mm-4339 === Subject: Polynomial Roots http://mypeoplepc.com/members/jon8338/polynomial/ === Subject: Re: Polynomial Roots > http://mypeoplepc.com/members/jon8338/polynomial/ Roots can only be found explicitly (as opposed to Newton's approximation)if there is a splitting field. The cubic and quartic can be decomposed into quadratics. Golois proved, and his proof is now universally accepted, that a general quintic and higer power is not analytically soluble. - Ian Parker === Subject: Re: Polynomial Roots >http://mypeoplepc.com/members/jon8338/polynomial/ It's wrong after just a few steps. You start with the equation to be solved (1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 You rewrite (1) as a dot product of vectors ... (2) P_1 . T = 0 where P = and T = <1, t, t^2, ..., t^n> So far, no problem. You then note that the 2-vector is perpendicular to the 2-vector <-a_p/a_0, 1>. Fine. You then make the false claim (3) <1, t^p>/(1 + t^(2p)) = <-a_p, a_0>/((a_0)^2) + (a_p)^2) Although you neglect to use _words_ to defend your claim, I can guess what your flawed reasoning probably was ... Since the vectors P_1, T are perpendicular, you concluded that they must be perpendicular when restricted to pairs of components. There's no way to justify that. Here's a counterexample ... Consider the equation -3 + t + t^2 + t^3 = 0 which has the obvious solution t=1. Thus, the vector equation is <-3, 1, 1, 1> . <1, 1, 1, 1> = 0 But there is no perpendicular pair of corresponding components, which is what you claimed. Thus, your \solution\ is simply bogus. Don't you think you should have checked a few examples before launching into the monstrosity you posted? quasi === Subject: Re: Polynomial Roots >On Sun, 19 Aug 2007 03:39:22 -0400, \Jon G.\ > >>http://mypeoplepc.com/members/jon8338/polynomial/ > >It's wrong after just a few steps. > >You start with the equation to be solved > >(1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 > >You rewrite (1) as a dot product of vectors ... > >(2) P_1 . T = 0 where > > P = > > and > > T = <1, t, t^2, ..., t^n> > That's right where I have the problem. P is a vector of scalars, and T is a vector of indeterminates. >So far, no problem. > >You then note that the 2-vector is perpendicular to the >2-vector <-a_p/a_0, 1>. Fine. > >You then make the false claim > >(3) <1, t^p>/(1 + t^(2p)) = <-a_p, a_0>/((a_0)^2) + (a_p)^2) > >Although you neglect to use _words_ to defend your claim, I can guess >what your flawed reasoning probably was ... > >Since the vectors P_1, T are perpendicular, you concluded that they >must be perpendicular when restricted to pairs of components. > >There's no way to justify that. > >Here's a counterexample ... > >Consider the equation > > -3 + t + t^2 + t^3 = 0 > >which has the obvious solution t=1. > >Thus, the vector equation is > > <-3, 1, 1, 1> . <1, 1, 1, 1> = 0 > >But there is no perpendicular pair of corresponding components, which >is what you claimed. > >Thus, your \solution\ is simply bogus. > >Don't you think you should have checked a few examples before >launching into the monstrosity you posted? > >quasi Brian === Subject: Re: Polynomial Roots On Sun, 19 Aug 2007 04:49:49 -0400, Brian VanPelt > >>On Sun, 19 Aug 2007 03:39:22 -0400, \Jon G.\ >> >>>http://mypeoplepc.com/members/jon8338/polynomial/ >> >>It's wrong after just a few steps. >> >>You start with the equation to be solved >> >>(1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 >> >>You rewrite (1) as a dot product of vectors ... >> >>(2) P_1 . T = 0 where >> >> P = >> >> and >> >> T = <1, t, t^2, ..., t^n> >> >That's right where I have the problem. P is a vector of scalars, and >T is a vector of indeterminates. So what? It's still can be regarded as a dot product which equals 0. The error is not there -- it's when the OP tries to assert that you can restrict the dot product to just two components and then claim that the restricted dot product is still 0. quasi === Subject: Re: Polynomial Roots > On Sun, 19 Aug 2007 04:49:49 -0400, Brian VanPelt > > >On Sun, 19 Aug 2007 04:40:13 -0400, quasi > > > >>On Sun, 19 Aug 2007 03:39:22 -0400, \Jon G.\ > > >> > >>>http://mypeoplepc.com/members/jon8338/polynomial/ > >> > >>It's wrong after just a few steps. > >> > >>You start with the equation to be solved > >> > >>(1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 > >> > >>You rewrite (1) as a dot product of vectors ... > >> > >>(2) P_1 . T = 0 where > >> > >> P = > >> > >> and > >> > >> T = <1, t, t^2, ..., t^n> > >> > >That's right where I have the problem. P is a > vector of scalars, and > >T is a vector of indeterminates. > > So what? > > It's still can be regarded as a dot product which > equals 0. > > The error is not there -- it's when the OP tries to > assert that you > can restrict the dot product to just two components > and then claim > that the restricted dot product is still 0. > > quasi Independently of your subsequent counterexample, I think that Brian comment is related to a formal question. If we are talking about scalar product, the only way to make sense to the first lines is: (a) Consider P = (a_0, a_1, a_2, ..., a_n) belonging to C^(n+1). (b) Consider S={T = (1, t, t^2, ..., t^n) in C^(n+1): t is a root of the given polynomial} (c) Consider the scalar product in C^(n+1): < x, y >= Sum{i=0 to n} x_i y_i* Then P is orthogonal to S. Fernando. === Subject: Re: Polynomial Roots On Sun, 19 Aug 2007 07:40:54 EDT, fernando revilla > >> On Sun, 19 Aug 2007 04:49:49 -0400, Brian VanPelt >> >> >On Sun, 19 Aug 2007 04:40:13 -0400, quasi >> > >> >>On Sun, 19 Aug 2007 03:39:22 -0400, \Jon G.\ >> >> >> >> >>>http://mypeoplepc.com/members/jon8338/polynomial/ >> >> >> >>It's wrong after just a few steps. >> >> >> >>You start with the equation to be solved >> >> >> >>(1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 >> >> >> >>You rewrite (1) as a dot product of vectors ... >> >> >> >>(2) P_1 . T = 0 where >> >> >> >> P = >> >> >> >> and >> >> >> >> T = <1, t, t^2, ..., t^n> >> >> >> >That's right where I have the problem. P is a >> vector of scalars, and >> >T is a vector of indeterminates. >> >> So what? >> >> It's still can be regarded as a dot product which >> equals 0. >> >> The error is not there -- it's when the OP tries to >> assert that you >> can restrict the dot product to just two components >> and then claim >> that the restricted dot product is still 0. >> >> quasi > >Independently of your subsequent counterexample, I think >that Brian comment is related to a formal question. > >If we are talking about scalar product, the only way to make >sense to the first lines is: > >(a) Consider P = (a_0, a_1, a_2, ..., a_n) belonging >to C^(n+1). > >(b) Consider S={T = (1, t, t^2, ..., t^n) in C^(n+1): t >is a root of the given polynomial} > >(c) Consider the scalar product in C^(n+1): > >< x, y >= Sum{i=0 to n} x_i y_i* > >Then P is orthogonal to S. Ok, although for vector spaces over C, this is not the standard inner product. But I agree, the vector <1, t, t^2, ..., t^n> needs to be in the same space, so by regarding t as a root in C, both vectors are in C^(n+1). Alternatively, the components of both vectors could be regarded as polynomials in the indeterminate t with real (or complex) coefficients. Thus, the essential error is not the misuse of inner product notation, which can easily be given a valid interpretation, but rather the claim that when a dot product is zero, it's still zero when the vectors are restricted to 2-vectors. By the way, there is at least one other very serious error. The OP later asserts that since is perpendicular to <1, t, t^2, ..., t^n> it follows that <1, t, t^2, ..., t^n> is in the same direction as a particular vector which is perpendicular to . Firstly, it could be the opposite direction, but that's minor. What's major is that it need not even be parallel, since the set of vectors which are perpendicular to is an n-dimensional hyperplane. Thus. finding a particular vector perpendicular to does not determine the vector <1, t, t^2, ..., t*n>. But such later errors are irrelevant. The first unrecoverable error already suffices to invalidate the claimed solution. quasi === Subject: Re: Polynomial Roots >On Sun, 19 Aug 2007 07:40:54 EDT, fernando revilla > >> >>> On Sun, 19 Aug 2007 04:49:49 -0400, Brian VanPelt >>> >>> >On Sun, 19 Aug 2007 04:40:13 -0400, quasi >>> > >>> >>On Sun, 19 Aug 2007 03:39:22 -0400, \Jon G.\ >>> >>> >> >>> >>>http://mypeoplepc.com/members/jon8338/polynomial/ >>> >> >>> >>It's wrong after just a few steps. >>> >> >>> >>You start with the equation to be solved >>> >> >>> >>(1) a_0 + a_1*t + a_2*t^2 + ... + a_n*t^n = 0 >>> >> >>> >>You rewrite (1) as a dot product of vectors ... >>> >> >>> >>(2) P_1 . T = 0 where >>> >> >>> >> P = >>> >> >>> >> and >>> >> >>> >> T = <1, t, t^2, ..., t^n> >>> >> >>> >That's right where I have the problem. P is a >>> vector of scalars, and >>> >T is a vector of indeterminates. >>> >>> So what? >>> >>> It's still can be regarded as a dot product which >>> equals 0. >>> >>> The error is not there -- it's when the OP tries to >>> assert that you >>> can restrict the dot product to just two components >>> and then claim >>> that the restricted dot product is still 0. >>> >>> quasi >> >>Independently of your subsequent counterexample, I think >>that Brian comment is related to a formal question. >> >>If we are talking about scalar product, the only way to make >>sense to the first lines is: >> >>(a) Consider P = (a_0, a_1, a_2, ..., a_n) belonging >>to C^(n+1). >> >>(b) Consider S={T = (1, t, t^2, ..., t^n) in C^(n+1): t >>is a root of the given polynomial} >> >>(c) Consider the scalar product in C^(n+1): >> >>< x, y >= Sum{i=0 to n} x_i y_i* >> >>Then P is orthogonal to S. > >Ok, although for vector spaces over C, this is not the standard inner >product. > >But I agree, the vector <1, t, t^2, ..., t^n> needs to be in the same >space, so by regarding t as a root in C, both vectors are in C^(n+1). > >Alternatively, the components of both vectors could be regarded as >polynomials in the indeterminate t with real (or complex) >coefficients. > >Thus, the essential error is not the misuse of inner product notation, >which can easily be given a valid interpretation, but rather the claim >that when a dot product is zero, it's still zero when the vectors are >restricted to 2-vectors. > >By the way, there is at least one other very serious error. > >The OP later asserts that since > > is perpendicular to <1, t, t^2, ..., t^n> > >it follows that <1, t, t^2, ..., t^n> is in the same direction as a >particular vector which is perpendicular to . >Firstly, it could be the opposite direction, but that's minor. What's >major is that it need not even be parallel, since the set of vectors >which are perpendicular to is an >n-dimensional hyperplane. Thus. finding a particular vector >perpendicular to does not determine the vector <1, t, >t^2, ..., t*n>. > >But such later errors are irrelevant. The first unrecoverable error >already suffices to invalidate the claimed solution. > >quasi Yeah, I did mean incompatibility of vector dots, and I should explain myself. I went to the sight without having much enthusiasm at the outset because I knew the result was false. Basically, when I saw his dot product, I pretty much just left the sight - reaffirmed that it had no substance. So, I have to admit, I didn't read it very carefully. I saw the 2-vector thing but really didn't look very carefully - like when you don't listen to someone who just rattles on, but you catch bits and pieces of what they are saying. If the algorithm works for any polynomials, I suspect that it is a very select collection! The conditions on such polynomials must be endless. Brian === Subject: Re: Polynomial Roots > http://mypeoplepc.com/members/jon8338/polynomial/ I don't believe it. What are the roots to x^7 + x - 1 = 0? --- Christopher Heckman === Subject: Re: Polynomial Roots On Sun, 19 Aug 2007 07:55:18 -0000, Proginoskes >> http://mypeoplepc.com/members/jon8338/polynomial/ > >I don't believe it. What are the roots to x^7 + x - 1 = 0? > > --- Christopher Heckman I'm with you. While the powers of t form a basis for a vector space - an infinite one, I wasn't aware that scalars did that, unless it was a vector space of dimension 1. Are you allowed to mix dimensions like that? Brian === Subject: An interesting thing Hi all: Reading Paulo Ribenboim's \13 Lectures on Fermat's Last Theorem\, I came across the following letter sent to him by F. Schlichting from G\.9attingen, in relation with the Wolfskehl Prize founded in 1908 by the Royal Society of Sciences in G\.9attingen for whoever could prove FLT. Please do pay special attention to the paragraph beginning with \ Nearly all solutions...\ . I think that explains particularly well the pool of cranks, anticantorians weirdos, infinite-finite cuckoos and G\.9attingen, March 23, 1974 Please excuse the delay in answering your letter. I enclose a copy of the original announcement, which gives the main regulations, and a note of the \Akademie\ which is usually sent to persons who are applying for the prize, now worth a little bit more than 10,000 DM. There is no count of the total number of \solutions\ submitted so far. In the first year (1907-1908), 621 solutions were registered in the files of the Akademie, and today they have stored about 3 meters of correspondence concerning the Fermat problem. In recent decades it was handled in the following way: the secretary of the Akademie divides the arriving manuscripts into (1) complete nonsense, which is sent back immediately, and into (2) material which looks like mathematics. The second part is given to the mathematical department and there, the work of reading, finding mistakes and answering is delegated to one of the scientific assistants (at german universities these are graduate individuals working for PhD or habilitation and helping the professors with teaching and supervision) - at the moment I am the victim. There are about 3 to 4 letters to answer per month, and there is a lot of funny and curious material arriving, e.g. like the one sending the first half of his solution and promising the second if we would pay 1,000 DM in advance; or another one, who promised me 10 per cent of his profits from publications, radio and TV interviews after he got famous, if only I would support him now; if not, he threatened to send it to a russian mathematics department to deprive us the glory insists on personal discussion. Nearly all solutions are written on a very elementary level (using the notions of high school mathematics and perhaps some indigested papers in number theory), BUT CAN NEVERTHELESS BE VERY COMPLICATED TO UINDERSTAND. -[the caps are NOT original] - Socially, the senders are often persons with a technical education but a failed career who try to find success with a proof of the Fermat problem. I gave some of the manuscripts to physicians who diagnosed heavy schyzophrenia. One condition of Wolfskehl's last will was that the Akademie had to publish the announcement of the prize yearly in the main mathematical periodicals. But already after the first years the pediodicals refused to print the announcement, because they were overflowed by letters and crazy manuscripts. So far, the best effect has been had by another regulation of the prize: namely, that the interest from the original 100,000 Mark could be used by the Akademie. For example, in the 1910's the heads of G\.9attingen mathematics department (Klein, Hilbert, Minkowski) used this money to invite Poincare to give six lectures in G\.9attingen. Since 1948 however the remainder of the money has not been used. I hope that you can use this information and would be glad to answer any further questions. F. Schlichting I hope you enjoyed this. Tonio === Subject: Re: is it true that the hindus educate the world i math? :>: you claimed Boole invented the idea of binary : >: representation of numbers. : > Read what I said: : > \The ingenious method of expressing every possible number using : > a set of TWO symbols (each symbol having a place value and an : > absolute value) emerged in England, with George Boole, father of : > the computer.\ -- Androcles. : You lost, two can play by your rules. === Subject: Re: Roman Numeral Odometer > In reference to: > A Decimal Roman Numeral Odometer > > following up on: > Mark's Reform of the Year AD XI > > which is expanded on at: > http://federation.g3z.com/Mathematics/index.htm#roman There used to be a Roman Numeral calculator at \ http://www.geocities.com/mdcluz/games/Romans/ , but it's not there any more. Anyone know where it moved to? --- Christopher Heckman === Subject: Re: #12 Great Attractor is Nucleus of Atom Totality and Great Wall \ & Sloan Great Wall Re: ATOM TOTALITY (Atom Universe) THEORY REPLACES BIG BANG \ THEORY IN PHYSICS > >It would be nice to see if Luminet's data places the Great Attractor as \ the center of his dodecahedron. If it places the Great Attractor anywhere \ near the center then we have confirmation of Atom Totality theory. > > In my travels on a plane flight I met a chemist and told him that the > radiation from atomic waste can be neutralized chemically. He thought > that was a most interesting idea indeed. Although I did not tell him > how to do it, he (or others that follow him) will figure it out. > > Now, in meeting you on this web forum, I'll do something similar for > you (and those that follow you)... > > Atom Totality theory is wrong, and Luminet's data won't place the > Great Attractor as the center of his dodecahedron, because the Great > Attractor is yet another local affair in our part of the universe > among many others to be discovered. Great Attractors are > nucleotides. Filaments are DNA strands. The universe is ALIVE. It is always better to bring logic into a science discussion, than ever to bring philosophy into a science discussion. Philosophy is too much \draped feelings\ posing as if it had intellectual content. So you want to have the Cosmos as a \living entity.