mm-357 === Subject: : Re: (n to the 5th power) mod 10> Is there an easy way to prove that, when a is an integer, > the equation `(a ^ 5) mod 10 = a mod 10' holds?> By the way, it also seems to work for powers of 9, 13, 17,> etc. That is: `(a ^ (4b + 1)) mod 10 = a mod 10'.> MauroFor a modulus as small as 10, one can simply calculate a^5 for each possible residue, 0,1,2,3,4,5,6,7,8 and 9, or even for 0, 1, 2, 3, 4 and 5 and note that 6 = 10-4, 7 = 10-5, 8 = 10-2 and 9 = 10-1. === Subject: : Re: [OT] de mortuis nil nisi bene (was: Armand Borel Finally dead)> There is a solution to everything except death> and death is the solution to everything!Yes, it is this sort of feeling, which forbids anystupid noise.I want to express great thanks to Steven E. Landsburg,who brought into our minds the wonderful story aboutSerre, walking with Steven's father.Isn't it really great how this turned the mind of thisreal-world-man? In such a way that he released hisvery son into the fancy world of mathematics. And notonly so: he showed real interest and having-been-touchedby going into the book-store and buying Hardy's apology.Touching - that is.With best regardsRainer Rosenthalr.rosenthal@web.de === Subject: : Re: A parking problem> Consider a parking lot with N spaces. You arrive at the lot at a random> instant and find all N spaces occupied. Assuming that the occupancy times> have uniform distribution in (0,1) hours what is the average waiting time> and its standard deviation until a space becomes vacant.Let T be the waiting time.of the N occupants leaves before time t.What is the probability that any given occupant will staylonger than t?What is the probability that all N do? The complement of thatis the probability that T <= t. - Randy === Subject: : Re: Calculus is irrational?Robin Chapman > I totally agree that there is rationality present in calculus, that is> why it isn't useless. But I'm not about to admit that its 100%> rational.> Would it be 99% rational? 74% rational? 31.4159...% rational? etc.There you mathematicians go, trying to quanitify everything :-)I don't know exactly. It's measure of rationality would be < 100% andthat is a good enough for the simple problems that I consider. (Forexample, is it rational to dismiss Peter Lynd's paper about Time onthe grounds that it largely ignores calculus? Well, if Time is theboundry of our rationality, and calculus does not lie completelyinside that boundry, then yes: it is irrational to dismiss a rationalnotion on the grounds that it conflicts with an irrational notion. Bydefintion, we should expect the irrational notion to contradict therational one... if it didn't it wouldn't be irrational.)For more complex problems there might be value in quantifying theirrationality of a system, but an inevitible result of thatquantification is the the solution leading to a result will containirrationality :-). === Subject: : Re: Calculus is irrational?> Hamish Reid > It's been quite well defined for centuries... which is one of the > reasons we keep asking you to use some existing definitions rather than > your own. And to keep using the *same* definitions....> But I'm saying that the defintion is lacking. Being able to reason is> not all there is to being rational.What else could there be? Being able to reason is almost exactly the definition for rational given by my dictionaries and by common and formal usage for centuries now. It's a very useful definition.> I say you have to be reasoning about things that are directly> observable to be rational.Again, *why*? Why restrict or add to an already common and useful definition? It doesn't gain you anything except confusion.> If you disagree, whats a more effective defintion?Well, I'm in favour of using the common definition(s) of rational and rationality, none of which touch on real world concerns. And why should they? There are other words and phrases for that...>Also, are there any> other words that already fit my defintion?How about experiential? A perfectly good word long used for what you're trying to say, but it has the disadvantage that it doesn't allow to say vaguely portentous things like caluculs is irrational. There are other words and phrases, but hey, you already knew that, right?>If rational isn't a good> word for what I'm talking about (and I happen to think it's the> perfect word), then I need to find a new word,Not a new word -- a word or phrase that's been around for centuries in your case.>but my point will still> stand that the solutions of calculus are not something observable we> can reason about.You're wrong here too, but it's evident that you don't know enough math to know how wrong you are, and I -- like almost everyone else in this thread -- will simply give up and not bother with that one...I'm postulating that> the differece is that we have to be reasoning about what we have> observed.> Again, *why*? If you want to talk about *human experience*, do so, and > if you want to talk about reasoning only from direct human experience, > do so, but don't pretend that that's all that human rationality can > mean. > Why not? I think it does a very good job of defining rationality.No one else does, and unless you want to just talk to yourself, you might make the effort to find out how your audience uses words and phrases. You're doing the equivalent of barging into (say) an aviation discussion and loudly claiming that the 737 you just took to JFK had no flaps -- and after a while, people start to realise that by flaps you meant forward-mounted canards or something just as odd. Then you stick to this definition while people point out that that's not what flaps are at all. For that matter, you're doing the equivalent of barging into a math group and claiming that infinity is defined as 1/0... (at which point most people will conclude that you either know little about math or you're horribly confused).> Someone asked me the question why does your rationality include> unicorns?> It doesn't. Because a unicorn has never been observed, it does not lie> in our rationality.Make that does not lie in our experiences of the 'real world' and I probably wouldn't disagree. But since we can reason about unicorns in detailed and coherent ways, it is -- *by definition* -- within our rationality.>We can theorize about one, suggest that it might> look like, but ONLY IF our defintion of rationality demands that what> we reason about must be observed is the unicorn excluded from> rationality.Do you think black holes are rational, using your bizarre sense of rational?> My defintion of rationality leaves out the fairy tales.It also leaves out the vast majority of stuff that's susceptible to reason, i.e. that the rest of the world defines as rational. Your loss, I guess.>Your defintion> of rationality, lacking the requirement of observation, does not> exclude these fairy tales.Nor does it exclude almost all of math, as yours does. And it's not *my* definition -- it's standard.> Do you disagree with my summary? If not, do you see why a more percise> defintion of rationality is in order?I have no reason to believe that the centuries-old definition of rationality is wrong or useless or particularly defective in any way. You keep trying to conflate rational and real world without seeming to realise that they're quite different concepts. You're being quite irrational about it all.> Because we cant' observe> something bouncing an infinite number of times the solution doesn't> even get off to a rational start.> You know, you just have to think less concretely about these things... > all math is abstraction, and this is a classic case of that. Just like > the number three is an abstraction.> It is an abstraction, yes. The number three is an abstraction of three> things. However, whether something is an abstraction or not does not> determine whether it is rational or irrational. The abstraction of 3> lies within our rationality. I observe it frequently in the real world> by abstracting observed things.Ditto for caluculus and most of math. You don't seem to be able to get even your own non-standard usage of the word rational straight, which would be funny if it weren't a little pathetic. Your three is as abstract and not-part-of-the-real-world as calculus.> The abstraction that leads to infinity must include everything,> including our rationality.*Why*? The above is just a string of non sequiturs unless you can precisely define what you're talking about. You've shown almost no knowledged of what infinity or calculus or rational mean in real-world and / or fomal usage, so there's no reason to dig much further here.>If so, rationality cannot be contained in> rationality. Incompleteness theorem and all that.I'm guessing that like the concept inifinity you don't have a coherent grasp on incompleteness theorem (or even the difference between theorem and theory, but never mind...).> The fact that it is math means > that some measure of rational thought went into it (well, more or less).> Do you understand my point that although calculus isn't completely> rational I'm not saying that it is completely irrational?I understand your point, it's just that that as presented it's simply and trivially wrong. If you'd amend it to something like calculus, although entirely rational, isn't entirely rooted in our real-world experiences, and can be both counterintuitive and / or abstract in ways I don't like then you'd get little disagreement here, since it's both trivially true and quite a commonplace observation.But that's *not* what you're saying, at least according to any standard usage, and that equivocation seems to be deliberate.> I totally agree that there is rationality present in calculus, that is> why it isn't useless. But I'm not about to admit that its 100%> rational.And your position is that it's, erm, 76% rational? How would you know? Is it rational to state something like that? Why? If calculus itself isn't part of rationality as you define it, how can you reason about it?> You agree with me that there are elements of calc that lie outside the> real world. If not irrational, what is the adjective you give to> something that lies outside the real world?I'd use the phrase outside the real world myself, but since you seem to dislike using existing usage and words in standard ways, and seem to want a single word, why not make something up? A word like juzastic or something? Then no one would complain that you keep irrationally and unreasonably using the word rational to mean something quite different from what it means in the real world.... Hamish === Subject: : Collatz ConjectureHere is a DC project attacking the Collatz Conjecture. I am sure they willget into more interesting projects but its not bad. They basically takeyour Program/Executable, make some small changes to it and run it as a dcproject (I think you also need to supply the work), very cool...Anyways:http://distributed.redirectme.net/ === Subject: : Re: Decidability in mathematics>> In sci.math, Michael Hochster>> :>>: I've noticed an interesting lack of understanding about ring and field>: operations from several posters who've replied to my posts where I>: prove that adding pi to the ring of integers results in a field.>Words you seem not to know the definitions of:>(1) Ring>(2) Field>(3) Decidable>> (4) Limit.>> (5) Definition.> (6) Proof.(7) Unwelcome.-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: : Re: Decidability in mathematicsI was taught what convergent and divergent series were in high school. Ofcourse I was in the honors mathematics program so I probably got more than alot of my classmates that weren't in that program.David Moran === Subject: : Direct sum = free product for non-abelian groups???I'm not much up on the current rage of category theory.I always thought that the direct sum in group theory occurs when wehave a group G, with subgroups H and K; and(i) H intersect K = {1}(ii) HK = G(iii) H and K are both normal subgroups of Gin which case we write G = H + K, and say that G is the direct sum ofsubgroups H and K.Someone has written me that in current mathematical usage, the aboveterminology is incorrect; and that direct sums with this definitionare _only_ used in the category of _abelian_ groups.Furthermore, the direct sum of non-abelian groups H and K isapparently now considered to be what was formerly termed the freeproduct of H and K.Is this correct? How is the situation I define above as direct sumcurrently named? Surely not that G equals the direct product of H andK (although it is obviously isomorphic to such a product)?TIA.Cheers - Chas === Subject: : Re: expectation maximization> some additional mathematics and no statistics background. Could anyone> please kindly advise me on what should I read into before I can understand> this topic??I assume that you mean the Expectation Maximization Algortihm (EM Algorithm)for parameter estimation. I have used it in the past in the context of Gaussianmixture models.A Gentle Tutorial of the EM Algorithm and its Application to Parameterhttp://www.cvgpr.uni-mannheim.de/hornegge/MEDBV/ handouts/em-tutorial.pdf === Subject: : Re: Factorial/Exponential Identity, Infinity> I guessWe've noticed. === Subject: : Re: How do you discuss mathematics with a person who thinks mathematics is a part of physics?>> ... He calls the mathematics that is done without any connection to the>> physical world imaginary.> Like it or not, every major development in mathematics ( or in any> science for that matter} > Mathematics is not a science.> It was not explicitly stated that mathematics is a science. You areperhaps assuming that mathematics or any *other* science was meant. But the words can be read as meaning mathematics, equally with anyscience.David Ames === Subject: : Re: How do you solve this?2^x = a*x=> ln_2 (2^x) = ln_2 (a*x)=> x = ln_2 (a*x)I am not exactly sure where Cantrell is going?Lurch> I have no idea how to solve this, please help.> 2^x = a*x> It can't be solved in terms of functions with which you're probably> familiar, but it can be solved in terms of the Lambert W function.Briefly,> it's the multivalued inverse of the function x*e^x; for more informationon> the Lambert W function, see> . > I'm guessing that you're interested in real solutions for x (and that a is> also real). If that is the case, then it's easy to see (by graphing both> 2^x and a*x) that there can be, depending on the value of a, either no> solution, one solution, or two solutions. When there are real solutions,> they are given by> x = - LambertW(ln(2)/a) / ln(2)> using one of, or both of, the real branches of the Lambert W function.> David Cantrell === Subject: : Re: How do you solve this?No. Log[2,ax] is not a constant...> 2^x = a*x> => ln_2 (2^x) = ln_2 (a*x)> => x = ln_2 (a*x)> I am not exactly sure where Cantrell is going?> Lurch> I have no idea how to solve this, please help.>> 2^x = a*x> It can't be solved in terms of functions with which you're probably> familiar, but it can be solved in terms of the Lambert W function.> Briefly,> it's the multivalued inverse of the function x*e^x; for more information> on> the Lambert W function, see> . > I'm guessing that you're interested in real solutions for x (and that ais> also real). If that is the case, then it's easy to see (by graphing both> 2^x and a*x) that there can be, depending on the value of a, either no> solution, one solution, or two solutions. When there are real solutions,> they are given by> x = - LambertW(ln(2)/a) / ln(2)> using one of, or both of, the real branches of the Lambert W function.> David Cantrell === Subject: : Re: how to calculate a paralell line> hi everybody.> i know this is pretty basic. i know i had this at school. i know there> should be an answer allready formulated to this question somewhere in> the web, but i can't find it. so i come to you folks. i have following> question. i need a formula which allways calculates in a cart.> coordinate system a second line in a certin given distance to a given> first line. i know where the first line begins, where it ends, so what> it gradient is. i know the distance between those two lines. i know> that a line between those two lines, beeing in a right angle two both,> will be as long as the distance. i know i only need one point of the> second line, because it will have the same gradient as the first. i> know it has to do something with the further going formulas of> pythagoras, but i just can't get them reconfigured. so if anybody can> set me on the right path, i would be very thankfull.> thanks> n> ps: by the way this is for a visual programming task i am sitting on. If you're doing this in a *plane*, then it's not too hard. Suppose that the given line has equation ax + by = r , that (x0,y0) is a point that lies on that line and that you want to find the equation of one of the two lines parallel to the given line and at a distance d > 0 from it. Let e1, e2 denote the standard basis for R^2. Then x0 e1 + y0 e2 represents a point on the given line and a e1 + b e2 is a vector perpendicular to that line. Normalize that vector (dividing its components by sqrt(a^2+b^2)) -- then it's easy to find a point that lies on one of the lines that you're seeking - a vector from the origin to the point is just x0 e1 + y0 e2 + d( a/sqrt(a^2+b^2) e1 + b/sqrt(a^2+b^2) e2) = (x0 + [ad/sqrt(a^2+b^2)]) e1 + (y0 + [bd/sqrt(a^2+b^2)]) e2 . And the equation of the line parallel to your original line that passes through the point that we just found is ax + by = a(x0 + [ad/sqrt(a^2+b^2)]) + b(y0 + [bd/sqrt(a^2+b^2)]) which simplifies pretty readily to become just ax + by = r + d sqrt(a^2+b^2) . (To get the _other_ line that solves your problem, replace a e1 + b e2 by its negative and run through the above steps again -- you should end up with the equation ax + by = r - d sqrt(a^2+b^2) -- no doubt, that's obvious to you on the basis of the geometry ...) === Subject: : Re: how to measure the orthogonality of a matrix?> Dear all,> I am facing with the following question:> If a matrix T is orthogonal/orthonormal, we can have T*T'=T'*T=I...> Now I have matrix A which is known to be non-orthonogonal/orthonormal...> but it is nearly orthogonal/orthonormal, that's to say, A*A' and A'*A are> nearly identity matrix I(all ones on diagonal line and very few non-zeros> off the diagonal line...)> Is there any mathematical method/term that I can used to measure the> degree> of such kind of non-orthogonality? How much orthogonality does a matrix> have?> Can anybody give me some hints?[...]Theoretically, by polar decomposition, one can produce anorthogonal matrixB = A * (A' * A)^(-1/2)The distance from B to A in the operator norm has aconditionally explicit inequality formula:denote I - A'*A = E and norm(E)=r, then for r < 1, norm(B-A) <= r / (1 + sqrt(1 - r))A lazy proof can be conducted using Singular ValueDecomposition, plus some elementary facts, such as thatthe Maclaurin series for the right-hand side as afunction of r has non-negative coefficients.What I do not recall (it's been a long time, and our officesare in the process of moving, so finding references ismission impossible):(1) Is this estimate actually an equality?(2) Is the estimate valid for all unitarily invariant norms? (Frobenius norm is here of interest.)(3) The polar factor B is known to be the closest orthogonal matrix to A in Frobenius norm. How about other norms?Remark: The l-infinity norm of a matrix (maximum of absoluterow sums) is not unitarily invariant, but within a moderatefactor away from the operator norm; I vaguely remember afactor of sqrt(n) or n, the order of the matrix.Cheers, ZVK(Slavek). === Subject: : Hubble spheres as type II superconductor vorticesPPS The Hubble horizon of radius c/H(t) of our local universe section on the large scale spatially flat expanding accelerating post-inflationary bubble is analogous to the core of the superconducting vortex singularity where the local macro-quantum coherent order parameter PSI vanishes. That is where the micro -> macro quantum vacuum phase transition happened after all in which the non-classical entropy (log of volume of phase space of the vacuum) is lowered to set the cosmic clock for the irreversible arrow of time in which the initial singularity big bang starts out with low entropy compared to the later universe. This Hubble horizon where PSI = 0 is a 2D surface out of which our 3D space of Einstein's curved 4D spacetime emerges like the projected holographic image in Lenny Susskind's sense. That's whyLp* = Lp^2/3(c/Ho)^1/3 ~ 1 fermi (10^-13 cm)where Lp ~ 10^-33 cm and c/Ho ~ 10^28 cmYou can think of the infinity of Tegmark's Level I parallel universes on a single post-inflationary bubble as a kind of type II superconductor lattice of vortices. Each vortex line maps to a Hubble horizon. Then there are, in the chaotic inflationary model of Tegmark's Level II) an infinity of such bubbles floating in hyperspace (in Kip Thorne's sense) like a glass of champagne (brewed by Rashi de Troyes 1000 years ago ;-)) These are not the QUBIT parallel worlds of David Deutsch's multiverse. The QUBIT multiverse is a landscape for the ensemble of IT system points of geometrodynamic post-inflationary bubbles to roll on like invs = (h/m)Grad(arg psi) in deBroglie-Bohm-Vigier theory in 3D, which for 4D geometrodynamics becomesdu = Lp*^2 argPSI,u,u is the ordinary partial derivative of the Goldstone phase argPSI, u = 0,1,2,3PSI = <0|e+e-|o> is the inflation complex scalar fielddu = Hagen Kleinert's world crystal lattice distortion field in which curvature and torsion are disclination and dislocation defect densities as shown for ordinary crystals by Bullough, Bilby et-al at Harwell AERE Theoretical Physics Division in studies of breakdown of nuclear reactor control rods. I was at Harwell in that division in Summer of 1966 after spending time at Heisenberg's Max Planck Institute in Munich at a Wheeler seminar. === Subject: : Re: Is there a simple way to see if a matrix can be factorized>The original post is to seek A= BC where each element of A, B, C >is positive integer.>Is it a simple way to do this ?> Maybe not really simple, but it's a finite search.That posting of mine assumed det(C) <> 0. The case where det(A) = 0is ratherspecial and should be dealt with separately. In fact, this case iseasy.Again I'll write [ a b ] [ s t ] [ w x ]A = [ c d ], B = [ u v ], C = [ y z ], A = B C,where all entries are positive integers, and I'm assuming det(A) = 0.So det(B) = 0 or det(C) = 0. WLOG assume it's det(C) = 0 (otherwisedo the same analysis on A^T = C^T B^T). Then I claim the necessaryand sufficient condition to factor A is gcd(a,b) > 1 and gcd(c,d) > 1.The columns of C are parallel. Writing w/y = x/z = m/n in lowest terms, we have [w, y] = e [m, n] and [x, z] = f [m, n] where e,f,m,n are positive integers. Then [a,c]^T = B [w,y]^T = e B [m,n]^T and similarly [b,d]^T = f B [m,n]^T. Let [p,q]^T = B [m,n]^T, so p and q are positive integers with a = p e, c = q e, b = p f, d = q f. Now [p, q] = m [s, u] + n [t, v]. In particular p >= 2 and q >= 2, sogcd(a,b) >= 2 and gcd(c,d) >= 2.Conversely, suppose gcd(a,b) = p > 1 and gcd(c,d) = q > 1. Then(using the fact that [a,b] and [c,d] are parallel) we can write [a,b] = p [e,f] and [c,d] = q [e,f]. Then we can write [p,q] = [1,1] + [p-1,q-1], and we have the factorization[ 1 p-1 ] [ e f ] [ pe pf ][ 1 q-1 ] [ e f ] = [ qe qf ] = A 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: Is there a simple way to see if a matrix can be factorized> Perhaps it will be true that:> If A is a matrix of unique primes or is pairwise coprime,> and B and C contain no factors of any number in A, and with special> consideration for 1, which is not a prime, but can be a factor, and cannot be> coprime to anything;> Then A # B*C, and we have a new cryptographic method with public key one of A,> B, or C, and private key one of the others? [3 8] [3 5] [73 127] [5 8] [8 14] = [79 137]Note that all entries of A are prime.