mm-2919 ==== Subject: Re: Nilpotent matrix yes, that's right . And sorry I missed some additional assumption, it should assume that char K= 0 and Trace A^r = 0 for all r >= 0. Then A ==== Subject: Re: Nilpotent matrix es, that's right . And sorry I missed some additional assumption, it > should assume that char K= 0 and Trace A^r = 0 for all r >= 0. Then A The trace of A*k is the sum of k-th powers of the eigenvalues of A (in an appropriate extension of K), multiplicity repsected. So, symbolically they are symmetric polynomials in the eigenvalues, of all degrees. (We only need k from 1 to n where A is n-by-n.) On the other hand, the coefficients of the characteristic polynomial of A are elementary symmetric polynomials in the eigenvalues. There is a theorem relating these two types of symmetric polynomials, and you will get the result. (char K = 0 is perhaps needed to divide by some positive integers.) Look up symmetric polynomials. (There may be an easier way, using Vandermonde matrices.) ==== Subject: Re: Nilpotent matrix >> yes, that's right . And sorry I missed some additional assumption, it >> should assume that char K= 0 and Trace A^r = 0 for all r >= 0. Then A >The trace of A*k is the sum of k-th powers of the eigenvalues of A (in an >appropriate extension of K), multiplicity repsected. So, symbolically they >are symmetric polynomials in the eigenvalues, of all degrees. (We only >need k from 1 to n where A is n-by-n.) >On the other hand, the coefficients of the characteristic polynomial of A >are elementary symmetric polynomials in the eigenvalues. There is a >theorem relating these two types of symmetric polynomials, and you will >get the result. (char K = 0 is perhaps needed to divide by some positive >integers.) Look up symmetric polynomials. Indeed: note that in characteristic p <> 0, the p x p identity matrix I has trace(I^r) = 0 for all r, but it certainly isn't nilpotent. >(There may be an easier way, using Vandermonde matrices.) I don't know if it's easier, but: Let K* be an extension of K in which the characteristic polynomial P(X) of A splits. Then there is a basis of (K*)^n in which A is upper triangular, with its eigenvalues on the diagonal. Note that the choice of basis doesn't affect the trace. Now if A is not nilpotent, at least one of these diagonal elements, say t, is nonzero. Let Q(X) be the product of X - r for r = 0 and all eigenvalues of A except t. Note that Q(A)_{ii} = Q(A_{ii}), which is 0 if A_{ii} <> t, and nonzero if A_{ii} = t. Thus trace(Q(A)) = m Q(t), where m is the number of t's on the diagonal of A. But since Q(A) is a linear combination of A^j for positive integers j, trace Q(A) = 0, contradiction. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: MOVING DIMENSIONS THEORY CATCHING ON AT STANFORD I so agree. nouns are neccesary to keep every thing from verbing at once. thus quoth: Space is an emergent property of time, mass and the laws of physics. Space has no separate existence. Any higher time integrals than velocity are nigh meaningless... only good for avoiding collisions. ;>) thus: in practice, no-one has a problem with this part of your expose, regardless of any claims of einsteinmaniaologists (cosmic reverberaration of the Sound of one hand, Clapping .-)... in other words, it's really some thing that stands on its own, without need of a new cosmogyny, or New Age storytelling. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush8.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html ==== Subject: Re: constructible numbers >> Rene Decarte showed that 2^(1/3) was not a constructible number around >> 1640 (while working on the cube duplication problem). I don't know >> that this was the first such proof, but I think it probably was. >The usual proof involves a little bit of field theory. >In 1640, field theory was still nearly 200 years into the future. >I'd be surprised if Descartes had the tools to prove >that you can't construct cube root 2 with ruler and compass. Pierre Laurent Wantzel proved the impossibility of both angle trisection and cube duplication in 1837: see e.g. . I think Descartes considered the problem, but I don't know what progress he made. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Cofinite topology on the reals >>This definition of open is not useful in general topological spaces - in >>fact it is circular. A neighbourhood of a point x in a topological space >>X is any subset N of X such that there exists an open set U with x in U >>subset N. Neighbourhood is defined in terms of open, rather than the >>other way round. >>If U is any open set and x in U then U is a neighbourhood of x and so >>trivially U contains a neighbourhood of each of its points (viz. U itself). >> There are oodles of ways to characterize a topology, and >> one of them is to use neighborhoods, in which a set can be >> a neighborhood of one point, and not of another. >To define open sets in terms of neighborhoods requires introduction via >axioms for neighborhoods. This was the traditional intuitive analysis >inspired approach likely of Hausdorff but as time modernized topology, a >topology was easier defined not by neighborhoods, but by open sets and the >scant axioms for them. >> This can even be useful, as it enables one to speak of compact >> neighborhoods in locally compact spaces. >The discrete topology is the unique topology >for which every compact set is open. >The discrete topology is the unique topology >for which every compact _nhood_ is open???? Not if there are no compact neighbourhoods, e.g. in infinite-dimensional normed linear spaces. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: ISO example of a concrete non-arithmetic ring >> I know of examples of groups that can be readily understood by a >> lay person without requiring any use of numbers and arithmetic >> (e.g. the group of permutations of a finite set). But I can't >> think of any non-arithmetic examples of rings. Does anyone know >> of one? >Yes, a Boolean ring defined as subsets of a set S, ie P(S), >with + being symmetric difference and * being intersection. >A + B = (A / B) - (A / B) = AB / BA >A * B = A / B interesting example. kj -- NOTE: In my address everything before the first period is backwards; and the last period, and everything after it, should be discarded. ==== Subject: Re: half page proof of 4 Color Mapping Theorem; using Moebius 4 adjacency maximum So you made it more complicated, hoping it would get by. When all the irrelevancies are removed, your proof looks like: I get the feeling you are getting back at me for declaring your irrelevant counterexamples as irrelevant. But my above proof example was aimed at making a proof where I have a Reductio ad Absurdum Implication comprise of only raising a 5th color and nothing about adjacency which is treated by entering dimension into the argument. So a question arises. I notice that 3rd dimension has infinite adjacency and infinite coloring required. I notice that in Eucl 2nd dimension that 4 colors are required and that adjacency has a maximum of 4. So Chris, is there a geometry where coloring and adjacency have a variance. A configuration that say requires 5 colors but maximum adjacency is say 6. A configuration where 5 colors are required but maximum adjacency is 3. What is the 4 Color Mapping like in Loba geom instead of Eucl geom plane. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Subject: Re: half page proof of 4 Color Mapping Theorem; using Moebius 4 adjacency maximum > [...] > So a question arises. I notice that 3rd dimension has infinite > adjacency and infinite coloring required. I notice that in Eucl 2nd > dimension that 4 colors are required and that adjacency has a maximum > of 4. So Chris, is there a geometry where coloring and adjacency have a > variance. A configuration that say requires 5 colors but maximum > adjacency is say 6. A configuration where 5 colors are required but > maximum adjacency is 3. The number of colors will always be at least the adjacency. Now, by configuration do you mean a geometry, or a map? Because I've been telling you that there are maps where the coloring and adjacency numbers are different, for a couple of months now. > What is the 4 Color Mapping like in Loba geom instead of Eucl geom > plane. I suspect that, since the 4CT can be expressed without reference to geometry, that it will be the same. ==== Subject: Re: half page proof of 4 Color Mapping Theorem; using Moebius 4 adjacency maximum The number of colors will always be at least the adjacency. Now, by configuration do you mean a geometry, or a map? Because I've been telling you that there are maps where the coloring and adjacency numbers are different, for a couple of months now. > What is the 4 Color Mapping like in Loba geom instead of Eucl geom > plane. I suspect that, since the 4CT can be expressed without reference to geometry, that it will be the same. (and) > I am waiting for Chris to get back with an answer as to whether there > is any model of any geometry that has a *Variance* of its adjacency > with its Color requirement. Any model, even the torus. Is the torus a geometry? Your terminology is non-standard. The torus, double torus, etc., are really called surfaces (2-manifolds). If you mean surfaces, then no, on every surface, adjacency is the same as the number of colors. However, pseudosurfaces, like the spindle surface might be different .... (Couldn't find a picture online.) Imagine taking a (hollow) sphere, and saying the north and south poles are the same point. Or take the circle (x-1)^2 + y^2 = 1^2 and rotating it around the y-axis. > We notice that in 3rd dimension Euclid geom that adjacency maximum is > infinite and Color requirement is also infinite so there is no variance > between adjacency and color requirement. We notice that in Eucl 2nd > dimension that adjacency maximum is 4 and that color requirement is > also 4, so there is no variance here either. > So is there any variance anywhere in geometry of maximum adjacency to > the number of colors required. If there is no variance between > adjacency and color requirement then my proof of 4 Color Mapping in one > sentence holds. Actually, that's true, but it would require a proof. And since the 4CT would follow from it, I don't expect the proof to be simple. --- Christopher Heckman Since every surface is the same for the coloring, I suspect there is no variance. And perhaps your last paragraph maybe an easy proof. Consider the alternative that there is a mapping requiring more colors than the maximum mutual-adjacency. What havoc would that create? I think the answer is simple in that more colors than the maximum mutual-adjacency forces the map from a surface or 2nd dimension into a third dimensional map where colors required are infinite and mutual adjacency is also infinite. So I think the proof that in 2nd dimension all planar maps have a maximum mutual adjacency equal to the number of colors required. If false, then it has to be in the 3rd dimension, contradiction. And from this theorem, 4CM is a corollary. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Subject: Re: half page proof of 4 Color Mapping Theorem; using Moebius 4 adjacency maximum ==== Subject: Re: half page proof of 4 Color Mapping Theorem; using Moebius 4 adjacency maximum <11178081.1139374692104.JavaMail.jakarta@nitrogen.mathforum.org> No I do not believe it is slippery. I just think it has a optical like illusion to its proving. Remember those optical illusions we see in books, thinking that a distance is shorter but the same size. So I think the 4 Color Mapping has this optical illusion type of logic. A.P. ==== Subject: Timothy P. Daly Today I found some comments from a correspondent who, as a person belonging to the Axiom team, influenced my life in July of 1993. ISSAC'93 was held at Kiev, organized by Glushkov Institute of Cybernetics... A live whirligig Manuel Bronstein, staid Gaston Gonnet... Stars! http://www-sop.inria.fr/cafe/Manuel.Bronstein/ http://www.inf.ethz.ch/personal/gonnet/ Dan Richardson in the flesh (envy me, boys!) http://www.bath.ac.uk/~masdr/ Axiom was shown publicly then... Having seeing my eyes glowing with crazy interest, benevolent and noble Steve Hague of NAG http://www.nag.co.uk/about/shague.asp lent me his Axiom book for a night. Needless to say, I kept reading the Bible overnight, and I got really stunned with prospects it presumably shows! What an inviting prospect!, I thought delightedly (in 1993). Before answering his comments, I'd like to let you an opportunity to feel about this person. http://www.amazon.com/exec/obidos/tg/detail/-/0387978550/103-8872015-5381429 ?v=glance What we read in the Axiom: The Scientific Computation System by Richard D. Jenks and Robert S. Sutor, Hardcover, 742 pages Springer-Verlag (July 1, 1992), ISBN 0387978550 Timothy P. Daly (IBM, Yorktown) is pursuing a Ph.D. in computer science at Brooklyn Polytechnic Institute and is responsible for porting, testing, performance, and system support work for AXIOM. Now here comes a terrific list of his skills in development (honestly, I take my hat off to his, I emphasize, skills in software _development_) http://daly.axiom-developer.org/ http://daly.axiom-developer.org/resume.txt ==== Subject: Quantized Indexing (References update) http://www.1stworks.com/ref/RefLib.htm Quantized Indexing Home: http://www.1stworks.com/ref/qi.htm * Local copies for several external papers (which had broken links) were added combinatorics 8a. R. Brak Enumerative Combinatorics (lattice paths) Lectures, http://www.1stworks.com/ref/combin_P1.PDF 28. L.D. Davisson Universal noiseless coding IEEE Trans. Inform. Theory IT-19 (6), 783-795, 1973 http://www.1stworks.com/ref/Davisson1973Universal.pdf 29. J. Rissanen, G.G. Langdon Universal Modeling and Coding IEEE Trans. Inform. Theory IT-19 (1), 12-23, 1981 http://www.1stworks.com/ref/Rissanen1981Universal.pdf 30. A. Barron, J. Rissanen, Bin Yu The minimum description length principle in coding and modeling IEEE Trans. Inform. Theory IT-44 (6), 2743-2760, 1998 http://www.1stworks.com/ref/Barron1998minimum.pdf 33. R. Krichevsky, V. Trofimov The performance of universal encoding IEEE Trans. Inform. Theory IT-27 (2), 199-207, 1981 http://www.1stworks.com/ref/Krichevsky1981performance.pdf 53. M. Feder, N. Merhav, M. Gutman Universal prediction of individual sequences IEEE Trans. Inform. Theory IT-38 (4), 1258-1270, 1992 http://www.1stworks.com/ref/Feder1992Universal.pdf 54. J. Rissanen Universal coding, information, prediction, and estimation IEEE Trans. Inform. Theory IT-30 (4), 629-636, 1984 http://www.1stworks.com/ref/Rissanen1984Universal.pdf 55. M.J. Weinberger, J. Rissanen, M. Feder A universal finite memory source IEEE Trans. Inform. Theory IT-41 (3), 643-652, 1995 http://www.1stworks.com/ref/Weinberger1995universal.pdf ==== Subject: Re: infinity sci.math_20050522: So, the first, second, ..., n'th derivatives of the equation of motion exists and the universe is infinite and infinite sets are equivalent, is that so? sci.math_20050523: Please don't read this as discouragement. On the contrary, you have some solid and correct intuitive notions about things infinite and infinite sets. sci.math_20050523: One consideration you have is to compare infinite sets of integers. As numbers, that is directly addressed by number theory. The asymptotic density of the even integers within the integers, for example, is one half, half of the integers are even. There are vagaries in the number theoretic definition, for example the standard number theory does not address comparable infinitesimal ratios. ==== Subject: finding intrinsic dimensionality hi, as part of my project i need to find the intrinsic dimensionality of surfaces whose parametric forms are available.can someone help me out on this. ==== Subject: Re: ? eigenbasis and other similar matrices > For self-adjoint matrix, we know much about its eigen decomposition. > But are there general theory that assues certain types of matrices > to have eigen decomposition? What at least, except the self-adjoint matrix, > what are other types of matrix that we know they have eigen decomposition? There is a more general result comprising this you have given above: A complex square matrix A is orthogonally diagonalizable iff A is normal (i.e. AA*=A*A where A* denotes the self-adjoint matrix of A). HTH J. ==== Subject: Re: ? eigenbasis and other similar matrices > For self-adjoint matrix, we know much about its eigen decomposition. > But are there general theory that assues certain types of matrices > to have eigen decomposition? What at least, except the self-adjoint matrix, > what are other types of matrix that we know they have eigen decomposition? There is a more general result comprising this you have given above: A complex square matrix A is orthogonally diagonalizable iff A is normal (i.e. AA*=A*A where A* denotes the self-adjoint matrix of A). HTH J. ==== Subject: Re: Question about operator theory >Could anybody give me an example of linear operator such that >>annihilator of its kernel does not conside with closure of image of >>its adjoint ? >>annihilator(kerA) != closure(Im(A')) >Well, linear operator and closure can mean >various things. Assuming that you're talking about >bounded operators on Banach spaces, it seems to me >after a bit of scribbling that annihilator(kerA) >is always equal to the weak*-closure of Im(A'), >so you need an example where the norm closure >of Im(A') is not weak* closed. By closed range theorem ( see for instance Taylor Introduction to functional analysis) closure of Im(A') always belongs to annihilator of ker(A). I know example of operator when annihilator(kerA) != closure(Im(A')) - it is differential operator on L2(a,b) , but this operator is not bounded . Sorry , but I didn't understand yoyr example ... > Assuming you knew all that (or rather, assuming > that that's all correct and you knew all that) > here's an example where annihilator(kerA) > is not equal to the norm closure of Im(A*): > Let X be L^1(T), the space of 2pi-periodic > integrable functions. Fix phi in X such that > all the Fourier coefficients of phi are non-zero, > and define A : X -> X by the convolution > Af = f*phi (f convolved with phi). > It's not hard to show that if g is in L^infinity > then A*(g) is the convolution of g with psi, > where psi(x) = phi(-x). So Im(A*) is the space > of all g*psi for g in L^infinity. > Hence Im(A*) is contained in C(T), the continuous > functions on T. The fact that all the Fourier > coefficients of psi are non-zero shows that > Im(A*) contains every character (every function > of the form e(t) = exp(int), n an integer). > Hence Im(A*) is norm-dense in C(T). So its > norm closure is C(T), and hence its weak*closure > is all of L^infinity(T). So the two closures > are not equal. >If that's right, then if you're talking about >bounded operators on a Hilbert space (or reflexive >spaces in general) then there is no such example, >because a norm-closed subspace is always weakly >closed. > >David C. Ullrich David C. Ullrich ==== Subject: Re: Question about operator theory >> On Tue, 07 Feb 2006 10:25:55 -0600, David C. Ullrich >Could anybody give me an example of linear operator such that >annihilator of its kernel does not conside with closure of image of >its adjoint ? >>annihilator(kerA) != closure(Im(A')) >>Well, linear operator and closure can mean >>various things. Assuming that you're talking about >>bounded operators on Banach spaces, it seems to me >>after a bit of scribbling that annihilator(kerA) >>is always equal to the weak*-closure of Im(A'), >>so you need an example where the norm closure >>of Im(A') is not weak* closed. >By closed range theorem ( see for instance Taylor Introduction to >functional analysis) >closure of Im(A') always belongs to annihilator of ker(A). >I know example of operator when annihilator(kerA) != closure(Im(A')) - >it is differential operator on L2(a,b) , but this operator is not >bounded . >Sorry , but I didn't understand yoyr example ... Well, it's a simple example of a bounded operator on a Banach space such that annihilator(kerA) is not equal to the _norm_ closure of Im(A*). I'd explain if I knew _what_ you didn't understand about it. (Do you know basic things about Fourier series? Do you know what a convolution is, and what convolutions have to do with Fourier coefficients?) >> Assuming you knew all that (or rather, assuming >> that that's all correct and you knew all that) >> here's an example where annihilator(kerA) >> is not equal to the norm closure of Im(A*): >> Let X be L^1(T), the space of 2pi-periodic >> integrable functions. Fix phi in X such that >> all the Fourier coefficients of phi are non-zero, >> and define A : X -> X by the convolution >> Af = f*phi (f convolved with phi). >> It's not hard to show that if g is in L^infinity >> then A*(g) is the convolution of g with psi, >> where psi(x) = phi(-x). So Im(A*) is the space >> of all g*psi for g in L^infinity. >> Hence Im(A*) is contained in C(T), the continuous >> functions on T. The fact that all the Fourier >> coefficients of psi are non-zero shows that >> Im(A*) contains every character (every function >> of the form e(t) = exp(int), n an integer). >> Hence Im(A*) is norm-dense in C(T). So its >> norm closure is C(T), and hence its weak*closure >> is all of L^infinity(T). So the two closures >> are not equal. >>If that's right, then if you're talking about >>bounded operators on a Hilbert space (or reflexive >>spaces in general) then there is no such example, >>because a norm-closed subspace is always weakly >>closed. >> >>David C. Ullrich >> David C. Ullrich David C. Ullrich ==== Subject: Re: Question about operator theory Here things which I understand and which I doesn't 1) Domain of A , is it defined on all L1(T) ? Domain of A' ? 2) Ker(A) all measurable functions which are equal to zero almost everywhere . So annihilator of Ker(A) is whole L_inf(T) . 3) Image of A' contains anly absolutely continuous functions so it contains only continous functions. 4) It contains 1 , exp(j*n*t) and any continous function on T could be approximated by harmonic polynomial => Im(A') = C(T) because for continous functions on compact infinity norm is equal to supprenum norm . 5) C(T)!= L_inf(T) >> On Tue, 07 Feb 2006 10:25:55 -0600, David C. Ullrich >Could anybody give me an example of linear operator such that >annihilator of its kernel does not conside with closure of image of >its adjoint ? >>annihilator(kerA) != closure(Im(A')) >>Well, linear operator and closure can mean >>various things. Assuming that you're talking about >>bounded operators on Banach spaces, it seems to me >>after a bit of scribbling that annihilator(kerA) >>is always equal to the weak*-closure of Im(A'), >>so you need an example where the norm closure >>of Im(A') is not weak* closed. >By closed range theorem ( see for instance Taylor Introduction to >functional analysis) >closure of Im(A') always belongs to annihilator of ker(A). >I know example of operator when annihilator(kerA) != closure(Im(A')) - >it is differential operator on L2(a,b) , but this operator is not >bounded . >Sorry , but I didn't understand yoyr example ... > Well, it's a simple example of a bounded operator > on a Banach space such that annihilator(kerA) > is not equal to the _norm_ closure of Im(A*). > I'd explain if I knew _what_ you didn't understand > about it. (Do you know basic things about > Fourier series? Do you know what a convolution > is, and what convolutions have to do with > Fourier coefficients?) >> Assuming you knew all that (or rather, assuming >> that that's all correct and you knew all that) >> here's an example where annihilator(kerA) >> is not equal to the norm closure of Im(A*): >> Let X be L^1(T), the space of 2pi-periodic >> integrable functions. Fix phi in X such that >> all the Fourier coefficients of phi are non-zero, >> and define A : X -> X by the convolution >> Af = f*phi (f convolved with phi). >> It's not hard to show that if g is in L^infinity >> then A*(g) is the convolution of g with psi, >> where psi(x) = phi(-x). So Im(A*) is the space >> of all g*psi for g in L^infinity. >> Hence Im(A*) is contained in C(T), the continuous >> functions on T. The fact that all the Fourier >> coefficients of psi are non-zero shows that >> Im(A*) contains every character (every function >> of the form e(t) = exp(int), n an integer). >> Hence Im(A*) is norm-dense in C(T). So its >> norm closure is C(T), and hence its weak*closure >> is all of L^infinity(T). So the two closures >> are not equal. >>If that's right, then if you're talking about >>bounded operators on a Hilbert space (or reflexive >>spaces in general) then there is no such example, >>because a norm-closed subspace is always weakly >>closed. >> >>David C. Ullrich >> David C. Ullrich David C. Ullrich ==== Subject: Re: Dihedral subgroup of a symmetric group On 4-Feb-2006, magidin@math.berkeley.edu (Arturo Magidin) > [.snip.] > [.The dihedral group of order 24.] >However, I was wondering if there is some other way to create a >dihedral group (by squeezing or twisting or applying whatever other >action to the n-gon) if you do not need to keep the symmetries of the >n-gon. > As an addendum, note that S_{12} contains other copies of the dihedral > group of order 24 which do not contain any 12-cycle. OK, I can't resist the challenge... :) > For example, you > can start with an element of order 12, such as > r = (1,2,3)(4,5,6,7) > and find an element s of order 2 such that srs=r^{-1}; for instance, > s = (2,3)(5,7) will give > srs = (1,3,2)(4,7,6,5) = r^{-1} > so the subgroup of S_{12} generated by r and s is isomorphic to the > dihedral group of order 24; since it fixes 8, 9, 10, 11, and 12 > pointwise, it clearly cannot contain any 12-cycle. Up to conjugacy, I count 31 subgroups of S_{12} isomorphic to the dihedral group of order 24. I've included rs as well as r and s in the tables below, so the cycle structures of the three conjugacy classes of involutions {r^6, s, rs} can be compared and contrasted. o 1 of degree 4+3 = 7 (yours above): r = (1,2,3,4) (5,6,7), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) o 3 of degree 4+3+2 = 9: r = (1,2,3,4) (5,6,7), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) (8,9) r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) (8,9) r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) o 3 of degree 6+4 = 10: r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (7,8)(9,10), rs = (1,6)(2,5)(3,4) (8,10) r = (1,2,3)(4,5,6) (7,8,9,10), s = (1,4)(2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) o 1 of degree 4+3+3 = 10: r = (1,2,3,4) (5,6,7) (8,9,10), s = (2,4) (6,7) (9,10), rs = (1,4)(2,3) (5,7) (8,10) o 1 of degree 8+3 = 11: r = (1,2,3,4)(5,6,7,8) (9,10,11), s = (1,5)(2,8)(3,7)(4,6) (10,11), rs = (1,8)(2,7)(3,6)(4,5) (9,11) o 3 of degree 4+4+3 = 11: r = (1,2,3,4) (5,6,7,8) (9,10,11), s = (2,4) (6,8) (10,11), rs = (1,4)(2,3) (5,8)(6,7) (9,11) r = (1,2,3,4) (5,6,7,8) (9,10,11), s = (2,4) (5,6)(7,8) (10,11), rs = (1,4)(2,3) (6,8) (9,11) r = (1,2,3,4) (5,6)(7,8) (9,10,11), s = (2,4) (5,7)(6,8) (10,11), rs = (1,4)(2,3) (5,8)(6,7) (9,11) o 6 of degree 4+3+2+2 = 11: r = (1,2,3,4) (5,6,7), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) (8,9) (10,11) r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (10,11), rs = (1,4)(2,3) (5,7) (8,9) (10,11) r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) (10,11) r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) (8,9) (10,11) r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) (10,11) r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) o 1 of degree 12 (the one that started this all): r = (1,2,3,4,5,6,7,8,9,10,11,12), s = (2,12)(3,11)(4,10)(5,9)(6,8), rs = (1,12)(2,11)(3,10)(4,9)(5,8)(6,7) o 9 of degree 6+4+2 = 12: r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (7,8)(9,10) (11,12), rs = (1,6)(2,5)(3,4) (8,10) (11,12) r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (7,8)(9,10), rs = (1,6)(2,5)(3,4) (8,10) (11,12) r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (7,8)(9,10) (11,12), rs = (1,6)(2,5)(3,4) (8,10) r = (1,2,3)(4,5,6) (7,8,9,10), s = (1,4)(2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) r = (1,2,3)(4,5,6) (7,8,9,10) (11,12), s = (1,4)(2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) r = (1,2,3)(4,5,6) (7,8,9,10) (11,12), s = (1,4)(2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) o 3 of degree 4+3+3+2 = 12: r = (1,2,3,4) (5,6,7) (8,9,10), s = (2,4) (6,7) (9,10) (11,12), rs = (1,4)(2,3) (5,7) (8,10) (11,12) r = (1,2,3,4) (5,6,7) (8,9,10) (11,12), s = (2,4) (6,7) (9,10), rs = (1,4)(2,3) (5,7) (8,10) (11,12) r = (1,2,3,4) (5,6,7) (8,9,10) (11,12), s = (2,4) (6,7) (9,10) (11,12), rs = (1,4)(2,3) (5,7) (8,10) -- Jim Heckman ==== Subject: Re: Dihedral subgroup of a symmetric group >On 4-Feb-2006, magidin@math.berkeley.edu (Arturo Magidin) >> [.snip.] >> [.The dihedral group of order 24.] >>However, I was wondering if there is some other way to create a >>dihedral group (by squeezing or twisting or applying whatever other >>action to the n-gon) if you do not need to keep the symmetries of the >>n-gon. >> As an addendum, note that S_{12} contains other copies of the dihedral >> group of order 24 which do not contain any 12-cycle. >OK, I can't resist the challenge... :) >> For example, you >> can start with an element of order 12, such as >> r = (1,2,3)(4,5,6,7) >> and find an element s of order 2 such that srs=r^{-1}; for instance, >> s = (2,3)(5,7) will give >> srs = (1,3,2)(4,7,6,5) = r^{-1} >> so the subgroup of S_{12} generated by r and s is isomorphic to the >> dihedral group of order 24; since it fixes 8, 9, 10, 11, and 12 >> pointwise, it clearly cannot contain any 12-cycle. >Up to conjugacy, I count 31 subgroups of S_{12} isomorphic to the >dihedral group of order 24. I've included rs as well as r and s in >the tables below, so the cycle structures of the three conjugacy >classes of involutions {r^6, s, rs} can be compared and contrasted. I just checked this with a computer calculation, and it also came up with 31 conjugacy classes of subgroups of S_{12} isomorphic to dihedral group of order 24, so I expect you have it right! Derek Holt. >o 1 of degree 4+3 = 7 (yours above): > r = (1,2,3,4) (5,6,7), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) >o 3 of degree 4+3+2 = 9: > r = (1,2,3,4) (5,6,7), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) (8,9) > r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) (8,9) > r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) >o 3 of degree 6+4 = 10: > r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) > r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (7,8)(9,10), rs = (1,6)(2,5)(3,4) (8,10) > r = (1,2,3)(4,5,6) (7,8,9,10), s = (1,4)(2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) >o 1 of degree 4+3+3 = 10: > r = (1,2,3,4) (5,6,7) (8,9,10), s = (2,4) (6,7) (9,10), rs = (1,4)(2,3) (5,7) (8,10) >o 1 of degree 8+3 = 11: > r = (1,2,3,4)(5,6,7,8) (9,10,11), s = (1,5)(2,8)(3,7)(4,6) (10,11), rs = (1,8)(2,7)(3,6)(4,5) (9,11) >o 3 of degree 4+4+3 = 11: > r = (1,2,3,4) (5,6,7,8) (9,10,11), s = (2,4) (6,8) (10,11), rs = (1,4)(2,3) (5,8)(6,7) (9,11) > r = (1,2,3,4) (5,6,7,8) (9,10,11), s = (2,4) (5,6)(7,8) (10,11), rs = (1,4)(2,3) (6,8) (9,11) > r = (1,2,3,4) (5,6)(7,8) (9,10,11), s = (2,4) (5,7)(6,8) (10,11), rs = (1,4)(2,3) (5,8)(6,7) (9,11) >o 6 of degree 4+3+2+2 = 11: > r = (1,2,3,4) (5,6,7), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) (8,9) (10,11) > r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (10,11), rs = (1,4)(2,3) (5,7) (8,9) (10,11) > r = (1,2,3,4) (5,6,7) (8,9), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) (10,11) > r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7), rs = (1,4)(2,3) (5,7) (8,9) (10,11) > r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7) (8,9), rs = (1,4)(2,3) (5,7) (10,11) > r = (1,2,3,4) (5,6,7) (8,9) (10,11), s = (2,4) (6,7) (8,9) (10,11), rs = (1,4)(2,3) (5,7) >o 1 of degree 12 (the one that started this all): > r = (1,2,3,4,5,6,7,8,9,10,11,12), s = (2,12)(3,11)(4,10)(5,9)(6,8), rs = (1,12)(2,11)(3,10)(4,9)(5,8)(6,7) >o 9 of degree 6+4+2 = 12: > r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) > r = (1,2,3,4,5,6) (7,8,9,10), s = (2,6)(3,5) (7,8)(9,10) (11,12), rs = (1,6)(2,5)(3,4) (8,10) (11,12) > r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) > r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (7,8)(9,10), rs = (1,6)(2,5)(3,4) (8,10) (11,12) > r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) > r = (1,2,3,4,5,6) (7,8,9,10) (11,12), s = (2,6)(3,5) (7,8)(9,10) (11,12), rs = (1,6)(2,5)(3,4) (8,10) > r = (1,2,3)(4,5,6) (7,8,9,10), s = (1,4)(2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) > r = (1,2,3)(4,5,6) (7,8,9,10) (11,12), s = (1,4)(2,6)(3,5) (8,10), rs = (1,6)(2,5)(3,4) (7,10)(8,9) (11,12) > r = (1,2,3)(4,5,6) (7,8,9,10) (11,12), s = (1,4)(2,6)(3,5) (8,10) (11,12), rs = (1,6)(2,5)(3,4) (7,10)(8,9) >o 3 of degree 4+3+3+2 = 12: > r = (1,2,3,4) (5,6,7) (8,9,10), s = (2,4) (6,7) (9,10) (11,12), rs = (1,4)(2,3) (5,7) (8,10) (11,12) > r = (1,2,3,4) (5,6,7) (8,9,10) (11,12), s = (2,4) (6,7) (9,10), rs = (1,4)(2,3) (5,7) (8,10) (11,12) > r = (1,2,3,4) (5,6,7) (8,9,10) (11,12), s = (2,4) (6,7) (9,10) (11,12), rs = (1,4)(2,3) (5,7) (8,10) >-- >Jim Heckman ==== Subject: Re: Poisson Probability > dead period, or is it just ignored? I do not think this is the case, > but it does bring up a good point that I over looked. > I am a little confused as to where you get the idea It sounds as if you are doubtful, and somehow imagine that I am making this up. It's not my idea, but is a well-known property of the Poisson process, to be found in hundreds of books. Dimensionally, it is OK. If r is a rate (number of events per second, say) its reciprocal is a time, in seconds. Note: I said the mean TIME to the next event is 1/r, not the mean number of events. The mean number of events in an interval of length T is rT; the mean time between the separate events is 1/r; for example, if the mean time between events is 1/2 second, then there are 2 events per second on average. RGV > that the mean is equal to 1/r if the rate is r. > Also, the book is Data Reducation and Error Analysis by Bevington. ==== Subject: How to reduce fractions? Is there any way to reduce this fraction: n^2/k^2 + n/k I have tried something like: 2n^2/k^2 but that is not correct. if its not possible to reduce further what is the reason? ==== Subject: Re: How to reduce fractions? >Is there any way to reduce this fraction: >n^2/k^2 + n/k That's a sum of 2 fractions. First you need to combine them into 1 fraction. You know k/k=1, so n^2/k^2 + n/k*(k/k) = n^2/k^2 + n*k/k^2 = (n^2+n*k)/k^2 = n*(n+k)/k^2 Then you might be able to reduce it if you know the values of n and k. --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. ==== Subject: Re: How to reduce fractions? > Is there any way to reduce this fraction: > n^2/k^2 + n/k > I have tried something like: > 2n^2/k^2 but that is not correct. > if its not possible to reduce further what is the > reason? You could rewrite it several ways, like n/k * (n/k + 1) ..or... (n^2 + n * k) / k^2. However, to *reduce* it I believe you'd need some further relationship between n and k. For instance, if you knew that n = r * k or something similar, then you could reduce it to r^2 + r. Kyle ==== Subject: Re: How to reduce fractions? >Is there any way to reduce this fraction: >n^2/k^2 + n/k >I have tried something like: >2n^2/k^2 but that is not correct. >if its not possible to reduce further what is the reason? n^2 n n(n+k) --- + - = ------ k^2 k k^2 Suppose that n and k are relatively prime, yet a prime p divides both n(n+k) and k^2. Since p divides k^2, it must divide k. Since n and k are relatively prime p cannot divide n, so p must divide n+k. However, since p divides k and p divides n+k, p must divide n, but then n and k are not relatively prime. Thus, if n and k are relatively prime, then so is n(n+k)/k^2. Thus, if n/k cannot be reduced, neither can n^2/k^2 + n/k. Rob Johnson take out the trash before replying ==== Subject: need help to find a counter example In the context of the evolutionary stable strategy, I need to find an example of function for my these. The problem is: to find a function W(x,y):D-->R+, where D is an open domain of the plane containning [0,1]^2, differentiable in D, and such that there exists a point x* with 00 for all 0<=x, > Answer to the World Wide Wade: > Your function W depend on x*, which is not the hypothesis. > Michael. You should learn how to use the Reply feature of your newsreader. Your response above has no context; this can make it difficult for readers to understand. Here's my post: ------- <32245430.1139392734018.JavaMail.jakarta@nitrogen.mathforum.org>, > In the context of the evolutionary stable strategy, I need to find an example > of function for my these. > The problem is: to find a function W(x,y):D-->R+, where D is an open domain > of the plane containning [0,1]^2, differentiable in D, and such that there > exists a point x* with 0 1) dW(x,x)/dy>0 for all 0<=x 2) dW(x*,x*)/dy=0 > 3) dW(x,x)/dy<0 for all x* 4) there exists 0 0 for all x < x* 2) dW(x*,x*)/dy = 0 3) dW(x,x)/dy < 0 for all x > x* 4) W(x*,x*) < W(x*,y) for all y not equal to x*. ==== Subject: Re: need help to find a counter example <32245430.1139392734018.JavaMail.jakarta@nitrogen.mathforum.org>, > In the context of the evolutionary stable strategy, I need to find an example > of function for my these. > The problem is: to find a function W(x,y):D-->R+, where D is an open domain > of the plane containning [0,1]^2, differentiable in D, and such that there > exists a point x* with 0 1) dW(x,x)/dy>0 for all 0<=x 2) dW(x*,x*)/dy=0 > 3) dW(x,x)/dy<0 for all x* 4) there exists 0 In the context of the evolutionary stable strategy, I need to find an example of function for my these. > The problem is: to find a function W(x,y):D-->R+, where D is an open domain of the plane containning [0,1]^2, differentiable in D, and such that there exists a point x* with 0 1) dW(x,x)/dy>0 for all 0<=x 2) dW(x*,x*)/dy=0 > 3) dW(x,x)/dy<0 for all x* 4) there exists 0 please, help. I see no problem here. Your conditions are only about the partial derivatives of W on the diagonal; away from it W can behave as it will ( or you). So just take any function satisfying 1-4 on the diagonal and then bend it downwards (in some distance from the diagonal) on the line from x* on the y-axis. Ciao Karl ==== Subject: how to inverse a highly symmetric matrix I would be grateful to anyone who could give me some insight into solving the following problem. I have a matrix with the form: 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 i.e. it is just filled with 0 and 1 and, basically, you just have to know one quadrant to define the whole matrix. The size of my matrix is extremely large and I wish to take advantage of its simple form to reduce the computation cost of inversing it (or calculating a pseudo-inverse). Unfortunately I do no see how to do it. So is there really a way to simplify the calculus or should I just use an algorithm such as SVD and let it run for 2 days? Maybe this problem has already been treated in a book or paper? Borg. ==== Subject: Re: how to inverse a highly symmetric matrix >I would be grateful to anyone who could give me some insight into >solving the following problem. >I have a matrix with the form: >0 0 0 0 0 0 >0 0 1 1 0 0 >0 1 1 1 1 0 >0 1 1 1 1 0 >0 0 1 1 0 0 >0 0 0 0 0 0 >i.e. it is just filled with 0 and 1 and, basically, you just have to >know one quadrant to define the whole matrix. It's not clear exactly what the form is. Do you just mean you have a 2n x 2n matrix A whose entries are 0's and 1's and it has the symmetries A[i,j] = A[2n+1-i,j] and A[i,j] = A[i,2n+1-j], or is there more to it? >The size of my matrix is extremely large and I wish to take advantage of >its simple form to reduce the computation cost of inversing it (or >calculating a pseudo-inverse). Obviously a pseudo-inverse, since it has repeated rows and columns. >Unfortunately I do no see how to do it. So is there really a way to >simplify the calculus or should I just use an algorithm such as SVD and >let it run for 2 days? The fact that the entries are 0's and 1's is not likely to be particularly useful for the pseudo-inverse. On the other hand, the symmetry should help. By reversing the order of the second half of the indices for rows and columns, you get a matrix of the form [ A A ] [ A A ] and if I'm not mistaken the pseudo-inverse of this should be [ B/4 B/4 ] [ B/4 B/4 ] where B is the pseudo-inverse of A. Reverse the order of the second half of the indices for rows and columns of this, and you get the pseudo-inverse of your original matrix. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: Finite groups exercise >> On Tue, 07 Feb 2006 07:32:08 EST, Larry Hammick >If G is a finite group of odd order, and y is any element of G, then the >mapping >x mapsto xy >of G to G is an _even_ permutation of G. >Hint: This is a very easy problem :) >> Yes, a question about algebra that _I_ see a solution to more >> or less as soon as I start thinking about it is very easy. >> Are you asking for a hint or wondering whether we can do it or what? >Er, I responded but the response seems to have vanished from Google. >Again then, assume that some homothety xmapsto xy is an odd permutation, >deduce that the odd homotheties are then equinumersous with the even, >and notice the contradiction: the total number of homotheties is odd by >hypothesis. I saw your reply and replied to it. In case you missed that, what I said was this: Heh-heh. I don't see exactly why your obvious proof is right - why are the odd translations (sorry, I know how to spell that) equinumerous with the even ones? My obvious argument: If P is one of those permutations and n is the order of the group then P^n is the identity; otoh if P is odd and n is odd then P^n is odd, hence not the identity. >Why did I post this thing? I was hoping to sucker somebody into talking >about >decomposition series or orbit-counting or some such thing :) David C. Ullrich ==== Subject: Re: Finite groups exercise > On Tue, 07 Feb 2006 07:32:08 EST, Larry Hammick >>If G is a finite group of odd order, and y is any element of G, then the >>mapping >>x mapsto xy >>of G to G is an _even_ permutation of G. >>Hint: This is a very easy problem :) > Yes, a question about algebra that _I_ see a solution to more > or less as soon as I start thinking about it is very easy. > Are you asking for a hint or wondering whether we can do it or what? >>Er, I responded but the response seems to have vanished from Google. >>Again then, assume that some homothety xmapsto xy is an odd permutation, >>deduce that the odd homotheties are then equinumersous with the even, >>and notice the contradiction: the total number of homotheties is odd by >>hypothesis. > I saw your reply and replied to it. In case you missed that, what I > said was this: > Heh-heh. I don't see exactly why your obvious proof is right - why > are the odd translations (sorry, I know how to spell that) > equinumerous with the even ones? > My obvious argument: If P is one of those permutations and n is > the order of the group then P^n is the identity; otoh if P is > odd and n is odd then P^n is odd, hence not the identity. Quite so. My way was something like this: Prop.: If G is a finite group (of any order, commutative or not) and the translation (thank you) x mapsto vx is odd for some v in G, then the odd translations are equinumerous with the even (and therefore the order of G is even). Proof: x mapsto bx and x mapsto ax. So if a is odd then b is even iff ab is odd. We get a map odds-to-evens and evens-to-odds. If one part is not surjective then the other is not injective, which is impossible. So both parts are bijective. Okay, maybe deduce in a flash was a bit of an overclaim on my part :) LH ==== Subject: similarity over a field Let A,B be two matrices over an infinite field F, such that they are similar(not necessarily over F)-i mean there exists the matrix U(not neccesary over F) such that UAU^{-1}=B. Prove that there exists U over F such that UAU^{-1}=B. ==== Subject: Re: similarity over a field days. My association with the Department is that of an alumnus. >Let A,B be two matrices over an infinite field F, such that they are >similar(not necessarily over F)-i mean there exists the matrix U(not >neccesary over F) such that UAU^{-1}=B. Prove that there exists U over >F such that UAU^{-1}=B. Rather than prove this, I will prove the closely related proposition: This is a homework exercise for eugene, who decided to have someone do it for him/her. Moreover, he/she was too lazy to even ask for help, simply copying the statement and hoping someone would answer it for him. The proof is left as an easy exercise for the reader. Now, as for a hint, consider the system of linear equations XA-BX=0. Does it have a solution? Why? -- ' ==== Subject: Re: similarity over a field > <1867713.1139395736595.JavaMail.jakarta@nitrogen.mathf > orum.org>, >Let A,B be two matrices over an infinite field F, > such that they are >similar(not necessarily over F)-i mean there exists > the matrix U(not >neccesary over F) such that UAU^{-1}=B. Prove that > there exists U over >F such that UAU^{-1}=B. > Rather than prove this, I will prove the closely > related proposition: > This is a homework exercise for eugene, who > who decided to have someone > do it for him/her. Moreover, he/she was too lazy > azy to even ask for > help, simply copying the statement and hoping > ing someone would answer > it for him. > The proof is left as an easy exercise for the reader. I just wanted to figure out the problem that interests me and didn't mind to copy my homework : you would laugh if you see my homework-it's much more easier... I thought one of the purposes of the such a forum is to discuss problem rather then blaming other users in copying-Sorry, if my comprehension isn't correct, i'll try to reform. > Now, as for a hint, consider the system of linear > equations > XA-BX=0. Does it have a solution? Why? For example it has a solution-the matrix U. You must have meant the solution over F, supposing that A and B are over F ? ==== Subject: Re: similarity over a field days. My association with the Department is that of an alumnus. >> <1867713.1139395736595.JavaMail.jakarta@nitrogen.mathf >> orum.org>, >>Let A,B be two matrices over an infinite field F, >> such that they are >>similar(not necessarily over F)-i mean there exists >> the matrix U(not >>neccesary over F) such that UAU^{-1}=B. Prove that >> there exists U over >>F such that UAU^{-1}=B. >> Rather than prove this, I will prove the closely >> related proposition: >> This is a homework exercise for eugene, who >> who decided to have someone >> do it for him/her. Moreover, he/she was too lazy >> azy to even ask for >> help, simply copying the statement and hoping >> ing someone would answer >> it for him. >> The proof is left as an easy exercise for the reader. >I just wanted to figure out the problem that interests me and didn't >mind to copy my homework : you would laugh if you see my >homework-it's much more easier... I thought one of the purposes of >the such a forum is to discuss problem rather then blaming other >users in copying-Sorry, if my comprehension isn't correct, i'll try >to reform. You want help? Ask for help. You know, as in I am trying to solve this problem. This is what I've done so far. This is where I am stuck. Can someone give me a hint? Rather than just writing down the problem. You are not giving an assignement to the readers of this newsgroup. You are asking for help. > Now, as for a hint, consider the system of linear equations > XA-BX=0. Does it have a solution? Why? >For example it >has a solution-the matrix U. You must have meant the solution over >F, supposing that A and B are over F ? What do you know about systems of linear equations? What are the conditions for having a solution. Do they depend on the base field? -- ' ==== Subject: Re: similarity over a field <29190119.1139412116714.JavaMail.jakarta@nitrogen.mathforum.org> <1867713.1139395736595.JavaMail.jakarta@nitrogen.mathf >> orum.org>, >>Let A,B be two matrices over an infinite field F, >> such that they are >>similar(not necessarily over F)-i mean there exists >> the matrix U(not >>neccesary over F) such that UAU^{-1}=B. Prove that >> there exists U over >>F such that UAU^{-1}=B. >> Rather than prove this, I will prove the closely >> related proposition: >> This is a homework exercise for eugene, who >> who decided to have someone >> do it for him/her. Moreover, he/she was too lazy >> azy to even ask for >> help, simply copying the statement and hoping >> ing someone would answer >> it for him. >> The proof is left as an easy exercise for the reader. >I just wanted to figure out the problem that interests me and didn't >mind to copy my homework : you would laugh if you see my >homework-it's much more easier... I thought one of the purposes of >the such a forum is to discuss problem rather then blaming other >users in copying-Sorry, if my comprehension isn't correct, i'll try >to reform. > You want help? Ask for help. You know, as in I am trying to solve > this problem. This is what I've done so far. This is where I am > stuck. Can someone give me a hint? Rather than just writing down the > problem. You are not giving an assignement to the readers of this > newsgroup. You are asking for help. > Now, as for a hint, consider the system of linear equations > XA-BX=0. Does it have a solution? Why? >For example it >has a solution-the matrix U. You must have meant the solution over >F, supposing that A and B are over F ? > What do you know about systems of linear equations? What are the > conditions for having a solution. Do they depend on the base field? > -- > 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 As an alternative method to those already recommended, you can prove it using the rational canonical form of the matrices (assuming you know what this is). In fact, in this case you don't even need to assume your field is infinite. As a hint for how to do this, suppose that F is a subfield of K and that A and B are similar over K. Then prove that a rational canonical form over K is a rational canonical form over F, and use the uniqueness of rational canonical forms. Mike ==== Subject: Re: similarity over a field > As an alternative method to those already recommended, you can prove it > using the rational canonical form of the matrices (assuming you know > what this is). In fact, in this case you don't even need to assume > your field is infinite. As a hint for how to do this, suppose that F > is a subfield of K and that A and B are similar over K. Then prove > that a rational canonical form over K is a rational canonical form over > F, and use the uniqueness of rational canonical forms. What _is_ the rational canonical form? