goedel meaning the non-logical or proper axioms,) the
theory can never be complete!
Don't you think there can be at least one theory that can be complete,
where completeness means at least consistency?
If not, your theory is inconsistent:  Goedel says so.
You must start from nothing and assume what truth there is from that.
Ross
X-Spam-This: SpamCopies@YahooGroups.Com
Maybe, maybe not. It depends on your specific algorithm for choosing at
random. For example, you might generate a denominator with probability
90% of 1, 9% of 2, 0.9% of 3, etc., except you skip 5 as denominator,
and numerator always exactly 5. You don't cover all the rationals, but
you're still choosing only rational numbers so who cares if you cover
them all. With that algorithm, you're very likely to pick an integer.
Compare to this question: Pick an integer at random. What's the
probability that integer is in the range 1 thru 9? That question can't
be answered until you specify the algorithm. For example, if the
algorithm is flip coin, heads you pick zero, tails you pick nonzero as
follows: Heads the number is positive, tails it's negative, in both
cases you get the magnitude as follows: Heads it's 1, tails you flip
again, heads it's 2, tails you flip again, heads it's 3, tails you flip
again, etc., heads you stop at the current magnitude, tails you
increment by 1 and flip again, eventually you are sure to flip tails at
which point you are done picking the integer.
That conclusion doesn't follow, for the same reason as above. For
example, suppose your algorithm is to run the algorithm in the
paragraph above, then add 3274873843 to the result. Slightly more than
50% chance of the result being divisible by 3274873843.
Same rebuttal as before.
Note that it is *impossible* to pick an integer, or rational, at random
such that you cover them all and the probability of each is the same as
any other. So you need to have some biassed algorithm that picks some
more often than others, and once you get into biassed algorithms the
grossly biassed ones I gave as examples above are just as reasonable as
any other algorithms.
X-RFC2646:  Original
[Robert Maas, to Bruce Stephens]
... [skipping to the end] ...
They're unreasonable in this context, though.  For example, do you think 
Dirichlet was deluded in some way when he proved that 6/pi^2 is the 
probability that two integers chosen at random are coprime?  I sure hope you 

In number theory, a claim that this-or-that probability holds for an integer 

chosen at random has a conventional, rigorous meaning:  you consider the 
claim applied to an appropriate finite set of integers (1 through N is most 

common), and compute the probability wrt the uniform distribution on that 
finite set.  Then you look at what happens as N (the size of appropriate 
sets) goes to infinity.  If a limit (in the ordinary sense of the word) 
exists, then the probability for an integer chosen at random is, by 
agreement, that limit.
For example, the probability that a strictly positive integer chosen at 
random is even = 0.5, and neither expert nor neophyte raises an eyebrow at 
    P(N) = floor(N/2) / N
exists and is 0.5, where P(N) is the probability that an integer chosen at 
random uniformly from {1, 2, ..., N} is even.
... [going back] ...
By the conventional meaning in this field, it's 0 (the limit of min(9, N)/N 

