mm-368 === >>Not where I went to school. A perfectly valid prrof would be marked >>down if it was not elegant; quite properly, IMHO. >> I see no reason for this. This is not the way mathematics is >> done; first get the proof, then MAYBE look for elegance. >Taking away the elegance from ANY proof, transforms the proof into a >ious, boring and mechanical procedure, which should interest no true >mathematician. >> Also, >> it is quite common for elegant proofs to hide the concepts; >I'd rather have an elegant proof which hides some of the concepts yet gives >the overall idea, than a mechanistic and non-interesting proofwhich boggles >the mind with inane details which are of no interest to the reader. Many of the so-called elegant proofs completely hide the concepts. Proving the Central Limit Theorem for sums of independent identical distributions is certainly the most elegant, but it hides virtually everything; it has no probability in it at all. This is also the case when the distributions are not identical. However, the somewhat clumsy Lindeberg proof extends to the martingale case, and gives an idea of what is happening. I can give lots of somewhat unrela proofs, but does anyone really understand the theorem? Similarly, Cramer's proof of the Levy-Cramer theorem that the sum of two independent random variables is not normal unless they both are is probably the only easy proof, but it likewise obscures the ideas. Any time characteristic functions are used to prove a probability theorem, the concepts are not even present. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Why is math so difficult for some people? === > Similarly, Cramer's proof of the Levy-Cramer theorem that > the sum of two independent random variables is not normal > unless they both are is probably the only easy proof, but > it likewise obscures the ideas. Any time characteristic > functions are used to prove a probability theorem, the > concepts are not even present. Do you happen to know of a probabilistic proof of the Cram.8er-L.8evy theorem? -- A. === Subject: Re: Why is math so difficult for some people? > Many of the so-called elegant proofs completely hide the > concepts. Proving the Central Limit Theorem for sums of > independent identical distributions is certainly the most > elegant, but it hides virtually everything; it has no > probability in it at all. This is also the case when the > distributions are not identical. But it shows the power of the Fourier transform. > However, the somewhat clumsy Lindeberg proof extends to > the martingale case, and gives an idea of what is happening. > I can give lots of somewhat unrela proofs, but does anyone > really understand the theorem? > Similarly, Cramer's proof of the Levy-Cramer theorem that > the sum of two independent random variables is not normal > unless they both are is probably the only easy proof, but > it likewise obscures the ideas. Any time characteristic > functions are used to prove a probability theorem, the > concepts are not even present. Would you say that the elementary proofs of say the prime number theorem due to Selberg-Erdos etc, give more insight than methods using complex analysis and Fourier analysis as done by Wiener, Ikehara and Newman? -- === Subject: Re: Why is math so difficult for some people? >> Many of the so-called elegant proofs completely hide the >> concepts. Proving the Central Limit Theorem for sums of >> independent identical distributions is certainly the most >> elegant, but it hides virtually everything; it has no >> probability in it at all. This is also the case when the >> distributions are not identical. >But it shows the power of the Fourier transform. I do not disparage the power of the Fourier transform as a method of getting numerical answers, but in most cases in which it is used, it conceals the important ideas. In some cases, such as the Central Limit Problem, the concealment is great, as the representation, which is admitly not easy to write down without characteristic functions, has an easier probability interpretation than the characteristic function admits. I am responsible for the current procedures based on the characteristic function for the generation of random variables with stable distributions, as the method I saw based on the Ibragimov-Chernin representation of the density of extremal stable random variables was carried over to the Zolotarev extension of their formula. Alas, I do not know of any other situation where the procedure yields anything useful. >> However, the somewhat clumsy Lindeberg proof extends to >> the martingale case, and gives an idea of what is happening. >> I can give lots of somewhat unrela proofs, but does anyone >> really understand the theorem? >> Similarly, Cramer's proof of the Levy-Cramer theorem that >> the sum of two independent random variables is not normal >> unless they both are is probably the only easy proof, but >> it likewise obscures the ideas. Any time characteristic >> functions are used to prove a probability theorem, the >> concepts are not even present. >Would you say that the elementary proofs of say the prime number >theorem due to Selberg-Erdos etc, give more insight than methods >using complex analysis and Fourier analysis as done by Wiener, >Ikehara and Newman? I will have to pass on these, not having enough familiarity with them. The occurrence of prime numbers is sufficiently erratic that I do not feel that any of the approaches has achieved what Erdos would call appropriate for THE BOOK. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Homological algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VELis06821; >Can someone suggest a good book on Homological algebra?? The standrad modern text for a first course in H.A. is Weibel's An Introduction to Homological Algebra (Cambridge Studies in Advanced Mathematics, 38) Its not that expensive, and it does things in the correct order. If you want to go furhter, you can try Gelfand & Manin HTH DL === Subject: Re: Was: Convergence on a space with no topology by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VIOTn25039; >Convergence on a space with no topology >It was mentioned that: >Hey, how about including the context? > (S,d) is a metric space and > for all n in N, A_n subset S >The following does appear to be equivilant... >1. A_n -> A >What's your definition of A_n -> A ? Am I to presume it is #2? >2. A_n subset A forall n and for all p > 0, p in R for all a in A >exists q in N for all n > q exists b in A_n: d(a,b) < p If you believe me, you may assume the definition of your choice, since I believe them to be equivilant (for me, that means I have an outline lazy boy proof of this). The earlier definition, which I presume us both to have been using throughout our previous discussions, was A_n -> A <-> A_n subset A forall n and for all a' in A exists (a'_n) subset X such that a'_n -> a and a'_n in A_n for all n. >A stricter form of convergence: >Ya, uniform convergence. Agreed, uniform convergence is the way I've been refering to it in my notes, too, although I was yet to label it as such. >1. A_n ->* A >2. A_n subset A forall n and for all p > 0, p in R exists q in N >for all a in A for all n > q exists b in A_n: d(a,b) < p >So you work out a definiton of convergence for P(S). >A question then to ask is there a metric or topology for P(S) >based upon d of (S,d) that will give you that convergence. >Do you need to restrict S? >Consider if compact S helps. Ok, but lets not forget my all important B*, either. >Anyway the question has already be answered, the Hausdorff metric. >As it turns out, it's a metric for bounded closed subsets of S. >Thus for that metric, P(S) could be restriced to compact sets. see below >Now the inverse question, as the Hausdorff metric does give >convergence of A_n for compact subsets of P(S), can you convert >Hausdorff metric convergence into something more akin to what >you're wanting? >Whenever you work out a definiton of A_n -> A, will it give >convergence for more that compact subsets? If not, then you've >done nothing more that what Hausdorff did many decades ago. I think we agree here a lot, but, as for me, I remain unconvinced on this point. Take the example given on this page: N_n -> N. We know (no proof as of yet) that from this follows convergence in the Hausdorff metric. Since !(N_n ->* N), uniform convergence and convergence in the Hausdorff metric appear quite different, to me at least. see below >-- dh >Here's the extend of my notes on Hausdorff distance > dh(A,B) = max(sup{ d(a,B) | a in A }, sup{ d(b,A) | b in B }) >metric for nonnul closed bounded subsets of metric space (S,d) > dh(A,A) = 0. If dh(A,B) = 0: for all a in A, d(a,B) = 0 > A subset { x | d(x,B) = 0 } = cl B; similarly B subset cl A > dh(A,C) <= dh(A,B) + dh(B,C). Restricting a,b,c to A,B,C resp. > for all a,b, d(a,C) <= d(a,b) + d(b,C) <= d(a,b) + sup_b d(b,C) > for all a, d(a,C) <= d(a,B) + sup_b d(b,C) > sup_a d(a,C) <= sup_a d(a,B) + sup_b d(b,C) > sup_c d(c,A) <= sup_b d(b,A) + sup_c d(c,B) similarily > conclude with max(u+v, x+y) <= max(u,x) + max(v,y) >It'll show why closed bounded is needed just for the metric to be >defined. I prefer to think compact, for compact sets are indeed >closed bounded and I've hunch that compact's needed for convergence. >-- set theory >As you want convergence without a topology, why base A_n -> A upon >an underlying metric? Here's a simple definition of A_n -> A >For A_n subset S, > limsup A_n = /{ { / Aj | j > k } | k in N } > liminf A_n = /{ { / Aj | j > k } | k in N } >Yes, analogous to limsup, liminf of analysis. >Exercise: liminf A_n subset limsup A_n >A_n converges when limsup A_n = liminf A_n > (A definition of Cauchy sequence?) >and we say A_n -> A when > A = limsup A_n = liminf A_n >I think it a better way to go. It's not limi to limit from above or >from below as is yours nor will it haunt you with metrics or topologies. >In addition as it's an already established definition, we won't have the >problems your definitions are presenting. It's a more fruitful approach. Thoughts such as these have been in the back of my mind for a few days, (which, btw., is just a little bit longer than my thoughts on this subject as a whole have been.). Especially concerning non-uniform convergence -> as apposed to uniform convergence. I wondered if I could find example for A_n -> A with A != {x | x in A_n for infinitely many n}. which believe is equivilant either your limsup or liminf def. (forgive me for not knowing this off the top of my head) Just a thought: If B(0, r) subset X is a closed ball of radius r at 0 in a metric space X, and r_n -> r with r_n != r for all n, then B(0,r_n) -> B(0, r) Proof? (I've got another lazy boy proof of it). Three points to consider: 1. B(0,r) != {x | x in B(0,r_n) for infinitely many n} 2. Since B(0,r_n) -> B(0, r), B(0,r_n) should converge to B(0,r) in the Hausdorff metric. over the reals...) does it also follow that B(0,r_n) ->* B(0, r). For the moment I am entirely concetrating on point 3, i.e. trying understand more about uniform convergence. The way I see it as of now, it is inherently unlike set theoretic convergence described above. And note again, as was mentioned in a previous message, it has pleasant property A_n ->* A, then (A_n) is a *Cauchy-sequence. >Projects: >Revamp the defintion of A_n converges into something more > Cauchy sequence like >Determine when or if a topology can be introduced onto P(S) to make > A_n -> A equivalent topologically and set theoretically. >Does convergence by Hausdorff metric give set theory convergence? >Does your definition of convergence give set theory convergence? > Can you make it equivalent to set theory convergence? > I don't think so as it's based upon a metric while set theory > is more general and flexible that that. >Can set theoretic convergence in P(S) trickle down to > convergence in S? Havent had the time yet to go through your points, above. I will look at them again, tomorrow. (but still see below) >-- A simple example regarding both types of convergence: >Let N be the natural numbers (no zero) and N_n = {1,2, ...,n}. >It is easy to see that N_n -> N. However, !(N_n ->* N). >A_n subset A forall n and for all p > 0, p in R for all a in A >exists q in N for all n > q exists b in A_n: d(a,b) < p >Indeed, trivially > A_n -> /{ A_n | n in N } for any ascending sequence >do you have, using the dual superset definition for A_n -> A > A_n -> /{ A_n | n in N } for any descending sequence? >Both are so for the set theory definition. >Proof: >We have to check: >Every sequence of subsets of N convergent to N, isn't >uniformly convergent to N unless the sequence is eventually N. My notes, exactly. >Every sequence of subsets of R convergent to R, isn't uniformly >convergent to N unless the sequence is eventually dense. Yes, I agree. >You claim A_n ->* A iff >A_n subset A forall n and for all p > 0, p in R exists q in N >for all a in A for all n > q exists b in A_n: d(a,b) < p >For all n, let A_n = D be a dense subset of R. >Let A be any set with D subset A subset R. >Then A_n ->* A for all such A's. >Your uniform convergence, and hence your convergence doesn't have unique >limits. Just want to reiterate a statement from before: [snip] Under these conditions, it appears the limit is almost unique, i.e., it is unique if the set A' subset A your are refering to above is closed in the metric topology of X; (a proof followed) >Moreover, a sequence can even converge to different sets with >different cardinalities. Will have to look at this, later, I've got to run for now. >for all p > 0, p in R exists q in N for >all a in N for all n > q exists b in N_n: |a-b| < p >This is equivilant to (choose p < 1) >exists q in N for >all a in N for all n > q exists b in N_n: a = b >which, in turn, is equivilant to >exists q in N for all a in N for all n > q: a < n + 1 >This is not valid for a = q + 2 and n = q + 1, q.e.d. >Whatever math I dream up is already old hat. > -- William's Metatheorem >---- === Subject: Re: White Noise Dilemma by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA13UrL31982; >Here is a question that has had me in a quandary for several >years now. >Let X(t) be a zero-mean IID process with variance of >sigma^2. Now we know that since this function is >IID, it has a white PSD and therefore its autocorrelation >function at lag 0, R_XX(0), should be b*delta(tau), where >delta(tau) is the usual Dirac delta function and b is some >constant. >However, we also know that the autocorrelation function R_XX(tau) >is defined to be E[X(t)*X(t+tau)], and therefore that R_XX(0) >is E[X^2] (where we have dropped the dependence on t since the >process is IID). Therefore in this sense R_XX(0) = E[X^2] = sigma^2, >which is not infinite. >How do you resolve this discrepancy? Check again. The autocorrelation function is never infinite. Try normalizing === Subject: find (x,y) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1D8Zo02174; so: x+y=127008 x and y have 45 divisors in common. x,y natural === Subject: hypercube/tesseract help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1Jo9b26896; I am not at all mathematical, please keep in mind when replying. How could I find hypercube images for hebrew letters? Does anyone know how to create them? Is it even possible? === Subject: Re: someone told me ... by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1Jo7B26872; >That, somewhere in the decimal expansion of pi, >there is ASCII code for the date of my birth, the >date of my death, a thumb-scale sketch of my >life on this earth, and even a few JPEG's of me >at various activies during my life. >This is sci.math, and that statement must be >true, or false, or Godel unknowable. >Which is it? If you don't know the answer and >must speculate, please don't. If you do know >the answer, please post. >Ben Well, if normality is unknown for pi, what are some numbers that are proven normal? Every one of them would satisfy Ben's requirement. === Subject: Re: new way of describing ellipse for math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248C325624; >Playing around with stringtheory today I was in need of a more convenient method of >describing an ellipse. Instead of the two focal points and the major >and minor axis. I need a method to tie in directly to that of square and rectangle. >That the circle is to the ellipse what a square is to a rectangle and dispense with >the notion of focal points and axes. >I want this because in a Coulomb Unification the perfectness of a Coulomb force in >that it is so perfect it comes as a solo force whereas the StrongNuclear pairs with >the WeakNuclear to restore the nuclear region as a NuclearCoulomb and for the space >of electrons where the Gravity pairs with Antigravity restores a Coulomb force there. >Thus all the forces are Unifyed by being Coulomb forces of a particular region. And >it is stringtheory that can give math data as to the StrongNuclear force from Euler's >Gamma Function. >So I need a better way of going from circle to ellipse without foci and axes. >We can consider an ellipse as a squashed circle. We can consider a rectangle as >a squashed evenly square. >Proposal: If I start with a given rectangle and its 4 corners, I cannot construct a >circle by using those 4 corners. However, I can construct a ellipse using those >4 corners, but can I get a unique ellipse from a given rectangle? I think not. But I >want a unique ellipse given a particular rectangle. So, if I determine the diagonal >of that given rectangle and use it as the diameter of a circle, I wonder if I can >thus >use that circle to determine a unique ellipse that inscribes the rectangle. >What I want when finished is an easier way of describing ellipses by saying there is >a unique rectangle and a unique circle given a particular ellipse. So to speak the >circle that is squashed to get the ellipse. >The old pedantics was to fiddle around with major axis and minor axis and F1 >and F2. Remember the old pedantic of string attached to F1 and F2 and that >a pencil in the string will draw the ellipse. >Well, instead of that what I want in this new pedantic is that given a rectangle, >then what is the unique ellipse associa to that rectangle? And it is got from >the diagonals of the rectangle conver to the diameter of a circle and this circle >is then made to fit the 4 corners of the rectangle forming an ellipse. I get rid of >focal points and string. >The reason I want this for stringtheory is that the photon is perfect since the >Coulomb force is perfect and thus the photon would be a circle as a string. But the >Neutrino would be almost a circle since its restmass is tiny but not zero. So the >neutrino string would be almost a circle and almost a square instead of a >rectangle. But in the Strongnuclear force these string-ellipses would be very much >elonga especially when you get to heavy atomic nuclei such as uranium where the >nuclei are the most elonga. >I suspect that in Euler's Gamma Function that a string of stringtheory would rarely >become square or rectangle. That the photon would be the only circular >string and that most every other case of a string would be elliptical. >Can someone in mathematics tell me if my above method of getting ellipses >is a new method or whether some other mathematician has gone down that >path of describing ellipses? I would be awfully surprized if this is a new >method since so much of mathematics is trampled over and picked clean. >Archimedes Plutonium, a_plutonium@hotmail.com >whole entire Universe is just one big atom where dots >of the electron-dot-cloud are galaxies Haven't you ever used a drawing program? To draw an ellipse, you typically start at one corner of a rectangle and end at the diagonally opposite corner. The ellipse fits inside with its axes parallel to the sides of the rectangle. Old Hat. === Subject: Re: Repeat: White Noise Dilemma by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248Cw25630; >I find it odd (2n+1) that no one has responded to this. I'm guessing >that either a) everyone thinks it is a homework problem (it is not), >or b) no one knows the answer. Come on, all you big strong mathematicians - >certainly you can answer a little ol' question like this, can't you? ... >Here is a question that has had me in a quandary for several >years now. >Let X(t) be a zero-mean IID process with variance of >sigma^2. Now we know that since this function is >IID, it has a white PSD and therefore its autocorrelation >function at lag 0, R_XX(0), should be b*delta(tau), where >delta(tau) is the usual Dirac delta function and b is some >constant. >However, we also know that the autocorrelation function R_XX(tau) >is defined to be E[X(t)*X(t+tau)], and therefore that R_XX(0) >is E[X^2] (where we have dropped the dependence on t since the >process is IID). Therefore in this sense R_XX(0) = E[X^2] = sigma^2, >which is not infinite. >How do you resolve this discrepancy? >-- >% Randy Yates % ...the answer lies within your soul >%% Fuquay-Varina, NC % 'cause no one knows which side >%%% 919-577-9882 % the coin will fall. >%%%% % 'Big Wheels', *Out of the Blue*, ELO >http:// home.earthlink.net/~yatesc r You got the wrong delta function. You want the one that is zero everywhere except at the origin, where it is one. No infinities please. The autocorrelation you seek is equal to the variance when the two functions overlap and equal to aero otherwise. === Subject: Re: Something about standard deviation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248CE25643; >Standard deviation is calcula by taking the square root of the >average of the square of the deviations from mean...Isn't it? >Why is it like that...?? >A more meaningful approach is taking the average of the magnitudes of >the deviations from the mean..something which is more obvious ? >Is it for the ease of calculations ?? >Please clarify me... >Jean In addition to moments of inertia, electrical power is also best described by variance in an alternating current situation. So variance is useful in the real world while your function isn't === Subject: Re: Greek Alphebet by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1Jo8V26876; >
 In my studies on triginometry, and calculus, I am comming
across many
>> symbols which appear to be the Greek Alphebet, as there was
a chart of
>> the Alphebet in the front of the book. I am wondering if
someone could
>> tell me what they all mean, or point me to a site where I
could find
>> out.
>> Thank you,
>> Tom
>You need this rule: The symbols mean whatever the author says
they mean.
>Take each one in the context where it is discussed. Somewhere
else, the
>same symbol is very likely to mean something different.
>Even the venerable PI does not necessarily mean the constant
3.1415...,
>although that is probably the meaning in the context of the
subjects you
>mentioned.
>Lynn Killingbeck
===
Subject: Re: Why is math so difficult for some people?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA2ChuR27568;
>> I'm a perfect zero in art. Some people have no trouble
>> with it. Here is my vision of hell (something that has
>> actually happened at various personal-enrichment seminars):
>> somebody sits you down with a pile of magazines and a
>> pair of scissors and says make a collage representing
>> yourself. After all, anybody can make a collage,
>> right?
>Well, it could be worse. They could be coming in with a
>pile of bricks and a bunch of mortar and then asking you
>to make a college representing yourself. Then you'll
>have to be good at architecture too.
Brick are very lucky.
===
Subject: find (x,y)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA2Chu427562;
so:
x+y=127008
x and y have 45 divisors in common.
x,y natural
127008=(2^5)(3^4)(7^2)
===
Subject: Who contribu most to mathematics?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA2ChvA27582;
This is somewhat of a political statement but here it is
anyway.
I am a patriotic American who does not cheat on his taxes,does
not
drive drunk, does not carry a gun or knife, obeys traffic laws,
obeys the rights of others, is courteous to other people, does
not
litter, does not smoke, tries to be courteous to telemarketers
(sometimes you really can't), sees the other guy's point of
view,
even right wing republican's and who is sick and tired of the
French
bashing by the republican party.
I am tired of freedom fries, freedom cuffs, freedom ja vou,
freedom
bread, freedom curves, freedom connection, freedom kissing and
all
other bull the right wing disgustingly bestow on France, our
ally!
I think France has made more contributions to mathematics than
any
other nation. There is no Nobel prize for math but the French
out
rank the U.S there too or are very close.
If we list, up to now, all the mathematicians professional and
amateur including those who immigra to a country and gave each
points 1-100, I would bet France would top the list in total
points.
My SWAG of some of the others' rank.
Italy
Germany
Russia
India
Greece
England
Japan
China
Uni States
Egypt
Hungary
Austria
Serbia
.....
A good historian could probably do this in his head.
Why are we bashing France because they did not want to Invade
Iraq
on the basis of the lies of the Asses of evil BCR?
This nation should wake up and not forget who put them here
and keep
in damn good mind who is taking them away every day.
Again, I love my country, just not how is running it.
Cino Hilliard
===
Subject: Re: Who contribu most to mathematics?
> I am a patriotic American who does not cheat on his
taxes,does not
> drive drunk, does not carry a gun or knife, obeys traffic
laws,
> obeys the rights of others, is courteous to other people,
How about having the courtesy not to post this dreck to
sci.math?
--
===
Subject: Re: Why is math so difficult for some people?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA2ChuW27578;
> at 04:56 PM, leah_lidtorf@yahoo.com.au (Leah Lidtorf) said:
>>I dont get it.Im a perfect 0 at math.
>It could be you, it could be your textbooks, it could be your
>teachers. There is a cultural bias against intellectual
subjects, and
>there is a cultural bias against females in the sciences, so
right off
>the bat you're facing a double whammy. The public schools are
full of
>teachers who hate the subjects that they are teaching, and
children
>pick up on that. Studies have shown that teachers, even female
>teachers, tend to be less encouraging to females than to
males.