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland ==== Subject: Re: similarity over a field You can look for example here http://mathworld.wolfram.com/RationalCanonicalForm.html to find out what is the rational canonical form Message was edited by: eugene ==== Subject: Re: similarity over a field <29190119.1139412116714.JavaMail.jakarta@nitrogen.mathforum.org> This doesn't quite do it for you need a non-singular U. I know I once did prove this, but I cannot now recall the proof. But it is why you may actually require the field to be infinite. Perhaps it goes something like this. If the determinant vanishes identically on the space of solutions over the small field, then on that linear space it is identically 0 and hence vanishes on any extension field. ==== Subject: Re: similarity over a field >This doesn't quite do it for you need a non-singular U. I know I once >did prove this, but I cannot now recall the proof. But it is why you >may actually require the field to be infinite. Perhaps it goes >something like this. If the determinant vanishes identically on the >space of solutions over the small field, then on that linear space it >is identically 0 and hence vanishes on any extension field. Yes, that's exactly it: if {U_i} are a basis of the space of solutions for the system of equations, det(sum_i x_i U_i) is a polynomial in the variables x_i, which can't be the 0 polynomial because it has a nonzero value for some choice of the x_i in the extension field. If the field F is infinite, such a polynomial can't be identically 0 on F^n. But the result is true even over finite fields: if two matrices over any field F are similar over an extension field, then they are similar over F. This can be done using Rational Canonical Form. See e.g. Herstein, Topics in Algebra, sec. 6.7. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: Re: S^n is simply connected, n>=2 <%h_Ef.17650$_S7.16145@newssvr14.news.prodigy.com> <5c5Gf.42317$dW3.31467@newssvr21.news.prodigy.com> to prove that q doesn't hit all points of S^2? > That's correct. However, there do exist continuous maps from S^1 onto > S^2 (that is, maps that pass through all the points on S^2). The purpose > of the exercise is to show that any map (such as one of these surjective > maps) can be homotoped to one that doesn't hit all the points of S^2. > Once that's done, then you can retract to a point. > Dale. ==== Subject: Re: S^n is simply connected, n>=2 > to prove that q doesn't hit all points of S^2? >>That's correct. However, there do exist continuous maps from S^1 onto >>S^2 (that is, maps that pass through all the points on S^2). The purpose >>of the exercise is to show that any map (such as one of these surjective >>maps) can be homotoped to one that doesn't hit all the points of S^2. >>Once that's done, then you can retract to a point. >>Dale. Let's recall the original problem: Take S^n be the unit sphere surface of R^{n+1}. For any loop p in S^n, find a loop q in R^{n+1} such that (1) p and q have the same basepoint; (2) q consists of finitely many line segments; (3) ||p(s)-q(s)||<=1 for all 0<=s<=1, from which deduce that S^n is simply connected for n>=2. (I note with surpise that I had forgotten that we were talking about any n>=2, rather than S^2 alone ... not that it matters for the proof. I'll return to the usage of S^n rather than S^2 and apologize for my lapse of memory) Note that q consists of finitely many line segments (projected radially to the sphere S^n, of course). Thus, q consists of the union of a finite number of geodesic (i.e., great-circle, in the case n=2) arcs: q = A_1 union A_2 union ... union A_k The complement of q in S^n then consists of the intersection: (S^n A_1) intersect (S^n A_2) intersect ... intersect (S^n A_k). Each of the sets S^n A_j is open and dense in S^n, which is a complete metric space. The Baire category theorem says that a countable intersection of such sets is nonempty, so we're done. Dale ==== Subject: Clocks and calendars Since this thread on measuring time passage has deteriorated into a discussion of relativity; it's time to rethread: Clocks and pendulums measure _periods_ of time: Calendars, and diaries measure the _passage_ of time. Don ==== Subject: Re: Clocks and calendars > Since this thread on measuring time passage has deteriorated into a > discussion of relativity; it's time to rethread: > Clocks and pendulums measure _periods_ of time: Calendars, and diaries > measure the _passage_ of time. > Don That's right. Different devices for different user features. Clocks with repeating representations give a greater dynamic range -- with one device you can easily tell the difference between any of the 86,400 seconds in a day, but aren't good for logging one-time events. Calendars and diaries are good for storing information pertaining to unique events but have got crappy resolution. PD ==== Subject: Re: Clocks and calendars > Since this thread on measuring time passage has > deteriorated into a discussion of relativity; it's > time to rethread: Pun intended? > Clocks and pendulums measure _periods_ of time: > Calendars, and diaries measure the _passage_ of time. > Don ==== Subject: Re: Clocks and calendars message : Since this thread on measuring time passage has deteriorated into a : discussion of relativity; it's time to rethread: : : Clocks and pendulums measure _periods_ of time: Calendars, and diaries : measure the _passage_ of time. : : Don : Nothing measures time, but the passage of time is indicated using arbitary units applied to a start and stop event. Smolley ==== Subject: Re: Clocks and calendars : Since this thread on measuring time > passage has deteriorated into a > : discussion of relativity; it's time to > rethread: > : Clocks and pendulums measure _periods_ > of time: Calendars, and diaries > : measure the _passage_ of time. > : Don > Nothing measures time, but the passage > of time is indicated using arbitary > units applied to a start and stop event. Starting and stopping of what? > Smolley ==== Subject: Re: Clocks and calendars > Since this thread on measuring time passage has deteriorated into a > discussion of relativity; it's time to rethread: > Clocks and pendulums measure _periods_ of time: Calendars, and diaries > measure the _passage_ of time. > Don *Plonk* ==== Subject: Re: JSH: My mistakes I think it's a problem with your presentation. When you think you have an idea how about posting it saying I found something that looks interesting but it seems to contradict established mathematics. Where is my mistake? If you did that you might get a friendlier response. ==== Subject: Re: JSH: My mistakes um, what in Hell was the mistake? is this some sort of plausible deniability? was there a missing link to some prior exposition? > I think it's a problem with your presentation. When you think you have > an idea how about posting it saying I found something that looks > interesting but it seems to contradict established mathematics. Where > is my mistake? If you did that you might get a friendlier response. thus: can't make heads or tails of it. either a) you're becoming dystraught, or b) you've got an interesting way of constructing the dodecaXIasteron (or cuboctahedron .-)... I mean, a 6D point is constructed as thirty tetrahedra !?!... I couldn't get the URL to work, initially, but maybe it will, now. http://mathforum.org/kb/thread.jspa?forumID=130&threadID=357139&messageID=10 92489#1092489 > Six dimensions to three dimensions with Synergetics Coordinates is shown > in the MathForum geometry.research URL above if you can get it to work. > A six-dimensional point is shown as thirty tetrahedrons. thus: I so agree.... nouns are neccesary to keep every thing from verbing at once. thus quoth: Space is an emergent property of time, mass and the laws of physics. Space has no separate existence. Any higher time integrals than velocity are nigh meaningless... only good for avoiding collisions. ;>) thus: in practice, no-one has a problem with this part of your expose, regardless of any claims of einsteinmaniaologists (cosmic reverberaration of the Sound of one hand, Clapping .-)... in other words, it's really some thing that stands on its own, without need of a new cosmogyny, or New Age storytelling. --Give Earth a Trickier Dick Cheeny -- out of office, after gigayears! http://tarpley.net/bush8.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rand.org/publications/randreview/issues/rr.12.00/ http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html ==== Subject: Re: JSH: My mistakes > Yes, I did it again, went off with claims of easy proofs which turned out > to be wrong, as I fought it out over an approach that just didn't work. > Big deal. Well, that's all that you have been doing for the past ten years or so: claim that you have a proof for some mathematical problem, being told over and over again that you are wrong, only to acknowledge, in the end, that you have been wrong all along. It will no doubt happen again several times, for you are either mentally disturbed, and therefore incapable to realize how stupid you look adopting such behavior, or else you are a complete asshole who gets a kick out of trying people's patience. ==== Subject: Re: JSH: My mistakes >Yes, I did it again, went off with claims of easy proofs which turned >out to be wrong, as I fought it out over an approach that just didn't >work. Big deal. >I don't doubt that as they've done in the past there will be posters >who will go on and on about me screwing up and boldly claiming I had >these dramatic proofs when I was just wrong. No. Would people do that? I mean really, that's simply not appropriate. Just because everything you've said for the last 10 years has been either not-even-wrong, wrong, or trivial... >And admitting I am wrong! >Wow, you should see the heat that flows over my just coming out and >saying that I screwed up. Hint. The heat doesn't come from your admitting you screwed up. It comes from your _continually_ screwing up, over and over, all the while insisting that _this_ time you're right - no matter how many times in the past it's turned out you were wrong, it's always true that anyone disputing your _current_ claims is just lying to protect the status quo. >I get made fun of for that as well, as if you people are perfect. >Nope. You just don't try hard enough. Sure, if you want to never have >to back-pedal and admit mistakes, you can work very hard to never make >them, look very carefully over every single argument you come up >with--and then just sit on them--as try as hard as you can, you will >fail, and screw-up if you dare to put it out there. >I do dare. I love it. And I make a lot of mistakes along the way. >So yeah, the other side is to say though that I screw up a lot. And I >do. Lots of claims of proofs where I've had to back-track as I was >wrong. >But that's extreme mathematics. That's one word for it. >You push the limit. Forget about social crap so you don't care so much >about the little people who will get all hot and bothered about your >mistakes. Those are the same as the little people who are all going to be fired for claiming you were wrong, right? So now it turns out you _were_ wrong. Are they still going to be fired? (More to the point: You were wrong _again_. Are we all going to be fired the _next_ time we say you're wrong? Just curious.) >Also, when you're arguing with people, emotion can get the best of you >and you stop thinking straight as you get defensive and look for any >kind of explanation that you think defends your case. Speak for yourself. This is not the way most of us behave. >In any event, I had some claims of proof over the weekend that were >just wrong. Went really wild Sunday with quite a lot of crap and a lot >of crap arguments where I was arguing positions that just didn't work >like I thought as I was happy as hell, telling myself how brilliant >these little proofs were. >And they weren't proofs. >James Harris David C. Ullrich ==== Subject: Math Teacher Arrested at JFK!!! They say that at New York's Kennedy airport today, an individual later discovered to be a public school teacher was arrested trying to board a flight while in possession of a ruler, a protractor, a setsquare, a slide rule, and a calculator. At a morning press conference, Attorney General Alberto Gonzalez said he believes the man is a member of the notorious Al-gebra movement. The FBI is charging him with carrying weapons of math instruction. Al-gebra is a fearsome cult, Gonzalez said. They desire average solutions by means and extremes, and sometimes go off on tangents in a search for absolute value. They use secret code names like 'x' and 'y' and refer to themselves as 'unknowns,' but we have determined they belong to a common denominator of the axis of medieval with coordinates in every country. As the Greek philanderer Isosceles used to say, 'there are 3 sides to every triangle.' When asked to comment on the arrest, President Bush said, If God had wanted us to have better weapons of math instruction, he would have given us more fingers and toes. ==== Subject: Re: Math Teacher Arrested at JFK!!! Are you really so antilogarithmic? Do you really need to fly off on such hyperbolic tangents? ==== Subject: Inclusion or Exclusion of numbers from a list... I have written a computer program to extract a list classes from patent documents, returned from a search of all patents. Basically each patent fits into a class which describes the patent. E.G. class 414 might be digital cameras etc. So basically I have a list of classes, from the search, and how often each of them occurs. like: 234 -> 30 times 543 -> 28 times 676 -> 21 times 656 -> 17 times 343 -> 12 times 124 -> 4 times 455 -> 3 times So I was wondering if there is a mathematically sound way of calculating a cut off value, that the patents from classes below that value get excluded from the search results. Above for example, the calculated value might be 10, which would mean that patents from classes with below 10 hits, (124 and 455 in this case) are deemed irrelevant and exluded from the search results. It is basically a system for filtering out irrelevant documents. Maybe something involving the mean and/or standard deviation? Any help would be greatly appreciated. Sophie. ==== Subject: Re: Inclusion or Exclusion of numbers from a list... > I have written a computer program to extract a list classes from > patent documents, returned from a search of all patents. Basically each > patent fits into a class which describes the patent. E.G. class 414 > might be digital cameras etc. > So basically I have a list of classes, from the search, and how often > each of them occurs. like: > 234 -> 30 times > 543 -> 28 times > 676 -> 21 times > 656 -> 17 times > 343 -> 12 times > 124 -> 4 times > 455 -> 3 times > So I was wondering if there is a mathematically sound way of > calculating a cut off value, that the patents from classes below that > value get excluded from the search results. > Above for example, the calculated value might be 10, which would mean > that patents from classes with below 10 hits, (124 and 455 in this > case) are deemed irrelevant and exluded from the search results. It is > basically a system for filtering out irrelevant documents. You haven't stated your objective very clearly. Irrelevant to what? I thought this was a sort of the entire patent database, so what does irrelevant mean? - Randy ==== Subject: Re: Inclusion or Exclusion of numbers from a list... Its not a sort of the entire database, but of results of a search. For example if a user searches for patents containing the search string digital camera, a list of all patents containing the string will be returned, many of which are in classes relating to digital cameras, but some of which just happen to contain that search string and have nothing to do with what the user is searching for, ie Digital Cameras. What I need is a means by which to filter out patents from these irrelevant classes. As these irrelevant classes will occur less often, they will tend to be at the bottom of a list like the one above, so I was wondering if there is a sound robust way of calculating this cutoff? ==== Subject: Re: Inclusion or Exclusion of numbers from a list... > Its not a sort of the entire database, but of results of a search. For > example if a user searches for patents containing the search string > digital camera, a list of all patents containing the string will be > returned, many of which are in classes relating to digital cameras, but > some of which just happen to contain that search string and have > nothing to do with what the user is searching for, ie Digital Cameras. > What I need is a means by which to filter out patents from these > irrelevant classes. > As these irrelevant classes will occur less often, they will tend to be > at the bottom of a list like the one above, so I was wondering if there > is a sound robust way of calculating this cutoff? Probably. What you want is something like choose a list of classes that contains X% of the matches, say 90%. A heuristic I'd try would be to look at how much each new value represents of the total so far. 234 -> 30 times (change = 30/30 = 1.0) 543 -> 28 times (change = 28/58 = 0.48) 676 -> 21 times (21/79 = 0.27) 656 -> 17 times (17/96 = 0.18) 343 -> 12 times (12/108 = 0.11) 124 -> 4 times (4/112 = 0.036) 455 -> 3 times (3/115 = 0.026) So you could if you set a cutoff somewhere below 10% (stop adding classes if they change the total so far by less than 10%, or maybe 5%), then you eliminate those last two classes. You probably want to experiment with a bunch of different searches to find a cutoff that seems right to you. As to sound, I think the theoretical basis would be found in the field of statistics called sequential estimation, but I'm not very knowledgable about that field. What you're trying to find is the sequential cutoff that ensures (with high probability) that you have the top X% of the classes. - Randy ==== Subject: Re: Well Ordering the Reals David R Tribble said: > David R Tribble said: >> See, Brian, Tony thinks any set of finite naturals must be finite >> itself. My analogy was that any box containing red apples must be >> red itself. >> Needless to say, Tony does not like this logical analogy. > That's because the analogy is not logical. It's an attempt to paint me with > quantifier dyslexia, when my argument is entirely different from what you > describe. If I thought anything like that, then I would probably say that the > set of reals in [0,1] is finite, since all values are finite, but I have never > made any such claim, because it's not true. > Which brings up an interesting question. Why is it that [0,1] is an > infinite set containing only finite reals, but N={0,1,2,...}, which > contains only finite naturals, is only a finite set? > Both sets have similar properties, i.e., any member of each set implies > the existence of another member in the set: for any real r in the set > (0,1], another real r/2 also exists in the set; likewise, for any > natural n in N, 2n also exists in the set. So while it's obvious that > the first set is infinite, you claim that the second set is only > finite, which is obviously false. If it were obviously false, then I'd agree, but it's not obvious at all, and in fact, seems obviously true to me. But, it's not good enough to say it's obvious. You want a rationale, and I'll give it again, for the nth time. The set of reals is dense in the reals (obviously), so there is no measurable difference between successive reals. The set of naturals is sparse, with a constant unit difference between successive members. The range of a set, divided by the average difference between successive members, give the size of the set. So, when you have an interval of 1, divided by an average increment of 0 in the reals, you have 1/0=oo and an infinite set. But, if you have an average difference of 1, as you do in the naturals, then the size of the set is the range divided by 1, which is the range of the set. If the range is the greatest possible difference between elements, and the greatest possible difference between any two naturals can only be finite, then one has a finite range, and given the equality between range and set size, one has a finite set. If you start at 1 and count up, at any given point the natural you just added is the size of the set so far. So the size of the set is no infinite, until you have started adding infinite values to the set. Could this be stated any more clearly? > I have given several different > proofs of the necessary finiteness of any set of finite naturals, ... > None of which hold up under the rigors of logic. Like you would know. You don't even seem to remember what I have said dozens of times, much less have any deep grasp of logic. You just keep all your red apples in your red box, and enjoy your red set theory. I'll study quantity, -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > David R Tribble said: > David R Tribble said: >> See, Brian, Tony thinks any set of finite naturals must be >> finite itself. My analogy was that any box containing red >> apples must be red itself. >> Needless to say, Tony does not like this logical analogy. > > That's because the analogy is not logical. It's an attempt to > paint me with quantifier dyslexia, when my argument is entirely > different from what you describe. If I thought anything like > that, then I would probably say that the set of reals in [0,1] is > finite, since all values are finite, but I have never made any > such claim, because it's not true. Which brings up an interesting question. Why is it that [0,1] is > an infinite set containing only finite reals, but N={0,1,2,...}, > which contains only finite naturals, is only a finite set? Both sets have similar properties, i.e., any member of each set > implies the existence of another member in the set: for any real r > in the set (0,1], another real r/2 also exists in the set; > likewise, for any natural n in N, 2n also exists in the set. So > while it's obvious that the first set is infinite, you claim that > the second set is only finite, which is obviously false. If it were obviously false, then I'd agree, but it's not obvious at > all, and in fact, seems obviously true to me. So TO claims that it is obviously true that one can have a bijection between what he claims is a finite set and what he calims is an an inifinite set! And TO claims says standard math has problems! But, it's not good > enough to say it's obvious. You want a rationale, and I'll give it > again, for the nth time. The set of reals is dense in the reals > (obviously), so there is no measurable difference between successive > reals. The set of naturals is sparse, with a constant unit difference > between successive members. The range of a set Does not exist for the sets in questin, since, as TO defines range, only finite sets can have ranges. I have given several different proofs of the necessary finiteness > of any set of finite naturals, ... None of which hold up under the rigors of logic. > Like you would know. Like many of us would know. While we do not all claim to be supermathematicians such as those few who really undersand Wiles proof of FLT, it does not take anywhere near that much to see the gaping holes in TOmatics, and most of us qualify for that. > You don't even seem to remember what I have said > dozens of times, much less have any deep grasp of logic. When TO keeps repeating stale arguments that have been shown false many times, there is not much point in remembering them, as he is sure to post them again soon. > You just > keep all your red apples in your red box, and enjoy your red set > theory. I'll study quantity, truth value, and symbolic It would be a refreshing change if TO would do some serious studying instead of merely spoutng over and over again the same tired falsehoods. ==== Subject: Re: Well Ordering the Reals David R Tribble said: > David R Tribble said: >> Your T-riffic 0:111...111 has two ends, yet you claim it somehow also >> has an infinite number of digits, which is a contradiction in terms. > There is no contradiction. If the index of the leftmost bit is infinite, then > the string represents an infinite value. That should be obvious. > That's exactly like saying an apple sits at the leftmost edge of an > endless table. If the table really is endless, there is no leftmost > edge. Likewise, if the number of bits really is infinite, there is no > leftmost bit. > That's the contradiction; there is no last bit in an infinite string of > bits. That should be obvious. That is true, and as I've said, the T-riffics either have an infinite string of all 1's or all 0's to the left of the leftmost limit point. But, if the table is truly infinite, you can pick points which are infinitely far apart, and if you define the pattern between them as a repeating pattern, it is quite feasible to work with the strings, without a clear concept of exactly how long they are, whether it be finite or infinite. Do you have a problem with saying that 0:1010...1010.1010... is 1/3 of N, where N=2^x, and x is the limit point? If I say the limit point is at 100, then N=2^100, and the T-riffic number is 1/3 of that. Does that finite example give you any grief, or does it just become a problem when someone says the limit point is at some infinite index? I don't see that it matters what the limit points are. Why do you? -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > That is true, and as I've said, the T-riffics either have an infinite > string of all 1's or all 0's to the left of the leftmost limit point. So how does one add 1 to the string of ALL 1's? > I don't see that it matters what the > limit points are. Why do you? TO-numbers cannot support a standard arithmetic as described. ==== Subject: Re: Well Ordering the Reals Virgil said: > imaginatorium@despammed.com said: > Pure Box Theory: >> Let's say I have a box. This box holds red boxes. Now I claim that >> since the first box contains only red boxes, the first box itself must >> be red. > > Brian Chandler >> I think your box theory would be much more convincing with large >> instead of red. Everyone (including Tony) can see that a shed of >> green apples isn't necessarily green*. But obviously, only a large box >> can contain a large box. > > Trying to use color as a property in this analogy is David's attempt to > make me > look stupid and obscure the situation. It's typical false QD > accusation, used > in desperation. > > See, Brian, Tony thinks any set of finite naturals must be finite > itself. My analogy was that any box containing red apples must be > red itself. > > Needless to say, Tony does not like this logical analogy. Well, I think he has a point. We can all see that a box containing red > things does not have to be red. Equally, we can see that a box > containing Large things needs to be Large; a box containing only Tiny > things may be Tiny, but a box containing imponderably enormous things > needs to be imponderably enormous itself. Tony's basic problem is that he thinks that infinite somehow means > something in the latter category, rather than something like endless > - which is more clearly in the same category as red. Brian Chandler > http://imaginatorium.org > More clearly? So, you don't think infinite is a term which can be applied > to a quantity, and therefore to a mathemtical expression? I am not sure what > you think is infinite about the set of reals in a finite interval, then, > besides the quantity of entities included in the set. Red is a property > which > is not generally considered to be quantitative, although everything can be > quantified, and red really means an interval in the electromagnetic > frequency > spectrum. The color of the box has no effect on the contents, whereas the > size > of the box, its internal volume, determines directly what it can hold. No > finite box can hold an infinite number of finite equal boxes, since the > quotient of two finites is always finite. Redness is irrelevant to the > naturals. They have no color, but they DO have values, and those values are > separated by a constant finite difference. You can't have an infinite number > of > constant finite differences in a finite range, and you can't have an infinite > range without infinite differences. > But you do not need an infinite range to have an infinte set. > The set of values {1/(x+1): x in N} is an infinite set of finite values. > And unbounded sets do not have ranges in TO's sense anyway. If x must always be a finite natural, then there will always be a finite difference between successive members of this set. Only when x is infinite can 1/(x+1)-1/(x+2) be infinitesimal. Within the range of this function, from 0 to 1, there can only fit a finite number of finite differences between successive elements. When you truly have infinite x, then and only then do you have a condensation point at 0, and only then do you have an infinite set within that finite range. You DON'T need an infinite range to have an infinite set. Within [0,1], a finite range, there are truly an infinite number of reals, but the salient difference between the reals and the naturals is that of density. The reals are everywhere dense, whereas the naturals are everywhere sparse. Without SOME point of real density within the set, it is impossible to have an infinite set, within a finite value range, because the range divided by the average difference between successive values is equal to the number of values in the set. Unless that average difference is 0, the size of a set witha finite range will be some finite value, and the average difference can only be 0, if there are some finite differences, when an infinite number of 0 differences are added to the mix to bring the average down to an infinitesimal level. -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > Virgil said: imaginatorium@despammed.com said: > Pure Box Theory: Let's say I have a box. This box holds >> red boxes. Now I claim that since the first box contains >> only red boxes, the first box itself must be red. > > > Brian Chandler >> I think your box theory would be much more convincing with >> large instead of red. Everyone (including Tony) can >> see that a shed of green apples isn't necessarily green*. >> But obviously, only a large box can contain a large box. > > > Trying to use color as a property in this analogy is > David's attempt to make me look stupid and obscure the > situation. It's typical false QD accusation, used in > desperation. > > See, Brian, Tony thinks any set of finite naturals must be > finite itself. My analogy was that any box containing red > apples must be red itself. > > Needless to say, Tony does not like this logical analogy. Well, I think he has a point. We can all see that a box > containing red things does not have to be red. Equally, we can > see that a box containing Large things needs to be Large; a box > containing only Tiny things may be Tiny, but a box containing > imponderably enormous things needs to be imponderably enormous > itself. Tony's basic problem is that he thinks that infinite somehow > means something in the latter category, rather than something > like endless - which is more clearly in the same category as > red. Brian Chandler http://imaginatorium.org > More clearly? So, you don't think infinite is a term which > can be applied to a quantity, and therefore to a mathemtical > expression? I am not sure what you think is infinite about the > set of reals in a finite interval, then, besides the quantity of > entities included in the set. Red is a property which is not > generally considered to be quantitative, although everything can > be quantified, and red really means an interval in the > electromagnetic frequency spectrum. The color of the box has no > effect on the contents, whereas the size of the box, its internal > volume, determines directly what it can hold. No finite box can > hold an infinite number of finite equal boxes, since the quotient > of two finites is always finite. Redness is irrelevant to the > naturals. They have no color, but they DO have values, and those > values are separated by a constant finite difference. You can't > have an infinite number of constant finite differences in a > finite range, and you can't have an infinite range without > infinite differences. But you do not need an infinite range to have an infinte set. The > set of values {1/(x+1): x in N} is an infinite set of finite > values. And unbounded sets do not have ranges in TO's sense > anyway. If x must always be a finite natural, then there will always be a > finite difference between successive members of this set. Only when x > is infinite can 1/(x+1)-1/(x+2) be infinitesimal. Within the range of > this function, from 0 to 1, there can only fit a finite number of > finite differences between successive elements. How strange that TOmatics allows so many things impossible in standard mathematics but the prohibits things that are so trivially true in standard mathematics as the endlessness of the set of finite naturals. In sequences, endless = infinite and the sequence of naturals is endless. Or does TO again claim to have last value for that sequence? ==== Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: > David R Tribble said: >> Uh, no. I was describing your binary infinite naturals as >> p-adics. > > Despite the fact that they are distinctly NOT the p-adics? What is >> the point of that obfuscation? > Virgil said: >> HOW are they different form p-adics? You keep claiming a difference that >> is not immediately obvious. > It's not immediately obvious that there are limit points which define > digits at > infinite positions? Do the T-riffics have ellipses to the left or in the > middle, or both. Where are the ellipses in the p-adics? Define > immediately > obvious, and then explain how 1:000...000 and ....1111 look the same to > you. Well, ...111 is an infinite number because it has only one end, and > does not have a last digit at the other end. Your T-riffic 0:111...111 has two ends, yet you claim it somehow also > has an infinite number of digits, which is a contradiction in terms. There is no contradiction. > If one requires any sort of consistent arithmetic, there are oodles of > contradictions. > For example, TO insists on having a least significant digit position and > a most significant digit position and uncountably many in between. Uncountably many, with no first or last, as I said in my last post. There are implied digits, even if insignificant, forever in both directions. The bits are indexed by hyperintegers, if that's the right term. > If one starts out with the least significant digit being 1 and > alternating 1's and 0's with increasing significance, TO cannot > determine whether the most significant digit will be a 0 or a 1. That's right, but you're describing a 2-adic. A T-riffic must have some defined point where all bits to the left, forever, are either all 1's or all 0's. > Similarly, one can have an unending sequence of ever more significant > 1's ascending from the least significant position with all uncountably > many other digits zero, and there cannot be any next number after such > a number, or immediate successor to such a number, as there can not > be any least significant 0 digit position to change into a 1. Here, again, you are talking about using the finite bit positions in the definition of your string. Yeah, yeah, there is no largest finite. We all know that. One would think you'd stop trying to construct sets around a concept that doesn't work. Does the standard theory address this? Sure, it says the index set must be countable, so there's nothing beyond those bits. But, this is a copout of a way to deal with that problem. If one cannot define the point between finite and infinite, so what? That doesn't mean the infinite doesn't exist, and it doesn't mean bits can't exist in infinite positions. Sure, there is an issue with any set defined this way. So, don't declare limit points at aleph_0, and you won't run into a contradiction. Aleph_0 isn't a value in my theory anyway. It's the Twilight Zone, where there is no largest finite, or smallest infinite. It's not allowed to be used as a limit point, since all infinite limit points must be declared as a formula on N. Do you know of a formula on N that describes aleph_0? If so, use that, and you won't have a problem. -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > Virgil said: David R Tribble said: > Your T-riffic 0:111...111 has two ends, yet you claim it > somehow also has an infinite number of digits, which is a > contradiction in terms. There is no contradiction. > > If one requires any sort of consistent arithmetic, there are oodles > of contradictions. For example, TO insists on having a least significant digit > position and a most significant digit position and uncountably many > in between. > Uncountably many, with no first or last, as I said in my last post. TO claims his TO-numbers are of form sum(x=-oo->oo: 2^x a_x) for some index set, so let us restrict our attention to those TO-numbers for which a_N = 0 for all x < 0, and simply ignore all zeros to the right of the radix position. Each of this restricted set of numbers is representable in the form sum(x=0->oo: 2^x a_x). Now let us consider one of these numbers which has countably many 1's in the least significant postions (and uncountable sequence of them) and zero's elsewhere and we want to add 1 to it. What does this new number look like? If our set of indices were well ordered and uncountable, we can have a number with 1's up to the first limit ordinal position, and adding 1 puts zeros in their places and a 1 in the limit ordinal position with no problems, and similarly for other situations. But unless there is something like a well ordering, such endless strings of 1's cannot have successors. If one starts out with the least significant digit being 1 and > alternating 1's and 0's with increasing significance, TO cannot > determine whether the most significant digit will be a 0 or a 1. > That's right, but you're describing a 2-adic. WRONG! 