as N goes to infinity).
Ahd that's reasonable when it's unreasonable to assume the conventional 
meaning of probability for integer chosen at random is intended.  At 
least 
to me, Bruce clearly intended the conventional meaning.
The other examples work out similarly, and what Bruce said all made good 
sense with that meaning in mind. 
X-Spam-This: SpamCopies@YahooGroups.Com
That's not a good way of saying it, because it makes it seem like 1/2
and 3/6 are different rational numbers when in fact they represent the
same rational number. You need to say clearly that rational numbers are
equivalence classes of such x/y representations modulo the equivalence
relation generated by the equation x/y = (kx)/(ky) for any k, or better
say that rational numbers are equivalence classes of ordered pairs
(n,d) where d isn't zero, modulo the relation generated by (n,d) =
(kn,kd) and then introduce the notation n/d to represent the
equivalence class containing (n,d). Then 1/2 = 3/6 because 1/2 =
{(1,2), (2,4), (3,6), (4,8), etc.} = 3/6.
   posting-account=mmiamAwAAAATmtxVYi9NGy7tls83IKO3
What do you mean by specialized choices? Sure, if you choose a factor
of j^2 T, you get a factor of M^2 T. If you don't pay any attention to
what you're doing, you may get a trivial factor of M^2 T, or you may
get a non-trivial factor of M^2 T.
But if you plan ahead, then you can figure out which factors (f_1, f_2,
..., f_N) of M^2 T will give you non-trivial factors of M^2 T.
Then, to factor an integer, all you need to do is use your method with
ONE of f_1, f_2, .., or f_N. If you know that using these factors will
give you a non-trivial factor, then you solve the factorization problem
(the RSA version, where M = p * q) with 100% SUCCESS ON THE FIRST TRY.
This even beats the 50% success rate that you've been claiming all
along.
The mathematics behind a real SF Theorem would guarantee it. Period.
If you still haven't understood this vital point, consider solving the
following problem:
PROBLEM. Given r, a non-zero rational number, find an non-zero integer
N so that r * N is an integer.
What _you're_ saying is that it's okay to just try N's at random. Maybe
you solve the problem, maybe you don't. Does it really matter? The
mathematics doesn't mind, after all ... (This has been your whole
approach to the factoring problem.)
What _I'm_ saying is that if you choose your N correctly, with some
knowledge about r, you can solve the problem with ONLY ONE GUESS. In
short, I would say Let N be the denominator of r. Then r * N is the
numerator of r, which is (by definition) an integer. Period. I'm done.
I get to go outside and play while you're stuck inside, choosing
numbers at random and hoping you get lucky.
It is a fact that an infinite number of N's that will NOT solve the
problem. So, if you're unlucky enough to choose N's from that set of
numbers, you'll never finish the problem.
With my theorem, I only have to test one; with your theorem, you may
have to test infinitely many. Whose theorem has more content?
That's because I don't have a definite choice. But, conceivably, I
could discover a theorem which would tell me how to do it. Then I
_would_ have a reason for a special choice; the theorem says that it
works, that I get a non-trivial factor of M.
That's not the case. Proofs exist outside of the material world; they
still imply the same things.
Time to get The Shovel.
You made a claim earlier that mathematics doesn't care about factoring.
Well, there is a finite set of mathematical facts, all of which are
true, which do say that an integer has been factored:
(1) 20 = 2 * 2 * 5.
(2) For all positive integers m and n, (m+1)(n+1) is not 2.
(3) For all positive integers m and n, (m+1)(n+1) is not 5.
These are sufficient facts to prove the statement 20 can be factored
into a product of prime numbers. And they're all written in
Mathematics.