>>Am I too dumb for math?
>Second, don't be mislead by the popular prejudices. Women can
do
>Mathematics, some of them well enough that most of us would
give our
>right arms to have the same talent. Emmy N.9ather comes to
mind, but
>she's not the only one.
This lady was great
http://www.agnesscott.edu/lriddle/women/germain.htm
Cino
===
Subject: can't show that .....
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA3DAJZ21959;
Hi All,
I dont know where is mistake:
a and r are vectors, a=a1*i+a2*j+a3*k, a1,a2,a3-const,
r=x*i+y*j+k*z
fi=fi(r)
rot(fi*a x r))=(grad(fi).r).a-(grad(fi).a).r+2*fi*a
I use rot(fi*v)=grad(fi)xv+fi*rot(v) and can't get 2*fi*a,
always have
3*fi*a.
What am I missing???
Tnx
Ihor
===
Subject: Re: congruence arithmatic/modular arithmatic
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA3DB4Z22197;
>
 Are there any books available that deal exclusively or
almost exclusively
with
>> congruence/modular/clock arithmatic? Most number theory
books briefly
discuss
>> this subject, but I want more information and information
about
>applications of
>> this kind of math. Thank you very much for your help. James
D. Hirsch,
Ph.D.
>me too!
>any websites while we're at it?
>Jesse Samuels
>jcs.ng@asu.edu
===
Subject: Re: find (x,y)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA3DAna22183;
>so:
>x+y=127008
>x and y have 45 divisors in common.
>x,y natural
>127008=(2^5)(3^4)(7^2)
HINT: If x & y have 45 divisors in common, then
127008 has those same divisors (and maybe more).
===
Subject: Re: Mathematicians' quotes about !math
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA3DAQl22014;
When can a theorem be wrong?
===
Subject: Re: Mathematicians' quotes about !math
> When can a theorem be wrong?
When it's not the theorem they asked for.
--
===
Subject: Re: Union of hyperplanes
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA3H3nN06697;
>Does anyone know convenient references to the following
results?
>(1) Let V be a d-dimensional vector space over a finite field
F. If
>d>1, then V is a union of |F|+1 hyperplanes but not |F|
hyperplanes.
>(2) If F is an infinite field, or even a division ring, then
V is
>not a union of finitely many hyperplanes.
>S. Glasby
A quick proof for (2): we can assume that V=F^d.
A hyperplane H in V is given by a linear polynomial f in d
variables. A
union of finitely many hyperplanes is given as the set of
zeros in V of the
product of finitely many linear polynomials. Therefore if V is
the union of
finitely many hyperplanes, then there exists a non-zero
polynomial
f(X_1,...,X_d) in the polynomial ring F[X_1,...,X_d] such that
f(c_1,...,c_d)=0 for every (c_1,...,c_d) in F^d.
However if F is infinite, such a polynomial does not exist as
one can prove
by induction on d: for d=1 the assertion is clear because a
polynomial in one
variable of degree n has at most n zeros.
For d>1 write f=f_0+f_1X_d+f_2X_d^2+...+f_nX_d^n , where the
f_i are
polynomials in the variables X_1,...,X_(d-1).
For every (c_1,...,c_(d-1)) in F^(d-1) the polynomial
f(c_1,...,c_(d-1),X_d)
in one variable has infinitely many zeros by assumption. Thus
by induction
hypothesis it is the zero polynomial. Thus for all i and all
(c_1,...,c_(d-1)) the equations f_i(c_1,...,c_(d-1))=0 hold.
By induction
hypothesis this implies that all polynomials f_i are equal to
zero. So f is
equal to zero.
H
===
Subject: Re: Proving the Cubic Formula in college Calc.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA448dj21873;
>I have been trying to find a way to prove the Cubic Formula
for some
>extra credit in my class. If someone knew how to do it and
showed
>me... it would be greatly apprecia. Thank You!
Now if we gave this to you, would you deserve the extra credit?
Look up Cardan's formula and derivation
===
Subject: Re: minimum foam
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA42HCY14115;
>Hi Stephen
>>Does anyone disagree that space fills with irregular
tetrahedra
>>such that each vertex is shared by 20 tetradrahedra? NO?
Okay.
>>Then....
Place a vertex at the center of each tetrahedron in that
maximum
>>foam and connect the dots. Each vertex has 4 edges connec.
>>Voila! Space fills with irregular pentagonal dodecahedra.
>>Minimum foam.
Yes? No?
>>Does anyone know if space fills with irregular pentagonal
>>dodecahedra? Why or why not?
>> If you go to 4 dimensions, then the tetrahedra can be
regular.
>If I 3 go with dimensions, can I fill space with irregular
tetrahedra
>such that all edges share 5 tetrahedra and all vertexes share
20
>tetrahedra?
>> If you try to do this with distortion in Euclidean 3-space,
you will
>> get a distor version of the same result. It is analogous to
>> trying to tile the plane with triangles, while allowing
only five
>> around each vertex, or tiling with three pentagons around
each vertex.
>So your saying, yes, irregular tetrahedra and irregular
dodecahedra
>are both space fillers?
>Thanks
>Dick
Space can be filled with exactly 600 distor tetrahedra or with
exactly 120 distor dodecahedra. If the 4-dimensional polytopes
are drawn on the hypersurface of a hypersphere, they can be
projec
from a point on the hypersphere to a 3-dimensional hyperplane
tangent
to the diametrically opposite point on the hypersphere.
The projec images of the polyhedra will have spherically curved
faces, and the angles between the faces will be the same as on
the
hypersphere.
Unfortunately, their volumes will not be the same, as required
by the
minimum foam problem. Further distortion cannot remedy the
volume
of the one polyhedron which has infinite volume.
Sorry about that,
Stephen
===
Subject: Maple question
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA4FWL901507;
Does anyone know of a way to extract coefficients of certain
expressions,
like if I have z = A*f(x) + B*g(x), how can I get A (and B)
without copying
it from the output? I want something like A = get_coeff(z,
f(x))
I guess it could be possible to divide z by f(x) and then
expand it in a 0th
order taylor series to keep only the constant (A). Is there a
better way?
regards,
Daniel
===
Subject: Re: Maple question
> Does anyone know of a way to extract coefficients of certain
expressions,
like if I have z = A*f(x) + B*g(x), how can I get A (and B)
without copying
it from the output? I want something like A = get_coeff(z,
f(x))
Try the op(.) command
===
Subject: Problems on elementary functions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA4NKGn03790;
Is the function
sin x, if x>=0
f(x)=
arc sin x, if 0>x>=-1
an elementary function?
Is the function
sin x / x, if x>0 or x<0
g(x)=
1, if x=0
an elementary function?
How to prove your conclusions?
===
Subject: Re: Problems on elementary functions
> Is the function
> sin x, if x>=0
> f(x)=
> arc sin x, if 0>x>=-1
> an elementary function?
> Is the function
> sin x / x, if x>0 or x<0
> g(x)=
> 1, if x=0
> an elementary function?
> How to prove your conclusions?
Perhaps you should first give us the precise definition of
elementary
function which you wish to use.
David
===
Subject: Re: JSH: My victories get lost
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA51QQX13640;

You should do the same as your victories -- get lost.
===
Subject: real analysis - absolute continuity & variation of
fcns
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA564Fn31804;
Using the notation of Royden,
1)
w.t.s the total variaton of an absolutely continuous function
on [a, b] satisfies:
T^a _b (f) = int{over [a,b]} |f'| dx.
What am I missing? Isn't this result obvious from the FTOC?
2)
Also looking for a function of bounded variation that satisfies
T^a _b (f) > int{over [a,b]} |f'| dx.
... obviously this function cannot be absolutely continuous by
earlier statement. I'm having difficulties thinking up a
function
of bounded variation that isn't absolutely continuous.
===
Subject: Re: real analysis - absolute continuity & variation
of fcns
> Using the notation of Royden,
> 1)
> w.t.s the total variaton of an absolutely continuous function
> on [a, b] satisfies:
> T^a _b (f) = int{over [a,b]} |f'| dx.
> What am I missing? Isn't this result obvious from the FTOC?
No, it's not. The FTOC applies to continuous integrands, not
L^1 functions.
And even when f is C^1, I don't see the FTOC coming in. I see
a proof via
the mean value theorem and convergence of Riemann sums.
That T(f) <= int |f'| is trivial. To go the other way, you
need to prove
something like this: If g is in L^1, then
sum (i=1,n) | int_[x(i-1), xi] g | -> int_[a,b] |g|
as n -> oo. Here a = x_0 < x_1 < ... < x_n = b and the mesh
size -> 0.
> 2)
> Also looking for a function of bounded variation that
satisfies
> T^a _b (f) > int{over [a,b]} |f'| dx.
If you allow noncontinuous functions, this is easy. (Recall
that every
monotone function is BV.) For a continuous example you could
look at the
Cantor function.
===
Subject: Problem with Euler's function
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5DMI127963;
Please can anyone help me?
How can i prove that phi(n) is even if and only if n>=3?
I am right in saying:
1) suppose n>=3, and want to prove phi(n) is even:
If n is a prime, then phi(n)=n-1=even, since prime numbers are
odd.
If n is not prime, then n is product of primes, n=p[1]^a[1] x
.... x
p[m]^a[m]. for p[i] not equal p[j] if i not equal j.
Then phi(n)= product(from i=1 to m) of (p[i]-1) *p[i]^(a[i]-1).
since (p[i]-1) is even, then phi(n) is also even.
Is this a reasonable proof?
2) suppose phi(n) is even,
we know phi(1)=1 and phi(2)=1, phi(3)=2. So n must be at least
3.
Is this enough to be a good proof?
Thank you ever so much for your help.
===
Subject: Re: Problem with Euler's function
> Please can anyone help me?
> How can i prove that phi(n) is even if and only if n>=3?
> I am right in saying:
> 1) suppose n>=3, and want to prove phi(n) is even:
> If n is a prime, then phi(n)=n-1=even, since prime numbers
are odd.
> If n is not prime, then n is product of primes, n=p[1]^a[1]
x .... x
p[m]^a[m]. for p[i] not equal p[j] if i not equal j.
> Then phi(n)= product(from i=1 to m) of (p[i]-1)
*p[i]^(a[i]-1).
> since (p[i]-1) is even, then phi(n) is also even.
> Is this a reasonable proof?
Add the case where the only prime factor is 2 (i.e. n=2^x).
> 2) suppose phi(n) is even,
> we know phi(1)=1 and phi(2)=1, phi(3)=2. So n must be at
least 3.
> Is this enough to be a good proof?
> Thank you ever so much for your help.
No problemo,
--
Unpatched IE vulnerability: DNSError folder disclosure
Description: Gaining access to local security zones
Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html
===
Subject: problem with euler's function phi(n)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5DMQb27968;
How can i find all positive integers n such that phi(n) is not
a multiple of
3? I have read from a book that there are infinitely many
integers n so that
3 does not divide phi(n), but no formal proof is given, it
just said that
some of the integers are of the form 5^k, k=1,2,.....
But this is not helping me to find all the positive integers n
though.
Can anyone help me?
===
Subject: Re: problem with euler's function phi(n)
> How can i find all positive integers n such that phi(n) is
not a multiple
of 3? I have read from a book that there are infinitely many
integers n so
that 3 does not divide phi(n), but no formal proof is given,
it just said
that some of the integers are of the form 5^k, k=1,2,.....
> But this is not helping me to find all the positive integers
n though.
> Can anyone help me?
Phi is a multiplicative ( phi(p^n.q^m) = phi(p^n) * phi(q^m) )
so only need
to consider prime powers.
What's Phi(p^n) for
a) n=1
b) n>1
?
have 3 as a divisor of phi(n).
--
Unpatched IE vulnerability: Web Archive buffer overflow
Description: Possible automa code execution.
Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html
===
Subject: Re: problem with euler's function phi(n)
Kathy  grava .88 la saucisse et au
marteau:
> How can i find all positive integers n such that phi(n) is
not a
> multiple of 3? I have read from a book that there are
infinitely many
> integers n so that 3 does not divide phi(n), but no formal
proof is
> given, it just said that some of the integers are of the
form 5^k,
> k=1,2,.....
> But this is not helping me to find all the positive integers
n
> though. Can anyone help me?
If n = p1^a1 + p2^a2 + ... + pk^ak, then:
phi(n) = n(1-1/p1)(1-1/p2)...(1-1/pk)
phi(n) is a multiple of 3 if n is a multiple of 3 at a power
strictly
greater than 1 or if one of its prime divisors is of the form
3i+1.
Otherwise:
- if n is a multiple of 3, then n(1-1/3) = 2n/3 will not be a
multiple of 3
- if n is not a multiple of 3 and has no prime divisors of
the form 3i+1, none of the (pk-1)/pk will be a multiple of 3.
- on the contrary, if one of the pk is of the form 3i+1, then
(pk-1)/pk
will be a multiple of 3.
--
Nicolas
===
Subject: Re: Gaussian Continued Fractions
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5DMYw27987;
>> Try this: Instead of rounding down, round to the nearest
Gaussian
>> integer, regardless of its direction.
>This wouldn't be the right thing to do, I think. If we
consider a
>complex number (that we are trying to find the CF of) which
is a
>negative real, then we would want to round towards zero.
>This is because:
>if x = [a(1),a(2),a(3),...],
>then -x = [-a(1),-a(2),-a(3),...].
Leroy Quet
Using round to nearest, e = (3, -4, 2, 5, -2, -7, 2, 9, ... ).
This can be conver to the usual form as follows:
1. At each sign change, insert a term equal to 1 between them
while
reducing their absolute values by one each.
2. Replace all terms by their absolute values.
This gives: e = (2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ... ).
(Of course, this is assuming that the original number is
positive.)
So there is no problem with using round to nearest.
Besides Gaussian integers, there are also Eisenstein integers,
(a + b w), where a & b are integers and 1 + w + w^2 = 0,
and Hurwitz integers which are quaternions (a + b i + c j + d
k),
where the a, b, c & d are all integers, or else they are all
odd
half integers. There are 6 Eisenstein units and 24 Hurwitz
units.
Continued fraction expansions can be formed from each of these
by
alternately subtracting out the nearest integer and then
taking the
reciprocal of the result.
Analogously, using geometric inversion instead of the
reciprocal,
there are lattices of any finite dimensionality that can be
trea
the same way. This relates to the sphere packing problem
because
this procedure only fails if there is room to pack more spheres
of unit diameter in the holes of the lattice.
The traditional continued fraction series can be rela to a
regular tiling of the hyperbolic plane with triangles which
have
all three angles equal to zero. These triangles can be paired
to
produce squares, or a triangle can be grouped with its 3
neighbors to
produce a regular hexagon. The groups need to be placed
symmetrically
as if the joining edge is a mirror to make a square or a
hexagonal
tiling pattern.
The Gaussian version is rela to a 3-dimensional hyperbolic
tiling
with octahedra which have 90 degree angles between faces and
vertex
angles equal to zero.
The Eisenstein CF expansion is rela to a tiling with tetrahedra
which have 60 degree angles between faces and vertex angles
equal
to zero. By surrounding one of these tetrahedra with four
others,
a cube is formed, thus making a cube tiling pattern.
===
Subject: Re: Colors of Infinity : Arbitary Maths
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5DLxO27949;
10002021415341453416335333134135354135415341534413579+
77974341321.1534324456
46543444145643
534146444324534534345345414545315345645345345345343546542258588
8856546546546
456465456746844+45646767435467464655753456745675669+++
45674567564534535364563
746878756453486756453456467465746545646545646787867875424546+
456445
456
54
5
5
53
4
4
(protons structure)
done by Dr/phisicist Dan
===
Subject: Factorizations
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5H43g11605;
can anyone help me
===
Subject: Re: Factorizations
> can anyone help me
1st - Algebraic (Cyclotomic and Aurefeuillian)
2nd - Sieve/trial-divide
3rd - Pollard's P-1 or Rho.
4th - Lenstra's ECM.
5th - If small (<100 digits) QS, else more ECM
And that's prettu much all you need to know.
Try google for more info.
--
Unpatched IE vulnerability: DNSError folder disclosure
Description: Gaining access to local security zones
Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html
===
Subject: Re: I can't stand it anymore
[snip]
Unc,
Is it racist to state that Pygmies are short, and Watusis are
tall?.
--
There are two things you must never attempt to prove: the
unprovable -- and
the
obvious.
--
--
http://www.crbond.com
===
Subject: Re: I can't stand it anymore
>I have been biting my tongue about the IQ test but I can't
any more.
>How reliable is a test that use the term Asian to represent
the most
>diverse of ethnic and cultural groups?
>If you want to know more about the diversity, ask me and I
will give
>you tons of specific examples. Here is one: An example of
cultural
>differences (that affects their educational goals) of the two
groups -
>ethnically the same and speak the same EXACT langauge - in
Burma who
>are the descendants of the people who came to Burma from a
town known
>as Surat, India during the British colonial days.
> Now, do not assume that they are racially the same as the
majority of
>northern Indian population, i.e Indo-Aryan. What I am aware
of is that
>they are descendants of those who came from Central Asia (to
india)
>and may be MIXED with the locals of India (I haven't checked
into
>that) like the majority of the current day Pakistanis.
>See the complexity yet?
>Add to that the Afghans, particularly the Pathan, which also
form a
>comunity in Burma.
>And then you have all these different native groups in Burma,
namely
>Burman, Mon, Shan, Karen, Kachin, Chin, etc. where the Karen
with a
>good size Christian population now claiming that they are one
of the
>lost tribes.
> The Mon are the first group who adop Bhuddism which made
them the
>most literate group *in the old old days* through the monastry
>education. Shans are the stauch Bhuddists too. I read
somewhere on
>the Internet that Mon origina from ancient India. They look
like
>typical oriental though some have darker skin while others
are very
>light-skinned) instead of the dravidians of ancient Inida.
>Shan is ethincally similar to some groups in Thialand from
the area
>that borders Burma. Talking about Thailand, not all Thias are
>ethincally the same.
>Now, the Chinese would claim that the high score of IQ tests
is
>because of them Chinese, i.e not all Asians are equal. And the
>Japanese thinks (may be not so much anymore) they are
superior to all
>other Asians.
>Does that term *Asian* include the Arabs?
>What about the people of Egypt (I avoid calling them
Egyptians to
>differentiate from the ancient Egyptians) some of whom are
Arabs, some
>of whom are descendants of anciant Egyptians, some are
descendnats of
>a group called Berber (my spellig may be wrong).
>And how many Ethiopians look or act like a typical African.
>Why are people using the racist IQ tests as God's given ruler
to
>measure intlligence?
http://metropolis.japantoday.com/tokyo/recent/feature.asp
In the off chance you are interes in whether or not the
Japanese
people came through the star gate or whether they crashed and
are decended from the ones who crashed in China, I am inclined
to believe that they came through the stargate. But that does
not
detract from the Chinese who have equal right to be here.
We aren't going to change anything at this point.
I have seen on the net, some very Japanese looking ET's
and one in particular that looked very Japanese. But for the
most part, the ones with the large cranium, look more
Chinese. They are scientists as far as I know. The ones
that were here last year were an advanced race. They beamed
my son home from school if you can image that. At lunch time.
The really odd thing about it was, that for about 20 minutes
after,
his eyes were slan down. Then his appearance returned to
normal.
His conscious self didn't know anything about it. I think he
just
assumed he walked home.
At the time a bunch of us had been discussing the future as per
The 5th Element. And I was arguing for rejuvenation machines.
Seems silly I know. But no more silly than this...
http://www.rense.com/general41/flying.htm
or
this...
http://dbarkertv.com/UPDATE.htm
As I understand it, this is a reptoid ship.
Were the ones who arrived first, and notified the glactic
government,
after Hiroshima, distant relatives of the Japanese people?
That is a nice notion, is it not?
Ignore the stupid pictures of the alien autopsy.
Some humans are very sick mentally and spiritually. Those are
the
kind of people who make mock ups like that to hide their own
evil
deeds and to take advantage of their fellow man.
Thats the price we pay for freedom. We have to tolerate some
people
who truly should not be allowed to exist.
So what is wrong with the Japanese people?
They are still suffering from the old, how could God, allow
this
to happen?
That is a very good question.
That will undoubly be answered one day.
But take it from me, there is nothing wrong with the Japanese
people.
They are beautiful people.
And do have a very valid inherent right to be here.
But try to think of people as individuals, not as a race.
Do not ask what is wrong with Japanese people, say what is
wrong with
some human beings. You cannot blame a victim for a crime.
And the answer again, is the price of freedom.
The freedom to be bad as well as good, makes people free.
That does not however mean, there are no consequences for
people's actions.
Be thankful for your good nature. That is a blessing. Some may
see that as a weakness, but then they usually find out the
hard way, that there is only one true power in the universe.
Trying to compete with him, is a very silly idea that will
only lead to misery.
What Japan has to offer humanity, is invaluable to the future
of mankind.
So don't go changing.
-*-
===
Subject: Re: I can't stand it anymore
I have been biting my tongue about the IQ test but I can't
any more.
How reliable is a test that use the term Asian to represent
the most
>diverse of ethnic and cultural groups?
If you want to know more about the diversity, ask me and I
will give
>you tons of specific examples. Here is one: An example of
cultural
>differences (that affects their educational goals) of the two
groups -
>ethnically the same and speak the same EXACT langauge - in
Burma who
>are the descendants of the people who came to Burma from a
town known
>as Surat, India during the British colonial days.
Now, do not assume that they are racially the same as the
majority of
>northern Indian population, i.e Indo-Aryan. What I am aware
of is that
>they are descendants of those who came from Central Asia (to
india)
>and may be MIXED with the locals of India (I haven't checked
into
>that) like the majority of the current day Pakistanis.
See the complexity yet?
Add to that the Afghans, particularly the Pathan, which also
form a
>comunity in Burma.
And then you have all these different native groups in Burma,
namely
>Burman, Mon, Shan, Karen, Kachin, Chin, etc. where the Karen
with a
>good size Christian population now claiming that they are one
of the
>lost tribes.
The Mon are the first group who adop Bhuddism which made them
the
>most literate group *in the old old days* through the monastry
>education. Shans are the stauch Bhuddists too. I read
somewhere on
>the Internet that Mon origina from ancient India. They look
like
>typical oriental though some have darker skin while others
are very
>light-skinned) instead of the dravidians of ancient Inida.
Shan is ethincally similar to some groups in Thialand from
the area
>that borders Burma. Talking about Thailand, not all Thias are
>ethincally the same.
Now, the Chinese would claim that the high score of IQ tests
is
>because of them Chinese, i.e not all Asians are equal. And the
>Japanese thinks (may be not so much anymore) they are
superior to all
>other Asians.