2-adics do not have most significant digits. > A T-riffic must have some defined point where all bits to the left, > forever, are either all 1's or all 0's. When did that rule come into being? And what happens if all bits are 1 and you want to add 1? Do you end up with zero? If so, you do not have an orderd set of numbers. Similarly, one can have an unending sequence of ever more > significant 1's ascending from the least significant position with > all uncountably many other digits zero, and there cannot be any > next number after such a number, or immediate successor to such > a number, as there can not be any least significant 0 digit > position to change into a 1. Here, again, you are talking about using the finite bit positions in > the definition of your string. I am talking about the requirement the for each bit position, except for end digits, there must be both a next larger and a next smaller, and even for end digitrs there must bd a next one in one orr the other direction. This requires that the digit positions have a serial order. Such serial orders with two ends are necessarily finite in the Dedekind sense, and are, in any case, no more than countably infinite. > Yeah, yeah, there is no largest > finite. While true, that is not relevant here. It seems to be something pointless that TO brings up whenever he has run out of other things to say. > Does the standard > theory address this? Sure, it says the index set must be countable, > so there's nothing beyond those bits. But, this is a copout of a way > to deal with that problem. Not when standard theory can PROVE that the index set, if sequentially ordered, must be no more than countable. > If one cannot define the point between > finite and infinite, so what? That doesn't mean the infinite doesn't > exist, and it doesn't mean bits can't exist in infinite positions. It does if one can prove that a sequentially ordered set is at most countably infinite. And that is eminently provable in standard mathematics. For a sequentially ordered set with an end value, either largest or smallest, simple induction works:end value is first, next one in the direstion away from it is successor, and that gets all of them. For a sequential ordering with neither end, one merely splits it into two oppositely directed sequences each with one end value. Thus we have proved: THEOREM: In any standard set theory any sequencially ordered set is at most countably infinite. ==== Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: > David R Tribble said: >> But you are still willing to say oo+1/2 > oo, right? Is oo+1/2 < oo+1? > > Under some circumstances. Like I said, there are two ways to view the >> number >> circle. > Virgil said: >> No one else is viewing a number circle, as such a representation >> destroys the order property of the standard reals. > http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html Apparently you did not notice that oo+1=oo in R*. So adding oo to the > real number line does not change its behavior; it still does not act > like a finite number. The T-riffics do not depend on the number circle. With the compactification > at > oo, of course any finite change is relatively negligible, and leaves one at > the > same point on the diagram. That's kind of irrelevant. I just posted that to > refute Virgil's vacuous claim that no one looks at number circles. > Get it right, TO! > No one looks at number circles as representations of ordered sets. > The TO-numbers require an index set in which each element except the > firsts has an immediate predecessor, each element except the last has an > immediate successor, and has uncountably many members. Really, there is no last bit. All T-riffics either have an infinite string of 1's or of 0's, from some point leftward. Also, there is no first bit, since the T-riffics can represent not only infinite naturals, but infinite reals as well, and so there is an unending string of bits to the right as well. That is why I say the interpretation of a digital number should be sum(x=-oo->oo: 2^x a_x). I still don't see why you have a problem with variously defined limit points, and repeating strings between. Both you and Tribble seem to be entirely hug up on the index set being countable, but if that were the case, then there would be no bits in infinite positions, and there would be no infinite values that could be represented. So, it's crucial that the index set be uncountable. How else could one represent any exact infinite quantity? > TO has yet to provide any axiom system in which such a monstrosity can > exist and have consistent arithmetic. In any standard system it is > impossible. > Ergo, anything based on the existence of such a system which cannot > exist in standard mathematics should be ignored until an axiom system in > which it can exist has beep published. Did you curse, and then beep it out? Yes, I have a chunk of my head working on an axiomatic basis for my system, including a replacement for the axiom of choice, and possibly requiring a probabilistic logic system. So, you're right, until I have concocted such a system, you are under no obligation to accept my theorems, since they can't be conclusively proven. -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > Virgil said: David R Tribble said: > David R Tribble said: >> But you are still willing to say oo+1/2 > oo, right? Is >> oo+1/2 < oo+1? > > Under some circumstances. Like I said, there are two ways to >> view the number circle. > Virgil said: >> No one else is viewing a number circle, as such a >> representation destroys the order property of the standard >> reals. > http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.h > tml Apparently you did not notice that oo+1=oo in R*. So adding oo > to the real number line does not change its behavior; it still > does not act like a finite number. The T-riffics do not depend on the number circle. With the > compactification at oo, of course any finite change is relatively > negligible, and leaves one at the same point on the diagram. > That's kind of irrelevant. I just posted that to refute Virgil's > vacuous claim that no one looks at number circles. Get it right, TO! No one looks at number circles as representations of ordered sets. The TO-numbers require an index set in which each element except > the firsts has an immediate predecessor, each element except the > last has an immediate successor, and has uncountably many members. > Really, there is no last bit. All T-riffics either have an infinite > string of 1's or of 0's, from some point leftward. Then 0:111...000 is NOT a representatain of a TO numer? Also, there is no first bit, since > the T-riffics can represent not only infinite naturals, but infinite > reals as well, and so there is an unending string of bits to the > right as well. There must be a least significant bit used for non-zero digits in representing integer values. > That is why I say the interpretation of a digital > number should be sum(x=-oo->oo: 2^x a_x). That implies that there is a largest TO-number, the one for which ALL the a_x in your sum(x=-oo->oo: 2^x a_x) are equal 1. It also appears to imply that the range of x is the set of standard integers, which is only countably infinite. In which case, it is very like the 2-adics. > I still don't see why you > have a problem with variously defined limit points, and repeating > strings between. What TO manages not to see constitutes the majot part of mathematics. Unless each of TO's limit points is a limit ordinal, there can be no arithmetic, at least with the usual proprties of arithmetic, defineable on his sytem. > Both you and Tribble seem to be entirely hug up on > the index set being countable, Not at all. It need not be countable, but if there is to be anything like a self-consistent arithmetic on it, including anything like an extension of standard addition, the index set for integer digit positions must be at least well ordered. TO has yet to provide any axiom system in which such a monstrosity > can exist and have consistent arithmetic. In any standard system it > is impossible. Ergo, anything based on the existence of such a system which cannot > exist in standard mathematics should be ignored until an axiom > system in which it can exist has beep published. Did you curse, and then beep it out? Why should I bother to curse or beep out what fails of itself so miserably? > Yes, I have a chunk of my > head working > on an axiomatic basis for my system, including a replacement for the > axiom of choice, and possibly requiring a probabilistic logic system. > So, you're right, until I have concocted such a system, you are under > no obligation to accept my theorems, since they can't be conclusively > proven. Strictly speaking, they are not theorems until they have been proven! ==== Subject: Re: Well Ordering the Reals Virgil said: > Virgil said: David R Tribble said: > David R Tribble said: >> But you are still willing to say oo+1/2 > oo, right? Is >> oo+1/2 < oo+1? > > Under some circumstances. Like I said, there are two ways to >> view the number circle. > Virgil said: >> No one else is viewing a number circle, as such a >> representation destroys the order property of the standard >> reals. > http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.h > tml Apparently you did not notice that oo+1=oo in R*. So adding oo > to the real number line does not change its behavior; it still > does not act like a finite number. The T-riffics do not depend on the number circle. With the > compactification at oo, of course any finite change is relatively > negligible, and leaves one at the same point on the diagram. > That's kind of irrelevant. I just posted that to refute Virgil's > vacuous claim that no one looks at number circles. Get it right, TO! No one looks at number circles as representations of ordered sets. The TO-numbers require an index set in which each element except > the firsts has an immediate predecessor, each element except the > last has an immediate successor, and has uncountably many members. Really, there is no last bit. All T-riffics either have an infinite > string of 1's or of 0's, from some point leftward. > Then 0:111...000 is NOT a representatain of a TO numer? No it's not, but not for the reason I just mentioned. The leftmost 0 is really an implicit infinite string of 0's, so you could write it ....0:111...000.0...., but it's not a valid T-riffic number because the string between the zero point and the other limit point (the colon) is not a repeating string of bits. There is no telling how many 1's or 0's there are, until you define an intermediate limit point where the 0's become 1's. If you mean 0's in finite positions, I addressed the problem with that in another post today. Aleph_0 is not valid limit point, since it's not expressible in terms of any finite formula on N or 1. > Also, there is no first bit, since > the T-riffics can represent not only infinite naturals, but infinite > reals as well, and so there is an unending string of bits to the > right as well. > There must be a least significant bit used for non-zero digits in > representing integer values. Yes, for whole numbers the bits to the right of the zero point are either all 0's or all 1's. In the latter case, we can make them all 0's and add 1 to the number left of the zero point, maintaining the same value. Once you have all 0's to the right of the zero point, they are no longer significant, and the 2^0 bit is the least significant bit for a whole number. Notice I said that you have an infinite number of potentially significant bits when representing a REAL, not an integer, in which case you use negative bit indexes. > That is why I say the interpretation of a digital > number should be sum(x=-oo->oo: 2^x a_x). > That implies that there is a largest TO-number, the one for which ALL > the a_x in your sum(x=-oo->oo: 2^x a_x) are equal 1. How do you figure? The bits extend forever in either direction. There is no first or last bit, and no greatest or smallest value. > It also appears to imply that the range of x is the set of standard > integers, which is only countably infinite. In which case, it is very > like the 2-adics. In structure it is obviously different from the 2-adics, with the additional limit points allowing for specific infinite values and some basic arithmetic. Furthermore, while the adics may be defined as having countably many bits, this is not the case with the T-riffics, as bits may be defined to be at infinite positions, so we can represent infinite values witha finite representation. > I still don't see why you > have a problem with variously defined limit points, and repeating > strings between. > What TO manages not to see constitutes the majot part of mathematics. > Unless each of TO's limit points is a limit ordinal, there can be no > arithmetic, at least with the usual proprties of arithmetic, defineable > on his sytem. I have already shown you some basic arithmetic on these numbers, and it has absolutely nothing to do with the bogus limit ordinals. > Both you and Tribble seem to be entirely hug up on > the index set being countable, > Not at all. It need not be countable, but if there is to be anything > like a self-consistent arithmetic on it, including anything like an > extension of standard addition, the index set for integer digit > positions must be at least well ordered. And yet, the only explicitly well ordered uncountable set is the limit ordinals, and that is only because of its kludgy predecessor discontinuities. As far as I am concerned, the set of bits is all sequential, but due to standard definitions of countability, it is not considered well ordered, since it DOESN'T have such discontinuities. One can always identify the predecessor of any bit in a T-riffic, or it's not a T-riffic. And yet, bits exist in infinite positions. So, while well ordering may be considered an important criterion for an index set in standard theory, sequentiality is sufficient as far as I am concerned, and the problem of infinite numbers of predecessors is not a problem. What problem does that cause in this construction, specifically? TO has yet to provide any axiom system in which such a monstrosity > can exist and have consistent arithmetic. In any standard system it > is impossible. Ergo, anything based on the existence of such a system which cannot > exist in standard mathematics should be ignored until an axiom > system in which it can exist has beep published. Did you curse, and then beep it out? > Why should I bother to curse or beep out what fails of itself so > miserably? Do you know what the beep a joke is? ;) > Yes, I have a chunk of my > head working > on an axiomatic basis for my system, including a replacement for the > axiom of choice, and possibly requiring a probabilistic logic system. > So, you're right, until I have concocted such a system, you are under > no obligation to accept my theorems, since they can't be conclusively > proven. > Strictly speaking, they are not theorems until they have been proven! Strictly speaking, you are correct, but they follow from other concepts, so I consider them theorems, in the non-strict sense. I'll try to be more rigorous, but it'll take time to get an axiom set together, especially as I have to consider the logistical system itself in order to integrate the concepts of space, truth, quantity, and language. Did I leave anything out? Do I need an axiom of the kitchen sink? :) -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > Virgil said: > Really, there is no last bit. All T-riffics either have an > infinite string of 1's or of 0's, from some point leftward. Then 0:111...000 is NOT a representatain of a TO numer? No it's not, but not for the reason I just mentioned. The leftmost 0 > is really an implicit infinite string of 0's, so you could write it > ....0:111...000.0...., but it's not a valid T-riffic number because > the string between the zero point and the other limit point (the > colon) is not a repeating string of bits. Let us just consider the subset representing integral numbers, and that particular such number for which ALL bits are 1. What is its successor? There is no telling how > many 1's or 0's there are, until you define an intermediate limit > point where the 0's become 1's. If you mean 0's in finite positions, > I addressed the problem with that in another post today. There must be an integral TO-number with an endless but countable sequence of 1's for its least significant integral digits, but zeroes elsewhere. What is its successor? > Aleph_0 is > not valid limit point, since it's not expressible in terms of any > finite formula on N or 1. There are no valid limit points unless the integral indices are well ordered. Also, there is no first bit, since > the T-riffics can represent not only infinite naturals, but > infinite reals as well, and so there is an unending string of > bits to the right as well. > > There must be a least significant bit used for non-zero digits in > representing integer values. Yes, for whole numbers the bits to the right of the zero point are > either all 0's or all 1's. Then we can consider the possibilities of an arithmetic on those integer TO-numbers. > Notice I said that you have an > infinite number of potentially significant bits when representing a > REAL, not an integer, in which case you use negative bit indexes. Does this mean that there are only finitely many, in the Dedekind sense, significant (non-zero) bits for an integer? If so, how do they differ from standard binary integers? > That is why I say the interpretation of a digital number should > be sum(x=-oo->oo: 2^x a_x). > That implies that there is a largest TO-number, the one for which > ALL the a_x in your sum(x=-oo->oo: 2^x a_x) are equal 1. > How do you figure? The bits extend forever in either direction. There > is no first or last bit, and no greatest or smallest value. But then there can be no TO-whole number larger than sum(x=-oo->oo: 2^x), and no TO-real larger than sum(x=-oo->oo: 2^x), so that TO-numbers, unlike satndard numbers, HAVE a largest. It also appears to imply that the range of x is the set of standard > integers, which is only countably infinite. In which case, it is > very like the 2-adics. In structure it is obviously different from the 2-adics, with the > additional limit points allowing for specific infinite values and > some basic arithmetic. Except that any not-well ordered index set causes fatal problems as previously demonstrated. > Furthermore, while the adics may be defined as > having countably many bits, this is not the case with the > T-riffics, as bits may be defined to be at infinite positions, so we > can represent infinite values witha finite representation. The bits of TO's number system seem to be sequentially ordered in the sense that for each one except a possible end value ther is an immedeate next one in both directions, with at most one end value, and it has been proved several times that such a set of indices can be no more than countably infinite. > I still don't see why you have a problem with variously defined > limit points, and repeating strings between. What TO manages not to see constitutes the major part of > mathematics. Unless each of TO's limit points is a limit ordinal, there can be > no arithmetic, at least with the usual proprties of arithmetic, > defineable on his sytem. I have already shown you some basic arithmetic on these numbers, and > it has absolutely nothing to do with the bogus limit ordinals. TO has shown no such thing. For example, consider the TO-natural having digits of 1 in a endless sequence from the radix point upwards but being zero for all uncountably many other indices. How does one add 1 to such a number unless there is a next index after all those 1's which is a limit ordinal? Both you and Tribble seem to be entirely hug up on the index set > being countable, Not at all. It need not be countable, but if there is to be > anything like a self-consistent arithmetic on it, including > anything like an extension of standard addition, the index set for > integer digit positions must be at least well ordered. And yet, the only explicitly well ordered uncountable set is the > limit ordinals, and that is only because of its kludgy predecessor > discontinuities. As far as I am concerned, the set of bits is all > sequential, but due to standard definitions of countability, it is > not considered well ordered, since it DOESN'T have such > discontinuities. One can always identify the predecessor of any bit > in a T-riffic, or it's not a T-riffic. The point is can one always determine the next bit after a countable subset of indices, or more generally after any bounded above set of indices. Unless one can do that, one cannot define the successor of an arbitrary number. > And yet, bits exist in > infinite positions. So, while well ordering may be considered an > important criterion for an index set in standard theory, > sequentiality is sufficient as far as I am concerned, and the > problem of infinite numbers of predecessors is not a problem. What > problem does that cause in this construction, specifically? Suppose one can partition of the TO-number index set into an upper and lower subset, U and L respectively, both at least countably infinite, with each index in U greater than each index in L and every index in L l less than every index in U, and suppose further that this can be done so that U has no smallest member. There must be a TO number with 0 at every member of U and 1 at every member of L, and adding 1 to that number is impossible in TO-arithmetic, or anyone else's arithmetic as well. The only condition that would prohibit this sort of self-contradictory situation would be the well ordering of the index set. > TO has yet to provide any axiom system in which such a > monstrosity can exist and have consistent arithmetic. In any > standard system it is impossible. Ergo, anything based on the existence of such a system which > cannot exist in standard mathematics should be ignored until an > axiom system in which it can exist has beep published. Did you curse, and then beep it out? Why should I bother to curse or beep out what fails of itself so > miserably? > Do you know what the beep a joke is? ;) TO is enough of a joke for now! > Yes, I have a chunk of my > head working > on an axiomatic basis for my system, including a replacement for > the axiom of choice, and possibly requiring a probabilistic logic > system. So, you're right, until I have concocted such a system, > you are under no obligation to accept my theorems, since they > can't be conclusively proven. Strictly speaking, they are not theorems until they have been > proven! > Strictly speaking, you are correct, but they follow from other > concepts, so I consider them theorems, in the non-strict sense. I'll > try to be more rigorous, but it'll take time to get an axiom set > together, especially as I have to consider the logistical system > itself in order to integrate the concepts of space, truth, quantity, > and language. Did I leave anything out? Do I need an axiom of the > kitchen sink? :) If you feel compelled to include a kitchen sink, TO, there should be some sort of axiom stating how it is to fit into the rest of your theory. Kitchen sinks are no part of any standard mathematics, so do not appear in any standard axiom set. ==== Subject: Re: Well Ordering the Reals > Pure Box Theory: >> Let's say I have a box. This box holds red boxes. Now I claim that >> since the first box contains only red boxes, the first box itself must >> be red. > > Brian Chandler >> I think your box theory would be much more convincing with large >> instead of red. Everyone (including Tony) can see that a shed of >> green apples isn't necessarily green*. But obviously, only a large box >> can contain a large box. > > Trying to use color as a property in this analogy is David's attempt to make me > look stupid and obscure the situation. It's typical false QD accusation, used > in desperation. > > See, Brian, Tony thinks any set of finite naturals must be finite > itself. My analogy was that any box containing red apples must be > red itself. > > Needless to say, Tony does not like this logical analogy. > Well, I think he has a point. We can all see that a box containing red > things does not have to be red. Equally, we can see that a box > containing Large things needs to be Large; a box containing only Tiny > things may be Tiny, but a box containing imponderably enormous things > needs to be imponderably enormous itself. > Tony's basic problem is that he thinks that infinite somehow means > something in the latter category, rather than something like endless > - which is more clearly in the same category as red. > Brian Chandler > http://imaginatorium.org > More clearly? So, you don't think infinite is a term which can be applied > to a quantity, and therefore to a mathemtical expression? No, I think this is the root of your misunderstanding. In particular, you use infinite exactly as though it were a sort of quality that some (otherwise perfectly ordinary) numbers might have. This is exactly wrong. (I don't quite understand why you quote back More clearly - I just meant that it might be clearer to you that endless is not the sort of thing you seem to think infinite is.) > I am not sure what > you think is infinite about the set of reals in a finite interval, then, > besides the quantity of entities included in the set. It's really remarkably simple. I use a definition of what it means for a set to be infinite, and I test the set in question against the definition. The problem with waffle like ... > Red is a property which > is not generally considered to be quantitative, although everything can be > quantified, .... etc is that it depends on vagaries like being generally considered, which may be fine for religion or town planning, but is completely alien to mathematical thinking. For the record, (and the umpteenth time, I expect), here's how I would determine if the set of reals in the interval [0, 1] is finite or infinite. I use the (I think more intuitive) definition that a set is infinite if a 1-1 mapping from it into the natural numbers has no largest element in its range. (In other words, it can't be mapped 1-1 to an initial subset of the natural numbers, and of course I'm using mathematical natural numbers, none of your nonsense.) Um, suppose n were the largest element in the range; that means that all of the reals in [0, 1] have been mapped into {1, 2, 3, ... n }. Consider the subset of these reals that is {1, 1/2, 1/3, ... 1/n }; if not all of these have been mapped, immediate contradiction, but if all of these have been mapped, then no other numbers can have been mapped, since this bit is 1-1, and here's one number that hasn't: 1/pi - once again contradiction. constant finite differences in a finite range, and you can't have an infinite > range without infinite differences. So you have repeated ad nauseam. Many people have now tried to explain to you why your argument is wrong. Is there any chance you accept that you are not going to convince any mathematicians? (Try religionists or town planners...) But I think the basic problem is the I-word. You tell us that the set of finite natural numbers (which is actually what everyone else means by the natural numbers, but I'll play silly terminology for a moment) is finite but unbounded. Well, the F-word won't get us anywhere, either, but let me ask about the unbounded. Does this mean that somewhere in this unbounded set of finite natural numbers there is at least one individual finite natural number which is unbounded? Brian Chandler http://imaginatorium.org ==== Subject: Re: Well Ordering the Reals imaginatorium@despammed.com said: > imaginatorium@despammed.com said: > >> Pure Box Theory: >> Let's say I have a box. This box holds red boxes. Now I claim that >> since the first box contains only red boxes, the first box itself must >> be red. > > Brian Chandler >> I think your box theory would be much more convincing with large >> instead of red. Everyone (including Tony) can see that a shed of >> green apples isn't necessarily green*. But obviously, only a large box >> can contain a large box. > > Trying to use color as a property in this analogy is David's attempt to make me > look stupid and obscure the situation. It's typical false QD accusation, used > in desperation. > > See, Brian, Tony thinks any set of finite naturals must be finite > itself. My analogy was that any box containing red apples must be > red itself. > > Needless to say, Tony does not like this logical analogy. > > Well, I think he has a point. We can all see that a box containing red > things does not have to be red. Equally, we can see that a box > containing Large things needs to be Large; a box containing only Tiny > things may be Tiny, but a box containing imponderably enormous things > needs to be imponderably enormous itself. > > Tony's basic problem is that he thinks that infinite somehow means > something in the latter category, rather than something like endless > - which is more clearly in the same category as red. > > Brian Chandler > http://imaginatorium.org > More clearly? So, you don't think infinite is a term which can be applied > to a quantity, and therefore to a mathemtical expression? > No, I think this is the root of your misunderstanding. In particular, > you use infinite exactly as though it were a sort of quality that > some (otherwise perfectly ordinary) numbers might have. This is exactly > wrong. Infinite means larger than any finite. Finite is defined in terms of numbers. Therefore, infinite is defined in terms of numbers. You define it in terms of sets, and that's fine, as long as your definition does not contradict the numerical concept of infinity. Endless becomes rather vague in the standard treatment. It's like you are still stuck in Zeno's paradox, but that's been solved for a long time. It's time to apply infinite series to your concepts of infinity, and the infinite series of unit increments diverges to an infinite value. > (I don't quite understand why you quote back More clearly - I just > meant that it might be clearer to you that endless is not the sort of > thing you seem to think infinite is.) Because it is not at all clear that infinite is more like red than it is like large. Do you have infinitely red apples? Do you have infinitely large spaces? Which concept, largeness or redness, is more likely to be considered without end? Which concept, red or large, seems to go better with the concept of infinite? Infinity seems more clearly related to largeness than redness to me, so I don't know why you think that handwaving declarations of what's clear have any value in this discussion. It's not clear at all, but rather the opposite is clear. > I am not sure what > you think is infinite about the set of reals in a finite interval, then, > besides the quantity of entities included in the set. > It's really remarkably simple. I use a definition of what it means > for a set to be infinite, and I test the set in question against the > definition. The problem with waffle like ... waffle? not hardly. So, you don't consider an infinite set to contain an infinite quantity of elements? Whatever.....that's why you call the finite naturals an infinite set. > Red is a property which > is not generally considered to be quantitative, although everything can be > quantified, .... etc > is that it depends on vagaries like being generally considered, which > may be fine for religion or town planning, but is completely alien to > mathematical thinking. Uh, yeah, red is not a term generally applied mathemtically. That's why Tribble's comment was so useless. > For the record, (and the umpteenth time, I > expect), here's how I would determine if the set of reals in the > interval [0, 1] is finite or infinite. I use the (I think more > intuitive) definition that a set is infinite if a 1-1 mapping from it > into the natural numbers has no largest element in its range. (In other > words, it can't be mapped 1-1 to an initial subset of the natural > numbers, and of course I'm using mathematical natural numbers, none of > your nonsense.) > Um, suppose n were the largest element in the range; that means that > all of the reals in [0, 1] have been mapped into {1, 2, 3, ... n }. > Consider the subset of these reals that is {1, 1/2, 1/3, ... 1/n }; if > not all of these have been mapped, immediate contradiction, but if all > of these have been mapped, then no other numbers can have been mapped, > since this bit is 1-1, and here's one number that hasn't: 1/pi - once > again contradiction. Or, you could use concepts like the set size is the range divided by the average difference between successive members, which in the case of the reals is 0, giving an infinite set size for any finite range. But, maybe that's too quantitative for you, when considering quantitative sets? > constant finite differences in a finite range, and you can't have an infinite > range without infinite differences. > So you have repeated ad nauseam. Many people have now tried to explain > to you why your argument is wrong. Is there any chance you accept that > you are not going to convince any mathematicians? (Try religionists or > town planners...) Yeah, or apple vendors with red carts full of infinitely red apples. (snore) > But I think the basic problem is the I-word. You tell us that the set > of finite natural numbers (which is actually what everyone else means > by the natural numbers, but I'll play silly terminology for a moment) > is finite but unbounded. Well, the F-word won't get us anywhere, > either, but let me ask about the unbounded. > Does this mean that somewhere in this unbounded set of finite > natural numbers there is at least one individual finite natural > number which is unbounded? Oh, come on! Of course not. Any natural you identified IS a bound on the interval from 0 to itself. Stop trying to play QD with me. That's bull, and you know it. It's almost as bad as Tribble's apples. The set is unbounded because it is defined inductively and monotonically increasing. The set is finite due to the restriction of finiteness on the elements and the identity relationship between element count and value. That couldn't be simpler. The set theoretic approach conflicts with the quantitative approach in this instance, due to the inability of the set theoretic approach to deal properly with the relationships between element value and position in the set. I am working on a solution. It will probably take a few months more, as it is starting to involve questions of the basic underlying logical system, which needs to be expanded paorbabilistically to be fully integrated with quantitative and symbolic concepts. I am considering using P(x), a property of an element x, as a primitive to replace set membership, and defining sets in terms of properties. So, it's not a minor tweak to ZFC. It's a more integrative approach. Thoughts? > Brian Chandler > http://imaginatorium.org -- Smiles, Tony ==== Subject: Re: Well Ordering the Reals > imaginatorium@despammed.com said: > More clearly? So, you don't think infinite is a term which can be > applied > to a quantity, and therefore to a mathemtical expression? No, I think this is the root of your misunderstanding. In particular, > you use infinite exactly as though it were a sort of quality that > some (otherwise perfectly ordinary) numbers might have. This is exactly > wrong. > Infinite means larger than any finite. Finite is defined in terms of numbers. > Therefore, infinite is defined in terms of numbers. You define it in terms of > sets, and that's fine, as long as your definition does not contradict the > numerical concept of infinity. Endless becomes rather vague in the standard > treatment. Endless is exact and precise in the sense that a set is endless if and only if there is an injection from the set to a proper subset, as any such injection determines an endless sequence of values of that set. If TO does not understand endless in this precise sense, then he does not understand it at all. (I don't quite understand why you quote back More clearly - I just > meant that it might be clearer to you that endless is not the sort of > thing you seem to think infinite is.) In either case, endless or infinite, it implies and is implied by the existence of an injection form the entire set to a proper subset. I am not sure what > you think is infinite about the set of reals in a finite interval, then, > besides the quantity of entities included in the set. It's really remarkably simple. I use a definition of what it means > for a set to be infinite, and I test the set in question against the > definition. The problem with waffle like ... > waffle? not hardly. TO is an expert at waffling when presented with clear questions for which he has no satisfactory answers. > So, you don't consider an infinite set to contain an infinite quantity of > elements? Whatever.....that's why you call the finite naturals an infinite > set. TO has the cart before the horse again. Containing an infinite quantity of elements is not what we use to define either endless or infinite. We do it the other way round, so that when a set is infintie, or a sequence is endless, THEN we can say that it has an infinite 'number' of elements. > Or, you could use concepts like the set size is the range divided by the > average difference between successive members Why would any one waqnt to do anything so stupid as to base number of members on a non-existant quality like the range of an unbounded set? > You can't have an infinite number of > constant finite differences in a finite range, and you can't have an > infinite > range without infinite differences. So you have repeated ad nauseam. Many people have now tried to explain > to you why your argument is wrong. Is there any chance you accept that > you are not going to convince any mathematicians? (Try religionists or > town planners...) > Yeah, or apple vendors with red carts full of infinitely red apples. (snore) TO asleeep at the switch again? But I think the basic problem is the I-word. You tell us that the set > of finite natural numbers (which is actually what everyone else means > by the natural numbers, but I'll play silly terminology for a moment) > is finite but unbounded. Well, the F-word won't get us anywhere, > either, but let me ask about the unbounded. Does this mean that somewhere in this unbounded set of finite > natural numbers there is at least one individual finite natural > number which is unbounded? > Oh, come on! Of course not. If no natural is unbounded why must any natural be infinite? Any natural you identified IS a bound on the > interval from 0 to itself. Stop trying to play QD with me. That's bull, and > you > know it. It's almost as bad as Tribble's apples. The set is unbounded because > it is defined inductively and monotonically increasing. The set is finite Maybe TO-finite, but Dedekind infinite, which is the only infinite there is in standard mathematics. TO-finite and TO-infinite only apply in TO-matics, not in mathematics. ==== Subject: Re: Well Ordering the Reals It was a lie. And the more I saw, the more I hated lies. -- Captain Willard ... we can usually breathe a sigh of relief when a logarithm appears inside O-notation, because O ignores multiplicative constants. There is no difference between O(lg n), O(ln n), and O(log n), as n -> oo; similarly, there is no difference between O(lg lg n), O(ln ln n), O(log log n). -- Graham, Knuth, and Patashnik, _Concrete Mathematics_. Sound is motion. Motion, is sound. (It's not really sound.) Ross ==== Subject: Re: Well Ordering the Reals imaginatorium@despammed.com said: > >> Pure Box Theory: >> Let's say I have a box. This box holds red boxes. Now I claim that >> since the first box contains only red boxes, the first box itself must >> be red. > > > Brian Chandler >> I think your box theory would be much more convincing with large >> instead of red. Everyone (including Tony) can see that a shed of >> green apples isn't necessarily green*. But obviously, only a large box >> can contain a large box. > > > Trying to use color as a property in this analogy is David's attempt to make me > look stupid and obscure the situation. It's typical false QD accusation, used > in desperation. > > See, Brian, Tony thinks any set of finite naturals must be finite > itself. My analogy was that any box containing red apples must be > red itself. > > Needless to say, Tony does not like this logical analogy. > > Well, I think he has a point. We can all see that a box containing red > things does not have to be red. Equally, we can see that a box > containing Large things needs to be Large; a box containing only Tiny > things may be Tiny, but a box containing imponderably enormous things > needs to be imponderably enormous itself. > > Tony's basic problem is that he thinks that infinite somehow means > something in the latter category, rather than something like endless > - which is more clearly in the same category as red. > > Brian Chandler > http://imaginatorium.org > > More clearly? So, you don't think infinite is a term which can be applied > to a quantity, and therefore to a mathemtical expression? > No, I think this is the root of your misunderstanding. In particular, > you use infinite exactly as though it were a sort of quality that > some (otherwise perfectly ordinary) numbers might have. This is exactly > wrong. > Infinite means larger than any finite. So you say. I don't think that's a mathematically coherent definition. Even supposing you had a proper definition of finite (e.g. a set is finite if counting the elements against a ditty stops at a number name in the ditty), then the way to define infinite is not finite. By saying larger than you assume there is some a priori way you can compare some magnitude of sets both finite and infinite before you have actually defined these terms. You might notice, btw, that the above definition of finite (about stopping) is such that a set cannot be slightly infinite, or in some twilight zone between finite and infinite (etc etc) because a ditty either stops, or does not stop. Anyway, sorry, but this is all increasingly pointless. You refuse to read an elementary textbook about set theory, so you might at least know something about that which you wish to prove wrong... of finite natural numbers (which is actually what everyone else means > by the natural numbers, but I'll play silly terminology for a moment) > is finite but unbounded. Well, the F-word won't get us anywhere, > either, but let me ask about the unbounded. > Does this mean that somewhere in this unbounded set of finite > natural numbers there is at least one individual finite natural > number which is unbounded? > Oh, come on! Of course not. Any natural you identified IS a bound on the > interval from 0 to itself. Stop trying to play QD with me. That's bull, and you > know it. It's almost as bad as Tribble's apples. The set is unbounded because > it is defined inductively and monotonically increasing. So when you say it's unbounded, would an attempt to recite its elements against a ditty stop at a number name in the ditty or not? > The set is finite due > to the restriction of finiteness on the elements and the identity relationship > between element count and value. That couldn't be simpler. The set theoretic > approach conflicts with the quantitative approach in this instance, due to the > inability of the set theoretic approach to deal properly with the relationships > between element value and position in the set. I am working on a solution. It > will probably take a few months more, as it is starting to involve questions of > the basic underlying logical system, which needs to be expanded > paorbabilistically to be fully integrated with quantitative and symbolic > concepts. I am considering using P(x), a property of an element x, as a > primitive to replace set membership, and defining sets in terms of properties. > So, it's not a minor tweak to ZFC. It's a more integrative approach. Thoughts? You want honest thoughts? (I don't think so.) You are not doing mathematics, because you know in advance what answer you hope to find. Cross between religion and bad engineering - well good luck! Brian Chandler http://imaginatorium.org ==== Subject: Naturals Construction Set This is what Leopold Kronecker said (1886): 'Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk' (the whole numbers have been made by God, all the rest is work of humans) Though we feel sympathy for Kronecker's constructive stand, we do not agree with his statement that the naturals would be the work of a supernatural being. There are at least five methods to construct the natural numbers out of well ... something or nothing, but all by ourselves, as human beings. Method 1 -------- Assume the existence of the empty set {} (: well, it _exists_). Then define 0 := {} , 1 := {0} , 2 := {0,1} , 3 := {0,1,2} ... This is the common definition of the finite ordinals. Right? Observe that it's impossible to have 3 without 2, or 2 without 1, or 1 without 0 . In general, if you have n , then you have gone all the way through 0,1,2,3,4,5, ... ,n . And you have constructed all the numbers between 0 and n . Method 2 -------- There is nothing new in the following approach, which I could look up in the notes of an intuitionist mathematician (W. Peremans, Eindhoven). http://www.math.leidenuniv.nl/~naw/serie5/deel01/jun2000/pdf/peremans.pdf O is a symbol, / is a symbol. There are no other symbols. O is a string, / is a string. If a is a string then aO is a string. If a is a string then a/ is a string. Nothing else is a string. Definition of equality: O = O ; / = / . If a and b are strings and a = b then aO = bO and a/ = b/ . And there are no other rules. Definition of the naturals N. O is a formula. If the string a is a formula then a/ is a formula. O in N. If a in N then a/ in N. And there are no other rules for being a member of N. Then follows a bunch of theorems. At last we identify: O = 0 , O/ = 1 , O// = 2 , O/// = 3 , O//// = 4 , ... O//////////////// = 16 . And so on and so forth. Observe that it's impossible to have 3 without 2, or 2 without 1, or 1 without 0 . In general, if you have n , then you have gone all the way through 0,1,2,3,4,5, ... ,n . And you have constructed all the numbers between 0 and n . Method 3 -------- The constructivist's method 2 is quite similar to the barkeeper's method, at the time I was a regular visitor of the local pub. He kept a record of the beer mugs on my account by scribbling strokes '/' on a piece of paper, as follows: / , // , /// , //// , ///// , ... Such a procedure is called turven in Dutch (I suppose to score or keep a tally in English. It is noted that, with no special symbol for the empty string, the definition of a number zero is problematic) Method 4 -------- Suppose there is a finite set X . Then X has a finite cardinality n , where n is a natural number of course. Then it's impossible to have n without (n-1), being the cardinality of a subset of X , namely all of its members except one. Extending this observation gives rise to the following: if n is the cardinality of a finite set X, then there also exists a cardinality (n-1), and so on and so forth, until the process of having lower cardinalities is stopped at 0. Click on picture in: http://hdebruijn.soo.dto.tudelft.nl/fototjes/gefopt5.htm Method 5 -------- In our modern digital computers, the equivalent of the set of naturals (more precise: the materialization of it) is ranging from zero until, say 4294967295 (: unsigned 32 bit) or 2^63-1 (: signed 64-bit). Anyway, it's a closed range and there are no gaps in it. OK. Enough is enough. | With all five of the above constructive methods, its impossible to | have n without having (n-1), except when n = 0 . A set of naturals | where this is the case will be called: a Naturals Construction Set. ------------------------- Now let us count the number of descendants in a finite set of naturals { n,n-1, ... ,k, ... ,3,2,1,0 } , where each k has a descendant (k-1) except k = 0 . Obviously, this count turns out to be (n-1). With all of the above methods of creating the naturals, the finite set of n naturals obtained has the property that the number of descendants in that set is (n-1). The only member in the set that has no descendant is always the same element, namely zero. That remains so, irrespective of the question how large n may become. The following set has been proposed by Stephen@nomail.com as a set that would result in THE naturals. It is the union of two disjoint parts: O = { x | x is odd and x < n } E = { x | x is even and 2x < n } An example for n = 10 : {0,1,2,3,4,5,7,9} = {9,7,5,4,3,2,1,0} Whereby it is clear that the naturals 9 and 7 have no descendants. Thus the set by Stephen is a set of naturals, but does _not_ represent a set where the naturals can be _created_ in . Hence, according to the above definition, it is not a _construction_ set of naturals. There are holes in it: 1/3 of its members will _never_ have a descendant; it would have been impossible to construct these members from their descendants. See also the argument in: The crucial difference being that successor has been replaced now by descendant, which has the advantage that only zero has no descendant. Therefore this property is _independent_ of the magnitude of the set of all naturals in statu nascendi. But we know now that the construction set _must_ look like: {0,1,2,3,4,5,6,7,8,9, ... ,n} . And nothing else. The last hurdle is taking the limit for large n . What does it mean? Nothing. What we actually must define is: what it means that (k in N). Definition: (k in N) iff there can be constructed a natural n such that k in {0,1,2,3,4,5,6,7,8,9, ... ,n}, with other words such that: k <= n. Example: if (p in N), can it be concluded that (2.p in N) ? Yes. The worst case being that p = n . But then start making successors of n (: method 1-5) until we arrive at {0,1,2,3,4,5,6,7,8,9, ... ,2.n}. Any objections against this approach? Han de Bruijn ==== Subject: Re: Naturals Construction Set Han de Bruijn said: > This is what Leopold Kronecker said (1886): 'Die ganzen Zahlen hat der > liebe Gott gemacht, alles andere ist Menschenwerk' (the whole numbers > have been made by God, all the rest is work of humans) Though we feel > sympathy for Kronecker's constructive stand, we do not agree with his > statement that the naturals would be the work of a supernatural being. > There are at least five methods to construct the natural numbers out of > well ... something or nothing, but all by ourselves, as human beings. > Method 1 > -------- > Assume the existence of the empty set {} (: well, it _exists_). > Then define 0 := {} , 1 := {0} , 2 := {0,1} , 3 := {0,1,2} ... > This is the common definition of the finite ordinals. Right? > Observe that it's impossible to have 3 without 2, or 2 without 1, or 1 > without 0 . In general, if you have n , then you have gone all the way > through 0,1,2,3,4,5, ... ,n . And you have constructed all the numbers > between 0 and n . > Method 2 > -------- > There is nothing new in the following approach, which I could look up > in the notes of an intuitionist mathematician (W. Peremans, Eindhoven). > http://www.math.leidenuniv.nl/~naw/serie5/deel01/jun2000/pdf/peremans.pdf > O is a symbol, / is a symbol. There are no other symbols. > O is a string, / is a string. If a is a string then aO is a string. > If a is a string then a/ is a string. Nothing else is a string. > Definition of equality: O = O ; / = / . If a and b are strings and > a = b then aO = bO and a/ = b/ . And there are no other rules. > Definition of the naturals N. > O is a formula. If the string a is a formula then a/ is a formula. > O in N. If a in N then a/ in N. And there are no other rules for > being a member of N. Then follows a bunch of theorems. > At last we identify: O = 0 , O/ = 1 , O// = 2 , O/// = 3 , O//// = 4 , > ... O//////////////// = 16 . And so on and so forth. > Observe that it's impossible to have 3 without 2, or 2 without 1, or 1 > without 0 . In general, if you have n , then you have gone all the way > through 0,1,2,3,4,5, ... ,n . And you have constructed all the numbers > between 0 and n . > Method 3 > -------- > The constructivist's method 2 is quite similar to the barkeeper's > method, at the time I was a regular visitor of the local pub. He