\ Now let me pose logic over the main features of your above opinion of the Atom Totality theory. Supporting data and evidence is logic but let me put an overarching logic that would dismiss your \Live Cosmos.\ The logic is that \all matter is atoms and since the Cosmos is not a void itself but a entity of matter then the Cosmos is an atom, albeit a huge single atom\. Your Alive Cosmos is a wish or feeling or opinion. My \Atom Universe\ falls out of the logical constraint that since all matter is atoms, then the Universe itself is one big atom. So you see, logic dismisses wishes and desires and feelings and puts constraints on what can be true and what is fantasy feelings. Not only do I list mounting evidence that the Cosmos is a big atom but that Logic insists a theory is fully Consistent. The Big Bang is not consistent with the Atomic theory that all matter is composed of atoms, for it leaves out the entity of matter that is the Cosmos itself. So maybe you have learned something here, that your feelings and opinions are not as good as logical analysis of a situation. Oh, and did the airplane rider you were yakking with over chemistry intervening in nuclear force radioactivity, figure out that you are a nonscientist. And since you are a nonscientist that you must be exceedingly arrogant and thus stupid. AP not worth saving in the archive === Subject: Re: two norms are equivalent ... <46c68da3$0$434$426a74cc@news.free.fr> <46c6c413$0$36440$4fafbaef@reader5.news.tin.it> > > I think that the OP is quoting from a book and that the definition of > > equivalence there is probably in terms of the topologies defined by the > > norms being the same. > > Yes, you are quite right. You should have said so then. --- Christopher Heckman === Subject: Re: approximation by series expansion > > 2/(-1+a)^(1/2)/a^(1/2)*((2*a-1)*((-1+a)*a)^(1/2)+2*a^2-2*a)^j* > > ((-8*a+1+8*a^2)*((-1+a)*a)^(1/2)+8*a^3+4*a-12*a^2)^(-j) > > Yes, my fault again. Actually I tried the integral in Maple 10, but > c_j:= 1/Pi*int(2/(2*a - 1 - cos(t))*cos(j*t), t=-Pi..Pi); > did not evaluate imediately. (What did You do to make it evaluate? > Substitute the cosine?) Here's what I did (or more precisely, what I should have done the first time...). The generating function of c_j is g(s) = 1/pi int_{-pi}^pi sum_{j=0}^infty 2 s^j/(2 a - 1 - cos(t)) cos(j t) dt. > sum(s^j*2*cos(j*t)/Pi/(2*a-1-cos(t)),j=0..infinity); int(%, t=-Pi..Pi) assuming a>0, s>0, s<1; g:= factor(%,sqrt((a-1)*a)); 2 2 1/2 2 1/2 2 (2 a - 2 a - (a - a) + 2 a (a - a) ) g := - ---------------------------------------------- 2 1/2 (a - 1) a (1 + s - 2 a - 2 (a - a) ) Note that this is of the form A/(s-B), so its Taylor series is of the form sum_{j=0}^infty -A s^j/B^(j+1), i.e. c_j = -A/B^(j+1). > R,A := selectremove(has,g,s); 1 R, A := ---------------------------, 2 1/2 1 + s - 2 a - 2 (a - a) 2 2 1/2 2 1/2 2 (2 a - 2 a - (a - a) + 2 a (a - a) ) - ---------------------------------------------- (a - 1) a > B := s-1/R; 2 1/2 B := -1 + 2 a + 2 (a - a) Thus a formula for c_j, somewhat nicer than what I posted before, is > -A/B^(j+1); 2 2 1/2 2 1/2 2 (2 a - 2 a - (a - a) + 2 a (a - a) ) ---------------------------------------------- 2 1/2 (j + 1) (a - 1) a (-1 + 2 a + 2 (a - a) ) Actually, an even more simplified form, which it takes a bit of Maple acrobatics to get, is 2 --------------------------------------- 2 1/2 2 1/2 j (a - a) (-1 + 2 a + 2 (a - a) ) -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Graph Theory: Cuts, Minimal Cuts and Bonds On Aug 17, 1:15 am, Narek Saribekyan > I'm confused in definitions of these terms. Can anyone give me some > explantions? > > These definitions are from Reinhard Diestel's \Graph Theory, 3rd > Edition\ book. > > Minimal with some property: > The books says \More generally, when we call a graph minimal or > maximal with some property, but have not specified any particular > ordering, we are reffering to the subgraph relation. When we speak of > minimal or maximal sets of vertices or edges, the reference is simply > to set inclusion.\ > > E(X,Y) set: > If x is from X and y is from Y (vertices), then we call the edge xy > from E an X-Y edge. We denote the set of this edges in E (edge set of > graph) by E(X, Y). Instead of E{{ x }, V) and E(X, {y}) we simply > write E(x,Y) and E(X,y). The set of all the edges in vertex v is > denoted by E(v). > > Cut: > \If {V1,V2} is a partition of V the set E(V1,V2) (already defined, see > above) of all the edges of G crossing(defines 'crossing') this > partition is called a cut (or cocycle). Recall that for V1={ v} this > cut is denoted by E(v).\ > > Bond: > \A minimal non-empty cut in G is a bond.\ > > By these definitions, I suppose that every cut is a minimal(with a > fixed partition). Well, yes, but when looking at minimality, you will be looking at different partitions. One partition = one cut. It may happen that E(V1,V2) properly contains E(W1,W2) (as subsets of the edges of G); in that case, E(V1,V2) cannot be a bond. > I mean, if we remove some edges from our cut, the > remaining edges cannot form a cut with the same partition(or, which is > the same, there's no subset of a cut that forms a cut with the same > partiton). Thats true for any cut, so any cut is a minimal. Thus every > non-empty cut is a bond??? Another way to define a cut (which makes \minimality\ clearer) is as a set of edges whose removal disconnects G. In this case, the minimal sets of edges [...] _are_ bonds, where minimal means in terms of set inclusion: That is, if S1 and S2 are [nonempty] cuts [viewed as sets of edges], and S1 properly contains S2, then S1 is not a bond. --- Christopher Heckman === Subject: Re: Ftn equation > On Aug 18, 2:53 am, Robert Israel > > > > > > > > > f : R -> R > > > > > > f(x + y/3) = f(x) + f(y)/2 > > > > > > (This problem is from an highschool in Korea) > > > > > > Is there anyone who can solve this equation?? > > > > > y:=0 => f(0)=0. > > > > > f(x/3)=f(x)/2. (x:=0, y:=x) (1) > > > > > f(4x/3)=3f(x)/2 (y:=x) (2) > > > > > f(x+y)=f(x)+f(3y)/2. (x:=x, y:=3y in the original eq) (3) > > > > > (1),(2) => f(4x/3)=3f(x/3) => f(4x)=3f(x) (4) > > > > > (1) => f(x)=f(3x)/2 (5) > > > > > f(x+x)=f(x)+f(3x)/2 (x:=x, y:=3x in the original eq) => using \ (5) > > > > > f(2x)=2f(x) (6) > > > > > However using (4) and (6) two times, > > > > > 3f(x)=f(4x)= 2f(2x)=4f(x) > > > > > which implies > > > > > f(x)=0 for all x. > > > > > Remark. This result is valid for f:V->W where V,W are arbitrary > > > > vector spaces. > > > > How about the case of f(ax + by) = Af(x) + Bf(y)? > > > This generalizing looks interesting... > > > Anyone help? > > > There are a number of special cases to look at. For example, > > f(x) = c x + d satisfies this equation in the following cases: > > 1) c = d = 0 > > 2) c = 0, B = 1 - A > > 3) d = 0, a = A, b = B > > 4) a = A, b = B = 1 - A > > > In some of these cases there are also more exotic solutions that can't \ be > > constructed explicitly but need some form of the Axiom of Choice. > > I'd be surprised to find a solution other than f(x) = 0 in any other \ case. > > -- > > Robert Israel isr...@math.MyUniversitysInitials.ca > > Department of Mathematics http://www.math.ubc.ca/~israel > > University of British Columbia Vancouver, BC, Canada > > 4) (i) c not 0, d=0 => a=A, b=B. > (ii) c,d not 0 => a=A, b=B, A+B=1. These are included in my 3) and 4). Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: JSH: Looking back, wondering > Or if I made a mistake you can point it out, but if I did not make a > mistake, and have a proof, then it must be true then, right? The adventures of Soupman - Helping people in soup-related situations Adventure number 7. Town about to be flooded by minestrone soup One day mild mannered Clart Bent was out jogging. As he jogged past by the minestrone soup factory he noticed that the roof was bulging upwards. Pausing only to put on his Soupman costume he rushed into the factory. It was one of the new totally automatic minestrone soup factories. Soupman immediately noticed that the Master Outlet Safety Override Emergency Backup Valve was corroded shut! At that moment the entire factory exploded, and an enormous ball of superheated minestrone soup was launched upwards. It would soon fall back and devastate Squallville!! Using his special Anti-minestrone Soup Atom Laser, Soupman was able to zap the deadly ball. Squallville was saved! *Soupman* === Subject: #15 deuterium water on Europa is it 1/2 of that of Earth? Errors in \ this journal report on Jupiter deuterium Re: ATOM TOTALITY (Atom Universe) \ THEORY REPLACES BIG BANG THEORY IN PHYSICS I have gone as far as I think I can on zircon aging of the Outer planets versus Inner planets and will just have to wait for further news and research. Perhaps someone who has zircon dated a meteorite found a anomaly of 8 billion years or greater but threw away his findings thinking it was a mistake. So I am going to attack this age problem from another angle; with the abundance of heavy water (deuterium) in comets more so than in Earth's ocean water. I suspect this journal report is illogical and gummed up: --- quoting from \ http://www.americanscientist.org/template/AssetDetail/assetid/14384/page/4;js\ essionid=aaa5LVF0 The deuterium atom is interesting because every natural bit of it was created in the Big Bang-about 20 to 30 parts per million (ppm) hydrogen atoms. It's not easy to make deuterium any more, and its abundance has been diminishing over the eons to the point where the local interstellar clouds only have about 5 to 15 ppm deuterium. When the solar system was formed 4.56 billion years ago, the deuterium abundance was still about 20 ppm, as demonstrated by its presence in the atmospheres of Jupiter and Saturn, which were captured from the gas of the solar nebula before it dissipated. Earthly seawater contains 160 ppm of deuterium, an enrichment of eight times relative to the deuterium in the solar nebula. So we would expect the comets to be similarly enriched, no? No. As it happens, the water in the three bright comets-Halley, Hyakutake and Hale-Bopp-that recently swung past the Earth averaged about 320 ppm deuterium, an enrichment of 16 times over the deuterium in the solar nebula, and twice that found in seawater! How do we account for these differences? --- end quoting that website --- It is late at night here and maybe my mind is too sleepy at the moment, but I suspect the is equal to the up the abundance of Jupiter's deuterium in gaseous form with the comet form of deuterium in water the form on Earth in ocean water. I think that is a bad and flawed assumption deuterium in water found on Europa would be very much different from the abundance of deuterium found in gaseous form on Jupiter or Saturn and to compare gaseous heavy water to ocean water is flawed, at least I suspect so. If my suspicion is true, then the entire apples to oranges and assuming they are the same thing. But I believe the heavy water of Earth's oceans compared to the heavy water of the three comets listed is a valid comparison since they are \water or ice\. So the entire Jupiter data is inappropriate since it is gaseous form. Perhaps the researchers recalibrated the gaseous form to accomodate Earth and Comets but I doubt it. I doubt water can exist on Jupiter in the form of water on Earth or Comets. the fact that Earth ocean water is 160 ppm deuterium and Comet water is 320 ppm or twice as much heavy water. I am going to guess that if the water on Europa can be measured that it will be found to be 80 ppm deuterium. Or water found on any satellites of the Outer Planets. Why do I guess this? Because I suspect Dirac radioactivities created our Solar System and not the Nebular Dust Cloud theory. And that the Sun and Inner Planets are 10 billion years old, and the Outer Planets are 5 billion years old and that the Comets are also 10 billion years old. So there must be a mechanism in which the deuterium on Earth has lost 1/2 of its deuterium. Perhaps it is because of the closer orbit to the Sun that 1/2 of the deuterium has been lost whereas the Comets contain their pristine original deuterium with little loss to the Sun. Maybe deuterium can be a measuring rod for the age of the components of the Solar System. Perhaps I should also look into lithium, beryllium and boron abundances of the inner as compared to outer planets. Instead of Earth acquiring its ocean waters from meteorite and comet impacts, a better scenario is that the inner planets and satellites were swallowed up by the Sun leaving behind a few satellites such as the protoMercury, protoVenus, proto Earth and which the water migrated to Earth. For example, what happens when a satellite the size of Moon slams into another satellite of the same size which has a huge oceans? Would not the water migrate? So that all the water on ancient Mercury, ancient Venus, ancient Mars ends up on Earth. If in 5 billion years from now, that there would not be a Mercury, Venus, Earth or Mars but swallowed up by Jupiter as well as all the other gas giants, and all that is remaining in our Solar System is Sun, the star-Jupiter and 4 new planets of what were formerly satellites of Jupiter and Saturn and Neptune. Two of these new planets are Europa and Titan and would not all the water remaining migrate to one of these 4 new planets? So instead of a Nebular Dust Cloud theory as the mechanism of Solar Systems, what we have is a growing of stars and planets via Dirac radioactivity, and that Solar Systems grow from a single star into a twin star system and when the twin star is borne the former inner planets were swallowed up and the satellites of the outer planets become the new planets. Do we have any evidence for this pattern? By all means we do. We have the recent discovery of exoplanets in that they usually have huge Jupiter sized planets orbiting nearby their star. 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: #16 Earth has 1/2 the heavy-water of Comets, but does Earth have 2X \ the abundance of lithium, beryllium, boron as does Jupiter?? Re: ATOM \ TOTALITY (Atom Universe) THEORY REPLACES BIG BANG THEORY IN PHYSICS Perhaps I can date Earth versus Jupiter as that of 10 billion years versus 5 billion years from the abundance of lithium, beryllium, boron or some other chemical element. So far I have found just one relationship of twice as much abundance. The relationship that Earth water is 160 ppm deuterium and Comets are 320 ppm. Can I find another data where a chemical element is twice as abundant between Earth and Jupiter? It is too late tonight. 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: #17 cores of planets and satellites tells us their ages and that \ Earth is twice as old as Jupiter Re: ATOM TOTALITY (Atom Universe) THEORY \ REPLACES BIG BANG THEORY IN PHYSICS I took a brief false turn in the roadway for evidence that inner- planets are 10 billion years old versus 5 billion years for outer-planets. A bad turn in the road because atmospheric science of the Solar System is not going to tell me much if anything about age. Instead of atmospheres, I need to get to the cores of these objects. Because the difference between a Nebular Dust Cloud theory versus Dirac Radioactivity as the creation of the Solar System would be registered in the core composition. It has been known for a very long time that the inner planets have dense cores and the outer planets have lighter cores. This is because Earth is 10 billion years old and Jupiter is a young object of 5 billion years old. Mars probably had a dense core as Earth but some collision, much like the Earth and Moon collision separated the core of Mars and we see it as the Asteroid belt. I was googling for evidence and data on the outer planetary cores: http://adsabs.harvard.edu/abs/2001Icar..151..204K Unless I am reading that harvard report wrong, it says the cores of the satellites of the outer-planets are similar to the core of the Moon and about 1/2 the density of the core of Earth. So that Europa, Titan, Io and others are about 1/2 the core density of Earth core. So, backtracking, I should focus purely on cores of planets as the age reckoning of planets. What is the age of our Sun? Is it 10 billion or 5 billion years old? Well what is its core composition? Does it have a dense core and does it have alot of thorium and uranium? I am going to need to have to reconcile the idea that most Solar Systems have twin stars. So that most Solar Systems have to be at least 10 billion years old, so that 5 billion to give birth to one of the stars and another 5 billion for the twin star. What is the age of the Milky Way Galaxy? Is it 10 billion or 15 billion years old? The key is to measure core abundance for thorium and uranium. Zircon crystals can be a very excellent measure also. Just knowing that Mercury, Venus, Earth and Mars+Asteroids have twice the density of their cores compared to the Outer Planets and their satellites tells us that Earth is 10 billion years old and Jupiter is 5 billion years old. 