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: : James HarrisThe more I view the postings of/from James Harris, the replies andcounterreplies, I have a certain thought that this is more than acompetition amongst mathemathicans and amateurs. I think this is a survey!No one can be as dumb as James. Everything is laid out to him, either asdefinitions or as proofs. He can't refute them. Still he disagrees,sometimes unpolitely, somtimes unscholardly.I think that this is some survey that some of the mathematisions hasstarted, and it has gone wild. Can they stop it before I die in laughs!Karl-Olav Nyberg === Subject: : Re: James Harris* Karl-Olav Nyberg> The more I view the postings of/from James Harris, the replies and> counterreplies, I have a certain thought that this is more than a> competition amongst mathemathicans and amateurs. I think this is a survey!> No one can be as dumb as James. Everything is laid out to him, either as> definitions or as proofs. He can't refute them. Still he disagrees,> sometimes unpolitely, somtimes unscholardly.> I think that this is some survey that some of the mathematisions has> started, and it has gone wild. Can they stop it before I die in laughs!I do think James harris is relatively stupid, but not _that_ stupid.He certainly lacks quite a lot of _social_ intelligence since hestubbornly repeats and repeats his random babble. However, I thinkhis _motive_ is not his self-proclaimed genious, but just to make alot of noise in the sci.math community. It really is about his needfor strokes or feedback. I guess he has a poor social life.Or perhaps everything is a gigantic theatrical spam.-- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52 electron-dot-cloud are galaxies === Subject: : math answers on a daily rise percentage Re: 100 free SBC; VonNeumann Gametheory how to play StockMarketToday I was faced with this problem once again. Math answered my lastquestion as to the proper percent to sell of a switching Cell in order tomaximizegains. The answer there was 1/e, and its compliment of 1 -(1/e) as in thecaseBut now math needs to answer for the case of a daily rise in stock. Onaverage,what is the percent rise of a stock when it exceeds rational, practicalbasis?Is a 8% rise in a day overexuberance? Is it at 5% or more?Today Qwest was up 35 cents to $4.50 for a 8% rise. And started to slipbefore the close. Should I have made a switch of 1/e shares of Qwest andboughtSBC today? Will tomorrow bring a further rise in Qwest and a slipping of SBCor something else?Math answered the problem of quantity to keep and quantity to sell.But now, math needs to answer as to what is Overexuberance for a daily riseina stock. I suspect the answer is going to look something like this:If company A rises 5% or more in a day then the chances of it falling backtobefore it rose is something about 90%. And only in 10% of the cases does astock continue to rise on consecutive days.I need a Probability Rule of Thumb to tell me what percent of a day rise inCompany A is it best to make a sale because the probability of Company Arising even more are slim.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies electron-dot-cloud are galaxies === Subject: : Re: math answers on a daily rise percentage Re: 100 free SBC; VonNeumann Gametheory how to play StockMarket> Today I was faced with this problem once again. Math answered my last> question as to the proper percent to sell of a switching Cell in order to> maximize> gains. The answer there was 1/e, and its compliment of 1 -(1/e) as in the> case> But now math needs to answer for the case of a daily rise in stock. On> average,> what is the percent rise of a stock when it exceeds rational, practical> basis?> Is a 8% rise in a day overexuberance? Is it at 5% or more?> Today Qwest was up 35 cents to $4.50 for a 8% rise. And started to slip> before the close. Should I have made a switch of 1/e shares of Qwest and> bought> SBC today? Will tomorrow bring a further rise in Qwest and a slipping of SBC> or something else?> Math answered the problem of quantity to keep and quantity to sell.> But now, math needs to answer as to what is Overexuberance for a daily rise> in> a stock. I suspect the answer is going to look something like this:> If company A rises 5% or more in a day then the chances of it falling back> to> before it rose is something about 90%. And only in 10% of the cases does a> stock continue to rise on consecutive days.> I need a Probability Rule of Thumb to tell me what percent of a day rise in> Company A is it best to make a sale because the probability of Company A> rising even more are slim.> Archimedes Plutonium, a_plutonium@hotmail.com> whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxiesI guess what I am asking is this sort of situation. On a pair of dice you havepossible numbers of 2 to 12 and average of say 6. So let us say that Qwestat 4.15 with a rise of .35 in a day, let us say Qwest is at 6. And so you rollthe dice and it comes up, let us say to number 9. Now, the important question,what is the probability on the next roll of the dice that the number 9 isexceeded?I am trying to frame a Mathematical Analogy to the concept of a daily rise instock as it overexceeds its intrinsic value due to exuberance.Long periods of dice rolls of 2 dice will be on average the number 6. What Iam looking for is what is the probability of successive rolls of 2 dice wherethe number that appears keeps getting higher and higher. Say you roll a 7, thenan 8, and then a 10 and then an 11 and then a 12. What is the probability ofgetting increasingly higher in number.This should, sort of copy the behaviour of a stock company that risesconsecutively.Archimedes Plutonium, a_plutonium@hotmail.comwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Subject: : Re: Most consistent prime function > If the time required to compute the result of an> algorithm is fixed and does not change as the> problem size (n in this case) varies, then it's> called constant time, not linear time. I tend> to doubt your algorithm is constant time, although> without knowing exactly what your algorithm is> supposed to do it's hard to say. Considering the original poster is bounding his input size(by log(100) ) then any algorithm running on such an inputwill trivially run in time bounded by a constant... For the original poster: talking about runtime in that wayis meaningless when you restrict the the size of the input.Runtime complexity is an asymptotic measure which is consideredas the input large grows large (i.e. if the size of the inputdoubles, how does it affect the runtime? Such a question makesno sense to ask when your input fits in one word of memory.)J === Subject: : Re: Most consistent prime function>>If the time required to compute the result of an>>algorithm is fixed and does not change as the>>problem size (n in this case) varies, then it's>>called constant time, not linear time. I tend>>to doubt your algorithm is constant time, although>>without knowing exactly what your algorithm is>>supposed to do it's hard to say.> Considering the original poster is bounding his input size> (by log(100) ) then any algorithm running on such an input> will trivially run in time bounded by a constant...Yep... Based on his description I'm uncertain abouthow his algorithm should behave. Although it may bethe case, it wasn't clear to me that the input sizeis bounded by log(100).In fact, I read this:i.e. the time it takes to execute the functiondoes not increase/decrease as the size of thevariable n changesto mean that n could get arbitrarily large...> For the original poster: talking about runtime in that way> is meaningless when you restrict the the size of the input.> Runtime complexity is an asymptotic measure which is considered> as the input large grows large (i.e. if the size of the input> doubles, how does it affect the runtime? Such a question makes> no sense to ask when your input fits in one word of memory.)> J === Subject: : Re: Most consistent prime function > Yep... Based on his description I'm uncertain about> how his algorithm should behave. Although it may be> the case, it wasn't clear to me that the input size> is bounded by log(100). The point I made was that the algorithm could do absolutely anythingand as long as it always terminates, and you restrict the size of the input, the algorithm runs in constant time since for any imput you can bound the runtime by some (maybe large) constant. I interpreted his description as bounding the input size to log(100) because he said his program runs on the numbers 1 to 100. And, yes, the poster needs to be much more specific about what he claims the algorithm does and about how it works for us to give any worthwhile feedback.i.e. the time it takes to execute the function> does not increase/decrease as the size of the> variable n changes Yes, from this one does assert that the algorithm is constant time (and analysis of the algorithm? I assumed he observed this while running his program on his inputs ranging from 1 to 100, which is pretty much nonsense.J === Subject: : Re: ordered n-tuples as sets> Since Kuratowski's definition works for ordered pairs, let's make an> ordered triple> (a,b,c) = (a,(b,c))> = {{a}, {a,(b,c)}} = {{a},{a,{{b},{b,c}}}}>> and similarly for an ordered n-tuple>> (a1,a2,...,an) = (a1, (a2,...,an))> Recursively nesting ordering pairs works, but, unfortunately, setnesting> level is unbounded. On the other hand, the alternative definition> (a,b,c,...) = {{a},{a,b},{a,b,c},...}> has nesting level 2, but clearly doesn't work. Is there an alternative> ordered n-tuple definition which require sets nesting no more than 3levels?> If 3 is not giong to cut it, some bigger number, maybe?<...>> the S-tuple (a, b, c, ...) can be considered as a set of ordered> pairs (s,d)<...>That was the key to the answer, thank you!An ordered n-tuple can be defined as(a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)}which clearly has a limit on curly brakets nesting.Alternatively, it could be defined transitively closed:-)(a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an),((a1,an-1)((a1,an),...}> I'm worring about set nesting because the alternative representationwith> bounded nesting level would be useful representing paths in DAG.<...>After thinking it over I take this statement back:-( === Subject: : Re: ordered n-tuples as sets> An ordered n-tuple can be defined as> (a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)}But that doesn't work. Consider:(1,2,1) = {(1,2),(2,1)} ??(2,1,2) = {(2,1),(1,2)} ??The right-hand sides are equal, but the left-hand sidesare not. === Subject: : Re: ordered n-tuples as sets> Since Kuratowski's definition works for ordered pairs, let's make an> ordered triple> (a,b,c) = (a,(b,c))> = {{a}, {a,(b,c)}} = {{a},{a,{{b},{b,c}}}}>> and similarly for an ordered n-tuple>> (a1,a2,...,an) = (a1, (a2,...,an))> Recursively nesting ordering pairs works, but, unfortunately, set nesting> level is unbounded. On the other hand, the alternative definition> (a,b,c,...) = {{a},{a,b},{a,b,c},...}> has nesting level 2, but clearly doesn't work. Is there an alternative> ordered n-tuple definition which require sets nesting no more than 3 levels?> If 3 is not giong to cut it, some bigger number, maybe?> Doesn't this all bypass the question of motive, i.e., the desire to> distinguish the second (or third, or i-th) element in the n-tuple?> In which case it seems to me that given an index set S, and a domain> D, the S-tuple (a, b, c, ...) can be considered as a set of ordered> pairs (s,d) (or if you prefer, a set of elements of the form {s, {{s},> d}}) where s in S and d in D. Each S-tuple x then becomes a function x> | S -> D; and cardinality of S is not (... as much of ...) a problem.But it's still a problem, if he's worried about unboundednesting levels, because the elements of S (where S = N) arethemselves nested sets. Or do I miss your point? === Subject: : Re: ordered n-tuples as sets > But it's still a problem, if he's worried about unbounded> nesting levels, because the elements of S (where S = N) are> themselves nested sets. Or do I miss your point?I guess the upshot must be that he shouldn't worry about unboundednesting. At least, not if he wants N as his index set, since ithappens already in N and there's nothing he can do about that.In no case does any member of any of these sets have an infinitenesting level of braces. So no problem, right? === Subject: : Re: ordered n-tuples as sets> An ordered n-tuple can be defined as> (a1,a2,a3,...,a_n-1,an) = {(a1,a2),(a2,a3),..., (a_n-1,an)}> But that doesn't work. Consider:> (1,2,1) = {(1,2),(2,1)} ??> (2,1,2) = {(2,1),(1,2)} ??Any counter example for transitevely closed definition as well?(1,2,1) = {(1,2),(2,1),(1,1)} !=!= (2,1,2) = {(2,1),(1,2),(2,2)} === Subject: : Product spaceDear all,Let X be a metric space. Let U, V be two subsets in X.Is the following statement true?The product space (U,V) is open in (X cross X) if and only if U, V are bothopen sets in X.If it is true, would kindly provide a concise roadmap of showing it ?Thank you,K. Yam === Subject: : Re: Product space> Dear all,> Let X be a metric space. Let U, V be two subsets in X.> Is the following statement true?> The product space (U,V) is open in (X cross X) if and only if U, V are both> open sets in X.> If it is true, would kindly provide a concise roadmap of showing it ?> Thank you,> K. YamU,V open implies UxV is open: actually for topological spaces, this follows by definition.UxV open implies U and V are open: show that the projections p,q:XxX->X, p(x,y)=x, q(x,y)=y are conitinuous.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen === Subject: : Re: Product space> Dear all,> Let X be a metric space. Let U, V be two subsets in X.> Is the following statement true?> The product space (U,V) is open in (X cross X) if and only if U, V are both> open sets in X.> If it is true, would kindly provide a concise roadmap of showing it ?It's helpful to recall the definition of the product topology. The base of the product topology is composed of all sets O_U x O_V where O_U is an open set in U, and O_V is an open set in V.<== If U and V are both open, let (u,v) be an element of U x V. Since U and V are each open in X, there are open subsets (of X) O_u and O_v containing u and v respectively, with O_u subset of U, and O_v subset of V. Then (u,v) is an element of O_u x O_v, which is open by the definition of the product topology; further, O_u x O_v is a subset of U x V. Therefore U x V is open.I'll leave the other direction to you. === Subject: : Re: Publishing>And technically speaking, I did not attack Serre or any other editor: I>described their editorial practises,...>Is this the cause of Aberg's animus against Serre?>Did Serre reject one of his papers?Serre is evidently considered the Immaculate, and those even hinting atsomething else will get their motives questioned even before any facts canbe presented. Would it not be better to get the facts before questioningthem?Serre, as editor, does not referee papers he does not want to seepublished, but instead quickly returns a letter stating that he willpretend that the already formally submitted paper has not been submittedto him, and if the submitter insists on having the paper refereed, it willnot become published anyhow. I think there must be quite a few such Serreletters around. (In the cases I know of, the papers where publishedelsewhere and eventually judged to be of utmost quality, but Serre'sunusual editorial practises caused a long delay.)This broke off from what the people I knew at the time thought were theethically correct procedures of scientific publishing, that paperssubmitted to a scientific journal should be registered and properlyrefereed. If the journal has some requirements that submitted papersshould be registered when submitted, then there is a formal clerical erroras well.I think that if Serre says that he knows that a paper will not bepublished even if refereed, prosecutors will say that it is circumstantialevidence that he has a way to bypass proper refereeing. Such methods wasdescribed to me in those days, but I have not made investigations as towho is doing what (as I do not really care, not working in that circuit).I think that Serre has made contributions not only to mathematics, butalso his very own contributions to scientific ethics. :-)Why don't you ask him about it? Hans Aberg * Anti-spam: remove remove. from email address. * Email: Hans Aberg * Home Page: === Subject: : Re: Push or Pull problem> Question:In a rectangle made up of 11*12 square tile,I can choose to> either push or pull a square.> But when pushing or pulling,ALL adjacent tile(one to the left,to the> right,above,and below) will also be affected.> For all affected square,if it is now up,it would be pushed to becomedown.If it is down,it would be pulled to become up> I have to make ALL square tile change from ALL down to ALL up> Diffculties:If I want to use equation to solve this,I would need a lot> of variables,if I use logic,there would be too many possibility.> Method tried:So far I've solved 3*3 and 4*4,but not 11*12.I think I> know the reason.For 3*3,press all 4 corner,and the centre one.For> 4*4,suppose the buttom-left square is called (1,1) ,and the> buttom-right is (4,1) ,the top left is (1,4) then the simplest is> (1,3) (3,4) (4,2) (2,1) , while pressing all square EXCEPT (1,2) (2,1)> (3,4) (4,3) is also ok.> I suppose all square start at 0,when a button is pressed,itself and> all adjacent will increase by 1,and obviously if the num is odd,then> it would be up,if it is even it would be down.I found that if all> square is pressed once,the four corner is 3,the side is 4,and others> are 5.If the edge of two side , i.e. (1,2) (2,1) is pressed, only the> side will change(but only effect 2 square at the side),that's how> solution of 3*3 and 4*4 came from.> Also,I found that no square would be needed to be pressed for more> then 1 time.If the same square is pressed 2 times,it would cause +2 to> all adjacent square,but both when odd num+ 2 or even num+2 would not> change that num from odd to even or even to odd,so the min. step> needed would not more than the total number of the square.> Finally,I found that all square must have a number of 1,3 or 5,and if> there are pressed square adjacent to or 2square to another square,or> it's just diagonally adjacent to that,then there will be at least one> square changed the state two times.(What a mess)> Can anyone solve this?I tried it for months and still no luck.Have a look at the 3x4 case. It may offer you more insight. Jumping from 3x3 and 4x4 to 11x12 (if that's what you did) is a significant leap in the complexity of the problem. The 3x4 can be solved, but its solution is somewhat different in nature from the 3x3 or 4x4 case.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Questions for James HarrisI am trying to understand this 'Ring of Objects' of which you speak somuch, but I haven't seen a definition. The closest thing that I haveseen to a definition is The Object Ring is a commutative ring thatincludes all numbers such that 1 and -1 are the only members that areboth a unit and an integer, and where no non-unit member is a factorof any two integers that are coprime.This raises more questions than it answers:1. What kinds of things are members of your object ring? complexnumbers? polynomials? functions? elements of some strange quotientring? vectors of some kind?2. What do you mean by 'coprime' in this context?3. You seem to merely (not counting the other issues with what you aresaying, like the ones I raise in my other questions) assert theexistence of a ring with these properties without proving uniquenessor uniqueness up to isomorphism of this ring, or proving somethinglike the existence of a maximal subring of C (the field of complexnumbers) with this property.4. What is the definition of integrality for members of this ring? Forsomething like Z[1/2], we can use the usual notion of integrality(and, by the way, this ring is not a field since 3 is in it, 3 is not0, and 1/3 isn't in it), and then things are all well and good. Butwhat does it mean to say that an object is an integer?These and other clarifications (NOT obfuscations) would beappreciated.---- David === Subject: : Set nomenclature questionF is a field and its elements are denoted by B_n_m. Does this meanthere is a one-to-one and onto function from the pairs of positiveintegers to F? Or, for example, could B_n_m = A (some element of F)forall pairs(n,m)?This is an effort to understand the concept of all possible countableunions of intersections (and intersections of unions) of B_n's. Asimple illustration involving a 2-element F would be useful indeed.Many thanks. === Subject: : Re: simple problem~hot-girl escribi.97 en elmensaje|nbi085l$15h$1@news.hananet.net:> hello~gentleman and lady~> f(x,y) = (x^2)(y^2) / (x^2)+(y^2)> lim f(x,y) = 0 [(x,y)->(0,0)]> -------------------------------> it is very simple.> i can understand that this limit go zero by the feel.> but i can't guarantee it.> help me~doctor.> i need your mild advice.Change to polar coordinates:x = r*cos(t)y = r*sin(t)(x, y) ---> (0, 0) <===> r ---> 0Then,Lim(r^2*cos^2(t)*r^2*sin^2(t)/r^2, r, 0) = Lim(r^2*cos^2(t)*sin^2(t), r, 0)= 0because r^2 --> 0 and cos^2(t)*sin^2(t) is bounded.-- Best regards,Ignacio Larrosa Ca.96estroA Coru.96a (Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: : Re: simple problem~> f(x,y) = (x^2)(y^2) / (x^2)+(y^2)f(x,y) = (x^2)*[(y^2)/(x^2 + y^2)] <= x^2. === Subject: : Re: solving algebraic equation> To solve an algebraic equation, like> sqrt{ 1 + 2x } = x (1)> one may proceed by squaring both sides and solve for x, however,> thereby introducing a sign ambiguity. > Then the (two) solutions come out as x = 1 +/- sqrt{ 2 }. But only the+ solution satisfies (1) (for the positive branch of the square root> function).> Question: is there a way to circumvent this ambiguity, i.e. solve such> that one doesn't need to re-substitute the obtained solutions and> check whether they actually satisfy the equation?I can't think of one, except by keeping an inequalityaround to make sure each step is completely equivalent.That is, what you are doing is a series of steps, someof which involve if and only if transformations(and are therefore reversible), and some of which involveimplies, i.e. only if transformations which arenot one way.sqrt(a) = b implies a = b^2, but a = b^2 does notimply sqrt(a) = b since sqrt(a) could equal -b.So here's how I'd transform your starting statement through<=> (if and only if) transformations:sqrt(1 + 2x) = x, <=> 1 + 2x = x^2 AND x >= 0 (because x is the right handside of the first equation, and sqrt(anything) >= 0 if real) <=> x^2 - 2x - 1 AND x >= 0 <=> x = 1+-sqrt(2) AND x >= 0 <=> x = 1+sqrt(2)Is that an improvement? - Randy === Subject: : Re: Synergetics coordinatesDear Clifford,Do You mean that I have to explain my reply more detailed to your earlierpost before You can consider, or do You accept my arguments as such?Of course - You are welcome to express counterarguments.I consider synergetics coordinates is nothing but a new name for old factsas a final goal to support somebody's effort to lift his profile as a greatinventor.