kept > a record of the beer mugs on my account by scribbling strokes '/' on > a piece of paper, as follows: / , // , /// , //// , ///// , ... > Such a procedure is called turven in Dutch (I suppose to score or > keep a tally in English. It is noted that, with no special symbol > for the empty string, the definition of a number zero is problematic) > Method 4 > -------- > Suppose there is a finite set X . Then X has a finite cardinality n , > where n is a natural number of course. Then it's impossible to have n > without (n-1), being the cardinality of a subset of X , namely all of > its members except one. Extending this observation gives rise to the > following: if n is the cardinality of a finite set X, then there also > exists a cardinality (n-1), and so on and so forth, until the process > of having lower cardinalities is stopped at 0. Click on picture in: > http://hdebruijn.soo.dto.tudelft.nl/fototjes/gefopt5.htm > Method 5 > -------- > In our modern digital computers, the equivalent of the set of naturals > (more precise: the materialization of it) is ranging from zero until, > say 4294967295 (: unsigned 32 bit) or 2^63-1 (: signed 64-bit). Anyway, > it's a closed range and there are no gaps in it. OK. Enough is enough. > | With all five of the above constructive methods, its impossible to > | have n without having (n-1), except when n = 0 . A set of naturals > | where this is the case will be called: a Naturals Construction Set. > ------------------------- > Now let us count the number of descendants in a finite set of naturals > { n,n-1, ... ,k, ... ,3,2,1,0 } , where each k has a descendant (k-1) > except k = 0 . Obviously, this count turns out to be (n-1). > With all of the above methods of creating the naturals, the finite set > of n naturals obtained has the property that the number of descendants > in that set is (n-1). The only member in the set that has no descendant > is always the same element, namely zero. That remains so, irrespective > of the question how large n may become. > The following set has been proposed by Stephen@nomail.com as a set that > would result in THE naturals. It is the union of two disjoint parts: > O = { x | x is odd and x < n } > E = { x | x is even and 2x < n } > An example for n = 10 : {0,1,2,3,4,5,7,9} = {9,7,5,4,3,2,1,0} > Whereby it is clear that the naturals 9 and 7 have no descendants. Thus > the set by Stephen is a set of naturals, but does _not_ represent a set > where the naturals can be _created_ in . Hence, according to the above > definition, it is not a _construction_ set of naturals. There are holes > in it: 1/3 of its members will _never_ have a descendant; it would have > been impossible to construct these members from their descendants. See > also the argument in: > The crucial difference being that successor has been replaced now by > descendant, which has the advantage that only zero has no descendant. > Therefore this property is _independent_ of the magnitude of the set of > all naturals in statu nascendi. But we know now that the construction > set _must_ look like: {0,1,2,3,4,5,6,7,8,9, ... ,n} . And nothing else. > The last hurdle is taking the limit for large n . What does it mean? > Nothing. What we actually must define is: what it means that (k in N). > Definition: (k in N) iff there can be constructed a natural n such that > k in {0,1,2,3,4,5,6,7,8,9, ... ,n}, with other words such that: k <= n. > Example: if (p in N), can it be concluded that (2.p in N) ? > Yes. The worst case being that p = n . But then start making successors > of n (: method 1-5) until we arrive at {0,1,2,3,4,5,6,7,8,9, ... ,2.n}. > Any objections against this approach? > Han de Bruijn It seems fine for finite naturals, which is undoubtedly all you are addressing. I am not sure I'd take this approach, when trying to define infinite naturals, since it rests on a notion of countability which I think constrains the concept somewhat. Personally, I think the naturals really should start at 1, since as you say, all naturals can be repesented as a string of tally marks, and the empty set of them is not considered, and so only one symbol is required. Plus, if you start at 1, it becomes immediately obvious that each n is at position n, and so if you want actually infinite positions in the set, and an actually infinite set, n must be infinite and you must have infinite values in the set. That's my thinking. We'll see what the standard-bearers have to say. :) -- Smiles, Tony ==== Subject: Re: Naturals Construction Set > Any objections against this approach? Han de Bruijn > It seems fine for finite naturals, which is undoubtedly all you are > addressing. I am not sure I'd take this approach, when trying to > define infinite naturals, since it rests on a notion of > countability which I think constrains the concept somewhat. As TO rejects all standard mathematics, including constructive notions, that does not conform to the dictates of his intuition, he has, so far, nothing to contribute to anything. > Personally, I think the naturals really should start at 1, since as > you say, all naturals can be repesented as a string of tally marks, > and the empty set of them is not considered, and so only one symbol > is required. Where they start is a matter of convention and definition, not of logic. TO-matic delusions snipped here > That's my thinking. We'll see what the standard-bearers have to say. > :) -- Smiles, > Tony ==== Subject: My old friend the SINC function > Real, physical quantites have uncertainties. That is > one of the fundamental properties of physics. And it > isn't just due to quantum considerations. Take an > average metal bar. It has no exact length, not down > to the precision of an atomic width or so. There are > temperature fluctuations and small forces from > Brownian motion which will cause that bar's length > to fluctuate. The atoms themselves are in motion which > is another cause of inexactness. As Randy describes himself, there are _fluctuations_ of all kind that will cause x = 0, when conceived as a physical quantity, to fluctuate. Meaning that x = 0 is actually to be _interpreted_ as x = 0 +/- delta, where delta is the uncertainity. But we don't need Randy's argument in order to see that there is kind of an uncertainity in _every_ realization of the real numbers. Take a look at the floating point quantities in a digital computer: 0.000000000000000000000000000000123597059137504570... |----------------------------||---------------------> oo material ideal Then, inside the computer at hand, _this_ value of x cannot possibly be distinguished from x = 0. But suppose we buy a somewhat better computer with extended precision, such that: 0.000000000000000000000000000000123597059137504570... |-------------------------------------||------------> oo material ideal Then suddenly the value which was supposedly equal to zero becomes just _close_ to 0 instead. We conclude that, in the material world, a real 0 cannot be distinguished from its (supposedly very small) environment. Now consider an old friend of mine, the function: sinc(x) = sin(x)/x for x <> 0 = 1 for x = 0 But NO ! We insist upon our Enduring Freedom and, as _mathematicians_, we like to define, instead, the following: S(x) = sin(x)/x for x <> 0 = 2 for x = 0 So I have the question: which of the above functions, S(x) or sinc(x), has an empirical counterpart? Is it possible that S(x) will _ever_ have an empirical counterpart? Just answer the question, please. Yes or No. Jean-Claude Arbaut, Jiri Lebl and Randy Poe all have been clever enough not to give a straight answer to this simple question. Who has the guts? Han de Bruijn ==== Subject: Re: My old friend the SINC function sinc(x) has a valid use: In Fourier optics, the far-field pattern of a square or rectangular aperture that projects a coherent light source is a product of squared sinc functions. For a reference, see Goodman Fourier Optics. ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >>Pluto would see a brighter sun if Jupiter nearly eclipses, eh? >>Gravitational lensing shows nothing but how a photon's path is >>permanently altered as per Isaac, when passing a massive object. It is >>quite obvious, that a photon following a line of equal gravitational >>potential (GR), returns to the path it had prior to the close encounter >>with the mass, and NO lensing (magnification or changed position) would >>be evident. > No, actually gravitational lensing can work according to Newton > (GR gives the same answer) >>Uh...wrong. Newton predicts 1+2gh/c^2; GR predicts 1+gh/c^2. >>Granted, the qualitative effect is similar. > The BaTh prediction for light falling from GPS clocks is exactly > equivalent to the fictitious 'GR correction'. > Work it out yourself Ghost, if you don't believe me. I already did, modeling the light as a ball of mass E/c^2 moving at lightspeed. The result is as I've stated. Granted, one can fudge a bit here; the ball could be mass 1/2 E/c^2 and have total energy E, and even the momentum would be right: E/c. However, this fix doesn't correlate all that well with other effects, some of which involve protons with energies of hundreds of GeV. > if a large object lies at certain distances between the observer > and the star. > But the main reason for the increase in brightness is > directly linked to c+v. >>Jim G >>c'=c+v > HW. > www.users.bigpond.com/hewn/index.htm > HW. > www.users.bigpond.com/hewn/index.htm -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <88apb3-et4.ln1@sirius.tg00suus7038.net In sci.math, HW@..(Henri Wilson) > : >>In sci.math, HW@..(Henri Wilson) >><2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com>: > On 5 Feb 2006 19:45:39 -0800, jgreenfield@seol.net.au >[snippage] >>Pluto would see a brighter sun if Jupiter nearly eclipses, eh? >>Gravitational lensing shows nothing but how a photon's path is >>permanently altered as per Isaac, when passing a massive object. It is >>quite obvious, that a photon following a line of equal gravitational >>potential (GR), returns to the path it had prior to the close encounter >>with the mass, and NO lensing (magnification or changed position) would >>be evident. No, actually gravitational lensing can work according to Newton > (GR gives the same answer) >>Uh...wrong. Newton predicts 1+2gh/c^2; GR predicts 1+gh/c^2. >>Granted, the qualitative effect is similar. > The BaTh prediction for light falling from GPS clocks is exactly > equivalent to the fictitious 'GR correction'. > Work it out yourself Ghost, if you don't believe me. > I already did, modeling the light as a ball of mass E/c^2 moving > at lightspeed. The result is as I've stated. Modeling light as being affected by gravity is a modification of Newtonian gravity. Consistent with the rest of Newton, that would also predict that light causes a gravitational effect on other masses. I don't buy the limit as m->0 argument. What that tells you is how arbitrarily small masses behave, not how massless objects behave. Reference: endless arguments in sci.math on the meaning of limits. I'm certainly willing to accept that if you postulate photons as behaving as if they have gravitational mass E/c^2, then you get lensing. But I wouldn't call that Newton's theory. By the way, doesn't that make the degree of deflection wavelength dependent? - Randy ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <88apb3-et4.ln1@sirius.tg00suus7038.net I don't buy the limit as m->0 argument. What that tells you is > how arbitrarily small masses behave, not how massless > objects behave. Reference: endless arguments in sci.math > on the meaning of limits. I am willing to bet that the folks who don't understand relativity won't be able to understand limits as the way modern calculus defines them. > I'm certainly willing to accept that if you postulate photons as > behaving as if they have gravitational mass E/c^2, then > you get lensing. But I wouldn't call that Newton's > theory. It is a kludge, and even then the actual photon mass - if any - is way less than E/c^2. > By the way, doesn't that make the degree of deflection > wavelength dependent? Yep. > - Randy ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> No, actually gravitational lensing can work according to Newton (GR gives the >> same answer) if a large object lies at certain distances between the observer >> and the star. >Really? Can you tell me which formula of Newton's applies to >the effect of gravity on light? > The Newtonian blueshift of light from a GPS satellite is exactly equivalent to > 45 us per day. > Work it out. With what equations? What equation describes Newtonian blueshift? How can you assert that an equation comes with a certain answer exactly if you've never done the calculation, never seen such a calculation, and don't even know what equation you'd use for that calculation? - Randy ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >>There is no Newtonian prediction of the deflection of rays of light. If >>you assume a point test mass, which light neither is nor is composed of, >>then you might get that result, but then what does GR get for the test mass? >> Light accelerates down a gravity well like anything else. >In Newtonian gravitation, the force on an object of mass m >in the presence of mass M is given by >F = GMm/r^2 >Please tell me how this applies to light to predict that light >accelerates down a gravity well like anything else. >Let's say M is the sun and I want to know the effect of >the sun on light from a He-Ne laser. Can you sketch out >the calculation for me? What's m? Actually, that can be done. If you assume that the relationship above remains valid at the limit of m -> 0, then you get radial acceleration of GM/r^2 and you can solve the equations of motion. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >>There is no Newtonian prediction of the deflection of rays of light. If >>you assume a point test mass, which light neither is nor is composed of, >>then you might get that result, but then what does GR get for the test mass? >> Light accelerates down a gravity well like anything else. >In Newtonian gravitation, the force on an object of mass m >in the presence of mass M is given by >F = GMm/r^2 >Please tell me how this applies to light to predict that light >accelerates down a gravity well like anything else. >Let's say M is the sun and I want to know the effect of >the sun on light from a He-Ne laser. Can you sketch out >the calculation for me? What's m? > Actually, that can be done. If you assume that the relationship above > remains valid at the limit of m -> 0, then you get radial acceleration > of GM/r^2 and you can solve the equations of motion. ..idiot. > Mati Meron | When you argue with a fool, > meron@cars.uchicago.edu | chances are he is doing just the same Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <828iu1dnh4km8e6f41gecqfd5o59eakso6@4ax.comThere is no Newtonian prediction of the deflection of rays of light. If >you assume a point test mass, which light neither is nor is composed of, >then you might get that result, but then what does GR get for the test mass? > Light accelerates down a gravity well like anything else. In Newtonian gravitation, the force on an object of mass m in the presence of mass M is given by F = GMm/r^2 Please tell me how this applies to light to predict that light accelerates down a gravity well like anything else. Let's say M is the sun and I want to know the effect of the sun on light from a He-Ne laser. Can you sketch out the calculation for me? What's m? - Randy ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <828iu1dnh4km8e6f41gecqfd5o59eakso6@4ax.com >There is no Newtonian prediction of the deflection of rays of light. If >you assume a point test mass, which light neither is nor is composed of, >then you might get that result, but then what does GR get for the test mass? > Light accelerates down a gravity well like anything else. > In Newtonian gravitation, the force on an object of mass m > in the presence of mass M is given by > F = GMm/r^2 > Please tell me how this applies to light to predict that light > accelerates down a gravity well like anything else. > Let's say M is the sun and I want to know the effect of > the sun on light from a He-Ne laser. Can you sketch out > the calculation for me? What's m? Newton doesn't work with light, explicitly. You can assume it has a nonzero mass and work out the equations of motion, for example, but the answer you get will be wrong. > - Randy ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <828iu1dnh4km8e6f41gecqfd5o59eakso6@4ax.com > On Tue, 07 Feb 2006 11:56:13 -0600, >There is no Newtonian prediction of the deflection of rays >of light. If you assume a point test mass, which light >neither is nor is composed of, then you might get that >result, but then what does GR get for the test mass? > > Light accelerates down a gravity well like anything else. > In Newtonian gravitation, the force on an object of mass m > in the presence of mass M is given by > F = GMm/r^2 > Please tell me how this applies to light to predict that light > accelerates down a gravity well like anything else. > Let's say M is the sun and I want to know the effect of > the sun on light from a He-Ne laser. Can you sketch out > the calculation for me? What's m? > Newton doesn't work with light, explicitly. > You can assume it has a nonzero mass and work out the equations of > motion, for example, but the answer you get will be wrong. [Eric Gisse]. | G*M*e | Force = |-- -- - -- --| |(n-1)*(r*c)^2| ..where, e/c^2 is TEST-REST-mass m1. ..mD=equivalent ambient DisCHARGEed. ..and, n=mD/m1=1, at weightlessness. ```Brian. > - Randy Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... <7pc2u19otvt62an7gjkkfdo6uv7ghpkgpv@4ax.com> <9v6au1lb47fnn61nktf7hlqo66848oe827@4ax.com> <2tafu1lre4bn5fp68h02gjmik37gguo7qb@4ax.com> <828iu1dnh4km8e6f41gecqfd5o59eakso6@4ax.com > On Tue, 07 Feb 2006 11:56:13 -0600, >There is no Newtonian prediction of the deflection of rays >of light. If you assume a point test mass, which light >neither is nor is composed of, then you might get that >result, but then what does GR get for the test mass? > > Light accelerates down a gravity well like anything else. > In Newtonian gravitation, the force on an object of mass m > in the presence of mass M is given by > F = GMm/r^2 > Please tell me how this applies to light to predict that light > accelerates down a gravity well like anything else. > Let's say M is the sun and I want to know the effect of > the sun on light from a He-Ne laser. Can you sketch out > the calculation for me? What's m? > Newton doesn't work with light, explicitly. > You can assume it has a nonzero mass and work out the equations of > motion, for example, but the answer you get will be wrong. [Eric Gisse]. | G*M*e | Force = |-- -- - -- --| |(n-1)*(r*c)^2| ..where, e/c^2 is TEST-REST-mass m1. ..mD=equivalent ambient DisCHARGEed. ..and, n=mD/m1=1, at weightlessness. ```Brian. > - Randy Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >In sci.math, HW@..(Henri Wilson) >: >>On 5 Feb 2006 19:45:39 -0800, jgreenfield@seol.net.au Pluto would see a brighter sun if Jupiter nearly eclipses, eh? Gravitational lensing shows nothing but how a photon's path is >permanently altered as per Isaac, when passing a massive object. It is >quite obvious, that a photon following a line of equal gravitational >potential (GR), returns to the path it had prior to the close encounter >with the mass, and NO lensing (magnification or changed position) would >be evident. >>No, actually gravitational lensing can work according to Newton >>(GR gives the same answer) >Uh...wrong. Newton predicts 1+2gh/c^2; GR predicts 1+gh/c^2. >>There is no Newtonian prediction of the deflection of rays of light. If >>you assume a point test mass, which light neither is nor is composed of, >>then you might get that result, but then what does GR get for the test mass? > Light accelerates down a gravity well like anything else. Yes it does, In GR, but Newton simply doesn't address an interaction between photons and masses, that is, not of this sort. Only absorption and/or emission of light by matter occurs. Light cannot interact with matter in any other way. Here's my personal interpretation of gravitational lensing: Firstly, it appears, by the above argument, that gravitational bending of light rays is literally impossible in the Newtonian context. Diffraction, OTOH is quite possible, and this is quite likely what is happening in gravitational lensing. Diffraction involves the production of secondary radiation that interferes constructively or destructively with the primary radiation, producing a resultant energy pattern that gives an illusion of a bent beam of light. In all cases, electron/electron interactions occur along the shortest possible worldline (*path* in QED). Putting a large mass between them doesn't change that worldline, though it might be interpreted to have done so when the details are omitted. The intervening mass only introduces a electrons can and will interact with, and there is thus a resultant interference effect. Feynman dubbed this effect sum over histories. Note that this approach is used to explain diffraction, and pretty much every other light related effect observed. The roadblock to making this connection between gravity and electromagnetism lies in the notion that photons are real, and are absorbed in their entirety. Time to get over that erroneous conclusion. The world of the electron is infinitely complex, yet from those complexities emerge such perfectly orderly macroscopic effects such as holographic imaging. Em waves are perfectly spherical, just a matter of retarded potentials, and the link between gravity and QM is that both are rather narrow-minded in their approaches. The unobservables are more numerous in those theories than their theorists are willing to admit, or to part with. The very fact that gravity can influence light behavior tells us that gravity has an underlying electromagnetic underpinning. Light it after all massless. Richard Perry > You can get the same answers by imagining that space contracts parabolically > and light retains its constant speed.....but that's a pretty dumb thing to > do..... >>Richard Perry > HW. > www.users.bigpond.com/hewn/index.htm ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... > The very fact that gravity can influence light behavior tells us that > gravity has an underlying electromagnetic underpinning. Light it after all > massless. Perhaps if I say it often enough someone may catch on. Ok, we've got the clue. Now where do we go next? The very fact that mass can influence e-m behavior tells us that mass has an underlying electromagnetic underpinning. Mass is after all lightive, m = E/c^2. Magnet --- Magnetic force. Electric --- Static force. Mass -- Gravitic force. Androcles. ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >>The very fact that gravity can influence light behavior tells us that >>gravity has an underlying electromagnetic underpinning. Light it after all >>massless. > Perhaps if I say it often enough someone may catch on. > Ok, we've got the clue. Now where do we go next? > The very fact that mass can influence e-m behavior tells us that > mass has an underlying electromagnetic underpinning. Mass is after > all lightive, m = E/c^2. Exactly. Richard Perry > Magnet --- Magnetic force. > Electric --- Static force. > Mass -- Gravitic force. > Androcles. ==== Subject: Re: Einstein was dumb. ahahaha... AHAHAHA... ahahaha... >The very fact that gravity can influence light behavior tells us that >gravity has an underlying electromagnetic underpinning. Light it after >all massless. >> Perhaps if I say it often enough someone may catch on. >> Ok, we've got the clue. Now where do we go next? >> The very fact that mass can influence e-m behavior tells us that >> mass has an underlying electromagnetic underpinning. Mass is after >> all lightive, m = E/c^2. > Exactly. > Richard Perry So the next step is Exactly... Very enlightening, I'm sure. Androcles >> Magnet --- Magnetic force. >> Electric --- Static force. >> Mass -- Gravitic force. >> Androcles. ==== Subject: JSH: Generalized Decker example, examples Decker's example can be generalized to help in finding rational solutions, where you will find that the mathematics follows the distributive property: f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) = f(25 x^2 + 30 x + 2) and f Q(x) = (5a_1(x) + f)(5a_2(x) + f) where the a's are defined by a^2 - (x - 1)a + f(x^2 + x) = 0 and you can let f be a rational number. Now there will be posters who will loudly declare coprimeness means nothing with rationals, but consider f(x^2 + 3x + 2) = (fx + f)(x + 2) and solutions with rationals, as guess what? You can STILL see that one factor is multiplied by f, even with rationals. If you don't think so, play with that example with some rational f's and rational x and see if the factor multiplied by f doesn't betray that it was. Now then, if I am wrong, some rational f can be found with a rational x that shows it. Like let f=32 with the generalized Decker example, and find some rational solutions and see if that f gets split up. Or let f=1024 or anything you want!!! You see, no counterexample exists, as the distributive property is right!!! So posters here at best can loudly proclaim that Galois Theory doesn't work with rationals but only with non-rationals, which is the dodge because it actually doesn't work, but you can't see that with non-rationals. The proof I've given relies on the distributive property. Even lower rung mathematicians cannot be incapable of quickly seeing it MUST be correct, but clearly as this impasse continues they are running. And them running means they are hoping that none of you who are not already established in careers, who are just learning as you're still in school, will stop protecting them by ignoring this result. They are in the weak positon of needing your protection so that they can teach you wrong mathematical ideas, as if some students start protesting, they will collapse like the cowards they are, running the other way, selling each other out to protect themselves. First mathematicians on the block will be the ones who are posters on sci.math, and their own will destroy their careers. People like Magidin and Ullrich will be out of their universities so fast your head will spin, as they are tossed to the wolves. So they sit and wait, checking each day to see if any of you are breaking out of the wall of silent acceptance, or irrational denial in the face of a simple proof that relies on the distributive property at the point of dispute. That check may be the best evidence against those mathematicians as they leave cyber clues to prove they knew, but were checking to see if they could keep getting away with lying. James Harris ==== Subject: Re: JSH: Generalized Decker example, examples > Decker's example can be generalized to help in finding rational > solutions, where you will find that the mathematics follows the > distributive property: > f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (x - 1)a + f(x^2 + x) = 0 And your claim is that one of the roots must always be divisible by f and one coprime to f. So take x=-2, f=-9 we get a^2 + 3a - 18 = (a+6)(a-3) The two roots are -6 and 3, neither is divisible by -9. -William Hughes ==== Subject: Re: JSH: Generalized Decker example, examples > Decker's example can be generalized to help in finding rational > solutions, where you will find that the mathematics follows the > distributive property: > f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (x - 1)a + f(x^2 + x) = 0 > And your claim is that one of the roots must always be divisible > by f and one coprime to f. > So take x=-2, f=-9 we get > a^2 + 3a - 18 = (a+6)(a-3) > The two roots are -6 and 3, neither is divisible by -9. > -William Hughes That's no different from the special cases with the original Decker example when x = 1 mod f, as then you have that sqrt(f) is a factor of both. Well, at least you're trying. You need x-1 coprime to f. Go look, you should still be able to find some rational examples, as remember, you can use rational f, where the requirement becomes with f = b/a b/a - 1 = (b-a)/a that b-a be coprime to f. James Harris ==== Subject: Re: JSH: Generalized Decker example, examples >Decker's example can be generalized to help in finding rational >solutions, where you will find that the mathematics follows the >distributive property: >f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) That's actually f(25 x^2 + 30 x + (f - 5)) >and >f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >where the a's are defined by >a^2 - (x - 1)a + f(x^2 + x) = 0 >>And your claim is that one of the roots must always be divisible >>by f and one coprime to f. >>So take x=-2, f=-9 we get >>a^2 + 3a - 18 = (a+6)(a-3) >>The two roots are -6 and 3, neither is divisible by -9. >> -William Hughes > That's no different from the special cases with the original Decker > example when x = 1 mod f, as then you have that sqrt(f) is a factor of > both. > Well, at least you're trying. > You need x-1 coprime to f. So how about x = 3, f = 7? Would you disallow that because the a polynomial doesn't have integral roots? Hint: for ANY f > 0, this polynomial only has integral roots for x = -1, 0. Rick ==== Subject: Re: JSH: Generalized Decker example, examples > Decker's example can be generalized to help in finding rational > solutions, where you will find that the mathematics follows the > distributive property: > > f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) > > and > > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > > where the a's are defined by > > a^2 - (x - 1)a + f(x^2 + x) = 0 > And your claim is that one of the roots must always be divisible > by f and one coprime to f. > So take x=-2, f=-9 we get > a^2 + 3a - 18 = (a+6)(a-3) > The two roots are -6 and 3, neither is divisible by -9. > -William Hughes > That's no different from the special cases with the original Decker > example when x = 1 mod f, as then you have that sqrt(f) is a factor of > both. > Well, at least you're trying. > You need x-1 coprime to f. Seems a bit arbitrary. Ok x=-2,f=-14 x-1=-3 is coprime to -14 a^2 + 3a - 28=(a+7)(a-4) roots are -7 and 4, neither is divisible by -14 -William Hughes ==== Subject: Re: JSH: Generalized Decker example, examples >>Decker's example can be generalized to help in finding rational >>solutions, where you will find that the mathematics follows the >>distributive property: >>f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) >> = f(25 x^2 + 30 x + 2) >> This should be f(25 x^2 + 30 x +(f - 5)) >>and >>f Q(x) = (5a_1(x) + f)(5a_2(x) + f) >>where the a's are defined by >>a^2 - (x - 1)a + f(x^2 + x) = 0 >And your claim is that one of the roots must always be divisible >by f and one coprime to f. >So take x=-2, f=-9 we get >a^2 + 3a - 18 = (a+6)(a-3) >The two roots are -6 and 3, neither is divisible by -9. > -William Hughes >>That's no different from the special cases with the original Decker >>example when x = 1 mod f, as then you have that sqrt(f) is a factor of >>both. James thinks this is a special case. As I explained in another thread, it's actually typical behavior, though James doesn't see that yet. >>Well, at least you're trying. >>You need x-1 coprime to f. > Seems a bit arbitrary. Ok > x=-2,f=-14 x-1=-3 is coprime to -14 > a^2 + 3a - 28=(a+7)(a-4) > roots are -7 and 4, neither is divisible by -14 > -William Hughes Or, how about x = -2, f = -77? We get a^2 + 3 a - 154 = (a - 11)(a + 14) with roots 11 and -14, neither of which are divisible by -77 and neither of which are coprime to -77. Not only that, but x - 1 is coprime to -77 (and, to forestall other ad hoc objections, x is coprime to -77). As with your example, we don' even need no steenkin' algebraic integers to refute James' conjecture. At least I'm trying (heh), Rick p.s. Not only that, but (5(11)-77)(5(-14)-77) = (-22)(-147) and neither of these factors are divisible by -77. Perhaps James might consider restricting f to primes, as long as he's throwing in more or less arbitrary restrictions :) ==== Subject: Re: JSH: Generalized Decker example, examples > At least I'm trying (heh), Very trying, for JSH. ==== Subject: Re: JSH: Generalized Decker example, examples > Decker's example can be generalized to help in finding rational > solutions, where you will find that the mathematics follows the > distributive property: > f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (x - 1)a + f(x^2 + x) = 0 > And your claim is that one of the roots must always be divisible > by f and one coprime to f. > So take x=-2, f=-9 we get > a^2 + 3a - 18 = (a+6)(a-3) > The two roots are -6 and 3, neither is divisible by -9. And if you want f coprime to 3, then take x=-2, f=-20, we get a^2 + 3a - 40 = (a+8)(a-5) The two roots are -8 and 5, neither is divisible by -20. -William Hughes ==== Subject: Re: JSH: Generalized Decker example, examples > First mathematicians on the block will be the ones who are posters on > sci.math, and their own will destroy their careers. People like > Magidin and Ullrich will be out of their universities so fast your head > will spin, as they are tossed to the wolves. And me? What about me? > So they sit and wait, checking each day to see if any of you are > breaking out of the wall of silent acceptance, or irrational denial in > the face of a simple proof that relies on the distributive property at > the point of dispute. > That check may be the best evidence against those mathematicians as > they leave cyber clues to prove they knew, but were checking to see if > they could keep getting away with lying. And you'll be famous for being the man who discovered all this. Tell me: do you think that Oprah will invite you to go to her show? Jose Carlos Santos ==== Subject: Re: JSH: Generalized Decker example, examples >> That check may be the best evidence against those mathematicians as >> they leave cyber clues to prove they knew, but were checking to see if >> they could keep getting away with lying. > And you'll be famous for being the man who discovered all this. Tell me: > do you think that Oprah will invite you to go to her show? He's said in the past that getting on Oprah's show is one of his goals. It's appropriate: Trivial aspirations for a trivial, insignificant little man. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: JSH: Generalized Decker example, examples days. My association with the Department is that of an alumnus. >> First mathematicians on the block will be the ones who are posters on >> sci.math, and their own will destroy their careers. People like >> Magidin and Ullrich will be out of their universities so fast your head >> will spin, as they are tossed to the wolves. >And me? What about me? You haven't been around long enough, Jose. Sorry. But to quote Stephen Colbert from _The Colbert Report_, I called it. Three years and two months ago: (last paragraph) Two years and three month ago: -- ' ==== Subject: Re: Generalized Decker example, examples > Decker's example can be generalized to help in finding rational > solutions, where you will find that the mathematics follows the > distributive property: > f Q(x) = f((x^2 + x)(5^2) + (-1 + x)(5) + f) > = f(25 x^2 + 30 x + 2) > and > f Q(x) = (5a_1(x) + f)(5a_2(x) + f) > where the a's are defined by > a^2 - (x - 1)a + f(x^2 + x) = 0 > and you can let f be a rational number. Now there will be posters who > will loudly declare coprimeness means nothing with rationals, but > consider > f(x^2 + 3x + 2) = (fx + f)(x + 2) > and solutions with rationals, as guess what? You can STILL see that > one factor is multiplied by f, even with rationals. If you don't think > so, play with that example with some rational f's and rational x and > see if the factor multiplied by f doesn't betray that it was. So What ? this is trivial. High School algebra @best. ==== Subject: Re: Complex Exponent 02/06/2006 >Ok, I can't seem to solve this one: >Let r be a real number. >Let z = a+bi be a complex number. >Write r^z as x+yi r^z = e^{z ln r} But if s is a natural log of r then so is s+n*i*2Pi, for integer n. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org ==== Subject: What property of compactness does this proof rely on? Hello Have a theorem that goes If b element R Q then the orbit of every point x element of S1 under rotation r_b is dense in S1. The proof goes Given b element RQ, x element S1 and epsilon >0 , the points (r_b)^n (x) .where n is element of the positive intergers, are distinct (easy part done by contradiction, assume (r_b)^n (x) = (r_b)^m (x) then (m-n)*b element of integers -contradiction) Then the proof goes on to say as S1 is compact, there exists n not equal to m, s.t 0 < d((r_b)^n (x) , (r_b)^m (x) ) < epsilon how does compactness imply that? ==== Subject: Re: What property of compactness does this proof rely on? days. My association with the Department is that of an alumnus. >Hello >Have a theorem that goes If b element R Q then the orbit of every >point x element of S1 under rotation r_b is dense in S1. >The proof goes >Given b element RQ, x element S1 and epsilon >0 , the points (r_b)^n >(x) .where n is element of the positive intergers, are distinct (easy >part done by contradiction, assume (r_b)^n (x) = (r_b)^m (x) then >(m-n)*b element of integers -contradiction) >Then the proof goes on to say >as S1 is compact, there exists n not equal to m, s.t 0 < d((r_b)^n (x) >, (r_b)^m (x) ) < epsilon >how does compactness imply that? I think it is the fact that every sequence has a converging subsequence; and then every converging sequence is Cauchy, which would give you the condition. -- ' ==== Subject: Recursion Made Simple? Is there a nonmathematical way to express the essential idea of recursion, or is it only meaningful within the context of formal mathematics? ==== Subject: Re: Recursion Made Simple? > Is there a nonmathematical way to express the essential idea of recursion, > or is it only meaningful within the context of formal mathematics? One of Murphy's Laws is: Whatever it is you have to do, you have to do something else first. Of course, the something else is just another thing you have to do. ==== Subject: Re: Recursion Made Simple? > Is there a nonmathematical way to express the essential idea of recursion, > or is it only meaningful within the context of formal mathematics? > One of Murphy's Laws is: > Whatever it is you have to do, you have to do something else first. Corollary: NOTHING gets ever done. -- Ioannis --- http://ioannis.virtualcomposer2000.com/ ==== Subject: Re: Recursion Made Simple? > Is there a nonmathematical way to express the essential idea of recursion, > or is it only meaningful within the context of formal mathematics? In the dictionary under recursion the entry is: See recursion ==== Subject: Re: Recursion Made Simple? > In the dictionary under recursion the entry is: > See recursion LOL ==== Subject: Re: Recursion Made Simple? Recursion is the process a procedure goes through when one of the steps of the procedure involves rerunning the entire same procedure. A procedure that goes through recursion is said to be recursive. Something is also said to be recursive when it is the result of a recursive procedure. Wikipedia For example if you want to find out all the possible decendants of some person, you would do this: a) find all children of person X b) for all children, find all decendants (this is the recursive part) Gerben ==== Subject: Re: Convergence of discrete newton > Hi all, > I am reading Dennis and Schnabel's Numerical Optimization book, and > there's a comment about convergence that I do not understand. On page > 103, it is stated that if we use Newton's method for unconstrained > minimization using a finite-diference approximation to the Hessian > matrix, then it is easy to see that the method displays quadratic > convergence, provided the finite difference step-size for the Hessian > decreases linearly with each iteration of the sequence. > I don't understand why this must be true, after re-reading a couple of > times. Most importantly, why is the rate of convergence NOT quadratic > if the step size is bounded? If the step size is bounded from below, the difference quotient doesn't converge to the true derivate. So, when it comes to the (asymptotic) oder of convergence, such a scheme is not better than the simplified Newton (which uses a fixed approximation for the Jacobian). -- Helmut Jarausch Lehrstuhl fuer Numerische Mathematik RWTH - Aachen University D 52056 Aachen, Germany ==== Subject: Re: Convergence of discrete newton >Hi all, >I am reading Dennis and Schnabel's Numerical Optimization book, and >there's a comment about convergence that I do not understand. On page >103, it is stated that if we use Newton's method for unconstrained >minimization using a finite-diference approximation to the Hessian >matrix, then it is easy to see that the method displays quadratic >convergence, provided the finite difference step-size for the Hessian >decreases linearly with each iteration of the sequence. >I don't understand why this must be true, after re-reading a couple of >times. Most importantly, why is the rate of convergence NOT quadratic >if the step size is bounded? Any suggestions for other references (or >a sketch of a proof even...) will be most appreciated. >SF problem F(x)=0 notation: JF(x) = Jacobian matrix at x JFD(x,h) = finite difference approximation with difference step assumption: we use a finite difference approximation to the partial derivatives of consistency order p, assuming that F is (p+1) times continuously differentiable. example (F(x+h*e_i)-F(x))/h = JF(x)*e_i + E(x,h)*e_i , ||E(x,h)||<= h*M2*n where M2 is an upper bound for any of F's second partial derivatives and e_i the i-th coordinate unit vector. finite difference Newtons method: x_new=x_old - inv(JFD(x_old,h))*F(x_old) = =x_old - inv(JF(x_old)+E(x_old,h))*F(x_old) = =x_old - inv(JF(x_old)*(I+inv(JF(x_old)*E(x_old,h)))*F(x_old) =x_old - inv(I+inv(JF(x_old)*E(x_old,h))*(inv(JF(x_old))*F(x_old) x^* solution, assuming F(x^*) invertible Banach perturbation lemma: inv(I+inv(JF(x_old)) * E(x_old,h) ) =I + E1(x_old,h) and ||E1(.,h)||<= Const*h^p x_new-x^* = x_old-x^* -(I+E1(.,.))*inv(JF(x_old))*(F(x_old)-F(x^*)) now by Taylors theorem F(x^*) = F(x_old) +JF(x_old)*(x^*-x_old) + O(||x^*-x_old||^2) inserting this gives x_new-x^* = x_old-x^* - (I+E1(.,.))*(x_old-x^*) + O(||x^*-x_old||^2) = O(||x^*-x_old||^2) - E1(.,.)*(x_old-x^*) taking norms and inserting the bound on ||E1(.,.)