That's because I can't devote all of my time looking for one. I have to
eat, eject bodily waste, sleep, teach, grade papers, turn on my
computer to check the latest news, etc.
But the reason itself could exist out there, even if no one is looking
for it. And if someone finds it, then they have the reason why the
problem can be solved in one step.
But I may be wrong. Maybe there isn't a reason out there, and maybe
someone can find it why there isn't.
But in either case, the answer requires more research, which you are
unwilling to do (refusing to think that the KEY is out there, after
all), and in some sense unable to do (not having had the training to
know when you do have a real reason, which can be proven). And the idea
is powerful enough that someone _should_ search for it, just in case it
exists.
Sure, it's the same set even.
But it's not really between factors of j^2 T and M^2 T, because you
also have k_1 and k_2 floating around in your equations (for Ax, in
particular). The factors you get for M^2 T depend on j, f_1, and k_1
(you say this in your SF Theorem, that all k_1 need to be considered).
These are all things which were not given to you. In the language of
mathematics, f_1 is a function of j, f_1, and k_1.
f_1 = F(j, f_1, k_1). (And technically, the function has to be 1-1 as
well. However, it still has to be a function.)
If you just know j and f_1, then there's no way to find Ax; the formula
for Ax uses k_1:
(This is from your Factoring Theorem post in the Surrogate Factoring
group, April 14.)
Yes, and a theorem which had actual content would show how to choose
the factors of the first number wisely.
What is the question, though? How many factors are there?
Or maybe a better question is How many factors do you have to test, to
find one that gives you a non-trivial factor?
Or maybe, related to that last question, there are an infinite number
of factors which give you trivial factors of M, so you could literally
spend forever checking factors and never finding any.
Is this the great advance you've been promising? At least someone
testing to see whether the integers between 2 and M-1 will finish the
task.
No, convincing YOU, James Steven Harris, the writer of the post I'm
responding to, is key. I've been talking to YOU this whole post.
Sometimes I talk to other people, but most of the time, I'm talking to
YOU.
I've been patient and explained, trying to give you analogies which
will help you despite your lack of formal training. But when you can't
(or won't) accept that, you turn to social rhetoric, while I have
remained on topic.
The simle fact is your theorem, as is, gives no help in solving the
factoring problem; a real theorem (provided it exists) would.
The Surrogate Factoring Theorem isn't important.
Surrogate factoring may be important; there may be a way to salvage the
basic idea. (If you have a factor of T, you get a non-trivial factor of
M.)
But nothing is happening because you refuse to work with
mathematicians. When people say your proof doesn't say anything, or
your analysis is wrong, you don't look at what they're saying. You
become anti-social ...
... a case in point.
You understand only the mathematics of your SF Theorem. If you knew the
complete mathematics of the factoring problem, you could solve the RSA
problems and prove that you're not a crank, a quack, a kook, or a
troll.
Using what I know about human behavior, I thus conclude that one of two
things must be true:
(1) You are unable to solve the factoring problem.
(2) You are able to solve the factoring problem, but won't, for some
deep psychological reason.
If you're not really a crank, then you want the Surrogate Factoring
idea to work. Why are you blocking the help that would make it work?
     --- Christopher Heckman
X-RFC2646:  Original
[JSH]
...
Since you said this part was most important, I spared everyone a repetition 