(Please notice this message is no way against You and the honest informationYou wanted to share with sci.math. community). My criticism was dedicatedagainst the original reference and against the inventor's ambitions.)Tapio> Keywords: fourth dimension, simplex, tetrahedron, coordinate geometry,> sixty degree coordinate system, sphere packing, linear algebra,> operations research, linear programming, Newton's method, roots of> unity, derivative, field, cyclotomic fields, B numbers, Bucky Numbers,> Synergetics, R. Buckminster Fuller.> I've read too much about most of the keywords from too many books. An> afterimage, persistence of vision, time lag, stops me from seeing simple> things from a new point of view for awhile.> Cliff Nelson> Fyi: The nice property that was mentioned in the given link, is based> on the Viviani's theorem.> http://mathworld.wolfram.com/VivianisTheorem.html> Ternary phase diagrams have been used during the last century -> perhaps longer. See for example:http://www.sv.vt.edu/classes/MSE2094_NoteBook/ 96ClassProj/experimental/terna> ry2.html> Btw: MS-Excel has radar plot property and, on the other hand,> xyz-cartesian plot, which can be viewed from any angle so that you> can result in synergetics view. (45-45 degrees rotation of the> point of view) Imho: I didn't know a certain angle of view in the xyz> cartesian coordination system is called synergetics coordinates.> ;-)> Tapio> What Eric W. Weisstein calls the x,y,z axes of the triangle are> rotated one position to the right from mine in his description,> but, it looks like at least some version of Synergetics coordinates> might become respectable in academia now that they are mentioned> at:> http://mathworld.wolfram.com/SynergeticsCoordinates.html> Cliff Nelson === Subject: : Trying to understand Languages and StructuresIn preparation for a graduate course in Logic that I will be takingthis fall, I am trying to get an overview of the course by goingthrough some notes on an undergraduate course in the subject.This is my first proof outside of Propositional Logic and I waswondering if someone would help me out by critiqueing it.Problem: Let L = Lab. Show that for every L-term t(x1, ..., xn) there areintegers k1, ..., kn such that for every abelian group A = (A, 0, -,+) and every a1, ..., an belonging to An we have tA(a1, ..., an) =k1a1 + ... + knan.Solution:Let x = (x1, ..., xn), and a = a1, ..., an. Case 1: t(x) is the variable xi (has arity 0)Then tA(a) = ai by the definition of tA. Define kj = 0 if j != i and ki = 1. Then tA(a) = k1a1 + ... + knan.Case 2: t(x) = - x (has arity 1)Then tA(a) = - ai by the definition of tA. Define kj = 0 if j != i and ki = -1.Then tA(a) = k1a1 + ... + knan.Case 3: t(x) = + xixj (has arity 2)Then tA(a) = + aiaj by the definition of tA. Define km = 0 if m != i and m != j and km = 1 if m = i or m = j.Then tA(a) = k1a1 + ... + knan.Case 4: t(x) = Ft1(x)t2(x)...tn(x) (has arity > 2)Then tA(a) = FAt1A(a)t2A(a)...tnA(a) by the definition of tA. Since each function of L is either nullary, unary, or binary(definition of Lab), each ti will fall under case 1, case 2, or case3. Hence each ki will be either 0, 1, or -1.Then tA(a) = k1a1 + ... + knan.Since every term falls into either case 1, 2, 3, or 4, it follows thatthe proposition is true.QED.I tried to put the proof in html, but I cant display html in thesegroups. The actual proof with superscripts, subscripts and such islisted at www.geocities.com/blakman211/lgprf1.htm === Subject: : Re: Typing Monkeys Don't Write Shakespeare>Typing Monkeys Don't Write Shakespeare>By JILL LAWLESS>.c The Associated Press>LONDON (AP) - Give an infinite number of monkeys an infinite number of >typewriters, the theory goes, and they will eventually produce the works >of Shakespeare.>Give six monkeys one computer for a month, and they will make a mess.>Researchers at Plymouth University in England reported this week that >primates left alone with a computer attacked the machine and failed to >produce a single word.>``They pressed a lot of S's,'' researcher Mike Phillips said Friday. >``Obviously, English isn't their first language.''>[...]>The notion that monkeys typing at random will eventually produce >literature is often attributed to Thomas Huxley, a 19th-century >scientist who supported Charles Darwin's theories of evolution. Typing Monkeys Don't Write Shakespeare -- anagrams****************************************************** Yes, Ess. Darwin kept on making potty here.Newton: Pigs take a merry . Donkys pee.******************************************************==== ===================== Endeavor to persevere========================= === Subject: : Re: Typing Monkeys Don't Write ShakespeareWhat the hell is the point of this? Is it just supposed to be really funny orsomething?Somebody needs to be explained the difference bewteen six and infinity.>>Typing Monkeys Don't Write Shakespeare>>By JILL LAWLESS>>.c The Associated Press>>LONDON (AP) - Give an infinite number of monkeys an infinite number of >>typewriters, the theory goes, and they will eventually produce the works >>of Shakespeare.>>Give six monkeys one computer for a month, and they will make a mess.>>Researchers at Plymouth University in England reported this week that >>primates left alone with a computer attacked the machine and failed to >>produce a single word.>>``They pressed a lot of S's,'' researcher Mike Phillips said Friday. >>``Obviously, English isn't their first language.''>>[...]>>The notion that monkeys typing at random will eventually produce >>literature is often attributed to Thomas Huxley, a 19th-century >>scientist who supported Charles Darwin's theories of evolution. >Typing Monkeys Don't Write Shakespeare -- anagrams>***************************************************** *>Yes, Ess. Darwin kept on making potty here.>Newton: Pigs take a merry . Donkys pee.>****************************************************** === Subject: : Re: Units in Z[pi], field forced REVISED>>In this post I use a limit approach to show that adding pi to the ring>>of integers forces you into a field.> [.snip.]>I note that in addition to all the many errors you are committing in>assuming that rings are closed under taking limits (or worse, taking a>limit of a sequence of rings!), you are still making the>categorical error of thinking that a ring with an infinite number of>units must be a field.>>No I'm not. You didn't read carefully. Each integer except 0, gets>>forced into being a unit if you add pi to the ring, which is why doing>>so gives you a field.>Even assuming you were correct in what your ring is, you failed to>show that EVERY NONZERO ELEMENT is a unit. That's what you need in>order to show that it is a field. Showing that every integer is a unit>is NOT enough. You are still making a categorical error.>>My assessment is that many of you don't quite understand what pi is,>>and thinking it's just some other number like 2, you think you can add>>it to the ring of integers without consequences.>My assessment is that you have no idea what a ring is (an assessment>amply confirmed by your definition in your webpage of > A commutative ring is a set of numbers where for any members a and b,> a+b = b+a and ab = ba.You're making that up. He actually gives that as the _definition_ ofcommutative ring?I mean that's just incredible - if he's going to give the definitionwhy wouldn't he look it up somewhere and give the real definition,even if he didn't understand it?>and therefore, that you in fact have no idea what you are talking>about when you talk about Z[pi], or rings, or fields.>Your entire discussion so far only reveals the true depths of your>ignorance, and how you have spent almost two years making statements>with terminology whose meaning you do not know. In short, you have>spent a year trying to bluff your way by apeing and parroting,>sprinkling your discussion with buzzwords you do not understand, and>hoping nobody would notice.>We had noticed, by the way, but your discussion now makes it>->painfully<- clear to just about everyone.>>However, pi is transendental, so it's not a number like 2. It has>>certain properties, which you can't escape if you follow the math.>Of course pi is not like 2. That's why the ring Z[2] is just the>integers, while the ring Z[pi] is isomorphic to the ring Z[x] of all>polynomials with integer coefficients. Different things.>====================================================== ================>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 critize. A great> many people are staggered to this extend, 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>======================================================= ===============>Arturo Magidin>magidin@math.berkeley.edu************************ === Subject: : Re: Units in Z[pi], field forced REVISED Visiting Assistant Professor at the University of Montana. [.snip.] [.about James.]>>My assessment is that you have no idea what a ring is (an assessment>>amply confirmed by your definition in your webpage of >> A commutative ring is a set of numbers where for any members a and b,>> a+b = b+a and ab = ba.>You're making that up. He actually gives that as the _definition_ of>commutative ring?No, I'm not making it up. Yes, he gives that as the definition ofcommutative ring.http://www.msnusers.com/AmateurMath/ objectmathematic.msnwIt showed up one or two days ago. I used his definition to show thatevery set is a commutative ring, and went one step further and evenmade sure that addition and multiplication were operations on theset. Amazing, huh?>I mean that's just incredible - if he's going to give the definition>why wouldn't he look it up somewhere and give the real definition,>even if he didn't understand it?Why bother? It would just be looking up a social fossil, wouldn't it?=========================================================== ===========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 critize. A great many people are staggered to this extend, 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======================================================== ==============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Units in Z[pi], field forced REVISED> Now then as to the social fossil, it is that the fielf of rationals is> NOT truly a field as elements are excluded to fit into a definition,> otherwise it would be reals.> There is no *field* of rationals, and the term is a social fossil.JSH you really are a creative guy. === Subject: : Re: Units in Z[pi], field forced REVISED>In this post I use a limit approach to show that adding pi to the ring>>of integers forces you into a field.>>No you don't. A ring is not required to be closed under taking limits.>The fact that you think it should be doesn't change that fact.>I didn't say it did. Your insinuation that I did, is false.>My guess is that you're lost, so you make up things.>>You are talking about pi^2/6 = 1+ 1/4 + 1/9 + ...>>This is formally written as pi^2/6 = lim k-> oo (1 + 1/4 + 1/9 + ... + >>1/(k^2) )>>Now, in order for this limit to lie in the rationals, the rationals must >>be closed under infinite sums.> Well, not quite. If the rationals were closed under converging> infinite sums, then you would be able to conclude that this limit lies> in the rationals. The point is that the rationals are not necessarily> closed under converging infinite sums, so the fact that this sum> converges is not sufficient to conclude that the limit must lie in thefield of rationals. > Because the operations in a field are finitary, so derived operations> are finitary, so there is no reason why one must consider infinite> sums when dealing with a structure we want to test for the field axioms.> I'm sure that's what you meant, just thought I would make the> precision explicit (since I committed the same sin of carelessness> earlier...)Unfortunately, there's a fine line between when the precision matters, and it's not always easy to see when that line has been crossed.I wonder if James understands what infinity is and is not.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Units in Z[pi], field forced REVISED>However,>that forces out elements that would result from field operations, like>pi.>>The axioms for a field do not require it to be closed under infinite>>sums. Taking an infinite sum is not a field operation.> I didn't say it was.> The actual rule is decidability, otherwise known as convergence.I thought decidability had to do with computability theory.> Infinite sums of differing elements are excluded from the ring of> integers because NONE of them can be decidable.Huh? What are you trying to say here?>Some might realize that in fact the ad hoc rule >>I'm not sure what you mean by ad hoc. The word rational meansexpressible as a ratio (of integers). That's why the first five>>letters if rational are R A T I O. The distinction between>>rationals and irrationals was known long before rings and fields were>>axiomatised, and it would be bizarre to use the term field of>>rationals for something that included irrationals.>>-- Richard> My point is that there is no *field* of rationals, and in fact, you> only have rationals, which are numbers written as the ratio of two> integers, where the denominator is non-zero.Look up field again. You are trying to change its definition.> That's it.> As if you consider a *field* of rationals then infinite sums that> converge, as they are decidable, would be included.That's not part of the definition.> Like 1/2 = 1/3 + 1/3^2 + 1/3^3+... is included because it fits the> definition.> While pi^2/6 = 1 + 1/2^2 + 1/3^2 +...+1/k^2+... is excluded because it> does not.> So it's the *definition* which defines rationals, which do not make up> a field.You got the first part right.> Mathematicians, by a definition, arbitrarily exclude certain elements> that *are* decidable in the field using field operations, so it's not> really a field.Decidable is not a term normally used when talking about fields.> The term field of rationals is a social fossil, as it doesn't exist> as an actual field in mathematics, as the field is broken by an ad hoc> rule.So stop making ad hoc rules and start using the real ones.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Units in Z[pi], field forced REVISED>> The axioms for a field do not require it to be closed under infinite>> sums. Taking an infinite sum is not a field operation.>I didn't say it was.>The actual rule is decidability, otherwise known as convergence.No, the axioms for a field do not require it to be closed under theoperation of convergent infinite sums either.>My point is that there is no *field* of rationals, and in fact, you>only have rationals, which are numbers written as the ratio of two>integers, where the denominator is non-zero.The rationals, under the usual operations of addition andmultiplication, form a field. >As if you consider a *field* of rationals then infinite sums that>converge, as they are decidable, would be included.Perhaps you have come across a different definition of field from theusual one. Can you tell me where you learnt that a field must beclosed under convergent infinite sums?-- Richard-- Spam filter: to mail me from a .com/.net site, put my surname in the headers.FreeBSD rules! === Subject: : Re: Units in Z[pi], field forced REVISED> ...> Why bother? It would just be looking up a social fossil, wouldn't it?There are probably characters in Henry James novels who look up socialfossils and are forever ostracized by polite society for it.-- G.C. === Subject: : Re: Units in Z[pi], field forced REVISED[...]> Now then as to the social fossil, it is that the fielf of rationals is> NOT truly a field as elements are excluded to fit into a definition,> otherwise it would be reals.> There is no *field* of rationals, and the term is a social fossil.While you're at it, why don't you make a list (even a short one is betterthan nothing) of what you call social fossils.David Bernier === Subject: : Re: Units in Z[pi], field forced REVISED> [.snip.]> [.about James.]>My assessment is that you have no idea what a ring is (an assessment>amply confirmed by your definition in your webpage of A commutative ring is a set of numbers where for any members a and b,> a+b = b+a and ab = ba.>>You're making that up. He actually gives that as the _definition_ ofcommutative ring?>No, I'm not making it up. Yes, he gives that as the definition of>commutative ring.>http://www.msnusers.com/AmateurMath/ objectmathematic.msnw>It showed up one or two days ago. Well that's incredible.>I used his definition to show that>every set is a commutative ring, and went one step further and even>made sure that addition and multiplication were operations on the>set. Amazing, huh?>>I mean that's just incredible - if he's going to give the definition>>why wouldn't he look it up somewhere and give the real definition,>>even if he didn't understand it?>Why bother? It would just be looking up a social fossil, wouldn't it?>========================================================== ============>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 critize. A great> many people are staggered to this extend, 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>======================================================= ===============>Arturo Magidin>magidin@math.berkeley.edu************************ === Subject: : Re: Units in Z[pi], field forced REVISED>> The logical conclusion, following standard use of limit arguments in>> mathematics, is that Z[pi] has an infinite number of units.>> Now then as to the social fossil, it is that the fielf of rationals is>> NOT truly a field as elements are excluded to fit into a definition,>> otherwise it would be reals.>> There is no *field* of rationals, and the term is a social fossil.I missed a couple points in the following...> A field R satisfies the following properties:The following is for all a,b,c in R:a+b is in Ra*b is in R> (a+b)+c=a+(b+c)> there exists 0 such that a+0=0+a=a> For all a there exists -a such that a+(-a)=(-a)+a=0 ^ in R> a+b=b+a> (ab)c=a(bc)> a(b+c)=ab+ac> (a+b)c=ac+bc> ab=ba> for all a<>0 there exists b such that ab=1 ^ in R> Which of these do the rationals not satify? A field is defined by its > properties, not by its elements.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Units in Z[pi], field forced REVISED Visiting Assistant Professor at the University of Montana.>> A field R satisfies the following properties:>>The following is for all a,b,c in R:>a+b is in R>a*b is in RAs usual, it depends; if you define '+' and '*' to be binaryoperations on R, then the fact that a+b and a*b are elements of R ispart of the fact that they are operations.=================================================== ===================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Units in Z[pi], field forced REVISED>A field R satisfies the following properties:>>The following is for all a,b,c in R:>>a+b is in R>>a*b is in R> As usual, it depends; if you define '+' and '*' to be binary> operations on R, then the fact that a+b and a*b are elements of R is> part of the fact that they are operations.Agreed, that is usually understood, but I don't trust James to claim that a+b can fall outside R.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Units in Z[pi], field forced REVISED> In this post I use a limit approach to show that adding pi to the > ring> of integers forces you into a field.>> No you don't. A ring is not required to be closed under taking limits.>> The fact that you think it should be doesn't change that fact.>> I didn't say it did. Your insinuation that I did, is false.>> My guess is that you're lost, so you make up things.>You are talking about pi^2/6 = 1+ 1/4 + 1/9 + ...>This is formally written as pi^2/6 = lim k-> oo (1 + 1/4 + 1/9 + ... >+ 1/(k^2) )>Now, in order for this limit to lie in the rationals, the rationals >must be closed under infinite sums.>> Well, not quite. If the rationals were closed under converging>> infinite sums, then you would be able to conclude that this limit lies>> in the rationals. The point is that the rationals are not necessarily>> closed under converging infinite sums, so the fact that this sum>> converges is not sufficient to conclude that the limit must lie in the>field of rationals.>> Because the operations in a field are finitary, so derived operations>> are finitary, so there is no reason why one must consider infinite>> sums when dealing with a structure we want to test for the field axioms.>> I'm sure that's what you meant, just thought I would make the>> precision explicit (since I committed the same sin of carelessness>> earlier...)> Unfortunately, there's a fine line between when the precision matters, > and it's not always easy to see when that line has been crossed.> I wonder if James understands what infinity is and is not.Should we start him off in a new research area by telling him about Zeno's paradox? Better not.Gib === Subject: : Re: Units in Z[pi], field forced REVISED>>Well, not quite. If the rationals were closed under converging>infinite sums, then you would be able to conclude that this limit lies>in the rationals. The point is that the rationals are not necessarily>closed under converging infinite sums, so the fact that this sum>converges is not sufficient to conclude that the limit must lie in the>field of rationals.>Because the operations in a field are finitary, so derived operations>are finitary, so there is no reason why one must consider infinite>sums when dealing with a structure we want to test for the field axioms.>I'm sure that's what you meant, just thought I would make the>precision explicit (since I committed the same sin of carelessness>earlier...)>> Unfortunately, there's a fine line between when the precision matters, >> and it's not always easy to see when that line has been crossed.>> I wonder if James understands what infinity is and is not.> Should we start him off in a new research area by telling him about > Zeno's paradox? Better not.> GibAs entertaining as that could be... I don't think it's a good idea. How about explaining that the Cantor set has an infinite number of elements of measure 0?-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Units in Z[pi], field forced REVISED> Should we start him off in a new research area by telling him about > Zeno's paradox? Better not.You see what a brilliant troll JSH is? For a long time in order to argue with JSH one needed to contemplate the algebraic integers, and not everyone wants to do that. So JSH brilliantly puts out a series of posts in which he claims that the rationals aren't a field; that rings are closed under limits; and that convergence is now called decidability. All of a sudden lots more people can argue with him, and so you have hundreds of followups to his latest round of trollery. And when he threatens to stop posting, it's like the local dealer running out of heroin. People start to go into JSH withdrawal. Do you realize that this entire newsgroup has become JSH codependant? === Subject: : Weakly dense in linear orderLet S be a linear order. Claim:S is order isomorphic to a subset of reals iffsome countable C subset S with for all x,y in S, if x < y, then some z in C with x <= z <= y. ie., C is weakly dense in S.To begin with, C need not be order dense, hence it may not be not orderisomorphic to the rationals Q, but a subset of Q. If an element of C hasa cover or is a cover of an element of C, then insert a copy of Q into Cbetween those two elements. The resulting C would then be orderisomorphic to Q and strictly between any two elements of the resulting Sis an element of C.If S is then a continuous order, it's order isomorphic to an open, closed,half open or half closed unit interval. What do I do when S isn't acontinuous order?