|| you see: for fixed (small) h the convergence will be fast but linear. (since p>=1) if h=O(||x_old-x^*||), for example h=min{delta,||F(x_old)||} then convergence will be quadratic hth peter x_new-x^* = x_old-x^* ==== Subject: Re: Request Algebra Proof Of Ruiz Identity Bullseye. I resisted the approach below because it is a visualization stretch for me. Although not as satisfying as direct algrebra, it is certainly algebra-only. After experimenting on paper, the logic does seem valid, so I'm forced to accept it. >I don't see a problem with using repeated differences of polynomials. >It is pretty easy to show that taking the difference of a degree n >polynomial yields a degree n-1 polynomial. Using the binomial >theorem, we get that x^n - (x-1)^n = n x^{n-1} + lower powers of x. >Taking n differences gives n! x^0 + lower powers of x, and this is >just n! To email me, remove theobvious. from my address. ==== Subject: Re: Algorithm for computing GL-orbits Content-Length: 952 Originator: rusin@vesuvius >Let K be a finite field. Suppose G:=GL(n,K) acts linearly and faithfully >over a finite dimensional K-vector space V. Does exist an algorithm for >finding the G-orbits in V ? Does exist a Schreir-Sims analog ? Yes, both GAP and Magma can compute orbits of any matrix group over a finite field acting on vectors or r-dimensional subspaces for some r. Tha main problem is that for large dimension the space of all vectors, on which the group is acting, is very large. Magma uses a version of the Schreier-Sims algorithm for computing the orders of matrix groups. Derek Holt. >The situation I'm particularly interested in is where V is the third >exterior power (or trivectors space) over a finite K-vector space E : if >dim(E) > 7 and as far I know, the orbits have not been determined yet. >Pascal ==== Subject: Dynamic programming approach to discrete optimization problem Problem statement (PS) J({x(k)}) -> min_{x(0),...,x(N)}, J({x(k)})=sum_{k=-1}^N { h_k(F(k+1)x(k+1)-C(k)x(k))+ q_k(y(k)-H(k)x(k))}, where F(k), C(k-1), H(k-1) - some real rectangular matrixes of approppriate dimensions, y(-1)=0, H(-1)=0, h_k(x) = (A(k)x,x), q_k(y) = (B(k)y,y), A(k),B(k) - symmetric positive-defined matrixes, (u,v) = sum u_i*v_i I need to solve a problem (PS) using a dynamic programming approach. Sergiy. ==== Subject: permutation statistics. The number of inversions of a permutation... Let x be a permutation of [n]={1,2,...,n}, let x^-1 denote the inverse permutation and let inv(x) denote the number of inversions in a permutation x. Then how do I show that inv(x)=inv(x^-1), that is the number of inversions in a permutation is equal to the number of inversions in its inverse. ==== Subject: Re: Cantorian pseudomathematics david petry said: > |But we can tell, just by looking at X, whether it is a constructively > |meaningful statement. Sometimes it is not so clear when the statement > |is made in informal natural language, though. > I think that when you just look at a statement and decide > whether it is constructively meaningful you do it in a > naive way. Going with your gut on these things is a bad idea. > You need to spend more time dealing with criteria that are > explicit enough that someone can tell whether a statement > qualifies or not without being you. > I believe I have stated very precisely what the criteria are. A > statement is meaningful iff it makes predictions about the results of > computational experiments. A computational experiment can be put in the > form Turing machine T, with input M, will halt within N steps. > I am also claiming that we can build a new formalism for mathematics > such that the axioms themselves are computationally meaningful, and > such that the laws of logic we use preserve computational meaning. Then > every grammatically valid sentence in our formalism will be meaningful. > In general, it is not so easy to take a statement written in the > formalism of ZFC (for example), and determine whether it has a > meaningful counterpart. David, it sounds like you want a statement to be a symbolic string that is manipulable through formal rules so that it produces a quantity. When you speak of a Turing machine, that reaffirms that idea. Now, you also seem to be talking about defining a grammar that sets down rules for such statements and their combinations. I think this sounds like a good idea. So, what primitive notions do you need to add to this grammar for that purpose? Have you identified specific problematic constructions that have no computational meaning? > Errett Bishop's constructive analysis textbook in at least > one edition contains a mistake regarding AC. He made a > mistake which led him to think it was true. So if there > is some way that one can just magically intuit its > unacceptability from a constructive point of view by just > looking at it, for the first time, it is at least a method > that had escaped his attention. > For constructivism, lists are a much more natural structure than sets, > and the axiom of choice for lists is trivially true. Indeed, it can be > difficult to take concepts from set theory and figure out how they fit > into the constructivist view. > Just a comment: I've always felt that Bishop tried too hard to make his > constructivism resemble classical mathematics. > |Proving theorems like AC->X is not doing constructive mathematics, > even > |in a broad sense. > So now you're ready to explain what core, essential aspect > of the concept of constructivity it violates? > AC is not part of constructive mathematics. > No. Of > course not. You derive too much enjoyment from glibly > tossing off such claims for stopping to examine the > principles in question to be attractive to you. > Glib nonsense. > Constructivity is typically explained in terms of the > meaning assigned to existential quantifiers. In particular, > each existence claim should be accompanied by a > construction, where it is often left rather vague as > to what should be counted as a construction. > As I have often pointed out, what I'm advocating is not exactly > conventional constructivism. There is nothing vague at all about > existence in what I am doing. Well, specifically, you wish an element to be the result of a symbolic manipulation according to specific rules which you consider computational. Are there any systems of symbolic manipulation which you don't consider computational, or any computations which don't involve symbolic tranformations? > One case, however, is more crucial than most people who've > heard of constructivism realize, and that's the case of > integers (and consequently other finite combinatorial > constructions). A constructive proof that an integer exists > must (by the nature of the concept of constructive) give > a way of computing it (in decimal form, say). > But that is already slightly vague. What I am advocating is that we > would have to derive from the proof an upper bound on how much > computation we need to do in order to compute the integer, in order to > say that we have a way to compute it. What are you computing the integer from? A predecessor? By how much computation, what do you mean, exactly. How do you measure computation in your scheme. Does that give a size to the set under consideration? >A second case > that is less crucial but generally fits people's intuitions > about what constructive should mean is that a construction > of a set should give us a way to express it as {x:P(x)} for > some explicit property P. > As I already pointed out, using lists would be preferable to using sets > here. A list being just a set with order, right? > So for an axiom system we have two natural criteria for > constructivity, known as the numeric existence property > and the set existence property. It has the numeric existence > property if there is a procedure for converting a proof of > a statement of the form (En)Q(n) where n ranges over the > integers into a proof of Q(n) for a specific n given in > decimal form. It has the set existence property if there is > a procedure for converting a proof of a statement of the > form (ES)Q(S) where S ranges over sets (of some kind) into > a proof of a statement of the form Q({x:R(x)}) for some > predicate R. > There are two fundamental axioms famous for creating > problems with the numeric and set existence principles. > Most mathematically trained people are aware of proofs > using the axiom of choice of the existence of things like > a Vitali set (with one representative from each member > of R/Q) where one doesn't expect to be able to give an > explicit example. This is associated with ZFC's failure > of the set existence principle. Now a lot of people don't > realize this can be patched over by going to ZF+V=L, > where V=L implies AC. But set theorists usually seem to > prefer axioms that contradict V=L so it doesn't make much > difference. The second and more serious issue is with the > way the law of excluded middle destroys the numeric > existence principle. > Not every constructivist is happy with every formal system > satisfying the numeric and set existence properties. I don't > know of any additional requirement, however, for which there > is a consensus that it is a necessity for a system to be a > constructive system, and its proofs constructive proofs. > Among people who want proofs to be constructive, I suppose > there's often interest in their being predicative as well. > It strikes me as somewhat arbitrary and severe to say that > this is some kind of absolute requirement. > I believe that you are approaching constructivism from the point of > view that Cantorian set theory is the big picture, and constructive > mathematics is merely a mildly interesting small picture which must fit > somewhere within the big picture. That, I suspect, is the obstacle to > communication we seem to be having. Actually, I think Keith is addressing a core issue in this area of foundations. Set theory has this primitive is an element, but it seems to me that one can start with has a property, and define is an element and is equal based on the concept of properties. If one defines sets in terms of objects that share a property, and then goes on to define the properties in terms of a grammar with some primitive types as values, then perhaps that grammatical definition would satisfy what you are trying to achieve. This is what I am currently considering. > Then there's the issue of computational meaning. > Some famous judge said of pornography that he couldn't > define it precisely but could recognize it when he saw it. > If one wants to be able to say something more satisfying > than that about mathematics that is obscenely lacking in > computational meaning, then one has further work to do. > Whatever. > You seem not to believe me when I say it, but there's a > simple, straightforward way of applying the mathematical > theories must make predictions rule that tells us to dump > the same two axioms, and nothing else. Take computational > predictions to mean only the most obvious kind of thing, > namely, statements of the form (n)P(n) where n ranges over > the integers and P is a primitive recursive predicate. > Then say that an axiom system should be no more complicated > than necessary to entail all conclusions of that form that > it does. > So you want to eliminate statements like for all n, A(n) exists, > where 'A' is the Ackerman function. Why would you do such an arbitrary > thing? > By the work of Goedel on AC, we can show that removing it > has no effect on which arithmetic statements we can prove, > let alone the kind of computational prediction I'm > considering here. Likewise there are some basic results on > the law of excluded middle that show it also doesn't give > us any more predictions than we had to begin with. I don't > know whether it counts as a coincidence or not, but these > are the same two axioms as were nonconstructive in the > obvious way. > Again, you're starting with classical set theory, and working down to > constructivism. That's bound to lead to confusion. It seems like Keith is pointing out that AC and LEM are not necessary, which should be good news. Of course, he is putting things in terms of standard theory, but I think he's doing a good job of describing what differences there are in basic assumptions between the two schools of thought, no? > The rest of the axioms in ZFC do give new predictions. The > axiom of replacement is often written in a form implying > the law of excluded middle, but it can be written not to. > Without AC or LEM this leaves us with a system called IZF > that satisfies the numeric and set existence properties. > I guess it's not very well liked for practical use, but in > a theoretical way it nicely matches ZFC. One can prove the > existence of the same recursive functions in each for > instance. > IZF can prove the consistency of IZF with the axiom of > replacement removed, which IZF with the axiom of replacement > removed can't do (by Goedel's second incompleteness theorem). > Since a consistency statement is equivalent to one of the > form (n)P(n) this counts as some kind of additional > prediction. > IZF with the axiom of replacment removed can prove the > consistency of IZF with the power set axiom and the axiom > of replacement both removed. > IZF with both the power set axiom and the axiom of replacement > removed can prove the consistency of IZF with the axiom of > infinity removed. If the axiom of infinity is replaced with > its negation, we get a system essentially the same as elementary > arithmetic. There's a model of it consisting only of the sets > that can be built up in finitely many steps from the empty set. > The only way to get the full computational content is to > keep all the axioms in IZF. Presumably there's some way > to simplify the system by replacing axioms, but none of > the axioms can be just tossed out. > I'm surprised the mob hasn't bumped you off yet. Clearly, you know too > much. Keith apparently knows a lot. I found this explanation very interesting ans > I've pointed this out to you before that there is something magical > (deceptive) about consistency proofs. If you believe that the theorems > of some formalism are in fact truths, then you must believe (at a > minimum) that the formalism is consistent. So if you use a strong > formal system to prove the consistency of a weaker system, you are > already assuming that the stronger system, and hence the weaker system, > is consistent. So you are proving nothing, except to true believers > who already believe the conclusion. > The criteria I described above give us a principled way to > designate certain kinds of mathematics as nonconstructive. > Is there a principled way to designate some of the > mathematics that counts as constructive by those criteria, > and say that, nevertheless, it lacks computational meaning? > Again, you are implicitly making the claim that in order to talk about > constructive mathematics, we must be able to talk about classical > mathematics. There's something fundamentally dishonest about that; you > are stacking the deck in your favor. > As far as I can tell, you just go on your own gut reaction, > without any principle that someone not sharing your > intuitions could apply in a consistent way. It's hard to see > how one could in a principled way dispose of all the > Cantorian mathematics you have such a problem with > Here's what I am claiming: we can build up constructive mathematics > independently of classical mathematics. Then, we can notice that > classical mathematics implies the existence of a universe much bigger > than the universe of constructive mathematics, and hence there is > necessarily something about classical mathematics that is fantasy, > where fantasy is defined to be anything beyond the world we can > observe (i.e. the constructive universe). > There is no unprincipled gut reaction involved. Hmmmm..... How do you know that what you view as non-constructive has just not had its construction discovered/defined yet? How will you know when all possible constructions have been exhausted? Doesn't that depend largely on how you define the grammar for the statements you allow? In a sense, this seems to be pushing the problem back a level, but not necessarily solving the ultimate problem of consistency, but I am no expert. > |I know we've discussed this a few times before. The result that is > not > |disputed by either constructivists or classical mathematicians is that > |if we are given a well defined list of well defined real numbers (so > |that every digit of every number can be computed), then the diagonal > |method gives us a new number not on that list. The classical > |mathematicians claim, essentially, that the argument is still valid > |when the list and the numbers on the list may not be well defined. > You're equivocating between computable, which is a property > of certain objects, and definedness which is really a > property of the description of the object that needs to be > satisfied before one can ask whether the object described > by the description has some property or not. > You describe constructivists as if they automatically > assume that reals are computable. They don't. The argument > makes no use of any such assumption. > Constructively, it makes no sense to say something exists unless it can > be computed. At least, that's what *this* constructivist says. And by computed, you mean that it can be produced as a state reached by Turing machine processing a symbolic input string, right? But then, doesn't its computability depend on the prior definition of a state in the Turing machine which represents the constructed value? I think what you need to do, to define what you mean by computationally meaningful, is design your Turing machine and thus the proposed grammar that you envision as restricting the kinds of statements that are allowed to define set membership. Does that make any sense? -- Smiles, Tony ==== Subject: Re: Cantorian pseudomathematics I believe I have stated very precisely what the criteria are. A > statement is meaningful iff it makes predictions about the results of > computational experiments. A computational experiment can be put in the > form Turing machine T, with input M, will halt within N steps. > I am also claiming that we can build a new formalism for mathematics > such that the axioms themselves are computationally meaningful, and > such that the laws of logic we use preserve computational meaning. Then > every grammatically valid sentence in our formalism will be meaningful. > David, it sounds like you want a statement to be a symbolic string that is > manipulable through formal rules so that it produces a quantity. I don't actually understand that statement of yours. > So, what primitive notions > do you need to add to this grammar for that purpose? Have you identified > specific problematic constructions that have no computational meaning? The use of the unrestricted existential quantifier quickly leads to statements which have no computational meaning. In other words, if you tell me that something exists, I want you to tell me precisely how I can find it. If you can't do that, I will claim that you don't really know what you are talking about. > Are there any systems of symbolic manipulation which you don't consider > computational, or any computations which don't involve symbolic tranformations? I have trouble grasping what you are really asking here. > What I am advocating is that we > would have to derive from the proof an upper bound on how much > computation we need to do in order to compute the integer, in order to > say that we have a way to compute it. > By how much > computation, what do you mean, exactly. How do you measure computation in your > scheme. A Turing machine takes one step at a time, and we can count those steps. > As I already pointed out, using lists would be preferable to using sets > here. > A list being just a set with order, right? No. A list can have elements repeated. > Here's what I am claiming: we can build up constructive mathematics > independently of classical mathematics. Then, we can notice that > classical mathematics implies the existence of a universe much bigger > than the universe of constructive mathematics, and hence there is > necessarily something about classical mathematics that is fantasy, > where fantasy is defined to be anything beyond the world we can > observe (i.e. the constructive universe). > Hmmmm..... How do you know that what you view as non-constructive has just not > had its construction discovered/defined yet? Well, constructively speaking, the burden is on someone who claims something exists to prove that it exists, and not on me to prove it doesn't. ==== Subject: Re: Cantorian pseudomathematics > [ ... ] Cantorian [ ... ] It may have become kind of common belief that the nomer Cantorian has been invented by David Petry. But this is not true. Just search for the string Cantorian in the web-version of the book by L.E.J. Brouwer: http://www.ru.nl/w-en-s/gmfw/bronnen/brouwer4-III.html Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics > It's pretty clear that there is at least a small handful of prominent > mathematicians, applied mathematicians, and logicians who agree with me > that Georg Cantor introduced an element of make-believe into > mathematics, and that this element should be expunged. Look at what > they say: Do you happen to know Allan Tarp? Just discovered him via Google. He seems to be one more critic against the modern style of writing and thinking in (modern) pure mathematics, especially in connection with mathematics education. There may be some differences between the two of you, but i thought you might be interested anyway. He also had somewhat of a struggle, apparantly, getting himself heard and understood. -- Herman Jurjus ==== Subject: Re: Cantorian pseudomathematics <43e9b474$0$2027$ba620dc5@text.nova.planet.nl It's pretty clear that there is at least a small handful of prominent > mathematicians, applied mathematicians, and logicians who agree with me > that Georg Cantor introduced an element of make-believe into > mathematics, and that this element should be expunged. Look at what > they say: > Do you happen to know Allan Tarp? > There may be some differences between the two of you, but i thought you > might be interested anyway. I never heard of him until now. I can't see anything but differences between him and me, from the web pages I looked at. Why do you see a similarity? ==== Subject: Re: Cantorian pseudomathematics On Wed, 08 Feb 2006 10:05:55 +0100, Herman Jurjus He seems to be one more critic against the modern style of writing and >thinking in (modern) pure mathematics, especially in connection with >mathematics education. >There may be some differences between the two of you, but i thought you >might be interested anyway. He also had somewhat of a struggle, >apparantly, getting himself heard and understood. I wonder why? http://www.mai.liu.se/SMDF/madif5/papers/Tarp.pdf ==== Subject: Re: Cantorian pseudomathematics >>Dik T. Winter schreef: >Indeed, it is in violation of the Dutch Wet op het auteursrecht. >Re-distribution is only permitted if there is permission from the >copyright holder. Whether there is a copyright notice or not is >irrelevant. >>OK. See you in the courtroom. > Han, you really are an asshole. Here we go again ... Still no *real* arguments, huh, Hughes? The _truth_ is that CopyRights stem from medieval times. CopyRights are violated a zillion times a day all over the world. Computers & internet have made Copyrights: _oldfashioned, obsolete, outdated_. Look at this: http://huizen.dto.tudelft.nl/deBruijn/muziek/download.htm : bottom http://huizen.dto.tudelft.nl/deBruijn/programs/fortran/Contour.FOR : top I am not breaking the law. The law has been broken by modern technology. And there is nothing you or Dik T. Winter or any other dienstklopper can do about it. It's dweilen met de kraan open. Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics ... >>Dik T. Winter schreef: >Indeed, it is in violation of the Dutch Wet op het auteursrecht. >Re-distribution is only permitted if there is permission from the >copyright holder. Whether there is a copyright notice or not is >irrelevant. >> >>OK. See you in the courtroom. Apparently there is some rogue-cancelling going on. However, it is not me you will see in court, when it comes to court. > Here we go again ... Still no *real* arguments, huh, Hughes? > The _truth_ is that CopyRights stem from medieval times. CopyRights are > violated a zillion times a day all over the world. Computers & internet > have made Copyrights: _oldfashioned, obsolete, outdated_. Look at this: So because something is easy to do it should be permitted? of it. Otherwise it would remain on the server on which you did post it and nobody would be able to read it. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== Subject: Re: Cantorian pseudomathematics > So because something is easy to do it should be permitted? Permitted by _who_? Somebody is imposing his standards on other people. According to Operation Survival Earth, money and private property are key indications of a primitive level of civilization. The Western world is continuously trying to impose its standards upon the rest of humanity and one can't even imagine that making a sustainable future could be of an entirely different kind. Johann Sebastian Bach felt himself much _honored_ when somebody stole a musical theme from him. There is no Law of Nature about CopyRights. Han de Bruijn ==== Subject: Re: Cantorian pseudomathematics >> So because something is easy to do it should be permitted? > Permitted by _who_? Somebody is imposing his standards on other people. > According to Operation Survival Earth, money and private property are > key indications of a primitive level of civilization. That of course translates to According to someone who claims to have had encounters with aliens, money and private property are key indications of primitive level of civilization. :) Stephen ==== Subject: Re: Cantorian pseudomathematics Discussion, linux) <43087$43e32c1e$82a1e2b0$1563@news1.tudelft.nl> <7b71c$43e37a9b$82a1e2b0$7242@news2.tudelft.nl> <43e37b44$0$19705$8fcfb975@news.wanadoo.fr> <23eau15cuv8ekjpm1t46348g7snjp83vnq@no.spam> <87zml35qbw.fsf@phiwumbda.org> <87hd7b5l1y.fsf@phiwumbda.org If you think Han is violating copyright law, then take a look at this > website: http://www.archive.org/index.php > They are trying to archive (and hence distribute) the whole web! > Sounds felonious to me! > IANAL (I am not a lawyer) I'm not a lawyer either, but I honestly don't see *how* the Wayback Machine *isn't* violating copyright (which is usually a civil matter, not a criminal matter). That doesn't mean I don't like it. Nonetheless, it seems to be re-distributing work without the permission of the copyright holder. -- The math doesn't care about their mortgages. It doesn't care about their political needs. [...] Today's mathematicians have to hate mathematics because mathematics doesn't look out for them. It doesn't pay attention to their needs. --- JSH analyzes mathematicians ==== Subject: Re: Cantorian pseudomathematics <43087$43e32c1e$82a1e2b0$1563@news1.tudelft.nl> <7b71c$43e37a9b$82a1e2b0$7242@news2.tudelft.nl> <43e37b44$0$19705$8fcfb975@news.wanadoo.fr> <23eau15cuv8ekjpm1t46348g7snjp83vnq@no.spam> <87zml35qbw.fsf@phiwumbda.org> <87hd7b5l1y.fsf@phiwumbda.org> <87slqugx4c.fsf@phiwumbda.org If you think Han is violating copyright law, then take a look at this > website: http://www.archive.org/index.php > They are trying to archive (and hence distribute) the whole web! > Sounds felonious to me! > IANAL (I am not a lawyer) > I'm not a lawyer either, but I honestly don't see *how* the Wayback > Machine *isn't* violating copyright (which is usually a civil matter, > not a criminal matter). > That doesn't mean I don't like it. Nonetheless, it seems to be > re-distributing work without the permission of the copyright holder. As Han points out, laws are changing, and they have to. For example, to speed up service, many internet providers cache frequently used pages (that is, they store the page on the local server). Could you seriously claim that that is a copyright violation? Anyway, this is quickly becoming an off topic discussion by people who have no idea what they are talking about (myself included), so that's all from me on this. ==== Subject: Re: Is there a name for MatLab matrix op .* ? Alex, ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] On Tue, 07 Feb 2006 14:54:46 +0000, Duncan Muirhead >> Does anyone have an idea how to prove that the following inequality holds for the coefficients of any positive definite matrix A: >> [inv(A)](k,k) >= 1/[A(k,k)] >k=0: >If L is the cholesky factor of A, we can write >L = (l 0 ) where l = 1/sqrt(A(0,0) > (x L~) >and >inv(L) = (1/l 0 ) > (y M~) >so inv( A) = inv(L)' * inv( L) = (1/(l*l) + y'*y z' ) > ( z N ) >and inv(A)(0,0) = 1/A(0,0) + y'*y >= 1/A(0,0) >k>0 >Apply a permutation matrix and wave your hands about As noted by Rob Johnson, this proof implicitly assumes A to be Hermitian, since there are positive definite non-Hermitian matrices for which there are no Cholesky decompositions. ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] > I have one question: > Shouldn't be l=sqrt(A(0,0))? Then we get: > inv(A)(0,0) = 1/A(0,0) + y'*y + z'*z .... > Am I getting it right? Oops, yes. Sorry. But I think its just Inv(A)(0,0) = 1/A(0,0) + y'*y ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] >Does anyone have an idea how to prove that the following inequality >holds for the coefficients of any positive definite matrix A: >[inv(A)](k,k) >= 1/[A(k,k)] > This is the same as asking if > -1 > (u,Au) (u,A u) >= 1 [1] > for any unit vector u. For the question you ask, let u be the k^th > coordinate vector. > Suppose that { v_k } are eigenvectors of A with eigenvalues { w_k }. > Since A is positive definite, w_k > 0. Suppose that > --- > u = > a v [2] > --- k k > Then > --- > (u,Au) = > a a (v ,v ) w [3] > --- j k j k k > -1 --- > (u,A u) = > a a (v ,v ) 1/w [4] > --- j k j k k > Since u is a unit vector, we have > --- > > a a (v ,v ) > --- j k j k > = (u,u) > = 1 [5] > and since 1/x is convex for x > 0, we can apply Jensen's inequality > to [3] and use [4] to get > -1 > 1/(u,Au) <= (u,A u) [6] > Inequality [6] is a restatement of [1] as long as u can be written as > a linear combination of eigenvectors of A. All vectors in R^n can be > written as linear combinations of eigenvectors if and only if A is > diagonalizable. I tried to remove the need for diagonalizability, but > then I realized that the matrix > +- -+ > | 1 1 | > A = | -1 1 | > +- -+ > whose inverse is > +- -+ > -1 | 1/2 -1/2 | > A = | 1/2 1/2 | > +- -+ > is positive definite, but not diagonalizable, and fails to satisfy the > conclusion above. So the condition of diagonalizability can not be > removed completely. I suspect he means SYMMETRIC positive-definite. --Ron Bruck ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] >>Does anyone have an idea how to prove that the following inequality >>holds for the coefficients of any positive definite matrix A: >>[inv(A)](k,k) >= 1/[A(k,k)] >> This is the same as asking if >> -1 >> (u,Au) (u,A u) >= 1 [1] >> for any unit vector u. For the question you ask, let u be the k^th >> coordinate vector. >> Suppose that { v_k } are eigenvectors of A with eigenvalues { w_k }. >> Since A is positive definite, w_k > 0. Suppose that >> --- >> u = > a v [2] >> --- k k >> Then >> --- >> (u,Au) = > a a (v ,v ) w [3] >> --- j k j k k >> -1 --- >> (u,A u) = > a a (v ,v ) 1/w [4] >> --- j k j k k >> Since u is a unit vector, we have >> --- >> > a a (v ,v ) >> --- j k j k >> = (u,u) >> = 1 [5] >> and since 1/x is convex for x > 0, we can apply Jensen's inequality >> to [3] and use [4] to get >> -1 >> 1/(u,Au) <= (u,A u) [6] >> Inequality [6] is a restatement of [1] as long as u can be written as >> a linear combination of eigenvectors of A. All vectors in R^n can be >> written as linear combinations of eigenvectors if and only if A is >> diagonalizable. I tried to remove the need for diagonalizability, but >> then I realized that the matrix >> +- -+ >> | 1 1 | >> A = | -1 1 | >> +- -+ >> whose inverse is >> +- -+ >> -1 | 1/2 -1/2 | >> A = | 1/2 1/2 | >> +- -+ >> is positive definite, but not diagonalizable, and fails to satisfy the >> conclusion above. So the condition of diagonalizability can not be >> removed completely. >I suspect he means SYMMETRIC positive-definite. Whether intended or not, that would have insured the diagonalizability of A and saved me the time I spent before I realized the necessity of diagonalizability. Rob Johnson take out the trash before replying ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] >I suspect he means SYMMETRIC positive-definite. > Whether intended or not, that would have insured the > diagonalizability > of A and saved me the time I spent before I realized > the necessity of > diagonalizability. In the problem formulation (which is not mine) there were no other assumptions except that the matrix is positive definite (not necessarily real). I guess that we also have to assume that the matrix is hermitian. My impression is that, when they say complex positive definite, some people implicitly assume hermitianity. Maybe that was the case with the person who formulated the problem. ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] >>I suspect he means SYMMETRIC positive-definite. >> Whether intended or not, that would have insured the >> diagonalizability >> of A and saved me the time I spent before I realized >> the necessity of >> diagonalizability. >In the problem formulation (which is not mine) there were no other >assumptions except that the matrix is positive definite (not >necessarily real). I guess that we also have to assume that the matrix >is hermitian. My impression is that, when they say complex positive >definite, some people implicitly assume hermitianity. Maybe that was >the case with the person who formulated the problem. I looked up positive definite in each of MathWorld, Wikipedia, and PlanetMath; all three say that positive definite assumes hermitian. So it appears that the statement of the problem is complete under that assumption. I apologize for my confusion. Rob Johnson take out the trash before replying ==== Subject: Re: Proving that [inv(A)](k,k) >= 1/[A(k,k)] >>I suspect he means SYMMETRIC positive-definite. >> Whether intended or not, that would have insured the >> diagonalizability >> of A and saved me the time I spent before I realized >> the necessity of >> diagonalizability. >In the problem formulation (which is not mine) there were no other >assumptions except that the matrix is positive definite (not >necessarily real). I guess that we also have to assume that the matrix >is hermitian. My impression is that, when they say complex positive >definite, some people implicitly assume hermitianity. Maybe that was >the case with the person who formulated the problem. > I looked up positive definite in each of MathWorld, Wikipedia, and > PlanetMath; all three say that positive definite assumes hermitian. > So it appears that the statement of the problem is complete under that > assumption. > = 0 implicitly says it's REAL and >= 0. And the statement that it's real IMPLIES that A is Hermitian. In the real case, it DOESN'T imply A is symmetric. This is deduced by expanding (basically, the complex polarization identity allows you to conclude this, but the real version doesn't quite do it). --Ron Bruck ==== Subject: Complex Analysis Question I am curious if there is a general way to understand the solution to the following (simple) complex analysis problem. Suppose we have an n-dimensional vector space, and a fixed (i.e. known) vector x in C^n. The question is whether there is a way to describe the set of all vectors z such that: (i) z' * x = 1 (where ' denotes conjugate transpose). If you sketch out this problem, it is not difficult to see that the conditions: (ii.a) Re(z' * x) = 1 and (ii.b) Im(z' * x) = 0 give you two linear equations in 2n unknowns, which you could then solve (in a least squares sense). This method of analysis, however, does little to describe what the solution set is like. I am curious if anyone else has another way of thinking about this problem. More general pictures for how to visualize this condition (like as a projection for example) do not seem that easy to conjur up. Juno ==== Subject: Re: Complex Analysis Question Notice that if z_1 and z_2 are both solutions of z' * x = 1, then z_1 - z_2 is a solution to the equation z' * x = 0. What can you say about the set of solutions z to the equation z' * x = 0? _____ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 > I am curious if there is a general way to understand the solution to > the following (simple) complex analysis problem. > Suppose we have an n-dimensional vector space, and a fixed (i.e. known) > vector x in C^n. > The question is whether there is a way to describe the set of all > vectors z such that: > (i) z' * x = 1 > (where ' denotes conjugate transpose). > If you sketch out this problem, it is not difficult to see that the > conditions: > (ii.a) Re(z' * x) = 1 > and > (ii.b) Im(z' * x) = 0 > give you two linear equations in 2n unknowns, which you could then > solve (in a least squares sense). > This method of analysis, however, does little to describe what the > solution set is like. I am curious if anyone else has another way of > thinking about this problem. More general pictures for how to visualize > this condition (like as a projection for example) do not seem that easy > to conjur up. > Juno ==== Subject: A question about surfaces Let S be a compact orientable surface. Prove that the Gauss map of S is surjective. How to prove it? Do you need the classification theorem or something? ==== Subject: Re: A question about surfaces What exactly the Gauss map is? ==== Subject: a linear algebra question Hi all, Can anyone help me with the following question? Suppose B is an nxn matrix (I know it is symetric but that may not be important but fill free to use it if needed). Any 2x2 minor of B is known to be positive definite (not semidefinite! but strictly positive definite). Is it true that also B is positive definite? (again not just semidefinite I need the strict positive definite). NCDude P.S. I know that the oposite is not true (i.e. for a matrix to be positive definite it is not necessary to have all its minors being positive definite) what I'm looking for is the direction that says that if the minors are positive definite then also the big matrix is positive definite. ==== Subject: Re: a linear algebra question >Can anyone help me with the following question? >Suppose B is an nxn matrix (I know it is symetric but that may not be >important but fill free to use it if needed). Any 2x2 minor of B is >known to be positive definite (not semidefinite! but strictly positive >definite). Is it true that also B is positive definite? (again not just >semidefinite I need the strict positive definite). The business with minors checks to see whether a quadratic form is positive definite on the coordinate planes. So what you want is a quadratic form that's positive definite on planes but not on the whole space. Very well: make an indefinite form whose primary negative direction is far from any coordinate plane, e.g. q(x,y,z) = x^2 + y^2 + z^2 - k (x+y+z)^2 If you make k too small, Poppa Bear will exclaim, This quadratic form is too definite!. If you make k too large, Momma Bear will say, This quadratic form is not definite enough! Choose an intermediate k so Baby Bear will say But this quadratic form is just right! dave ==== Subject: Re: a linear algebra question Suppose B is an nxn matrix (I know it is symetric but that may not be >important but fill free to use it if needed). Any 2x2 minor of B is >known to be positive definite (not semidefinite! but strictly positive >definite). Is it true that also B is positive definite? (again not just >semidefinite I need the strict positive definite). > The business with minors checks to see whether a quadratic form is > positive definite on the coordinate planes. So what you want is a > quadratic form that's positive definite on planes but not on the > whole space. Very well: make an indefinite form whose primary > negative direction is far from any coordinate plane, e.g. > q(x,y,z) = x^2 + y^2 + z^2 - k (x+y+z)^2 > If you make k too small, Poppa Bear will exclaim, This quadratic > form is too definite!. If you make k too large, Momma Bear will > say, This quadratic form is not definite enough! Choose an intermediate > k so Baby Bear will say But this quadratic form is just right! So, proving that math is a fairy tale? R.G. Vickson > dave ==== Subject: Re: a linear algebra question > Hi all, > Can anyone help me with the following question? > Suppose B is an nxn matrix (I know it is symetric but that may not be > important but fill free to use it if needed). Any 2x2 minor of B is > known to be positive definite (not semidefinite! but strictly positive > definite). Is it true that also B is positive definite? (again not just > semidefinite I need the strict positive definite). > NCDude > P.S. I know that the oposite is not true (i.e. for a matrix to be > positive definite it is not necessary to have all its minors being > positive definite) what I'm looking for is the direction that says that > if the minors are positive definite then also the big matrix is > positive definite. If the minors in question are the matrices of the form B(i,i) B(i,j) B(j,i) B(j,j) then the matrix ( 4 3 3 ) ( 3 4 -3 ) ( 3 -3 4 ) has all its 2 x 2 minors positive definite, but is not positive definite itself, as its determinant is -98 On the other hand if B is positive definite then all its minors are also positive definite. For example if M is the minor above, and v is the vector (x,y)', then v'*M*v = u'*B*u where u is a vector with x in the i'th slot and y in the j'th. So v'*M*v >= 0, and is only 0 if u is zero, ie if both x and y are 0. Duncan ==== Subject: how much to rotate by? (NEED HELP!!) I have x and y coordinates that need to be rotated in such a way with any method, so that the results I get from the rotation, when atan-1(y/x) must equal as close to 60 degrees as possible. eg. (257,381) when rotated by 184.10 degrees gives (229,398) now atan-1(398/229) = 60.08. The 184.10 rotation angle I had to guess..I am trying to obtain a method that would prevent me from guessing how many degrees to rotate by so when i do the atan-1 function i get 60 degrees or as close to it as possible.. When responding with a solution to my problem kindly show the math workings also so i can find the solution easier to understand and so can replicate it without difficulty. ==== Subject: Re: how much to rotate by? (NEED HELP!!) > I have x and y coordinates that need to be rotated in such a way with any method, so that the results I get from the rotation, when atan-1(y/x) must equal as close to 60 degrees as possible. > eg. (257,381) when rotated by 184.10 degrees gives (229,398) now atan-1(398/229) = 60.08. This doesn't make sense. A rotation of a point in the first quadrant by 180 degrees should have put it into the third quadrant, with x and y both negative. > The 184.10 rotation angle I had to guess..I am trying to obtain a method that would prevent me from guessing how many degrees to rotate by so when i do the atan-1 function i get 60 degrees or as close to it as possible.. > When responding with a solution to my problem kindly show the math workings also so i can find the solution easier to understand and so can replicate it without difficulty. The formula for a general counterclockwise is: x' = x*cos(theta) - y*sin(theta) y' = x*sin(theta) + y*cos(theta) So I'd do the following: (1) Find the current angle alpha using a two-argument arctan or cartesian-to-polar conversion. (For your example, the point (257,381) is at alpha = 55.999 degrees). (2) Rotate by theta = 60 - alpha. For your example, that means theta = 4.001 degrees, cos(theta) = .9976, sin(theta) = 0.0698, and x' = 229.8, y' = 398.0, and atan(y'/x') = 60.00. Note that you apparently mean something different by rotate than I do. I'm rotating about the origin, in which case a rotation of 184 degrees does not give me the values you got. - Randy ==== Subject: Re: how much to rotate by? (NEED HELP!!) Randy, thank you very much the solution is good. but just to clarify something with you by means of an example. Using (176,316) and the same formula as mentioned in your number 2 paragraph. 60-alpha now what if alpha is greater than 60? Do we reverse the subtraction or we still minus 60-alpha no matter the case? in this case atan(316/176)=60.88 so 60-alpha would give -.88 reversing the sum would simply give .88 with the -, but 2 different rotation results nevertheless. just want to make very sure. thank you! ==== Subject: Re: how much to rotate by? (NEED HELP!!) <6874585.1139422859215.JavaMail.jakarta@nitrogen.mathforum.org Randy, thank you very much the solution is good. but just to clarify something with you by means of an example. > Using (176,316) and the same formula as mentioned in your number 2 > paragraph. 