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: #18 yes! the cores of Sun, Earth, Jupiter, Io indicate 10 billion \ and 5 billion year ages Re: ATOM TOTALITY (Atom Universe) THEORY REPLACES BIG \ BANG THEORY IN PHYSICS I have got to move on to finishing this edition of this book since I have only 10 days remaining in this month and find myself on a topic that should be in a different book altogether \Growing Solar System from Dirac Radioactivity theory replaces Nebular Dust Cloud theory\ The reason I ventured into this topic is because the Cosmos has layered ages where some stars are older than the recent Plutonium Atom MiniBang Accretion, so that where the oldest stars of 20 billion years old are of the Uranium Atom Totality and the newest accretion layer is only 8 billion years old (the Freedman vs. Sandage debate). So if the Universe is layered ages, then the Solar System is probably layered ages where the Inner Planets and Sun are older than Outer Planets. I was looking for zircon crystals to prove there exists such crystals which register a age of 8 to 10 billion years old, of perhaps a crystal found inside some eucrite meteorites. But I believe I have found the very best way of determining the age of Sun and planets and satellites in our Solar System-- their cores. Cores of stars become dense in iron as they age. So if you see a star with a large iron core, it is an old star. Likewise we can age Earth and Jupiter and Io if we know their core data. Now I am using this website for information on cores: http://www.nineplanets.org/sol.html And from my Harvard source which says that Jupiter's satellites are about 40% the size of Earth's core. Io core is huge compared to Earth. Earth core is a factor of 300,000 in size whereas Jupiter core (even though much is unknown and says about 10-15 Earth masses) is only a factor of 30 in size to Io. What that tells me is the Nebular Dust Cloud theory cannot cope with those figures. That if the Nebular Dust Cloud theory were true then the ratios of cores of Earth with Sun and Jupiter with Io should have been in somewhat agreement. About the only agreement we see in cores of the Solar System is that the Inner-planets cores are relatively the same, and that the Outer-planet cores are relatively the same amoung one another, and that the cores of the Outer-planet satellites are approx 40% the size of Earth's Core. So what the core data suggests is the age of the Inner planets and Sun are of the same age and twice the age of the Outer Planets and their satellites. That Sun and Mercury, Venus, Earth and Mars are 10 billion years old and Jupiter, Saturn, Uranus, Neptune and their satellites are only 5 billion years old. Also, I want to say something that is pretty neat about the Growing Solar System theory in that it allows for the Sun to be younger, say 5 billion years old, yet Earth being 10 billion years old. That is a remarkable feature of that theory and where the Nebular Dust Cloud could never have accomodated. The reason Growing Solar System can have such a feature is because of Dirac Radioactivity. And I do not remember if I called the concept as \seed-dot\ of the electron-dot-cloud. How our Solar System started was that Sun and Mercury and the other planets and their satellites were borne of a \seed-dot\ which taps directly into the Nucleus of the Atom Totality and from which cosmic rays or gamma bursts from the Nucleus end up at this \seed-dot\ making it grow. So it starts growing from a few rays and bursts and more are added to that seed dot. I called this concept in the 1990s as \neutron materialization\ and later called it \dirac radioactive materialization\. Our planet Earth, every day is bombarded from cosmic rays and gamma ray bursts. and created our planet and continues to grow our planet. But some planets like Jupiter seem to have a fountain of growth where Jupiter receives even more Dirac radioactivity and grows Jupiter faster than even the Sun. So in this vision of solar system dynamics, one planet can accelerate in growth while another has tiny growth. So one can envision how in this theory, Earth could be twice as old as the Sun, and where Jupiter could be growing exponentially faster than the Sun. But I should save this for another book and not crowd this book with a theory of Solar System dynamics. So I am going to stop on this topic, in order to finish this 2nd edition on this book before the end of August. And sadly, I see now that I need a 3rd edition just to straighten-out what I have cluttered into the 2nd edition. So this is an added benefit to writing and publishing a Internet book, in that the reader can follow my thought processes. In the 3rd edition of this book, I should include this age of the Solar System of its layered age in the chapter of the layered age of the Cosmos as 20 billion year old stars inside the newest accretion of 8 billion year old Minibang of the Plutonium Atom Totality. I do not anticipate the 3rd edition until late 2008, unless some big new discovery appears before then. But why talk about 3rd edition when I am not even half way through the 2nd edition. Summary: Yes! indeed! the pattern of the cores of the Sun and Inner Planets compared to the cores of the Outer-Planets and their satellites suggests that the Inner Planets are twice as old as the Outer-Planets, so that Sun and Earth are 8-10 billion years old and Jupiter and Europa are 4-5 billion years old. Also, the cores should trashcan the Nebular Dust Cloud theory because proto-Jupiter as it was sweeping up the gases of the primordial Dust Cloud would not have a physics that allows for Europa and Io to have such a huge sized metal core. The metal in the dust-swath of the proto-Jupiter would have sunk into Jupiter, leaving any satellites that formed as impoverished of a dense core. 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: Unitary oiperaor and countable spectrum > Let T be a unitary operator on a Hilbert space H such that its spectrum > sigma(T) is a countable set. > > How is it possbile to prove that there exists a sequence n_j in N such \ that > T^{n_j} -> I (the identity operator) in the strong operator norm \ topology. > > Any ideas or hints are welcome. By the Lebesgue dominated convergence theorem applied to the spectral \ measures for T, it suffices to find a sequence such that t^(n_j) -> 1 for every t in sigma(T). Do you know a theorem of Dirichlet about simultaneous Diophantine \ approximation? -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Unitary oiperaor and countable spectrum > >> Let T be a unitary operator on a Hilbert space H such that its spectrum >> sigma(T) is a countable set. >> >> How is it possbile to prove that there exists a sequence n_j in N such >> that T^{n_j} -> I (the identity operator) in the strong operator norm >> topology. >> >> Any ideas or hints are welcome. > > By the Lebesgue dominated convergence theorem applied to the spectral > measures for T, it suffices to find a sequence such that t^(n_j) -> 1 for > every t in > sigma(T). Do you know a theorem of Dirichlet about simultaneous > Diophantine approximation? The measure space is atomic, so there is no need for any advanced convergence theorems from measure theory. Why do you mention them? -- rusty === Subject: Re: Unitary oiperaor and countable spectrum > Let T be a unitary operator on a Hilbert space H such that its spectrum > sigma(T) is a countable set. > > How is it possbile to prove that there exists a sequence n_j in N such > that T^{n_j} -> I (the identity operator) in the strong operator norm > topology. > > Any ideas or hints are welcome. Since the spectrum is closed countable, its structure as a Borel set its just a countable set of atoms. So, by the spectral theorem for unitary operators, the operator is diagonalizable. Thus you just have to prove that for a countable closed subset S of the circle, there is a sequence n_j such that z^{n_j} -> 1 for each z in S. -- rusty === Subject: Re: Unitary oiperaor and countable spectrum > Let T be a unitary operator on a Hilbert space H such that its spectrum > sigma(T) is a countable set. > > How is it possbile to prove that there exists a sequence n_j in N such > that T^{n_j} -> I (the identity operator) in the strong operator norm > topology. > > Any ideas or hints are welcome. > Note that this result it not to hard to prove if H is finite dimensional and \ that all of the spectral values of T lie on the unit circle. -Bill === Subject: Re: cantorian algebra > this is how cantor or cantorians solve algebra > > 1) x^3 + y*(x^4) + 2x + 13 = 4y + 3xy - 7 > > solution x and y are aleph_0 or aleph_1 or aleph_2 > etc > > 2) x^2 + 2x - 7 = x > > aleph_1 is a solution > > 3) 14*x^3 -27 x^2 + root(5)x -1729 = x > > aleph_0 is a solution > > 4) x = x + 1 > > aleph_0 is a solution > > 5) x^2 = x^2 + 2x + 1 > > aleph_0 is a solution > > 6) 3x = 3x + 1 > > aleph_1 is a solution > > 7) x^1729 + x^19 + 5* x^17 + 23 x + 7 = 13 > > step 1 --> add x to both sides > > x^1729 + x^19 + 5* x^17 + 24 x + 7 = 13 + x > > step 2 --> solve > > aleph_1 is a solution > > > with these 7 example equations it is clear that using > cantor set theory is very usefull for solving > algebra > > even the counterintuitive 4) x = x+1 can be solved by > using cantor > > and even a high degree equation like > 7) x^1729 + x^19 + 5* x^17 + 23 x + 7 = 13 > > is no problem for cantor set theory... > > > cantor set theory is therefore recommended for > students who want to solve algebra , but dont > understand galois theory , roots or imaginary numbers > > > it is a very powerfull theory and very handy for > first year students > > arent cantorians brilliant !! > > > exercises: > > 1) x^79 + 3x = 79x give 1 solution > > 2) x+y = x + 4y give 1 solution > > 3) x/x = x give 1 solution > > 4) x = x + 384351 give 1 solution > > 5) x^19 + x^12 -3x^2 + 5x = 9 give 1 solution > > > hint : the solution to these 5 equations are quite > similar > > > for calculus we have similar results : > > integral x dx from 0 to aleph_0 = aleph_0 > > integral x^1729 + 3x dx from 0 to aleph_0 = aleph_0 > > > cantor seems to have constructed a very solvable > field extension of the reals > > however in a recent thread tommy1729 gave a bit > harder equation > > exercise: > > 2^x = aleph_0 > > keep in mind that 2^1729 is only finite > > and 2^ aleph_0 is aleph_1 > > .... > > > deadline for the papers is tomorrow at 9. > > thats for the students of course > > cantorians will have no problem solving that equation > of course .... > > > tommy1729 Does one likewise use algebra to \solve\ the calculus? Perhaps you also use a drill to open a can. Tom === Subject: Re: cantorian algebra > > this is how cantor or cantorians solve algebra > > > > 1) x^3 + y*(x^4) + 2x + 13 = 4y + 3xy - 7 > > > > solution x and y are aleph_0 or aleph_1 or aleph_2 > > etc > > > > 2) x^2 + 2x - 7 = x > > > > aleph_1 is a solution > > > > 3) 14*x^3 -27 x^2 + root(5)x -1729 = x > > > > aleph_0 is a solution > > > > 4) x = x + 1 > > > > aleph_0 is a solution > > > > 5) x^2 = x^2 + 2x + 1 > > > > aleph_0 is a solution > > > > 6) 3x = 3x + 1 > > > > aleph_1 is a solution > > > > 7) x^1729 + x^19 + 5* x^17 + 23 x + 7 = 13 > > > > step 1 --> add x to both sides > > > > x^1729 + x^19 + 5* x^17 + 24 x + 7 = 13 + x > > > > step 2 --> solve > > > > aleph_1 is a solution > > > > > > with these 7 example equations it is clear that > using > > cantor set theory is very usefull for solving > > algebra > > > > even the counterintuitive 4) x = x+1 can be solved > by > > using cantor > > > > and even a high degree equation like > > 7) x^1729 + x^19 + 5* x^17 + 23 x + 7 = 13 > > > > is no problem for cantor set theory... > > > > > > cantor set theory is therefore recommended for > > students who want to solve algebra , but dont > > understand galois theory , roots or imaginary > numbers > > > > > > it is a very powerfull theory and very handy for > > first year students > > > > arent cantorians brilliant !! > > > > > > exercises: > > > > 1) x^79 + 3x = 79x give 1 solution > > > > 2) x+y = x + 4y give 1 solution > > > > 3) x/x = x give 1 solution > > > > 4) x = x + 384351 give 1 solution > > > > 5) x^19 + x^12 -3x^2 + 5x = 9 give 1 solution > > > > > > hint : the solution to these 5 equations are quite > > similar > > > > > > for calculus we have similar results : > > > > integral x dx from 0 to aleph_0 = aleph_0 > > > > integral x^1729 + 3x dx from 0 to aleph_0 = > aleph_0 > > > > > > cantor seems to have constructed a very solvable > > field extension of the reals > > > > however in a recent thread tommy1729 gave a bit > > harder equation > > > > exercise: > > > > 2^x = aleph_0 > > > > keep in mind that 2^1729 is only finite > > > > and 2^ aleph_0 is aleph_1 > > > > .... > > > > > > deadline for the papers is tomorrow at 9. > > > > thats for the students of course > > > > cantorians will have no problem solving that > equation > > of course .... > > > > > > tommy1729 > > Does one likewise use algebra to \solve\ the > calculus? > Perhaps you also use a drill to open a can. > > Tom no offense but the comparison does not hold algebra is very closely related to calculus look up \L-algebra\ for instance ... or recall it ( depending on how familiar you are with that ) another example are certain types of analytic computations requiring \ algebra or even an integral of 1/polynomial(x) uses algebra and btw cantor set theory is not even a drill ... rather it is as usefull as a dirty wet towel with holes in it ... the post is a sarcastic note of course tommy1729 \ i dont believe in mathematics \ Albert Einstein === Subject: Finding the group of symmetry [Setting] Let S be a set, and G = SL_n(F), the special linear group on a field F. Now G act on S nontrivially. T is a subset of S. [Question] Is there a way that we can find the/a subgroup H of G leave T fixed? i.e. h*t belong to T for all h in H and t in T. [detail] The setting may be too vague. What I had in mind is that S is F[x,y,..]/~, the polynomial ring in several variables mod some equivalence relation. And G act on S by acting on the vector of variables. I think this should be a well studyed problem in invariant theory, I === Subject: Re: Finding the group of symmetry i donno... maybe look at the group action on sets... u want T(subset)S to be a G-set... try lookat the normalizer and/or stabalizer of the group G, Cauchy-Forbenius theorem can help with counting. good luck... if ur into torrents ... search for Math Complete. its the bible of maths === Subject: Re: Finding the group of symmetry I had no luck in Math Complete (mathcomplete.com?). Maybe my question is too vague. What if I rephrase the problem as: In Z^n (integer lattice of n dimension), I have a set of point T, that form a convex polytope (only the vertices). Now let S_n act on Z^n by permuting the basis. E.g., the transposition (1 2) act on (1 2 3 ...) by making it (2 1 3 ...). Now I want to know which subgroup of S_n fixes the polytope, i.e. the orbit of any x in T is in T. I think this should be a well known problem, but so far I had no luck finding anything. > i donno... maybe look at the group action on sets... > u want T(subset)S to be a G-set... try lookat > the normalizer and/or stabalizer of the group G, > Cauchy-Forbenius theorem can help with counting. > > good luck... > if ur into torrents ... search for Math Complete. its > the bible of maths === Subject: Re: Finding the group of symmetry I think this is a well known problem in polytopes or geometric group theory or something, but I don't know anything about it. On the other hand, as you describe it, it is just a stabilizer calculation in a permutation group and there are reasonably efficient ways to do this from a computational perspective. If you just have some points, then you can use computational group theory software to get an answer: gap> pnts:=[[1,2,3],[2,3,1],[3,1,2]]; [ [ 1, 2, 3 ], [ 2, 3, 1 ], [ 3, 1, 2 ] ] gap> h := Stabilizer( SymmetricGroup(3), Set(pnts), OnSetsTuples); Group([ (1,2,3) ]) gap> Orbit(SymmetricGroup(3),AsSet(pnts),OnSetsTuples); [ [ [ 1, 2, 3 ], [ 2, 3, 1 ], [ 3, 1, 2 ] ], [ [ 1, 3, 2 ], [ 2, 1, 3 ], [ 3, 2, 1 ] ] ] An easy to read introduction to the details of efficient but old methods to do this is in Butler's Algorithms for Permutation Groups, and a modern version is in Holt et al. Handbook of Computational Group Theory. You would likely not need anything in Serres's Permutation Group Algorithms. === Subject: Re: Hubble and singularity On Aug 18, 10:27 pm, Mark Nudelman > > > > > > > Hubble and singularity > > > According to Hubble's law, radius r of universe as a function of time > > t is: > > > dr/dt= H r > > > Integrating > > r=exp(Ht) +R or > > r=r0 exp(Ht) > > > Clearly at t=0, r<> 0 > > > So expansion of universe started from a finite non zero radius ie > > from > > a promordial ball of radius r0. > > Hubble's Law is not a law of nature. It's just a rough empirical rule > that fits the observations. There's no justification for extrapolating > it all the way back to the beginning, especially when you get close to > zero radius and quantum effects come into play. > > --Mark- Hide quoted text - > > - Show quoted text - Yeah. You are right that extrapolation back to zero size is not justified. Unfortunately that is EXACTLY what has been done. That extrapolation is responsible for the singularity. Near the minimum radius quantum effects will be prnounced. The very Uncertainty Principle might cause a band in the primordial ball of enrergy and matter. === Subject: Upper bound for problem hello everybody i was try to solve the problem \145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits\. here is problem link http://projecteuler.net/index.php?section=problems&id=34. although i solved the problem but one thing i want know how can i find the upper bound . === Subject: Re: Upper bound for problem > hello everybody > i was try to solve the problem \145 is a curious number, as 1! + 4! + > 5! = 1 + 24 + 120 = 145. > Find the sum of all numbers which are equal to the sum of the > factorial of their digits\. here is problem link > http://projecteuler.net/index.php?section=problems&id=34. although i > solved the problem but one thing i want know how can i find the upper > bound . --------------- I would say that 7*9! = 2540160 is surely an upper bound. The smallest 8-digit number is 10000000 and the largest value of 8 factorials of digits from 0 to 9 is only 8*9! = 2903040. The same holds for larger numbers of digits. I found only four numbers with this property, 1, 2, 145, and 40585. Does that agree with your findings? Roger Stafford === Subject: Re: Make Operating System Faster !! >> Make your Windows XP faster and smoother >> > > > me, look at me\ psychosis. No psychosis, Google encourages people to look for click revenue on Googles' blogspot. He/she is doing just that also with the help of === Subject: Re: Make Operating System Faster !! Keywords: X-Usenet-Spam: YES format=flowed; reply-type=response > >> >>> >>> Make your Windows XP faster and smoother >> >> \look at me, look at me\ psychosis. > > No psychosis, Google encourages people to look for click revenue on > Googles' blogspot. He/she is doing just that also with the help of Ah, so that's the intent of Google with its blogger site. However, I don't see anywhere that Google is overtly prompting their bloggers to private rather than stroke his ego in public. http://www.blogger.com/content.g transmitting malware and viruses.\ Yeah, right. Must be nice to live in a world where blinders and rose-colored glasses make for a pleasant and lazy view. I'll believe Google is interested in stopping their abuse contact that says they won't do anything. NOTE: Sorry, didn't notice the cross-posting to unrelated groups. FollowUp-To header added to move discussion to the 24hoursupport.helpdesk group, the only group where the original post === Subject: Re: Make Operating System Faster !! 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http://www.arabzwaj.com/welcome/viewtopic.php?t=1751 http://www.arabzwaj.com/welcome/viewtopic.php?t=1838 http://www.arabzwaj.com/welcome/viewforum.php?f=11 http://www.arabzwaj.com/welcome/viewtopic.php?t=234 www.arabzwaj.com :-arabzwaj@yahoo.com hide ip + : :- http://d.turboupload.com/d/839284/157615851606157516051580hide.rar.html :- http://www.megaupload.com/?d=BBCGC4J0 the best web site formarrige and frindship www.arabzwaj.com === Subject: Inverting the floor function (sort of) The OEIS sequence A097075 has an explicit formula: A097075(n) = (1+sqrt(2))^n/4 + (1-sqrt(2))^n/4 + (1) (-1)^n/2. The sub-sequence A097075(2n+1) is related to the sequence A053141(n) as I happened to find out: A097075(2n+1) = floor(A053141(n)*sqrt(2)+1/2) (2) There is no explicit formula for A053141 in the OEIS. But as A097075 does have such a formula, we \only\ have to /invert the floor-function/ to provide us with an explicit formula for A053141. Well I know that this is nonsense, in a way. But at least we can hope for a not so bad approximation due to equation (2). And this is indeed so. We know from (2): A097075(2n+1)-1/2 <= A053141(n)*sqrt(2) < A097075(2n+1)+1/2 and thus we can define two functions LowInv(n) = (A097075(2n+1)-1/2)/sqrt(2) and HigInv(n) = (A097075(2n+1)+1/2)/sqrt(2). A simple-and-stupid approach is to take the arithmetic mean: and see how well we did: : ----------------------------------------------- : 0 0 0 : 1 2 2 : 2 14 14 : 3 84 84 : 4 492 492 : 5 2870 2870 : 6 16730 16730 : 7 97512 97512 : 8 568344 568344 : 9 3312554 3312555 : 10 19306982 19306989 : 11 112529340 112529380 : 12 655869060 655869360 : 13 3822685022 3822686800 : 14 22280241074 22280252000 : 15 129858761424 129858820000 : :______________________________________________________________ : : Table 1: Comparison of exact but recursive sequence A053141 : I should have stopped that mad enterprise by now but I was still curious and asked Maple to compute the quotient 1., 1., 1., 1., 1., 1., 1., 1., 1.0000003, 1.0000004, 1.0000004, 1.0000005, 1.0000005, 1.0000005, 1.0000005, 1.0000006, 1.0000005, 1.0000006, 1.0000007, 1.0000007, 1.0000006, 1.0000008, 1.0000007, 1.0000007, 1.0000008, 1.0000008, 1.0000009, 1.0000009, 1.0000009, 1.0000010, 1.0000010, 1.0000010, 1.0000011, 1.0000011, 1.0000011, 1.0000012, 1.0000012, 1.0000012, 1.0000013, 1.0000012, 1.0000013, 1.0000014, 1.0000014, 1.0000013, 1.0000014, 1.0000014, 1.0000015, 1.0000016, 1.0000015, 1.0000016, 1.0000017, 1.0000017, 1.0000017, 1.0000018, 1.0000018, 1.0000018, 1.0000018, 1.0000018, 1.0000018, 1.0000019 There is a clear tendency. Are there any tricks to get closer to an explicit formula for A053141? I admit I tried the geometric mean instead of the arithmetic mean - but only with a neglectable effect to the better. Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Inverting the floor function (sort of) > The OEIS sequence A097075 has an explicit formula: > > A097075(n) = (1+sqrt(2))^n/4 + > (1-sqrt(2))^n/4 + (1) > (-1)^n/2. > > The sub-sequence A097075(2n+1) is related to the > sequence A053141(n) as I happened to find out: > > A097075(2n+1) = floor(A053141(n)*sqrt(2)+1/2) (2) > > There is no explicit formula for A053141 in the OEIS. There is; you just overlooked it. In 2002, Bruce Corrigan gave a(n) = (1/8)*(-4 + (2 + sqrt(2))*(3 + 2*sqrt(2))^n + (2 - sqrt(2))*(3 - \ 2*sqrt(2))^n). BTW, OEIS seems to be down now and so Rainer kindly supplied me, via email, a link to A053141: . David > But as A097075 does have such a formula, we \only\ have > to /invert the floor-function/ to provide us with an > explicit formula for A053141. > > Well I know that this is nonsense, in a way. But at > least we can hope for a not so bad approximation due to > equation (2). And this is indeed so. > > We know from (2): > A097075(2n+1)-1/2 <= A053141(n)*sqrt(2) < A097075(2n+1)+1/2 > and thus we can define two functions > LowInv(n) = (A097075(2n+1)-1/2)/sqrt(2) > and > HigInv(n) = (A097075(2n+1)+1/2)/sqrt(2). > > A simple-and-stupid approach is to take the arithmetic mean: > > > and see how well we did: > > : ----------------------------------------------- > : 0 0 0 > : 1 2 2 > : 2 14 14 > : 3 84 84 > : 4 492 492 > : 5 2870 2870 > : 6 16730 16730 > : 7 97512 97512 > : 8 568344 568344 > : 9 3312554 3312555 > : 10 19306982 19306989 > : 11 112529340 112529380 > : 12 655869060 655869360 > : 13 3822685022 3822686800 > : 14 22280241074 22280252000 > : 15 129858761424 129858820000 > : > :______________________________________________________________ > : > : Table 1: Comparison of exact but recursive sequence A053141 > : > > I should have stopped that mad enterprise by now but I was > still curious and asked Maple to compute the quotient > > 1., 1., 1., 1., 1., 1., 1., 1., > 1.0000003, > 1.0000004, 1.0000004, > 1.0000005, 1.0000005, 1.0000005, 1.0000005, > 1.0000006, 1.0000005, 1.0000006, 1.0000007, 1.0000007, 1.0000006, > 1.0000008, 1.0000007, 1.0000007, 1.0000008, 1.0000008, 1.0000009, > 1.0000009, 1.0000009, 1.0000010, 1.0000010, 1.0000010, 1.0000011, > 1.0000011, 1.0000011, 1.0000012, 1.0000012, 1.0000012, 1.0000013, > 1.0000012, 1.0000013, 1.0000014, 1.0000014, 1.0000013, 1.0000014, > 1.0000014, 1.0000015, 1.0000016, 1.0000015, 1.0000016, 1.0000017, > 1.0000017, 1.0000017, 1.0000018, 1.0000018, 1.0000018, 1.0000018, > 1.0000018, 1.0000018, 1.0000019 > > There is a clear tendency. Are there any tricks to get closer to > an explicit formula for A053141? I admit I tried the geometric mean > instead of the arithmetic mean - but only with a neglectable effect > to the better. > > Rainer Rosenthal > r.rosenthal@web.de === Subject: Re: Inverting the floor function (sort of) > >>A097075(2n+1) = floor(A053141(n)*sqrt(2)+1/2) (2) >>There is no explicit formula for A053141 in the OEIS. > There is; you just overlooked it. Oh well, thank you very much. It would have been nice to get an explicit formula with that weird inversion, though ;-) You may be interested in the way I found the connection (2). So I would like to demonstrate it. It's part of our fiddling around with two grids G and H', where G is the standard Z x Z = { (i,j) | i,j integers}. The other grid H' is rotated to the right by an angle of 45 and then \ shifted in such a way that the mesh with corners H'(0,0) and H'(1,1) is centered at G(0,0): : | S : G(0,3) + . . : | . . : | P R : | . . . : + . . . : | . Q : | . . : | . . : + . . +G(5,1) : .H'(0,1) . : H'(0,0) . | . . : \\ . | . . : -------<----+---->.+------+------+------+------+------+- : . | . \\ G(3,0) G(5,0) : . | . H'(1,1) : .H'(1,0) : | :______________________________________________________________ : : Figure 1: Two grids G and H' rotated and shifted : The fun thing is to observe the many patterns that show up an never ever are the same except for the point symmetry in this large pattern. When I tried to locate an H' square with corner P = H'(0,k) in such a way that the center of side SR (Figure 1) is very close to the center of as quare mesh of grid G, I happened to meet these very sequences A053141 and A097075. To explain this, let me zoom into figure 1: : | : | : | D. C : | . . : | . . : | P . : | . . A . B : | . . . : | . Q : | . . : | . . : | : | : | : | : | :______________________________________________________________ : : Figure 2: Two special meshs of G and H'. The center of : G-mesh ABCD is midpoint of side PQ of H'-mesh PQRS. : See figure 1: P = H'(0,k) : Point A can be written as A = G(m,m) : The explanation of figure 2 introduces integer coordinates k and m. It turns out that for better and better approximations of figure 2 you need to select m = A053141(n) and for n=1,2,3,... (3) k = A097075(2n+1) Calculating distances I found out that relation (2) must hold. ***** Oops, now I am not that sure anymore about relation (2). I will have to recalculate and maybe I am able to get rid of the floor- function then ... Anyhow - it has taken me quite a long time to ASCIIfy my grids. Now it's time to post my pictures :-) Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Inverting the floor function (sort of) > > : | S > : G(0,3) + . . > : | . . > : | P R > : | . . . > : + . . . > : | . Q > : | . . > : | . . > : + . . +G(5,1) > : .H'(0,1) . > : H'(0,0) . | . . > : \\ . | . . > : -------<----+---->.+------+------+------+------+------+- > : . | . \\ G(3,0) G(5,0) > : . | . H'(1,1) > : .H'(1,0) > : | > :______________________________________________________________ > : > : Figure 1: Two grids G and H' rotated and shifted > : That's OK. The midpoint of SR is marked . The next (zoom) \ figure contained a serious mistake, as I talked about being the center of \ PQ. Please read that figure as follows: > To explain this, let me zoom into figure 1: > > : | > : | > : | D.----- C > : | .| . | > : | . | . | > : | P | .| > : | . . A ---.- B > : | . . . > : | . Q > : | . . > : | . . > : | > : | > : | > : | > : | > :______________________________________________________________ > : > : Figure 2': Two special meshs of G and H'. The center of > : G-mesh ABCD is midpoint of side RS of H'-mesh PQRS. > : See figure 1: P = H'(0,k) > : Point A can be written as A = G(m,m) > : > Anyhow - it has taken me quite a long time to ASCIIfy my grids. Now > it's time to post my pictures :-) Sorry for the mistakes. I am quite sure you like this geometric thing, because you must have done this all the time in your amazing work with packing squares and the like. Rainer ' 'o ' . o. o. _________________________o___o___o__________________________________ Rainer Rosenthal ' o' o' r.rosenthal@web.de . .o . === Subject: Re: Inverting the floor function (sort of) Am 20.08.2007 09:27 schrieb Rainer Rosenthal: > > Sorry for the mistakes. I am quite sure you like this geometric thing, > because you must have done this all the time in your amazing work > with packing squares and the like. :-))) Go on, dear! -- --- Gottfried Helms, Kassel === Subject: Re: Cartan Forms in General Relativity cut and paste... cut and paste.. cut and paste.. hahaha === Subject: Re: Explaining my latest factoring research You're always explaining, but never successfully, and never factoring any damn thing. === Subject: Re: JSH: Explaining my latest factoring research >Here's a post to go over the latest with my factoring research and >explain a key breakthrough, as well as address the issue of >demonstration. > >First some background, since August of last year I have focused on >very simple congruences in an attempt to develop an integer >factorization method, where now I present as > >y^2 = x^2 mod T > >which is of course the difference of squares with T the target prime, Typo, I suspect that you meant \T the target composite\. :) >and the one addition I make is introducing a variable I call k, where > >k = 2x mod T > >which is the second congruence. > >The breakthrough came when I focused on a hypothetical T that is made >up of two primes: > >T = p_1*p_2 > >where the p's are differing primes, As you note elsewhere, this prevents your method being a solution to THE factoring problem since it can now only factor numbers of a certain form. It cannot factor 120 for instance. > >2x = c_1*p_1 + c_2*p_2 > >from which I have c_1 = (2x)*(p_1)^{-1} mod p_2, so you can pick ANY >non-zero x coprime to p_1 and p_2, That is a strange way to express it, since at this stage we do not know what p_1 and p_2 are as we have not yet factored T. Would it not be better to say something like: \Pick a non-zero x. If x is not coprime to T, then the problem is solved. 