-- Conversely if S is order isomorphic to a subset of the reals.How do I construct a countable C that's weakly dense in S?--------== Posted via Newsfeed.Com - Unlimited-Uncensored-Secure Usenet News==----http://www.newsfeed.com The #1 Newsgroup Service in the World! >100,000 Newsgroups---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === Subject: : Re: Weakly dense in linear order> Let S be a linear order. Claim:> S is order isomorphic to a subset of reals iff> some countable C subset S with for all x,y in S,> if x < y, then some z in C with x <= z <= y.> ie., C is weakly dense in S.Suppose C is weakly dense in S. Let C = {c_n: n = 1, 2, ...}. For x inS define f(x) = Sum(1/2^n: c_n < x) + Sum(1/2^n: c_n <= x). Then f isan order-embedding of S into R.Suppose S is a subset of R. For each pair (r,s) of rational numbers,if there exists an element a of S such that r < a < s, choose one suchelement. Let A be the set of elements so chosen. Let B be the set ofall elements b of S such that, for some positive number p, at leastone of the intervals (b-p, b) and (b, b+p) is disjoint from S. Let Cbe the union of A and B. Then C is a countable subset of S which isweakly dense in S. === Subject: : What cardinality may a ring have?Given a non-empty set S of cardinality kappa (finite or infinite) canone always define +, * and 0 so that is a ring?-- G.C. === Subject: : Re: What cardinality may a ring have? Visiting Assistant Professor at the University of Montana.>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring?Well, if you already know S has a cardinality kappa, then yes: bijectit with the direct sum of kappa copies of the integers. I ->think<-that there is no need to invoke choice to show that the latter set hascardinality kappa, so we can compose the two without invoking choice.Otherwise, you may have to invoke the Axiom of Choice, since inparticular you need to define a group structure, and without AC theremay be sets withoout a group structure. And you can always make anabelian group into a ring by defining multiplication a*b=0; but aslong as you have AC, you can use the direct sum of copies of any ringof smaller cardinality.================================================== ====================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What cardinality may a ring have?>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring?Certainly for any finite kappa, yes (Z_kappa); also certainly for aleph0 and c (= cardinality of P(N)).Under GCH, for all kappa, since set operations render P(X) a boolean ring.The masters will have to weigh in for the case not GCH.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: What cardinality may a ring have? Visiting Assistant Professor at the University of Montana.>>Given a non-empty set S of cardinality kappa (finite or infinite) can>>one always define +, * and 0 so that is a ring?>Certainly for any finite kappa, yes (Z_kappa); also certainly for aleph0 >and c (= cardinality of P(N)).>Under GCH, for all kappa, since set operations render P(X) a boolean >ring.>The masters will have to weigh in for the case not GCH.Well, you could always do the boolean ring (without complementation)defined on the finite subset of X, which has the same cardinality asX. But the real sticky point, I think, is AC. Even the bijectabilityof S with the collection of all finite subset of S is equivalent tochoice.http://groups.google.com/groups?selm=b1jtbp%241g04% 40odds.stat.purdue.eduHmmm... We've had group structure on any set; now ring structure. Iguess in a couple of months someone will ask if we can put a fieldstructure on any infinite set. (-:=========================================================== ===========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What cardinality may a ring have?>> Given a non-empty set S of cardinality kappa (finite or infinite) can>> one always define +, * and 0 so that is a ring?> [...]> Under GCH, for all kappa, since set operations render P(X) a > boolean ring.Oops. Even under GCH, I guess this argument does not cover limit cardinals.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: : Re: What cardinality may a ring have?>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring?> Well, if you already know S has a cardinality kappa, then yes: biject> it with the direct sum of kappa copies of the integers. I ->think<-> that there is no need to invoke choice to show that the latter set has> cardinality kappa, so we can compose the two without invoking choice.> Otherwise, you may have to invoke the Axiom of Choice, since in> particular you need to define a group structure, and without AC there> may be sets withoout a group structure. And you can always make an> abelian group into a ring by defining multiplication a*b=0; but as> long as you have AC, you can use the direct sum of copies of any ring> of smaller cardinality.Harris thread that I used his definition to show thatevery set is a commutative ring, and went one step further and evenmade sure that addition and multiplication were operations on theset.And I thought that if the set was **non-empty** you could always do it;a*b=0 ensuring commutativity. Just wanted to be sure. I'm happy withChoice.-- G.C. === Subject: : Re: What cardinality may a ring have? Visiting Assistant Professor at the University of Montana.>>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring? [.snip.]>Harris thread that >I used his definition to show that>every set is a commutative ring, and went one step further and even>made sure that addition and multiplication were operations on the>set.>And I thought that if the set was **non-empty** you could always do it;>a*b=0 ensuring commutativity. Just wanted to be sure. I'm happy with>Choice.Ah, but his definition was a set in which for all a,b, a+b=b+a anda*b=b*a. So if the set is empty, it satisfies it by vacuity; and ifthe set is nonempty, pick any element a in S, and define +:SxS->S and*:SxS->S by x+y=a, x*y=a. Surely you agree it meets the definition.(Perhaps I should have said ' I used his definition to show that everyset is a commutative ring', with scare quotes)======================================================= ===============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What cardinality may a ring have?>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring?>>Certainly for any finite kappa, yes (Z_kappa); also certainly for aleph0 >>and c (= cardinality of P(N)).>>Under GCH, for all kappa, since set operations render P(X) a boolean >>ring.For arbitrary cardinals, it is equivalent to AC.Hajnal and Kertesz showed that AC is equivalent to everyinfinite set being one of many things, from two sidedcancellative semigroup to field. To get a field ofcardinality kappa, take the rational functions overthe rationals with kappa indeterminates; making thisalgebraically closed does not increase the cardinality.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: : Re: What cardinality may a ring have? Visiting Assistant Professor at the University of Montana. [.snip.]>Given a non-empty set S of cardinality kappa (finite or infinite) can>one always define +, * and 0 so that is a ring?>For arbitrary cardinals, it is equivalent to AC.Now, I am just splitting hairs, but I am mildly curious. Say we do notassume AC, but interpret non-empty set of cardinality kappa asmeaning that we are given a bijection from S to the cardinal.Is there a constructive bijection from Z^(kappa) to kappa, whereZ^(kappa) is the set of almost always zero function from kappa to Z?============================================================ ==========It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What cardinality may a ring have?> ...> Hmmm... We've had group structure on any set; now ring structure. I ^^^^^^^^^^^^^^^^^^^I should have googled for this since taking a*b = 0 to make a ring isobvious even to a mathematical illiterate like me.-- G.C. === Subject: : Re: What cardinality may a ring have?> Given a non-empty set S of cardinality kappa (finite or infinite) can> one always define +, * and 0 so that is a ring?It can certainly be done for any S having any natural number cardinality greater than one.Given that S is countably infinite, it can be done in a myriad of ways just as subrings of the ring of rationals.Similarly for cardinality of the reals (equal to that of the complexes), there are lots of (uncountable) such subrings of the complex number field. === Subject: : Re: What cardinality may a ring have?> Given a non-empty set S of cardinality kappa (finite or infinite) can> one always define +, * and 0 so that is a ring?> It can certainly be done for any S having any natural number> cardinality greater than one.Why _greater_ than one? Surely S = {0}, with +, *, 0 defined in theonly way possible, is a ring? Rather perversely such an S is even aring with a multiplicative identity!> Given that S is countably infinite, it can be done in a myriad of> ways just as subrings of the ring of rationals.> Similarly for cardinality of the reals (equal to that of the> complexes), there are lots of (uncountable) such subrings of the> complex number field.-- G.C. === Subject: : Re: What cardinality may a ring have? Visiting Assistant Professor at the University of Montana.> Given a non-empty set S of cardinality kappa (finite or infinite) can> one always define +, * and 0 so that is a ring?> It can certainly be done for any S having any natural number>> cardinality greater than one.>Why _greater_ than one? Surely S = {0}, with +, *, 0 defined in the>only way possible, is a ring? Rather perversely such an S is even a>ring with a multiplicative identity!Indeed, it is the terminal object of the category of rings. But some people demand that their rings have a 1, and moreover that 1be different from 0, so it is possible for some not to consider thezero ring as a ring.========================================================= =============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What cardinality may a ring have?> [.snip.]>>Given a non-empty set S of cardinality kappa (finite or infinite) can>>one always define +, * and 0 so that is a ring?>>For arbitrary cardinals, it is equivalent to AC.>Now, I am just splitting hairs, but I am mildly curious. Say we do not>assume AC, but interpret non-empty set of cardinality kappa as>meaning that we are given a bijection from S to the cardinal.>Is there a constructive bijection from Z^(kappa) to kappa, where>Z^(kappa) is the set of almost always zero function from kappa to Z?This will follow if and only if kappa^2 = kappa andaleph_0 <= kappa. That the condition is necessary isthat the non-negative functions which add to 3 take have cardinality kappa^2.However, kappa^2 = kappa for all kappa is equivalent to AC.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: : Re: What cardinality may a ring have?> Given a non-empty set S of cardinality kappa (finite or infinite) can> one always define +, * and 0 so that is a ring?> It can certainly be done for any S having any natural number>> cardinality greater than one.>Why _greater_ than one? Surely S = {0}, with +, *, 0 defined in the>only way possible, is a ring? Rather perversely such an S is even a>ring with a multiplicative identity!In my old-fashioned world, all rings had at least two members, an additive identity and a distinct multipicative identity.> Indeed, it is the terminal object of the category of rings. > But some people demand that their rings have a 1, and moreover that 1> be different from 0, so it is possible for some not to consider the> zero ring as a ring.> ============================================================== ========It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> ============================================================== ========> Arturo Magidin> magidin@math.berkeley.edu === Subject: : What do you call...Something that's almost completely like a ring in every respect exceptthat it has a couple of different rules? Something more detailed thana set, because it involves operators, but where the operatorsthemselves may not actually be operatable between two numbers orn-dimensional sets of numbers, but rather between any particularn-dimensional number and any number of predetermined, pregivenoperators? But, which by applying the operators in any orderwhatsoever to any number in the set one also gets a number in the set.What do you call that?As an example of one of the things I'm talking about, take the set ofdifferent positions on the unit sphere, or radial vectors, along withthe orientation vector, such as that of an abserver on the sphere, tobe be the possible values of the set.Then take two operations:One operation rotates the sphere a given fixxed angle about the axisorthogonal to both the current radial vector and to the currentorientation vector, and which passes through the center of the sphere,and therefore changes your position, essentially moving you thedirection you're pointing.The other operation rotates the sphere another given fixxed angleabout the axis corresponding to your current radial vector, and whichalso passes through the center of the sphere, and therefore changesyour orientation, or changing the direction you're pointing.And most of all, take the set of points such that if you take anypoint on the set, and apply the given operations, you can only getanother point on the set. And maybe I should add that you can get fromany point to any other point using only the given operations, ifthat's necessary.What would that be called?I know that at least five sets of this sort exist, each correspondingto the platonic solids.Does anybody know if this sort of thing has been thought of already?And if so, what its given name is? Or am I the first to consider suchthings?(...Starblade Riven Darksquall...) === Subject: : Re: What do you call...> Something that's almost completely like a ring in every respect except> that it has a couple of different rules? Something more detailed than> a set, because it involves operators, but where the operators> themselves may not actually be operatable between two numbers or> n-dimensional sets of numbers, but rather between any particular> n-dimensional number and any number of predetermined, pregiven> operators? But, which by applying the operators in any order> whatsoever to any number in the set one also gets a number in the set.> What do you call that?No clue.> As an example of one of the things I'm talking about, take the set of> different positions on the unit sphere, or radial vectors, along with> the orientation vector, such as that of an abserver on the sphere, to> be be the possible values of the set.> Then take two operations:> One operation rotates the sphere a given fixxed angle about the axis> orthogonal to both the current radial vector and to the current> orientation vector, and which passes through the center of the sphere,> and therefore changes your position, essentially moving you the> direction you're pointing.> The other operation rotates the sphere another given fixxed angle> about the axis corresponding to your current radial vector, and which> also passes through the center of the sphere, and therefore changes> your orientation, or changing the direction you're pointing.> And most of all, take the set of points such that if you take any> point on the set, and apply the given operations, you can only get> another point on the set. And maybe I should add that you can get from> any point to any other point using only the given operations, if> that's necessary.> What would that be called?This sounds to me like a type of permutation group with an infinite number of elements.> I know that at least five sets of this sort exist, each corresponding> to the platonic solids.Those are permutation groups, I believe. I'm not completely sure I'm understanding your example. Are you thinking of rotations, for example, of a cube?> Does anybody know if this sort of thing has been thought of already?> And if so, what its given name is? Or am I the first to consider such> things?> (...Starblade Riven Darksquall...)-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: What do you call... Visiting Assistant Professor at the University of Montana.>Something that's almost completely like a ring in every respect except>that it has a couple of different rules? Something more detailed than>a set, because it involves operators, but where the operators>themselves may not actually be operatable between two numbers or>n-dimensional sets of numbers,By n-dimensional sets of numbers, do you mean n-tuples? Or do youmean sets of n-tuples?> but rather between any particular>n-dimensional number and any number of predetermined, pregiven>operators? But, which by applying the operators in any order>whatsoever to any number in the set one also gets a number in the set.>What do you call that?Are you sure they don't qualify under algebra in the General Algebrasense? A set with a collection of (possibly partial) finitaryoperations and relations?>As an example of one of the things I'm talking about, take the set of>different positions on the unit sphere, or radial vectors, along with>the orientation vector, such as that of an abserver on the sphere, to>be be the possible values of the set.>Then take two operations:>One operation rotates the sphere a given fixxed angle about the axis>orthogonal to both the current radial vector and to the current>orientation vector, and which passes through the center of the sphere,>and therefore changes your position, essentially moving you the>direction you're pointing.>The other operation rotates the sphere another given fixxed angle>about the axis corresponding to your current radial vector, and which>also passes through the center of the sphere, and therefore changes>your orientation, or changing the direction you're pointing.>And most of all, take the set of points such that if you take any>point on the set, and apply the given operations, you can only get>another point on the set. And maybe I should add that you can get from>any point to any other point using only the given operations, if>that's necessary.>What would that be called?They sound like plain old operations to me. This would be an algebrain the general algebra sense.======================================================== ==============It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes)======================================================= ===============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: What do you call...> Something that's almost completely like a ring in every respect except> that it has a couple of different rules? Something more detailed than> a set, because it involves operators, but where the operators> themselves may not actually be operatable between two numbers or> n-dimensional sets of numbers, but rather between any particular> n-dimensional number and any number of predetermined, pregiven> operators? But, which by applying the operators in any order> whatsoever to any number in the set one also gets a number in the set.> What do you call that?> No clue.> As an example of one of the things I'm talking about, take the set of> different positions on the unit sphere, or radial vectors, along with> the orientation vector, such as that of an abserver on the sphere, to> be be the possible values of the set.> Then take two operations:> One operation rotates the sphere a given fixxed angle about the axis> orthogonal to both the current radial vector and to the current> orientation vector, and which passes through the center of the sphere,> and therefore changes your position, essentially moving you the> direction you're pointing.> The other operation rotates the sphere another given fixxed angle> about the axis corresponding to your current radial vector, and which> also passes through the center of the sphere, and therefore changes> your orientation, or changing the direction you're pointing.> And most of all, take the set of points such that if you take any> point on the set, and apply the given operations, you can only get> another point on the set. And maybe I should add that you can get from> any point to any other point using only the given operations, if> that's necessary.> What would that be called?> This sounds to me like a type of permutation group with an infinite > number of elements.I'm looking only for ones with a finite number of elements. I'mwondering if there's a special name for THAT.> I know that at least five sets of this sort exist, each corresponding> to the platonic solids.> Those are permutation groups, I believe. I'm not completely sure I'm > understanding your example. Are you thinking of rotations, for example, > of a cube?I meant, if you put each one of them on a sphere, then you'll have aset of points corresponding to one of the type of sets I'm talkingabout.> Does anybody know if this sort of thing has been thought of already?> And if so, what its given name is? Or am I the first to consider such> things?(...Starblade Riven Darksquall...) === Subject: : Re: What do you call...> Something that's almost completely like a ring in every respect except> that it has a couple of different rules? Something more detailed than> a set, because it involves operators, but where the operators> themselves may not actually be operatable between two numbers or> n-dimensional sets of numbers, but rather between any particular> n-dimensional number and any number of predetermined, pregiven> operators? But, which by applying the operators in any order> whatsoever to any number in the set one also gets a number in the set.> What do you call that?> No clue.> As an example of one of the things I'm talking about, take the set of> different positions on the unit sphere, or radial vectors, along with> the orientation vector, such as that of an abserver on the sphere, to> be be the possible values of the set.> Then take two operations:> One operation rotates the sphere a given fixxed angle about the axis> orthogonal to both the current radial vector and to the current> orientation vector, and which passes through the center of the sphere,> and therefore changes your position, essentially moving you the> direction you're pointing.> The other operation rotates the sphere another given fixxed angle> about the axis corresponding to your current radial vector, and which> also passes through the center of the sphere, and therefore changes> your orientation, or changing the direction you're pointing.> And most of all, take the set of points such that if you take any> point on the set, and apply the given operations, you can only get> another point on the set. And maybe I should add that you can get from> any point to any other point using only the given operations, if> that's necessary.> What would that be called?> This sounds to me like a type of permutation group with an infinite > number of elements.I'm looking only for ones with a finite number of elements. I'mwondering if there's a special name for THAT.> I know that at least five sets of this sort exist, each corresponding> to the platonic solids.> Those are permutation groups, I believe. I'm not completely sure I'm > understanding your example. Are you thinking of rotations, for example, > of a cube?I meant, if you put each one of them on a sphere, then you'll have aset of points corresponding to one of the type of sets I'm talkingabout.> Does anybody know if this sort of thing has been thought of already?> And if so, what its given name is? Or am I the first to consider such> things?