60-alpha now what if alpha is greater than 60? Do we > reverse the subtraction or we still minus 60-alpha no matter the case? Use 60 - alpha. You can take sine and cosine of negative numbers. > in this case atan(316/176)=60.88 so 60-alpha would give -.88 > reversing the sum would simply give .88 with the -, but 2 > different rotation results nevertheless. cos(-0.8839 deg) = 0.9999 sin(-0.8839 deg) = -0.0154 x' = 176*0.9999 - 316*(-0.0154) = 175.98 + 4.87 = 180.85 y' = 176*(-0.0154) + 316*(0.9999) = -2.71 + 315.96 = 313.25 - Randy ==== Subject: Re: how much to rotate by? (NEED HELP!!) randy, thank you very much!! =) ==== Subject: Re: Beyond Not Even Wrong: Beyond String Theory: New Theories on Time Rocking The Physics World http://physicsmathforums.com/showthread.php?p=1332#post1332 > sciencenews sciencenews is offline > Junior Member > Posts: 1 > Default Mayer's New Theories > Please see the Website www.stanford.edu/~afmayer and review the > lectures. It will take you about 10 minutes to make an initial > evaluation. I would like to see sober academic discussion of the new > ideas here. > Edit/Delete Message Reply With Quote > sciencenews > View Public Profile > Send a private message to sciencenews > Find all posts by sciencenews > Add sciencenews to Your Buddy List > #2 IP > Old Today, 01:15 PM > astro astro is online now > Administrator > Posts: 64 > Default Moving Dimensions Theory stipulates that time is an emergent > property of a fourth exp > Moving Dimensions Theory stipulates that time is an emergent property > of a > fourth expanding dimension. > Your diagrams seem to agree with what I've been saying: > http://www.stanford.edu/~afmayer/do...ture2Signed.pdf (PAGE 26) > http://www.stanford.edu/~afmayer/do...ture2Signed.pdf (PAGE 23) > The fourth expanding dimension appears as a spherically-symmetric > wavefront > in the three spatial dimensions. So this wavefront moves along the length, width, and height, dimensions but why is it expanding instead of contrating as it moves in these other expanding dimensions? > At every point in the universe, the fourth dimension is expanding, > manifesting itself as a spherically-symmetric wavefront in our three > dimensions. Hence the spercially symmetric propagation of a photon's > probability wave--a photon is matter that exists in the expanding > fourth > dimension. > Hence matter, when fully rotated into the fourth expanding dimension, > becomes orthogonal to the three spactial diemnsions and travels with > the > fourth expanding dimension. We see such matter--when it's rotated fully > into the fourth expanding dimension--as photons. > Hence the equivalence of matter and energy. Energy is but matter that > exists in the fourth expanding dimension. > Hence the length contraction associated with all motion. The more > something is rotated into the fourth dimension, the shorter it appears > to > us in the three dimensions, and the faster it moves. > Many other phenomena are accounted for throughout the MDT paper. > Your diagrams seem to agree with what I've been saying: > http://www.stanford.edu/~afmayer/do...ture2Signed.pdf (PAGE 23) > http://www.stanford.edu/~afmayer/docs/Lecture2Signed.pdf > Best, > Dr. Elliot McGucken > Hello All, > Please respect that Moving Dimensions Theory is just a theory. > I look forward to feedback and insights regarding its logic. > Moving Dimensions Theory > http://physicsmathforums.com > Questions Addressed by MDT: > Why is the speed of light constant in all frames? > Why are light and energy quantized? > Why are there non-local effects in quantum mechanics? > Why does time stop at the speed of light? > How come a photon does not age? > Why are inertial mass and gravitational mass the same thing? > Why do moving bodies exhibit length contraction? > Why are mass and energy equivalent? > Why does time's arrow point in the direction it points in? Why > entropy? > Why do photons appear as spherically-symmetric wavefronts traveling > with the velocity c? > Why is there a minus sign in the following metric? > x^2+y^2+z^2-c^2t^2=s^2 > What deeper reality underlies Einstein's postulates of relativity? > What deeper reality underlies Newton's laws? > What underlies the laws of Inertia? > Why does general relativity fail at short distances? Why does quantum > mechanics dominate at short distances? > Why have so many great minds, Einestin, Godel, Wheeler, Hawking, and > Penrose called for a new conception of time? > If at first the idea is not absurd, then there is no hope for it. > --Albert Einstein > If I have seen further it is by standing on the shoulders of giants. > --Isaac Newton > Max Planck, the father of quantum theory, felt that the pioneer > scientist must have a vivid intuitive imagination, for new ideas are > not generated by deduction, but by artistically creative imagination. > An important scientific innovation rarely makes its way by gradually > winning over and converting its opponents: What does happen is that the > opponents gradually die out. > --Max Planck > Moving Dimensions Theory (MDT) > Today I am writing regarding Moving Dimensions Theory-a deeper model > for explaining diverse phenomena in both quantum mechanics and > relativity. > The General Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions. > The Specific Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions at the rate of c in quantized units of the Planck length. > Relativistic, classical, and quantum mechanical phenomena, as well as > time itself, are emergent properties of this fundamental principle. > Newton's laws, the principle of Inertia, Einstein's postulates, and > model. > A FEW YEARS BACK > A few years back, while surfing a towering wave on the Outer Banks of > North Carolina, a beautiful thought occurred to me. Suppose the wave I > was riding represented a coordinate in a dimension. Then although I was > approaching shore, I was not moving in this dimension. > The dimension itself was moving with me-I was surfing the dimension. > In a flash I saw that that is why photons never age-they are moving > along with the fourth dimension, and thus stationary relative to it. In > another flash I saw that that is why a photon's space-time interval > is represented by a null vector, or a 0, no matter how far it travels. > Indeed Einstein stated that an object's velocity through space-time > was always c-even stationary objects are traveling at the velocity c > through time! How could this be, were it not for a fourth expanding > dimension, which matter could surf as photons, giving rise to our > notion of time? And so it is that Moving Dimensions Theory was born as > the wave crested and crashed about me, thundering on down, as I fought > to remain surfing amidst the foam, facing the setting sun silhouetting > the Hatteras light. > And the waves kept on crashing that night. The nonlocal EPR > paradox/effect could be explained by the underlying nonlocality of an > expanding fourth dimension. The equivalence of mass and energy, the > light-it could all be understood via a single principle of Moving > Dimensions Theory: the fourth dimension is expanding relative to the > three spatial dimensions. MDT reached back thousands of years to > resolve Zeno's paradox, then voyaged forth to ease Godel's, > Einstein's, Hawking's, and Penrose's concerns with the > paradoxical nature of a block universe, and arrived in the present, > quelling the oft exaggerated conflicts between relativity and quantum > mechanics, and pointing the way to the future by accounting for > time's arrow and entropy herself. At long last GR and QM could be > married in theory as harmoniously as they are in nature with Moving > Dimensions Theory's simple postulate: > The General Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions. > The Specific Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions at the rate of c in quantized units of the Planck length. > Classical physics, quantum mechanics, and relativity descend from this > simple postulate. Light, and thus all energy, is quantized as the > dimension which transports it expands in a quantized manner. Light > travels at a constant velocity in all frames because velocity is > measured relative to time which is measured relative to the light that > is transported by the fourth expanding dimension. Thus both fundamental > constants h and c emerge from the fundamental nature of the expansion > of the fourth dimension relative to the three spatial dimensions. And > thus MDT provides a simple, unifying postulate accounting for the > classical, relativistic, and quantum mechanical properties of this > universe. > And it's always been simple postulates, as opposed to abstruse > mathematics and mythologies, that have furthered physics. > Moving Dimensions Theory Can Unify GR & QM: > By offering an underlying reality from where both branches of physics > emerge-an underlying reality of a fourth dimension expanding relative > to three spatial dimensions, MDT unifies relativity and quantum > mechanics not with indecipherable mathematical mythologies, but with a > simple postulate. MDT explains quantum mechanical effects such as > and energy, as well as the two postulates of relativity: the speed of > light is constant in all inertial frames and the laws of physics are > the same for all inertial observers. MDT also explains relativistic > effects such as time dilation and length contraction. The beauty of > Moving Dimensions Theory is that it explains properties of quantum > mechanics and relativity in the deeper context of a unified framework, > opening a door to a deeper physical reality-the fourth dimension is > expanding relative to the three spatial dimensions. > The Purpose of Physics > The purpose of physics has ever been to unify diverse physical > phenomena with simple postulates, laws, and formulas reflecting the > deeper physical reality. MDT unifies relativity and quantum mechanics > by positing that they are both emergent properties of moving > dimensions. MDT's simple postulate-the fourth dimension is > expanding relative to the three spatial dimensions-offers the first > satisfactory explanation of the Einstein Podolsky Rosen (EPR) effect > and the nonlocal behavior inherent to the math and physical reality of > quantum mechanics. Time itself is viewed not as the fourth dimension, > but as an emergent phenomena arising from the expansion of the fourth > dimension relative to the three spatial dimensions. This logic > alleviates a confusion of time with an actual fourth dimension where > one can travel back and forth at will, thus addressing Godel's, > Einstein's, Hawking's, Barbour's, and Penrose's concerns about > frozen time, and accounting for time's relentless arrow, the second > law of thermodynamics, and entropy. > This is but a brief treatment of a much larger project. > The General Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions. > The Specific Postulate of Moving Dimensions Theory: > The fourth dimension is expanding relative to the three spatial > dimensions at the rate of c in quantized units of the Planck length. > and relativity, including the following: > The Constant Velocity of Light: > Light travels with constant velocity of c, because the fourth dimension > is expanding relative to the three spatial dimensions at the rate of c. > Light, or energy, is matter rotated completely into the expanding > fourth dimension, orthogonal to the three spatial dimensions. No matter > how fast a spaceship is traveling, when it turns its lights on, the > light can only propagate as fast as the expanding fourth dimension can > carry it. > The Constant Velocity of Light in All Inertial Frames: > The velocity of light is always measured relative to the velocity of > time, and the velocity of time is always measured relative to the > velocity of light. This tautology assures us that the velocity of light > will always be the same for all observers in all inertial frames, as > the velocity of light is being measured relative to the velocity of > light in that frame. However, as demonstrated by experiments, time and > light travel slower close to gravitational masses, when measured from > distant frames. > What is Time? > Time is an emergent property of the underlying reality that a fourth > dimension is expanding relative to the three spatial dimensions. All > our measurements of time are based on the emission and propagation of > photons, and all photons propagate by surfing the expanding fourth > dimension. So it is that time inherits properties of the fourth > dimension, but time is not the fourth dimension. > Too many physicists have extended dimensional properties to the notion > of time, rather than realizing that time is an emergent property tied > closely to a fourth expanding dimension. Because our notions of time > are linked ... > Edit/Delete Message Reply With Quote > astro > View Public Profile > Send a private message to astro > Find all posts by astro > Add astro to Your Buddy List > #3 Report Bad Post IP > Unread Today, 03:26 PM > Dr_Strangelove Dr_Strangelove is online now > Junior Member > Location: Michigan > Posts: 14 > Send a message via Yahoo to Dr_Strangelove > Default Mayers Lectures > Mayer uses some relativistic reasoning to propose what he calls a > transverse red shift. Using the equivalence principal he applies this > reasoning to gravity. It's a fair argument and deserves consideration. > He also develops a modified Schwarzschild metric; again form > relativistic reasoning. > I don't see where either of these proposals would lead one to assume > that time is structured any more then the present theory, and he > doesn't mention directional time until late in lecture 2. > He provides a plethora of empirical evidence to refute the validity of > present theory and subjectively discusses how his modifications might > explain the myriad of anomalies. > I was pleased to see that many of the anomalies I conjectured based on > directional time have in fact been observed. > http://members.triton.net/daveb It is surprising that such experiments > such as gravity probe B have been funded in light of the overwhelming > empirical evidence suggesting that satellites move in as of yet > unpredictable ways. > Only after he goes into cosmological effects does he specifically > propose structured time and his reasoning is basically similar to my > own; the apparent expansion of the Universe is a manifestation of the > direction of time. I postulated that idea several years ago, and > published it on my web site as well as many physics forums. > His time flows radically outward from some center point very much like > what I believe Proffessor Elliot McGucken has proposed. > In conclusion, I don't see how his transverse red shift and modified > Schwarzschild metric lead one to propose structured time anymore then > conventional relativity does. It is implicit in his arguments that red > shift is related to time direction; something that my work and Dr. > Elliopts explicitly state. It is possible that his idea of directional > time resulted form other then his imagination.The top 2 results of a > suggesting directional time. > Nonetheless, he provides a wealth of empirical evidence in favor of our > structured time ideas. http://members.triton.net/daveb > http://physicsmathforums.com/showthread.php?p=1332#post1332 ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, > but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. I need to fill in a few details to understand your argument. Please bear with me. Let's go back to basics. You will agree that a key to your argument is the constant terms. Your notation implies that a_1(x) and a_2(x) are functions of x. You know that a_1(0) = 0 and a_2(0) = -1. These are the constant terms of these two functions. Let f(x) be the maximal divisor of 7 which also divides a_1(x), in the algebraic integers. That is, f(x) is an algebraic integer and 7/f(x) is an algebraic integer and a_1(x)/f(x) is an algebraic integer. I am not saying anything here about what f(x) actually is. Define g(x) similarly for a_2(x): that is, g(x) is an algebraic integer and 7/g(x) is an algebraic integer and a_2(x)/g(x) is an algebraic integer. Further it must be the case that f(x) g(x) = 7, because 7 factors out of the whole expression only once. Now consider (5 a_1(x) + 7)/f(x) = 5 a_1(x)/f(x) + 7/f(x). This is an algebraic integer because each part is separately. Similarly, (5 a_2(x) + 7)/g(x) = 5 a_2(x)/g(x) + 7/g(x). Now, what you claim is that f(x) must equal 7 for all x and g(x) must equal 1 for all x, at least up to units. Here is how I think your reasoning goes, though you don't spell it all out here. The constant term of 5 a_2(x) + 7 must be 5 a_2(0) + 7 = 5*(-1) + 7 = 2. Therefore the constant term of (5 a_2(x) + 7)/g(x) must be 2/g(x). Now, since g(x) is a divisor of 7, it is coprime to 2. So the only possibility for g(x) is that it is a unit, i.e., it is essentially equal to 1. And then you can conclude f(x) must be 7. Is this a reasonable summary of your logic? Marcus > Note that the result that only one of the factors was multiplied by 7 > follows from the distributive property, and noting the values of the > constant terms. > That completes the proof. It's rather short as you can see. > It contradicts with standard ideas from Galois Theory because from that > theory, if the a's are non-rational and irreducible over Q, except for > the special case where both have sqrt(7) which I'll go over next, then > by those ideas NEITHER of the a's can have 7 as a factor. > And it can be shown that neither does have 7 as a factor in the ring of > algebraic integers. > But that then proves--as the simple proof cannot be wrong or it > wouldn't be a proof--that the ring of algebraic integers is incomplete, > and for those who don't understand why it shows that, consider the > analogy where considering evens as a ring, 2 and 6 are coprime in that > ring because 3 is not even. > Make sense? Similarly, 7 being coprime > integers, when it has been proven to be a factor by a simple algebraic > proof, > shows that it is excluded because its co-factor is not an > algebraic integer. > I would be interested in at least some posters actually trying to > answer the proof itself, and its steps, especially the key step where I > note that the constant factor, which is 7, proves that 7 was multiplied > through that particular factor. > Now to the sidepoint about special cases. It's not hard to explain > mathematically. > It's easy to determine that at x=1 mod 7, both of the a's have sqrt(7) > as a factor. > The explanation is simple enough, but it might help in understanding to > divide 7 from the factorization and solve for the solution where 7 is a > factor of only one, so > Q(x) = (x^2 + x)(5^2) + (-1 + x)(5) + 7 > = 25 x^2 + 30 x + 2 > and > Q(x) = (5b_1(x) + 1)(5a_2(x) + 7) > where a_1(x) = 7b_1(x), so I also have > b_1(x) a_2(x) = x^2 + x > and > 5a_2(x) + 7(5)b_1(x) = (-1 + x)(5) > so I can solve out a_2(x), to get > a_2(x) = (x^2 + x)/b_1(x), and substitute in the second to get > (x^2 + x)/b_1(x) + 7b_1(x) = -1 + x > which gives > 7 b_1(x)^2 - (x-1)b_1(x) + x^2 + x = 0 > and the solution that > b_1(x) = ((x-1) +/- sqrt((x-1)^2 - 28(x^2 + x)))/14 > and at x=1, I have have > b_1(1) = +/- sqrt(-14(5))/14 > so it's just that at that point and it turns out at all points where x > = 1 mod 7, that 7 b_1(x) has sqrt(7) as a factor. > So that case actually says nothing about the earlier argument, which > depends on the distributive property. > If you wish to dispute with only one factor having 7 as its factor, > then you might consider the more general case: > Q(x) = (5b_1(x) + f_1)(5b_2(x) + f_2) > where f_1 f_2 = 7, and > b_1(x) f_2 = a_1(x) > and > b_2(x) f_2 = a_2(x) > and answer the question of, what mathematical reason might there be for > any particular choice for the f's? > I've given a reason based on the distributive property and the constant > terms. > If you dispute that reason, you have an INFINITE number of possible > algebraic integer f's that are available, so how do you choose one > configuration? > Note again the rest here is a side discussion noting how the cases at x > = 1 mod 7 do not change the basic argument, which relies on the > distributive property. > That argument has gone through formal peer review and been published > with the more complicated expressions I used in my paper on > non-polynomial factorization, but then some social pressure was brought > to bear agains the journal and the paper was pulled, and a bit later > the journal keeled over and died. ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, > but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. > I need to fill in a few details to understand your argument. Please > bear with me. > Let's go back to basics. You will agree that a key to your argument > is the constant terms. Yes. And they are 7 and 2. > Your notation implies that a_1(x) and a_2(x) are functions of x. Yes, they are functions of x. > You know that a_1(0) = 0 and a_2(0) = -1. These are the constant > terms of these two functions. Those are their values at 0. The constant terms are the factors of the constant term of the polynomial--and I remind that the polynomial is 7(25 x^2 + 30 x + 2) where you can see that the constant term is 7(2). > Let f(x) be the maximal divisor of 7 which also divides a_1(x), > in the algebraic integers. That is, f(x) is an algebraic integer > and 7/f(x) is an algebraic integer and a_1(x)/f(x) is an algebraic > integer. I am not saying anything here about what f(x) actually > is. Ok. > Define g(x) similarly for a_2(x): that is, g(x) is an algebraic > integer and 7/g(x) is an algebraic integer and a_2(x)/g(x) is > an algebraic integer. Ok. > Further it must be the case that f(x) g(x) = 7, because 7 > factors out of the whole expression only once. Yup. > Now consider > (5 a_1(x) + 7)/f(x) = 5 a_1(x)/f(x) + 7/f(x). > This is an algebraic integer because each part is separately. Ok. > Similarly, > (5 a_2(x) + 7)/g(x) = 5 a_2(x)/g(x) + 7/g(x). > Now, what you claim is that f(x) must equal 7 for all x and > g(x) must equal 1 for all x, at least up to units. Huh? I claim that the distributive property proves that one factor is multiplied through by 7, while the other is not. And here it IS important to be extremely precise about what I claim, as what you just said I claim, is kind of all over the map in comparison, and notice, you make NO MENTION OF THE DISTRIBUTIVE PROPERTY. I wish to emphasize that while my actual claim relies on the distributive property. YOU MAKE NO MENTION OF THE DISTRIBUTIVE PROPERTY. Let's cut to the chase. You will note that you can find algebraic integers for f(x) and g(x), and I will say, so? The distributive property still works, so what is the other possible explanation? I am curious to know if you are willing to even concede that there is a possible explanation for how 7 can multiply through by the distributive property, and yet you can find f(x) and g(x) in the ring of algebraic integers. Can you think of a way? > Here is how I think your reasoning goes, though you don't spell it > all out here. > The constant term of 5 a_2(x) + 7 must be > 5 a_2(0) + 7 = 5*(-1) + 7 = 2. > Therefore the constant term of > (5 a_2(x) + 7)/g(x) > must be 2/g(x). No. The mathematics only tells you from the constant terms that one of the a's has been multiplied through by 7, while one has not, but not which one. By the distributive property, there is only one conclusion. Now then, how do you explain that f(x) and g(x) can be found where they are algebraic integers, when it can be shown that if f(x)=7, there are non-rational cases where they are not? I can explain it. If you wish, you can try to avoid my questions in this area, and then I can come back and explain exactly how. > Now, since g(x) is a divisor of 7, it is coprime to 2. So > the only possibility for g(x) is that it is a unit, i.e., it > is essentially equal to 1. And then you can conclude f(x) must > be 7. > Is this a reasonable summary of your logic? > Marcus You left in the rest of what I said, and I'll truncate it in a bit, but I want to leave in the next two paragraphs. > Note that the result that only one of the factors was multiplied by 7 > follows from the distributive property, and noting the values of the > constant terms. > That completes the proof. It's rather short as you can see. My position relies on a(b+c) = ab + ac and talk about units here or units there and what's true in the ring of algebraic integers do not change the reality of the key step in the proof. The ring of algebraic integers is a complex entity made more complex by special rules which some of you are acting like you don't understand, so you get results that contradict with what can be shown by proof. So with the ring of algebraic integers you can contradict the distributive property. Fun stuff, if you're a mathematician, a real one. But ultimately you have to resolve the contradiction to show that mathematics IS consistent...if you're a real mathematician. James Harris ==== Subject: Re: JSH: Decker example, proper answer >> Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. >> So both constant terms--those terms constant with respect to x--cannot >> be visible in the factorization. >> However, the actual constant terms can be determined by clearing x out, >> letting x=0, I have that one of the a's goes to 0, while the remaining >> one is -1, from >> a^2 + a = 0. >> And looking at >> 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) >> I have that the actual constant terms are 7 and 5(-1) + 7 = 2, >> but it's >> not mathematically determined which of the two factors was multiplied >> by 7, and which one actually is coprime to 7, but you know ONE must >> have been multiplied by 7, from the distributive property and the >> constant terms. >> I need to fill in a few details to understand your argument. Please >> bear with me. >> Let's go back to basics. You will agree that a key to your argument >> is the constant terms. >Yes. And they are 7 and 2. >> Your notation implies that a_1(x) and a_2(x) are functions of x. >Yes, they are functions of x. >> You know that a_1(0) = 0 and a_2(0) = -1. These are the constant >> terms of these two functions. >Those are their values at 0. The constant terms are the factors of the >constant term of the polynomial--and I remind that the polynomial is > 7(25 x^2 + 30 x + 2) >where you can see that the constant term is 7(2). >> Let f(x) be the maximal divisor of 7 which also divides a_1(x), >> in the algebraic integers. That is, f(x) is an algebraic integer >> and 7/f(x) is an algebraic integer and a_1(x)/f(x) is an algebraic >> integer. I am not saying anything here about what f(x) actually >> is. >Ok. >> Define g(x) similarly for a_2(x): that is, g(x) is an algebraic >> integer and 7/g(x) is an algebraic integer and a_2(x)/g(x) is >> an algebraic integer. >Ok. >> Further it must be the case that f(x) g(x) = 7, because 7 >> factors out of the whole expression only once. >Yup. >> Now consider >> (5 a_1(x) + 7)/f(x) = 5 a_1(x)/f(x) + 7/f(x). >> This is an algebraic integer because each part is separately. >Ok. >> Similarly, >> (5 a_2(x) + 7)/g(x) = 5 a_2(x)/g(x) + 7/g(x). >> Now, what you claim is that f(x) must equal 7 for all x and >> g(x) must equal 1 for all x, at least up to units. >Huh? I *think* that is your conclusion, no? You cannot assume that f(x) = 7 and g(x) = 1 at the outset. That would be assuming what you want to prove. You may ultimately CONCLUDE that f(x) = 7 and g(x) = 1 from your argument. That's all I am saying here. I just want to look at your argument in more detail. >I claim that the distributive property proves that one factor is >multiplied through by 7, while the other is not. Yes - I used the distributive property also - it's in the equation (5 a_1(x) + 7)/f(x) = 5 a_1(x)/f(x) + 7/f(x) Here I am using it in the form (a + b) / c = a/c + b/c, but it is equivalent to the multiplicative form. >And here it IS important to be extremely precise about what I claim, as >what you just said I claim, is kind of all over the map in comparison, >and notice, you make NO MENTION OF THE DISTRIBUTIVE PROPERTY. I used it, as I noted above. I didn't think it was significant enough to mention it. >I wish to emphasize that while my actual claim relies on the >distributive property. >YOU MAKE NO MENTION OF THE DISTRIBUTIVE PROPERTY. See above. >Let's cut to the chase. >You will note that you can find algebraic integers for f(x) and g(x), >and I will say, so? >The distributive property still works, so what is the other possible >explanation? You are using the distributive property in almost exactly the same way I am, except you use 7 where I am using f(x). That's OK. But I am not assuming at the outset that f(x) is 7. It is just some algebraic integer. It might be 7 and it might not, but it would be incorrect to assume it is 7 and then claim to prove it also. >I am curious to know if you are willing to even concede that there is a >possible explanation for how 7 can multiply through by the distributive >property, and yet you can find f(x) and g(x) in the ring of algebraic >integers. I am making fewer assumptions than you are. Saying that f(x) factors out of the first factor and g(x) factors out of the second factor is more general than saying 7 must factor out of the first factor and 1 must factor out of the second. I am trying to figure out how you can conclude that g(x) = 1 and f(x) = 7, so I don't want to assume it in the first place. >Can you think of a way? >> Here is how I think your reasoning goes, though you don't spell it >> all out here. >> The constant term of 5 a_2(x) + 7 must be >> 5 a_2(0) + 7 = 5*(-1) + 7 = 2. >> Therefore the constant term of >> (5 a_2(x) + 7)/g(x) >> must be 2/g(x). >No. No? Then what the heck is it ??? >The mathematics only tells you from the constant terms that one of the >a's has been multiplied through by 7, while one has not, but not which >one. No, it doesn't. It tells that you can factor 7 out of the whole thing. It is very possible that 7 is the product of two other numbers - I am calling them f(x) and g(x) - and that f(x) is a factor of a_1(x) and g(x) is a factor of a_2(x). The math doesn't tell you immediately that f(x) must be 7 and g(x) must be 1. It does tell you that f(0) must be 7 and g(0) must be 1, but it doesn't tell you their values for other x's. So I think it gets back to figuring out what the constant term of (5 a_2(x) + 7)/g(x) is. If it is 2/g(x), then you are right, because g(x) is posited to be a divisor of 7 and the only divisors of 7 that also divide 2 are units. So g(x) would have to be a unit, and you have your conclusion. Isn't that just what you are saying? >By the distributive property, there is only one conclusion. No - not a priori. You must assume a more general factorization and then prove that the one you want is the only one possible. think it is obvious, but the rest of us are probably a little slower. >Now then, how do you explain that f(x) and g(x) can be found where they >are algebraic integers, when it can be shown that if f(x)=7, there are >non-rational cases where they are not? No point in discussing non-rationals here as far as I can see. That just confuses things. >I can explain it. >If you wish, you can try to avoid my questions in this area, and then I >can come back and explain exactly how. >> Now, since g(x) is a divisor of 7, it is coprime to 2. That's just what I was saying when I was asking about 2/g(x) ! I just want to know if I am interpreting your argument correctly. I may be filling in some stuff that you thought was obvious. Marcus. >> So >> the only possibility for g(x) is that it is a unit, i.e., it >> is essentially equal to 1. And then you can conclude f(x) must >> be 7. >> Is this a reasonable summary of your logic? >> Marcus >You left in the rest of what I said, and I'll truncate it in a bit, but >I want to leave in the next two paragraphs. >> Note that the result that only one of the factors was multiplied by 7 >> follows from the distributive property, and noting the values of the >> constant terms. >> That completes the proof. It's rather short as you can see. >My position relies on >a(b+c) = ab + ac >and talk about units here or units there and what's true in the ring of >algebraic integers do not change the reality of the key step in the >proof. >The ring of algebraic integers is a complex entity made more complex by >special rules which some of you are acting like you don't understand, >so you get results that contradict with what can be shown by proof. >So with the ring of algebraic integers you can contradict the >distributive property. >Fun stuff, if you're a mathematician, a real one. >But ultimately you have to resolve the contradiction to show that >mathematics IS consistent...if you're a real mathematician. >James Harris ==== Subject: Re: JSH: Decker example, proper answer days. My association with the Department is that of an alumnus. [...] >Let's go back to basics. You will agree that a key to your argument >is the constant terms. >Your notation implies that a_1(x) and a_2(x) are functions of x. >You know that a_1(0) = 0 and a_2(0) = -1. These are the constant >terms of these two functions. >Let f(x) be the maximal divisor of 7 which also divides a_1(x), >in the algebraic integers. That is, f(x) is an algebraic integer >and 7/f(x) is an algebraic integer and a_1(x)/f(x) is an algebraic >integer. I am not saying anything here about what f(x) actually >is. >Define g(x) similarly for a_2(x): that is, g(x) is an algebraic >integer and 7/g(x) is an algebraic integer and a_2(x)/g(x) is >an algebraic integer. You are missing one condition each (to account for your maximal): (i) If y is any algebraic integer such that 7/y and a_1(x)/y are algebraic integers, then f(x)/y is an algebraic integer. (ii) If z is any algebraic integer such that 7/z and a_2(x)/z are algebraic integers, then g(x)/z is an algebraic integer. >Further it must be the case that f(x) g(x) = 7, because 7 >factors out of the whole expression only once. This would not follow without the two conditions above. (f(x)=g(x)=1 whenever x is nonzero satisfy all conditions you give, for instance) [.rest deleted.] -- ' ==== Subject: Re: JSH: Decker example, proper answer > What happened over the weekend is that I looked for another argument to > go along with the one that has been checked so thoroughly, as you > people keep fighting me over the distributive property, and I saw > mirages. > Big deal. > In this thread I went back to the tried-and-true where I guess people > will again argue with me over the freaking distributive property. > Oh well. I was having an email discussion a couple of days ago with one of the regulars here about your silence since the weekend. To quote from something I said in an email Monday: doubts he's having about his 'proof, though it's a toss-up whether he'll be depressed or just say , 'Maybe I was wrong, oh well, no big deal.' You're *so* predictable. -- Wayne Brown (HPCC #1104) | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) + 1 = 0 -- Euler | -- John Myers Myers, Silverlock ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. No. You keep tripping up on this point. Just because the product of two numbers, pq, is divisible by 7 in the algebraic integers, you cannot conclude that either p or q must be divisible by 7. That's true in the integers, but is not true in the algebraic integers. > Note that the result that only one of the factors was multiplied by 7 > follows from the distributive property, and noting the values of the > constant terms. > That completes the proof. It's rather short as you can see. It isn't a proof. Unsurprisingly, you've been seduced by the rational integers. There are only two rational integers x that give rational integer roots for r(a, x) = a^2 + (1 - x) a + 7(x^2 + x) namely, x = 0 r(a, x) = a^2 + a x = -1 r(a, -1) = a^2 + 2 a In these two cases, your conjecture is true, namely that one of the roots is dividible by 7 and the other is coprime to 7. Unfortunately, these are the ONLY values of x for which the roots behave as you want them to. For example, let x = 3. We see that 7 Q(3) = 7(25(9) + 30(3) + 2) = 7(317) and r(a, 3) = a^2 - 2 a + 7(12) so a_i = (2 +/- sqrt(4 - 4(84)) / 2 = 1 +/- sqrt(-83) and [5(1 + sqrt(-83)) + 7][5(1 - sqrt(-83)) + 7] = [12 + 5 sqrt(-83)][12 - 5 sqrt(-83)] = 144 + 25(83) = 2219 = 7(317), as expected. However, neither of 1 +/- sqrt(-83) is divisible by 7. If one were, then a = 7b for some algebraic integer b, but then we'd have (7b)^2 - 2(7b) + 7(12) = 0 from [1] with x = 3, so 49 b^2 - 2(7) b + 7(12) = 0 and hence b would satisfy the monic irreducible polynomial equation 7 b^2 - 2 b + 12 = 0. In addition, we can find explicit nonunit common factors of a and 7, although it takes a bit of work. We can show that (4627 + 423 a)(4627 - 423 a) = 7^9 and (4627 + 423 a)(72192 - 512 a) = a^9 And since t = (4627 + 423 a)^{1/9) satisfies t^{18} - 10100 t^{9} + 40353607 = 0 we have that t is an algebraic integer nonunit common factor of a and 7, establishing the fact that a and 7 are not coprime in the ring of algebraic integers. > It contradicts with standard ideas from Galois Theory because from that > theory, if the a's are non-rational and irreducible over Q, [You mean if r(a, x), considered as a polynomial in a, is irreducible over Q.] > except for > the special case where both have sqrt(7) which I'll go over next, then > by those ideas NEITHER of the a's can have 7 as a factor. That's right, but you don't need Galois Theory to establish that. In this example, except for the two cases mentioned above (x = 0, -1), either both roots will have 7 as a factor, or neither will. BTW, your special case isn't. > And it can be shown that neither does have 7 as a factor in the ring of > algebraic integers. Yes. > But that then proves--as the simple proof cannot be wrong or it > wouldn't be a proof--that the ring of algebraic integers is incomplete, > and for those who don't understand why it shows that, consider the > analogy where considering evens as a ring, 2 and 6 are coprime in that > ring because 3 is not even. > Make sense? Similarly, 7 being coprime in the ring of algebraic > integers, when it has been proven to be a factor by a simple algebraic > proof, shows that it is excluded because its co-factor is not an > algebraic integer. Go back to what I said above. If p and q are algebraic integers with the product pq divisible by 7, you cannot conclude that either of p or q must be divisible by 7. That's true in the rational numbers but not true in the algebraic integers. > I would be interested in at least some posters actually trying to > answer the proof itself, and its steps, especially the key step where I > note that the constant factor, which is 7, proves that 7 was multiplied > through that particular factor. 7 is not a prime in the ring of algebraic integers. That's where your explanation fails. Rick ==== Subject: Re: JSH: Decker example, proper answer Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. > No. You keep tripping up on this point. Just because the product of > two numbers, pq, is divisible by 7 in the algebraic integers, you > cannot conclude that either p or q must be divisible by 7. That's > true in the integers, but is not true in the algebraic integers. I conclude nothing from the ring of algebraic integers, as I simply use the distributive property, and often do that without specifying a ring, which often evokes howls and questions like, in what ring? a(b+c) = ab +ac so knowing that the term constant with respect to x, for ONE of the factors is 7, while the other is 2, when 7 was multiplied across, I know that ONE of the functions has a factor of 7, from the distributive property, without need of mention of a ring. An example to explain the concept is 7(x^2 + 3x + 2) = (7x + 7)(x + 2) where the constant terms are 7 and 2, and the same argument proves that 7 was multiplied through the first factor. The only difference with your more complex equations is that since the a's are roots of a quadratic generator, their value is obscured by radicals EXCEPT when you have rational solutions. But my point is that the distributive property still holds whether you can see the answer directly or not. > Note that the result that only one of the factors was multiplied by 7 > follows from the distributive property, and noting the values of the > constant terms. > That completes the proof. It's rather short as you can see. > It isn't a proof. Then point out an error. The steps are clearly outlined, and the most important step relies on the distributive property. Refuting that position requires refuting the distributive property. I think you are wrong and the distributive property is right. > Unsurprisingly, you've been seduced by the rational integers. > There are only two rational integers x that give rational integer > roots for > r(a, x) = a^2 + (1 - x) a + 7(x^2 + x) I have been seduced by algebra. And I LIKE the distributive property and refuse to give it up just because a lot of people argue with me, attacking it, because the key step in the proof relies on it. > namely, > x = 0 r(a, x) = a^2 + a > x = -1 r(a, -1) = a^2 + 2 a > In these two cases, your conjecture is true, namely that one of the > roots is dividible by 7 and the other is coprime to 7. That's easy enough with 7 as the factor, but you can generalize to f, and use ANY rational f to be able to see the same result. The DISTRIBUTIVE PROPERY holds in all cases. > Unfortunately, these are the ONLY values of x for which the roots > behave as you want them to. That is irrelevant as your example can be generalized. But it's also irrelevant because I have a short proof. > For example, let x = 3. We see that 7 Q(3) = 7(25(9) + 30(3) + 2) = > 7(317) and r(a, 3) = a^2 - 2 a + 7(12) so > a_i = (2 +/- sqrt(4 - 4(84)) / 2 = 1 +/- sqrt(-83) > and > [5(1 + sqrt(-83)) + 7][5(1 - sqrt(-83)) + 7] > = [12 + 5 sqrt(-83)][12 - 5 sqrt(-83)] > = 144 + 25(83) > = 2219 = 7(317), as expected. > However, neither of 1 +/- sqrt(-83) is divisible by 7. If one were, then > a = 7b for some algebraic integer b, but then we'd have > (7b)^2 - 2(7b) + 7(12) = 0 > from [1] with x = 3, so > 49 b^2 - 2(7) b + 7(12) = 0 > and hence b would satisfy the monic irreducible polynomial equation > 7 b^2 - 2 b + 12 = 0. > In addition, we can find explicit nonunit common factors of a and 7, > although it takes a bit of work. We can show that > (4627 + 423 a)(4627 - 423 a) = 7^9 > and > (4627 + 423 a)(72192 - 512 a) = a^9 > And since t = (4627 + 423 a)^{1/9) satisfies > t^{18} - 10100 t^{9} + 40353607 = 0 > we have that t is an algebraic integer nonunit common factor of a and 7, > establishing the fact that a and 7 are not coprime in the ring of > algebraic integers. Irrelevant to the question of proof. I focus on a short algebraic proof for a reason, which is that proof is supposed to be paramount. And I counter that the ring of algebraic integers is flawed in that it is not complete, so you can get some weird results that CONTRADICT the algebra. The only way to answer that claim is to go to the proof, and show that it is not a proof. > It contradicts with standard ideas from Galois Theory because from that > theory, if the a's are non-rational and irreducible over Q, > [You mean if r(a, x), considered as a polynomial in a, is irreducible > over Q.] > except for > the special case where both have sqrt(7) which I'll go over next, then > by those ideas NEITHER of the a's can have 7 as a factor. > That's right, but you don't need Galois Theory to establish that. In > this example, except for the two cases mentioned above (x = 0, -1), > either both roots will have 7 as a factor, or neither will. > BTW, your special case isn't. You may not need Galois Theory but the result sinks current usage. It also brings into question the theory of ideals. > And it can be shown that neither does have 7 as a factor in the ring of > algebraic integers. > Yes. > But that then proves--as the simple proof cannot be wrong or it > wouldn't be a proof--that the ring of algebraic integers is incomplete, > and for those who don't understand why it shows that, consider the > analogy where considering evens as a ring, 2 and 6 are coprime in that > ring because 3 is not even. > Make sense? Similarly, 7 being coprime in the ring of algebraic > integers, when it has been proven to be a factor by a simple algebraic > proof, shows that it is excluded because its co-factor is not an > algebraic integer. > Go back to what I said above. If p and q are algebraic integers with > the product pq divisible by 7, you cannot conclude that either of > p or q must be divisible by 7. That's true in the rational numbers > but not true in the algebraic integers. > I would be interested in at least some posters actually trying to > answer the proof itself, and its steps, especially the key step where I > note that the constant factor, which is 7, proves that 7 was multiplied > through that particular factor. > 7 is not a prime in the ring of algebraic integers. That's where your > explanation fails. > It's not an explanation, it's a proof. I wish to remind people that there is a PROOF and that proof is usually mostly ignored so that people can start going on about the ring of algebraic integers and results in that ring. BUT my point is that the the proof shows a problem with the ring of algebraic integers, so why just keep going back to where I claim there's a problem? I say you do that because you can't answer the proof, refuse to acknowledge the truth, and have no where else to go. James Harris ==== Subject: Re: JSH: Decker example, proper answer >Oh, over the weekend I had a lot of trouble handling an example from a >Rick Decker of Hamilton College that paralleled my own non-polynomial >factorization argument with simpler quadratics where one had a special >feature. >I made a lot of posts claiming proof where the arguments did not prove >what I thought they did when I posted them, and I was wrong. >So yes, those previous claims of proof made a few days ago, were crap. >Here is the correct answer to the Decker example. >As a synoposis up front as I've had at least one poster ask for >something beforehand that explains what's going to happen, I am going >to use the Decker equations with the same approach I've used with my >own more complex cubic generating equations to show a complete proof, >which just so happens to shoot down standard usage of Galois Theory, >and the theory of ideals. >The approach is basic, relies on simple algebraic concepts, where the >key step depends on the use of the distributive property. >In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) >and >7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) >where the a's are defined by >a^2 - (x - 1)a + 7(x^2 + x) = 0. ------------------ ERROR HERE ------------------ >Given >7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) >it cannot be the case that the 7's visible in the factorization both >result from the 7 multiplied on the left side, as the constant term of >Q(x) is coprime to 7, since it is 2. ------------------ END OF ERROR ----------------- >So both constant terms--those terms constant with respect to x--cannot >be visible in the factorization. >However, the actual constant terms can be determined by clearing x out, >letting x=0, I have that one of the a's goes to 0, while the remaining >one is -1, from >a^2 + a = 0. >And looking at >7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) >I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's >not mathematically determined which of the two factors was multiplied >by 7, and which one actually is coprime to 7, but you know ONE must >have been multiplied by 7, from the distributive property and the >constant terms. >>No. You keep tripping up on this point. Just because the product of >>two numbers, pq, is divisible by 7 in the algebraic integers, you >>cannot conclude that either p or q must be divisible by 7. That's >>true in the integers, but is not true in the algebraic integers. > I conclude nothing from the ring of algebraic integers, as I simply use > the distributive property, and often do that without specifying a ring, > which often evokes howls and questions like, in what ring? > a(b+c) = ab +ac > so knowing that the term constant with respect to x, for ONE of the > factors is 7, while the other is 2, when 7 was multiplied across, I > know that ONE of the functions has a factor of 7, from the distributive > property, without need of mention of a ring. You cannot talk about factors, divisors, coprime, and the like without specifying the ring in which you're working. Even disregarding this, how do you explain that when x = 3, as I used below, neither of the a_i are divisible by 7? Are you saying that your conjecture (one of a_i is divisible by 7 and the other is coprime to 7) is true only when the a polynomial has rational roots? If that's your position, I agree. I note, though, that there are only two x values for which a^2 - (x - 1)a + 7(x^2 + x) has integral roots. > An example to explain the concept is > 7(x^2 + 3x + 2) = (7x + 7)(x + 2) > where the constant terms are 7 and 2, and the same argument proves that > 7 was multiplied through the first factor. > The only difference with your more complex equations is that since the > a's are roots of a quadratic generator, their value is obscured by > radicals EXCEPT when you have rational solutions. > But my point is that the distributive property still holds whether you > can see the answer directly or not. >Note that the result that only one of the factors was multiplied by 7 >follows from the distributive property, and noting the values of the >constant terms. >That completes the proof. It's rather short as you can see. >>It isn't a proof. > Then point out an error. I did. Perhaps you missed it. In case you did, I highlighted it above. To see a counterexample, try x = 3. > The steps are clearly outlined, and the most important step relies on > the distributive property. > Refuting that position requires refuting the distributive property. > I think you are wrong and the distributive property is right. >>Unsurprisingly, you've been seduced by the rational integers. >>There are only two rational integers x that give rational integer >>roots for >> r(a, x) = a^2 + (1 - x) a + 7(x^2 + x) > I have been seduced by algebra. And I LIKE the distributive property > and refuse to give it up just because a lot of people argue with me, > attacking it, because the key step in the proof relies on it. No, in fact it doesn't. If your key step is the one I highlighted, then questions of relies on are meaningless, since the inference you make is wrong. >>namely, >> x = 0 r(a, x) = a^2 + a >> x = -1 r(a, -1) = a^2 + 2 a >>In these two cases, your conjecture is true, namely that one of the >>roots is dividible by 7 and the other is coprime to 7. > That's easy enough with 7 as the factor, but you can generalize to f, > and use ANY rational f to be able to see the same result. Yup, and the same thing will hold. Except for a few special cases with integral roots for r, your conjecture will be wrong. > The DISTRIBUTIVE PROPERY holds in all cases. >>Unfortunately, these are the ONLY values of x for which the roots >>behave as you want them to. > That is irrelevant as your example can be generalized. And, as I said, you'll face the same problems. > But it's also irrelevant because I have a short proof. You have a short non-proof. >>For example, let x = 3. We see that 7 Q(3) = 7(25(9) + 30(3) + 2) = >>7(317) and r(a, 3) = a^2 - 2 a + 7(12) so >> a_i = (2 +/- sqrt(4 - 4(84)) / 2 = 1 +/- sqrt(-83) >>and >> [5(1 + sqrt(-83)) + 7][5(1 - sqrt(-83)) + 7] >> = [12 + 5 sqrt(-83)][12 - 5 sqrt(-83)] >> = 144 + 25(83) >> = 2219 = 7(317), as expected. >>However, neither of 1 +/- sqrt(-83) is divisible by 7. If one were, then >>a = 7b for some algebraic integer b, but then we'd have >> (7b)^2 - 2(7b) + 7(12) = 0 >>from [1] with x = 3, so >> 49 b^2 - 2(7) b + 7(12) = 0 >>and hence b would satisfy the monic irreducible polynomial equation >> 7 b^2 - 2 b + 12 = 0. >>In addition, we can find explicit nonunit common factors of a and 7, >>although it takes a bit of work. We can show that >> (4627 + 423 a)(4627 - 423 a) = 7^9 >>and >> (4627 + 423 a)(72192 - 512 a) = a^9 >>And since t = (4627 + 423 a)^{1/9) satisfies >> t^{18} - 10100 t^{9} + 40353607 = 0 >>we have that t is an algebraic integer nonunit common factor of a and 7, >>establishing the fact that a and 7 are not coprime in the ring of >>algebraic integers. > Irrelevant to the question of proof. Not so. It's relevant since it provides a counterexample to your claim. > I focus on a short algebraic proof for a reason, which is that proof is > supposed to be paramount. Only when it's correct. When it's not, a counterexample serves as a proof (of refutation). > And I counter that the ring of algebraic integers is flawed in that it > is not complete, so you can get some weird results that CONTRADICT the > algebra. > The only way to answer that claim is to go to the proof, and show that > it is not a proof. Done. Several times, in fact, in this thread and others. Rick ==== Subject: Re: JSH: Decker example, proper answer >Oh, over the weekend I had a lot of trouble handling an example from a >Rick Decker of Hamilton College that paralleled my own non-polynomial >factorization argument with simpler quadratics where one had a special >feature. >I made a lot of posts claiming proof where the arguments did not prove >what I thought they did when I posted them, and I was wrong. >So yes, those previous claims of proof made a few days ago, were crap. Really? That's a surprise. I