of the rest:
There's an easy way to find out:  program it and measure.
Everyone else already sees this, so they won't bother:  given a rational 
factor r, gcd(numerator(r), M) = gcd(numerator(r) mod M, M):  all 
numerators fall into one of exactly M distinct congruence classes, and 
exactly 2 of those M classes yield a non-trivial gcd if M is the product of 

distinct primes.  So the probability is 2/M across all r in 0 thru M-1, 
across all r in 0 thru 2*M-1, across all r in 0 thru 3*M-1, ..., across all 

numerator _congruence class_ is equally likely, and (b) M is the product of 

two distinct primes.  The number of distinct congruence classes remains M no 

matter how large i grows, and the number of congruence classes is all that 
matters to the chance of finding a non-trivial factor.
Of course it does, and the above is a decent start at a proof (it needs 
generalization to upper limits not just of the special form i*M-1, and then 

proof that the limit exists and is in fact 2/M as the upper bound approaches 

infinity -- & that's all easy to do too).
But you're so determined to let hope triumph over sense here that your 
thinking appears impaired, so just program it and see it mostly thrashing 
uselessly for yourself.  Then maybe you'll be motivated enough to learn 
something new.  Or keep stewing in your juices -- still your choice whether 

to make progress or just keep ranting.
... [the usual] ... 
X-RFC2646:  Response
[JSH]
[Tim Peters]
Very careless of me there, and I apologize for that.  All instances of 2 
above, and in the rest except for in 0 thru 2*M-1, should be
    p_1 + p_2 - 2
instead.  Or see any of the previous 579 times the same thing was explained 

   posting-account=0oMoGgwAAAA45r-W1MV-8N0QRNWHaDRI
Lets say you have a number that is the product of 2 large primes. There
are 2 non-trivial factors, the 2 primes. Then there are trillions and
trillions of trivial rational factors. Why don't you demonstrate how
your SFT will find those 2 non-trivial factors. You have said that your
theorem has no bias towards either, so why would it come up with one of
those 2 factors as often as one of the trillions of trivial ones?
Doesn't quite seem like there is a 50% chance of coming up with one of
the two prime factors. A demonstration sure would put a rest to all the
arguing.
But you are not mathematicians.  I am a mathematician.
I guess you must be a liar then, since you have stated time and time
again that mathematicians are liars. And since everyone else here is
not a mathematician, then they must all be telling the truth.
Conversations among mathematicians are so elaborate!
M1. I'm famished! Let's get something to eat. Have you got any money?
M2. Yes, I have 50 cents.
M1. OK. Have you got any more?
Me, I cut to the chase -- when I'm M1 the conversation is more accurate:
M1'. What is the least upper bound of the set of numbers N to which you
     can truthfully say I have N cents?
M2.  Oh, fercryinoutloud. Give me a break!
M1'. Do you mean you would like a _unique_ break, or are you simply asking
     for a non-empty set of breaks?
(M1' now dines alone.)
dave
LOL
   posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
There
your
of
There are an infinity of rational factors, not just trillions and
trillions of them.
James Harris
   posting-account=0oMoGgwAAAA45r-W1MV-8N0QRNWHaDRI
There are an infinity of rational factors, not just trillions and
trillions of them.
James Harris
That doesn't answer any of the questions I asked. But I will reply to
your statement. Lets say I have 100 coins. Does it not also stand to
reason that I have 50 coins, or 10 coins. I did not say that all there
was was trillions of rational factors, but it is true that there are
thrillions of them, as well as billions, as well as millions. Two
non-trivial, an infinity of trivial factors. Please demonstrate how
your theorem finds either one of the 2 prime factors of my example 50%
of the time.
Hi all  
Help please! How does one solve for D in the following?
 Z=(A+jC)+(B+jD)
   posting-account=BjC-YAwAAADQ91Zm3XkS3aGs3XlaqZ4X
An example:
(13, 9) =(3,4)+(10,D)=(3+10, 4+D).
In addition the components are independent.
So D= ?
Have fun 
Hero
OK, that one was too easy. Here's a more 'serious' one. When the ducks
fly south for the winter, they always form the vee formation. But one
leg of the vee is always longer than the other. Why is that?
--Lynn
   posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK
ducks
Because the number of ducks is even?
On 16 Apr 2005 11:04:25 -0700, mensanator@aol.compost
Nope.
--Lynn
There are 10 kinds of people in this world - - those that understand
binary and those that don't.
--Lynn
X-RFC2646:  Original
Hi all,
I am confused while reading different books.
For stochastic processes, the autocorrelation or autocovariance must be 
positive semi-definite, but which one?
After reading several books, my conclusion is that both autocovariance and 
autocorrelation are positive semi-definite, am I right?
X-RFC2646:  Original
Number 1 is obvious from the fundamental theorem of arithmetic.
Each number has a unique representation of powers of primes eg 2^n1 * 3^n2 * 