(...Starblade Riven Darksquall...) === Subject: : Re: What do you call...>As an example of one of the things I'm talking about, take the set of>different positions on the unit sphere, or radial vectors, along with>the orientation vector, such as that of an abserver on the sphere, to>be be the possible values of the set.>Then take two operations:>One operation rotates the sphere a given fixxed angle about the axis>orthogonal to both the current radial vector and to the current>orientation vector, and which passes through the center of the sphere,>and therefore changes your position, essentially moving you the>direction you're pointing.>The other operation rotates the sphere another given fixxed angle>about the axis corresponding to your current radial vector, and which>also passes through the center of the sphere, and therefore changes>your orientation, or changing the direction you're pointing.>And most of all, take the set of points such that if you take any>point on the set, and apply the given operations, you can only get>another point on the set. And maybe I should add that you can get from>any point to any other point using only the given operations, if>that's necessary.>What would that be called?>>This sounds to me like a type of permutation group with an infinite >>number of elements.> I'm looking only for ones with a finite number of elements. I'm> wondering if there's a special name for THAT.>I know that at least five sets of this sort exist, each corresponding>to the platonic solids.>>Those are permutation groups, I believe. I'm not completely sure I'm >>understanding your example. Are you thinking of rotations, for example, >>of a cube?> I meant, if you put each one of them on a sphere, then you'll have a> set of points corresponding to one of the type of sets I'm talking> about.>Does anybody know if this sort of thing has been thought of already?>And if so, what its given name is? Or am I the first to consider such>things?> (...Starblade Riven Darksquall...)Ok, the image I'm getting from your description is (for instance) a cube. You are looking at it with a sense of up and can either rotate the cube so that the face you are looking at travels up, or right an specified number of degrees/radians. If O is the set of orientations, then up:RxO -> O, right:RxO -> O.As stated this seems like an interesting variation on scalar multiplication that could be restated as a permutation group on O. Am I understanding correctly?-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Which are the fastest algorithms to solve large polynomial systems?Hello.Which are the momentary fastest (serial) algorithms for the solution oflarge polynomial systems with several variables (rational coefficients, realsolutions, up to 50 variables, polynomial degrees up to 50, arbitrarily manypolynomials)? === Subject: : Re: Who here believes maths is all there is?>>As a teenager, I think that I really used to think this way. I thought that>>the fundamental building blocks of the universe would be some mathematical> WOW! You really *were* a mathematophile! > I doubt most professional mathies would be so devout.>> Then at the age of 23, I became a Christian,> Oh, you poor fellow. I sincerely hope you'll recover soon.I hope he doesn't. I've been a Christian for about 4.5 years now.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Who here believes maths is all there is?>>As a teenager, I think that I really used to think this way. I thoughtthat>>the fundamental building blocks of the universe would be somemathematical> WOW! You really *were* a mathematophile!> I doubt most professional mathies would be so devout.>> Then at the age of 23, I became a Christian,> Oh, you poor fellow. I sincerely hope you'll recover soon.> I hope he doesn't. I've been a Christian for about 4.5 years now.Oh it's contagious? Is there vaccine or cure for that terrible illness?> --> Will Twentyman> email: wtwentyman at copper dot net === Subject: : Re: Who here believes maths is all there is?>As a teenager, I think that I really used to think this way. I thought> that>the fundamental building blocks of the universe would be some> mathematical>WOW! You really *were* a mathematophile!>I doubt most professional mathies would be so devout.>Then at the age of 23, I became a Christian,>Oh, you poor fellow. I sincerely hope you'll recover soon.>>I hope he doesn't. I've been a Christian for about 4.5 years now.> Oh it's contagious? Is there vaccine or cure for that terrible illness?I was vaccinated, but apparently it wore off.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Why not anonymous publication?>Why can't, just for once, a math paper be published anonymously. > It happens from time to time, usually under the name of X> or the like.>Let>people focus on the paper, not the author. Peer reviews are already>done anonymously, right?> Wrong, if you mean what I think youmean.> About 20 or 25 years ago, the American Mathematical Society > briefly experimented with having peer reviewers see only a> redacted and anonymized version of the manuscript. The > experiment must have been the failure it was predicted to be,> universal (as far as I know) in mathematics: the author's name> is known to the reviewer (but not vice versa). Apparently there> are academic fields in which it is standard to anonymize manuscripts> for review, and maybe in some fields that's not completely laughable,> but I think it would be in mathematics (and evidently that was the> better or worse, many mathematicians--if they write more than> one paper--cite their own earlier papers, and usualy in ways > which make it clear that they *are* their own earlier papers.> Also for better or worse, many subfields of mathematics are> small enough that if an editor wants a competent referee,> then the referee will probably know the author and may well> have seen the paper in pre-print version. Finally (and this> is new), with the arXiv being there, in many subjects a> person who's likely to be a competent referee is now likely> to have seen an e-print even if he or she doesn't know the> author from Adam or Eve. Anonymity would be a farce.> By all means make arguments for your case if you care to> refute my various assertions. (Let people focus on the > paper, not the author isn't an argument; it's just an> assertion, too.)> Lee RudolphI would think math would be the best fit for anonymous publication. You don't have to refer to some physical lab of where experiments havetaken place. Even if it is reviewed non-anonymously, why not exposeit anonymously? As for citing earlier works, why must it be obviousthat they are the author's own--that should just be a matter of aquick edit? Let each paper stand on its own.True, in some circles the author will be obvious simply because ofdiscussions. But outside of that, it would allow people to focus onideas and not the people behind them. === Subject: : Re: Why not anonymous publication?>Why can't, just for once, a math paper be published anonymously. > Some have.> But for most papers, nobody knows more about the topic than the writer. And> for most writers and readers, nobody is more interested in communication on> that topic than the writer. That's why the writer's name and adress are> published, rather than the name of a waitress in Chelsea and the address of a> tobacconist in Hooveringham.OK, if you must, have a mailbox associated with every single paper,where people can send correspondence anonymously to the author. Abetter solution would be just a group discussion where everyone isanonymous. The author would then be likely to see the messages. === Subject: : Re: Why not anonymous publication?>I would think math would be the best fit for anonymous publication. >You don't have to refer to some physical lab of where experiments have>taken place. Even if it is reviewed non-anonymously, why not expose>it anonymously? As for citing earlier works, why must it be obvious>that they are the author's own--that should just be a matter of a>quick edit? Let each paper stand on its own.>True, in some circles the author will be obvious simply because of>discussions. But outside of that, it would allow people to focus on>ideas and not the people behind them.After this past week of sci.math, I might conjecture that Hans Abergotherwise be interested in) because he can't find it within himself to focus on the ideas and not the people behind them. With thatpossible exception, I can't think of *any* example in my experienceof real mathematics and real mathematicians (by those phrases I mean to exclude, for example, the crank posts and crank posters promulgating anti-Cantorian balderdash and anti-Wiles bafflegab) in which thefocus is *not* on ideas rather than on the people behind them.Obviously I'm a pawn of the establishment, willfully blind, etc.,etc.; but some specific examples of the problem you intend to fixwould be good to see, if you propose to persuade people to implementyour solution to it. Lee Rudolph === Subject: : Re: Why not anonymous publication?>I would think math would be the best fit for anonymous publication. >You don't have to refer to some physical lab of where experiments have>taken place. Even if it is reviewed non-anonymously, why not expose>it anonymously? As for citing earlier works, why must it be obvious>that they are the author's own--that should just be a matter of a>quick edit? Let each paper stand on its own.>True, in some circles the author will be obvious simply because of>discussions. But outside of that, it would allow people to focus on>ideas and not the people behind them.> After this past week of sci.math, I might conjecture that Hans Aberg> otherwise be interested in) because he can't find it within himself > to focus on the ideas and not the people behind them. With that> possible exception, I can't think of *any* example in my experience> of real mathematics and real mathematicians (by those phrases I mean > to exclude, for example, the crank posts and crank posters promulgating anti-Cantorian balderdash and anti-Wiles bafflegab) in which the> focus is *not* on ideas rather than on the people behind them.> Obviously I'm a pawn of the establishment, willfully blind, etc.,> etc.; but some specific examples of the problem you intend to fix> would be good to see, if you propose to persuade people to implement> your solution to it. > Lee RudolphA real problem? Elitism. Referring to concepts by names raises thebarrier of entry, and makes it seem like a closed group of snobs. Whynot commutative group instead of Abelian? Boolean algebra? What'swrong with binary algebra? === Subject: : Re: Why not anonymous publication?...>> Obviously I'm a pawn of the establishment, willfully blind, etc.,>> etc.; but some specific examples of the problem you intend to fix>> would be good to see, if you propose to persuade people to implement>> your solution to it. > Lee Rudolph>A real problem? Elitism. Referring to concepts by names raises the>barrier of entry, and makes it seem like a closed group of snobs. Why>not commutative group instead of Abelian? Boolean algebra? What's>wrong with binary algebra?I never, ever, would have guessed that *this* was what you were on about. And it's reasonably unclear to me how publishing anonymously will have much of an effect on this, in the shortterm. However, ignoring the connection to publication, I'lltake the opportunity to blather on for a bit. There are plenty of examples of opaque jargon, of whicheponyms like Abelian group and Boolean algebra are only a small minority. Arguably jargon *does* raise the barrier of entry; I can't, myself, see that eponymic jargon raises it anyfurther, and it may actually lower it, in that jargon derived fromcommon English may deceive the novice by making a technicaldiscussion seem to be saying something it isn't. Comparethe synonyms B-space (little used), Banach space, andcomplete normed topological vector space. A novice who'snever heard of Banach won't get anything but a convenient label(for an idea) out of either of the first two, and depending oncontext, ability, capacity to learn and read, etc., may or may not be able to continue on with the rest of the sentence,paragraph, chapter, book...--but certainly won't come away(from reading B-space or Banach space) with a terribly*wrong* idea. By contrast, I can imagine some novices takingthe words complete and normed (and maybe vector) andgoing absolutely ape, into a world of terrible miscomprehension.There are real examples like the one I've just pastiched, that have come up from time to time in sci.math, and--I'd bet--inthe classroom experience of many posters here.Lee Rudolph === Subject: : Re: Why not anonymous publication?>...<>> Obviously I'm a pawn of the establishment, willfully blind, etc.,<>> etc.; but some specific examples of the problem you intend to fix<>> would be good to see, if you propose to persuade people to implement<>> your solution to it. <>> Lee Rudolph>>A real problem? Elitism. Referring to concepts by names raises the>>barrier of entry, and makes it seem like a closed group of snobs. Why>>not commutative group instead of Abelian? Boolean algebra? What's>>wrong with binary algebra?>I never, ever, would have guessed that *this* was what you were on >about. And it's reasonably unclear to me how publishing >anonymously will have much of an effect on this, in the short>term. However, ignoring the connection to publication, I'll>take the opportunity to blather on for a bit. >There are plenty of examples of opaque jargon, of which>eponyms like Abelian group and Boolean algebra are only a >small minority. Arguably jargon *does* raise the barrier of >entry; I can't, myself, see that eponymic jargon raises it any>further, and it may actually lower it, in that jargon derived from>common English may deceive the novice by making a technical>discussion seem to be saying something it isn't. Compare>the synonyms B-space (little used), Banach space, and>complete normed topological vector space. A novice who's>never heard of Banach won't get anything but a convenient label>(for an idea) out of either of the first two, and depending on>context, ability, capacity to learn and read, etc., may or may >not be able to continue on with the rest of the sentence,>paragraph, chapter, book...--but certainly won't come away>(from reading B-space or Banach space) with a terribly>*wrong* idea.However, if we carried this idea out for all definitions,would be a complete normed topological vector space withthe norm given by an inner product, and these terms couldbe expanded. Would you like to use topological space forwhich all open coverings have countable refinementsinstead of Lindelof spaces. Or how about Stone-Cechcompactification? Or the Stone space of a Boolean algebra?In probability, we use the term Polish space and cadlag process. An infinitely divisible distributionhas a Levy measure. There are many others.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: : Re: Why not anonymous publication?...> <>> Obviously I'm a pawn of the establishment, willfully blind, etc.,> <>> etc.; but some specific examples of the problem you intend to fix> <>> would be good to see, if you propose to persuade people to implement> <>> your solution to it. > <>> Lee Rudolph>>A real problem? Elitism. Referring to concepts by names raises the>>barrier of entry, and makes it seem like a closed group of snobs. Why>>not commutative group instead of Abelian? Boolean algebra? What's>>wrong with binary algebra?>I never, ever, would have guessed that *this* was what you were on >about. And it's reasonably unclear to me how publishing >anonymously will have much of an effect on this, in the short>term. However, ignoring the connection to publication, I'll>take the opportunity to blather on for a bit. >There are plenty of examples of opaque jargon, of which>eponyms like Abelian group and Boolean algebra are only a >small minority. Arguably jargon *does* raise the barrier of >entry; I can't, myself, see that eponymic jargon raises it any>further, and it may actually lower it, in that jargon derived from>common English may deceive the novice by making a technical>discussion seem to be saying something it isn't. Compare>the synonyms B-space (little used), Banach space, and>complete normed topological vector space. A novice who's>never heard of Banach won't get anything but a convenient label>(for an idea) out of either of the first two, and depending on>context, ability, capacity to learn and read, etc., may or may >not be able to continue on with the rest of the sentence,>paragraph, chapter, book...--but certainly won't come away>(from reading B-space or Banach space) with a terribly>*wrong* idea.> However, if we carried this idea out for all definitions,> would be a complete normed topological vector space with> the norm given by an inner product, and these terms could> be expanded. Would you like to use topological space for> which all open coverings have countable refinements> instead of Lindelof spaces. Or how about Stone-Cech> compactification? Or the Stone space of a Boolean algebra?> In probability, we use the term Polish space and cadlag process. An infinitely divisible distribution> has a Levy measure. There are many others.Well, why not type-A space? I think the example of a commutativegroup being called Abelian is at least one example of something thatis far more useful with a descriptive label. Certainly that only goesso far (as systematic chemical naming has found). But the problemremains of honoring people when nobody creates math, they justdiscover it. What to do when cultures meet up (what happens whenaliens come?). If you name things after people, you'll argue that youwant your culture's name, and it's a pointless battle. === Subject: : Re: WOW! Is this known.>WOW! Is this known?> Yes, it is known. If you define _any_ sequence using the rule that > a(n+1) = a(n)/2 if a(n) is even plus any other rules then you will get > exactly the same result. I sincerely hope you can find out yourself why that is the case.> I do not believe you are interpreting my table in the right way.> If you were interpreting it right then you could see that this table> would be invalid for any large n that was a counter example to the> Collatz conjecture.> A Collatz counterexample would be an odd number withoo in column 2 in your table. Nothing you've done> proves that such a number exists or doesn't exist.> All you've done is discover that if the Collatz iteration> takes m steps starting at n, then it takes (m+1) steps> starting at 2n. As has been said, that's obvious.> It MUST be true, because the first step takes you from> 2n to n.> I believe that this table just by its mathematical implications is > true----> oo making any counter example impossible.> Nope. Sorry. Which mathematical implications are those?> - RandyDuh!Then came the light. Sorry to all for being so dense! Then again whatcan you expect from a laymen.I will leave with one question. I believe I have asked about thisbefore with no answer.If there could possibly be a counter example to this conjecture whereall n was tested up to and including the counter example, then thiscounter example would have to be odd and the termination point of thisnew tree. My question, because of the density factor in the range ofintegers beyond all tested (n) that are members of the Collatz treebe able to support a second tree? In other words is there enough roomfor two growing and branching trees? I can't invision a counterexample (n) without it being the trunk of a new tree. === Subject: : Re: WOW! Is this known.> I will leave with one question. I believe I have asked about this> before with no answer.Now THIS is an interesting question. Putting all the naturalnumbers into a graph-theoretical representation based on theiroccurrence in repeated iterations of the Collatz conjecturefunction (odd n->3n+1, even n -> n/2). Then asking graph-theoretic questions about that tree.> If there could possibly be a counter example to this conjecture where> all n was tested up to and including the counter example, then this> counter example would have to be odd and the termination point of this> new tree.The first counterexample couldn't link to any m=n.I guess one possibility is a cycle, isn't it? Has anyoneproved that cycles can't exist?Yes, the first counterexample must be odd.And everything on that branch must be a counterexample.> My question, because of the density factor in the range of> integers beyond all tested (n) that are members of the Collatz tree> be able to support a second tree? In other words is there enough room> for two growing and branching trees? I can't invision a counter> example (n) without it being the trunk of a new tree.I'm not sure which way your tree grows. I'm thinking thatyou start from n (at the bottom), then 3n+1 is just abovethat, then either (3n+1)/2 or (9n+4), etc. But I can't seehow you can get branches.If your tree grows in the other direction (stuff aboven are numbers which when iterated eventually end upat n) then yes I guess there can be an entire treestructure above n. And, hmmm, every number in thatdirection must be >n as well. Certainly it containsat least the chain 2n, 4n, 8n, ...So I don't think it's right to describe n as the rootof a tree. But it is indeed located on an entire graphwhich is not connected to the graph of finite chains. - Randy === Subject: : Re: WOW! Is this known.> WOW! Is this known?>>Yes, it is known. If you define _any_ sequence using the rule that >a(n+1) = a(n)/2 if a(n) is even plus any other rules then you will get >exactly the same result. I sincerely hope you can find out yourself why >that is the case.> I do not believe you are interpreting my table in the right way.> If you were interpreting it right then you could see that this table> would be invalid for any large n that was a counter example to the> Collatz conjecture.> A Collatz counterexample would be an odd number withoo in column 2 in your table. Nothing you've done> proves that such a number exists or doesn't exist.> All you've done is discover that if the Collatz iteration> takes m steps starting at n, then it takes (m+1) steps> starting at 2n. As has been said, that's obvious.> It MUST be true, because the first step takes you from> 2n to n.> I believe that this table just by its mathematical implications is > true----> oo making any counter example impossible.> Nope. Sorry. Which mathematical implications are those?> - Randy> Duh!> Then came the light. Sorry to all for being so dense! Then again what> can you expect from a laymen.> I will leave with one question. I believe I have asked about this> before with no answer.> If there could possibly be a counter example to this conjecture where> all n was tested up to and including the counter example, then this> counter example would have to be odd and the termination point of this> new tree. My question, because of the density factor in the range of> integers beyond all tested (n) that are members of the Collatz tree> be able to support a second tree? In other words is there enough room> for two growing and branching trees? I can't invision a counter> example (n) without it being the trunk of a new tree.To get an idea what that may look like, try using 3x+5 instead of3x+1. Every system 3x+b has a trunk whose root is b. But in additionto the trunk320__425160 80__25 40 20__5 10 53x+5 also has positive loops at 1, 19 and 23. And, yes, they formseperate, unconnected trees:128__41 64 32__9 16 8__1 4 2 1608_201 62__19304 98__31152______49 76 38__11 19So the Collatz Conjecture is known to be false for 3x+5. _If_ 3x+1were false, then the root would be a really, really large number andthe loop length would be at least a 275,000 iterations. Hard toenvision, but not impossible. === Subject: : Re: WOW! Is this known.