Does that term *Asian* include the Arabs?
What about the people of Egypt (I avoid calling them
Egyptians to
>differentiate from the ancient Egyptians) some of whom are
Arabs, some
>of whom are descendants of anciant Egyptians, some are
descendnats of
>a group called Berber (my spellig may be wrong).
And how many Ethiopians look or act like a typical African.
Why are people using the racist IQ tests as God's given ruler
to
>measure intlligence?
> http://metropolis.japantoday.com/tokyo/recent/feature.asp
> In the off chance you are interes in whether or not the
Japanese
> people came through the star gate or whether they crashed and
> are decended from the ones who crashed in China, I am
inclined
> to believe that they came through the stargate. But that
does not
> detract from the Chinese who have equal right to be here.
> We aren't going to change anything at this point.
> I have seen on the net, some very Japanese looking ET's
> and one in particular that looked very Japanese. But for the
> most part, the ones with the large cranium, look more
> Chinese. They are scientists as far as I know. The ones
> that were here last year were an advanced race. They beamed
> my son home from school if you can image that. At lunch time.
> The really odd thing about it was, that for about 20 minutes
after,
> his eyes were slan down. Then his appearance returned to
normal.
> His conscious self didn't know anything about it. I think he
just
> assumed he walked home.
> At the time a bunch of us had been discussing the future as
per
> The 5th Element. And I was arguing for rejuvenation machines.
> Seems silly I know. But no more silly than this...
> http://www.rense.com/general41/flying.htm
> or
> this...
> http://dbarkertv.com/UPDATE.htm
> As I understand it, this is a reptoid ship.
> Were the ones who arrived first, and notified the glactic
government,
> after Hiroshima, distant relatives of the Japanese people?
> That is a nice notion, is it not?
> Ignore the stupid pictures of the alien autopsy.
> Some humans are very sick mentally and spiritually. Those
are the
> kind of people who make mock ups like that to hide their own
evil
> deeds and to take advantage of their fellow man.
> Thats the price we pay for freedom. We have to tolerate some
people
> who truly should not be allowed to exist.
> So what is wrong with the Japanese people?
> They are still suffering from the old, how could God, allow
this
> to happen?
> That is a very good question.
> That will undoubly be answered one day.
> But take it from me, there is nothing wrong with the
Japanese people.
> They are beautiful people.
> And do have a very valid inherent right to be here.
> But try to think of people as individuals, not as a race.
> Do not ask what is wrong with Japanese people, say what is
wrong with
> some human beings. You cannot blame a victim for a crime.
> And the answer again, is the price of freedom.
> The freedom to be bad as well as good, makes people free.
> That does not however mean, there are no consequences for
> people's actions.
> Be thankful for your good nature.
> That is a blessing.
> Some may see that as a weakness, but then they usually find
out the
> hard way, that there is only one true power in the universe.
> Trying to compete with him, is a very silly idea that will
> only lead to misery.
> What Japan has to offer humanity, is invaluable to the future
> of mankind.
Except what they have offered during their fascist days. They
never
apologized to us victims. My grandfather, a successful
business man
was murdered by the facist thugs at the prime of his life - he
was
barely 40 - for having close business associations with the
British
subjects). Don't tell me that the thugs did not torture him.
These beautiful Japanese didn't tell their offsprings the
atrocities
their government commit in Burma, Korea, and China.
> So don't go changing.
Nope. We just sit and watch the Chinese takeover.
Heck, Burma (Myanmar) should now be referred to as a Chinese
colony.
Burmese military dictatorhip is made up of those with Chinese
blood.
Of course, they are fully Burmanized.
===
Subject: Re: I can't stand it anymore
And yes, as far as we know, the same thing happened in India
years before.
But that is not as bad as mass extinctions in prehistoric
times.
So why do these things happen?
In the course of evolution, it must be necessary at times
to test certain limits.
Abuse of power is one that humans keep getting wrong.
And by power I don't necessarily mean strength.
But any abuse of power of power over another is wrong. To
properly handle power you need to have a lot of other
qualities. Like compassion. And the ability to cook small
fish. ;)
Look at the legends of Atlantis. Abuse of power.
Mu, same. Only totalitarian power.
Look at George Orwell. He addresses the abuse of power in 1984
and looks at the crappy part of human nature in Animal Farm.
Another part of being free to make mistakes, is to not be
sure that there is someone watching you decide.
Maybe there is, maybe there isn't.
People who act in good conscience, don't concern themselves
with
it at all. They are more free than one who tries to hide
trickery or deceit.
I think it was because of the treatment of women. That the
people had gone too far beyond normal sensuality into
fetishes. They had lost touch with nature. Lost touch with
their own humanity and were going too far with process
and order and formality, and subservience.
People need to be free. Japan is a very free nation today,
because of the sacrifices of their parents. Good or bad,
right or wrong, the problem appears to be fixed.
Japan is far better off today than India, or Pakistan, or
any dinosaur, or Neanderthal.
In fact it is one of the best countries in the world today.
Does it still have social problems? What country doesn't?
Would a person from America, rather go to Japan for a visit,
or to South Africa?
Is South Africa better off now, that the western world has
left them to their own devices?
Better for them perhaps, but I would not go there or take
my family there. I would however, love to visit Japan.
-*-
===
Subject: Re: I can't stand it anymore
: Nope. You favor the excellent and tolerate the mediocre.
Success and
: failure are not indistinguishable.
Well, it's pretty obvious to us which side of that line
*you're* on.
For how many years have you been promising us kg-sized
diamonds?
Considering
that you cannot even figure out how to post to usenet, it
doesn't come as
a big surprise to me that you are a failure in the more
important things
in life, e.g., being able to tell the difference between the
truth and a
lie.
-----
Richard Schultz schultr@mail.biu.ac.il
Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel
Opinions expressed are mine alone, and not those of Bar-Ilan
University
-----
. . . for while he was not dumber than an ox, he was not any
smarter.
-- James Thurber, _My Life and Hard Times_
===
Subject: Re: I can't stand it anymore
> amanda replied:
> I have been biting my tongue about the IQ test but I can't
any more.
> How reliable is a test that use the term Asian to represent
the
most
> diverse of ethnic and cultural groups?
> Are you Asian?
Yes. Not oriental though but grew up in that region.
> Rich
===
Subject: Re: I can't stand it anymore
amanda replied:
>>amanda replied:
>I have been biting my tongue about the IQ test but I can't
any more.
>How reliable is a test that use the term Asian to represent
the most
>diverse of ethnic and cultural groups?
>>Are you Asian?
> Yes. Not oriental though but grew up in that region.
So you recognize and understand the word. It has utility and
meaning.
Why do you get so upset about it?
Many hate the word American and raise all manner is idiotic
arguments
as to why it is wrong (and most of them defend the term native
American).
Do you have similar issues with the word American, or issues
like the
ones you have with the word Asian?
Rich
>>Rich
===
Subject: Re: I can't stand it anymore
> amanda replied:
>>amanda replied:
>I have been biting my tongue about the IQ test but I can't
any more.
How reliable is a test that use the term Asian to represent
the
most
>diverse of ethnic and cultural groups?
>>Are you Asian?
> Yes. Not oriental though but grew up in that region.
> So you recognize and understand the word. It has utility and
meaning.
> Why do you get so upset about it?
> Many hate the word American and raise all manner is idiotic
arguments
> as to why it is wrong (and most of them defend the term
native
American).
> Do you have similar issues with the word American, or issues
like the
> ones you have with the word Asian?
What gave you the idea that I was upset about the term Asian?
My only point was that IQ tests to determine people's
intelligence are bs.
> Rich
>>Rich
>
===
Subject: Re: I can't stand it anymore
> What gave you the idea that I was upset about the term Asian?
> My only point was that IQ tests to determine people's
intelligence are
bs.
Ah, of course. It would be far better indeed to administer
intelligence tests rather than IQ tests to determine people's
intelligence, right?
Really, you sound like someone who is disappoin by their
own results on such a test.
===
Subject: Re: I can't stand it anymore
amanda replied:
>>amanda replied:
>>amanda replied:
>I have been biting my tongue about the IQ test but I can't
any more.
How reliable is a test that use the term Asian to represent
the
most
>diverse of ethnic and cultural groups?
>>Are you Asian?
>Yes. Not oriental though but grew up in that region.
>>So you recognize and understand the word. It has utility and
meaning.
>>Why do you get so upset about it?
>>Many hate the word American and raise all manner is idiotic
arguments
>>as to why it is wrong (and most of them defend the term
native
American).
>>Do you have similar issues with the word American, or issues
like the
>>ones you have with the word Asian?
> What gave you the idea that I was upset about the term Asian?
You're entire original post, other than the first sentence.
> My only point was that IQ tests to determine people's
intelligence are
bs.
Their inventor, years ago, inven them not to determine
intelligence,
but as a test for slow learners. He had cautions against using
the test
as it's used today.
Nevertheless, the notion that they are totally worthless is
also worth
examining. The poster 'Uncle Al' (I think it was him) said
that they
measure
your ability to take IQ tests but that's naught but a
tautology. IQ tests
don't tale place in a vacuum and students should have learned
something
from
all those years in school.
Do you think that all tests are worthless? some are worthless?
Somewhere
in-between?
Rich
===
Subject: Re: I can't stand it anymore
> I have been biting my tongue about the IQ test but I can't
any more.
How reliable is a test that use the term Asian to represent the
most
> diverse of ethnic and cultural groups?
> [off-topic crapsnip]
> Apparently you've just failed another test -- the one
> that measures your ability to choose suitable newgroups
> for a given topic, and vice versa.
> Or may be math, physics, and chemistry are the subjects
Amanda got her
C-'s.
I was referring to people mentioning IQ tests in discussion
about
determinging intelligence of different race in all these 3
groups.
But, how could you possibly realized that? After all, you got
F in
all these 3 subjects.
===
Subject: Re: I can't stand it anymore
>But, how could you possibly realized that? After all, you got
F in
>all these 3 subjects.
Oh, I see. You're trolling - why didn't you just say that?
===
Subject: Re: I can't stand it anymore
> I was referring to people mentioning IQ tests in discussion
about
> determinging intelligence of different race in all these 3
groups.
> But, how could you possibly realized that? After all, you
got F in
> all these 3 subjects.
Seems to me that *you're* the one that brought it up.
So are you a Troll or just clueless?
===
Subject: Re: I can't stand it anymore
> I have been biting my tongue about the IQ test but I can't
any more.
> Which IQ test is this?
Whichever one people were referrign to in sci.chem, sci.phy,
and
sci.math.
I never checked into the those tetss. All I heard about IQ
test is
from the media when that book bell curve came out and from
people
using the term Asian.
> Martin Hogbin
===
Subject: Re: I can't stand it anymore
charset=iso-8859-1
> not all Asians are equal. And the
> Japanese thinks (may be not so much anymore)
> they are superior to all other Asians.
AHahahhaah.....ahahahaha.....easy babe, easy. This is a
hilarious chapter in the annals of human behavior.....
Serious Japanese insistence about their superiority reaches
back into the mist of time.........ever since the two peoples
diverged. According to the Japanese there was that Japanese
Princess who ed with and got impregna by a monkey.
Their offspring became the m/patriarch of all the Chinese
to come.............meanwhile on the Asian mainland the lore
of yore insists that in the grey mist of time there exis a
horny Chinese Princess that ed with and got impregna
by a monkey. Their off spring became the m/patriarch of
all the Japanese to come.............Meanwhile back in those
times of bygone wonders the Jewish proboscis superiority
was established by their claim to be God's chosen people,
to which the Pope's and his troops later on objec to
and retor that God had given this torch to Jesus and his
Xian Church was the superior gig now, which left the Yiddles
in the dust.
So, while in the West the good folks established exclusive
superiority over others by telling their God(s) what to do, the
much more practically inclined Asians achieved the same goal
by the insistence on some gross bestiality, a threat that was
suffiecnt for them to get to the top of the pecking order.
Ain't it a wonderful
world..........AHhahahahaha.....ahahaahnson
PS: somebody bring forth such lores about other cultures'
orginal attempts and explanations which were needed to give
them their raison d'etre at the top.
===
Subject: Re: I can't stand it anymore
>Why are people using the racist IQ tests as God's given ruler
to
>measure intlligence?
Probably for the same reason that people feel the need to
generalize
about all IQ tests.
The only surprising thing about this post was that you didn't
also
cross-post it to alt.flame, alt.fan.rush-limbaugh, and a dozen
other
unrela newsgroups.
===
Subject: Re: I can't stand it anymore
In sci.chem Norris  writ:
:>Why are people using the racist IQ tests as God's given
ruler to
:>measure intlligence?
: Probably for the same reason that people feel the need to
generalize
: about all IQ tests.
,
Why does everyone in the US insist on generalising about
people's need to
generalise about IQ tests?
Just curious...
Gavin
===
Subject: Re: I can't stand it anymore
>In sci.chem Norris  writ:
>:>Why are people using the racist IQ tests as God's given
ruler to
>:>measure intlligence?
>: Probably for the same reason that people feel the need to
generalize
>: about all IQ tests.
>,
> Why does everyone in the US insist on generalising about
people's need
to
>generalise about IQ tests?
> Just curious...
Is that a general curiousity, or something more specific?
But mostly I would say because they are seeking justification
for
social attitudes, generally or course. It has been ever thus,
just
that some of the target groups have changed.
The three grains of salt that I keep in mind wrt IQ tests are
the
Flynn effect (average IQ score increses with time), the fact
that in
1917 african americans from the four highest scoring northern
states
scored above whites from the nine southern states and that jews
were about as far below the white average as blacks are today.
Finally, a lot of people who read this web site are good little
gerbils, put any kind of test in front of them and they will
try to get 100%. There are a whole lot of other people
who if you ask them to take an IQ test instead of playing
golf, going shopping or whatever, say screw that, check
all the a's and let's get out of here.
josh halpern
josh halpern
> Gavin
>
===
Subject: Re: Math dependency logic REVISED


<3c65f87.0311020542.61653c2a@posting.google.com>
<2tncqv4kb8gf7chslv93qjbr34aca7h2qa@4ax.com>
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<3fa5b2a8_3@newsfeed.slurp.net>
<3c65f87.0311040738.6d7f37ec@posting.google.com>

<63dgqvcuool9gmchue32p952rpatmsm05l@4ax.com>

A8EwTYfhf*u~,Eu,tf6
$HN*MY&)u0G
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J6m5.EN?>Zh
Xh;Y
V|',x(js'Jfq02joVpj|#x
linux)
>> I kinda like this when you get all het up - you make a
bigger
>> fool of yourself than usual.
>> Here for example. All I said was that the disputes were not
over
>> the fact that if you start with something true and apply
correct
>> reasoning then the conclusion must be true, the disputes
>> were over the correctness of his reasoning. You say that
>> of course I'm lying? Then you must think that the people
>> who have been saying James is wrong have been disputing
>> the statement if you start with something true and apply
>> correct reasoning then the conclusion must be true.
>> You should point out where anyone has dispu that fact.
>> .
> You yourself dispu that fact. Starting from C1-C4,
> C1 AxAy[x=y -> Az(z in x <-> z in y)]
> C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in
> A)Classification
> C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
> Extensionality)
> by applying correct reasoning, one deduces Ex~(x=x).
> Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
point out
> the error.
So C1-C4 |- Ex~(x = x)? I assume that's right -- I won't check.
Ullrich said it didn't? Maybe, though if so, it's likely just
a minor
error (perhaps caused by context?).
Where does this entail a denial of if you start with something
true
and apply correct reasoning then the conclusion must be true?
Are you just lying? Or, shall we be charitable and assume that
you're
stupid?
--
Well *supposedly* a correct and profound math paper can get
published
in a 'reputable journal' which means that the journals I've
faced so
far may lose a lot of their luster once the full story comes
out.
--- , on the quality of math journals rejecting his paper
===
Subject: Re: Math dependency logic REVISED
I kinda like this when you get all het up - you make a bigger
> fool of yourself than usual.
Here for example. All I said was that the disputes were not
over
> the fact that if you start with something true and apply
correct
> reasoning then the conclusion must be true, the disputes
> were over the correctness of his reasoning. You say that
> of course I'm lying? Then you must think that the people
> who have been saying James is wrong have been disputing
> the statement if you start with something true and apply
> correct reasoning then the conclusion must be true.
You should point out where anyone has dispu that fact.
> .
>> You yourself dispu that fact. Starting from C1-C4,
>> C1 AxAy[x=y -> Az(z in x <-> z in y)]
>> C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)]
>> C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in
>> A)Classification
>> C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <->
x=y}] (Weak
>> Extensionality)
>> by applying correct reasoning, one deduces Ex~(x=x).
>> Exhibit of proof of Ex~(x=x) from C1-C4 and someone will
point out
>> the error.
>So C1-C4 |- Ex~(x = x)? I assume that's right -- I won't
check.
>Ullrich said it didn't? Maybe, though if so, it's likely just
a minor
>error (perhaps caused by context?).
Caused by context indeed. He quotes that thing I said really
a lot, but he never quotes the post where I actually _said_ it,
it's always a followup to a reply to a response. I'd like to
see
the original, where I make that statement with no > at the
start of the lines...
>Where does this entail a denial of if you start with
something true
>and apply correct reasoning then the conclusion must be true?
>Are you just lying? Or, shall we be charitable and assume
that you're
>stupid?
It _is_ quite remarkable how slow he's being about this point,
how it is that calling a proof incorrect, (correctly or
incorrectly)
does not involve denying that if you start with something true
and apply correct reasoning the conclusion must be true.
************************
===
Subject: Re: Math dependency logic REVISED
Adjunct Assistant Professor at the University of Montana.
[.snip.]
>>So C1-C4 |- Ex~(x = x)? I assume that's right -- I won't
check.
>>Ullrich said it didn't? Maybe, though if so, it's likely
just a minor
>>error (perhaps caused by context?).
>Caused by context indeed. He quotes that thing I said really
>a lot, but he never quotes the post where I actually _said_
it,
>it's always a followup to a reply to a response. I'd like to
see
>the original, where I make that statement with no > at the
>start of the lines...
http://groups.google.com/groups?safe=images&ie=UTF-8&oe=UTF-8&
as_umsgid=0e14
nu484feb15mpa21t5p0drsh9jbr94f%404ax.com
It is clear the context has been compromised. Here is that
part of the
exchange:
-- Begin Insert --
>>C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] LL2
>> And this follows from
>> AxAy[Az(z in x <-> z in y) -> x=y].
>Dull David hasn't noticed that C3-C4 prove
>~AxAy[Az(z in x <-> z in y) -> x=y].
You said set theory, which I took to mean ZF.
In ZF they do not prove this.
And in NBG they do not prove that Ex~(x=x). You
say that I'm too stupid to realize that they do
prove this. Exhibit of proof of Ex~(x=x) from
C1-C4 and someone will point out the error.
-- End Insert --
Exhibit a proof of Ex~(x=x) from C1-C4 and someone will point
out the
error is clearly meant to be
Exhibit a proof of Ex~(x=x) from C1-C4 [in ZF or in NBG] and
someone
will point out the error.
Prior comments were:
-- Begin Insert --
>Leave it to a brain-dead analysis teacher to
>say something like this. Not only
>haven't you the faintest idea why Ex~(x=x)
>follows from C1-C4, but you are far too
>lazy to attempt to figure this out.
>You may be good at something other than
>stiffing JSH, but nothing you have ever
>written regarding either logic or set
>theory gives any indication of this.
When you said set theory I assumed you meant
ZF. That's what set theory with no qualification
means these days.
If you meant NGB set theory then no, C1-C4 are
not inconsistent with set theory. It does _not_
follow that C1-C4 give an example of something
which is not equal to itself, or an example of
something which does not exist.
It is correct that I have no idea why Ex~(x=x)
follows from C1-C4. This is because (assuming
that NBG is consistent) NBG has a model in
FOL=. In that model everything is equal to
itself.
-- End Insert --
once again clearly circumscribing the discussion to specific
contexts
which John Correy is now misrepresenting.
===
Subject: Re: JSH: Attacking a proof is attacking yourselves
<3c65f87.0311051238.755f8789@posting.google.com>
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J6m5.EN?>Zh
Xh;Y
V|',x(js'Jfq02joVpj|#x
linux)
> See what I mean? I called David Ullrich a ing piece of dog,
> but here he is, yet again.
> The math professor trapped by his need to confuse others.
You're not trapped by a need to confuse others, right?
So, if we call you a ing piece of dog, you'll go away?
Just kidding. I love most of your material (though the ing
piece
of dog bits really aren't so clever --- think over this latest
raging asshole without a clue persona).
--
Jesse Hughes
By definition m is a variable. By definition all then (sic)
numbers
represen by letters are variables--that's algebra[,] Magidin.
-- shows deep understanding of algebra
===
Subject: Re: JSH: Attacking a proof is attacking yourselves
> Now I've found out that even though I've stepped out my
proof and
> explained in detail there are posters who just keep trotting
out the
> same things despite getting refu.
Maybe it's because _you_ keep trotting out the same thing
despite being
shown the errors in PAINFUL DETAIL!
> 7 and 22 are NUMBERS
Are they. Blimey! Didn't realise that. I don't think that 22
was covered
in my analytic number theory course...7 was, obviously, since
it's a
prime variable...er....number.
> Now I'm sick of it.
Good. So am I. I've kept fairly quiet about all this ridiculous
goings-on for over 7 years (that's a number, not a variable!)
but I
really have had enough. Shut up and go and do something
useful. I've
spent many years studying mathematics, number theory in
particular, and
have bumped into numerous people like you. None quite as
determined
though. Most reasonable people will admit that they have made
an
error and go away and work on it. Please do so.
>You people are going to play by some rules.
Er...how about the 'usual rules of mathematics'?
> ...anti-math behavior on display...
Hmm, I wonder where most of this comes from?
> After all, you people are attacking or sitting by while
people attack
> the idea that 7 and 22 are constant!!!
Not at all! I am quite happy with your assertion that 7 and 22
are
constants.
> And how can any of you not understand what it means for 22
not to have
> 7 as a factor in a ring where the only integer units are -1
and 1?
A very good question. Do _you_ understand what it would mean?
> These are *basic* points people...
Certainly are!
> YOU ATTACK THE VALIDITY OF BASIC THINGS THAT PEOPLE CAN BE
CERTAIN
> MUST BE TRUE!!!
Hmm. 'Goddamn proof'? Best not to bring religion into a
discussion about
mathematics. Always a bad mixture!
> Oh yeah, make no mistake, what I've given is a mathematical
proof.