5^n3 ... etc, in M none will contain 2  eg n1 =0. Now map to the number in P 

2^n2 * 3^n3 ... etc its a bijection with the correct properties.
 I guess all answers will be of this form (swaping primes powers around)
#2  I'm not really sure what you want or why?
   posting-account=WZMvOwwAAAC_B1TaayrVN99BJgDWQkUc
This means you will have an infinite number of people
flipping an infinite number of times each, if t ever
gets to 1.
Forever, basically.  At any given finite time,
you have only flipped a finite number of times.
But there are infinitely many people flipping before
you, and the expected number of them that match you,
for a prefix-length equal to
the-number-of-times-you-have-flipped-so-far,
is ALSO INFINITE.  In order to get the number of expected
matches-so-far down to something finite,
you have to determine some finite subset of the flippers
that are worth thinking about, at any given time.
Of course it is.  The reason for logic and set theory being studied so 
heavily around the time of Cantor's work was *precisely* to determine 
whether our tools are properly built.  The result of that work was to 
have tools that are reliable, consistent, and serve as a framework of 
thought for the rest of mathematics.
Would you want to design a bridge, not knowing if your calculations were 
correct?  Would you want to build a house over a sinkhole?  Would you 
knowingly live in such a house?
I rely on the soundness of the arguments presented.  In the beginning of 
the _Courage?_ thread you said Spontaneously, Zermelo came up with the 
blunt idea to just define without any proof his axiom of choice.  I 
understand my tools well enough to know that an *axiom* requires no 
proof.  That is why it is called an axiom, and not a theorem.  I wish 
you would consider the possibility that you are attempting to critique 
something without understanding the thinking and terminology behind it. 
  To suggest that an axiom requires proof is nonsense.  To suggest that 
an axiom should be proven consistent with the other axioms is 
reasonable, but no more can be asked.
-- 
Will Twentyman
email: wtwentyman at copper dot net
I believe I was responding to a theme you have of calling *a* real 
number *uncountable* or alluding that it might be infinite.
numerical identity is irrelevant to whether a set is uncountable.
But at the moment, you do not even use the words of mathematics with the 
same meaning as the rest of us.  You need to start with a clear 
explanation of your concepts.  The ideas you are working with are likely 
to be covered already, just not in Cantor's work.  Numerical Analysis 
comes to mind.
-- 
Will Twentyman
email: wtwentyman at copper dot net
And if a problem has a unique solution, it serves as a definition and 
you can round the answer to whatever precision you desire.  The rounded 
answer is not the answer, however.
I suppose by this you are suggesting that something being absolutely 
compelling is related to infinity.  On the other hand, it could be tied 
to probability and something that is absolutely compelling could be 
associated with 1.
-- 
Will Twentyman
email: wtwentyman at copper dot net
You misunderstood.  Cantor defined what it means for the set of real 
numbers to be uncountable.  His proof simply showed that the real 
numbers satisfy that definition.
In the mathematics we are discussing, numbers are not dealt with up to a 
certain number of digits.  What you have in mind is a topic for 
Numerical Analysis, which is a very different branch of mathematics.
Please understand, you will make no headway in your arguments because 
you are starting with radically different assumptions about the meaning 
of numbers, sets, etc.  If you wish to critique Cantor, you *must* deal 
with his meaning.  If you do not accept his meaning, you can only say 
that his results cannot be reliably applied to your (different) system.
-- 
Will Twentyman
email: wtwentyman at copper dot net
On Fri, 15 Apr 2005 17:35:03 +0200, Eckard Blumschein
How in the world could anyone possibly think that was an answer to my
question?
************************
David C. Ullrich
Cantor used whole infinite numbers in order to distinguish them from
fractions. Not only in his early works.
In his final presentation of set theory: Beitr.8age zur Begr.9fndung der
transfiniten Mengenlehre. [Math. Annalen Bd. 46, S. 481-512 (1895);
Es ist zweckm.8a¤ig, sich zun.8achst mit denjenigen Zahlen von 
Z(aeph_0)
vertraut zu machen, welche ganze (rationale) Funktionen endlichen
Grades von omega sind. Jede derartige Zahl l.8a¤t sich, und dies 
nur auf
eine Weise, in die Form bringen ...
ganze Zahlen von Z(aleph_0) = whole numbers of second number class
(infinite numbers)
ganze rationale Funktionen von omega = whole rational function of
omega
One chapter is named: Û 15.  Die Zahlen der zweiten 
Zahlenklasse
Z(aleph_0).
Die zweite Zahlenklasse hat eine kleinste Zahl. = The second number
class has a smallest number.
Small wonder that you firmly believe in Cantor's theory. You should
study it!
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
Die umbegrentzheit (unendliche) Zahlen, wieder die Kardinalen, Sie aber
wird die Negativzahlen zuwirden.