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If u r eagerly seeking for Job Guidance, than u have found the perfect place on web. http://UltraJob.blogspot.com/ Comments r welcomed. === Subject: Re: Math Trek: Calculating the Word Spurt >> People who insist that \baby talk\ is detrimental >> might just be speaking /a priori/ and without >> evidence. It's at least plausible that basics are >> laid down faster by using simpler constructions. >> You don't start by feeding a baby on steak, after >> all. > > Yes, but you do feed the baby real food. Yes, sorry, my fault, I should have known not to use an analogy in this group. In other places they are enlightening. Here they are merely distracting, attacked as if they were isomorphic, and not illustrative. The main point is that research has shown that \baby talk\ seems to be beneficial. The suggestion is that it's because it is better suited to the current state of development. The proposed theories suggest that using \adult talk\ to babies does not provide the optimal stimulus for learning sound formation, turn taking, and phoneme recognition. As with most things, extremes and excesses are generally bad, but people who denounce \baby talk\ as inane without having any reason other than to say it's baby talk, may be missing the point. YMMV, do what you like with the information provided. Reject it if you will, argue with the researchers if you care to, or spout unresearched and ill-informed opinions here. Or accept that perhaps \baby talk\ has benefits that the researchers in the field suggest. === Subject: Re: Math Trek: Calculating the Word Spurt >>> People who insist that \baby talk\ is detrimental >>> might just be speaking /a priori/ and without >>> evidence. It's at least plausible that basics are >>> laid down faster by using simpler constructions. >>> You don't start by feeding a baby on steak, after >>> all. >> >> Yes, but you do feed the baby real food. > >Yes, sorry, my fault, I should have known not >to use an analogy in this group. In other >places they are enlightening. Here they are >merely distracting, attacked as if they were >isomorphic, and not illustrative. Chill, please. >The main point is that research has shown that >\baby talk\ seems to be beneficial. The suggestion >is that it's because it is better suited to the >current state of development. The proposed >theories suggest that using \adult talk\ to >babies does not provide the optimal stimulus >for learning sound formation, turn taking, and >phoneme recognition. Define baby talk. Is it the words or the intonation? The Maybe, it is the way that the words are said. What if one spoke real words in baby talk intonation? Would that be any better than the nonsense that some use? That is something I would like to hear about. >As with most things, extremes and excesses are >generally bad, but people who denounce \baby talk\ >as inane without having any reason other than to >say it's baby talk, may be missing the point. Or might not be. Is the words or the intonation? >YMMV, do what you like with the information >provided. Reject it if you will, argue with >the researchers if you care to, or spout >unresearched and ill-informed opinions here. Or criticise possible omissions in the research and suggest something to cover the point. > >Or accept that perhaps \baby talk\ has benefits >that the researchers in the field suggest. Accept a maybe? Gene Wirchenko Computerese Irregular Verb Conjugation: I have preferences. You have biases. He/She has prejudices. === Subject: Re: Powers and logic On Sat, 18 Aug 2007 16:36:47 -0400, quasi >To dramatize the issue, here's a challenge problem ... > >Find the sum of the digits (base 10) of 9^(9^(9^9)). I spent a fruitless half hour banging my head against this last night. Please reassure me that there is a solution. Then I'll bang my head against the thing some more, because it looks like a very nice problem. -- Angus Rodgers Contains mild peril === Subject: Re: Powers and logic On Sun, 19 Aug 2007 15:07:27 +0100, Angus Rodgers >On Sat, 18 Aug 2007 16:36:47 -0400, quasi > >>To dramatize the issue, here's a challenge problem ... >> >>Find the sum of the digits (base 10) of 9^(9^(9^9)). > >I spent a fruitless half hour banging my head against >this last night. Please reassure me that there is a >solution. Then I'll bang my head against the thing >some more, because it looks like a very nice problem. I apologize -- it's not a nice problem -- it wasn't intended to be. I have no idea how to solve it. While the answer is provably within range of physical storage (it's easy to show that it's less than 10 gigabytes), I know of no tractable way of finding it. I'm not saying that there is no tractable method -- just that if there is, I don't see it. Let me propose a seemingly easier challenge ... Let x = 9^(9^(9^9)). What is the leading digit (base 10) of x? Even for this problem, while it might be doable, I have no idea how to do it. quasi === Subject: Re: Powers and logic >On Sun, 19 Aug 2007 15:07:27 +0100, Angus Rodgers >>On Sat, 18 Aug 2007 16:36:47 -0400, quasi >>>To dramatize the issue, here's a challenge problem ... >>>Find the sum of the digits (base 10) of 9^(9^(9^9)). >>I spent a fruitless half hour banging my head against >>this last night. Please reassure me that there is a >>solution. Then I'll bang my head against the thing >>some more, because it looks like a very nice problem. >I apologize -- it's not a nice problem -- it wasn't intended to be. >I have no idea how to solve it. >While the answer is provably within range of physical storage (it's >easy to show that it's less than 10 gigabytes), I know of no tractable >way of finding it. I'm not saying that there is no tractable method -- >just that if there is, I don't see it. No, 9^(9^9) takes about 30 megabytes of storage. Raising 9 to such a power is beyond what most believe to be the capabilities of the universe. >Let me propose a seemingly easier challenge ... >Let x = 9^(9^(9^9)). >What is the leading digit (base 10) of x? >Even for this problem, while it might be doable, I have no idea how to >do it. It is easy to give an algorithm which could be calculated in a terabyte or so cycle times. Hint: how would you calculate the leading digit of 9^(9^9) rather easily on a computer with sufficient accuracy, well within reach. Unless one happens to be unlucky, the usual \double precision\ in which most floating arithmetic is done is quite adequate. >quasi -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Powers and logic On 19 Aug 2007 22:35:21 -0400, hrubin@odds.stat.purdue.edu (Herman >>On Sun, 19 Aug 2007 15:07:27 +0100, Angus Rodgers > >>>On Sat, 18 Aug 2007 16:36:47 -0400, quasi > >>>>To dramatize the issue, here's a challenge problem ... > >>>>Find the sum of the digits (base 10) of 9^(9^(9^9)). > >>>I spent a fruitless half hour banging my head against >>>this last night. Please reassure me that there is a >>>solution. Then I'll bang my head against the thing >>>some more, because it looks like a very nice problem. > >>I apologize -- it's not a nice problem -- it wasn't intended to be. > >>I have no idea how to solve it. > >>While the answer is provably within range of physical storage (it's >>easy to show that it's less than 10 gigabytes), I know of no tractable >>way of finding it. I'm not saying that there is no tractable method -- >>just that if there is, I don't see it. > >No, 9^(9^9) takes about 30 megabytes of storage. Raising >9 to such a power is beyond what most believe to be the >capabilities of the universe. You didn't read the problem The problem was not to represent 9^(9^(9^9)) base 10. I was well aware that such a result would be out of range. The problem was to find the sum of its digits. Unless I made a mistake, the sum of the decimal digits of 9^(9^(9^9)) is _not_ out of range -- I estimated that the space requirement to be less than 10 gigabytes. >>Let me propose a seemingly easier challenge ... > >>Let x = 9^(9^(9^9)). > >>What is the leading digit (base 10) of x? > >>Even for this problem, while it might be doable, I have no idea how to >>do it. > >It is easy to give an algorithm which could be calculated >in a terabyte or so cycle times. Well, a terabyte of cycles is what -- half an hour of computing time? Ok, out of curiosity, what's the answer? I'm not suggesting that you do it -- the question is for anyone who is also curious and who feels like fooling around with the required algorithms (and who was a terabyte of cycles to spare). >Hint: how would you calculate the leading digit of 9^(9^9) rather >easily on a computer with sufficient accuracy, well within reach. Since 9^(9^9) can be multiplied out in full, with successive squaring to reduce the number of multiplications, and with the result stored in a file, one can always find the leading digit by inspection. However, such an almost brute force method can't possibly work for 9^(9^(9^9)) since the space requirement for the full representation is out of bounds. Perhaps there's a way to truncate the partial products to some kind of floating point representation, keeping both upper and lower bounds, but as far as I can see, such truncation would lose too much accuracy to be useful in finding the leading digit. I'll have to think about it. >Unless one happens to be unlucky, the usual \double >precision\ in which most floating arithmetic is done >is quite adequate. I'm not saying I don't believe it -- but I do find that claim very surprising. quasi === Subject: Re: Powers and logic >On Sun, 19 Aug 2007 15:07:27 +0100, Angus Rodgers > >>On Sat, 18 Aug 2007 16:36:47 -0400, quasi >> >>>To dramatize the issue, here's a challenge problem ... >>> >>>Find the sum of the digits (base 10) of 9^(9^(9^9)). >> >>I spent a fruitless half hour banging my head against >>this last night. Please reassure me that there is a >>solution. Then I'll bang my head against the thing >>some more, because it looks like a very nice problem. > >I apologize -- it's not a nice problem -- it wasn't intended to be. > >I have no idea how to solve it. > >While the answer is provably within range of physical storage (it's >easy to show that it's less than 10 gigabytes), I know of no tractable >way of finding it. I'm not saying that there is no tractable method -- >just that if there is, I don't see it. > >Let me propose a seemingly easier challenge ... > >Let x = 9^(9^(9^9)). > >What is the leading digit (base 10) of x? > >Even for this problem, while it might be doable, I have no idea how to >do it. Of course, a symbolic answer can be given by the expression floor( x / (10^(floor(log[10](x))) ) But that's not acceptable. The required answer must be one of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9. quasi === Subject: Re: Powers and logic > On Sat, 18 Aug 2007 16:36:47 -0400, quasi > > >To dramatize the issue, here's a challenge problem > ... > > > >Find the sum of the digits (base 10) of 9^(9^(9^9)). > > I spent a fruitless half hour banging my head against > > this last night. Please reassure me that there is a > solution. Then I'll bang my head against the thing > some more, because it looks like a very nice problem. > -- > Angus Rodgers > Contains mild peril its a very big number !! if most digits are not zero then digitsum(A) is quite large too !! what reasons do we have to assume that number A even fits on this mathforum \ ?? and if A is quite big , then probably the digits cannot be expressed in \ closed form ... ( complexity theory , compression theory , statistics ) tommy1729 === Subject: Re: Powers and logic On Sun, 19 Aug 2007 17:57:16 EDT, tommy1729 >> On Sat, 18 Aug 2007 16:36:47 -0400, quasi >> >> >To dramatize the issue, here's a challenge problem >> ... >> > >> >Find the sum of the digits (base 10) of 9^(9^(9^9)). >> >> I spent a fruitless half hour banging my head against >> >> this last night. Please reassure me that there is a >> solution. Then I'll bang my head against the thing >> some more, because it looks like a very nice problem. >> -- >> Angus Rodgers >> Contains mild peril > >its a very big number !! > >if most digits are not zero > >then digitsum(A) is quite large too !! > >what reasons do we have to assume that number A even fits on this mathforum \ ?? > >and if A is quite big , then probably the digits cannot be expressed in \ closed form ... > >( complexity theory , compression theory , statistics ) Right. The number itself, if written out (base 10), probably doesn't fit on any medium (not even the whole universe). However the sum of the digits is provably less than 10 gigabytes. Although that would surely exceed the maximum allowable post on mathforum, it would still fit on a typical hard drive. But just because it would fit on a disk doesn't mean there's any practical way of finding it. quasi === Subject: look this page about mathematic http://www.pennergame.de/ref.php?uid=5572 === Subject: Re: JSH: Solving the factoring problem Entire approach from yesterday is flawed. Guess I was reaching. I'll be deleting just about everything I posted yesterday from Google. James Harris > After the explanation of my current factoring research that I just > posted I realized that wrapped up in this latest analysis is solution > to the factoring problem. > > It's fairly easy but as usual I'll post what I think it is to see if I > made any mistakes: > > Starting as usual with the key congruences: > > y^2 = x^2 mod T > > and > > k = 2x mod T > > I note that a solution must exist for those equations for any non-zero > x coprime to T, as I worked out in detail in my prior post. > > Using them I easily get > > (x+k)^2 = y^2 + 2k^2 + n*T > > where n is some non-zero integer, which I've also explained and now I > can address the value of n--and solve the factoring problem. > > As we know a solution to the difference of square that non-trivially > factors T must exist, so assume you have that x, and some arbitrary n, > can it be reached? > > The answer is yes if it is even, because > > k = 2x mod T > > explicitly is k = 2x + m*T, where m is some non-zero integer, so I can > always use > > (k+/-a*T) = 2x + ( m+/-a)*T > > and letting k* = k+/-a*T, I then have > > (x+k*)^2 = y^2 + 2k*^2 + n*T > > so if k=2, I can add or subtract as needed to reach ANY even n, so n > needs to be even. > > Notice that knowing how much is being added or subtracted from k to > get k* is unimportant as you're looking at everything mod T, so with > even n, and k=2, the size limit discussed in my prior post is the only > block and over that size limit you are 100% guaranteed to factor T non- > trivially, which solves the factoring problem. > > Note: You are moving in multiples of k*T, as remember k = 2x mod T is > first multiplied by k and then added back to the congruence of > squares. > > The factoring problem is done. > > James Harris === Subject: JSH: Musing on failure Yesterday I started with a long and detailed post talking about the latest insight with my factoring research, but after that post I had several more claiming that I had now solved the factoring problem which all were flawed. So there was the one detailed and correct post and then a leap to believing I was done, instead of in a position where a lot more work might be ahead, so it was just a fantasy reaching thing. I WISH I could just get a stupendously simple solution that could end all the debate and finally corner mathematicians and show them for what I know them to be, but its mathematics. You have to prove. Oh, so does it really worry me about what people might think? No, because if I DO figure out the factoring problem none of this will matter and everything will be over in a blink, as in overnight. Immediately everything would change and my past failures would mean nothing. That is the burden on the math world, hope I fail but know that if I do succeed you lose everything--overnight. So feel secure with that if you want, as I'm still thinking about this research and did get a key insight recently. It boggles my mind that I have to find something practical to break the math world because it has left mathematical proof behind to live in its own world of pretend. And factoring is it. I have my problems with wishing but I admit them. You people do NOTHING right in huge areas but hold on to the fantasy that you are correct for the social and ego benefits. I can admit failure, even huge failure. You are not strong enough to live in the real world. So I have to find a mathematical solution that would make obsolete the system that underpins the world's Internet security to prove beyond anyone's doubt that the current math world is full of cons and liars. Gauss and Newton in comparison had it easy. But they did live in a simpler world. James Harris === Subject: Re: JSH: Musing on failure > Yesterday I started with a long and detailed post talking about the > latest insight with my factoring research, but after that post I had > several more claiming that I had now solved the factoring problem > which all were flawed. > > So there was the one detailed and correct post and then a leap to > believing I was done, instead of in a position where a lot more work > might be ahead, so it was just a fantasy reaching thing. > > I WISH I could just get a stupendously simple solution that could end > all the debate and finally corner mathematicians and show them for > what I know them to be, but its mathematics. You have to prove. > > Oh, so does it really worry me about what people might think? > > No, because if I DO figure out the factoring problem none of this will > matter and everything will be over in a blink, as in overnight. > > Immediately everything would change and my past failures would mean > nothing. > > That is the burden on the math world, hope I fail but know that if I > do succeed you lose everything--overnight. > > So feel secure with that if you want, as I'm still thinking about this > research and did get a key insight recently. > > It boggles my mind that I have to find something practical to break > the math world because it has left mathematical proof behind to live > in its own world of pretend. And factoring is it. > > I have my problems with wishing but I admit them. You people do > NOTHING right in huge areas but hold on to the fantasy that you are > correct for the social and ego benefits. > > I can admit failure, even huge failure. > > You are not strong enough to live in the real world. > > So I have to find a mathematical solution that would make obsolete the > system that underpins the world's Internet security to prove beyond > anyone's doubt that the current math world is full of cons and liars. > > Gauss and Newton in comparison had it easy. But they did live in a > simpler world. > > James Harris Well you can do whatever you want. I'm just never going to hold my breath, considering that the failures just keep coming. Does this mean you concede that your factoring algorithm you just posted is bunk? === Subject: Re: JSH: Musing on failure >Yesterday I started with a long and detailed post talking about the >latest insight with my factoring research, but after that post I had >several more claiming that I had now solved the factoring problem >which all were flawed. You do sometimes read too much into what you have done. You need to check and double check your work before making claims. The bigger the claim the more checking - and you do tend to make some very big claims. Alternatively just post it as unchecked and drop the big claims. Making incorrect claims only annoys people. > >So there was the one detailed and correct post and then a leap to >believing I was done, instead of in a position where a lot more work >might be ahead, so it was just a fantasy reaching thing. > >I WISH I could just get a stupendously simple solution that could end >all the debate and finally corner mathematicians and show them for >what I know them to be, but its mathematics. What you wish is irrelevant. Reality is reality irrespective of your wishes. A lot of very good mathematicians: Gauss, Euler, Fermat have looked at the factoring problem and did not find a solution. If there is a solution out there then it is probably going to be difficult to find. If is was easy then it would already have been found. > >Oh, so does it really worry me about what people might think? > >No, because if I DO figure out the factoring problem none of this will >matter and everything will be over in a blink, as in overnight. You are getting back into what you wish the world to be like. Better to stick with the world as it is. You will be happier if you do not expect the world to be diferent to what it is. rossum > > >James Harris === Subject: Re: JSH: Musing on failure > You are getting back into what you wish the world to be like. Better > to stick with the world as it is. You will be happier if you do not > expect the world to be diferent to what it is. > Comment, and this has nothing to do with James or the validity of his stuff, it's just a response to what you said, in general: I think though that one should try to _change_ the world as much as possible, for the better. > rossum > === Subject: Re: JSH: Musing on failure > >> You are getting back into what you wish the world to be like. Better >> to stick with the world as it is. You will be happier if you do not >> expect the world to be diferent to what it is. >> > >Comment, and this has nothing to do with James or the validity >of his stuff, it's just a response to what you said, in general: > >I think though that one should try to _change_ the world as much >as possible, for the better. \Lord grant me the strength to change what can be changed, The patience to suffer what cannot be changed, And above all the wisdom to know the difference.