Indeed it is, albeit wrong. Don't feel bad though, even the
best
mathematicians have made some corking errors!
> I've proven that *as a group* mathematicians can be so
irrational as
> to question that 22 and 7 are constant.
Er, no. I don't think that even 'modern math' teaching asserts
that 7
and 22 are anything other than constant. I could be wrong
though.
> You are attacking your own interests by attacking
mathematical proof.
Ooo no, you seem to misunderstand the idea of publishing work
on a forum
such as this...only by getting others to examine our work do
we find
errors that, by definition, we cannot see ourselves. In an
on-topic
aside, look at the progress of Wiles' own proof of FLT; when
colleagues
spot error(s) he took a nice walk, went back to his desk, and
worked
on fixing what was broken.
What he did _not_ do is proclaim that his detractors (read:
peer
reviewers) did not understand what he was doing (doubtless,
many of them
did not understand _all_ of his work but they examined what
they _did_
understand) and that they were all ganging up on him!
Having been in a similar situation (albeit on a much smaller
problem)
and, in fact I'm still working on it after 3 years (I probably
will
never get there but I'm learning a lot in the process of not
getting
there!)
In short, whilst mathematical proofs are wonderfully powerful,
the
process of producing them can be very difficult and strewn
with errors!
>
ttfn (no doubt)
JasonG
===
Subject: Re: JSH: Attacking a proof is attacking yourselves
> Now I've found out that even though I've stepped out my
proof and
> explained in detail there are posters who just keep trotting
out the
> same things despite getting refu.
> Maybe it's because _you_ keep trotting out the same thing
despite being
> shown the errors in PAINFUL DETAIL!
That's not true. I've stepped out the proof into 7 main steps.
If you think there's an error, point it out.
Math society needs to either play by rules or accept the
consequences
of not playing by the rules, which is that the world not
believe what
you say!!!
If mathematicians can't accept a proof as simple as mine,
where not
accepting it means questioning basic algebra, the fact that 22
and 7
are constants, and that 22 does not have 7 as a factor in the
ring of
algebraic integers, then why should the world believe you on
anything?
Maybe you're confident that you can fool the world, like
magicians
with illusions, so that you can keep going confidently suppor
by a
deluded world.
But can you really fool yourselves?
Think about the corrosion upon math society as student after
student
learns that even a math proof can be questioned if enough
mathematicians don't care for the result. What good will a
math prize
be then, eh?
What will it matter when everyone knows the game is rigged, so
who
knows what's really true, and what's just a truth that
mathematicians have decided they want?
Attack the proof and you attack the foundations that make your
work
meaningful.
===
Subject: Re: Algebraic Closure
>What is the algebraic closure of a finite field?
I'd be happy just to know the algebraic closure of Z/<2>.
> Let K be a finite field and L the algebraic closure of K.
Then we know
that for every integer n there exists exactly one field M
within L that has
degree n over K.
> The field M looks like K[X]/pK[X], where p is an arbitrary
irreducibel
> polynomial of degree n.
> There are algorithms to determine the irreducibel
polynomials over the
> field K (for example Berlekamps algorithm).
> The field L is the union over all the field M, and these
fields form
> an ascending chain.
> Every algebraic computation within L typically involves only
finitely
> many elements. So you can perform the calculation in one of
the M's,
> that are under your control.
> So in a sense one can say that the algebraic closure of K is
>known<.
First off, it is easy to show that if F has p^a elements and E
has
p^b, then F is (isomorphic to) a subfield of E iff a divides
b. Armed
with this, you can form the tower of fields GF(p), GF(p^2),
GF(p^6),
GF(p^24),...,GF(p^n!) and their union is the algebraic closure
of
GF(p). The point is that any algebraic extension of a finite
field is
a root extraction and one sees that all roots of the elements
of GF(p)
appear somewhere in that tower. This is about as explicit as
you are
going to get.
===
Subject: Re: Algebraic Closure
> What is the algebraic closure of a finite field?
> I'd be happy just to know the algebraic closure of Z/<2>.
On the algebraic closure of two, by H.W. Lenstra, Jr. (Nederl.
Akad.
Wet.,
Proc., Ser. A 80, 389-396 (1977)).
===
Subject: Maximizing and minimizing in optimization
Is it possible to convert a problem where we are trying to
maximize a
quantity (profit) into a problem where we are trying to
minimize a
(different) quantity (loss)? If yes, can we do the transform
in linear
time? In particular, is it possible to transform a Network flow
problem where we try to maximize the flow between the source
and sink
into a shortest path problem where we are trying to find the
shortest
between two points (say source and sink)?
Thanks and regards,
--
C P Pradip
===
Subject: Re: Maximizing and minimizing in optimization
>Is it possible to convert a problem where we are trying to
maximize a
>quantity (profit) into a problem where we are trying to
minimize a
>(different) quantity (loss)?
Maximizing f is the same as minimizing -f. Or perhaps what you
have in
mind is the equivalence in Linear Programming between solving
the primal
problem (a max problem) and the dual problem (a min problem).
> If yes, can we do the transform in linear
>time? In particular, is it possible to transform a Network
flow
>problem where we try to maximize the flow between the source
and sink
>into a shortest path problem where we are trying to find the
shortest
>between two points (say source and sink)?
Maximum flow is equivalent to minimal cut. If it's on a planar
graph,
and the source and sink are on the same face, then this is
equivalent
to a shortest-path problem in the dual graph.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: naive geometry questions
> Here are two geometry questions from a curious layperson.
> I think these are wonderful questions; such a curious
layperson is
> welcome into my classes any time! Let met take them in
opposite order.
[snip]
> Cool, huh?
Very
I wish having had you as a diffgeo professor,
in stead of that brilliant but boring man we had.
Dirk Vdm
===
Subject: Re: naive geometry questions
> Here are two geometry questions from a curious layperson. If
they're
unclear,
> please tell me. Ditto if they're addressed to the wrong
forum.
1) Are there surfaces that can be bent to each other (the way
a flat
piece of
> paper is bent to a cylinder) but not to anything flat?
> I'm not sure I understand very well... it depends on what
> kind of bending would be allowed. For instance, would
> a cube and a sphere be 'bendable' into each other?
> Spheres and eggs?
> If you think about it, I guess you will have a hard time
> explaining (and even defining) what you mean exactly
> with 'bending'.
How about 'preserving Gaussian curvature'? The surfaces with
Gaussian
curvature = 0 are developable. E.g. cones, cylinders, the
plane.
--
G.C.
===
Subject: Re: naive geometry questions
>The other surface is the catenoid, formed by rotating a
catenary,
Q: How do you make a catenoid?
A: Pull its tail.
Lee Rudolph
===
Subject: Re: naive geometry questions
>The other surface is the catenoid, formed by rotating a
catenary,
> Q: How do you make a catenoid?
> A: Pull its tail.
How do you make a hormone?
--
Subject: Re: naive geometry questions
===
The other surface is the catenoid, formed by rotating a
catenary,
Q: How do you make a catenoid?
A: Pull its tail.
> How do you make a hormone?
:-))
Who says Geometry! when he has grown up?
Dirk Vdm
===
Subject: VECTORS AND SCIENCE?
One of my students has given me a chemistry math problem which
he
found and I can't see how to solve it.
It reads like this:
Consider a molecule such as cyclohexane in the trans
configuration in
which as the bonds are the same length
|>
c | | b
d | |
<---------------< |
| |
| >---------------->
e | | a
| | f
< |
(Sorry for the bad diagram but the arrows chase each other
round
a->b->c->d->e->f)
and the angle theta between two consecutive bonds is constant
throughout the molecule. (Notice also that a and d are
parallel). The
angle of skew alpha between two bonds are separate by a single
intermediate one is defined as the angle which these two bonds
appear
to form when viewed along the intermedite one. Show that this
is given
by
cos alpha = - cos theta / 2 (cos (theta/2))^2
Does any one know how to solve this? I would really appreciate
it!
Sarah
===
Subject: Some topological questions
I think I have solved a problem in my recent dimnsion theory
homework.
It relies on these claims:
A set that separates a space also separates a subspace of that
space.
The open halfspace R^n+, where the n:th coordinate is positive
(and not
merely non-negative), is homeomorphic with R^n.
If both of thse are true then my answer should be sufficient.
Are they
true?
Thanks in advance for any answers.
--
/-- Joona Palaste (palaste@cc.helsinki.fi) -------------
Finland --------
-- http://www.helsinki.fi/~palaste ---------------------
rules! --------/
It's time, it's time, it's time to dump the slime!
- Dr. Dante
===
Subject: Re: Some topological questions
Adjunct Assistant Professor at the University of Montana.
>I think I have solved a problem in my recent dimnsion theory
homework.
>It relies on these claims:
>A set that separates a space also separates a subspace of
that space.
So: the set {(x,0) : x in R} separates R^2. Therefore, it also
separates the subspace {(a,b): a,b>0} of R^2...
Surely you have more hypothesis? You need something like this,
I
think:
Suppose that X is a space, and A is a subset that separates X,
so
X-A = Y_1 disjoint union Y_2. Let B be a subspace of X, and
assume
that B intersect Y_1 and B intersect Y_2 are both nonempty.
Then
Acap B separates B.
Is that what you were thinking?
>The open halfspace R^n+, where the n:th coordinate is
positive (and not
>merely non-negative), is homeomorphic with R^n.
Let f:(0,infty) -> R be your favorite homeomorphism.
Then map R^n+ to R^n by the identity on each coordinate but
the n-th,
and by f on the nth coordinate. Is that a homeomorphism?
===
Subject: Re: Some topological questions
 scribbled the following:
>>I think I have solved a problem in my recent dimnsion theory
homework.
>>It relies on these claims:
>>A set that separates a space also separates a subspace of
that space.
> So: the set {(x,0) : x in R} separates R^2. Therefore, it
also
> separates the subspace {(a,b): a,b>0} of R^2...
> Surely you have more hypothesis? You need something like
this, I
> think:
> Suppose that X is a space, and A is a subset that separates
X, so
> X-A = Y_1 disjoint union Y_2. Let B be a subspace of X, and
assume
> that B intersect Y_1 and B intersect Y_2 are both nonempty.
Then
> Acap B separates B.
> Is that what you were thinking?
Yes, more or less. I understand that a set that separates a
space
does not separate a subspace if the set and the subspace are
disjoint. However such is not the case in my homework. In my
homework the set and the subspace intersect each other. So
therefore
the subspace is separa by a non-empty set.
>>The open halfspace R^n+, where the n:th coordinate is
positive (and not
>>merely non-negative), is homeomorphic with R^n.
> Let f:(0,infty) -> R be your favorite homeomorphism.
> Then map R^n+ to R^n by the identity on each coordinate but
the n-th,
> and by f on the nth coordinate. Is that a homeomorphism?
It should be. I used the exact same reasoning in my homework.
As my
favourite homeomorphism I used f(x) = ln x.
Thanks for your answer!
--
/-- Joona Palaste (palaste@cc.helsinki.fi) -------------
Finland --------
-- http://www.helsinki.fi/~palaste ---------------------
rules! --------/
Ice cream sales somehow cause drownings: both happen in summer.
- Antti Voipio & Arto Wikla
===
Subject: Re: Some topological questions
>  scribbled the following:
>I think I have solved a problem in my recent dimnsion theory
homework.
>It relies on these claims:
>A set that separates a space also separates a subspace of
that space.
>> So: the set {(x,0) : x in R} separates R^2. Therefore, it
also
>> separates the subspace {(a,b): a,b>0} of R^2...
>> Surely you have more hypothesis? You need something like
this, I
>> think:
>> Suppose that X is a space, and A is a subset that separates
X, so
>> X-A = Y_1 disjoint union Y_2. Let B be a subspace of X, and
assume
>> that B intersect Y_1 and B intersect Y_2 are both nonempty.
Then
>> Acap B separates B.
>> Is that what you were thinking?
>Yes, more or less. I understand that a set that separates a
space
>does not separate a subspace if the set and the subspace are
>disjoint. However such is not the case in my homework. In my
>homework the set and the subspace intersect each other. So
therefore
>the subspace is separa by a non-empty set.
Again, not enough: take two circles that are tangent at a
point;
this space is separa by the point of tangency, but the point of
tangency does not separate each of the circles separately,
even though
the intersection is nonempty (in fact, even though they each
CONTAIN
the entire separating subset). You really need the subspace
not just to
intersect the separating set, but to intersect both parts that
the big
space gets separa into. But if you have that, then your
conclusion
holds.
>The open halfspace R^n+, where the n:th coordinate is
positive (and not
>merely non-negative), is homeomorphic with R^n.
>> Let f:(0,infty) -> R be your favorite homeomorphism.
>> Then map R^n+ to R^n by the identity on each coordinate but
the n-th,
>> and by f on the nth coordinate. Is that a homeomorphism?
>It should be.
Well, prove it! Show that the inverse image of an open ball is
a
product of open intervals, and that will do it.
===
Subject: Stone's Representation Theorem(s?): help!
I'm not english, so I dearly hope none of my terms turns out
to be
misleading or unaccurate.
I'm a math student, I'm going to have an exam that has a part
on Stone's
Representation Theorem, and I miss, on my notes, a part of it,
I'd wish to
know if anyone can either help me to complete it or address me
to a place
where I can find it. Let me clarify the situation:
In my course, Stone's Representation theorem has been divided
in two parts,
called Stone's First Theorem and Stone's second Theorem.
The first one is the one i've seen talked about somewhere on
this
newsgroup,
which says that if we take a Boole algebra B, then build the
set of the
ultrafilters on this Boole algebra, U(B), and then consider
the set of
parts
of U(B), P(U(B)), there is a function psi: B -> P(U(B)) which
allows us to
both structure U(B) as a topological space and then consider
the Boole
algebra of its clopen, B(U(B)), subset of P(U(B)), and,
considered only
between the two structures B and B(U(B)), that is, narrowing
the codomain,
psi turns out to be an isomorphism between Boole algebras.
This First
Theorem I know, I have completely, and it's quite simple to
demonstrate.
The Second Theorem is, it seems, far more difficult, and I
definitely miss
a
part of it. Beyond this, I've never heard about it on
sci.math, and I
wonder
if it exists under some other name, since when the Theorem of
Stone is
named
everyone immediately talks only about the aforementioned
isomorphism.
The First Theorem ended with finding an isomorphism between
two structures
(two Boole algebras); the Second theorem aims to find a
_homeomorphism_
between two _topological spaces_, passing through a Boole
algebra, exactly
as the first theorem arrived from a Boole algebra to another
one by passing
through a topological space.
We start from X, topological space, and build the Boole
algebra of its
clopens, B(X). Then we build the set of the ultrafilters of
this Boole
algebra, U(B(X)). Then we define a function h: X -> U(B(X))
defined like
this:
for every x belonging to X, h(x) is {A in B(X) | x is in A}.
It's shown quite easily that this subset of B(X) is indeed an
ultrafilter,
and therefore is indeed an element of U(B(X)). We structure
U(B(X)) as a
topological space using the topology induced from the old
psi-function,
which makes U(B(X)) compact and totally disconnec (it was
proved in the
First Theorem).
We then suppose that X is totally disconnec (this hypothesis
is always
taken as gran , from now on), and from that we deduce that h is
injective.
Then, some properties get demonstra, which use in parallel h
and psi,
and
precisely:
1) For every A in B(X), h[X] / psi(A) = h[A]
2) h is a continuous function
3) For every A in B(X), the closure in U(B(X)) of h[A] is psi
(A).
4) in particular, the closure of h[X] in U(B(X)) is U(B(X))
itself, and
therefore h[X] is dense in U(B(X)).
So we have demonstra that h is continuous, injective, and also
that its
image is dense in U(B(X)).
Here ends the part of the theorem that I have.
I have the feeling that we're very near to demonstrate that h
is also
surjective (maybe, it could be enough to demonstrate that h[X]
is closed in
U(B(X)), therefore its closure, U(B(X)), is h[X] itself, but I
don't know
if
it's possible to prove it), and that it has an inverse
function that is
continuous (maybe it could be shown that h is open, or that
it's closed, I
don't know...), to prove that h is a homeomorphism and finish
the theorem.
Can anybody either help me to fill the blanks or give me an
advice of the
right text to look at in my university's library (we have tons
of books and
publications in english, here) to complete the proof? A big
big thank you
to
anyone who'll want to help me.
Federico
===
Subject: Technical Name Of Equatorially Concentric Rings?
I know that polar concentric rings on a sphere/spheroid are
known as
lines of latitude or parallels, but what are equatorially
concentric
rings called?--e.g., the 90 equatorially concentric ring
would be a
line of longitude or meridian, but what about the other,
semi-circle
rings **parallel TO A MERIDIAN**?
In terms of arcradius/radius of curvature, the perpendicular
meridian value is known as the normal: Is it that a meridian is
the prime normal, equals the 90 normal; the parallel
semi-circle/ellipse 1 away is the 89
normal, 2 away is the 88
normal, 3 away is the 87 normal, etc.,
in the same way that the
equator is the 0 latitude, 1 away is the
1 latitude, etc.?
Or, as an annulus is a band bounded by two concentric rings,
could
all of the rings comprising the annulus be something like
annulobes?
~Kaimbridge~
-----
WanKaimbridge (w/mugshot!):
http://www.angelfire.com/ma2/digitology/Wan_KMGC.html
----------
DigitologyThe Grand Theory Of The Universe:
http://www.angelfire.com/ma2/digitology/index.html
***** Void Where Permit; Limit 0 Per Customer. *****
===
Subject: Re: Limsup, Liminf
> Since there are finitely many natural numbers m <=N, for the
> rearrangement (and possible subset), there must be a K for
which all the
> s_n's <=N are in the s_k's <=K. You compute the new boundary
K from the
> old boundary N. The new boundary may be higher, but it does
exist.
I see it now, many thanks for your patience.
===
Subject: Re: c-number
> I'm reading Peskin and Schroeder's Intro to QFT and in this
> book the author constantly refers to c-numbers... without
> saying what a c-number is. I couldn't find c-number
> on mathworld.wolfram.com or physicsworld.wolfram.com.
> Can someone please tell me what the hell a c-number is?
 adam
>>I haven't read Peskin and Schroeder but if their usage of
the term
>>c-number is the same as everyone else's usage then they are
referring
>>to commuting variables. For example, Bryce DeWitt introduces
the
>>concepts of c-numbers (commuting variables) and a-numbers
(anticommuting
>>variables) in his book Supermanifolds. If a,b are c-numbers
then
>>a*b = b*a
>This is correct. The problem is that Peskin and Schroeder
don't go
>into 'grassman numbers' in any detail at all. DeWitt's book
is a good
>intro.
>--Dan Grubb
okay, thanks. I think you're right about the definition of a
c-number,
I asked my professor about it and she said some crap about
commuting and
stuff, but then she said that I could just think of a c-number
as
a complex number.
thanks alot,
adam
===
Subject: Re: Key core error argument, stepped out
> Mathematicians are *still* apparently hell-bent on running
from the
> result which is key in showing an over hundred year old
problem, but
> at least arguing with them gives me a sense of places where
people get
> confused. So here's the argument again with slight changes
based on
> what I've gathered.
One other thought on this occurred to me - see after 7., below.
> I also number out the main steps, so if anyone thinks
there's a
> problem and wishes to reply back, they need to give at least
*some*
> numbers.
> 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078,
where x is
> in the ring of algebraic integers, notice that P(x) has a
constant
> term that is 1078.
> 2. It can be shown that
> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
> where the *same* polynomial has been put in a form which
allows a
> factorization into non-polynomial factors so that I have
> P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
> where the a's are roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
> 3. Now let x=0, so
> P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)
> as the cubic defining the a's at x=0 is
> a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked
a_1(0) and
> a_2(0) to equal 0, which leaves a_3(0) with a value of 3.
> 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched.
Then I
> have
> P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
> P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
> 5. Now P(x) has a factor of 49 as
> P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22
> which means that
> (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)
> has a factor of 49.
> Most posters seem ok with the steps up to the final two. In
what
> follows it's important to understand the word coprime.
> In integers, 2 and 3 are coprime as are 12 and 13, as it
simply means
> they don't share non-unit factors.
> However, in reals, NO numbers are coprime, as for instance,
2(3/2) =
> 3, so 2 is a factor of 3.
> That's an important point to consider going forward.
> 6. However, the constant term of P(x)/49 is 22, which is
verified by
> again setting x=0, which gives P(0)/49 = 22.
> But for two of the factors of P(x), the constant terms is 7,
which is
> coprime to 22. Therefore, *none* of the constant terms can
have 7 as
> a factor.
> (By saying that 7 is coprime to 22, I'm making a choice as
to where
> the proof is going. Since I've been talking about algebraic
integers,
> where 7 is coprime to 22, it's natural to go with a choice
where 7 is
> coprime to 22.)
> Given that the constant terms are independent of x's value,
it must be
> the case that dividing P(x) by 49 divides the two constant
terms equal
> to 7, by 7.
> 7. But to divide 7 from those constant terms requires
dividing
> through two of the factors, so
> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
 300125 x^3 - 18375 x^2 - 360 x + 22
> from reverse use of the distributive property, which gives a
constant
> term coprime to 7, as required.
> For some odd reason I've STILL had mathematicians refusing to
> acknowledge the truth.
The peculiar thing about this argument is that it doesn't
have a conclusion. JSH says that 7 must divide through two
of the factors, but what does divide mean?
No doubt JSH will say this is a silly question.
But there is content. In a previous version of this
argument, JSH was concluding that a_1/7 (or a_1/f, previously)
must be an algebraic integer. That conclusion is no longer
sta.
In fact it is not sta WHAT a_1/7 is.
Is it an algebraic integer? Is it an object ? Is it
just an algebraic *number* ?
What does divide mean? In what ring does the result of
the division exist?
We know that it is not the ring of algebraic integers. There
are counterexamples, which JSH now seems to accept, that show
that
is not possible. If the above proof were valid with the
conclusion
that a_1/7 is an algebraic integer, we would have a proof that
mathematics is inconsistent. JSH seems to reject this idea.
If all one can say is that a_1/7 is an algebraic number, there
is just no point in even mentioning it. That would be obvious
without any proof at all.
If a_1/7 is an object, that too seems rather vapid. No one
seems to know exactly what the ring of objects includes. JSH
has refused to answer questions about it. In any case he is not
saying that in this version of the proof. Plus properties of
the ring of objects are not established. Does it make any sense
to announce to the world, a_1/7 is an object! ?? Why should
anyone care? An object which is not an algebraic integer is
nonetheless an algebraic number or a transcendental, i.e., a
real
number. Thus the conclusion might be a_1/7 is a real number
which
is not an algebraic integer! By far the most likely response to
that would be, SO WHAT ??? We all accept that anyway.