(The transfinite cardinals are actually the infinite ordinals, and the
negative integers, via dual representation as the cumulative
hierarchy.)
Es is nicht fortunat, wo es wirde nur eins Theorem zuwirden, aber est
so.  Das Eins ist Vertig.
Goedel sagt so!
Ich sage so.
Ross
That are merely names or prescriptions how to find that number.
But we know the exact relation 2/1 of 2 and 1. Therefore two *is* a
number.
In order to call x a number, we must know exactly the representation
in the basic unit, x/1. This is impossible for irrational numbers.
In fact, because we know the exact relation between them and the unit.
Irrational numbers are nothing else but the aim of approximations.
These cannot be arbitrarily precise, because of physical constraints,
which cannot be avoided and must be taken into account even by pure
mathematics.
   posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
The infinite series is merely a prescription for how
to find that number.
The definition is that pi is the ratio of the circumference
to the diameter of a circle. The other prescriptions are
provably equal to that quantity. They are exact. What is
limited is how many steps of these exact algorithms we
have followed. Just because we haven't done the 10^20-th
step doesn't mean there isn't one.
              - Randy
On 16 Apr 2005 03:32:35 -0700, mueckenh@rz.fh-augsburg.de (W.
Irrationals can be pointed out exactly on straight lines.
Transcendentals cannot. But transcendentals can be pointed out on
curves. The problem is curves do not have any exact commensuration in
terms of straight lines. But that does not make the concepts involved
any the less numeric.
   posting-account=alQKkAwAAAA82xoCXcVIo1q-o-rv2IW-
Lester, any chance you could share your definitions for irrational,
transcendental, and can be pointed out? Quick hint: in the normal
mathematical definitions, transcendental numbers are a subset of the
irrationals, so your claim couldn't possibly make sense. I might guess
that by 'irrational' you meant to say 'algebraic' (the reals that are
not transcendental), but then I still don't know what 'can be pointed
out' means.
Brian Chandler
http://imaginatorium.org
On 16 Apr 2005 08:59:43 -0700, imaginatorium@despammed.com in
My usage was pretty thoroughly picked over a little while back on the
thread Epistemology 201: The Science of Science.
The idea of irrationality used I believe by Euler gives irrationality
as convergent infinite series which would include both non rationals
and transcendentals. I don't like the term algebraic because algebra
defines both rational, irrational, and transcendental concepts.
The idea of rationality, irrationality, and transcendentality I use
gives rationality as integral ratios which can be pointed out on a
straight line with right angles; non rationals or irrationals as those
numbers which can be pointed out exactly on straight lines using right
angles but cannot be represented by any finite series of cardinal
numbers or integral ratios; and transcendentals as those numbers which
cannot be pointed out on straight lines using right angles because
they exist on curves of various descriptions and not straight lines.
This means that rationals and non or irrationals coexist on straight
lines and transcendentals are a different numeric class altogether
which involves the ratio between curves and straight lines. I know
this is nonstandard usage. But it represents a strict interpretation
of the basic concepts involved based on geometry that doesn't just
rely on cardinal arithmetic and infinite series for their analysis.
The phrase pointed out just refers to the ability to use right
angles to define the distance between points on a straight line with
the Pythagorean theorem.
   posting-account=alQKkAwAAAA82xoCXcVIo1q-o-rv2IW-
irrational,
normal
guess
are
pointed
Hmm, I wonder if you mean Cauchy sequences, or something? Well, never
mind, I understand this isn't *your* definition, so I'll skip over it.
a
Up to the [*] this looks like the usual definition (well, with a bit of
rewriting); but I don't understand the last bit, nor do I understand
its relation to the first bit. Is this a further qualification
(implying that there are integral ratios which cannot be pointed
out), or is it just a comment, in which case there should surely be a
comma at the [*] position?
non rationals or irrationals as those
right
Can you give some examples? sqrt(2) ? fifth root of 5 ? solution of a
quintic polynomial not solvable by radicals? e+pi ? e*pi ?
Are there any results in your mathematics, such as that all
Zirrationals are algebraic numbers (I hope you don't mind me
identifying your concepts with your initial, and unless otherwise
stated, using the normal maths terms for the normal maths definitions),
or the other way round?
 and transcendentals as those numbers which
Bit baffled by this. Is pi a Ztranscendental? In the Cartesian plane,
doesn't the point (pi, 0) lie on the x-axis?
Could you actually draw a picture of these right angles being used for
the pointing out?
Or at least explain how a right angle helps??
Brian Chandler
http://imaginatorium.org
X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal 