\ James is not a Stoic philosopher, hence I doubt that he has the wisdom to know the difference. Given that assumption, it is probably better for James if he leaves well alone. rossum === Subject: Re: JSH: Musing on failure > Yesterday I started with a long and detailed post talking about the > latest insight with my factoring research, but after that post I had > several more claiming that I had now solved the factoring problem > which all were flawed. > And somehow this comes as a surprise? To whom? > So there was the one detailed and correct post and then a leap to > believing I was done, instead of in a position where a lot more work > might be ahead, so it was just a fantasy reaching thing. > And the sun came up this morning. Coincidence? Perhaps not. > I WISH I could just get a stupendously simple solution that could end > all the debate and finally corner mathematicians and show them for > what I know them to be, but its mathematics. You have to prove. > Therein lies the root of your problem, in my estimation. It seems that you are not motivated by anything more profound than the desire to prove that mathematicians are bad guys. You came into this quest a decade ago with this as your preordained conclusion, and have never done anything original that was not dedicated to that singular goal. All your claims of having a position in history comparable to Gauss, all the assertions that you're only after the truth, all those \you can't handle the truth\ gauntlets you throw down: they all come down to your incessant gorging on the sweets of self-congratulation. You want all the glory without having any of the guts; you want to play the blues without having paid your dues; in short, you want to reap the rewards without having put in any of the requisite effort. While you imagine that your decade of fruitless labor constitutes \the requisite effort\, it doesn't matter how hard you *think* you're working, it only matters that you actually get somewhere. As has been pointed out countless times, your near-total ignorance has made it less than likely that you'll achieve anything of significance. > Oh, so does it really worry me about what people might think? > > No, because if I DO figure out the factoring problem none of this > will matter and everything will be over in a blink, as in overnight. > IF. As in IF pigs could fly. As if. > Immediately everything would change and my past failures would mean > nothing. > Umm, IF pigs could fly, you meant to say. Of course, those past failures would still be sitting there for all to see. > That is the burden on the math world, hope I fail but know that if I > do succeed you lose everything--overnight. > I have neither hopes nor expectations on this account. Nothing in mathematics will be affected in the slightest amount by anything that you accomplish. That much is certain. > So feel secure with that if you want, as I'm still thinking about > this research and did get a key insight recently. > Ooooh. A key insight. > It boggles my mind that I have to find something practical to break > the math world because it has left mathematical proof behind to live > in its own world of pretend. And factoring is it. > Yes. You have to find something. Why it is that you \have to break the math world\, we're back to your obsession with fighting that same old devil. It's pretty amusing that every single time you've claimed to \have the goods\, that you've been shown to be as wrong as wrong can be; yet you continue with the claim that mathematicians are living in some \world of pretend\. > I have my problems with wishing but I admit them. You people do > NOTHING right in huge areas but hold on to the fantasy that you are > correct for the social and ego benefits. > Huge areas. Name one. Can there be a bigger fool than the one who fools himself? > I can admit failure, even huge failure. > Yes, and you retract those admissions as soon as anyone turns his back. Witness the FLT fiasco, fool-boy. > You are not strong enough to live in the real world. > Again, YOU are the one with the bogus claims, the 100% track record of failure over the span of a decade, the never-ending rant about how the world is really in for it because you haven't been awarded the big prize, and a wont to ascribe every event in terms that suggest that you are in control. Yet it is the rest of the world that is \not strong enough to live in the real world\. > So I have to find a mathematical solution that would make obsolete > the system that underpins the world's Internet security to prove > beyond anyone's doubt that the current math world is full of cons and > liars. > There's a term for people who have their minds made up before they have gotten any actual experience. I'm imagining that it applies to you. > Gauss and Newton in comparison had it easy. But they did live in a > simpler world. > Well, in particular, they weren't in the practice of spending decades attempting to show something that was patently untrue. > > James Harris > It's nice to be able to play \Spot the Loony\ on sci.math Dale === Subject: Re: JSH: Musing on failure On Sun, 19 Aug 2007 14:34:20 -0700, \W. Dale Hall\ >> >> So feel secure with that if you want, as I'm still thinking about >> this research and did get a key insight recently. [JSH] >> > Ooooh. A key insight. > You know, the hammer will arrive soon! F. === Subject: Re: Musing on failure > Yesterday I started with a long and detailed post talking about the > latest insight with my factoring research, but after that post I had > several more claiming that I had now solved the factoring problem > which all were flawed. > you owe sci.math an apology. === Subject: Re: JSH: Musing on failure > Yesterday I started with a long and detailed post talking about the > latest insight with my factoring research, but after that post I had > several more claiming that I had now solved the factoring problem > which all were flawed. > > So there was the one detailed and correct post and then a leap to > believing I was done, instead of in a position where a lot more work > might be ahead, so it was just a fantasy reaching thing. One cannot help but note you would have saved yourself a lot of embarrassment this weekend if your work was presented with a bit more humility. You had an idea that might lead to a faster way to factor; nothing wrong with that. You wanted to brainstorm that idea on the Math groups. Again, there is nothing wrong with that. After all, that is one of the major purposes of these groups. The problem is *how* you presented the idea. The right way to brainstorm a new, untested idea would be to write a post something like this: \Folks, I had an idea that may lead to a faster algorithm for factoring. Here is the theory ... Here is the algorithm ... Here is an example ... Fell free to comment or critique this idea\. If your posts had that tone, you would have gotten much friendlier replies, and there would be nothing so embarrassing that you would feel as obligation to delete them from the Google archives. history with the Wright brothers first flight and the Moon landing, as something so solidly proved that it is demeaning to ask for an example, as being so obviously right that the only people who could deny it your imagined cabal of lying Mathematicians who conspire to maintain the status quo. With this kind of over-the-top build-up, it is no wonder your failure has caused you so much angst. So why was there such excessive hubris in your your initial posts? This problem, like many of your problems, can be traced back to NPD. The longer you leave this condition untreated, the more pain it will cause you. > I WISH I could just get a stupendously simple solution that could end > all the debate and finally corner mathematicians and show them for > what I know them to be, but its mathematics. You have to prove. But doesn't the experience of this weekend cast any doubt on what you believe about mathematicians? In the past, you have attributed the failure of mathematicians to find a faster way of factoring on their vested interests. Now that you have discovered finding a quicker factoring method is harder than you originally thought, isn't it possible that mathematicians haven't found a faster way to factor is that it is a legitimately hard problem? > Oh, so does it really worry me about what people might think? Be honest with yourself: of course you worry about people might thing, as you should. > No, because if I DO figure out the factoring problem none of this will > matter and everything will be over in a blink, as in overnight. Apparently, you made no contingency plans in case you DO NOT figure out the factoring problem, hence the predicament you are in now. Again, you have become the victim of NPD thinking. > Immediately everything would change and my past failures would mean > nothing. > > That is the burden on the math world, hope I fail but know that if I > do succeed you lose everything--overnight. > > So feel secure with that if you want, as I'm still thinking about this > research and did get a key insight recently. > > It boggles my mind that I have to find something practical to break > the math world because it has left mathematical proof behind to live > in its own world of pretend. And factoring is it. > > I have my problems with wishing but I admit them. You people do > NOTHING right in huge areas but hold on to the fantasy that you are > correct for the social and ego benefits. > > I can admit failure, even huge failure. > > You are not strong enough to live in the real world. A bit of projection, perhaps? > So I have to find a mathematical solution that would make obsolete the > system that underpins the world's Internet security to prove beyond > anyone's doubt that the current math world is full of cons and liars. > > Gauss and Newton in comparison had it easy. But they did live in a > simpler world. > > > James Harris > -- \All things extant in this world, Gods of Heaven, gods of Earth, Let everything be as it should be; Thus shall it be!\ - Magical chant from \Magical Shopping Arcade Abenobashi\ \Drizzle, Drazzle, Drozzle, Drome, Time for this one to come home!\ - Mr. Wizard from \Tooter Turtle\ === Subject: Re: JSH: Musing on failure You touch yourself at night over geese. > Yesterday I started with a long and detailed post talking about the > latest insight with my factoring research, but after that post I had > several more claiming that I had now solved the factoring problem > which all were flawed. > > So there was the one detailed and correct post and then a leap to > believing I was done, instead of in a position where a lot more work > might be ahead, so it was just a fantasy reaching thing. > > I WISH I could just get a stupendously simple solution that could end > all the debate and finally corner mathematicians and show them for > what I know them to be, but its mathematics. You have to prove. > > Oh, so does it really worry me about what people might think? > > No, because if I DO figure out the factoring problem none of this will > matter and everything will be over in a blink, as in overnight. > > Immediately everything would change and my past failures would mean > nothing. > > That is the burden on the math world, hope I fail but know that if I > do succeed you lose everything--overnight. > > So feel secure with that if you want, as I'm still thinking about this > research and did get a key insight recently. > > It boggles my mind that I have to find something practical to break > the math world because it has left mathematical proof behind to live > in its own world of pretend. And factoring is it. > > I have my problems with wishing but I admit them. You people do > NOTHING right in huge areas but hold on to the fantasy that you are > correct for the social and ego benefits. > > I can admit failure, even huge failure. > > You are not strong enough to live in the real world. > > So I have to find a mathematical solution that would make obsolete the > system that underpins the world's Internet security to prove beyond > anyone's doubt that the current math world is full of cons and liars. > > Gauss and Newton in comparison had it easy. But they did live in a > simpler world. > > James Harris === Subject: Re: JSH: Musing on failure Listen bro You cant come to a forum like this,.. talking down to everyone and then expecting us to let you down easy when your wrong.. Arrogance is a luxury of the intelligent === Subject: Re: integral zeta > the integral of x > > it is x^2 / 2 + C > > and a primitive integral means C = 0 See: http://mathforum.org/kb/message.jspa?messageID=3881287&tstart=0 Fernando. === Subject: Re: integral zeta > > > the integral of x > > > > it is x^2 / 2 + C > > > > and a primitive integral means C = 0 > > See: > > http://mathforum.org/kb/message.jspa?messageID=3881287 > &tstart=0 > > Fernando. but i fear David C.Ullrich is a hopeless case ... tommy1729 === Subject: Re: integral zeta On Sun, 19 Aug 2007 18:04:07 EDT, tommy1729 > >> >> > the integral of x >> > >> > it is x^2 / 2 + C >> > >> > and a primitive integral means C = 0 >> >> See: >> >> http://mathforum.org/kb/message.jspa?messageID=3881287 >> &tstart=0 >> >> Fernando. > > >but i fear David C.Ullrich is a hopeless case ... Answer the question: What is the primitive integral of 2 sin(t) cos(t)? If it sin^2(t), is it -cos^2(t), or something else? And then of course explain exactly how your answer follows from your definition. (The name of a book that contains the definition that you say is in books would also be good.) >tommy1729 ************************ David C. Ullrich === Subject: Re: integral zeta On Sat, 18 Aug 2007 16:18:55 EDT, tommy1729 > > > >> On Thu, 16 Aug 2007 17:51:34 EDT, tommy1729 >> >> >> >David C.Ullrich >> > >> >> On Thu, 16 Aug 2007 07:16:32 EDT, tommy1729 >> >> >> >> >> >> >> tommy1729 a \.8ecrit : >> >> >> >>> tommy1729 a \.8ecrit : >> >> >> >>>> where do the zero's lie of integral zeta ? >> >> >> >>> Not to mention the problem of Riemann >> >> surfaces, >> >> >> >> what do you take as >> >> >> >>> integration constant? >> >> >> >>> >> >> >> >>> >> >> >> >>> Always getinf your questions out of your >> hat, >> >> do >> >> >> >> you? >> >> >> >> >> >> >> >> I though that he got them from somewhere >> >> >> else....... >> >> >> >> >> >> >> > >> >> >> > haha very funny guys >> >> >> > >> >> >> > *rolleyes* >> >> >> > >> >> >> > how about actually answering a question and >> >> doing >> >> >> or explaining math , apart from just making fun >> of >> >> >> the OP ?? >> >> >> > >> >> >> > tommy1729 >> >> >> >> >> >> >> >> >> But this was an answer : what about that >> constant? >> >> >> (and Riemann surfaces) >> >> > >> >> >a primitive integral ... >> >> >> >> If you'd think about what you're saying instead >> >> of trying so hard to be a smartass you'd >> understand >> >> when people explain why the question makes no >> sense. >> >> >> >> we understood that by \integral zeta\ you meant a >> >> primitive integral. >> > >> >yet at the end of this post you seem to have >> forgotten it hmm ? >> > >> > >> > There are several problems with >> >> that. First, the fact that zeta has a