Another possible statement of the conclusion: a_1/7 is an
algebraic integer divided by the number 7! True! We all accept
that a_1 itself is an algebraic integer. But to claim as a
startling
conclusion the fact that a_1/7 is an algebraic integer divided
by 7 - the response to *that* is going to be, Well, DUH!
So here we have these 7 steps about which much arguing has
occurred in the last few days, but there isn't any conclusion.
At least it is not sta here. What have we been arguing
about? What truth are we supposed to be acknowledging?
What is JSH now claiming to have proved ?
> But, surprisingly, the proof here, from what I read in a
link pos
> further down, is tighter than what mathematicians usually
claim is a
> proof, and represents perfection at a level most
mathematicians, even
> professional mathematicians, don't even attempt.
> To understand what I mean, you might want to see
> When is a proof? http://www.maa.org/devlin/devlin_06_03.html
> Here's an excerpt:
>  (right-or-wrong, rule-of-law) definition is that a proof is a
> logically correct argument that establishes the truth of a
given
> statement. The left wing answer (fuzzy, democratic, and human
> centered) is that a proof is an argument that convinces a
typical
> mathematician of the truth of a given statement.
> While valid in an idealistic sense, the right wing
definition of a
> proof has the problem that, except for trivial examples, it
is not
> clear that anyone has ever seen such a thing.
>  mathematicians seem to think they can just ignore it, while
they
> themselves have far vaguer works, which they expect the
world to
> celebrate.
> They're cheating.
 http://mathforprofit.blogspot.com
===
Subject: Re: Key core error argument, stepped out
<3c65f87.0311031501.63257981@posting.google.com>

<3c65f87.0311040730.5cddbe2c@posting.google.com>
A8EwTYfhf*u~,Eu,tf6
$HN*MY&)u0G
=N'
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
J6m5.EN?>Zh
Xh;Y
V|',x(js'Jfq02joVpj|#x
linux)
>> Fascinating perspective.
>> Isn't it?
> So you have coprime with one clear meaning in a ring like
integers,
> but things are different in a field.
No, it has the same meaning in the field, but the special
properties
of a field make that meaning trivial.
> Notice then that 3 and 6 are coprime, but 6 still has 3 as a
factor,
> so in fact, coprime simply loses any relevance.
> I think that's telling.
> Part of my point here is that the mathematics you've taken
for
> gran, with lots of definitions that seem ok, goes off into
some
> interesting places.
But everyone else here seems to already *know* that the
standard
definition of coprime yields a trivial concept for fields.
Mathematicians have *not* taken their definitions for gran.
They
accept that a concept may reduce to triviality in some
settings.
A function f:X -> Y is 1-1 if, for all x,x' in X, if f(x) =
f(x'),
then x = x'. If X is a singleton or empty, then every function
is
1-1. So, for certain sets X, this definition yields a trivial
concept. (Similarly, if Y is empty, then all functions with
codomain Y
are 1-1.) So what? That's the price of consistency (a price,
apparently, that JSH is reluctant to pay).
>> Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE
>> COPRIME. That's because, given any two nonzero real numbers
x and
y,
>> any common divisor of x and y is a unit.
Well, then every real number but 0 is a unit follows from that
>> position.
>> As that is the definition of a field, it looks about right.
> So mathematicians take these positions, which not only go
against
> common sense, they don't make sense in general, pushing
definitions.
> So you have this broken word coprime which has no use at all
if
> you're in the field of real numbers.
What is your preferred definition of coprime for fields? For
what
elements x and y of, say, R ought the sentence x is coprime to
y be
true? For which ought it be false? If you think that coprime
*shouldn't* be trivial for fields, suggest how you would
define it to
avoid triviality.
--
Jesse Hughes
[I]t's the damndest thing. There's something wrong with every
last
one of you, and I *never* thought that was a possibility. But
now I
feel it's the only reasonable conclusion. --JSH sees some
sorta light
===
Subject: Re: Key core error argument, stepped out
<3c65f87.0311031501.63257981@posting.google.com>

A8EwTYfhf*u~,Eu,tf6
$HN*MY&)u0G
=N'
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
J6m5.EN?>Zh
Xh;Y
V|',x(js'Jfq02joVpj|#x
linux)
> Sorry to butt in, but we untold milllions listening in can
take this
> tripe in silence no longer. The poster *should* be
> apologizing to this newsgroup and others for leading so many
of us
> astray for YEARS, but instead, he's quibbling as he appears
> hell-bent on continuing to obfuscate.
James S. Harris has led so many of us astray for YEARS? Who?
If you're one of those victims, I wouldn't be so quick to
publicly
proclaim so. Good thing that you've obfusca your email address,
because I think comments like this will draw Nigerian scams
and penis
enlarging spam like honey draws flies.
Perhaps you mean that JSH has *attemp* to lead so many of us
astray for years -- although, I'm not sure that's accurate
either.
Instead, he's attemp to convince us of his mathematical
arguments,
which he probably believes are valid but which evidently are
not --
not that I care to enter the discussion with him on this point.
--
Jesse F. Hughes
Contrariwise, continued Tweedledee, if it was so, it might be,
and
if it were so, it would be; but as it isn't, it ain't. That's
logic!
-- Lewis Carroll
===
Subject: Re: Key core error argument, stepped out
> 7. But to divide 7 from those constant terms requires
dividing
> through two of the factors, so
> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
 300125 x^3 - 18375 x^2 - 360 x + 22
> from reverse use of the distributive property, which gives a
constant
> term coprime to 7, as required.
Let's take a simpler example and see how your argument works.
Let
Q(x) = 7(25x^2 + 30x + 2) [1]
= 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2
and write
Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2]
Now we mirror your construction and observe that we may take
a_1(x) and a_2(x) to be roots of
r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3]
We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots
a_1(0) = 0, a_2(0) = -1. We see that this assignment is
consistent with your factorization [2], since
Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4]
and we see that we can rewrite [4] as
Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2)
or, emphasizing the constant term in [1] we may write,
letting b_2(0) = a_2(0) + 1,
Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2)
Now, however, you would have us write each factorization
over the algebraic integers as
Q(x)/7 = (5(a_1(x)/7) + 1)(5b_2(x) + 2) [5]
regardless of the (rational integer) value of x.
This constraint in [5] does indeed work in some cases. For
example, if we take x = -1 we see that
Q(-1) = -21 = (7)(-3)
r(-1) = a^2 + 2a
and we see that we may take a_1(-1) = 0, a_2(-1) = -2
and that doing so gives us the factorization
Q(-1) = (5(0) + 7)(5(-2) + 7)
and we see that
Q(-1)/7 = (5(a_1(-1)/7) + 1)(5a_2(-1) + 7) = (1)(-3)
The problem is that your restriction in [5] depends strongly
on the polynomial r(x) used to define the terms a_1(x)
and a_2(x). For x = 0, -1 the factorization in [5] worked
because of the way r(x) splits into particular linear factors,
but consider the case when x = 1. Now we have
Q(1) = 399 = (7)(57) = (7)(3 * 19)
r(1) = a^2 + 14
which gives us a_1(1) = sqrt(-14), a_2(1) = -sqrt(-14)
and so we have the factorization
Q(1) = (5 sqrt(-14) + 7)(-5 sqrt(-14) + 7)
which is indeed equal to 399, as expec. However, it's
immediately obvious that sqrt(-14) isn't divisible by 7
in the algebraic integers, so the factorization in [5]
does not hold. Instead of distributing the value 7
as 7 in the first factor and 1 in the second, we have
the distribution sqrt(7) in both factors, so we have
the factorization in algebraic integers
Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] *
[(-5 sqrt(-14) + 7)/sqrt(7)]
In fact, your constraint on the way 7 (or 49, in the example
you have been using) is split among the terms in your
nonpolynomial factorization is valid only in a small number
of cases for x.
Rick
===
Subject: Re: Key core error argument, stepped out
 7. But to divide 7 from those constant terms requires
dividing
> through two of the factors, so
> (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
 300125 x^3 - 18375 x^2 - 360 x + 22
> from reverse use of the distributive property, which gives a
constant
> term coprime to 7, as required.
 Let's take a simpler example and see how your argument works.
> Let
> Q(x) = 7(25x^2 + 30x + 2) [1]
> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2
That's dumb. The factorization I use isn't just pulled out of
the
blue.
More importantly, examples are a waste of time when the proof
is in
front of you, and the proof itself is so simple.
What I hate about posters who decide to go off on a tangent is
their
need to waste time. Obviously they have an agenda which is to
cast
doubt on my work, but rather than face it directly, they start
chattering about something else, claiming it's rela.
But over time it's clear that there's some key difference, or
worse,
even when their own examples support my case they simply fudge.
So their purpose is simply to obfuscate, which is why they
hide from
the clear and direct argument in front of them to send people
on wild
goose chases.
Rick Decker is a particularly abhorrent type as he pretends at
times
to be reasonable, but he probably realizes by now that he's
screwed as
my work is correct, and he's had a hand in hiding that fact.
Rather than show some common decency, he simply looks for
another
route to continue hiding the truth.
I point that out because I don't think he should remain in a
position
based on society's faith in his common decency, when he so
clearly
lacks any.
If he had any, he'd be posting retractions, rather than
working still
to hide the truth.
===
Subject: Re: Key core error argument, stepped out
>7. But to divide 7 from those constant terms requires dividing
>through two of the factors, so
>(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) =
 300125 x^3 - 18375 x^2 - 360 x + 22
>from reverse use of the distributive property, which gives a
constant
>term coprime to 7, as required.
>>Let's take a simpler example and see how your argument works.
>>Let
>> Q(x) = 7(25x^2 + 30x + 2) [1]
>> = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2
> That's [...]
> stuff. In fact, you should stop cutting and pasting,
period. Why not
>> post a link to your original post, instead, if all you are
going to do
>> is repeat it verbatim, ERRORS INCLUDED?
[.snip.]
Why don't you off ?
> Three times I offered to do so if you told me to stop posting
> replies. You declined to do so.
> Are you doing so now, in your oh-so-courteous way?
> Just say so: Do you want me to stop posting with comments
about your
> statements? Yes or no?
You're ing kidding me!!! You mean you're asking *me* whether or
not I want you to stop posting in my threads?
Quit posting in my threads .
off.
===
Subject: Re: Key core error argument, stepped out
>> Three times I offered to do so if you told me to stop
posting
>> replies. You declined to do so.
>> Are you doing so now, in your oh-so-courteous way?
>> Just say so: Do you want me to stop posting with comments
about your
>> statements? Yes or no?
> You're ******* kidding me!!! You mean you're asking *me*
whether or
> not I want you to stop posting in my threads?
We've all seen him asking you if you wan him to stop replying
to you.
(Actually, I thought it was more than three times, but I
haven't really
been counting.) I've always assumed that you ignored the offer
because
the attention you get from him (and others like him) is the
only thing
that lends you even the *appearance* of any credibility.
> Quit posting in my threads .
So you've decided to accept the offer. Either you've finally
learned
to read, or else you've finally realized you can't stay backed
into this
particular corner forever. (I'd bet on the latter.)
> **** off.
Please try at least to *pretend* that you are fit for decent
company.
--
rs,
Subject: Re: Key core error argument, stepped out
===
>>For crying out loud. You should stop cutting and pasting
incorrect
>>stuff. In fact, you should stop cutting and pasting, period.
Why not
>>post a link to your original post, instead, if all you are
going to do
>>is repeat it verbatim, ERRORS INCLUDED?
>> [.snip.]
>Why don't you off ?
>>Three times I offered to do so if you told me to stop posting
>>replies. You declined to do so.
>>Are you doing so now, in your oh-so-courteous way?
>>Just say so: Do you want me to stop posting with comments
about your
>>statements? Yes or no?
> You're ing kidding me!!! You mean you're asking *me* whether
or
> not I want you to stop posting in my threads?
> Quit posting in my threads .
> off.
I'm amazed you finally said that. He made the same offer about
a month
ago, and you didn't answer.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key core error argument, stepped out
Adjunct Assistant Professor at the University of Montana.
For crying out loud. You should stop cutting and pasting
incorrect
> stuff. In fact, you should stop cutting and pasting, period.
Why not
> post a link to your original post, instead, if all you are
going to
do
> is repeat it verbatim, ERRORS INCLUDED?
[.snip.]
>>Why don't you off ?
>> Three times I offered to do so if you told me to stop
posting
>> replies. You declined to do so.
>> Are you doing so now, in your oh-so-courteous way?
>> Just say so: Do you want me to stop posting with comments
about your
>> statements? Yes or no?
>You're ing kidding me!!! You mean you're asking *me* whether
or
>not I want you to stop posting in my threads?
I asked you three times before and you decline to ask. I was
wondering
if you had a martyr complex of some sort.
>Quit posting in my threads .
I will not stop posting in your threads (no such thing, this
is a
public forum); but I will keep my word from the offer before.
I will
not follow up on anything else you post, not even by
piggybacking. I
may reply to others who ask specific mathematical questions.
Just remember, next time you try to get on your high horse
about being interes in truth, that you are a hypocrite. Next
time
you claim that the fact that people don't reply means you are
correct,
remember what you just did. And next time you complain about
people
being rude, remember your post here, and you oh-so-courteous
parting
shot:
> off.
==============================================================
========
[Gabriele Rossetti] has left a vast body of writings... in
which
he has attemp to prove the truth of his unorthodox interpre-
tation of medieval literature. They present a formidable
record of unsystematic research in which we see an enthusiast
plunging farther and farther and farther from the logic of
facts
and good sense until truth is lost in the dreadful nightmare
of an idee fixe. There is no real evolution of the Theory
although it grows and expands until it embraces ever wider
horizons. The numerous inaccuracies of deduction,
mis-statements
of historical fact, and self-contradictions...have caused
critics
to turn away from them in disgust... [...] It is impossible to
read far... without realizing that we have to deal with a work
of
faith and imagination rather than of reasoning. There is an
appearance of reason, for the author is set on proving by logic
the truth of what he already believes by intuition. The truth
is plain to him and he cannot comprehend why others do not
immediately accept it, but as they desire demonstration he has
multiplied his proofs. It is the redundancy and confusion of a
prophet expounding by a familiar method the truth revealed to
his
own simple soul in a flash of inspiration... In such work as
this... it is idle to look for the calm reasoning of a scholar;
we do not find it, and there is little or no advantage in
attacking the obvious inconsistencies and absurdities that
abound.
-- E.R. Vincent, _Gabriele Rossetti in England_, quo in
_The Shakespearan Ciphers Examined_, by William F.
Friedman and Elizebeth S. Friedman
==============================================================
========
Subject: Re: Key core error argument, stepped out
===
>
>For crying out loud. You should stop cutting and pasting
incorrect
>stuff. In fact, you should stop cutting and pasting,
period. Why not
>post a link to your original post, instead, if all you are
going to do
>is repeat it verbatim, ERRORS INCLUDED?
 [.snip.]
>>Why don't you off ?
>Three times I offered to do so if you told me to stop posting
>replies. You declined to do so.
>Are you doing so now, in your oh-so-courteous way?
>Just say so: Do you want me to stop posting with comments
about your
>statements? Yes or no?
>>You're ing kidding me!!! You mean you're asking *me* whether
or
>>not I want you to stop posting in my threads?
> I asked you three times before and you decline to ask. I was
wondering
> if you had a martyr complex of some sort.
>>Quit posting in my threads .
> I will not stop posting in your threads (no such thing, this
is a
> public forum); but I will keep my word from the offer
before. I will
> not follow up on anything else you post, not even by
piggybacking. I
> may reply to others who ask specific mathematical questions.
It's good to know you'll still be paying attention to these.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key core error argument, stepped out
> ...
>In integers, 2 and 3 are coprime as are 12 and 13, as it
simply
means
>they don't share non-unit factors.
However, in reals, NO numbers are coprime, as for instance,
2(3/2) =
>3, so 2 is a factor of 3.
 But 2 is a unit, so your claim here is false.
 Hmmm...now that's interesting, every real but 0 is a unit,
eh?
 Yup. Every element of a field, except 0, is a unit.
 Fascinating perspective.
 Isn't it?
 So you have coprime with one clear meaning in a ring like
integers,
> but things are different in a field.
Is it? The definition of coprime in a ring with unit (like the
integers
and the reals) is equivalent to: two elements a and b are
coprime if there
are elements c and d in the ring such that a*c + b*d = 1.
(Things get a
bit more complex if the ring has no unit.) You use
(apparently) the
definition that a and b are coprime if every number that
divides both
a and b is a unit. Well, that again is equivalent for both the
integers
and the reals. So where is the difference in meaning?
> Notice then that 3 and 6 are coprime, but 6 still has 3 as a
factor,
> so in fact, coprime simply loses any relevance.
Yup in the reals 3 and 6 are coprime, but 6 has 3 as a factor.
4 also
has 3 as a factor. 4/3 is a real. Coprimeness indeed loses any
relevance in a field, as does having a factor of. Or would you
say
that 4 does *not* have 3 as a factor in the reals? If so, how
would
you define have a factor of? (I assume your meaning is is
divisible
by.)
> I think that's telling.
Is it? But that is the way reals work!
> Part of my point here is that the mathematics you've taken
for
> gran, with lots of definitions that seem ok, goes off into
some
> interesting places.
?
> Actually, in the real numbers, ANY TWO NONZERO NUMBERS ARE
> COPRIME. That's because, given any two nonzero real numbers
x and
y,
> any common divisor of x and y is a unit.
 Well, then every real number but 0 is a unit follows from
that
> position.
 As that is the definition of a field, it looks about right.
 So mathematicians take these positions, which not only go
against
> common sense, they don't make sense in general, pushing
definitions.
The definition of a field is that it is a ring (commutative)
where each
non-zero element has a multiplicative inverse. As having a
multiplicative
inverse is the definition of being a unit, I see no problem.
It is
easily seen that the reals form a field...
> So you have this broken word coprime which has no use at all
if
> you're in the field of real numbers.
Right. In a field coprime, divisor of are of no use. Why does
that
surprise you?
> I guess you could say that and it doesn't change things in
any
> meaningful way.
 But it's *fascinating* that pushed that position!
 Are you contradicting it?
 Nope. I'm just highlighting how screwed up things get when
> mathematicians are left to their own devices.
What is screwed up about it? How would *you* define
coprimeness for
numbers
a and b in the reals?
> Well consider then, he's saying that 3 is not a factor of 6
in
reals
> because they're coprime!
 Hmmm...that's my statement. It looks off in retrospect, as
instead
> coprime is broken, so that what it means in integers isn't
what it
> means in reals.
It means exactly the same thing. Namely in the ring you are
working in
there are numbers c and d such that a*c + b*d = 1.
> So you have 3 is coprime to 6, 3 is still a factor of 6, but
it's a
> unit, or trivial factor, which in one sense is ok, as every
real but 0
> is a factor of every other real, but the word coprime is now
a
> liability.
Eh? Chose a = 3, b = 6, c = 1/3 and d = 0 and see that a*c +
b*d = 1.
How would *you* define coprimeness in a field?
> And you know what? I think the way mathematicians usually
go, he's
> right!!!
 And indeed. But apparently you have no idea about the
mathematical
> meaning of coprime.
 For those who wonder, coprimeness is usually defined to
ignore
> *trivial* factors, where unit factors are, of course,
trivial.
Nope. That is *not* the usual definition. As always you do not
know
the mathematical definitions.
> So 2 is coprime to 3 in integers, but they both have 1 as a
trivial
> factor, of course.
 But notice how the word coprime gets broken when you end up
where
> *every* factor is trivial!
 Then you can say 3 is coprime to 6, in reals.
 Most of you do a quick switch in your heads when you're
operating in
> the real world, so that you handle the problem, and operate
in a
> particular ring based on your particular needs at the time.
 Mathematicians, well, they basically do the same thing, but
*claim* to
> be more precise.
 Fun stuff, eh?
 Yeah.
 HELL YEAH!!! Mathematicians have broken or screwed up terms
all over
> the place, but tend to make ad hoc rules to handle them.
 For instance, with coprime an ad hoc rule would be NOT to
use the
> word with a field!!!
Why? The definition is in a ring. As a field is also a ring,
why should
it not be used in a field?
> The math world is spectacularly illogical and quirky, and
uses lots of
> end-runs and ad hoc rules to cover up messes, which is why I
say the
> situation is like Ptolemy's circles.
 The physics community should take note of the reality versus
just
> ignoring it as we usually do.
Tsk.
--
Subject: Re: Key core error argument, stepped out
===
As a stopped clock is right twice a day...
: > HELL YEAH!!! Mathematicians have broken or screwed up
terms all over
: > the place, but tend to make ad hoc rules to handle them.
That is overstating the case. Math is a very broad thing.
There are a great many subfields. It should not surprise anyone
that terms and concepts that are central in one subfield (no
pun intended)
are sufficiently analogous to concepts in some other subfield
as to make
the same word appropriate, but to leave the applicability less
central.
: > For instance, with coprime an ad hoc rule would be NOT to
use the
: > word with a field!!!
Well, yes, precisely, THAT rule WOULD be ad hoc -- that's
PRECISELY WHY
nobody has adop that rule.
: Why? The definition is in a ring. As a field is also a ring,
why
should
: it not be used in a field?
This is a stupid question, Dik. As ANYbody who knows what YOU
know knows,
IN A FIELD, THE CONCEPT IS DEGENERATE. Almost ANY pair of
things is
coprime. It doesn't SAY ANYTHING ABOUT a pair of elements of a
field
to say they are coprime. Even if you decide to KEEP using the
term,
you wind up HAVING no use for it.
But even when Dik is wrong, the person he is arguing against is
much more wrong.
: > The math world is spectacularly illogical and quirky, and
uses lots
of
: > end-runs and ad hoc rules to cover up messes, which is why
I say the
: > situation is like Ptolemy's circles.
No, really, it isn't. By definition, it isn't.
There is nothing messy about the fact that the notion of
coprimeness, defined for rings in general, becomes degenerate
in certain special kinds of rings (e.g. fields).
For crying out loud, ADDITION becomes degnerate when one of the
addends is 0. That doesn't mean that we start disallowing
expressions like 7+0. If 7 was what you wan to say, then it's
true, you wouldn't ever need 7+0 to do THAT, but you MIGHT need
7+0 as an intermediate step in a proof or calculation.
Somebody just needs to stick to the math and just ing
quit it with all these lame generalizations about the content
of mathematicians' character. It is the content of their proofs
that matters.