listed above.
X-Abuse-Notes2: Reports sent via any other method will not be processed.
I thought you were claiming that pi was not really a number,
for precisely the reason that it could not be exactly calculated.
Why is it necessary to know this relationship?
Why is this so?
Matt
[...]
[...]
This is the elementary school notion of number formalised. It should be 
obvious that number is more subtle concept than  what is learned in 
Grade 1.
The notion that a formula is an recipe for producing a number is also a 
  school-based notion. School does not teach mathematics, it teaches 
arithmetic, so that it's no surprise that many people think that a 
number is the result of a calculation. Hence Mueckenheim's claim that 
4*arctan(1) does not represent a number. If this representation is 
rejected, then so should x/1 be rejected, since both are algebraic 
representations of numbers. Anyhow, x/1 means exactly the same for any 
x: that is, x/1 = x for any x whatsoever. M's claim that it's 
meaningless for irrational numbers makes it meaningless for any number, 
which contradicts his claim that it's meaningful only for natural numbers.
The mistake IMO arises in part from the confusion of numeral and number. 
Eg, the numerals 1, 2, 3 ... are not numbers, but representations of 
numbers. This is somewhat more obvious when they are replaced by their 
ordinary language equivalents (one, two, three,... or eins, zwei, drei, 
...). A numeral is not a number. (Eine Ziffer ist nicht eine Zahl. Im 
Deutschen sowie wie im Englischen bedeutet Nummer/number Zahl und 
Ziffer.)
A number, most abstractly considered, seems to me to be any element of a 
set that is defined in terms of arithmetic operations. An arithmetic 
operation * is one that produces (or relates?) elements of the set. Eg, 
consider {a, b, c...}, an operation * such that a*b = c, and some 
inverse operation ! such that c!b = a. Then a, b, c are numbers. (IIRC, 
at least two operations are required; is this correct?)
Thus the members of {Vehicles} = {truck, car, tractor,...} would be 
numbers IFF truck*tractor is defined such that the result is some 
vehicle. Ie,  truck*tractor = trucktor, and trucktor!truck = tractor. 
(This is, oddly enough, possible for some members of the set {Vehicles}, 
but not for all. Eg, in N. American usage, tractor + trailer = 
transport; transport - tractor = trailer, transport - trailer = tractor, 
where + is connect and - is disconnect. Fun, eh?)
On Sat, 16 Apr 2005 09:32:31 -0400, Wolf Kirchmeir
So if a formula is not a recipe for producing a number, then what are
the operations that produce members of the set of numbers that you
outline below if they aren't recipes for producing numbers?
Why doesn't school teach mathematics? If numbers are not the result
of calculations then what are the results of the operations you use to
define the set of numbers below?
So what we have here is your opinion that a number is any member of a
set that is defined in terms of operations that produce members of the
set. This may be true but is certainly not very enlightening. What
produces a set? What produces members of a set. You will undoubtedly
then enlighten us by saying that members of this set are what children
do with their fingers. But children do lots of things with their
two is not a number. Why not just say it's magic?
This is a rather curious essay in which you reject someones opinion of
what a number is and substitute your own opinion of what a number is.
What makes your opinion anything but an opinion?
   posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
In N. American usage, tractor is cab, rig, semi, truck, diesel,
tractor, and with the trailer sometimes it's hauler, 18-wheeler, or
road rig, the tractor-trailer.
Probablement ces sont les memes mots dans les autres langues.
Infantilisement, ces ne sont pas du choses infinites.  En l'autre cas,
on peut se voir comme un separacion du contexte, tous les choses minus
le rig, tous qui se sont et tou qui ne se sont pas, ou se qui ne sont
pas.
C'est tous ma parlance, j'espere et vous remercierais a parler en
anglais, je suis presque de l'illiterate.  Pardon mon francais.  Mais
en cours, beaucoup des gens vous comprennents.
Le contexte, c'est son seule contradiction, mais, ce ne c'est pas!
C'est tout, mais ce n'est rien.  Il y a du traits philosophiques qui se
trouvents dans le zero, l'infinite, ou on peut realiser du vraiisme, un
mot particuliere.  Es ne peurraient pas etre autre-meme, ainsi c'est
ca.
En concours, oui.  Le contexte, ou, la contexte, le contexte, c'est
essentialement tous les choses est mais rien.
Ross
   posting-account=jEbKGQ0AAADZF1UpkDsHa5gkWBqABnUE
unit.
The idiot Muckenschleim cannot distinguish between a number and its
representation.
Idiot Muckenschleim, Seven, IX, PI, e, 1001, 3 are but symbols. The
shape and syntax of the symbol are of absolutely no import to the
attributes or the nature of that which they stand for.
You are a malicious and fallacious bugger and should be barred from
teaching to unsuspecting students.
Another tiny yappy dog in the back of the pack.
-- 
I know that most men, including those at ease with
problems of the greatest complexity, can seldom
accept even the simplest and most obvious truth
if it be such as would oblige them to admit the falsity
of conclusions which they have delighted in explaining
to colleagues, which they have proudly taught to others,
and which they have woven, thread by thread,
into the fabric of their lives. -
        -- Tolstoy
   posting-account=alQKkAwAAAA82xoCXcVIo1q-o-rv2IW-
unit.
You speak authoritatively, as though you knew what pure maths is.
Unfortunately, you have actually no clue what mathematics is, so your
writings are nonsensical. At best, you seem to be writing about what we
might call applied calculation, which is a sort of branch of
engineering (spiced with the odd bit of philosophical waffle here and
there).
I think it would be extremely counterproductive to attempt to argue
with your statements above. If you wanted to know what mathematics is,
someone might suggest some reading matter.
Brian Chandler
http://imaginatorium.org
Hey, one crank per thread, please...
Apparently, it has never occurred to you that many people know 
/exactly/ what 'mainstream' mathematics is and simply disagree. 
You consider it 'counterproductive', whatever /that/ means, to 
build an argument against a point of view.  One must assume, 
based on your post, that you consider it more 'productive' to 
insult without justification.  You remind me of a little yappy 
dog in the back of the pack, urging your more courageous fellows 
to attack.
-- 
I know that most men, including those at ease with
problems of the greatest complexity, can seldom
accept even the simplest and most obvious truth
if it be such as would oblige them to admit the falsity
of conclusions which they have delighted in explaining
to colleagues, which they have proudly taught to others,
and which they have woven, thread by thread,
into the fabric of their lives. -
        -- Tolstoy