thought that the people denying the validity of your work were disputing the Truth of Mathematics itself! No way to refute anything you've said because after all a proof is a proof - Decker has been disputing the distributive property, etc etc etc. >Here is the correct answer to the Decker example. Ah, great. _This_ one is _correct_, eh? That's good to know... David C. Ullrich ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. No, you don't. . Your have k (5a_1(x) +7) = (b(x) + 1) with b(0)=0 and you conclude that k must be 1/7 and b(x) must be 5a_1(x)/7. However, this is just one possibility. Try k = 1/(7(1+x^2)) then k(a(x) +7) = (5a_1(x)/(7(1+x^2) + 1/(1+x^2)) =( [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] + 1) so k = 1/(7(1+x^2)) and b(x) = [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] is another possibility. (There are in fact an infinite number of possibilities.) The constant term in both cases is 7 before multiplication by k and 1 after multiplication by k. So looking at the constant term cannot tell you which k was used. -William Hughes ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > So yes, those previous claims of proof made a few days ago, were crap. > Here is the correct answer to the Decker example. > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > In the ring of algebraic integers, I have from Decker that > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > and > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > where the a's are defined by > a^2 - (x - 1)a + 7(x^2 + x) = 0. > Given > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > a^2 + a = 0. > And looking at > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. > No, you don't. . > Your have > k (5a_1(x) +7) = (b(x) + 1) > with b(0)=0 > and you conclude that k must be 1/7 and b(x) must be 5a_1(x)/7. That's not correct. Key here is that it is not determinable WHICH of the b's has been multiplied by 7, and which one has not. So you have 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) where you only know that one has 7 as a factor, and one does not, but there's no mathematical determination which one. But given that 7 is clearly a factor of only one, based on the constant terms, which is 7, for one and 2 for the other, it's shown that ONE of the a's in general has 7 as a factor, though, as has been pointed out repeatedly, an infinity of cases can be found where that is not true in the ring of algebraic integers. > However, this > is just one possibility. > Try k = 1/(7(1+x^2)) then > k(a(x) +7) = (5a_1(x)/(7(1+x^2) + 1/(1+x^2)) > =( [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] + 1) > so k = 1/(7(1+x^2)) and b(x) = [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] > is another possibility. (There are in fact an infinite number > of possibilities.) The constant term in both cases is 7 before That's actually a good thing to consider when thinking about your objections--an infinite number of possibilities--so how does the math choose? The answer is that it is forced to go one way by the distributive property. The functions here are only different from the functions in the example 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) where a_1(x) = 7x and a_2(x) = x - 5, in that it is obscured WHICH function has 7 as a factor, except for rational cases. Now then, apply your exact same objection to the example I've shown, and you'll see that it still appears to work!!! Or do you disagree? If so, why can't you find a k with my simple example just like before? > multiplication > by k and 1 after multiplication by k. So looking at the constant term > cannot tell you which k was used. > -William Hughes But your argument if true would also prove the same thing about 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) where a_1(x) = 7x and a_2(x) = x - 5. I suggest to you that your argument is flawed. James Harris ==== Subject: Re: JSH: Decker example, proper answer > Oh, over the weekend I had a lot of trouble handling an example from a > Rick Decker of Hamilton College that paralleled my own non-polynomial > factorization argument with simpler quadratics where one had a special > feature. > > I made a lot of posts claiming proof where the arguments did not prove > what I thought they did when I posted them, and I was wrong. > > So yes, those previous claims of proof made a few days ago, were crap. > > Here is the correct answer to the Decker example. > > As a synoposis up front as I've had at least one poster ask for > something beforehand that explains what's going to happen, I am going > to use the Decker equations with the same approach I've used with my > own more complex cubic generating equations to show a complete proof, > which just so happens to shoot down standard usage of Galois Theory, > and the theory of ideals. > > The approach is basic, relies on simple algebraic concepts, where the > key step depends on the use of the distributive property. > > In the ring of algebraic integers, I have from Decker that > > 7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) > = 7(25 x^2 + 30 x + 2) > > and > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > where the a's are defined by > > a^2 - (x - 1)a + 7(x^2 + x) = 0. > > Given > > 7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) > > it cannot be the case that the 7's visible in the factorization both > result from the 7 multiplied on the left side, as the constant term of > Q(x) is coprime to 7, since it is 2. > > So both constant terms--those terms constant with respect to x--cannot > be visible in the factorization. > > However, the actual constant terms can be determined by clearing x out, > letting x=0, I have that one of the a's goes to 0, while the remaining > one is -1, from > > a^2 + a = 0. > > And looking at > > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > > I have that the actual constant terms are 7 and 5(-1) + 7 = 2, but it's > not mathematically determined which of the two factors was multiplied > by 7, and which one actually is coprime to 7, but you know ONE must > have been multiplied by 7, from the distributive property and the > constant terms. > No, you don't. . > Your have > k (5a_1(x) +7) = (b(x) + 1) > with b(0)=0 > and you conclude that k must be 1/7 and b(x) must be 5a_1(x)/7. > That's not correct. > Key here is that it is not determinable WHICH of the b's has been > multiplied by 7, and which one has not. > So you have > 7 Q(0) = (5a_1(0) + 7)(5a_2(0) + 7) > where you only know that one has 7 as a factor, and one does not, but > there's no mathematical determination which one. > But given that 7 is clearly a factor of only one, based on the constant > terms, which is 7, for one and 2 for the other, it's shown that ONE of > the a's in general has 7 as a factor, though, as has been pointed out > repeatedly, an infinity of cases can be found where that is not true in > the ring of algebraic integers. > However, this > is just one possibility. > Try k = 1/(7(1+x^2)) then > k(a(x) +7) = (5a_1(x)/(7(1+x^2) + 1/(1+x^2)) > =( [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] + 1) > so k = 1/(7(1+x^2)) and b(x) = [5a_1(x)/(7(1+x^2)) + 1/(1+x^2) -1] > is another possibility. (There are in fact an infinite number > of possibilities.) The constant term in both cases is 7 before > That's actually a good thing to consider when thinking about your > objections--an infinite number of possibilities--so how does the math > choose? > The answer is that it is forced to go one way by the distributive > property. > The functions here are only different from the functions in the example > 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) > where a_1(x) = 7x and a_2(x) = x - 5, in that it is obscured WHICH > function has 7 as a factor, except for rational cases. > Now then, apply your exact same objection to the example I've shown, > and you'll see that it still appears to work!!! > Or do you disagree? > If so, why can't you find a k with my simple example just like before? > multiplication > by k and 1 after multiplication by k. So looking at the constant term > cannot tell you which k was used. > -William Hughes > But your argument if true would also prove the same thing about > 7(x^2 + 3x + 2) = (a_1(x) + 7)(a_2(x) + 7) > where a_1(x) = 7x and a_2(x) = x - 5. and indeed it does. There are an infinite number of functions k_1(x) and k_2(x) such that 7(x^2 + 3x + 2) = (7x + 7)((x-5) + 7) = k_1(x)(b_1(x) + 1)k_2(x)(b_2(x) + 1) (E.g. k_1(x) = 7(1+x^2), k_2(x) = 1/(x^2 +1), b_1(x) = [x/(1+x^2) + 1/(1+x^2) -1] b_2(x) = [(x-5)(x^2+1) + (1+x^2) -1] [there is only one choice, k_1 = 7, k_2 = 1 which gives polynomials] -William Hughes ==== Subject: Re: Measuring time passage Mail-To-News-Contact: abuse@dizum.com >Wouldn't time be measured best with a 'time line', along which Now >travels; rather than with a series of arrows or clocks? I've never measured time with a series of arrows. In fact, I've never measured time with a series of clocks, only with one at any epoch. So it's hard to say if measuring it with a line would be better. How does one use a line to measure time? -- Michael F. Stemper #include No animals were harmed in the composition of this message. ==== Subject: Re: Measuring time passage <1VQFf.11348$1e5.273902@news20.bellglobal.com> <57qdnTc427szeXreRVn-hA@comcast.com> <6URFf.11840$1e5.285666@news20.bellglobal.com> <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> | their metabolism, Does paint dry slower in the cockpit of the space shuttle? > Prove it! > and you don't have any metabolism slowing > proof yet either so i have no clue why you say such crap. > Do you know of an experiment that actually tested > human cell growth while traveling at super high speeds? > I would love to read about such. > Of course. > I can't.. because there is no such data that you are saying exists. > :) Let's stitch together a bunch of observations: 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. This average lifetime is identical for every collection of muons. 2. A muon traveling very fast with respect to the Earth travels farther than expected from it's 2.2 microseconds, and when we look at it's lifetime with our clock at rest on the Earth (which we've done nothing to), we see that it's lived 4.8 microseconds. 3. A clock traveling just as fast as the muon and alongside it measures its lifetime to be 2.2 microseconds. Now: Is the clock in (3) broken because it measures the *same* thing as the clock in (1)? PD ==== Subject: Re: Measuring time passage | Let's stitch together a bunch of observations: | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. | This average lifetime is identical for every collection of muons. | 2. A muon traveling very fast with respect to the Earth travels farther | than expected from it's 2.2 microseconds, and when we look at it's | lifetime with our clock at rest on the Earth (which we've done nothing | to), we see that it's lived 4.8 microseconds. could be FTL and slowing down at end of flight could be time dilation could be objects in motion have more energy so they can live longer anyway. (Are you thinking the energy a muon has when moving at lightspeed is the same energy as when it is sitting stationery on Earth?) that would be real silly (and against relativity) :) | 3. A clock traveling just as fast as the muon and alongside it measures | its lifetime to be 2.2 microseconds. no proof of such at all. see above for other realities that could be happening. including a relative velocity energy thought you are missing completely. | Now: Is the clock in (3) broken because it measures the *same* thing | as the clock in (1)? you have no clock doing such. you have data that is not even physical proof at all. you have a moving a c object supposedly having the same energy as a statinery to Earth object. you have problems to work out with such. LOL ==== Subject: Re: Measuring time passage > | Let's stitch together a bunch of observations: > | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. > | This average lifetime is identical for every collection of muons. > | 2. A muon traveling very fast with respect to the Earth travels farther > | than expected from it's 2.2 microseconds, and when we look at it's > | lifetime with our clock at rest on the Earth (which we've done nothing > | to), we see that it's lived 4.8 microseconds. > could be FTL and slowing down at end of flight No mechanism to speed them up after they're created. No mechanism to slow them down at the end of their flight. No object has ever been observed to exceed c. > could be time dilation Doesn't apply to speeds measured in the frame of reference of the observer. > could be objects in motion have more energy so they > can live longer anyway. Kinetic energy is entirely relative. > (Are you thinking the energy a muon has when moving at lightspeed > is the same energy as when it is sitting stationery on Earth?) > that would be real silly (and against relativity) Kinetic energy is relative. > :) > | 3. A clock traveling just as fast as the muon and alongside it measures > | its lifetime to be 2.2 microseconds. > no proof of such at all. Au contraire. We're moving along with lab muons at an enormous clip compared to other things all around us in the universe. ==== Subject: Re: Measuring time passage | No mechanism to speed them up after they're created. | No mechanism to slow them down at the end of their | flight. | No object has ever been observed to exceed c. how can you observe it at FTL how far away would you have to be to observer a muon traveling at FTL anyway? | Au contraire. We're moving along with lab muons at an | enormous clip compared to other things all around us | in the universe. Still not close to the speed they catch up to Earth at.. LOL ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> | No mechanism to slow them down at the end of their > | flight. > | No object has ever been observed to exceed c. > how can you observe it at FTL > how far away would you have to be > to observer a muon traveling at FTL anyway? Why do you have to be far away? The way you measure the speed of something is to record its passage at two different places and measure the time in between. If you have timing accuracy down to a picosecond, your two marks only need to be 3 mm apart to time something going 10c. - Randy ==== Subject: Re: Measuring time passage | | > | No mechanism to speed them up after they're created. | > | No mechanism to slow them down at the end of their | > | flight. | > | No object has ever been observed to exceed c. | > how can you observe it at FTL | > how far away would you have to be | > to observer a muon traveling at FTL anyway? | | Why do you have to be far away? FTL causes a problem with light being able to measure it. :) | The way you measure the speed of something is to | record its passage at two different places and measure | the time in between. Correct, so where is the proof you are actually measuring the same muon in this case? Hint: there is no proof. It is merely assumed from the group motion. If you have timing accuracy down | to a picosecond, your two marks only need to be | 3 mm apart to time something going 10c. If something moved passed your detector at 10c how would the detector be able to detect such when it uses a 1c detection rate? :) ==== Subject: Re: Measuring time passage > | > | No mechanism to speed them up after they're created. > | > | No mechanism to slow them down at the end of their > | > | flight. > | > | No object has ever been observed to exceed c. > | | > how can you observe it at FTL > | > how far away would you have to be > | > to observer a muon traveling at FTL anyway? > | Why do you have to be far away? > FTL causes a problem with light being able to measure it. > :) two different places while passing by in order to detect the time of passage. > | The way you measure the speed of something is to > | record its passage at two different places and measure > | the time in between. > Correct, > so where is the proof you are actually measuring the same > muon in this case? > Hint: there is no proof. > It is merely assumed from the group motion. You don't need to measure the same muon. You only have to know ehre it came from, where it ended up, and the average speed it had in the interrim. If the muons are created within a tight range of energies (which they are by dint of the generation mechanism), the maximum speed is known. > If you have timing accuracy down > | to a picosecond, your two marks only need to be > | 3 mm apart to time something going 10c. > If something moved passed your detector at 10c > how would the detector be able to detect such when it uses > a 1c detection rate? Light is not the detecting rate. The rate is given by the difference in time between two events (interactions). It doesn't matter how slow the messenger is, as long as their speeds are the same they will arrive with their message separated by the same interval. ==== Subject: Re: Measuring time passage | > | | > | > | No mechanism to speed them up after they're created. | > | > | No mechanism to slow them down at the end of their | > | > | flight. | > | > | No object has ever been observed to exceed c. | > | > | > | > how can you observe it at FTL | > | > how far away would you have to be | > | > to observer a muon traveling at FTL anyway? | > | | > | Why do you have to be far away? | > FTL causes a problem with light being able to measure it. | > :) | | two different places while passing by in order to | detect the time of passage. LOL | You don't need to measure the same muon. To get a physical proof that a single muon does such ... you do to. sheesh! | Light is not the detecting rate. The rate is given | by the difference in time between two events | (interactions). lol what a bunch of bull covered in chocolate! lol ==== Subject: Re: Measuring time passage > | | > | > | > | > | No mechanism to speed them up after they're created. > | > | > | No mechanism to slow them down at the end of their > | > | > | flight. > | > | > | No object has ever been observed to exceed c. > | > | | > | > how can you observe it at FTL > | > | > how far away would you have to be > | > | > to observer a muon traveling at FTL anyway? > | > | > | > | Why do you have to be far away? > | | > FTL causes a problem with light being able to measure it. > | > :) > | two different places while passing by in order to > | detect the time of passage. stationary atoms in the detector. > LOL > | You don't need to measure the same muon. > To get a physical proof that a single muon > does such ... you do to. No. Statistical measuring methods apply. Or don't you know anything about measurement science? > sheesh! > | Light is not the detecting rate. The rate is given > | by the difference in time between two events > | (interactions). > lol > what a bunch of bull covered in chocolate! You're ranting again. ==== Subject: Re: Measuring time passage | > | > | > | > | | > | > | > | No mechanism to speed them up after they're created. | > | > | > | No mechanism to slow them down at the end of their | > | > | > | flight. | > | > | > | No object has ever been observed to exceed c. | > | > | > | > | > | > how can you observe it at FTL | > | > | > how far away would you have to be | > | > | > to observer a muon traveling at FTL anyway? | > | > | | > | > | Why do you have to be far away? | > | > | > | > FTL causes a problem with light being able to measure it. | > | > :) | > | | > | two different places while passing by in order to | > | detect the time of passage. | | stationary atoms in the detector. It did not even hit any, (just like you muon does not hit other atoms.) LOL | No. Statistical measuring methods apply. Or don't | you know anything about measurement science? ROFLOL this from a guy that has no clue about clock malfunctions! LOL Wrong, not proven at all. ==== Subject: Re: Measuring time passage <_buGf.29246$1e5.537696@news20.bellglobal.com> It did not even hit any, (just like you muon does not hit other atoms.) It has to essentially hit an atom, or at least get very close to it, to ionize it. Why do you think a stationary atom can't be hit piece of paper can't be hit by a bullet? - Randy ==== Subject: Re: Measuring time passage | It has to essentially hit an atom, or at least get very | close to it, to ionize it. Ok fine Randy. so, now that you know it passed detector #1 how do you know it is the same one passing detector #2? Did you mark it with a sharpie as it passed detector #1? :) ==== Subject: Re: Measuring time passage > | It has to essentially hit an atom, or at least get very > | close to it, to ionize it. > Ok fine Randy. > so, now that you know it passed detector #1 > how do you know it is the same one passing detector #2? > Did you mark it with a sharpie as it passed detector #1? 3D track geometry and coincidence statistics tell the tale. If you're really worried about this you can use a single bubble chamber type detector and through multiple collisions. The speed can be obtained by measuring the energy deposited (the track length will be correlated with the initial kinietic energy). ==== Subject: Re: Measuring time passage <_buGf.29246$1e5.537696@news20.bellglobal.com> | close to it, to ionize it. > Ok fine Randy. > so, now that you know it passed detector #1 > how do you know it is the same one passing detector #2? > Did you mark it with a sharpie as it passed detector #1? When I see a hole in my first sheet of paper, and a hole in my second sheet of paper, how do I know the same bullet made both of them? - Randy ==== Subject: Re: Measuring time passage | When I see a hole in my first sheet of paper, and a hole | in my second sheet of paper, how do I know the same | bullet made both of them? How close are these detectors for the muons Randy? ==== Subject: Re: Measuring time passage <7J4Gf.14520$1e5.384272@news20.bellglobal.com> | detect the time of passage. sitting still. Do you think it can't do that? Look. Imagine I fire a bullet through two pieces of paper. Do you think that for some reason a fast-moving bullet won't make a hole in a slow-moving piece of paper? Does something in the paper have to catch up to the bullet for it to make a hole? THAT's a muon going through a detector. - Randy ==== Subject: Re: Measuring time passage | sitting still. Do you think it can't do that? The detector uses something to detect what speed does the detection devices detector use. sheesh what don't you get about that? | THAT's a muon going through a detector. so the detector is made of paper? Is it the rubber ruler type paper too? Is the detector slowing down the muon? (hint: it must if it is physically detecting it.) LOL ==== Subject: Re: Measuring time passage The detector uses something to detect through. It could be sound waves, making their leisurely way to a microphone in the setup I described earlier. It doesn't matter how long it takes the signal device. You could send it by trained earthworm. > what speed does the detection devices detector use. Zero. The atoms of the detector are sitting still. > sheesh > what don't you get about that? Your fixed idea that a detector needs to chase something down in order to detect it, like a piece of paper needs to chase down a bullet or else the bullet will have no effect on it. > | THAT's a muon going through a detector. > Is the detector slowing down the muon? In the same sense as the paper is slowing down the bullet. > (hint: it must if it is physically detecting it.) Congratulations. You have made a correct statement. When you build a good detector, you design it so that the energy of detection is a miniscule fraction of the energy of the thing you're trying to detect. Hint: The detector doesn't make a noticeable difference - Randy ==== Subject: Re: Measuring time passage | through. Randy, (what form of detection are you using) sheesh! you are really ignoring that basic question huh? | When you build a good detector, you design it so | that the energy of detection is a miniscule fraction | of the energy of the thing you're trying to detect. | | Hint: The detector doesn't make a noticeable difference You don't know they are moving at c. is beign detected at both places so you don't have any proof of the speed at all.. Dang.. you are so silly. ==== Subject: Re: Measuring time passage Randy, > (what form of detection are you using) > sheesh! > you are really ignoring that basic question huh? I'm wrong. Not light. Electrons. Here's one experiment using homemade equipment: http://www.cosmicrays.org/ We detect the muons by utilizing a homebrew Geiger-M.9fller detector. The Geiger counters are supplied by high voltage, which creates a very high electric field near the anode detector, it strips off some electrons of some atoms. These electrons move towards the positively charged wires, are accelerated by the huge electric field and have enough energy to strip more electrons from other gas molecules. These electrons are accelerated too in order to strip more and more electrons. Avalanche effect in a Geiger-M.9fller tube detector This electric avalanche consisting of more than a billion negative charges rains down on the positively charged wire, causing a current Got that? You have a bunch of gas molecules, milling around at low velocity the way gas molecules do. A some of them. The detector is designed to amplify one free electron into many. All that avalanching and moving toward the positive electrode happens long first couple electrons, and left. Now, what question have I not answered? - Randy ==== Subject: Re: Measuring time passage I'm wrong. Not light. Electrons. AHA! You see, I had a nice conversation about muons hitting electrons a little bit ago. and according to Greg, the muons don't move electrons around. Free electrons and even atmospheric gases can not slow the muon at all according to greg. yet somehow, the muon can be stopped by water? hmm? Water can stop a muon traveling at c? (or even .5c for that matter?) ==== Subject: Re: Measuring time passage > I'm wrong. Not light. Electrons. > AHA! > You see, > I had a nice conversation about muons hitting electrons > a little bit ago. > and according to Greg, the muons don't move electrons > around. > Free electrons and even atmospheric gases can not > slow the muon at all according to greg. The interaction rate in air is comparatively low. The muons that make it to ground level have enjoyed a relatively uneventful trip. > yet somehow, the muon can be stopped by water? > hmm? > Water can stop a muon traveling at c? > (or even .5c for that matter?) Some will pass through without being detected. Some will be stopped. The water has a much greater density (number of targets per cubic centimeter) than air does. Water also consists of polar molecules (electric dipoles) which may contribute to increasing the reaction cross section, but I'm not certain about that. I'll have to look that up. ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> | to a picosecond, your two marks only need to be > | 3 mm apart to time something going 10c. > If something moved passed your detector at 10c > how would the detector be able to detect such when it uses > a 1c detection rate? What does a 1c detection rate mean? It goes through detector 1. Detector 1 marks a time. It doesn't matter how fast gets a mark. A little way away, the muon goes through detector 2. However fast it goes, detector 2 marks the time when it was detected. Where does this 1c detection rate come in? You have two things that mark a time when they detected the muon. You can send the data by carrier pigeon and subtract the time with quill pens as slowly as you like. But the by the time between recorded events. What the heck is a 1c detection rate? - Randy ==== Subject: Re: Measuring time passage | What does a 1c detection rate mean? It goes through | detector 1. Detector 1 marks a time. detector 1 marks the time of the muon passing by, so what.. what proof do you have that the same muon is passing by the second detector? Do you time race cars the same way? all race cars were doing 200 mph. because we detected one went by a detector and then another one went by the second detector so we just call it the same car. It is just so silly. lol ==== Subject: Re: Measuring time passage > | What does a 1c detection rate mean? It goes through > | detector 1. Detector 1 marks a time. > detector 1 marks the time of the muon passing by, > so what.. > what proof do you have that the same muon is > passing by the second detector? Coincidence detection. Straight line path between hits (you can stack detectors, too). The frequency between detections is low enough to statistically weed out false positives. > Do you time race cars the same way? > all race cars were doing 200 mph. > because we detected one went by a detector > and then another one went by the second detector > so we just call it the same car. > It is just so silly. Presumably they should both pass both detectors. ==== Subject: Re: Measuring time passage | Coincidence detection. Straight line path | between hits (you can stack detectors, too). | The frequency between detections is low | enough to statistically weed out false positives. So you can prove that the detectors can detect FTL, but you have no FTL things you can prove such with? LOL | | > Do you time race cars the same way? | > all race cars were doing 200 mph. | > because we detected one went by a detector | > and then another one went by the second detector | > so we just call it the same car. | > It is just so silly. | | Presumably they should both pass both detectors. You don't know what cars though. no marking on the muons. average does not prove a single lifetime. If so. no human should live past the average human in your physical-less proof method. ==== Subject: Re: Measuring time passage > | Coincidence detection. Straight line path > | between hits (you can stack detectors, too). > | The frequency between detections is low > | enough to statistically weed out false positives. > So you can prove that the detectors can > detect FTL, but you have no FTL > things you can prove such with? More or less. The detectors are up to detecting have ever been observed. Nothing is known about the properties of tachyons, purely hupothetical than c. They couldn't have a great interaction cross section or they would have been detected by now. > LOL > | | > Do you time race cars the same way? > | > all race cars were doing 200 mph. > | > because we detected one went by a detector > | > and then another one went by the second detector > | > so we just call it the same car. > | > It is just so silly. > | Presumably they should both pass both detectors. > You don't know what cars though. > no marking on the muons. > average does not prove a single lifetime. If the density of cars on the track is known (by count rate) and the average time between cars passing is then known, then detector events that occur in significantly shorter time intervals are statistically more likely to be caused by a single vehicle. If the density is very low, so that one car passes every minute, say, then count pairs that are a second apart are almost surely caused by an individual car. > If so. > no human should live past the average human > in your physical-less proof method. Do you know what an average is? How about correlation? Standard deviation? ==== Subject: Re: Measuring time passage | More or less. The detectors are up to detecting | have ever been observed. Ya sure, and i have a lilipution universe detector that works perfect too, but none have been observed. LOL ==== Subject: Re: Measuring time passage > | More or less. The detectors are up to detecting > | have ever been observed. > Ya sure, and i have a lilipution universe detector that works perfect too, > but none have been observed. You should patent that. You'll have to explain why the detectors would be unable to measure velocities greater than c. Simply making a snide comment doesn't cut it. ==== Subject: Re: Measuring time passage <57qdnTc427szeXreRVn-hA@comcast.com> <6URFf.11840$1e5.285666@news20.bellglobal.com> <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. > | This average lifetime is identical for every collection of muons. > | 2. A muon traveling very fast with respect to the Earth travels farther > | than expected from it's 2.2 microseconds, and when we look at it's > | lifetime with our clock at rest on the Earth (which we've done nothing > | to), we see that it's lived 4.8 microseconds. > could be FTL and slowing down at end of flight Could be, but it's not. The speed is easily measured in beamlines and it is not FTL. > could be time dilation Yes, exactly. > could be objects in motion have more energy so they > can live longer anyway. Could be, but there's no evidence that increasing the energy increases the lifetime. Moreover, isn't it then odd that the clock moving with the muon (that according to you is broken) measures *exactly* the same value as the muons at rest? In other words, you'd have to explain why the muon's lifetime *actually* gets longer, and why the malfunctioning of the clock just happens to remove *exactly* the same amount of time, so that the two completely different effects exactly cancel. > (Are you thinking the energy a muon has when moving at lightspeed > is the same energy as when it is sitting stationery on Earth?) > that would be real silly (and against relativity) > :) > | 3. A clock traveling just as fast as the muon and alongside it measures > | its lifetime to be 2.2 microseconds. > no proof of such at all. Not so. > see above for other realities that could be happening. > including a relative velocity energy thought you are missing > completely. > | Now: Is the clock in (3) broken because it measures the *same* thing > | as the clock in (1)? > you have no clock doing such. > you have data that is not even physical proof at all. > you have a moving a c object No, the muon is not moving at c. > supposedly having the same > energy as a statinery to Earth object. No, I didn't say that either. > you have problems to work out with such. > LOL ==== Subject: Re: Measuring time passage | | > | Let's stitch together a bunch of observations: | > | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. | > | This average lifetime is identical for every collection of muons. | > | 2. A muon traveling very fast with respect to the Earth travels farther | > | than expected from it's 2.2 microseconds, and when we look at it's | > | lifetime with our clock at rest on the Earth (which we've done nothing | > | to), we see that it's lived 4.8 microseconds. | > could be FTL and slowing down at end of flight | | Could be, but it's not. The speed is easily measured in beamlines and | it is not FTL. Complete utter bull. nobody has measured one muons speed in flight. and many things could be slowing it down and also keeping it alive longer as it travels through such. | > could be time dilation | | Yes, exactly. no, just could be no clocks on muons so it is real hard to physically prove such bull. | > could be objects in motion have more energy so they | > can live longer anyway. | | Could be, It is a relativistic fact that objects in motion have more energy. (isn't it?) such energy is keeping the needed conditions for longer life of the muon. too simple, no silly time dilation needed at all. still funny that 2 hydrogen and 1 oxygen can catch a muon and yet hyrogen and oxygen alone could never be slowing it down.. lol ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> <9YudnT93XMnrrHfenZ2dnUVZ_tydnZ2d@comcast.com | > | Let's stitch together a bunch of observations: > | > | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. > | > | This average lifetime is identical for every collection of muons. > | > | 2. A muon traveling very fast with respect to the Earth travels > farther > | > | than expected from it's 2.2 microseconds, and when we look at it's > | > | lifetime with our clock at rest on the Earth (which we've done nothing > | > | to), we see that it's lived 4.8 microseconds. > | | > could be FTL and slowing down at end of flight > | Could be, but it's not. The speed is easily measured in beamlines and > | it is not FTL. > Complete utter bull. > nobody has measured one muons speed in flight. Your ignorance of beamline instrumentation is not my problem. Try googling time-of-flight counter. > and many things could be slowing it down and also > keeping it alive longer as it travels through such. Such as? > | > could be time dilation > | Yes, exactly. > no, > just could be > no clocks on muons so it is real hard to > physically prove such bull. > | > could be objects in motion have more energy so they > | > can live longer anyway. > | Could be, > It is a relativistic fact that objects in motion have > more energy. (isn't it?) > such energy is keeping the needed conditions for > longer life of the muon. What needed conditions would those be? > too simple, > no silly time dilation needed at all. > still funny that 2 hydrogen and 1 oxygen can catch > a muon and yet hyrogen and oxygen alone could never be slowing it down.. I'm sorry, I don't get your point. > lol ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> <9YudnT93XMnrrHfenZ2dnUVZ_tydnZ2d@comcast.com | > | Let's stitch together a bunch of observations: > | > | 1. A muon at rest on Earth decays with a lifetime of 2.2 microseconds. > | > | This average lifetime is identical for every collection of muons. > | > | 2. A muon traveling very fast with respect to the Earth travels > farther > | > | than expected from it's 2.2 microseconds, and when we look at it's > | > | lifetime with our clock at rest on the Earth (which we've done nothing > | > | to), we see that it's lived 4.8 microseconds. > | | > could be FTL and slowing down at end of flight > | Could be, but it's not. The speed is easily measured in beamlines and > | it is not FTL. > Complete utter bull. > nobody has measured one muons speed in flight. Your ignorance of beamline instrumentation is not my problem. Try googling time-of-flight counter. > and many things could be slowing it down and also > keeping it alive longer as it travels through such. Such as? > | > could be time dilation > | Yes, exactly. > no, > just could be > no clocks on muons so it is real hard to > physically prove such bull. > | > could be objects in motion have more energy so they > | > can live longer anyway. > | Could be, > It is a relativistic fact that objects in motion have > more energy. (isn't it?) > such energy is keeping the needed conditions for > longer life of the muon. What needed conditions would those be? > too simple, > no silly time dilation needed at all. > still funny that 2 hydrogen and 1 oxygen can catch > a muon and yet hyrogen and oxygen alone could never be slowing it down.. I'm sorry, I don't get your point. > lol ==== Subject: Re: Measuring time passage | Your ignorance of beamline instrumentation is not my problem. Try | googling time-of-flight counter. Do you not get any such thought that light can not measure speeds greater than itself. Do you think a soundwave detector can measure something traveling at lightspeed? | What needed conditions would those be? Whatever a muon needs to decay slower. What makes muons decay to begin with? Stability and energy loss of such stability? | | > too simple, | > no silly time dilation needed at all. | > still funny that 2 hydrogen and 1 oxygen can catch | > a muon and yet hyrogen and oxygen alone could never be slowing it down.. | | I'm sorry, I don't get your point. What catches a muon? ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> <9YudnT93XMnrrHfenZ2dnUVZ_tydnZ2d@comcast.com> | googling time-of-flight counter. > Do you not get any such thought that > light can not measure speeds greater than itself. I'm not using light to measure its speed. I have two detectors that blip when the muon passes by and passes that information to a nearby, time-stamped recorders. I know how far apart the detectors are, and I know how much time elapsed between the blips. That tells me the speed. often they pass a particular place on the circle. This isn't rocket science, spaceman. This is pretty close to how they measure the speed of racecars at dragstrips and oval tracks. You don't have to be faster than the race cars to know how fast they're going. > Do you think a soundwave detector can > measure something traveling at lightspeed? > | What needed conditions would those be? > Whatever a muon needs to decay slower. > What makes muons decay to begin with? > Stability and energy loss of such stability? Nope. Try again. > | > too simple, > | > no silly time dilation needed at all. > | | > still funny that 2 hydrogen and 1 oxygen can catch > | > a muon and yet hyrogen and oxygen alone could never be slowing it down.. > | I'm sorry, I don't get your point. > What catches a muon? I don't have to catch a muon to see how fast it's going, any more than I have to catch a horse to see how fast it's going around the track. PD ==== Subject: Re: Measuring time passage | I'm not using light to measure its speed. I have two detectors that | blip when the muon passes by and passes that information to a nearby, | time-stamped recorders. I know how far apart the detectors are, and I | know how much time elapsed between the blips. That tells me the speed. PD, you forgot to mark the muon so you know it is the same one. ==== Subject: Re: Measuring time passage <5XWFf.13477$1e5.345331@news20.bellglobal.com> <7J4Gf.14520$1e5.384272@news20.bellglobal.com> <9YudnT93XMnrrHfenZ2dnUVZ_tydnZ2d@comcast.com> | googling time-of-flight counter. > Do you not get any such thought that > light can not measure speeds greater than itself. In what sense do you think light is measuring this? > Do you think a soundwave detector can > measure something traveling at lightspeed? What would a soundwave detector be? I can imagine comes through. In that case it certainly could be used to measure something at light speed. Detector 1: sound wave is triggered, makes its slow way to recording device, arrives eventually. Detector 2: sound wave is triggered, makes its slow way to recording device, arrives eventually. Observer measures that sound#2 arrived 1 us later than sound#1, 1 us apart. If they are separated by 300 m, the observer can thus tell that the triggering partlcle was moving at c. - Randy ==== Subject: Re: Measuring time passage | | > | Your ignorance of beamline instrumentation is not my problem. Try | > | googling time-of-flight counter. | > Do you not get any such thought that | > light can not measure speeds greater than itself. | | In what sense do you think light is measuring this? What could you be measuring it with? a rubber ruler? | | > Do you think a soundwave detector can | > measure something traveling at lightspeed? | | What would a soundwave detector be? I can imagine | comes through. In that case it certainly could be used | to measure something at light speed. | | Detector 1: sound wave is triggered, makes its slow way | to recording device, arrives eventually. | | Detector 2: sound wave is triggered, makes its slow way | to recording device, arrives eventually. Observer | measures that sound#2 arrived 1 us later than sound#1, | 1 us apart. | | If they are separated by 300 m, the observer can thus | tell that the triggering partlcle was moving at c. You are not using sound waves themselves to detect in the above example you made up. I said use sound to detect a lightspeed object have the detectors listen for sound and pass the photon by it at lightspeed and see how it registers it. :) ==== Subject: Re: Measuring time passage > You are not using sound waves themselves to detect > in the above example you made up. > I said use sound to detect a lightspeed object > have the detectors listen for sound and pass the photon > by it at lightspeed and see how it registers it. Sound is a pressure wave in a medium. The light has to interact with something in the medium in order to create a sound wave. ==== Subject: Re: Measuring time passage > | Your ignorance of beamline instrumentation is not my problem. Try > | googling time-of-flight counter. > Do you not get any such thought that > light can not measure speeds greater than itself. > Do you think a soundwave detector can > measure something traveling at lightspeed? A simple rotating toothed-wheel gizmo can measure the speed of light. Nothing in the contraption is moving faster than light. You could time the speed of a bullet with sound by detecting when it passed through two sheets of paper, the sound from each being picked up by microphones. > | What needed conditions would those be? > Whatever a muon needs to decay slower. > What makes muons decay to begin with? > Stability and energy loss of such stability? and have it born out through experiment, the world will applaud. > | > too simple, > | > no silly time dilation needed at all. > | | > still funny that 2 hydrogen and 1 oxygen can catch > | > a muon and yet hyrogen and oxygen alone could never be slowing it down.. > | I'm sorry, I don't get your point. > What catches a muon? Dumping energy by a head-on collision with a massive low interaction cross section, but happens in significant numbers with a high flux of muons and a dense medium. It is fortunate for us that the flux of muons created by individual cosmic ray events is quite high. ==== Subject: Re: Measuring time passage | A simple rotating toothed-wheel gizmo can measure the | speed of light. Nothing in the contraption is moving | faster than light. aha, so, you spin it and the frequency of spin etc.. .. no way the spin could be a multiplied frequency instead of the real frequency. | You could time the speed of a bullet with sound by | detecting when it passed through two sheets of paper, | the sound from each being picked up by microphones. not according to observers 10 miles away! LOL | and have it born out through experiment, the | world will applaud. Yup, and more cooler, I know you won't be doing such since you have already closed your box. :) | Dumping energy by a head-on collision with a massive | low interaction cross section, but happens in significant | numbers with a high flux of muons and a dense medium. | It is fortunate for us that the flux of muons created | by individual cosmic ray events is quite high. More cooler is that they only happen on Earth huh? and apparently water can catch them, but the 2 gases water is made up of do not effect them at all according to the stupid theory. lol ==== Subject: Re: Measuring time passage > More cooler is that they only happen on Earth huh? > and apparently water can catch them, but the 2 gases > water is made up of do not effect them at all > according to the stupid theory. > lol It's a matter of relative density and the interaction cross section. ==== Subject: Re: Measuring time passage | | > More cooler is that they only happen on Earth huh? | > and apparently water can catch them, but the 2 gases | > water is made up of do not effect them at all | > according to the stupid theory. | > lol | | It's a matter of relative density and the interaction | cross section. It is a matter of complete bull that the air can not effect the muon. ==== Subject: Re: Measuring time passage > | | > More