--
---
It's difficult ... you need to be uni to have any
strength, but internal issues have to be addressed.
--- E. Ray Lewis, on liberalism in America
===
Subject: Re: Key core error argument, stepped out
[cut]
> : Why? The definition is in a ring. As a field is also a
ring, why
should
> : it not be used in a field?
> This is a stupid question, Dik. As ANYbody who knows what
YOU know
knows,
> IN A FIELD, THE CONCEPT IS DEGENERATE. Almost ANY pair of
things is
> coprime. It doesn't SAY ANYTHING ABOUT a pair of elements of
a field
> to say they are coprime. Even if you decide to KEEP using
the term,
> you wind up HAVING no use for it.
I don't agree with your position here.
(Actually, from your other comments that I cut, you don't seem
to
be consistant maintaining the above position.)
I don't agree with your position (and others maintaining your
position also) because there may be some general theorem proved
already that has one of the pre-conditions that a certain pair
of elements are coprime, which theorem you wish to apply to
a certain type of field that also meets the other conditions
of the given theorem, so that then you can use the conclusion
of
that theorem in some proof you are working on for that type of
field.
I can't think off-hand of such a theorem for coprime,
but here follows a similar example.
The concept of unit is degenerate in a field, as already
mentioned. But, you can use the concept of integral
over a ring applied to algebraic over field to obtain
the less general result that the sum and product of
two algebraic numbers are algebraic numbers.
A second example might be a polynomial ring over a UFD
is a UFD, but applied to a field. This may not be a
good example: first, the result applied to a field
might be already proved first; and second, I believe
the proof for UFD works in quotient field anyway.
Hale
===
Subject: Re: Key core error argument, stepped out
...
> : Why? The definition is in a ring. As a field is also a
ring, why
should
> : it not be used in a field?
 This is a stupid question, Dik. As ANYbody who knows what
YOU know
knows,
> IN A FIELD, THE CONCEPT IS DEGENERATE. Almost ANY pair of
things is
> coprime. It doesn't SAY ANYTHING ABOUT a pair of elements of
a field
> to say they are coprime. Even if you decide to KEEP using
the term,
> you wind up HAVING no use for it.
 But even when Dik is wrong, the person he is arguing against
is
> much more wrong.
Exactly what is wrong about what I said? (And pray, keep the
caps-lock key
in the unpressed position.) Did I say that every use would be
degenerate?
Is it wrong to use a degenrerate concept? if so, what is wrong
about it?
 : > The math world is spectacularly illogical and quirky,
and uses lots
of
> : > end-runs and ad hoc rules to cover up messes, which is
why I say
the
> : > situation is like Ptolemy's circles.
 No, really, it isn't. By definition, it isn't.
> There is nothing messy about the fact that the notion of
> coprimeness, defined for rings in general, becomes degenerate
> in certain special kinds of rings (e.g. fields).
> For crying out loud, ADDITION becomes degnerate when one of
the
> addends is 0. That doesn't mean that we start disallowing
> expressions like 7+0. If 7 was what you wan to say, then it's
> true, you wouldn't ever need 7+0 to do THAT, but you MIGHT
need
> 7+0 as an intermediate step in a proof or calculation.
 Somebody just needs to stick to the math and just ing
> quit it with all these lame generalizations about the content
> of mathematicians' character. It is the content of their
proofs
> that matters.
> --
> ---
> It's difficult ... you need to be uni to have any
> strength, but internal issues have to be addressed.
> --- E. Ray Lewis, on liberalism in America
--
===
Subject: Re: Key core error argument, stepped out
> [.snip.]
>> But for two of the factors of P(x), the constant terms is
7, which is
>> coprime to 22. Therefore, *none* of the constant terms can
have 7 as
>> a factor.
(By saying that 7 is coprime to 22, I'm making a choice as to
where
>> the proof is going. Since I've been talking about algebraic
integers,
>> where 7 is coprime to 22, it's natural to go with a choice
where 7 is
>> coprime to 22.)
>
This is the heart of it. JSH says
[1] 5*b3 + 22
has constant term 22, and that constant term is coprime to 7.
>Correct.
Here, I believe, is how the thinking goes. If you have a
>polynomial with integer coefficients, say,
F(x) = a*x^3 + b*x^2 + c*x + 22,
then that *polynomial* is coprime to 7, because one of the
coefficients
>itself is coprime to 7. That is true whether the coefficients
are
>integers or algebraic integers. There is no possible common
factor
>of F(x) with 7. The constant term, F(0), is 22, and that is
all
>you need to know to say that the polynomial is coprime to 7.
This statement seems to me to be correct. One says that a
polynomial
>with integer coefficients is coprime to an integer m if *any
one* of
>its coefficients is coprime to m. Here the constant
coefficient happens
>to be coprime to 7, and that is enough.
> The statement is true IF by coprime you mean coprime in
Z[x], and
> you mean have no common divisors other than units. However,
it is
> not true if you mean coprime in Z^Z; the example of x^2+x
and 2
> comes to mind.
Right. The definition I intended was, all the coefficients are
coprime
to 7. I am trying to give JSH as much benefit of the doubt as
possible on this. Why? Not sure. Perhaps I shouldn't. It
doesn't matter
anyway.
> Using Dot's proof that the values of a polynomial with
algebraic
> integer coefficients are always divisible by an integer if
and only if
> each coefficient is a multiple of that integer (in the
algebraic
> integers) gives you that the statement is correct for
polynomials in
> A[x].
I had not noticed Dot's theorem. The x^2 + x example seems to
belie
it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
are not multiples of 2, and both of the coefficients are
algebraic
integers. You must therefore be thinking that the domain is
algebraic
integers as well.
> Of course in [1] above, b3 is not a polynomial. It is a
function
>of x. It takes values in the algebraic integers. But perhaps
the
>same reasoning that applies for polynomials applies for b3(x).
> But it does not.
Yes, of course I know that. Again I am trying to guess at JSH's
thinking.
One could make up a definition:
Definition. Two functions f(x) and g(x) which take on integer
values when x is an integer are COPRIME if there exists a
number
x0 such that f(x0) and g(x0) are coprime in the integers.
Whether such a definition is worth considering, I don't know.
By
this definition, f(x) = 2 and g(x) = x^2 + x are not coprime.
> [.snip.]
> Figuring out how JSH is thinking at any given point is not
>necessarily much of a reason to celebrate.
> One possiblity that occurred to me yesterday is that James
has not yet
> realized that any function f(x) from A to C can be written as
> f(x) = g(x) + c
> where c is a constant, and g is a function that takes any
specified
> value at any specified point. That is, if a is in A and b is
in C,
> then there always exists a function g(x) from A to C and a
constant c
> in A such that f(x) = g(x) + c, and g(a)=b.
> Just take c = f(a)-b, and let g(x) = f(x)-f(a)+b.
> So, given ANY f(x) function from A to C, there is a function
g(x) such
> that g(0)=0 and f(x) = g(x)+c, c a constant.
> He has found ONE function that has the right c, namely,
> (5a_1(x)+7)/7; so he thinks this is the ONLY function that
has the
> right c. Likewise, (5a_2(x)+7)/7 works, so it must be the
only
one
> that works; and (5b_3+22)/1 works, so it must be the only one
that
> does. He does not realize that for ANY complex-valued
functions
> w_1(x), w_2(x), w_3(x), such that w_1(0)=w_2(0)=7, w_3(0)=1,
and
> w_1(x)*w_2(x)*w_3(x)=49, one can write (5a_1(x)+7)/w_1(x),
> (5a_2(x)+7)/w_2(x), and (5b_3(x)+22)/w_3(x) as a function
which is 0
> at 0, plus 1, a function which is 0 at 0, plus 1, and a
function
> which is 0 at 0, plus 22. He found one possibility, he
thinks there
> is only one possibility. And that could be the mistake.
As no in my post, I think his problem is at a lower level.
He might say that 7*x^3 + 6*x^2 - 7*x + 7 is coprime to 7
because 6 is coprime to 7. I think he is going by the
definition
I described above: if a polynomial Q(x) has *any coefficient*
which is
coprime to 7, then that polynomial is coprime to 7. Therefore
if
the constant term is coprime to 7, the whole thing must be.
Then
he somehow generalizes this idea to all other functions.
It is just *verbal* reasoning, not mathematical reasoning,
and as such it misses the point. He doesn't even need that the
polynomial is coprime to 7. What he needs is that specific
*evaluations* of the polynomial are coprime to 7. He is
clearly not
understanding this, so he keeps repeating the bit about
constant
terms like a shaman singing a chant. I liked what you said
about
it yesterday: he assigns almost magical powers to the value of
his
functions at x = 0.
>
==============================================================
========
> Why do you take so much trouble to expose such a reasoner as
> Mr. Smith? I answer as a deceased friend of mine used to
answer
> on like occasions - A man's capacity is no measure of his
power
> to do mischief. Mr. Smith has untiring energy, which does
> something; self-evident honesty of conviction, which does
more;
> and a long purse, which does most of all. He has made at
least
> ten publications, full of figures few readers can criticize.
A great
> many people are staggered to this extent, that they imagine
there
> must be the indefinite something in the mysterious all this.
> They are brought to the point of suspicion that the
mathematicians
> ought not to treat all this with such undisguised contempt,
> at least.
> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
>
==============================================================
========
magidin@math.berkeley.edu
===
Subject: Re: Key core error argument, stepped out
Adjunct Assistant Professor at the University of Montana.
[.snip.]
>> Using Dot's proof that the values of a polynomial with
algebraic
>> integer coefficients are always divisible by an integer if
and only if
>> each coefficient is a multiple of that integer (in the
algebraic
>> integers) gives you that the statement is correct for
polynomials in
>> A[x].
> I had not noticed Dot's theorem. The x^2 + x example seems
to belie
>it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
>are not multiples of 2, and both of the coefficients are
algebraic
>integers.
But the values of x^2+x are not multiples of 2 for every
algebraic
integer value of x: if x=i, then you i-1, which is not a
multiple of
2.
> You must therefore be thinking that the domain is algebraic
>integers as well.
Indeed.
[.snip.]
===
Subject: Re: Key core error argument, stepped out
> Using Dot's proof that the values of a polynomial with
algebraic
> integer coefficients are always divisible by an integer if
and only if
> each coefficient is a multiple of that integer (in the
algebraic
> integers) gives you that the statement is correct for
polynomials in
> A[x].
 I had not noticed Dot's theorem. The x^2 + x example seems
to belie
> it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
> are not multiples of 2, and both of the coefficients are
algebraic
> integers. You must therefore be thinking that the domain is
algebraic
> integers as well.
I think you are reasoning on cross-purposes. As I remember,
Dot's theorem
was about polynomials over the algebraic integers. As a
polynomial over
the integers, x^2 + x is always divisible by 2, but not as a
polynomial
over the algebraic integers.
--
===
Subject: Re: Key core error argument, stepped out
Adjunct Assistant Professor at the University of Montana.
> Using Dot's proof that the values of a polynomial with
algebraic
> integer coefficients are always divisible by an integer if
and only
if
> each coefficient is a multiple of that integer (in the
algebraic
> integers) gives you that the statement is correct for
polynomials in
> A[x].
> I had not noticed Dot's theorem. The x^2 + x example seems
to belie
> it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
> are not multiples of 2, and both of the coefficients are
algebraic
> integers. You must therefore be thinking that the domain is
algebraic
> integers as well.
>I think you are reasoning on cross-purposes. As I remember,
Dot's theorem
>was about polynomials over the algebraic integers. As a
polynomial over
>the integers, x^2 + x is always divisible by 2,
I know what you meant, but no, as polynomial over the integers,
x^2+x is not always divisible by 2. In fact, it is simply not
divisible by 2 It is always divisible by 2 as a FUNCTION from
the
integers to the integers, meaning that the value of x^2+x is
divisibly
by 2 for every integer value x. That's part of the entire
point and
what has long caused James confusion, really: the difference
between
factoring polynomials and factoring values of polynomials.
> but not as a polynomial over the algebraic integers.
In case anyone is interes, here's a link to Dot's post:
The statement of her result is:
Theorem 1. Suppose A is the ring of integers of a number field
K.
Let f and g be two elements of A[x] that are coprime in K[x].
If
g_eval divides f_eval in A^A then g is a constant.
Theorem 2. Suppose A is the ring of integers of an algebraic
closure K of Q. Let f and g be two elements of A[x] that are
coprime in K[x]. If g_eval divides f_eval in A^A then g divides
f in A[x].
g_eval means the image of the polynomial in A^A, i.e., the
polynomial considered as a function.
So, if a polynomial with algebraic integer coefficients f(x)
has the
property that for every algebraic integer a, f(a) is divisible
by a
fixed algebraic integer b, then every coefficient of f(a) is a
multiple of b.
In addition, it is worth noting that James's conclusion would
be true
->if<- the factors of P(x) were polynomials. Then it would
follow by
applying Dedekind's Prague Theorem (aka Kummer's Lemma), as I
no
some little time ago, and then the fact that one of the
polynomials
has constant term coprime to f would imply that the each of
the other
2 has all coefficients divisibly by f. James's ->argument<-,
however,
would still be invalid.
===
Subject: Re: Key core error argument, stepped out
> Using Dot's proof that the values of a polynomial with
algebraic
> integer coefficients are always divisible by an integer if
and only
if
> each coefficient is a multiple of that integer (in the
algebraic
> integers) gives you that the statement is correct for
polynomials in
> A[x].
I had not noticed Dot's theorem. The x^2 + x example seems to
belie
> it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
> are not multiples of 2, and both of the coefficients are
algebraic
> integers. You must therefore be thinking that the domain is
algebraic
> integers as well.
> I think you are reasoning on cross-purposes. As I remember,
Dot's
theorem
> was about polynomials over the algebraic integers. As a
polynomial over
> the integers, x^2 + x is always divisible by 2, but not as a
polynomial
> over the algebraic integers.
Oh yeah, I finally realized where you were cheating Dik
Winter, as
using v=-1+49x my polynomial is simply enough
P(x) = (v^3+1)5^3 - 3v(5)7^2 + 7^3
which I'm sure you conveniently forgot when you came up with
your
examples.
Nevertheless my explanations before were correct, but maybe
now you
can start to see more of the particulars of why your hacks put
you in
a field.
What I hate about posters like you is that you don't seem to
care
about the truth, but only about convincing sci.math and other
newsgroup readers.
Subject: Re: Key core error argument, stepped out
===
>> Using Dot's proof that the values of a polynomial with
algebraic
>> integer coefficients are always divisible by an integer if
and only
if
>> each coefficient is a multiple of that integer (in the
algebraic
>> integers) gives you that the statement is correct for
polynomials in
>> A[x].
I had not noticed Dot's theorem. The x^2 + x example seems to
belie
>> it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
>> are not multiples of 2, and both of the coefficients are
algebraic
>> integers. You must therefore be thinking that the domain is
algebraic
>> integers as well.
>>I think you are reasoning on cross-purposes. As I remember,
Dot's
theorem
>>was about polynomials over the algebraic integers. As a
polynomial over
>>the integers, x^2 + x is always divisible by 2, but not as a
polynomial
>>over the algebraic integers.
> Oh yeah, I finally realized where you were cheating Dik
Winter, as
> using v=-1+49x my polynomial is simply enough
> P(x) = (v^3+1)5^3 - 3v(5)7^2 + 7^3
When you star this thread you said:
1. Let P(x) = 14706125 x3 - 900375 x2 - 17640 x + 1078, where
x is
in the ring of algebraic integers, notice that P(x) has a
constant
term that is 1078.
Where did the v's come frome?
> which I'm sure you conveniently forgot when you came up with
your
> examples.
Why would he think it's there?
> Nevertheless my explanations before were correct, but maybe
now you
> can start to see more of the particulars of why your hacks
put you in
> a field.
Or not.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key core error argument, stepped out
...
> Using Dot's proof that the values of a polynomial with
algebraic
> integer coefficients are always divisible by an integer if
and
only if
> each coefficient is a multiple of that integer (in the
algebraic
> integers) gives you that the statement is correct for
polynomials
in
> A[x].
 I had not noticed Dot's theorem. The x^2 + x example seems to
belie
> it: for any integer x, x^2 + x is divisible by 2, but the
coefficients
> are not multiples of 2, and both of the coefficients are
algebraic
> integers. You must therefore be thinking that the domain is
algebraic
> integers as well.
 I think you are reasoning on cross-purposes. As I remember,
Dot's
theorem
> was about polynomials over the algebraic integers. As a
polynomial
over
> the integers, x^2 + x is always divisible by 2, but not as a
polynomial
> over the algebraic integers.
 Oh yeah, I finally realized where you were cheating Dik
Winter, as
> using v=-1+49x my polynomial is simply enough
 P(x) = (v^3+1)5^3 - 3v(5)7^2 + 7^3
 which I'm sure you conveniently forgot when you came up with
your
> examples.
--
Subject: Re: Key core error argument, stepped out
===
>sci.physics snipped
>Newsgroups trimmed. Again.

[deletia]
>>4. Further let a_3(x) = b_3(x) + 3, to keep indices matched.
Then I
>>have
>>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
>>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
Note that all you have done is add and subtract 3 to define
b_3(x);
>that is, you are writing
b_3(x) = (a_3(x) - 3)
so
5a_3(x) + 7 = 5(a_3(x)-3+3) + 7
> = 5(a_3(x)-3) + 15 + 22
> = 5b_3(x) + 22.
The exact same process that you decried when I used it. You
claimed
>that doing this ma[de] no sense mathematically. Do you
still make
>that claim? Just curious.
>>That is a lie from as in fact he just subtrac and
>>added 3 on the same line. I'm focusing on constant terms,
not trying
>>to hide a correct argument with meaningless operations like
>>subtracting and adding 3.
>No it's not, James. You might want to check your facts before
you
>accuse others of lying.
>>You obviously are the one who didn't check facts as what I
said IS
correct.
> Actually, no. You said three things, and you implied a
fourth. Of the
> three things you said, one is false. Therefore, what you
said is NOT
> correct.
> The three things you said are:
> (a) I lied;
I had interpret the lie to refer to you coming up with the idea
first. In either case, you accurately described what he did,
and did it
first.
> (b) I just subtrac and added 3 on the same line;
> (c) You are focusing on constant terms.
> The thing you implied was that I was trying to hide a correct
> argument with meaningless operations.
> Statements (b) and (c) are correct. Statement (a) is false.
Your
> definition of b_3 MEANS that you've added and subtrac 3 to
go from
> 5a_3(x)+7 to 5b_3(x) + 22. Just because you did not say so
explicitly
> You said three things. One of them is false. Therefore, you
claim that
> what you said IS correct is false.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key core error argument, stepped out
Adjunct Assistant Professor at the University of Montana.
>>[deletia]
>4. Further let a_3(x) = b_3(x) + 3, to keep indices
matched. Then
I
>have
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
>>Note that all you have done is add and subtract 3 to define
b_3(x);
>>that is, you are writing
>>b_3(x) = (a_3(x) - 3)
>>so
>>5a_3(x) + 7 = 5(a_3(x)-3+3) + 7
>> = 5(a_3(x)-3) + 15 + 22
>> = 5b_3(x) + 22.
>>The exact same process that you decried when I used it. You
claimed
>>that doing this ma[de] no sense mathematically. Do you still
make
>>that claim? Just curious.
That is a lie from as in fact he just subtrac and
>added 3 on the same line. I'm focusing on constant terms,
not trying
>to hide a correct argument with meaningless operations like
>subtracting and adding 3.
>>No it's not, James. You might want to check your facts
before you
>>accuse others of lying.
>You obviously are the one who didn't check facts as what I
said IS
correct.
>> Actually, no. You said three things, and you implied a
fourth. Of the
>> three things you said, one is false. Therefore, what you
said is NOT
>> correct.
>> The three things you said are:
>> (a) I lied;
>I had interpret the lie to refer to you coming up with the
idea
>first.
I don't remember (and I do not see in the original post) any
claim of
priority. In fact, nowhere did I say first. All I said is what
james
had done, in this case, and no that it was the exact same thing
James had attacked recently.
In fact, when he attacked it I poin out that it was the same
thing
James had done in his Advanced Polynomial Factorization first
Lemma
(that was done in a different thread), so I certainly did not
claim
priority in any way! Not that James first came up with the
idea of
writing a function f as f(x) = g(x) + f(a) for some specific
value a;
it's just that it's not terribly interesting for the vast
majority of
functions...
==============================================================
========
Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can criticize. A
great
many people are staggered to this extent, that they imagine
there
must be the indefinite something in the mysterious all this.
They are brought to the point of suspicion that the
mathematicians
ought not to treat all this with such undisguised contempt,
at least.
-- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan
==============================================================
========
Subject: Re: Key core error argument, stepped out
===
>[deletia]

>4. Further let a_3(x) = b_3(x) + 3, to keep indices
matched. Then
I
>>have
>>P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)
>>P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).
Note that all you have done is add and subtract 3 to define
b_3(x);
>that is, you are writing
b_3(x) = (a_3(x) - 3)
so
5a_3(x) + 7 = 5(a_3(x)-3+3) + 7
> = 5(a_3(x)-3) + 15 + 22
> = 5b_3(x) + 22.
The exact same process that you decried when I used it.
This is what I interpret as priority---^^^^^^^^^^^^^^^
You claimed
>that doing this ma[de] no sense mathematically. Do you still
make
>that claim? Just curious.
>>That is a lie from as in fact he just subtrac and
>>added 3 on the same line. I'm focusing on constant terms,
not trying
>>to hide a correct argument with meaningless operations like
>>subtracting and adding 3.
No it's not, James. You might want to check your facts
before you
>accuse others of lying.
>>You obviously are the one who didn't check facts as what I
said IS
correct.
>Actually, no. You said three things, and you implied a
fourth. Of the
>three things you said, one is false. Therefore, what you said
is NOT
>correct.
>The three things you said are:
>(a) I lied;
>>I had interpret the lie to refer to you coming up with the
idea
>>first.
> I don't remember (and I do not see in the original post) any
claim of
> priority. In fact, nowhere did I say first. All I said is
what james
> had done, in this case, and no that it was the exact same
thing
> James had attacked recently.
> In fact, when he attacked it I poin out that it was the same
thing
> James had done in his Advanced Polynomial Factorization
first Lemma
> (that was done in a different thread), so I certainly did
not claim
> priority in any way! Not that James first came up with the
idea of
> writing a function f as f(x) = g(x) + f(a) for some specific
value a;
> it's just that it's not terribly interesting for the vast
majority of
> functions...