> I will leave with one question. I believe I have asked about this> before with no answer.> Now THIS is an interesting question. Putting all the natural> numbers into a graph-theoretical representation based on their> occurrence in repeated iterations of the Collatz conjecture> function (odd n->3n+1, even n -> n/2). Then asking graph-> theoretic questions about that tree.> If there could possibly be a counter example to this conjecture where> all n was tested up to and including the counter example, then this> counter example would have to be odd and the termination point of this> new tree.> The first counterexample couldn't link to any m those all have finite branches. Thus everything linked> to n must be >=n.> I guess one possibility is a cycle, isn't it? Has anyone> proved that cycles can't exist?> Yes, the first counterexample must be odd.> And everything on that branch must be a counterexample.> My question, because of the density factor in the range of> integers beyond all tested (n) that are members of the Collatz tree> be able to support a second tree? In other words is there enough room> for two growing and branching trees? I can't invision a counter> example (n) without it being the trunk of a new tree.> I'm not sure which way your tree grows. I'm thinking that> you start from n (at the bottom), then 3n+1 is just above> that, then either (3n+1)/2 or (9n+4), etc. But I can't see> how you can get branches.tree to, the terminating point of a new tree. Like the Collatztrees terminating point is one. Where this new trees terminating pointwould be the lowest valued (n) in this loop however large or smallthis loop might be! > If your tree grows in the other direction (stuff above> n are numbers which when iterated eventually end up> at n) then yes I guess there can be an entire tree> structure above n. And, hmmm, every number in that> direction must be >n as well. Certainly it contains> at least the chain 2n, 4n, 8n, ...Yes, stuff above n that (eventually) terminates (loops) where n is thelowest value in the loop thus terminating on the counter example n. > So I don't think it's right to describe n as the root> of a tree. But it is indeed located on an entire graph> which is not connected to the graph of finite chains.I agree with the first half but don't understand when you say is notconnected to the graph of finite chains> - Randy Dan === Subject: : Re: WOW! Is this known.> WOW! Is this known?> Is this known about the Collatz conjecture?> Where if either integer is odd or even then this rule applies,> 3n+1 if n is odd and n/2 if n is even.> In the table below, read the 2 left columns in each row as one> set and likewise the 2 right columns in the corresponding> row as the other set. > You will see the important relationship between the 2 sets.No, I don't. Could you please explain in more detail how the columnsare related to eachother? In your description above you did NOTmention A THING about chilren.How do you get children? By doing n(a+1) = (3n(a)+1) and n(b+1) =(n(b))/2? Where does the concept of the number of 'children' come into play anywhere?> Left # of Right col. # of > column children start # (n) children> start # (n) including double of left +1 of col.2 > start # col. start # children> 1 1 2 2> 2 2 4 3> 3 8 6 9> 4 3 8 4> 5 6 10 7> 6 9 12 10> 7 17 14 18> 8 4 16 5> 9 20 18 21> 10 7 20 8> 11 15 22 16> 12 10 24 11> 13 10 26 11 > 14 18 28 19> 15 18 30 19> 16 5 32 6> 17 13 34 14> 18 21 36 22> 19 21 38 22> 20 8 40 9> 21 8 42 9> 22 16 44 17> n... Etc. I believe this has to happen -----> oo> Why doesn't this prove the conjecture? ;-)> Really, think about how this pattern could be broken!> No way! > The 2 set comparison is interesting also because the path> going back to 4,2,1 is the same thus all the same children.> Where doubling (n) in the second set just adds 1 more term > to the path.> Also the second set shows up later in another row in the > first set.> More supporting evidence that the conjecture is true!> DanYou're not making any sense whatsoever.(...Starblade Riven Darksquall...) === Subject: : Re: WOW! Is this known.> tree to, the terminating point of a new tree. Like the Collatz> trees terminating point is one. Where this new trees terminating point> would be the lowest valued (n) in this loop however large or small> this loop might be!> If your tree grows in the other direction (stuff above> n are numbers which when iterated eventually end up> at n) then yes I guess there can be an entire tree> structure above n. And, hmmm, every number in that> direction must be >n as well. Certainly it contains> at least the chain 2n, 4n, 8n, ...> Yes, stuff above n that (eventually) terminates (loops) where n is the> lowest value in the loop thus terminating on the counter example n.No, it can't terminate on n. Are you claiming there's ann with the property that it is equal to either 3n+1 orn/2?Every iteration will take you from one number to adistinct number. If there is a Collatz counterexampleit can't terminate anywhere. Two possibilities are toincrease without bound or to loop.> So I don't think it's right to describe n as the root> of a tree. But it is indeed located on an entire graph> which is not connected to the graph of finite chains.> I agree with the first half but don't understand when you say is not> connected to the graph of finite chainsBecause if somewhere in your iteration you end up withm, and m has the property that eventually it iteratesto 1, then n can't be a counterexample. It's got afinite chain.For instance, here's the sequence starting at m=37. 37 112 56 28 14 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1Now clearly, if somewhere along the path while iteratingfrom starting point m, if I get to 37, then my nextiterations will follow exactly this sequence.If I get to 5, then my next iterations will be 16, 8,4, 2, 1, regardless of my starting point. So a Collatzcounterexample can't possibly ever end up on anynumber with a finite chain. That's why all the numbersin the infinite chain/loop must be greater than thefirst counterexample. We already know everything To get an idea what that may look like, try using 3x+5 instead of> 3x+1. Every system 3x+b has a trunk whose root is b. But in addition> to the trunk> 320__425> 160> 80__25> 40> 20__5> 10> 5> 3x+5 also has positive loops at 1, 19 and 23. And, yes, they form> seperate, unconnected trees:> 128__41> 64> 32__9> 16> 8__1> 4> 2> 1> 608_201 62__19> 304 98__31> 152______49> 76> 38__11> 19OK. 19 appears as one of the leaves, too. Hence everythinghere eventually ends up in the loop 19-62-31-98-49-152-76-38-19.> So the Collatz Conjecture is known to be false for 3x+5. _If_ 3x+1> were false, then the root would be a really, really large number and> the loop length would be at least a 275,000 iterations. Hard to> envision, but not impossible.Can you explain where that number (275,000) came from?Also how is it known that the root would be a reallyreally large number? Analysis or exhaustive search? - Randy === Subject: : Re: WOW! Is this known.> WOW! Is this known?> Is this known about the Collatz conjecture?> Where if either integer is odd or even then this rule applies,> 3n+1 if n is odd and n/2 if n is even.> In the table below, read the 2 left columns in each row as one> set and likewise the 2 right columns in the corresponding> row as the other set. > You will see the important relationship between the 2 sets.> No, I don't. Could you please explain in more detail how the columns> are related to eachother? In your description above you did NOT> mention A THING about chilren.> How do you get children? By doing n(a+1) = (3n(a)+1) and n(b+1) => (n(b))/2? Where does the concept of the number of 'children' come in> to play anywhere?> Left # of Right col. # of > column children start # (n) children> start # (n) including double of left +1 of col.2 > start # col. start # children> 1 1 2 2> 2 2 4 3> 3 8 6 9> 4 3 8 4> 5 6 10 7> 6 9 12 10> 7 17 14 18> 8 4 16 5> 9 20 18 21> 10 7 20 8> 11 15 22 16> 12 10 24 11> 13 10 26 11 > 14 18 28 19> 15 18 30 19> 16 5 32 6> 17 13 34 14> 18 21 36 22> 19 21 38 22> 20 8 40 9> 21 8 42 9> 22 16 44 17> n... Etc. I believe this has to happen -----> oo> Why doesn't this prove the conjecture? ;-)> Really, think about how this pattern could be broken!> No way! > The 2 set comparison is interesting also because the path> going back to 4,2,1 is the same thus all the same children.> Where doubling (n) in the second set just adds 1 more term > to the path.> Also the second set shows up later in another row in the > first set. More supporting evidence that the conjecture is true!> Dan> You're not making any sense whatsoever.> (...Starblade Riven Darksquall...)The table is right, it is just the logic I had about the table whichwas wrong. See later post.The term children is often used which only means the number ofintegers created from the parent start number.Where all parents are in column one and three and their childrenin columns two and four.BTW. The total count of children includes the parent also. ;-)Sorry about the confusion because of my dumb logic.Dan === Subject: : Re: WOW! Is this known.> To get an idea what that may look like, try using 3x+5 instead of> 3x+1. Every system 3x+b has a trunk whose root is b. But in addition> to the trunk> 320__425> 160> 80__25> 40> 20__5> 10> 5> 3x+5 also has positive loops at 1, 19 and 23. And, yes, they form> seperate, unconnected trees:> 128__41> 64> 32__9> 16> 8__1> 4> 2> 1> 608_201 62__19> 304 98__31> 152______49> 76> 38__11> 19> OK. 19 appears as one of the leaves, too. Hence everything> here eventually ends up in the loop 19-62-31-98-49-152-76-> 38-19.> So the Collatz Conjecture is known to be false for 3x+5. _If_ 3x+1> were false, then the root would be a really, really large number and> the loop length would be at least a 275,000 iterations. Hard to> envision, but not impossible.> Can you explain where that number (275,000) came from?According to MathWorld, Lagarias proved that there are no non-trivialloops <275,000 iterations. Don't ask me how he did it.> Also how is it known that the root would be a really> really large number? Analysis or exhaustive search?Also from Mathworld, it's been exhaustively tested to 2.7x10^18 (andsomeone is running a GIMPS-like distributed testing program to pushthat range higher). So maybe that's not really, really big dependingon how you define big. If it's beyond the range of your computer'smath packages, it's big. OTOH, 53-bit binary numbers are smallcompared to some of the patterns I've been studying, such as the 6thGeneration Quadra-Cycle Pulsar. With 4-bit rep units and a 2-bitresonator, this requires 4098 bits. So the exhaustive search willnever even come close to where the interesting numbers are found.> - Randy === Subject: : Re: WOW! Is this known.> To get an idea what that may look like, try using 3x+5 instead of> 3x+1. Every system 3x+b has a trunk whose root is b. But in addition> to the trunk> 320__425> 160> 80__25> 40> 20__5> 10> 5> 3x+5 also has positive loops at 1, 19 and 23. And, yes, they form> seperate, unconnected trees:> 128__41> 64> 32__9> 16> 8__1> 4> 2> 1> 608_201 62__19> 304 98__31> 152______49> 76> 38__11> 19> OK. 19 appears as one of the leaves, too. Hence everything> here eventually ends up in the loop 19-62-31-98-49-152-76-> 38-19.> So the Collatz Conjecture is known to be false for 3x+5. _If_ 3x+1> were false, then the root would be a really, really large number and> the loop length would be at least a 275,000 iterations. Hard to> envision, but not impossible.> Can you explain where that number (275,000) came from?> Also how is it known that the root would be a really> really large number? I forgot to mention in my previous post that a loop of 275,000iterations would probably require a number in the range of 20,000 to33,000 bits. So it would actually be a really, really large number.That's based on stopping times (S) for m-bit binary numbers, whereS = m * 13.457 for Mersenne Numbers (2^m - 1)S = m * 8.228 for Fermat Numbers (2^m + 1)S = m * 8.228 for random binary patterns (same as Fermat but for a different reason)> Analysis or exhaustive search?> - Randy === Subject: : Z[1/2] is reals, come one, come allI've been somewhat surprised from the discussion on Z[pi] so I thoughtI'd test the math newsgroup with something simpler: Z[1/2].Here the result that Z[1/2] is the field of reals follows simplyenough from 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...where here k is a non-zero integer, other than 1 or -1, and theexclusion of -1 is interesting, and goes to decidability, as the sumis NOT decidable for k=-1.Remember decidability is key, and many of you have learned of it asconvergence.Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so toget 1/7 in the ring, you can just use k=8, as you have 1/8, from1/2(1/2)(1/2).Similarly, you can get every integer for k, as now, for instance, with1/7, you can get 1/6, which puts 1/3 in the ring.Though I do wonder if the full proof is at all difficult. I thinknot, so I'll leave it as an exercise.Then again, it occurs to me that it would be strange for that to benew. So I'm VERY curious as to whether or not it is a standardresult, taught to math students.Of course, 1/2 is not special, as in fact any non-integer rationalincluded with the ring of integers, that is, Z[a/b] where 'a' and 'b'are integers and 'b' is nonzero, will give you reals.Notice that the field of rationals is skipped here, as*decidability* plays a key role, and without the title rationalthere is no definition to keep out other numbers like pi and e.Interestingly enough, introducing *any* non-integer to the ring ofintegers gives you reals. But that brings me back to what some peoplehave argued with me about when I talked about Z[pi], and I'm curiousto know if any of you think you can prove or disprove that statement.That is, I'm asking an important question to me:Are any of you math geniuses?Replies to this post will give me a much better understanding both ofhow you're all taught, how much mathematics you actually know in ameaningful way, and how amenable you are to learning mathematics,especially math at a deep level.James Harris === Subject: : Re: Z[1/2] is reals, come one, come all>I've been somewhat surprised from the discussion on Z[pi] so I thought>I'd test the math newsgroup with something simpler: Z[1/2].You're going to test us, eh? tee-hee.Yes, talking about Z[1/2] might be a better idea because it's simpler.>Here the result that Z[1/2] is the field of reals follows simply>enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...>where here k is a non-zero integer, other than 1 or -1, and the>exclusion of -1 is interesting, and goes to decidability, as the sum>is NOT decidable for k=-1.>Remember decidability is key, and many of you have learned of it as>convergence.Many of us have learned of it as convergence? Convergence is_not_ called decidability, except by _you_.>Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to>get 1/7 in the ring, you can just use k=8, as you have 1/8, from>1/2(1/2)(1/2).Yes, 1/8 is in Z[1/2]. And yes,1/7 = 1/8 + (1/8)^2 + ...And _no_, that does _not_ show that 1/7 is in Z[1/2]. BecauseZ[1/2] need not contain the limit of a sequence of elements ofZ[1/2].The other day you insisted you were not using the non-factthat a ring must be closed under the operation of taking limits.It's a good thing you decided to talk about a simpler example;here you _are_ definitely using that non-fact.The argument does show that 1/7 is in the (topological/metric)_closure_ of Z[1/2]. This is not big news; Z[1/2] is a well-knownring, and it's well-known that it's dense in the reals (which is tosay that its _closure_ is equal to the reals.)>Similarly, you can get every integer for k, as now, for instance, with>1/7, you can get 1/6, which puts 1/3 in the ring.>Though I do wonder if the full proof is at all difficult. I think>not, so I'll leave it as an exercise.>Then again, it occurs to me that it would be strange for that to be>new. So I'm VERY curious as to whether or not it is a standard>result, taught to math students.Me, I'm curious whether the fact that the moon is made ofgreen cheese is a standard result, taught to astronomy students.>Of course, 1/2 is not special, as in fact any non-integer rational>included with the ring of integers, that is, Z[a/b] where 'a' and 'b'>are integers and 'b' is nonzero, will give you reals.>Notice that the field of rationals is skipped here, as>*decidability* plays a key role, and without the title rational>there is no definition to keep out other numbers like pi and e.>Interestingly enough, introducing *any* non-integer to the ring of>integers gives you reals. But that brings me back to what some people>have argued with me about when I talked about Z[pi], and I'm curious>to know if any of you think you can prove or disprove that statement.One-line disproof, not even requiring the fact that pi istranscendental: Z[pi] is obviously countable and thereals are not countable.>That is, I'm asking an important question to me:>Are any of you math geniuses?None of us are the sort of math genius you are, thank heaven.>Replies to this post will give me a much better understanding both of>how you're all taught, how much mathematics you actually know in a>meaningful way, and how amenable you are to learning mathematics,>especially math at a deep level._Exactly_ as some time ago you were curious whether anyonebut you realized that integers were irrational.>James Harris************************ === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...The definition of a ring, which Z[1/2] is, does NOT require closure under infinite sums. So, which two polynomials P and Q in Z[x] have the property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When you find them, I'll be interested. Until then, you are not talking about Z[1/2].> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1> Remember decidability is key, and many of you have learned of it asconvergence.Then why not call it convergence? Decidability means something completely different.> Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.Using the definition of Z[1/2], it is far more than merely difficult.> Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students.It is not a standard result, as it is not true.> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.Which means you believe the rationals are the reals.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e> Interestingly enough, introducing *any* non-integer to the ring of> integers gives you reals. But that brings me back to what some people> have argued with me about when I talked about Z[pi], and I'm curious> to know if any of you think you can prove or disprove that statement.Z[1/2] is closed under finite addition, not infinite. Therefor 1/3 is not in Z[1/2] by the argument you propose.> That is, I'm asking an important question to me:> Are any of you math geniuses?> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level.If you want to know how we're taught, go get a book on Abstract Algebra from the library. It would explain a lot to you.> James Harris-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].You are probably the only one surprised by the responses. Isn't that aclue that YOU are off the beam? Your continued and consistent pattern offailures would be an embarrassment to anyone but you. And now you feelqualified to test others? Isn't that like letting the monkeys run thezoo?> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1.> Remember decidability is key, and many of you have learned of it asconvergence.We learned that because that is the term which has been accepted for theconcept. It has nothing to do with decidability which refers to anotherconcept entirely. Given your history of resisting standard definitions, Isuppose you will continue using the wrong term, however. Why not use theapproriate terminology? Your insistence on using inappropriate terminologymarks you as a militant anti-intellectual.> Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.Read: I have no idea how to do it. The real question is: Do you, or canyou, think at all? You state that you are dealing with rings. That termhas an accepted usage with specific requirements on its use. If you intendto hijack it for use in another context, you have an obligation to specifyYOUR definition of ring.> Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students.If things are very strange to you, and they makes you VERY curious,why not study math? Then you would have the benefit of knowing what istaught, and maybe even learning something.> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e.Decidability has nothing to do with it. That term refers to anotherconcept entirely.> That is, I'm asking an important question to me:> Are any of you math geniuses?Why is that important? You don't have to be a chicken to identify a badegg, and it certainly doesn't take a genius to determine that you are anidiot.> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level.It is your understanding of math that is in question here. You haverepeatedly exposed yourself as arrogant and uneducatable moron. Arrogantis too mild a term. You have exhibited continuous motion along thefollowing path:self-confident --> arrogant --> insolent --> insufferable --> JHS> James Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: : Re: Z[1/2] is reals, come one, come allIn sci.math, James Harris<3c65f87.0308181935.297fb314@ posting.google.com>:> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1.> Remember decidability is key, and many of you have learned of it asconvergence.> Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.There's a simpler proof: express every number in binary.That expansion is unique and results in a polynomial of x,where x = 1/2, for the fractional part. (Z covers theintegral part.)The main problem (as usual) is the endlessly repeatingbinaral and of course irrational numbers. It is notclear to me in Z[x] whether an infinite number ofoperations in +, -, and * is allowed, or not, inthe generation of the elements of Z[x]. If one doesallow an infinite number of operations, one might aswell define Z[1/2] = R and be done with it. :-)> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.> Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students.> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e.> Interestingly enough, introducing *any* non-integer to the ring of> integers gives you reals. But that brings me back to what some people> have argued with me about when I talked about Z[pi], and I'm curious> to know if any of you think you can prove or disprove that statement.> That is, I'm asking an important question to me:> Are any of you math geniuses?Obviously not.> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level.