cooler is that they only happen on Earth huh? > | > and apparently water can catch them, but the 2 gases > | > water is made up of do not effect them at all > | > according to the stupid theory. > | > lol > | It's a matter of relative density and the interaction > | cross section. > It is a matter of complete bull that the air > can not effect the muon. It can and does affect them. Many don't make the trip all the way down. Some are slowed significantly, some are not slowed at all. It's a matter of chance for individual muons. We only measure the ones that make it to the finish line. ==== Subject: Re: Measuring time passage > | A simple rotating toothed-wheel gizmo can measure the > | speed of light. Nothing in the contraption is moving > | faster than light. > aha, > so, > you spin it and the frequency of spin etc.. > .. > no way the spin could be a multiplied frequency > instead of the real frequency. > | You could time the speed of a bullet with sound by > | detecting when it passed through two sheets of paper, > | the sound from each being picked up by microphones. > not according to observers 10 miles away! > LOL > | and have it born out through experiment, the > | world will applaud. > Yup, > and more cooler, I know you won't be doing such > since you have already closed your box. > :) > | Dumping energy by a head-on collision with a massive > | low interaction cross section, but happens in significant > | numbers with a high flux of muons and a dense medium. > | It is fortunate for us that the flux of muons created > | by individual cosmic ray events is quite high. > More cooler is that they only happen on Earth huh? > and apparently water can catch them, but the 2 gases > water is made up of do not effect them at all > according to the stupid theory. > lol ==== Subject: Re: Measuring time passage > | A simple rotating toothed-wheel gizmo can measure the > | speed of light. Nothing in the contraption is moving > | faster than light. > aha, > so, > you spin it and the frequency of spin etc.. > .. > no way the spin could be a multiplied frequency > instead of the real frequency. The idea is to gradually speed up the rotation rate until the light passing between the teeth and being reflected back is just cut off (the light doesn't make it back in time and hits the tooth instead of the vacency. Then speed it up some more until the light reappears, passing through one vacency in the outbound direction and through the next vacancy on the return. Then do the math using the rotation rates, path lengths, etc. Fizeau did this experiment back in 1849, I think. > | You could time the speed of a bullet with sound by > | detecting when it passed through two sheets of paper, > | the sound from each being picked up by microphones. > not according to observers 10 miles away! > LOL > | and have it born out through experiment, the > | world will applaud. > Yup, > and more cooler, I know you won't be doing such > since you have already closed your box. > :) > | Dumping energy by a head-on collision with a massive > | low interaction cross section, but happens in significant > | numbers with a high flux of muons and a dense medium. > | It is fortunate for us that the flux of muons created > | by individual cosmic ray events is quite high. > More cooler is that they only happen on Earth huh? > and apparently water can catch them, but the 2 gases > water is made up of do not effect them at all > according to the stupid theory. > lol ==== Subject: Re: Measuring time passage | The idea is to gradually speed up the rotation | rate until the light passing between the teeth | and being reflected back is just cut off (the | light doesn't make it back in time and hits | the tooth instead of the vacency. Then speed it | up some more until the light reappears, passing | through one vacency in the outbound direction and | through the next vacancy on the return. Then do | the math using the rotation rates, path lengths, | etc. So what is the accuracy of speeding up the disk? can you speed up the disk by 0.00001 rotations back in the pre 1900's? ==== Subject: Re: Measuring time passage > | The idea is to gradually speed up the rotation > | rate until the light passing between the teeth > | and being reflected back is just cut off (the > | light doesn't make it back in time and hits > | the tooth instead of the vacency. Then speed it > | up some more until the light reappears, passing > | through one vacency in the outbound direction and > | through the next vacancy on the return. Then do > | the math using the rotation rates, path lengths, > | etc. > So what is the accuracy of speeding up the disk? > can you speed up the disk by 0.00001 rotations > back in the pre 1900's? Fizeau was able to achieve a measurement within about 5% of today's value. Foucult used a rotating mirror setup to get within about half a percent. I'm not certain what mechanisms were used to time the rotation rates, but it could have been as simple as a mechanical gearing mechanism. The required rotation rates can be kept low by the simple expediency of increasing the light path length. ==== Subject: Re: Measuring time passage <1VQFf.11348$1e5.273902@news20.bellglobal.com> <57qdnTc427szeXreRVn-hA@comcast.com> <6URFf.11840$1e5.285666@news20.bellglobal.com> <5XWFf.13477$1e5.345331@news20.bellglobal.com> <46CdnQro5YC-lnTeRVn-oQ@comcast.com> What a freakin joke! > So, you are now stating the science of measurement > is wrong. I'm sure saying it's changed over the years, yes. You have a problem with that? PD ==== Subject: Re: Measuring time passage | | > | Then we invented it wrong, | > What a freakin joke! | > So, you are now stating the science of measurement | > is wrong. | | I'm sure saying it's changed over the years, yes. You have a problem | with that? LOL so the standards have changed but only in your SR world, because they surely have not changed anywhere else. Rocket scientists do not include time dilation when sending things to get slingshotted by another moving object. :) If you think they do.. you are truly lost because We see the objects using Earth time and we have all the data of such objects using Earth time. If we start giving the sats there own time, we will be in real trouble if we ever get stuff moving real fast out there.. LOL ==== Subject: Re: Measuring time passage > | > | Then we invented it wrong, > | | > What a freakin joke! > | > So, you are now stating the science of measurement > | > is wrong. > | I'm sure saying it's changed over the years, yes. You have a problem > | with that? > LOL > so the standards have changed but only in your SR world, > because they surely have not changed anywhere else. > Rocket scientists do not include time dilation when sending > things to get slingshotted by another moving object. > :) > If you think they do.. > you are truly lost because We see the objects using Earth > time and we have all the data of such objects using Earth time. > If we start giving the sats there own time, we will be in real > trouble if we ever get stuff moving real fast out there.. > LOL GR corrections are used in the calculation of the position of Mercury. It's the only body that's deep enough the Sun's gravity well to make a significant difference from the Newtonian model for the spacecraft speeds currently in use. ==== Subject: Re: Measuring time passage <6URFf.11840$1e5.285666@news20.bellglobal.com> <5XWFf.13477$1e5.345331@news20.bellglobal.com> <46CdnQro5YC-lnTeRVn-oQ@comcast.com> | Then we invented it wrong, > | | > What a freakin joke! > | > So, you are now stating the science of measurement > | > is wrong. > | I'm sure saying it's changed over the years, yes. You have a problem > | with that? > LOL > so the standards have changed but only in your SR world, > because they surely have not changed anywhere else. Not so. Want to look at the history of the definition of the day? > Rocket scientists do not include time dilation when sending > things to get slingshotted by another moving object. > :) > If you think they do.. > you are truly lost because We see the objects using Earth > time and we have all the data of such objects using Earth time. That's where you're wrong, spaceman. We *do* take into account the slowing of transmission encoding due to the motion. Sorry, but we do. > If we start giving the sats there own time, we will be in real > trouble if we ever get stuff moving real fast out there.. > LOL ==== Subject: Re: Measuring time passage | Not so. Want to look at the history of the definition of the day? Yup, my bad, I forgot the standards changed and the change even helps the length contraction bull. a length that changed length with lightspeed.. lol no wonder they use that silly meter = time light travels such and such distance.. lol It is a perfect standard warping that leaves proof of such being wrong.. impossible.. LOL ==== Subject: Re: Measuring time passage <6URFf.11840$1e5.285666@news20.bellglobal.com> <5XWFf.13477$1e5.345331@news20.bellglobal.com> <46CdnQro5YC-lnTeRVn-oQ@comcast.com> | Then we invented it wrong, > | | > What a freakin joke! > | > So, you are now stating the science of measurement > | > is wrong. > | I'm sure saying it's changed over the years, yes. You have a problem > | with that? > LOL > so the standards have changed but only in your SR world, > because they surely have not changed anywhere else. You think we're still using cubits, or using the king's foot or somebody's thumb as a standard? Are you sure you want to say that standards have never changed? - Randy ==== Subject: Re: Measuring time passage | You think we're still using cubits, or using the king's foot | or somebody's thumb as a standard? You could do such, you can't use such and combine with other standards but you could use such standards for basic things. you don't get that fact huh? | Are you sure you want to say that standards have never | changed? I know they have changed. they even screwed up the meter. (making it equal to a distance that light travels) that is the whole problem.. DUH! ==== Subject: Re: Measuring time passage > | You think we're still using cubits, or using the king's foot > | or somebody's thumb as a standard? > You could do such, > you can't use such and combine with other standards > but you could use such standards for basic things. > you don't get that fact huh? > | Are you sure you want to say that standards have never > | changed? > I know they have changed. > they even screwed up the meter. (making it equal to a distance > that light travels) > that is the whole problem.. Oh, *THAT'S* the problem... Riiight. ==== Subject: Re: Measuring time passage | > | You think we're still using cubits, or using the king's foot | > | or somebody's thumb as a standard? | > You could do such, | > you can't use such and combine with other standards | > but you could use such standards for basic things. | > you don't get that fact huh? | > | Are you sure you want to say that standards have never | > | changed? | > I know they have changed. | > they even screwed up the meter. (making it equal to a distance | > that light travels) | > that is the whole problem.. | | Oh, *THAT'S* the problem... Riiight. That and the consistancy of lightspeed to all observers bull. It is funny that you can eat up such bull without even questioning what the food is. LOL ==== Subject: Re: Measuring time passage > | Why is it malfunctioning if it keeps pace with > | their metabolism, Does paint dry slower in the cockpit of the space shuttle? According to accepted theory, all processes are affected equally by time dilation. Nature doesn't single out clocks for special treatment. > Prove it! No. You prove the contrary. It's your idea that paint is somehow magically unaffected by the laws of physics while the rest of creation appears to follow them to the letter. So go ahead and show us how paint dries at Earth clock rate as measured by Earth-based clocks while in Earth orbit. > and you don't have any metabolism slowing > proof yet either so i have no clue why you say such crap. Plenty of evidence that SR and GR are correct (they've neither ever been contradicted in any empirical test) says that all process are affected equally. > Do you know of an experiment that actually tested > human cell growth while traveling at super high speeds? > I would love to read about such. So would I. And look forward to doing so when the technology permits. Although I suspect they'll probably start with something with a more rapid growth rate and simple to count, like bacterial cultures. > Of course. > I can't.. because there is no such data that you are saying exists. Here's an easy one for you James. Find one, just one! example where predictions made by either GR or SR have been contradicted by empirical evidence. ==== Subject: Re: Measuring time passage | According to accepted theory, all processes are affected | equally by time dilation. Nature doesn't single out clocks | for special treatment. You have no proof of such. You have never tested a human's metabolism in such. | No. You prove the contrary. It's your idea that | paint is somehow magically unaffected by the laws | of physics while the rest of creation appears to | follow them to the letter. So go ahead and show us | how paint dries at Earth clock rate as measured by | Earth-based clocks while in Earth orbit. You are the one saying motion is creating some new law? (the law that time slows the faster you move..) LOL (You have no proof for this law) You have malfunctioning clocks and nothing more. | Plenty of evidence that SR and GR are correct (they've | neither ever been contradicted in any empirical test) | says that all process are affected equally. predictions without an actual cause does not mean . the predictions are backing up my thoughts too. the clock malfunctioned and they can even predict such a malfunction using your model. :) | So would I. And look forward to doing so when the | technology permits. Although I suspect they'll | probably start with something with a more rapid | growth rate and simple to count, like bacterial | cultures. Let's hope they do such soon. :) | Here's an easy one for you James. Find one, just one! example | where predictions made by either GR or SR have been | contradicted by empirical evidence. Greg, You are very ignorant when you come to this statement. I have never said the predictions themselves are wrong. I am merely stating the given cause is lacking physical proof. and with such lacking proof, my clock malfunction stands as solid as the stupid time changing thought does. It is predicted correctly and everything. I am usign the SR theory and it agrees with my conclusion. The clock malfunctioned. ==== Subject: Re: Measuring time passage > | According to accepted theory, all processes are affected > | equally by time dilation. Nature doesn't single out clocks > | for special treatment. > You have no proof of such. > You have never tested a human's metabolism in such. You provide the funding to test everything in the universe and we'll get started. But why bother? All material things are made of in the way predicted. > | No. You prove the contrary. It's your idea that > | paint is somehow magically unaffected by the laws > | of physics while the rest of creation appears to > | follow them to the letter. So go ahead and show us > | how paint dries at Earth clock rate as measured by > | Earth-based clocks while in Earth orbit. > You are the one saying motion is creating some new > law? (the law that time slows the faster you move..) > LOL > (You have no proof for this law) > You have malfunctioning clocks and nothing more. You have not and cannot prove that the clocks malfunctioned so as to behave any differently from the way that nature intended a good clock to behave, as evidenced by the fact that the clocks are obsereved to run according to standard unit definition in their frames of reference at all times. > | Plenty of evidence that SR and GR are correct (they've > | neither ever been contradicted in any empirical test) > | says that all process are affected equally. > predictions without an actual cause does not > mean . Predictions are predictions; they are either correct or incorrect. A 100% success rate is hard to argue with. > the predictions are backing up my thoughts too. > the clock malfunctioned and they can even predict > such a malfunction using your model. You need to provide a definition of 'malfunction' then. If it just means behaving not the way you personally would like things to behave, then you're good to go. > :) > | So would I. And look forward to doing so when the > | technology permits. Although I suspect they'll > | probably start with something with a more rapid > | growth rate and simple to count, like bacterial > | cultures. > Let's hope they do such soon. > :) > | Here's an easy one for you James. Find one, just one! example > | where predictions made by either GR or SR have been > | contradicted by empirical evidence. > Greg, > You are very ignorant when you come to this statement. > I have never said the predictions themselves are wrong. > I am merely stating the given cause is lacking physical proof. > and with such lacking proof, my clock malfunction stands > as solid as the stupid time changing thought does. > It is predicted correctly and everything. > I am usign the SR theory and it agrees with my conclusion. > The clock malfunctioned. Not really, as you propose no mechanism within the standard theory to account for any malfunction, nor any way to identify such, nor any way to extend the theory to include such. You're simply expressing your discontent with the theory without providing grounds other than I don't like it. ==== Subject: Re: Measuring time passage | You provide the funding to test everything in the universe | and we'll get started. | | But why bother? All material things are made of | in the way predicted. LOL you are lost now! LOL | You have not and cannot prove that the clocks | malfunctioned SR proves it for me. too bad you don't get that fact. | Predictions are predictions; they are either | correct or incorrect. A 100% success rate is | hard to argue with. 100 percent success rate of the prediction of a clock malfunction! Sweet! SR really love me! :) | You need to provide a definition of 'malfunction' | then. If it just means behaving not the way you | personally would like things to behave, then you're | good to go. Oh, here we go again. I have already given such definition to you. A clock malfunction = rate change of clock when compared to another clock. | Not really, as you propose no mechanism within the | standard theory to account for any malfunction, sheesh. and SR provides no mechanism for time changing so we are even, and SR likes my clock malfunction the same amount it likes the time changing bull you eat up as if it were chocolate.. but of course... it is just chocolate colored bull. LOL ==== Subject: Re: Measuring time passage > | You provide the funding to test everything in the universe > | and we'll get started. > | But why bother? All material things are made of > | in the way predicted. > LOL > you are lost now! > LOL > | You have not and cannot prove that the clocks > | malfunctioned > SR proves it for me. > too bad you don't get that fact. You didn't define malfunction as I asked. > | Predictions are predictions; they are either > | correct or incorrect. A 100% success rate is > | hard to argue with. > 100 percent success rate of the prediction > of a clock malfunction! > Sweet! > SR really love me! > :) > | You need to provide a definition of 'malfunction' > | then. If it just means behaving not the way you > | personally would like things to behave, then you're > | good to go. > Oh, here we go again. > I have already given such definition to you. > A clock malfunction = rate change of clock when compared to > another clock. Then nature malfunctions, and your argument is only a semantic one. > | Not really, as you propose no mechanism within the > | standard theory to account for any malfunction, > sheesh. > and SR provides no mechanism for time changing > so we are even, and SR likes my clock malfunction the > same amount it likes the time changing bull you eat > up as if it were chocolate.. but of course... it is just > chocolate colored bull. SR provides a mechanism based upon the postulates of the theory. > LOL ==== Subject: Re: Measuring time passage | You didn't define malfunction as I asked. malfunction = to fail to operate(funtion) in the designed operation parameters. | Then nature malfunctions, and your argument is | only a semantic one. nope, you are just lost about the function of a clock. If it fails to function as desinged, it has malfunctioned. | SR provides a mechanism based upon the postulates of the | theory. It does no such thing, It has the cause as time changing rate It is the same cause that a clock malfunction would fit into. ==== Subject: Re: Measuring time passage > | You didn't define malfunction as I asked. > malfunction = to fail to operate(funtion) in the designed operation > parameters. So the clocks did not malfunction, because they behaved precicely as their designed operation parameters specified in their FoR. > | Then nature malfunctions, and your argument is > | only a semantic one. > nope, > you are just lost about the function of a clock. > If it fails to function as desinged, it has malfunctioned. As I said, so it didn't malfunction. It ran at precicely the designed rate in its FoR. > | SR provides a mechanism based upon the postulates of the > | theory. > It does no such thing, > It has the cause as time changing rate > It is the same cause that a clock malfunction would fit into. No, you have it backwards. Time dilation is a consequence predicted by application of the postulates. ==== Subject: Re: Measuring time passage | > | You didn't define malfunction as I asked. | > malfunction = to fail to operate(funtion) in the designed operation | > parameters. | | So the clocks did not malfunction, because they | behaved precicely as their designed operation | parameters specified in their FoR. no they did not. the proper function of the clock was to keep at sync with the Earth based clock. It failed to do such. End result proven. | As I said, so it didn't malfunction. It ran at precicely | the designed rate in its FoR. no, it did not. damn broken record here.. It failed to keep the same rate, end result proves it. SR agrees with my clock malfunction 100% If it did not agree, it would not be saying time changed rate.. lol | No, you have it backwards. Time dilation is a consequence | predicted by application of the postulates. No you have it in a pile of bull. time dilation is a consequence of clock malfunction. SR even proves my postulate 100 percent. :) ==== Subject: Re: Measuring time passage > | | > | You didn't define malfunction as I asked. > | | > malfunction = to fail to operate(funtion) in the designed operation > | > parameters. > | So the clocks did not malfunction, because they > | behaved precicely as their designed operation > | parameters specified in their FoR. > no they did not. > the proper function of the clock was to keep at sync > with the Earth based clock. Why? What makes an Earth-based clock special, except perhaps an overdevelopped parochial attitude? > It failed to do such. > End result proven. > | As I said, so it didn't malfunction. It ran at precicely > | the designed rate in its FoR. > no, it did not. > damn broken record here.. > It failed to keep the same rate, It kept the same rate in its own FoR. > end result proves it. > SR agrees with my clock malfunction 100% > If it did not agree, it would not be saying time > changed rate.. Right. It changed rate as measured from the ground-based clock, and it changed with respect to the rate of another clock moving at a different speed, which also disagreed with the ground-based clock. Those are the facts as stated. > lol > | No, you have it backwards. Time dilation is a consequence > | predicted by application of the postulates. > No you have it in a pile of bull. > time dilation is a consequence of clock malfunction. > SR even proves my postulate 100 percent. So if we destroy all the clocks there will be no time dilation? What about natural clocks, like all physical processes? ==== Subject: Re: Measuring time passage | Why? What makes an Earth-based clock special, except perhaps | an overdevelopped parochial attitude? Nothing makes it special, It is just where the standard we use was setup. :) | It kept the same rate in its own FoR. Nope, proven wrong in end result. | Right. It changed rate as measured from the ground-based | clock, and it changed with respect to the rate of another | clock moving at a different speed, which also disagreed | with the ground-based clock. Those are the facts as | stated. Correct, end result proves the clocks in motion did not keep the standard Earth based second. | So if we destroy all the clocks there will be | no time dilation? What about natural clocks, | like all physical processes? What about them, they are affected by temperature and all sorts of different surrounding enviromentals. We are talking about clocks. not physical processes. try sticking with the clocks for once. ==== Subject: Re: Measuring time passage > | Why? What makes an Earth-based clock special, except perhaps > | an overdevelopped parochial attitude? > Nothing makes it special, > It is just where the standard we use was setup. Well the standard specifies a way to duplicate the standard anywhere. > :) > | It kept the same rate in its own FoR. > Nope, > proven wrong in end result. Nope, proven byb empirical fact. > | Right. It changed rate as measured from the ground-based > | clock, and it changed with respect to the rate of another > | clock moving at a different speed, which also disagreed > | with the ground-based clock. Those are the facts as > | stated. > Correct, > end result proves the clocks in motion did not > keep the standard Earth based second. Correct; they did not keep the same rate of ticking as measured from the Earth. They did keep the standard Earth-based standard as measured in their own FoR. > | So if we destroy all the clocks there will be > | no time dilation? What about natural clocks, > | like all physical processes? > What about them, > they are affected by temperature and all sorts > of different surrounding enviromentals. > We are talking about clocks. > not physical processes. > try sticking with the clocks for once. Nothing to do with it. Clocks can be kept at standard temperatures and pressures in any FoR. ==== Subject: Re: Standard Deviation of PSIA >The latest Census survey reported that there are 3,916 niggers in North >Dakota, or 0.7% of the population. > North Dakota isn't a city, in case you never noticed, nincompoop. > The largest city in North Dakota has a population less than 100,000 > and hence does not even appear on the lists comparing crimes rates and > other data among the largest cities in the country. >Do you know of any reason that the niggers in North Dakota wouldn't >kill each other (and others) at the same rate as they do in New >Orleans, Detroit, DC, and lil ol' Gary, Indiana (pop. 100k, 83% >nigger), which is 100 murders per 100k niggers? > No motive? > Temperature? >So even though the justice department REFUSES to tell us how many >murders are committed by niggers in North Dakota, we can figure it out >anyway--they commit ALL of them. > Your unsubstantiated opinion is noted. > I looked for recent ND murders > http://www.prisontalk.com/forums/archive/index.php/t-98629.html > < Minn., had moved in drug circles and had testified against a drug > < dealer in a federal trial last fall. > < Hinojosa Jr., 22, of Dilworth, had been arrested for drug possession. > < told police Hinojosa was taken to a gravel road north of Glyndon and > < beaten on the head with an ax or hatchet. Sheriff Bill Bergquist said > < investigators believe Hinojosa owed the men $50 to $60 for drugs. A > < body found in eastern North Dakota is believed to be Hinojosa's. > <... > < motivation that we can see. > < > < drug ringleader, a former area American Idol finalist, and the mother > < of a man killed last summer, in a case involving drugs and murder. > < conspiracy to bring drugs into the United States and Mexico, is > < accused of killing Lee Avila, 28, of East Grand Forks, Minn. > < American Idol competition; Anthony Valdez, 22, of Turlock, Calif; > < and Lacey Kathryn Johnson, 18, and Jonathan Olbay Meyer, 21, both of > < Fargo. > http://crime.about.com/od/news/a/dru_sjodin.htm > < Sjodin left her job at the Columbia Mall in Grand Forks, North Dakota > < talking with her boyfriend on her cell phone when she said Oh my > < god. Her phone then went dead. Chris Lang received another call from > < Dru's phone a few hours later, but there was only static on the other > < end of the line. > < Rodriguz Jr, a registered sex offender from Minnesota who had > < recently completed a 23-year prison term, was charged with abducting > < Rodriquz lived, a federal grand jury charged Rodriguez with > < kidnapping and murder. > A total of 10 people charged with 3 murders. 2 were out-of-state > Hispanics, two were ND Hispanics. One was an out-of-state white, 5 > were ND whites (3 were explicitly stated to be white; the other three > that were not Hispanic did not have their race mentioned, but 2 of > them have German surnames, and the third was a girl, presumably one of > their girlfriends) So even though the justice department REFUSES to tell us how many murders are committed by niggers in North Dakota, we can figure it out anyway--they commit ALL of them. Allow me to repeat myself, if I may: Sometimes, almost, you make me feel sorry for John. Almost. -- cary ==== Subject: Re: Standard Deviation of PSIA faggotcarynation, Do you know how many murders there would have been in Washington, DC, in 1991 had the murder rate been equivalent to North Dakota, or Singapore, or Qatar, or Saudi Arabia? TWO!! Instead, there were 484. WHO killed the other 482? NIGGERS! Where is the news about these FOUR HUNDRED AND EIGHTY TWO NIGGER MURDERS, more than 160 TIMES as many murders as lobmamzer cited in North Dakota. Let me repeat: There was NO news coverage for 160 TIMES as many niggers who murdered someone in ONE city, in ONE year, while the above White murderers were headline news? Clearly you cannot comprehend this simple statistical comparison, can you? You read stories like the one lobmamzer quoted above, and think that this represents total reality. You ignore that there ARE almost compute in your mamzer brain. Because you're so easily beguiled that you can't comprehend this simple comparison, you must be a girl, not a man. You undoubtedly never will comprehend, no matter how many times it's posted, that when there's one WHITE murderer, it makes the news for a decade, while nigger murderers, even those killing Whites, are NEVER in the news. It's already been posted that, according to Yahoo, the number of web pages dedicated to ONE White putative murderer Scott Peterson was 11,276 TIMES as many as are dedicated to ALL of the 600,000 nigger murderers in the last 4 decades, combined. Does that compute? Of course it doesn't, and it never will. When measured PER MURDER, there were 577 MILLION times as many news nigger murderers, COMBINED. http://blackexile.com John Knight ==== Subject: Re: Closedness of Level Set <44kjruF2knidU3@individual.net Oh sorry, I did not state my question correctly. Here is the correct > one: > If a function f is continuous over R^n and suppose C is a *closed* > subset of R^n, is the function f:C->R at the point of the boundry of C > discontinuous? > If it's continuous on a larger set, then it's continuous on a smaller > set. Really in the above case it is far harder to be a continuous > function in R^n then in C. Just apply the sequence definition of > continuity and it should be totally obvious. Do you viewed f as a function over R^n? I thought f:C->R should only be defined within C and undefined everywhere else? In my opinion now, f should be discontinuous at the boundary of C. ==== Subject: Re: Closedness of Level Set >> Oh sorry, I did not state my question correctly. Here is the correct >> one: >> If a function f is continuous over R^n and suppose C is a *closed* >> subset of R^n, is the function f:C->R at the point of the boundry of C >> discontinuous? >> If it's continuous on a larger set, then it's continuous on a smaller >> set. Really in the above case it is far harder to be a continuous >> function in R^n then in C. Just apply the sequence definition of >> continuity and it should be totally obvious. >Do you viewed f as a function over R^n? I thought f:C->R should only be >defined within C and undefined everywhere else? Yes. And the definition of continuity only talks about the function on its domain. > In my opinion now, f >should be discontinuous at the boundary of C. The definition of continuity at a point can be stated as: f:C -> R is continuous at x, a member of C, if for every epsilon > 0 there is delta > 0 such that for all y _in C_ with |y - x| < delta, |f(y) - f(x)| < epsilon. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada ==== Subject: integral of the product of bessel functions Can someone help me in solving the following definite integral Integrate[BesselJ[n, x] BesselJ[k, x] (Cosh[h x] Sinh[h x] - h x)/(Sinh[h x]^2 - (h x)^2), {x, 0, Infinity}] when h > 0, n is even and k is odd. Note that the ratio Cosh[h x] Sinh[h x] - h x)/(Sinh[h x]^2 - (h x)^2) = D[Log[Sinh[h x]^2 - (h x)^2]/2, x] and it behaves as 2/x as x -> 0 and as Coth[h x] as x -> Infinity ==== Subject: Numerical Analysis I in Houston I am looking for the solutions of the excercices of this course: http://www.math.uh.edu/~rohop/fall_05/ (especially solution 1 to 5) (numerical analysis I in houston) or do someone know anyone, who saved the files?? I need them to prepare a test. andi ==== Subject: Re: Numerical Analysis I in Houston It's my understanding that solutions would be very hard to come by since it takes some privileged status to have access to full solutions to every exercise in any particular textbook and not everyone would be willing to share these with you. The text, your instructor, fellow students, and the internet should usually be enough to study for an exam, if you've given up on finding solutions. Good luck, M.M. ==== Subject: Re: pivot = leading entry > My Lin Alg prof uses pivot column and leading entry column > interchangeably all the time. She even says pivot, or leading entry > if you like vise versa a lot. > The text we use doesn't really explain pivots, and when it did lightly > touch on it, it was explained as a number used to zap zeros in all the > entries below it in the column. If that is the case, then the pivot > entries should be a subset (sometimes proper subset) of the set of > leading entries. > Am I right? Because sometimes when doing the pivotting, other entries > in other columns are zapped to zero, creating a leading entry > automatically with zeros below it (thus not needing to use this leading > entry at any time for pivotting) I detect loose use of terminology (or conventions). Leading often applies to the first non-zero entry in a list (the entry 2 in [0, 0, 2, 77, 338, 5]), and on other occasions it applies to the zeros at the beginning of the list (the expansion .00006303 has 4 leading zeros). Personally, I saw it best applied to leading ones in the echelon form of a matrix. Pivot is, in computational linear algebra, a chosen entry below which the matrix entries are to be zapped to zero, after a possible interchange of rows, either actually performed or symbolically stored. In some orthogonal transformation methods, one can talk about a pivot column, the one with largest magnitude within a submatrix to be processed. The distinction: A frequently recommended pivot is the largest entry in the column that is being processed (it may reduce the risk of large errors). But there can be other criteria (in linear programming, it is the result of other magnitude decisions). The distinction is obvious: if rounding errors are not an issue, a pivot may (or may not) be the leading non-zero entry. (And in introductory linear algebra, they are indeed not an issue.) As a fresh graduate, I listened to a lecture by a German professor about the pivoting strategies; he pronounced pivot with a silent t. In Slavic languages, pivo means beer, so we (Czechs and Slovaks in the audience) were delighted that he emphasized the importance of a proper choice of a pivo element. ==== Subject: Re: pivot = leading entry >My Lin Alg prof uses pivot column and leading entry column >interchangeably all the time. She even says pivot, or leading entry >if you like vise versa a lot. >The text we use doesn't really explain pivots, and when it did lightly >touch on it, it was explained as a number used to zap zeros in all the >entries below it in the column. If that is the case, then the pivot >entries should be a subset (sometimes proper subset) of the set of >leading entries. >Am I right? Because sometimes when doing the pivotting, other entries >in other columns are zapped to zero, creating a leading entry >automatically with zeros below it (thus not needing to use this leading >entry at any time for pivotting) >Is my reasoning correct? I'm not sure what your reasoning is trying to say. But in fact a pivot is exactly the same as the first non-zero entry in a non-zero row. What it's used for is not part of the definition. David C. Ullrich ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT > Not particularly relevant to the 4CT, but it shows how AP thinks. > THE TOTAL ATOM THEORY GUIDE TO WINNING GOLF > Total Atom Theory allows you to attain a perfect score in golf (18 on > an 18-hole course)! Here's how. > (1) Find a golf club which will launch the ball at a 45 degree angle. > (2) When you get to the first tee, measure the distance to the hole and > call it D. > (3) Hit the ball so that its initial velocity is D*k, where k is the > magic number. > Actually, it should be sqrt(9.8 * D), not D*k. And the ball should be > hit directly at the hole. How did you get sqrt(9.8*D)? D is measured in yards! > (4) The ball will land in the hole without even bouncing. > (5) Pick up the ball and go to the next hole. > Clearly golfers have ignored this procedure, which will guarantee a > perfect score. Tiger Woods and other players clearly have wasted their > time on their technique. > I forgot to add: This will guarantee me a spot in the annals of Golf > Theory. > --- Archimedes Plutnoium > The universe consists of one large Plutonium atom. > There is a Plutonium atom in my brain, which makes me a genius! > Bow down to me, Tiger Woods! ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT > Not particularly relevant to the 4CT, but it shows how AP thinks. > > THE TOTAL ATOM THEORY GUIDE TO WINNING GOLF > > Total Atom Theory allows you to attain a perfect score in golf (18 on > an 18-hole course)! Here's how. > > (1) Find a golf club which will launch the ball at a 45 degree angle. > > (2) When you get to the first tee, measure the distance to the hole and > call it D. > > (3) Hit the ball so that its initial velocity is D*k, where k is the > magic number. > Actually, it should be sqrt(9.8 * D), not D*k. And the ball should be > hit directly at the hole. > How did you get sqrt(9.8*D)? D is measured in yards! No, you need to measure D in the Total Atom Theory units. Yards are not included in them. > (4) The ball will land in the hole without even bouncing. > > (5) Pick up the ball and go to the next hole. > > Clearly golfers have ignored this procedure, which will guarantee a > perfect score. Tiger Woods and other players clearly have wasted their > time on their technique. > I forgot to add: This will guarantee me a spot in the annals of Golf > Theory. > --- Archimedes Plutnoium > The universe consists of one large Plutonium atom. > There is a Plutonium atom in my brain, which makes me a genius! > Bow down to me, Tiger Woods! ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT I don't think any journal would take a one-page proof of the 4CT. What is the minimum length proof necessary for acceptibility? >(One mathematician received so many 4CT proofs that he had a form letter > written up, which went something like: > ___, on line ___. > and he would fill in the blanks.) > --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT If any journal would even referee a 4CT proof it would be a miracle! > I don't think any journal would take a one-page proof of the 4CT. > What is the minimum length proof necessary for acceptibility? Ten pages maybe. There is a result by Lorenz Friess that said that if you have a near-triangulation (exactly one region has N sides, where N >= 4), then the graph is 4-colorable, and that there are at least 2^(N-2) colorings. (No good for triangulations -- as Maxwell Smart said, Missed it by THAT MUCH!) I'm in the middle of translating this paper to English, and it's about 100 pages long. --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT >[...] > Here are the two basic facts that AP is ignoring: > > (1) If you can color every triangulated map with 4 colors, that does > not mean you can color any map with 4 colors. > What is a 'triangulated' map? > Every region is a triangle, and you are coloring regions. (Note that if > you are coloring the VERTICES, you CAN assume that every region is a > triangle.) Actually, you can 3-color a triangulated map. > BTW, the error in ignoring (2) is called arguing from ignorance, > which is basically, I can't imagine any other possibility, so my > possibility must be true. The same error is present, in the same form, > in the following proof: > What is frustrating is; > I can't imagine any other possibility, therefore, I can't evaluate > the other possibilities! > Well, then you don't have a proof then, and you need to do some more > thinking. Only complex computer analysis can address the improbable possibilities since they cannot be addressed directly. That I have no desire to do! > BTW, a list of countries and a list of which countries are adjacent to > which is all that is needed in order to color a map. The following > scenario should make it clear, even to AP: > > The leaders of the various countries got together and decided they > would like to have high-speed roads (highways) connecting the capital > cities of their countries. To reduce the cost and eliminate routing > conflicts, it was decided that two capitals would have a highway > between them only if those two countries were adjacent (but not at a > point), and that the route between the two capitals should pass through > their common boundary. > I imagine that the citizens of Denver and Phoenix might be disappointed > with this plan? [...] > Colorado and Arizona aren't adjacent, because they only have a single > point in common. To be adjacent, the regions have to have a line > segment in common. It's a joke , man! > --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT >[...] > Here are the two basic facts that AP is ignoring: > > (1) If you can color every triangulated map with 4 colors, that does > not mean you can color any map with 4 colors. > What is a 'triangulated' map? > Every region is a triangle, and you are coloring regions. (Note that if > you are coloring the VERTICES, you CAN assume that every region is a > triangle.) Actually, you can 3-color a triangulated map. > BTW, the error in ignoring (2) is called arguing from ignorance, > which is basically, I can't imagine any other possibility, so my > possibility must be true. The same error is present, in the same form, > in the following proof: > What is frustrating is; > I can't imagine any other possibility, therefore, I can't evaluate > the other possibilities! > Well, then you don't have a proof then, and you need to do some more > thinking. Only complex computer analysis can address the improbable possibilities since they cannot be addressed directly. That I have no desire to do! > BTW, a list of countries and a list of which countries are adjacent to > which is all that is needed in order to color a map. The following > scenario should make it clear, even to AP: > > The leaders of the various countries got together and decided they > would like to have high-speed roads (highways) connecting the capital > cities of their countries. To reduce the cost and eliminate routing > conflicts, it was decided that two capitals would have a highway > between them only if those two countries were adjacent (but not at a > point), and that the route between the two capitals should pass through > their common boundary. > I imagine that the citizens of Denver and Phoenix might be disappointed > with this plan? [...] > Colorado and Arizona aren't adjacent, because they only have a single > point in common. To be adjacent, the regions have to have a line > segment in common. It's a joke , man! > --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT >[...] > Here are the two basic facts that AP is ignoring: > > (1) If you can color every triangulated map with 4 colors, that does > not mean you can color any map with 4 colors. > > What is a 'triangulated' map? > Every region is a triangle, and you are coloring regions. (Note that if > you are coloring the VERTICES, you CAN assume that every region is a > triangle.) > Actually, you can 3-color a triangulated map. Not if it has 4 countries, all adjacent. (This is a triangulated map.) But this is the only exception. (Brooks's Theorem.) --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT <25094907.1139173401165.JavaMail.jakarta@nitrogen.mathforum.org> An even simpler map where 4 non mutually adjacent countries require 4-colors is A B C B A D ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT <25094907.1139173401165.JavaMail.jakarta@nitrogen.mathforum.org An even simpler map where 4 non mutually adjacent countries require > 4-colors is > A > B C B > A D Yeah, Nevada and its neighbors. I realized this after I posted last night. (This may also be a minimal map that requires 4 colors and has an adjacency of 3.) --- Christopher Heckman ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT <25094907.1139173401165.JavaMail.jakarta@nitrogen.mathforum.org> Yeah, Nevada and its neighbors. I realized this after I posted last night. (This may also be a minimal map that requires 4 colors and has an adjacency of 3.) Well this is convoluted logic, for Nevada has 5 adjacencies, California, Oregon, Idaho, Utah, Arizona and of no surprize 4 colors would be needed, even though 3 of those 5 are Mutual Adjacencies. Same as your example, Chris, which has 3 Mutual Adjacencies but 2 or 3 more other adjacencies would not be surprizing to require 4 colors. Proof of the 4 CM would say that there does not exist 5 Mutually Adjacent countries. If instead that Nevada had in total 3 adjacencies and required 4 colors, then you have something to boast about. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== Subject: Re: Why Archimedes Plutonium will never proof the 4CT I'm not real sure what's connected to what there, but if it's how I think it is, then you can color that with 3. A A C B B A