Ah. Ok. In any case, I do recall him attacking that procedure
before
he adop it here.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Key core error argument, stepped out
good move. now, when are you going to finish the short proof
of Fermat's so-called Last Theorem
--is it NP-soluble or just Googol undecidable?--
and how long will it have become, then?
maybe it's just Shakespearean Sonnet Primate Team Time
solvable.
I think it's been sta, before,
that you can use examples with much smaller numbers, but
having the same mathematical concept content,
what ever in Hell that's supposed to be.
is there really any point in numbering a proof,
unless it involves several lemmas?
now, let me be less kind to Devlin's sophistry
--and I wish I'd gone to B&N, when he was there, to say it--
he's just lying, because the Pythagorean Theorem, and
its extension to Pythagorean Triples is anything but trivial,
although it is *simple*, or elementary.
but, then, you're just using him to beg the question,
only in a more direct way then you always do; eh?... or,
do you consider yourself a rightwinger or leftwinger
in all things, including proofs?
> And I'm adding in more changes as the word coprime that I
used
> before has become a flashpoint, so I'm taking it out.
> I also number out the main steps, so if anyone thinks
there's a
> problem and wishes to reply back, they need to give at least
*some*
> numbers.
> When is a proof? http://www.maa.org/devlin/devlin_06_03.html
> While valid in an idealistic sense, the right wing
definition of a
> proof has the problem that, except for trivial examples, it
is not
> clear that anyone has ever seen such a thing.
> http://mathforprofit.blogspot.com
--ils duces d'Enron!
http://tarpley.net/bush8.htm
http://www.wlym.com/PDF-SpReps/SPRP13.pdf
===
Subject: Re: How to prove that sqrt(2) exist?
1> So, I'm curious: how can one prove that sqrt(2)
> exists and it is a real number?
> Thanks!
> Carlos
> --
One possible proof is the following:
Accepting the Cantor's axiom: The reals between 0 - 1 are the
numbers
expresed by one point and then all the different infinite
arrays with
repetition of the ten digits.
The sqrt(2)/10 is one of these arrays. Say
0.14142135623730950.......
Sqrt(2) can be calcula iterating x = (2/x + x)/2 infinitely
times
begining with x = 1 . Taking in each iteration the x with
double of
decimals than in the preceding value.
L. Rodriguez
===
Subject: Re: can't show that .....
>Hi All,
>I dont know where is mistake:
>a and r are vectors, a=a1*i+a2*j+a3*k, a1,a2,a3-const,
>r=x*i+y*j+k*z
>fi=fi(r)
>rot(fi*a x r))=(grad(fi).r).a-(grad(fi).a).r+2*fi*a
>I use rot(fi*v)=grad(fi)xv+fi*rot(v) and can't get 2*fi*a,
always have
3*fi*a.
>What am I missing???
>Tnx
>Ihor
hmm.. it's been a while since I've done this stuff... but in
coordinate
free
form:
(Del x ((f a) x r))_i
=eps_ijk (Del_j ((f a) x r))_k)
=eps_ijk (Del_j (eps_klm (f a_l) (r_m))
=eps_kij eps_klm (Del_j) (f a_l) (r_m)
=(delta_il delta_jm - delta_im delta_jl) (Del_j) (f a_l) (r_m)
=( (Del_m) (f a_i) (r_m) ) - ( (del_l) (f a_l) (r_i) ) )
=((f a_i)(3) + (r_m) (a_i)(df/x_m))-((delta_il)(f a_l) +
(r_i)(a_l)(df/dx_l))
^^^^^^^^^ ^^^^^^^^^^^^^^^^^
notice:t3(f a_i) - (f a_i)
gives:
Del x ((f a) x r)
= 2(f a) + a(r.Del(f)) + r(a.Del(f))
right?
adam
===
Subject: A Question for
James:
Something's been bothering me. The world population
is now approximately 6.3 billion. In seven years you
have not convinced even ONE of them about ANYTHING.
The thing that is a source of puzzlement to me is that
in view of this obvious fact, is why have you not just
once sat down and said to yourself, Hmmm. Maybe.
Just possibly. Something's wrong here.
===
Subject: {Group theory} Number of units?
I have asked a question like this before and the answer was
that no
set could be big enough.
Here is a slightly different question.
Is it possible for a set to exist which contains all the units
of the
various isomorphically distinct groups that exist?
In other words, a set S such that:
--> for any group g with unit g_e, then either g_e is in S, or
there
exists some group h, isomorphic to g and with unit h_e,
such
that h_e is in S
--> if g,h are isomorphic but unequal groups with units g_e,
h_e,
then either g_e is in S or h_e are in S, but not both
(unless
they're equal)
--> if x is in S then x is the unit of some group g and for any
group
h that is isomorphic to g, with unit h_e, either h_e = x or
h_e is not in S
I understand this may very well depend whether you allow
choice.
If this set is possible, what is the cardinality of it?
Thanks in advance... and no, this ain't homework :)
===
Subject: Re: simple question about subspaces
<3f9d2f6a$16$fuzhry+tra$mr2ice@news.patriot.net>
<17a4a089.0311030148.792339b3@posting.google.com>
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>Not necessarily ordered.
There are other ways to do it besides using ordered, sets, but
they
are more complica and have the same effect.
It required introducing an additional operator, negation. It
still
requires an additional axiom, just not the same additional
axiom.
>This helps me get rid of the references to the identity in the
>axioms how?
Why would I defend a statement that I didn't write?
>Imagine, if you will, that someone has been giving a brief
>explanation of how it's possible to build a car without using
spark
>plugs to ignite the fuel.
Somebody has. So the analogy is even worse than it looked at
first
glance.
>Who gives a !
Thank you for your incisive comment. Will you now continue to
rebut
things that I didn't right, or will you instead continue to
tell me
things that I already knew and that cast no light? My guess is
both.
>You don't mean ordered.
Nu, tonto, so now you believe that you're Uri Geller. Perhaps
you are,
but that doesn't help, because his mind reading was just an
act. I
do mean ordered.
Since you have descended from rebutting statements that exist
only in
your head to asserting that I don't mean what I obviously do
mean, I
see no point in continuing this conversation with you. Or any
other. I
suggest that you trade in your Rhine cards for a functioning
brain.
*PLONK*
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolici bulk E-mail will be 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
===
Subject: Re: Graduate algebra book




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at 12:43 PM, toolshed37@yahoo.com (Nobody) said:
>If 3) were the case, you would not respond with such a snotty
remark
Snotty, perhaps, but warran. Introductory Abstract Algebra
should
be a required undergraduate course, IMHO. Obviously fewer
topics would
be covered, at in less depth, than for graduate studies.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolici bulk E-mail will be 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
===
Subject: Re: Graduate algebra book
> What's a good book for a first year graduate algebra course?
> Something with alot of emphasis on factorization, polynomial
rings,
> fields, PIDs, Galois Theory, and of course all the more
basic topics
> of algebra as well like groups, ideals, integral domains,
etc.
> Something that has lots and lots of _good_ exercises?
Hungerford is
> good, but I'll have finished most of the exercises by the
time I want
> to start doin the exercises in this new book. Lang is
probably good,
> but there don't seem to be very many exercises.
A severely critical opinion of Hungerford may be found at
http://www.math.hawaii.edu/~lee/algebra together with some
professional remarks as to what sort of algebra is commendable
at the
graduate level.
An on-line book of first-year graduate mathematics may be
found at
http://www.math.uiuc.edu/~r/
and this may be a direct answer to your question.
Both these references may be seen at
http://hilbert.dartmouth.edu/~leejstem/freebook.html#al
David Ames
===
Subject: Re: Graduate algebra book
> Perhaps the use of snotty is different in the UK than it is
here in
> the US. Chez nous, snotty is often used by the person who
feels
> abused. It is used to indicate that one's counterpart is
eleva by
> airs of superiority, unable to maintain a tone of equality.
Among us,
> it would be called a put-down, rather than abuse.
> But an unwarran use of a perjorative term is surely abuse
> in anyone's lexicon?
A fair point to raise. I first say that I am not party to the
dispute, nor do I intend to be. I gradua in math over 35 years
ago
and have hardly used it since then. I read this newsgroup only
for
interest. (And, if I ever get to retire, I may want to study
algebra
and other topics again.)
That said, let me observe that I have seen contentious remarks
in this
thread. Rather than raise the level of contention, it may be
preferable to make remarks that are dismissive put-downs. We
cannot
all be like Victoria, whose famous dismissive put-down was We
are not
amused. She had position, whereby her remark carried some
weight.
Frankly I think the best answer to the original post, wanting
an
appropriate graduate-level text, might come from faculty
rather than
from those who have comple a certain level of studies. I do not
have time during the week to search but I am certain I have
seen,
someplace on the Web, a complete book which is intended to be
an
appropriate first-year graduate text in algebra. Perhaps I may
have
time on the weekend (the 11th is a holiday and I am taking the
10th
off) to research that point.
David Ames
===
Subject: Re: Graduate algebra book
> You still haven't said what topics in algebra you consider
advanced
> enough to belong in a graduate algebra course. I'd be
curious to see
> the syllabus for an undergraduate introduction to abstract
algebra
> course which covered:
> The thing here is that you seem to be equating undergraduate
algebra
> with introductory algebra. I would say that for *most*
universities,
> these two are not the same. The university I went to for
undergrad (after
> 3 courses in linear algebra) have SIX upper-year
undergradute courses in
> abstract algebra (gran, two of these are not intended for
pure/applied
> math majors, but still the other four courses cover all the
topics you
> lis earlier.)
Well, I think the U.S. and other countries are wildly
different in
their structure of an undergraduate curriculum. I don't have a
problem admitting that my university is not so terrific. We
have 2
algebra courses which are available to undergraduates at my
university, the more advanced of which is almost identical to
the
introductory course, albeit slightly more fast paced. It so
happens
that I _am_ an undergraduate and I am trying to get a better
education
than what my university offers to the typical student. What I
have a
problem with, is listening to the egos of those people who
feel like
they are superior to everyone else for no particular reason.
All Mr.
Chapman had to say was generally those topics are covered at
the
undergraduate level, instead of the snide remark 'thats
graduate
algebra?' Or no remark about it at all would have been fine
too. I
also clearly sta in my original post that I would have been
finished with Hungerford by the time I star reading this new
post,
so another helpful response would have been if you're going to
have
finished reading hungerford, then perhaps what you want is not
a first
year graduate book, but something more advanced such as
__________________.
Actually, upon checking a few of the top universities'
websites, I
find that while the undergraduate curriculum generally does
offer
algebra courses which talk about galois theory and the topics I
mentioned, they are generally at the near-graduate level.
Thus, for
any AVERAGE university, they will not be offered until the
graduate
level. And even at these better universities, not everyone
takes
every single course, and I imagine not everyone takes 5 or 6
semesters
of algebra as an undergraduate. Should they take 5 semesters of
analysis and 5 more of topology as well? how is anyone ever
supposed
to graduate?
===
Subject: Re: Graduate algebra book
> Well, I think the U.S. and other countries are wildly
different in
> their structure of an undergraduate curriculum. I don't have
a
> problem admitting that my university is not so terrific. We
have 2
> algebra courses which are available to undergraduates at my
> university, the more advanced of which is almost identical
to the
> introductory course, albeit slightly more fast paced. It so
happens
> that I _am_ an undergraduate and I am trying to get a better
education
> than what my university offers to the typical student.
Good for you.
So you really want a decent undergraduate text, rather than
a postgrad text, to compensate for the impoverished curriculum
at your university.
> What I have a
> problem with, is listening to the egos of those people who
feel like
> they are superior to everyone else for no particular reason.
Who's that then?
> All Mr.
> Chapman had to say was generally those topics are covered at
the
> undergraduate level, instead of the snide remark 'thats
graduate
> algebra?'
Aaaaaaaah!
> Actually, upon checking a few of the top universities'
websites, I
> find that while the undergraduate curriculum generally does
offer
> algebra courses which talk about galois theory and the
topics I
> mentioned, they are generally at the near-graduate level.
Thus, for
> any AVERAGE university, they will not be offered until the
graduate
> level.
What is an AVERAGE university?
> And even at these better universities, not everyone takes
> every single course, and I imagine not everyone takes 5 or 6
semesters
> of algebra as an undergraduate. Should they take 5 semesters
of
> analysis and 5 more of topology as well? how is anyone ever
supposed
> to graduate?
It's quite possible to study a reasonable amount of algebra,
a reasonable amount of analysis and a reasonable amount of
topology
all in three years. (And also a reasonable amount of applied
maths
and statistics if that's your thing.) I'm a bit baffled about
your
reference
to semesters of algebra --- do you only study one topic per
semester
at your university?
Anyway, you'd be better off studying some maths rather than
flaunting
the chip on your shoulder. I'd second the suggestion of
Dummit/Foote
as a suitable book, and also re-iterate mine of Rotman's recent
Advanced Modern Algebra (this does some topics, e.g., the
homological
theory of local rings, not often found in texts at this level).
--
===
Subject: Re: Graduate algebra book
> Well, I think the U.S. and other countries are wildly
different in
> their structure of an undergraduate curriculum. I don't have
a
> problem admitting that my university is not so terrific. We
have 2
> algebra courses which are available to undergraduates at my
> university, the more advanced of which is almost identical
to the
> introductory course, albeit slightly more fast paced. It so
happens
> that I _am_ an undergraduate and I am trying to get a better
education
> than what my university offers to the typical student.
> Good for you.
> So you really want a decent undergraduate text, rather than
> a postgrad text, to compensate for the impoverished
curriculum
> at your university.
> What I have a
> problem with, is listening to the egos of those people who
feel like
> they are superior to everyone else for no particular reason.
> Who's that then?
> All Mr.
> Chapman had to say was generally those topics are covered at
the
> undergraduate level, instead of the snide remark 'thats
graduate
> algebra?'
> Aaaaaaaah!
> Actually, upon checking a few of the top universities'
websites, I
> find that while the undergraduate curriculum generally does
offer
> algebra courses which talk about galois theory and the
topics I
> mentioned, they are generally at the near-graduate level.
Thus, for
> any AVERAGE university, they will not be offered until the
graduate
> level.
> What is an AVERAGE university?
> And even at these better universities, not everyone takes
> every single course, and I imagine not everyone takes 5 or 6
semesters
> of algebra as an undergraduate. Should they take 5 semesters
of
> analysis and 5 more of topology as well? how is anyone ever
supposed
> to graduate?
> It's quite possible to study a reasonable amount of algebra,
> a reasonable amount of analysis and a reasonable amount of
topology
> all in three years. (And also a reasonable amount of applied
maths
> and statistics if that's your thing.) I'm a bit baffled
about your
reference
> to semesters of algebra --- do you only study one topic per
semester
> at your university?
No, but at my university (and in the Uni States in general) we
don't usually take more than one course of the same discipline
per
semester. If you go to a university's website and find that
they have
5 algebra courses, each of them is a semester long, usually
with one
being a prerequisite for the next (but not always). But when I
say
semesters of algebra i mean algebra courses, since each course
is
one semester long. So, for example, if there are 5 algebra
courses,
the only people who actually end up taking all of them are
those
people who have a pretty good idea that they are going to
specialize
in algebra. Because there simply isn't enough time to take so
many
courses and they aren't generally required anyway. Remember
that in
the states we have to take lots of courses which are completely
irrelevant to anything we plan to do in the future, such as 2
political science courses, 2 history courses, 3 or 4 english
courses,
some other courses such as psychology, art history, economics,
etc,
PLUS get a minor in another subject. I realize this is not the
case
in most european countries, so there is more time to focus on
studying
math and it's only natural that your undergraduate education
will take
you farther.
===
Subject: Re: Graduate algebra book
>> It's quite possible to study a reasonable amount of algebra,
>> a reasonable amount of analysis and a reasonable amount of
topology
>> all in three years. (And also a reasonable amount of
applied maths
>> and statistics if that's your thing.) I'm a bit baffled
about your
reference
>> to semesters of algebra --- do you only study one topic per
semester
>> at your university?
>No, but at my university (and in the Uni States in general) we
>don't usually take more than one course of the same
discipline per
>semester.
Oh dear, oh dear. I do hope by discipline you mean, for
instance, subfield of a subject. If you mean to imply
that taking more than one course per semester in all of
mathematics is unusual (for someone studying mathematics
seriously, e.g., majoring--or even minoring--in it),
then either you're wildly wrong or I am living in a dreamworld.
How could someone graduate from university with a degree in
mathematics having taken only eight courses in mathematics?
Lee Rudolph
===
Subject: Re: Graduate algebra book
>> It's quite possible to study a reasonable amount of algebra,
>> a reasonable amount of analysis and a reasonable amount of
topology
>> all in three years. (And also a reasonable amount of
applied maths
>> and statistics if that's your thing.) I'm a bit baffled
about your
reference
>> to semesters of algebra --- do you only study one topic per
semester
>> at your university?
>No, but at my university (and in the Uni States in general) we
>don't usually take more than one course of the same
discipline per
>semester.
> Oh dear, oh dear. I do hope by discipline you mean, for
> instance, subfield of a subject. If you mean to imply
> that taking more than one course per semester in all of
> mathematics is unusual (for someone studying mathematics
> seriously, e.g., majoring--or even minoring--in it),
> then either you're wildly wrong or I am living in a
dreamworld.
> How could someone graduate from university with a degree in
> mathematics having taken only eight courses in mathematics?

Sorry, I should have been more clear. Certainly people take
multiple
courses of the same subject every semester. I'm taking 4 math
courses
this semester. What I meant to say was that it's not common to
take,
say, 2 algebra courses the same semester. The closest one
might get
is an algebra course and a number theory course. Obviously
there are
exceptions. I imagine it's quite common at the graduate level
to take
multiple courses in the same subfield, but I think it's fairly
rare at
the undergraduate level.
===
Subject: Re: Graduate algebra book
Remember that in
> the states we have to take lots of courses which are
completely
> irrelevant to anything we plan to do in the future, such as 2
> political science courses, 2 history courses, 3 or 4 english
courses,
> some other courses such as psychology, art history,
economics, etc,
> PLUS get a minor in another subject. I realize this is not
the case
> in most european countries, so there is more time to focus
on studying
> math and it's only natural that your undergraduate education
will take
> you farther.
In France undergraduate students (who plan to graduate in
mathematics or
physics) study maths (eheh), physics (electronics, mechanics,
thermodynamics), chemy, computer sciences, and (a little)
English. We don't
have to minor in anything, but we can chose either maths or
physics as
a
prevalent subject for our two first years. That means you'll
have more
maths, or more physics, depending on what you'll do later.
Moreover, we have only two years of undergraduate school (ie
your BchS
is
one year more than ours) so we can specialize earlier (this
system is going
to change in two years in order to follow other european and
american
systems).
===
Subject: Ed Witten on NOVA's Elegant Universe
Look at Elegant Universe NOVA tonight.
http://www.pbs.org/wgbh/nova/elegant/
Brian Greene telephoned an ET Gray on brane universe next door
a
millimeter away using presumably Ray Chiao's gravity radio
transducer
of off-brane world gravity waves to EM waves. ;-) Also long
part on Star
Gate time and space-travel. Ed Witten & Co
http://superstringtheory.com/people/witten.html would be
considered real
kooks by the SI CSICOPS if their actual words were given to
the CSICOPS
without identifying who they were in a blind fold Turing Test.
The
tidy classical universe of James Oberg & Co is not the
Universe of Ed
Witten & Co.
BTW
Q to Ed Witten: How can the cosmological constant be so close
to zero
but not zero?
Ed Witten's answer: I really don't know. It's very perplexing
that
astronomical observations seem to show that there is a
cosmological
constant. It's definitely the most troublesome, for my
interests,
definitely the most troublesome, observation in physics in my
lifetime.
In my career that is.
My answer to the same question is at
http://qedcorp.com/APS/StarGate1.mov
PS
In my theory
Vacuum Coherence Field = (Higgs Amplitude Field)e^i(Goldstone
Phase Field)
This may correspond to locally gauging the one-parameter
dilation
subgroup of the 15 parameter conformal Penrose Massless
Twistor group
of space-time. I am not sure of that as yet. It does come from
a
dynamical instability in the Dirac electron vacuum, i.e.
virtual photon
exchange between virtual electrons and virtual positron holes
near the
-mc^2 Fermi surface well inside the 2mc^2 gap giving
non-perturbative
Bose-Einstein condensation of bound virtual electron-positron
pairs
into the complex scalar MACRO-QUANTUM local Vacuum Coherence
Field.
Einstein's c-number gravity field for curved space-time is then
guv = (Minkowski metric)u,v + (Planck Area*)(Goldstone
Phase)(,u,v)
torsion potential is
suv = (Planck Area*)(Goldstone Phase)[,u,v]
, is ordinary partial derivative.
/ zpf = Exotic Vacuum Random Zero Point Fluctuation Field =
(Planck
Area*)^-1[(Planck Volume*)|Vacuum Coherence Field|^2 - 1)
The strongly attractive dark matter cores of positive pressure
inside
the vibrating strings have |Vacuum Coherence Field| = 0.
*Note change of sign from my earlier formulae to make the sign
conventions internally consistent where, in the weak field
limit
Laplacian of w = -1 Exotic Vacuum Gravity Potential per unit
test
where a negative RHS is attractive gravity of positive
pressure in the
more general and a positive RHS is repulsive anti-gravity of
negative
pressure equal but opposite in sign to the zero point
fluctuation energy
density,
Laplacian of Gravity Potential of any stuff = G(Mass
Density)(1 + 3w)
w = Pressure/Energy Density
The Vacuum Coherence Field is a function of 4 local space-time
coordinates + bosonic coordinates of extra space-dimensions +
fermionic
matrix dimensions). Therefore the 1-dim strings vibrate in the
Calabi-Yau fiber space to make the lepto-quarks and gauge
force bosons.
All topological defects of the complex scalar Vacuum Coherence
Field
with local O(2) symmetry in ordinary space time must be 1-dim
strings.
Branes mean order parameters with local symmetry larger than
O(2).
That is, the Vacuum Coherence Field becomes a hyper-complex
number or
matrix function of matrix space-time coordinates on the short
scale
inside the effective Planck scale Lp* ~ 1 fermi not 10^-33 cm.
My toy
model here is for large macro-scale primarily > 10^-13 cm.