> James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> You are probably the only one surprised by the responses. Isn't that a> clue that YOU are off the beam? Your continued and consistent pattern of> failures would be an embarrassment to anyone but you. And now you feel> qualified to test others? Isn't that like letting the monkeys run the> zoo?Your post fits a pattern of attempting to influence rather than relyon mathematics and logic.I've seen that the group has consistently accepted such behavior.My assessment is that people in the math community lack the ability toengage in rational debate, and when that is shown, they instead resortto personal attacks, innuendo, and appeal to the crowd.The math community is a group of people where democracy has run amok.James Harris === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> The definition of a ring, which Z[1/2] is, does NOT require closure > under infinite sums. So, which two polynomials P and Q in Z[x] have the > property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When > you find them, I'll be interested. Until then, you are not talking > about Z[1/2].Your demonstrate a lack of basic mathematical knowledge.I'll use MathWorld as a reference.Please consider the information at the linkhttp://mathworld.wolfram.com/Ring.htmland comment. > where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1> Remember decidability is key, and many of you have learned of it asconvergence.> Then why not call it convergence? Decidability means something > completely different.You demonstrate a lack of understanding of basic mathematicalknowledge.Again using MathWorld as a reference, I direct you now to theinformation at the linkshttp://mathworld.wolfram.com/Decidable.htmlhttp:// mathworld.wolfram.com/Theory.htmland request that you comment.> Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.> Using the definition of Z[1/2], it is far more than merely difficult.I've noted that posters rely on trying to convince rather thanmathematical logic.I've also noted that they're very successful.My evaluation of the math community continues, but most of the work iscomplete. > Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students.> It is not a standard result, as it is not true.It is true based on mathematical logic, and standard definitions.I've merely outlined a rather basic mathematical result, which I canuse to test the understanding of posters whom I've deemed do indeedreflect the math community.I've gone from very complex concepts and extremely powerfulmathematical techniques to basic ones.Consistently, people from the math community have failed todemonstrate understanding of even basic mathematics.> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Which means you believe the rationals are the reals.Mathematicians use the definition of rational to arbitrarily excludeelements from what they call the field of rationals though it is infact, not a field.Using Z[a/b] where 'a' and 'b' are integers, where 'b' is nonzero,non-unit, and 'a' and 'b' don't share prime factors always results inthe field of reals.Here there is no way to put in the ad hoc requirement that is usedwith the field of rationals.That's what makes it a useful test.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e> Interestingly enough, introducing *any* non-integer to the ring of> integers gives you reals. But that brings me back to what some people> have argued with me about when I talked about Z[pi], and I'm curious> to know if any of you think you can prove or disprove that statement.> Z[1/2] is closed under finite addition, not infinite. Therefor 1/3 is > not in Z[1/2] by the argument you propose.You're just making things up.Interestingly, such a strategy tends to be effective with the mathcommunity.The democratic process rules in that community.> That is, I'm asking an important question to me:> Are any of you math geniuses?> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level.> If you want to know how we're taught, go get a book on Abstract Algebra > from the library. It would explain a lot to you.I'm field testing with actual subjects representative of the mathcommunity. Using simple questions I've managed to determine how muchmathematics YOU and others actually learned from such books.The answer is surprisingly little.The *social* function of the math community appears to be basedprimarily on a rigid hierarchy based on a high degree of belief incertain figures, either alive or dead, with little emphasis onunderstanding mathematics.Issues within the community are decided democratically, and people gowith the majority opinion.As mathematics is actually rigid and logical, I call it democracy runamok.Most of you have no idea what the truth is, and I've determined thatmany of you don't even believe in it. You think it's just a word, andthat truth is a social function.That is, to most of you, mathematics is a fashion show.James Harris === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].How delicious: The village idiot decides to test the village elders.As I've said before, you can't buy this kind of entertainment.-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> You are probably the only one surprised by the responses. Isn't that a> clue that YOU are off the beam? Your continued and consistent pattern of> failures would be an embarrassment to anyone but you. And now you feel> qualified to test others? Isn't that like letting the monkeys run the> zoo?> > Your post fits a pattern of attempting to influence rather than rely> on mathematics and logic.> I've seen that the group has consistently accepted such behavior.> My assessment is that people in the math community lack the ability to> engage in rational debate, and when that is shown, they instead resort> to personal attacks, innuendo, and appeal to the crowd.> The math community is a group of people where democracy has run amok.> James HarrisWell, James. In the portion of my post which you deleted, I discovered an error. To wit, inthe following passage:-----It is your understanding of math that is in question here. You haverepeatedly exposed yourself as arrogant and uneducatable moron. Arrogantis too mild a term. You have exhibited continuous motion along thefollowing path:self-confident --> arrogant --> insolent --> insufferable --> JHS-----should have concluded with:self-confident --> arrogant --> insolent --> insufferable --> JSHNote the corrected initials. And let this be a lesson to you: When an error has beenidentified, it is better to *correct* it than to *defend* it -- as is your practice.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Subject: : Re: Z[1/2] is reals, come one, come all Visiting Assistant Professor at the University of Montana.> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].>> Here the result that Z[1/2] is the field of reals follows simply> enough from>> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> The definition of a ring, which Z[1/2] is, does NOT require closure >> under infinite sums. So, which two polynomials P and Q in Z[x] have the >> property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When >> you find them, I'll be interested. Until then, you are not talking >> about Z[1/2].>Your demonstrate a lack of basic mathematical knowledge.Good thing I disconnected the hassiumy meter...>I'll use MathWorld as a reference.>Please consider the information at the link>http://mathworld.wolfram.com/Ring.htmlA ring is a set S together with two binary operators + and * (commonlyinterpreted as addition and multiplication, respectively) satisfyingthe following conditions:1. Additive associativity: For all a,b,c in S, (a+b)+c = a+(b+c)2. Additive commutativity: For all a,b in S, (a+b)=(b+a)3. Additive identity: There exists an element 0 in s such that for all a in S, 0+a=a+0=a.4. Additive inverse: For every a in S there exists -a in S such that a+(-a)=(-a)+a=0.5. Multiplicative associativity: For all a,b,c in S, (a*b)*c = a*(b*c)6. Left and right distributivity: For all a,b,c in S, a*(b+c)=(a*b) + (a*c); and (a+b)*c = (a*c) + (b*c).>and comment.The key here is binary operator. Addition and multiplication areassociated operations of n variables for every positive integer n, butyou cannot define an operation with an infinite number ofentries. Rings are not required to be closed under convergent infinitesums, according to the definition.> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1>> Remember decidability is key, and many of you have learned of it as>> convergence.> Then why not call it convergence? Decidability means something >> completely different.>You demonstrate a lack of understanding of basic mathematical>knowledge.>Again using MathWorld as a reference, I direct you now to the>information at the links>http://mathworld.wolfram.com/Decidable.html>http:// mathworld.wolfram.com/Theory.html>and request that you comment.You know, it looks like the problem is you don't understand what is inthose links and are desperately hoping someone will explain it toyou. But instead of asking for an explanation, you challenge. It makes you look like an even bigger idiot than usual.Decidable: A theory is decidable iff there is an algorithm which candetermine whether or not any sentence r is a member of the theory.As you can see, it is a term of art using the terms of artdecidable, algorithm, and sentence.Theory: A theory is a set of sentences which is closed under logicalimplication. Sentence: A sentence is a logic formula in which every variable isquantified. Now, explain to me again why you would apply the term decidable to aseries depending on its convergence, when decidable is an adjectivewhich is to be applied to a ->set of logic formulas in which everyvariable is quantified, closed under logical implication<-?That would mean that you think a series is, in particular, a set oflogic formulas in which every variable is quantified. But it is not. Aseries is a sequence of partial sums obtained from a sequence ofnumbers, and a sequence is a function with domain the naturalnumbers. You are misapplying the terms. Pure and simple. >> Using the definition of Z[1/2], it is far more than merely difficult.>I've noted that posters rely on trying to convince rather than>mathematical logic.By definition, Z[1/2] is the smallest subring of Q which contains Zand 1/2. But let's say that it is the smallest subring of the complexnumbers C which contains both Z and 1/2. They amount to the same thing.CLAIM: Z[1/2] = {a/b : a,b integers, b=2^n for some n>=0 integer}First, we verify that the set on the right is a ring. Since additionand multiplication are inherited from Q, we know they are associative,commutative, and that multiplication distributes over addition. So wejust need to check the other ring axioms.For simplicity, call the set on the right S.0 is in S, because 0=0/1 satisfies the membership condition for beingin S. If a/b is in S, then -a/b is also in S: b=2^n by hypothesis. So S hasan additive identity and inverses. Finally, we need to check that the sum of two elements of S lies in S,and that the product of two elements of S lies in S.If a/b, c/d lies in S, write a/b = a/2^n and c/d = c/2^m. Thena/b + c/d = a/2^n + c/2^m = (2^ma + 2^nc)/2^{n+m}. Therefore, it liesin S.(a/b)*(c/d) = (ac)/(bd) = (ac)/2^{n+m}, so the denominator is a powerof 2, hence it lies in S.Therefore, the set S is a ring.It contains the integers, for if a is an integer, then a = a/2^0 andlies in S. And it contains 1/2, clearly. Thus, S is a ring whichcontains Z and contains 1/2. Therefore, Z[1/2] is contained in S.To complete the claim, all we need to do is verify that S is containedin Z[1/2]. That is, that if R is ANY subring of C which contains bothZ and 1/2, then S is contained in R.Let a/b in S. Then a/b = a/2^{n} for some n>=0. Since R contains 1/2,and is a ring, it must contain (1/2)*...*(1/2) (n factors), so itcontains 1/2^n. It also contains a, since it contains allintegers. Being closed under multiplication, R must containa*(1/2^n)=a/2^n.Therefore, R contains every element of S. This proves that S iscontained in R. This proves that S must be contained in Z[1/2]. Sincethe other inclusion was already proven, this shows that S=Z[1/2].QED. [.snip.]> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Which means you believe the rationals are the reals.>Mathematicians use the definition of rational to arbitrarily exclude>elements from what they call the field of rationals though it is in>fact, not a field.You have things backwards. The rationals were first defined. Later,when the concept of field emerged, it was shown that the set ofrationals (which, naturally, exclude all reals which are notrationals) happen to form a field under the usual addition andmultiplication.Your complaint here is like the Monty Python sketch were John Cleesecomplains that his cat is excluded from getting a dog license.Did you look up field in Mathworld as well? http://mathworld.wolfram.com/Field.htmlhttp:// mathworld.wolfram.com/FieldAxioms.htmlhttp:// mathworld.wolfram.com/DivisionAlgebra.htmlThe latter states that a field is a division algebra which iscommutative under multiplication. The field axioms are listed on thesecond link.Please point out exactly which field axiom is not satisfied by the setof rational numbers under their usual addition and multiplication. [.snip.]====================================================== ================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 critize. A great many people are staggered to this extend, 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======================================================== ==============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Z[1/2] is reals, come one, come all>I've been somewhat surprised from the discussion on Z[pi] so I thought>I'd test the math newsgroup with something simpler: Z[1/2].>Here the result that Z[1/2] is the field of reals follows simply>enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...>>The definition of a ring, which Z[1/2] is, does NOT require closure >>under infinite sums. So, which two polynomials P and Q in Z[x] have the >>property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When >>you find them, I'll be interested. Until then, you are not talking >>about Z[1/2].> Your demonstrate a lack of basic mathematical knowledge.> I'll use MathWorld as a reference.> Please consider the information at the link> http://mathworld.wolfram.com/Ring.html> and comment.I do not see an infinite sum anywhere there. Where do you see it?I do see a polynomial evaluated at cuberoot(2) to demonstrate an element of Z[cuberoot(2)]. That would correspond to what I've written above. The information at mathworld is consistent with what I'm claiming. Now, where do you see that your infinite sum is supposed to be included?where here k is a non-zero integer, other than 1 or -1, and the>exclusion of -1 is interesting, and goes to decidability, as the sum>is NOT decidable for k=-1>Remember decidability is key, and many of you have learned of it as>convergence.>>Then why not call it convergence? Decidability means something >>completely different.> You demonstrate a lack of understanding of basic mathematical> knowledge.> Again using MathWorld as a reference, I direct you now to the> information at the links> http://mathworld.wolfram.com/Decidable.htmlA theory is decidable iff there is an algorithm which can determine whether or not any sentence r is a member of the theory. This does not apply to infinite sums or limits.> http://mathworld.wolfram.com/Theory.htmlA theory is a set of sentences which is closed under logical implication. That is, given any subset of sentences in the theory, if sentence r is a logical consequence of , then r must also be in the theory.This comes from my studies of formal logic in grad school (a bit rusty), it does not specifically talk about limits or infinite sums either.> and request that you comment.Neither of these terms applies to your infinite sum.Now, try this one:http://mathworld.wolfram.com/ConvergentSeries.html>Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to>get 1/7 in the ring, you can just use k=8, as you have 1/8, from>1/2(1/2)(1/2).>Similarly, you can get every integer for k, as now, for instance, with>1/7, you can get 1/6, which puts 1/3 in the ring.>Though I do wonder if the full proof is at all difficult. I think>not, so I'll leave it as an exercise.>>Using the definition of Z[1/2], it is far more than merely difficult.> I've noted that posters rely on trying to convince rather than> mathematical logic.I've shown above what you must do to find 1/3 in the ring Z[1/2]. You are trying to rely on infinite sums.> I've also noted that they're very successful.Because we use the definitions as stated, not as we wish them to be stated.> My evaluation of the math community continues, but most of the work is> complete.Who will you get to publish it?>Then again, it occurs to me that it would be strange for that to be>new. So I'm VERY curious as to whether or not it is a standard>result, taught to math students.>>It is not a standard result, as it is not true.> It is true based on mathematical logic, and standard definitions.No. You clearly do not understand the definitions.> I've merely outlined a rather basic mathematical result, which I can> use to test the understanding of posters whom I've deemed do indeed> reflect the math community.Please provide your statistical evidence that we reflect the mathematical community.> I've gone from very complex concepts and extremely powerful> mathematical techniques to basic ones.And only been able to correctly count primes, from all reports I've heard.> Consistently, people from the math community have failed to> demonstrate understanding of even basic mathematics.Well, someone hasn't, anyway.>Of course, 1/2 is not special, as in fact any non-integer rational>included with the ring of integers, that is, Z[a/b] where 'a' and 'b'>are integers and 'b' is nonzero, will give you reals.>>Which means you believe the rationals are the reals.> Mathematicians use the definition of rational to arbitrarily exclude> elements from what they call the field of rationals though it is in> fact, not a field.Which of the field axioms does it violate?http://mathworld.wolfram.com/FieldAxioms.htmlIf you can't find a violation, it's a field.> Using Z[a/b] where 'a' and 'b' are integers, where 'b' is nonzero,> non-unit, and 'a' and 'b' don't share prime factors always results in> the field of reals.No.> Here there is no way to put in the ad hoc requirement that is used> with the field of rationals.It's not an ad hoc requirement, it's the definition.http://mathworld.wolfram.com/RationalNumber.html> That's what makes it a useful test.>Notice that the field of rationals is skipped here, as>*decidability* plays a key role, and without the title rational>there is no definition to keep out other numbers like pi and e>Interestingly enough, introducing *any* non-integer to the ring of>integers gives you reals. But that brings me back to what some people>have argued with me about when I talked about Z[pi], and I'm curious>to know if any of you think you can prove or disprove that statement.>>Z[1/2] is closed under finite addition, not infinite. Therefor 1/3 is >>not in Z[1/2] by the argument you propose.> You're just making things up.Which part? Being closed under finite addition but not infinite, or 1/3 not being in Z[1/2]?> Interestingly, such a strategy tends to be effective with the math> community.> The democratic process rules in that community.I thought it was the process of defining terms and drawing conclusions from those definitions only.>That is, I'm asking an important question to me:>Are any of you math geniuses?>Replies to this post will give me a much better understanding both of>how you're all taught, how much mathematics you actually know in a>meaningful way, and how amenable you are to learning mathematics,>especially math at a deep level.>>If you want to know how we're taught, go get a book on Abstract Algebra >>from the library. It would explain a lot to you.> I'm field testing with actual subjects representative of the math> community. Using simple questions I've managed to determine how much> mathematics YOU and others actually learned from such books.> The answer is surprisingly little.How can you assess that when you don't know the content of the books? Perhaps we have learned a great deal and it simply doesn't correspond to what you believe it does.> The *social* function of the math community appears to be based> primarily on a rigid hierarchy based on a high degree of belief in> certain figures, either alive or dead, with little emphasis on> understanding mathematics.> Issues within the community are decided democratically, and people go> with the majority opinion.You can keep believing that, but you will find yourself losing more credibility if you try to push that argument. There are no meetings to vote on AC, or any of the other axioms. There are heated disagreements about what axioms *should* be used, and that tends to result in sub-branches of mathematics.> As mathematics is actually rigid and logical, I call it democracy run> amok.If you don't know how a system works from within, you are likely to misunderstand what you observe. This is one such case. The topics you are bringing up are not normally under dispute. You'll have to dig much deeper to reach those levels.> Most of you have no idea what the truth is, and I've determined that> many of you don't even believe in it. You think it's just a word, and> that truth is a social function.> That is, to most of you, mathematics is a fashion show.Then why do you keep trying to show up naked?-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Z[1/2] is reals, come one, come all>Using the definition of Z[1/2], it is far more than merely difficult.>>I've noted that posters rely on trying to convince rather than>>mathematical logic.> By definition, Z[1/2] is the smallest subring of Q which contains Z> and 1/2. But let's say that it is the smallest subring of the complex> numbers C which contains both Z and 1/2. They amount to the same thing.> CLAIM: Z[1/2] = {a/b : a,b integers, b=2^n for some n>=0 integer}> First, we verify that the set on the right is a ring. Since addition> and multiplication are inherited from Q, we know they are associative,> commutative, and that multiplication distributes over addition. So we> just need to check the other ring axioms.> For simplicity, call the set on the right S.> 0 is in S, because 0=0/1 satisfies the membership condition for being> in S. > If a/b is in S, then -a/b is also in S: b=2^n by hypothesis. So S has> an additive identity and inverses. > Finally, we need to check that the sum of two elements of S lies in S,> and that the product of two elements of S lies in S.> If a/b, c/d lies in S, write a/b = a/2^n and c/d = c/2^m. Then> a/b + c/d = a/2^n + c/2^m = (2^ma + 2^nc)/2^{n+m}. Therefore, it lies> in S.> (a/b)*(c/d) = (ac)/(bd) = (ac)/2^{n+m}, so the denominator is a power> of 2, hence it lies in S.> Therefore, the set S is a ring.> It contains the integers, for if a is an integer, then a = a/2^0 and> lies in S. And it contains 1/2, clearly. Thus, S is a ring which> contains Z and contains 1/2. Therefore, Z[1/2] is contained in S.> To complete the claim, all we need to do is verify that S is contained> in Z[1/2]. That is, that if R is ANY subring of C which contains both> Z and 1/2, then S is contained in R.> Let a/b in S. Then a/b = a/2^{n} for some n>=0. Since R contains 1/2,> and is a ring, it must contain (1/2)*...*(1/2) (n factors), so it> contains 1/2^n. It also contains a, since it contains all> integers. Being closed under multiplication, R must contain> a*(1/2^n)=a/2^n.> Therefore, R contains every element of S. This proves that S is> contained in R. This proves that S must be contained in Z[1/2]. Since> the other inclusion was already proven, this shows that S=Z[1/2].> QED.