The basic universal Regge slope is Ed Witten's alpha' where
Spin of hadronic resonance = alpha' (Energy)^2 + constant
alpha' = G*/hc^5
Where Lp*^2 = hG*/c^3
G* ~ 10^40 G(Newton)
G* = c^3Lp*^2/h
alpha' = (Lp*/hc)^2
Note that the center of the quantized vibrating Type II
superconductor
string in hyperspace has Vacuum Coherence Field = 0, therefore
/zpf(string core) = - 1/Lp*^2 = - (hc)^2alpha'
This, in the case of the electron, solves the problem of the
Abraham-Lorentz-Becker stresses that prevent the spatially
extended
electron from exploding due to its self-charge internal +
centrifugal
repulsion. The classical space-time singularity seems to be
removed as
well as the need for renormalization infinities - the latter
for reasons
poin out by Ed Witten & Co (e.g. April 1996 Physics Today
Reflections on the Fate of Space-Time
You can picture a bare electron either as string e^2/mc^2 ~ 1
fermi
(with ends glued to 3 dim brane world) or, via
complementarity, as a
tiny quasi rotating charged black hole. This bare electron is
surrounded
by a cloud of virtual electron-positron plasma reaching out to
a Compton
distance ~ 137(e^2/mc^2) = h/mc. That is for low energy
imaging. For
high energy imaging with deep small scale probes the same
electron
shrinks in apparent size to ~ 10^-18 cm for current technology
because
of the huge micro-gravity space warping.
===
Subject: Re: Ed Witten on NOVA's Elegant Universe
> Look at Elegant Universe NOVA tonight.
> http://www.pbs.org/wgbh/nova/elegant/
> Brian Greene telephoned an ET Gray on brane universe next
door a
> millimeter away using presumably Ray Chiao's gravity radio
transducer
> of off-brane world gravity waves to EM waves. ;-) Also long
part on Star
> Gate time and space-travel. Ed Witten & Co
> http://superstringtheory.com/people/witten.html would be
considered real
> kooks by the SI CSICOPS if their actual words were given to
the CSICOPS
> without identifying who they were in a blind fold Turing
Test. The
> tidy classical universe of James Oberg & Co is not the
Universe of Ed
> Witten & Co.
> BTW
> Q to Ed Witten: How can the cosmological constant be so
close to zero
> but not zero?
> Ed Witten's answer: I really don't know. It's very
perplexing that
> astronomical observations seem to show that there is a
cosmological
> constant. It's definitely the most troublesome, for my
interests,
> definitely the most troublesome, observation in physics in
my lifetime.
> In my career that is.
> My answer to the same question is at
http://qedcorp.com/APS/StarGate1.mov
> PS
> In my theory
> Vacuum Coherence Field = (Higgs Amplitude
Field)e^i(Goldstone Phase
Field)
> This may correspond to locally gauging the one-parameter
dilation
> subgroup of the 15 parameter conformal Penrose Massless
Twistor group
> of space-time. I am not sure of that as yet. It does come
from a
> dynamical instability in the Dirac electron vacuum, i.e.
virtual photon
> exchange between virtual electrons and virtual positron
holes near
the
> -mc^2 Fermi surface well inside the 2mc^2 gap giving
non-perturbative
> Bose-Einstein condensation of bound virtual
electron-positron pairs
> into the complex scalar MACRO-QUANTUM local Vacuum Coherence
Field.
> Einstein's c-number gravity field for curved space-time is
then
> guv = (Minkowski metric)u,v + (Planck Area*)(Goldstone
Phase)(,u,v)
> torsion potential is
> suv = (Planck Area*)(Goldstone Phase)[,u,v]
> , is ordinary partial derivative.
> / zpf = Exotic Vacuum Random Zero Point Fluctuation Field =
(Planck
> Area*)^-1[(Planck Volume*)|Vacuum Coherence Field|^2 - 1)
> The strongly attractive dark matter cores of positive
pressure inside
> the vibrating strings have |Vacuum Coherence Field| = 0.
> *Note change of sign from my earlier formulae to make the
sign
> conventions internally consistent where, in the weak field
limit
> Laplacian of w = -1 Exotic Vacuum Gravity Potential per unit
test
> where a negative RHS is attractive gravity of positive
pressure in the
> more general and a positive RHS is repulsive anti-gravity of
negative
> pressure equal but opposite in sign to the zero point
fluctuation energy
> density,
> Laplacian of Gravity Potential of any stuff = G(Mass
Density)(1 + 3w)
> w = Pressure/Energy Density
> The Vacuum Coherence Field is a function of 4 local
space-time
> coordinates + bosonic coordinates of extra space-dimensions
+ fermionic
> matrix dimensions). Therefore the 1-dim strings vibrate in
the
> Calabi-Yau fiber space to make the lepto-quarks and gauge
force bosons.
> All topological defects of the complex scalar Vacuum
Coherence Field
> with local O(2) symmetry in ordinary space time must be
1-dim strings.
> Branes mean order parameters with local symmetry larger than
O(2).
> That is, the Vacuum Coherence Field becomes a hyper-complex
number or
> matrix function of matrix space-time coordinates on the
short scale
> inside the effective Planck scale Lp* ~ 1 fermi not 10^-33
cm. My toy
> model here is for large macro-scale primarily > 10^-13 cm.
> The basic universal Regge slope is Ed Witten's alpha' where
> Spin of hadronic resonance = alpha' (Energy)^2 + constant
> alpha' = G*/hc^5
> Where Lp*^2 = hG*/c^3
> G* ~ 10^40 G(Newton)
> G* = c^3Lp*^2/h
> alpha' = (Lp*/hc)^2
> Note that the center of the quantized vibrating Type II
superconductor
> string in hyperspace has Vacuum Coherence Field = 0,
therefore
> /zpf(string core) = - 1/Lp*^2 = - (hc)^2alpha'
> This, in the case of the electron, solves the problem of the
> Abraham-Lorentz-Becker stresses that prevent the spatially
extended
> electron from exploding due to its self-charge internal +
centrifugal
> repulsion. The classical space-time singularity seems to be
removed as
> well as the need for renormalization infinities - the latter
for reasons
> poin out by Ed Witten & Co (e.g. April 1996 Physics Today
> Reflections on the Fate of Space-Time
> You can picture a bare electron either as string e^2/mc^2 ~
1 fermi
> (with ends glued to 3 dim brane world) or, via
complementarity, as a
> tiny quasi rotating charged black hole. This bare electron
is surrounded
> by a cloud of virtual electron-positron plasma reaching out
to a Compton
> distance ~ 137(e^2/mc^2) = h/mc. That is for low energy
imaging. For
> high energy imaging with deep small scale probes the same
electron
> shrinks in apparent size to ~ 10^-18 cm for current
technology because
> of the huge micro-gravity space warping.
ing beautiful! That's exactly what I always thought! Wow!
===
Subject: Re: cantor's infinity pt II
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hA5L0vw29046;
>set never have the same cardinality. All the members of N
are finite,
>)>but N itself is infinite.
>)
>)this sentence is difficult to understand. in order to be in
an infinite
>)set, each element must be unique. while i can visualize many
integers
with
>)finite values, i cannot understand how all integers have a
finite number
of
>)digits and still exist in an infinite set. it seems as
though in order
for
>)the set to be infinite, there has to exist some integer with
an infinite
>)number of digits. since the one-to-one function works on any
integer
>)without bounds, doesn't this prove that there exist at least
1 integer
>)without bounds?
>I believe the thing you have in mind is called w (lower case
greek
>omega). It is not an integer. The change from the finite to
the
>infinite cannot be made stepwise. Unlike adding more whiskers
to get a
>beard, one cannot add single elements to a finite set and get
an
>infinite set. One has to take a single giant leap to get from
merely
>very large finite sets to w. To put it another way, w is not a
>successor, it is a limit ordinal. There is no largest integer
is
>another way of putting it.
>--
>----
>char *p=char
*p=%c%s%c;main(){printf(p,34,p,34);};main(){printf(p,34,p,34);}
>This message made from 100% recycled bits.
>I don't speak for Alcatel <- They make me say that.
===
Subject: question about book on calculus in 7d
Is there a book on calculus in 7 dimensions?
Can it answer why calculus works best in 7 dimensions?
===
Subject: Ed Witten & Cosmological Constant Puzzle
The way Ed Witten might write my basic formulae in terms of the
fundamental string parameter alpha' with h = c = 1 so that
alpha' is
essentially the effective Planck area inside of which we need
fancy math
like non-commutative matrix space-time, i.e. Galois extension
of
space-time continuum of real numbers to the hypercomplex
numbers like
the fermionic extra space dimensions of supersymmetry. The
formulae
below are at scales larger than effective Planck area.
Einstein's geometrodynamic field = Minkowski metric + alpha'
(Goldstone
Phase)(,u,v)
,u are ordinary partial derivatives in 4D
Goldstone Phase = arg (Vacuum Coherence)
/zpf = Exotic Vacuum Unified Dark Energy/Matter Field =
(alpha')^-1[(alpha')^3/2|Vacuum Coherence|^2 - 1)
/zpf > 0 is w = -1 anti-gravity exotic vacuum dark energy
/zpf < 0 is w = -1 gravitating exotic vacuum dark matter which
mimics
w = 0 CDM for us distant observers.
Vacuum Coherence = (Higgs Field)e^(Goldstone Field)
O(2) symmetry implies 1D string quantized vortex core
topological
defects like in a Type II superconductor.
Look at Elegant Universe NOVA tonight.
http://www.pbs.org/wgbh/nova/elegant/
Brian Greene telephoned an ET Gray on brane universe next door
a
millimeter away using presumably Ray Chiao's gravity radio
transducer
of off-brane world gravity waves to EM waves. ;-) Also long
part on Star
Gate time and space-travel. Ed Witten & Co
http://superstringtheory.com/people/witten.html would be
considered real
kooks by the SI CSICOPS if their actual words were given to
the CSICOPS
without identifying who they were in a blind fold Turing Test.
The
tidy classical universe of James Oberg & Co is not the
Universe of Ed
Witten & Co.
BTW
Q to Ed Witten: How can the cosmological constant be so close
to zero
but not zero?
Ed Witten's answer: I really don't know. It's very perplexing
that
astronomical observations seem to show that there is a
cosmological
constant. It's definitely the most troublesome, for my
interests,
definitely the most troublesome, observation in physics in my
lifetime.
In my career that is.
My answer to the same question is at
http://qedcorp.com/APS/StarGate1.mov
PS
In my theory
Vacuum Coherence Field = (Higgs Amplitude Field)e^i(Goldstone
Phase Field)
This may correspond to locally gauging the one-parameter
dilation
subgroup of the 15 parameter conformal Penrose Massless
Twistor group
of space-time. I am not sure of that as yet. It does come from
a
dynamical instability in the Dirac electron vacuum, i.e.
virtual photon
exchange between virtual electrons and virtual positron holes
near the
-mc^2 Fermi surface well inside the 2mc^2 gap giving
non-perturbative
Bose-Einstein condensation of bound virtual electron-positron
pairs
into the complex scalar MACRO-QUANTUM local Vacuum Coherence
Field.
Einstein's c-number gravity field for curved space-time is then
guv = (Minkowski metric)u,v + (Planck Area*)(Goldstone
Phase)(,u,v)
torsion potential is
suv = (Planck Area*)(Goldstone Phase)[,u,v]
, is ordinary partial derivative.
/ zpf = Exotic Vacuum Random Zero Point Fluctuation Field =
(Planck
Area*)^-1[(Planck Volume*)|Vacuum Coherence Field|^2 - 1)
The strongly attractive dark matter cores of positive pressure
inside
the vibrating strings have |Vacuum Coherence Field| = 0.
*Note change of sign from my earlier formulae to make the sign
conventions internally consistent where, in the weak field
limit
Laplacian of w = -1 Exotic Vacuum Gravity Potential per unit
test
where a negative RHS is attractive gravity of positive
pressure in the
more general and a positive RHS is repulsive anti-gravity of
negative
pressure equal but opposite in sign to the zero point
fluctuation energy
density,
Laplacian of Gravity Potential of any stuff = G(Mass
Density)(1 + 3w)
w = Pressure/Energy Density
The Vacuum Coherence Field is a function of 4 local space-time
coordinates + bosonic coordinates of extra space-dimensions +
fermionic
matrix dimensions). Therefore the 1-dim strings vibrate in the
Calabi-Yau fiber space to make the lepto-quarks and gauge
force bosons.
All topological defects of the complex scalar Vacuum Coherence
Field
with local O(2) symmetry in ordinary space time must be 1-dim
strings.
Branes mean order parameters with local symmetry larger than
O(2).
That is, the Vacuum Coherence Field becomes a hyper-complex
number or
matrix function of matrix space-time coordinates on the short
scale
inside the effective Planck scale Lp* ~ 1 fermi not 10^-33 cm.
My toy
model here is for large macro-scale primarily > 10^-13 cm.
The basic universal Regge slope is Ed Witten's alpha' where
Spin of hadronic resonance = alpha' (Energy)^2 + constant
alpha' = G*/hc^5
Where Lp*^2 = hG*/c^3
G* ~ 10^40 G(Newton)
G* = c^3Lp*^2/h
alpha' = (Lp*/hc)^2
Note that the center of the quantized vibrating Type II
superconductor
string in hyperspace has Vacuum Coherence Field = 0, therefore
/zpf(string core) = - 1/Lp*^2 = - (hc)^2alpha'
This, in the case of the electron, solves the problem of the
Abraham-Lorentz-Becker stresses that prevent the spatially
extended
electron from exploding due to its self-charge internal +
centrifugal
repulsion. The classical space-time singularity seems to be
removed as
well as the need for renormalization infinities - the latter
for reasons
poin out by Ed Witten & Co (e.g. April 1996 Physics Today
Reflections on the Fate of Space-Time
You can picture a bare electron either as string e^2/mc^2 ~ 1
fermi
(with ends glued to 3 dim brane world) or, via
complementarity, as a
tiny quasi rotating charged black hole. This bare electron is
surrounded
by a cloud of virtual electron-positron plasma reaching out to
a Compton
distance ~ 137(e^2/mc^2) = h/mc. That is for low energy
imaging. For
high energy imaging with deep small scale probes the same
electron
shrinks in apparent size to ~ 10^-18 cm for current technology
because
of the huge micro-gravity space warping.
===
Subject: Re: More symmetry between derivative and
antiderivative?
> While dx/dt at x has only one parameter x,
> the integral from x1 to x2 has two parameters x1 and x2.
> Iff x2 was always equal to zero, then derivative and
antiderivative
> would be more similar to each other.
> Mathematicians might feel this idea somewhat cheeky.
> However, it has a physical background.
> Eckard
Congratulations. You just discovered the fundamental theorom of
calculus! I'd submit a paper if I were you.
Really - why all the debating over this? Flip open any calc
book and
read through the appendicies.
And, regarding the integeral of 1/x, it's solved via numerical
analysis. Again, check the same appendicies.
I suppose I should be prepared for the onslaught of pedants
now...
===
Subject: Re: More symmetry between derivative and
antiderivative?
> While dx/dt at x has only one parameter x,
> the integral from x1 to x2 has two parameters x1 and x2.
> Iff x2 was always equal to zero, then derivative and
antiderivative
> would be more similar to each other.
> Mathematicians might feel this idea somewhat cheeky.
> However, it has a physical background.
>> Actually, the fundamental theorem says that the former is
the
opposite
>> of the latter when the first bound remains constant and is
the point
>> where the function takes the value 0.
>> I hope I've been clear.
>I do not even understand what fundamental theorem you are
referring to.
>Please get more specific.
>Antiderivative is the same like integral. Therefore one could
expect
>some similarity.
>I blame Rn Descartes for the missing fix point of our
coordinates.
Antiderivative is not the same as integral; it gives a
means of calculating integrals.
Integral is FAR older than derivative. In fact, integrals
with respect to discrete measures were evalua
approximately 5000 years ago, and all measures (and most
integrals) are limits of these.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: More symmetry between derivative and
antiderivative?
> What about functions not defined at zero, such as f(x) = 1/x?
Just an addition: With ln(b)-ln(a) = ln(b/a) the number of
parameters
reduces from two to one in this case, too.
===
Subject: Re: More symmetry between derivative and
antiderivative?
> What about functions not defined at zero, such as f(x) = 1/x?
> Just an addition: With ln(b)-ln(a) = ln(b/a) the number of
parameters
> reduces from two to one in this case, too.
That only considers one out of infinitely many cases of
functions
with antiderivatives not defined at zero, so it hardley solves
the
problem.
===
Subject: Re: More symmetry between derivative and
antiderivative?
> What about functions not defined at zero, such as f(x) = 1/x?
> The antiderivatives of this function are too important to
ignore,
> but setting limits of integration to include zero is a no-no.
I consider the physically relevant space either according to
the
traditional arrow of time (-inf, 0) or matching to R+ (0, inf),
i.e. neither zero nor infinity are included.
You are correct, 1/x does not exist exactly at x.
Its antiderivative ln(x) + C contains anyway an infinite
constant C.
Only ln(x1)-ln(x0) = ln(x1/x0) makes sense in real-world
physics.
I presume, there is no reason not to hide C within x2=0.
The sugges leaning of integration on the limes x towards zero
seems
to fit quite naturally to a sperical rather than cubic notion
of the
world. It resolves several nasty problems. However, one cannot
expect it
to make the function 1/x more generally adequate to nature.
===
Subject: Re: More symmetry between derivative and
antiderivative?
> What about functions not defined at zero, such as f(x) = 1/x?
> The antiderivatives of this function are too important to
ignore,
> but setting limits of integration to include zero is a no-no.
> I consider the physically relevant space either according to
the
> traditional arrow of time (-inf, 0) or matching to R+ (0,
inf),
> i.e. neither zero nor infinity are included.
Physical relevance is irrelevant to the mathematics.
> You are correct, 1/x does not exist exactly at x.
> Its antiderivative ln(x) + C contains anyway an infinite
constant C.
Infinite constants are nonsense, at least in the mathematics of
Reimann integration.
> Only ln(x1)-ln(x0) = ln(x1/x0) makes sense in real-world
physics.
> I presume, there is no reason not to hide C within x2=0.
You will, no doubt, presume whatever nonsense you wish about
real-word physics, but that does not make it true or relevant
to
mathematics. The realities of mathematics are quite
independent of
your view of physics.
[garbage snipped]
===
Subject: Re: More symmetry between derivative and
antiderivative?
>What about functions not defined at zero, such as f(x) = 1/x?
>The antiderivatives of this function are too important to
ignore,
>but setting limits of integration to include zero is a no-no.
>>I consider the physically relevant space either according to
the
>>traditional arrow of time (-inf, 0) or matching to R+ (0,
inf),
>>i.e. neither zero nor infinity are included.
> Physical relevance is irrelevant to the mathematics.
Really? You caused me to crosspost this message.
Maybe, mathematicians believe they are independent. However,
they are
just not aware to what extent their discipline benefi not just
from
funding but also from people who dealt with practical
applications. I
will try and give the gist of an utterance by Oliver Heaviside:
Mathematicians said, this series does not converge. On that
condition
it will be of use. He ha unjustified rigor. Ironically, his
revenge was made as rigorous as possible.
>>You are correct, 1/x does not exist exactly at x.
>>Its antiderivative ln(x) + C contains anyway an infinite
constant C.
> Infinite constants are nonsense, at least in the mathematics
of
> Reimann integration.
Even if I am merely an old engineer with minimal training in
mathematics, I am aware of some facts concerning Riemann type
integration. Perhaps nobody defined integrals exclusively for
open
domains while the world excessively integrates from minus
infinite to
plus infinite. Is this correct?
>>Only ln(x1)-ln(x0) = ln(x1/x0) makes sense in real-world
physics.
>>I presume, there is no reason not to hide C within x2=0.
> You will, no doubt, presume whatever nonsense you wish about
> real-word physics, but that does not make it true or
relevant to
> mathematics. The realities of mathematics are quite
independent of
> your view of physics.
I was told, every paying costumer is the king at least in
America.
Doesn't physics pay for mathematics? Doesn't physics provide
tailor-made
models of nature?
===
Subject: Re: More symmetry between derivative and
antiderivative?
>What about functions not defined at zero, such as f(x) = 1/x?
The antiderivatives of this function are too important to
ignore,
>but setting limits of integration to include zero is a no-no.
>>I consider the physically relevant space either according to
the
>>traditional arrow of time (-inf, 0) or matching to R+ (0,
inf),
>>i.e. neither zero nor infinity are included.
> Physical relevance is irrelevant to the mathematics.
> Really?
Really!
> You caused me to crosspost this message.
How? Did I twist your arm?
> Maybe, mathematicians believe they are independent.
Do you claim that physics is the only use for mathematical
theories?
There are applications of mathematics outside of physics which
are
not in any way dependent on the theories of physics. In that
sense,
if no other, mathematics IS independent of physics.
> However, they are
> just not aware to what extent their discipline benefi not
just from
> funding but also from people who dealt with practical
applications.
Actuarial sciences and economics and other non-physics areas
fund
more math than physics does. Perhaps mathematicians should dump
physics and only go where the big money is?
> I will try and give the gist of an utterance by Oliver
Heaviside:
> Mathematicians said, this series does not converge. On that
> condition it will be of use. He ha unjustified rigor.
> Ironically, his revenge was made as rigorous as possible.
The whole mathematical theory of asymptotic series deals
usefully
and rigorously with series that diverge. So what is your point?
>>You are correct, 1/x does not exist exactly at x.
>>Its antiderivative ln(x) + C contains anyway an infinite
constant C.
> Infinite constants are nonsense, at least in the mathematics
of
> Reimann integration.
> Even if I am merely an old engineer with minimal training in
> mathematics, I am aware of some facts concerning Riemann type
> integration. Perhaps nobody defined integrals exclusively
for open
> domains while the world excessively integrates from minus
infinite to
> plus infinite. Is this correct?
Were you, as an old engineer or otherwise aware that Reimann
integrals are defined only on closed intervals?
To integrate over an open domain, including the set of all
reals,
requires an extension of the Reimann definition though some
sort of
limiting process.
I am not sure what you mean by excessively in this context.
>>Only ln(x1)-ln(x0) = ln(x1/x0) makes sense in real-world
physics.
>>I presume, there is no reason not to hide C within x2=0.
> You will, no doubt, presume whatever nonsense you wish about
real-word physics, but that does not make it true or relevant
to
> mathematics. The realities of mathematics are quite
independent of
> your view of physics.
> I was told, every paying costumer is the king at least in
America.
> Doesn't physics pay for mathematics?
No! See above!
Doesn't physics provide tailor-made
> models of nature?
Yes, but it is not the exclusive provider of models of
interest to
mathematicians. Very often it has been mathematics developed
without
physical models that has been later adap to the needs of
physics.
Group theory springs to mind as one example.
Thus one can produce a strong argument that mathematics does
not
need physics as much as physics needs mathematics.
You seem to want the tail (physics) to wag the dog
(mathematics).