> [.snip.]Which doesn't quite finish addressing the fact that 1/3 is NOT in Z[1/2].However, suppose 1/3 = a/b where b=2^n, n an integer >=0 and a an integer, then 2^n = 3a. This is not true since the integers are a unique factorization domain, which means 3 would have to appear on both sides of the equals sign. 2^n is written in factored form, 3 is not on the left side, therefor 1/3 <> a/b in the required form. Therefor, by contradiction, 1/3 is not in Z[1/2].-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Z[1/2] is reals, come one, come all> Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.> Read: I have no idea how to do it. The real question is: Do you, or can> you, think at all? You state that you are dealing with rings. That term> has an accepted usage with specific requirements on its use. If you intend> to hijack it for use in another context, you have an obligation to specify> YOUR definition of ring.There you go, appealing to *definitions* like those otherweaseling mathematicians. Obviously Z[1/2] should include1/7, so if the definition of ring demonstrates that it doesn't,then the definition is wrong.There, that was easy, wasn't it? - Randy === Subject: : Re: Z[1/2] is reals, come one, come all Visiting Assistant Professor at the University of Montana. [.snip.]>> CLAIM: Z[1/2] = {a/b : a,b integers, b=2^n for some n>=0 integer} [.snip.]>Which doesn't quite finish addressing the fact that 1/3 is NOT in Z[1/2].>However, suppose 1/3 = a/b where b=2^n, n an integer >=0 and a an >integer, then 2^n = 3a. This is not true since the integers are a >unique factorization domain, which means 3 would have to appear on both >sides of the equals sign. 2^n is written in factored form, 3 is not on >the left side, therefor 1/3 <> a/b in the required form. Therefor, by >contradiction, 1/3 is not in Z[1/2].You are absolutely right. My bad. At first I had defined the set S tobe{q in Q: if q is written in lowest terms, q=a/b, then b is a power of 2}which of course immediately shows that 1/3 is not in the set; butthere were some details that made the proof that the set is a ring abit more cumbersome. Then I changed the condition and forgot that Inow had to prove more stuff...====================================================== ================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 critize. A great many people are staggered to this extend, 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======================================================== ==============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Z[1/2] is reals, come one, come all> [.snip.]>CLAIM: Z[1/2] = {a/b : a,b integers, b=2^n for some n>=0 integer}> [.snip.]>>Which doesn't quite finish addressing the fact that 1/3 is NOT in Z[1/2].>>However, suppose 1/3 = a/b where b=2^n, n an integer >=0 and a an >>integer, then 2^n = 3a. This is not true since the integers are a >>unique factorization domain, which means 3 would have to appear on both >>sides of the equals sign. 2^n is written in factored form, 3 is not on >>the left side, therefor 1/3 <> a/b in the required form. Therefor, by >>contradiction, 1/3 is not in Z[1/2].> You are absolutely right. My bad. At first I had defined the set S to> be> {q in Q: if q is written in lowest terms, q=a/b, then b is a power of 2}> which of course immediately shows that 1/3 is not in the set; but> there were some details that made the proof that the set is a ring a> bit more cumbersome. Then I changed the condition and forgot that I> now had to prove more stuff...In any case, between the two 1/3 is not, never was, and never will be in Z[1/2] under the *standard definition* of the notation. If James doesn't like the way we define things, he can invent his own notation/definitions. It wouldn't be the first time people have done that, but it may not be earth-shaking either.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: : Re: Z[1/2] is reals, come one, come all...> Mathematicians use the definition of rational to arbitrarily exclude> elements from what they call the field of rationals though it is in> fact, not a field....If x is a rational number then there exists an integer adefinition of rational, and from simple definitions of ordinary addition and multiplication, it is easy to showthat rational numbers satisfy the field axioms of additiveand multiplicative Commutativity, Associativity,Distributivity, Identity, and Inverses. (See e.g.http://mathworld.wolfram.com/FieldAxioms.html for statements of field axioms.)Do you agree that the rationals satisfy the field axioms?If you don't agree, which axiom isn't satisfied?-jiw === Subject: : Re: Z[1/2] is reals, come one, come all>> I've been somewhat surprised from the discussion on Z[pi] so I thought>> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply>> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...>The definition of a ring, which Z[1/2] is, does NOT require closure >under infinite sums. So, which two polynomials P and Q in Z[x] have the >property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When >you find them, I'll be interested. Until then, you are not talking >about Z[1/2].>>Your demonstrate a lack of basic mathematical knowledge.>Good thing I disconnected the hassiumy meter...>>I'll use MathWorld as a reference.>>Please consider the information at the link>>http://mathworld.wolfram.com/Ring.html>A ring is a set S together with two binary operators + and * (commonly>interpreted as addition and multiplication, respectively) satisfying>the following conditions:>1. Additive associativity: For all a,b,c in S, (a+b)+c = a+(b+c)>2. Additive commutativity: For all a,b in S, (a+b)=(b+a)>3. Additive identity: There exists an element 0 in s such that for> all a in S, 0+a=a+0=a.>4. Additive inverse: For every a in S there exists -a in S such that a+(-a)=(-a)+a=0.>5. Multiplicative associativity: For all a,b,c in S, (a*b)*c = a*(b*c)>6. Left and right distributivity: For all a,b,c in S,> a*(b+c)=(a*b) + (a*c); and (a+b)*c = (a*c) + (b*c).>>and comment.>The key here is binary operator. Addition and multiplication are>associated operations of n variables for every positive integer n, but>you cannot define an operation with an infinite number of>entries. Rings are not required to be closed under convergent infinite>sums, according to the definition.>> where here k is a non-zero integer, other than 1 or -1, and the>> exclusion of -1 is interesting, and goes to decidability, as the sum>> is NOT decidable for k=-1> Remember decidability is key, and many of you have learned of it as>convergence.>Then why not call it convergence? Decidability means something >completely different.>>You demonstrate a lack of understanding of basic mathematical>>knowledge.>>Again using MathWorld as a reference, I direct you now to the>>information at the links>>http://mathworld.wolfram.com/Decidable.html>>http:// mathworld.wolfram.com/Theory.html>>and request that you comment.>You know, it looks like the problem is you don't understand what is in>those links and are desperately hoping someone will explain it to>you. But instead of asking for an explanation, you challenge. >It makes you look like an even bigger idiot than usual.Hard to believe that's possible, but it's true.>Decidable: A theory is decidable iff there is an algorithm which can>determine whether or not any sentence r is a member of the theory.>As you can see, it is a term of art using the terms of art>decidable, algorithm, and sentence.>Theory: A theory is a set of sentences which is closed under logical>implication. >Sentence: A sentence is a logic formula in which every variable is>quantified. >Now, explain to me again why you would apply the term decidable to a>series depending on its convergence, when decidable is an adjective>which is to be applied to a ->set of logic formulas in which every>variable is quantified, closed under logical implication<-?>That would mean that you think a series is, in particular, a set of>logic formulas in which every variable is quantified. But it is not. A>series is a sequence of partial sums obtained from a sequence of>numbers, and a sequence is a function with domain the natural>numbers. >You are misapplying the terms. Pure and simple. >[...]************************ === Subject: : Re: Z[1/2] is reals, come one, come all Visiting Assistant Professor at the University of Montana. [.snip.]>>You know, it looks like the problem is you don't understand what is in>>those links and are desperately hoping someone will explain it to>>you. But instead of asking for an explanation, you challenge. >>It makes you look like an even bigger idiot than usual.>Hard to believe that's possible, but it's true.Just goes to show. There ain't no such as thing as rock bottom.======================================================= ===============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 critize. A great many people are staggered to this extend, 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======================================================== ==============Arturo Magidinmagidin@math.berkeley.edu === Subject: : Re: Z[1/2] is reals, come one, come alltoo God-am much, dood! > That would mean that you think a series is, in particular, a set of> logic formulas in which every variable is quantified. But it is not. A> series is a sequence of partial sums obtained from a sequence of> numbers, and a sequence is a function with domain the natural> numbers. > Therefore, the set S is a ring. > To complete the claim, all we need to do is verify that S is contained> in Z[1/2]. That is, that if R is ANY subring of C which contains both> Z and 1/2, then S is contained in R.> Let a/b in S. Then a/b = a/2^{n} for some n>=0. Since R contains 1/2,> and is a ring, it must contain (1/2)*...*(1/2) (n factors), so it> contains 1/2^n. It also contains a, since it contains all> integers. Being closed under multiplication, R must contain> a*(1/2^n)=a/2^n. > Did you look up field in Mathworld as well? > http://mathworld.wolfram.com/Field.html> http://mathworld.wolfram.com/FieldAxioms.html> http://mathworld.wolfram.com/DivisionAlgebra.html--les ducs d'Enron!http://members.tripod.com/~american_almanac === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1.> Remember decidability is key, and many of you have learned of it asconvergence. Z[1/2] by definition is comprised of FINITE sums of integer multiples of powers of 1/2. Or, put another way, polynomials P(x) with integer coefficients evaluated at x = 1/2. An element of Z[1/2] is invariably a rational number withdenominator a power of 2. What you are describing is not Z[1/2]. It is instead all convergent INFINITE sums of integer multiples of powers of 1/2.Any real number can be expressed in this way. Let V[1/2] represent all such convergent series. Then it is truethat V[1/2] = R = all real numbers. This is not a new result! Also at this point not interestingor exciting. The distinction between finite sums [polynomials] and infinitesums [infinite series] is absolutely essential. You are badlyabusing current notation in implying that Z[1/2] = V[1/2].What you have done does not imply that Z[1/2] is a field. It is not. For example, 3 is in Z[1/2], but 1/3 is not. > Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2). In conventional notation, Z[1/2] is comprised only of FINITEsums of integer multiples of powers of 2. Thus 1/7 is not inZ[1/2].> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise. If you are in fact interested in INFINITE convergent sums,V[1/2], the proof that all real numbers are in it is trivial.> Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students. Students are taught that Z[x] = polynomials in the indeterminatex with integer coefficients. They are NOT taught that Z[x]is comprised of INFINITE sums, nor should they be. It is at bestan abuse of notation. Of course there is an enormous and old literature on functions represented as limits of finite sums.Math students are taught that not in algebra courses, but in analysis courses. It is NOT a neglected subject. You have notdiscovered something new and you have not added any new insightto something old. > Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e.> Interestingly enough, introducing *any* non-integer to the ring of> integers gives you reals. Only if you allow infinite sums. > But that brings me back to what some people> have argued with me about when I talked about Z[pi], and I'm curious> to know if any of you think you can prove or disprove that statement. Z[pi] is not a field, unless what you actually mean is V[pi] (see above).> That is, I'm asking an important question to me:> Are any of you math geniuses? Why is that important to you?> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level. Is this somehow related to 'objects' and your currently completely stymied non-proof of Fermat's Last Theorem? Or to your incorrect claims in Advanced Polynomial Factorization?Or are you hoping that we will forget that you have not provided satisfactory responses to the criticism of yourarguments in those areas? Nora B. > James Harris === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+... You _are_ aware that there's an *infinite* series on the righthand side of that equation, right ?? > The definition of a ring, which Z[1/2] is, does NOT require closure > under infinite sums. So, which two polynomials P and Q in Z[x] have the > property that P(1/2)+Q(1/2) = 1/3? How about P(1/2)*Q(1/2)=1/3? When > you find them, I'll be interested. Until then, you are not talking > about Z[1/2].> Your demonstrate a lack of basic mathematical knowledge. What a buffoon ... As others have noted, a ring need not be closed under the formation of infinite sums or products. It's a fact that (for instance) 1/3 is _not_ an element of Z[1/2] -- since the elements of Z[1/2] are precisely those rational numbers that can be written in the form m/2^n with m, n in Z and n >= 0, it's trivial that you'll not find 1/3 among the elements of Z[1/2] ...> I'll use MathWorld as a reference.> Please consider the information at the link> http://mathworld.wolfram.com/Ring.html> and comment. OK - what you find there is a perfectly adequate rendering of the definition of a ring -- note that there are two _binary_ operations required by the definition. Using induction, you can show that, once a ring R contains some element r, then R must contain the element P(r) where P is any polynomial with integer coefficients. What you _cannot_ do (in general) is make any sense of infinite sums of ring elements ... > where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1> Remember decidability is key, and many of you have learned of it as> convergence.> Then why not call it convergence? Decidability means something > completely different.> You demonstrate a lack of understanding of basic mathematical> knowledge. That's f***ing rich ... Here, we have a situation where there's a perfectly good, already-standard term to describe what you mean ... you _know_ the term already ... and yet you insist on misusing some _other_ term (which you apparently _don't_ understand) as you dribble on ...> Again using MathWorld as a reference, I direct you now to the> information at the links> http://mathworld.wolfram.com/Decidable.html> http://mathworld.wolfram.com/Theory.html> and request that you comment. Just where (in either of those pages ... or any of the ones they link to) do you see any mention of convergence ??? Of course, given 1/2 in the ring, you also get 1/4, 1/8, etc., so to> get 1/7 in the ring, you can just use k=8, as you have 1/8, from> 1/2(1/2)(1/2).> Similarly, you can get every integer for k, as now, for instance, with> 1/7, you can get 1/6, which puts 1/3 in the ring.> Though I do wonder if the full proof is at all difficult. I think> not, so I'll leave it as an exercise.> Using the definition of Z[1/2], it is far more than merely difficult.> I've noted that posters rely on trying to convince rather than> mathematical logic. Actually, the last is more _your_ stock in trade hereabouts ... It's easy enough to prove that Z[1/2] is what I said (up above) that it is -- given that description, what Will claims is *exactly* true: the full proof that you (evidently) can't come up with _is_ far more than merely difficult -- it's flat _impossible_ ...> I've also noted that they're very successful. The simplest explanation for that fact: the posters that you're whining about _are_ (in fact) shredding your various arguments using correct (and convincing) mathematical demonstrations. _That_ is why you can't find any supporters among even the lurkers to whom you so often address your amusing diatribes ...> My evaluation of the math community continues, but most of the work is> complete.> Then again, it occurs to me that it would be strange for that to be> new. So I'm VERY curious as to whether or not it is a standard> result, taught to math students.> It is not a standard result, as it is not true.> It is true based on mathematical logic, and standard definitions. Wow -- that's one for the Jesse F. Hughes quote collection -- JSH claiming to rely on standard definitions ... Boy! Just buried the needle on the ol' bullmeter ... Now that I've recovered ... your claim is _not_ true, you don't know or understand any of the standard definitions involved and you wouldn't recognize mathematical logic if it bit you on the ass ... > I've merely outlined a rather basic mathematical result, which I can> use to test the understanding of posters whom I've deemed do indeed> reflect the math community. No, you've blithered on incoherently, betraying your (seemingly invincible) ignorance of even basic mathematics (the stuff that freshman and sophomore math majors routinely master). > I've gone from very complex concepts and extremely powerful> mathematical techniques to basic ones. This may be a tactical error on your part -- at least with the Advanced Polynomial Factorization tripe or your alleged proof of FLT, you can grind out polynomials where it requires a good deal of dedication for anyone to show explicitly that the claims you make are simply *wrong*. With this current stuff, the corresponding demonstraton is just _too_ easy ...> Consistently, people from the math community have failed to> demonstrate understanding of even basic mathematics. <*teehee*>> Of course, 1/2 is not special, as in fact any non-integer rational> included with the ring of integers, that is, Z[a/b] where 'a' and 'b'> are integers and 'b' is nonzero, will give you reals.> Which means you believe the rationals are the reals.> Mathematicians use the definition of rational to arbitrarily exclude> elements from what they call the field of rationals though it is in> fact, not a field. Of course, it's field -- why don't you try to understand the definition of that concept ?? I'm sure you can find it at Mathworld. That definition (again) involves 2 binary operations (conveniently called addition and multiplication, by convention) which have a few properties. And ... lo ... the rational numbers come equipped with two such operations that (provably) have all of those properties. The use that mathematicians make of the definition of rational number ?? Why, we use it to exclude from the set of rational numbers things that are _not_ rational numbers -- most people would regard that as a perfectly reasonable thing to do ... > Using Z[a/b] where 'a' and 'b' are integers, where 'b' is nonzero,> non-unit, and 'a' and 'b' don't share prime factors always results in> the field of reals. It _never_ results in any such thing ... for starters, any such ring has only countably many elements (not nearly enough to get you R as a result) ...> Here there is no way to put in the ad hoc requirement that is used> with the field of rationals. In all of the cases that you're nattering on about, you've got a set that comes provided with a definition of just what it takes for something to be an element of that set ... and you regard the defintion as an _ad_hoc_ requirement ??? It must be really scary to live in _your_ world ...> That's what makes it a useful test.> Notice that the field of rationals is skipped here, as> *decidability* plays a key role, and without the title rational> there is no definition to keep out other numbers like pi and e> Interestingly enough, introducing *any* non-integer to the ring of> integers gives you reals. But that brings me back to what some people> have argued with me about when I talked about Z[pi], and I'm curious> to know if any of you think you can prove or disprove that statement.> Z[1/2] is closed under finite addition, not infinite. Therefor 1/3 is > not in Z[1/2] by the argument you propose.> You're just making things up. Nope -- _he_ isn't, but _you_ are. What _he_ said is *true* (and *provable*) -- you've said almost nothing in this entire collection of (bizarre) threads that can stand up to any sort of critical scrutiny (no matter how cursory). > Interestingly, such a strategy tends to be effective with the math> community.> The democratic process rules in that community. Actually, what counts (and convinces) is *mathematical proof*. > That is, I'm asking an important question to me:> Are any of you math geniuses?> Replies to this post will give me a much better understanding both of> how you're all taught, how much mathematics you actually know in a> meaningful way, and how amenable you are to learning mathematics,> especially math at a deep level.> If you want to know how we're taught, go get a book on Abstract Algebra > from the library. It would explain a lot to you.> I'm field testing with actual subjects representative of the math> community. Using simple questions I've managed to determine how much> mathematics YOU and others actually learned from such books.> The answer is surprisingly little. that you could get through the introductory chapter of any abstract algebra text and do the exercises ... and yet you feel competent to pompously denigrate people who could/can/have done so ... without even blushing (or realizing just how stupid you sound when you carry on in this manner). > The *social* function of the math community appears to be based> primarily on a rigid hierarchy based on a high degree of belief in> certain figures, either alive or dead, with little emphasis on> understanding mathematics.> Issues within the community are decided democratically, and people go> with the majority opinion.> As mathematics is actually rigid and logical, I call it democracy run> amok.> Most of you have no idea what the truth is, and I've determined that> many of you don't even believe in it. You think it's just a word, and> that truth is a social function.> That is, to most of you, mathematics is a fashion show. projection ?? Is that the term ?? Y'all know, right ?? To describe what the JSHster is doing when taking a dump in public (as we observe with the paragraphs just north of here) ??? > James Harris === Subject: : Re: Z[1/2] is reals, come one, come all> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1.I don't think you quite understand how important Harris' discovery is!Because any real can be written as the infinite sum of rational numbers i.e.an integer plus sigma a_n / 10^n, then James has shown that any real is inTake that Cantor-Tralfaz === Subject: : Re: Z[1/2] is reals, come one, come allJames Harris> I've been somewhat surprised from the discussion on Z[pi] so I thought> I'd test the math newsgroup with something simpler: Z[1/2].> Here the result that Z[1/2] is the field of reals follows simply> enough from> 1/(k-1) = 1/k + 1/k^2 + 1/k^3+...> where here k is a non-zero integer, other than 1 or -1, and the> exclusion of -1 is interesting, and goes to decidability, as the sum> is NOT decidable for k=-1.[barf]www.crank.net/harris.html