pole with >> >> non-vanishing residue at 1 means that it does not >> >> _have_ a primitive integral in various sets; >> >> the integral is \multivalued\, just like the >> >> integral of 1/z. (That's a simple-minded way >> >> of expressing what Denis is getting at when he >> >> mentions Riemann surfaces). >> > >> > >> >you dont completely understand what i mean >> > >> > >> >but i take almost full responsability for it >> > >> >i will specify below * >> > >> > >> >> >> >> Second, even if we ignore that problem, the >> question >> >> still makes no sense, because a given function >> >> has many different integrals. What are the zeroes >> >> of integral (x) on the line? Well, integral(x) >> >> could be x^2 + 1, in which case it has no zeroes. >> >> Or it could be x^2 - 4, which has two zeroes. Etc. >> > >> >like i said you seem to have forgotten here what a >> primitive integral is >> > >> > >> >the primitive integral of x is x^2 / 2 and only >> that. >> >> Well, then I didn't understand what you meant by >> \primitive integral\. >> >> Unfortunately you need to give a _definition_ of the >> concept instead of just an example. What does >> \primitive integral\ mean, exactly? >> >> >not x^2 - 4 or similar >> > >> >so primitive integral of zeta(z) = 0* >> > >> > >> > >> >> >> >> >> >> ************************ >> >> >> >> David C. Ullrich >> > >> > >> > >> >*the primitive integral of zeta the way i meant it : >> >> Ah. So you expected us to know what you _meant_, >> even though you didn't _say_ what you meant, and >> then you complain when people try to explain >> why what you actually _said_ is meaningless. >> >> Think about that next time you feel like >> shooting your mouth off about how stupid >> people are. >> >> >z + sum n=2 -> infinity [elog(n)^-1]*n^(-z) >> >> I imagine that \elog\ just means the logarithm? >> Inventing your own private notation is not the >> best way to communicate. >> >> Just curious, still: What _definition_ of >> \primitive integral\ allows us to conclude >> that _that_ is the primitive integral of zeta? >> >> >and the question is where do the zero's lie of >> > >> > >> >primitiveintegralzeta(z)=0 >> > >> > >> >you might worry about continuation and riemann >> surfaces , >> > >> >but that is not so important as it seems ... >> > >> > >> >as an example of that >> > >> > >> >the zero's of sum n=1 -> infinity n^(-z) >> > >> >lie in the critical line >> > >> >so it also converges there to zero. >> > >> >even if its neighbourhood is not analiticly >> continued; >> > >> >not even finite ! >> > >> >and not caring about riemann surfaces or similar ... >> >> Your comments here are just dripping with ignorance. >> >> You cannot talk about the zeroes of that sum except >> for Re(z) > 1 (where there are none) because >> elsewhere >> the sum simply does not converge. >> >> You _can_ talk about the zeroes of the zeta function >> elsewhere in the plane. The _reason_ you can talk >> about that is that the function defined by that >> sum _has_ a _unique_ analytic continuation to the >> plane >> minus {0}. That is not true for the \primitive >> integral\ of zeta, as defined by the sum above. >> >> This is not an example of why we don't need to >> worry about things, it's an illustration of >> exactly why we _do_. >> >> >that might ammaze you , since it is not often found >> in the books ... >> >> No, the fact that the sum defining zeta(z) for Re(z) >> > 1 >> converges elsewhere is indeed \not often found in >> books\. >> There'a a reason for that - it's not true. >> >> >or it might ammaze others ... >> > >> > >> >but if you consider that any function must have all >> of its zero's or at least an unempty subset of its >> zero's >> > >> >then we still have these zero's despite no analytic >> continuation or even continuation at all ... >> > >> >tommy1729 >> >> >> ************************ >> >> David C. Ullrich > >you are terribly wrong David > >thats the most polite way i can say it ... > >terribly wrong > > >apparantly you dont even know the integral of x > >it is x^2 / 2 + C > >and a primitive integral means C = 0 > >this is very basic math ... > >( and in the books ) What book can I find this definition in? I'm curious, because actually the definition makes no sense. Consider for example the function f(t) = 2 sin(t) cos(t). One integral is sin^2(t). Another integral is -cos^2(t). Those two integrals are different. Neither one contains a \+ C\. So. Which one is the \primitive integral\ of 2 sin(t) cos(t)? I _suspected_ that you were going to \define\ the \primitive integral\ to be \the integral with no +C included\. But that's no definition at all, because whether or not a given integral of f includes a \+C\ depends on _how_ we have decided to _write_ the integral, not on the integral itself. As is apparent elsewhere you seem to be confused about what a function is, as opposed to a formula representing a function. >and you are even more wrong about zeta and your other comments Really? Some explanation of exactly what's wrong about Giggle. >( outside of the books ) > >tommy1729 > >********************************************************** >\ we should be gratefull and feel honored that David C. Ullrich is here on \ sci.math \ David C. Ullrich 's lawyer in a threath that was a super ode to \ David ... ************************ David C. Ullrich === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b89482@news2.lightlink.com> <46b9d8eb@news2.lightlink.com> > > > > > In simple cases, indeed such a definition is possible, such as \ saying > > > > |E| is half of |N|. > > > > Why isn't it twice |N|? Don't you double each member of N to get E > > > Wanted to leave it, but can't let this go ... > > > First, I don't. I take every second element in N to get E. > > Then shouldn't the E you get one way 4 times as large as the E you get > the other way? Why? And which one is the 4 times as large? > > > > > Second, even if I did, are you asserting that multiplying every > > element in a set of natural numbers by a natural number produces a set > > that contains more elements than it did before? > > No. I am asking you why multiplying each member of N by 2 to get E is > any less valid than leaving out the odd members of N to get E. Don't the > two processes give the same resulting set? That's the question ... and the one I'm asking you, since taking them independently seems to imply different conclusions (ie size of N in the first case, size of half N in the other, if size isn't just cardinality). > > Or are you claiming that the resulting sets have different numbers of > elements? I'm trying to figure out what would happen. It's the others who have insisted without any real argumentation that they'd be the same. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > In simple cases, indeed such a definition is possible, such as \ saying > > > > > |E| is half of |N|. > > > > > > Why isn't it twice |N|? Don't you double each member of N to get E > > > > > Wanted to leave it, but can't let this go ... > > > > > First, I don't. I take every second element in N to get E. > > > > Then shouldn't the E you get one way be 4 times as large as the E you \ get > > the other way? > > Why? Why not? And which one is the 4 times as large? Both? > > > > > > > > > > Second, even if I did, are you asserting that multiplying every > > > element in a set of natural numbers by a natural number produces a \ set > > > that contains more elements than it did before? > > > > No. I am asking you why multiplying each member of N by 2 to get E is > > any less valid than leaving out the odd members of N to get E. Don't \ the > > two processes give the same resulting set? > > That's the question ... and the one I'm asking you, since taking them > independently seems to imply different conclusions (ie size of N in > the first case, size of half N in the other, if size isn't just > cardinality). If size IS cardinality, the problem vanishes, as all theses sets are of the same cardinality. > > > > > Or are you claiming that the resulting sets have different numbers of > > elements? > > I'm trying to figure out what would happen. It's the others who have > insisted without any real argumentation that they'd be the same. They have the same well defined cardinality, which is not quite the same as being the same sets. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46ac1376@news2.lightlink.com> <46b54e7f@news2.lightlink.com> <46b7513c@news2.lightlink.com> <46b85a87@news2.lightlink.com> <46b9a9e5@news2.lightlink.com> > > > The 'number' of all naturAls is naturally a non-natural, > > > Why? > > In the von Neumann naturals (which avoids the problem of a natural being > a member of itself), no natural as a set is a member of itself but every > naty=ural as a set is a memer of some other natural, so that the set of > all naturals not being a member of itself, or of any natural, is not a > natural. Um, this is only if you consider a natural number to be a set. But natural numbers could exist in a mathematical view that didn't contain sets at all. In fact, we could define a natural number so that it had all of the same qualities as current natural numbers in a mathematical \world\ that didn't contain sets at all. The only difference would be this notion about it being the size of the set of all naturals ... which wouldn't exist in that world. So why insist that a natural number be defined as a set of all natural numbers that lead up to it? That's the only way that this problem occurs. [snipped] > > > I'm not saying that it CAN'T be done, just that it isn't \naturally a > > non-natural\. It requires some axioms and/or definitions to get it > > that way [grin]. > > Since it requires some axioms and definitions to get any naturals at > all, that means without some axioms and definitions there aren't any > naturals at all. True, but yours requires a specific one or set of ones to claim that a natural number is a set. Or, at least, that the size of a set has to be a set itself ... === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > The 'number' of all naturAls is naturally a non-natural, > > > > > Why? > > > > In the von Neumann naturals (which avoids the problem of a natural \ being > > a member of itself), no natural as a set is a member of itself but \ every > > natural as a set is a member of some other natural, so that the set of > > all naturals not being a member of itself, or of any natural, is not a > > natural. > > Um, this is only if you consider a natural number to be a set. But > natural numbers could exist in a mathematical view that didn't contain > sets at all. In fact, we could define a natural number so that it had > all of the same qualities as current natural numbers in a mathematical > \world\ that didn't contain sets at all. The only difference would be > this notion about it being the size of the set of all naturals ... > which wouldn't exist in that world. So why insist that a natural > number be defined as a set of all natural numbers that lead up to it? > That's the only way that this problem occurs. As I was under the impression that we were considering alternates to cardinality, sets are required, so, using Occam's razor, one may as well let naturals be sets. > [snipped] > > > > > > I'm not saying that it CAN'T be done, just that it isn't \naturally \ a > > > non-natural\. It requires some axioms and/or definitions to get it > > > that way [grin]. > > > > Since it requires some axioms and definitions to get any naturals at > > all, that means without some axioms and definitions there aren't any > > naturals at all. > > True, but yours requires a specific one or set of ones to claim that a > natural number is a set. Or, at least, that the size of a set has to > be a set itself ... There are set theories, e.g., ZF, in which only sets are allowed, so if one is to have any naturals at all in such a system, they must be sets. And even in systems allowing non-set objects, there doesn't seem to be any advantage to insisting that naturals not be sets. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b88ed8@news2.lightlink.com> <46b9dad9$1@news2.lightlink.com> > On Aug 8, 10:30 am, Allan C Cybulskie [snipped] > > that > > for infinite sets a bijection between two sets necessarily means that > > they have the same \size\ (or number of elements, or count of > > elements, to me). > > That's not an AXIOM! It's not even a theorem! It's not even a > definition! So, it's not an axiom, it's not a definition, and it isn't a theorem. Since these are the three things that people have been going on about in this thread ... what is it then? What else do you have in mathematics that that statement can be? But, again, you miss the point. Is that statement accepted by standard Cantorian set theory? If yes, how is it justified? You nitpick over what that statement is, instead of dealing with its justification. > > Beyond that, the most you've ever been able to show about what I > > \don't know\ that matters is about axioms versus definitions. > > No, it started with my telling you that for any given set there is a > specific ordinal that is the cardinality of that set. Moreover, the > distinction between axiom and definition is utterly CRUCIAL to ALL of > this discussion. You simply can't talk meaningfully talk about this > subject without understanding the most basic distinctions between > axioms and definitions! Man oh man, wouldn't you think I were a fool > if I started shooting my mouth off with all kinds of stubborn opinions > about harmonic theory of music while not knowing the differences among > a note, a chord, and a scale?! If you commented that the second note sounded higher than the first, it would not matter if it was even a chord ... your claim would still hold. The same thing holds here; the distinction between axiom and definition only matters if that's what you use to appeal to a statement's justification, and thus simply stating \X is an axiom, Y is a definition\ would be sufficient ... and you have thus far only attacked me for not classifying them properly, not for using the wrong justifications directly. > > > I > > submit that since I'm basically a philosopher in these threads > > But you can't philosophize about that which you don't know anything > about! Sure I can. What I'm doing is watching how you guys justify things. If I considered you representative of all mathematicians, I'd have to conclude that mathematicians justify things on the basis of what they like better. Fortunately, I know better than that. > > > that > > the comment may be valid, but that axioms -- at least how you are > > using them -- BEHAVE like philosophical definitions, and so since I > > care a lot about philosophy and little about mathematics in general > > even if I DID look them all up I'd still use the terms incorrectly. > > But then this is just an issue of terminology, not substance. > > Any excuse for you not to accept the humility that you're like any > other person who walks the earth - to talk about mathematics > meaningfully, you have to learn about it, even if only to learn its > UTTERLY most BASIC considerations. You'd like to think that issues of terminology matter more than matters of substance, but fortunately that is not the case. Whether or not a statement is called an axiom or a definition is of little consequence to its consequences and what they mean. > > > Finally, I have no intention of providing an alternative theory. I'm > > basically doing what might be called \Philosophy of Mathematics\, > > analyzing why mathematics forms the theories and axioms and > > definitions it does and seeing if that's reasonable. Right now, I'm > > still standing by my original theory that the insistence on > > maintaining bijection as saying that two sets have the same size even > > for infinite sets is more for convenience and coolness than anything > > else > > You're no philosopher of mathematics. I am far more of a philosopher, in general, than you are. As for doing it about mathematics, I admit that I am not interested enough in mathematics to be one precisely, but that doesn't mean that I don't find it interesting when I come across it [grin]. [snipped] === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46ac1376@news2.lightlink.com> <46b54e7f@news2.lightlink.com> <46b7513c@news2.lightlink.com> <46b85a87@news2.lightlink.com> <46b9a9e5@news2.lightlink.com> <46b9eb45@news2.lightlink.com> > On Aug 8, 10:15 am, Allan C Cybulskie > > > This would of course explain my utter impatience with nitpicking over > > \definitions\. > > It's not nitpicking! If the definitions are not precise and properly > formed then nonsense results. The degree of precision in such matters > is one of the things that distinguishes mathematics from other > studies. It IS nitpicking, since if a definition is \imprecise\ on a USENET NEWSGROUP it is only relevant IF that imprecision DIRECTLY impacts the debate. Which would mean that first you determine what the discussion is about, and then ask specific questions about how that imprecision impacts the discussion. > > > Or, to sum it up, I want to explore YOUR or THE EXISTING definitions > > and what the consequences are, not invent my own > > Fine, then to explore them, you have to learn about them in a > SYSTEMATIC way just as they are FORMULATED systematically. That means > learning to work in the predicate calculus (some talented or > mathematically steeped people can get by without that step, but for > ordinary people, such as myself, I very strongly recommend first > learning the predicate calculus), then the language, primitives, > axioms, definitions, and proofs of the theorems, and some mathematical > logic helps too. There just is not a shortcut that obviates learning > the material from a class and/or textbook. Now, let me summarize this: I've taken university level mathematics through Abstract Algebra. In order to understand these issues even at a BASE level -- ie what happens when you take an element out of a set or add one to it -- you claim that we must understand the entire system, meaning that in order to understand mathematics you have to basically BE a mathematician (even intuitively). This makes mathematics a specialist field; only those who know it in full detail can do it (at least for some things; addition on normal numbers doesn't seem to fall into that category). There's nothing WRONG with mathematics being a specialist field, but it is something to consider. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > On Aug 8, 10:15 am, Allan C Cybulskie > > > > > This would of course explain my utter impatience with nitpicking over > > > \definitions\. > > > > It's not nitpicking! If the definitions are not precise and properly > > formed then nonsense results. The degree of precision in such matters > > is one of the things that distinguishes mathematics from other > > studies. > > It IS nitpicking, since if a definition is \imprecise\ on a USENET > NEWSGROUP it is only relevant IF that imprecision DIRECTLY impacts the > debate. Which would mean that first you determine what the discussion > is about, and then ask specific questions about how that imprecision > impacts the discussion. It is the duty, and sometimes the pleasure, of every mathematician to pits even the smallest of nits. The best math is nit-free. > > > > > > Or, to sum it up, I want to explore YOUR or THE EXISTING definitions > > > and what the consequences are, not invent my own > > > > Fine, then to explore them, you have to learn about them in a > > SYSTEMATIC way just as they are FORMULATED systematically. That means > > learning to work in the predicate calculus (some talented or > > mathematically steeped people can get by without that step, but for > > ordinary people, such as myself, I very strongly recommend first > > learning the predicate calculus), then the language, primitives, > > axioms, definitions, and proofs of the theorems, and some mathematical > > logic helps too. There just is not a shortcut that obviates learning > > the material from a class and/or textbook. > > Now, let me summarize this: > > I've taken university level mathematics through Abstract Algebra. In > order to understand these issues even at a BASE level -- ie what > happens when you take an element out of a set or add one to it -- you > claim that we must understand the entire system, meaning that in order > to understand mathematics you have to basically BE a mathematician > (even intuitively). This makes mathematics a specialist field; only > those who know it in full detail can do it (at least for some things; > addition on normal numbers doesn't seem to fall into that category). There was a time, now long gone, when a professional mathematician was expected to be reasonably familiar with /every/ area of mathematical study, however distant from his own. Nowadays, it is almost too much to expect one to be that familiar with all parts of one's own area. > > There's nothing WRONG with mathematics being a specialist field, but > it is something to consider. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b8986e$1@news2.lightlink.com> <46b89b46@news2.lightlink.com> <46b9a7e9@news2.lightlink.com> <1cb4e$46b9aff3$82a1e228$16601@news2.tudelft.nl> <1652c$46b9b58e$82a1e228$17511@news2.tudelft.nl> <704d0$46b9cbd3$82a1e228$26842@news1.tudelft.nl> > > > > > > > The truth is that you don't understand or don't care for the idea of > > > axiomatic mathematics. But this is a newsgroup about mathematical > > > pursuits. Waffling about how the universe \really\ is and whether > > > there are any \actual\ infinites in reality is the preserve of \ science > > > and philosophy. Not mathematics > > > Hmmmm. Quick check of the newsgroup headers: > > > sci(SCIENCE!).math, sci(SCIENCE!).logic, and comp.ai.PHILOSOPHY. > > > Based on what you said above, which newsgroup do you think that sort > > of discussion shouldn't apply to [grin]? > > > (Normally, I'd ignore this, but this is too good to pass up. And I > > added the obligatory Thomas Dolby on the SCIENCE! [grin]). > > Nevertheless, mathematics is _not_ a science. It does not follow the > scientific method. I never claimed it did. I claimed that the question that you claimed only has relevance to science and philosophy is perfectly relevant to two newsgroups about science and one newsgroup about philosophy ... despite your insistence to the contrary. Come on, man. Can you not ever admit your wrong, even to a jocular response about what is clearly a misstatement [grin]? > > Learned what a definition is, yet? Non sequitor. I know what it means to PHILOSOPHY, as would be relevant and appropriate for the newsgroup I'm in. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b8986e$1@news2.lightlink.com> <46b89b46@news2.lightlink.com> <46b9a7e9@news2.lightlink.com> <1cb4e$46b9aff3$82a1e228$16601@news2.tudelft.nl> <1652c$46b9b58e$82a1e228$17511@news2.tudelft.nl> <704d0$46b9cbd3$82a1e228$26842@news1.tudelft.nl> On 19 Aug, 15:17, Allan C Cybulskie > > > > > > > > > The truth is that you don't understand or don't care for the idea \ of > > > > axiomatic mathematics. But this is a newsgroup about mathematical > > > > pursuits. Waffling about how the universe \really\ is and whether > > > > there are any \actual\ infinites in reality is the preserve of \ science > > > > and philosophy. Not mathematics > > > > Hmmmm. Quick check of the newsgroup headers: > > > > sci(SCIENCE!).math, sci(SCIENCE!).logic, and comp.ai.PHILOSOPHY. > > > > Based on what you said above, which newsgroup do you think that sort > > > of discussion shouldn't apply to [grin]? > > > > (Normally, I'd ignore this, but this is too good to pass up. And I > > > added the obligatory Thomas Dolby on the SCIENCE! [grin]). > > > Nevertheless, mathematics is _not_ a science. It does not follow the > > scientific method. > > I never claimed it did. I claimed that the question that you claimed > only has relevance to science and philosophy is perfectly relevant to > two newsgroups about science and one newsgroup about philosophy ... > despite your insistence to the contrary. > > Come on, man. Can you not ever admit your wrong, even to a jocular > response about what is clearly a misstatement [grin]? Where does my wrong want admittance to? === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > We only say that cardinality, with the AoC, does what we claim for \ it. > > > > It is you who are trying to say that it has to do other things, as \ well. > > > What we say is that the cardinality does not work as a notion of > > \number of elements\ if it doesn't maintain the idea that removing an > > element from a set means that the set has less elements than it had > > before you do that. This isn't an insistence that it has to do other > > things, just that if it's gonna be about \number of elements\ at all > > it has to maintain that ESTABLISHED notion about sets. Now, you can > > insist that that's wrong, but YOU actually can't since you insisted > > that that was my misconception, and not what anyone argued. Yet that > > seems to be implied by saying that the proper subset doesn't have to > > have a smaller cardinality than the set itself ... if cardinality > > should be mapped to the \common\ notion of number of elements of > > course. > > Cardinality does everything you demand of \number\ for finite sets. Agreed. > If your notion of \number of elements\ demands that removing one \ element > from a set must always product a smaller \number of elements\, you have > yet to show that such a \number of elements\ definition is even \ possible. What do you mean? Why wouldn't it be possible? Just because I may not find a number to represent it that works that way? Or because you may not be able to do all the cool things you can do with cardinality? Or because you might not be able to find a number to represent it for all sets, or that you can determine it for all infinite sets? None of the above questions strike to \impossible\ for someone who isn't after a number to represent it, but is more concerned with how it behaves. > > It has, as I have said before, nothing to do with the number used to > > represent it, but just about what it means to have one element added > > or removed from a set (as a main push for it). So think less \ \number\ > > and think more \less or more\. > > There is a perfectly good partial ordering of sets that does precisely > what you claim to want. It is called the subset relation. But it is of > no use in comparing pairs of sets neither of which is a subset of the > other. I may look that up, since I am unconcerned about that last sentence [grin]. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > We only say that cardinality, with the AoC, does what we claim for \ it. > > > > > > It is you who are trying to say that it has to do other things, as \ well. > > > > > What we say is that the cardinality does not work as a notion of > > > \number of elements\ if it doesn't maintain the idea that removing \ an > > > element from a set means that the set has less elements than it had > > > before you do that. This isn't an insistence that it has to do other > > > things, just that if it's gonna be about \number of elements\ at \ all > > > it has to maintain that ESTABLISHED notion about sets. Now, you can > > > insist that that's wrong, but YOU actually can't since you insisted > > > that that was my misconception, and not what anyone argued. Yet that > > > seems to be implied by saying that the proper subset doesn't have to > > > have a smaller cardinality than the set itself ... if cardinality > > > should be mapped to the \common\ notion of number of elements of > > > course. > > > > Cardinality does everything you demand of \number\ for finite sets. > > Agreed. > > > If your notion of \number of elements\ demands that removing one \ element > > from a set must always product a smaller \number of elements\, you \ have > > yet to show that such a \number of elements\ definition is even \ possible. > > What do you mean? Why wouldn't it be possible? If you have a \number of elements\ (NOE) definition, which is to assign a \number\ to to every set, what properties should it have? 1. It should work like cardinality for finite sets. 2. If A is a PROPER subset of B then NOE(A) < NOE(B), and, of course, NOE(A) = NOE(A). 3. It should satisfy trichotomy, i.e., for every pair of sets A and B exactly one of NOE(A) < NOE(B), NOE(A) = NOE(B), NOE(A) > NOE(B) can hold true. Getting 1 and either 2 or 3 to hold does not seem too difficult, Cardinality does 1 and 3. but finding a way to make 1. 2 and 3 all work simultaneously has stumped mathematicians for quite some time now. I suspect that it will eventually found to be impossible in many current forms of set theory. > Just because I may > not find a number to represent it that works that way? Or because you > may not be able to do all the cool things you can do with > cardinality? Or because you might not be able to find a number to > represent it for all sets, or that you can determine it for all > infinite sets? If there were any easy way to do it, it would already have been done. > > None of the above questions strike to \impossible\ for someone who > isn't after a number to represent it, but is more concerned with how > it behaves. So prove y suspicions wrong. > > > > > It has, as I have said before, nothing to do with the number used to > > > represent it, but just about what it means to have one element added > > > or removed from a set (as a main push for it). So think less \ \number\ > > > and think more \less or more\. > > > > There is a perfectly good partial ordering of sets that does precisely > > what you claim to want. It is called the subset relation. But it is of > > no use in comparing pairs of sets neither of which is a subset of the > > other. > > I may look that up, since I am unconcerned about that last sentence > [grin]. But in order to have trichotomy (and a total, rather than merely a partial, ordering), one MUST be able to compare sets neither of which is a subset of the other. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > ... > > See, this is why we're in trouble. I argue -- for reasons that have > > never been addressed -- that if you take every second element from the > > set of all naturals you get the set of even naturals. This implies > > that the set of all even naturals should have exactly half the > > elements of the set of all natural numbers. > > Can you provide a *proof* of this surprising statement? How do you \ *define* > the number of elements, so that that is provable? Why do I need a proof, and why is this surprising? Take ANY finite set, and this WILL occur. Why would it be different for infinite sets? Heck, the \infinity/2 is infinity\ argument flat-out implies that this is indeed not all that surprising, and your fellow mathematicians have made hay with that claim. > > So the reply was that if you multiply > > every element in the set of natural numbers by 2 it would BECOME the > > set of even naturals. But this, to me, only holds true if there is an > > element in the set of even naturals for every element in the set of > > naturals. And the claim is that this is what the bijection > > demonstrates. But this seems to contradict the reasoning given > > above. And that leads me to question that bijection works for > > infinite sets, in that it really is the case that if there is a > > bijection between two infinite sets they have the same number of > > elements. The above reasoning implies that bijection gives a false > > positive for the even naturals and the set of all naturals. Now, am I > > wrong, or are you just ignoring it, or what? > > You are just using a different definition than is standard. In standard > mathematics \two sets have the same number of elements if there is a > bijection between the two sets\. This, however, disagrees with *your* > definition of \number of elements\. I'd reply to that, but I've done so twice now. Ah, hell, quick summary: if that was the case, then if we consider the counterfactual of two finite sets that were counted and gave a different number of sets, but a bijection could be proved to exist between them, you'd have to claim that they really did have the same number of elements, despite the results of the counting. But I suspect that few will accept that claim. And note, it's a COUNTERFACTUAL, which means it won't ever happen. But counterfactuals are INCREDIBLE tools for determining what a definition really does mean, since it separates the necessary from what simply happens to be the case. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > Implying that you can never finish actually doing it. See, by all \ the > > > > notions of \do to all\ that we've had with sets, it implies that \ if > > > > you do something to all elements in a set you do it to the last one \ as > > > > well. But for those infinite sets, that can't be done; the \ element, > > > > by your own reasoning, doesn't exist. So what does it mean to do > > > > something to every element in an infinite set? > > > > It means to do it for every element in that infinite set. > > > And you can't see why that's completely unhelpful? > > To you perhaps. But being helpful to you, and others as ignorant as you, > is not my goal in life. You mean people who actually know logic and reasoning? Read the section again. I said \What does it mean to do something to every element in an infinite set?\ You replied: \It means to do it for every element in that infinite set.\ Your reply merely restates the question, and thus is not an answer, useful or otherwise. It MIGHT be useful if you said \That's a primitive; it can't be defined by appealing to any other words\, but I'm afraid that that doesn't seem to be what you meant. > > > > It follows that N=1 and 2*n are defined (by induction) for all n in \ N. > > > Okay, so let's accept the inductive argument, and apply it to all the > > elements in the SET of all natural numbers. What's the number of > > elements of that set? > > Why must every set have a \number\ of elements? It is certainly not the > case that the set of natural numbers has a natural number of elements. Well, Stephen was the one that claimed it had to, not me. And if you want to say that \number of elements\ doesn't really apply to infinite sets, I'd disagree with you, but it would be a consistent and reasonable position for you to hold, and would end the debate for my part; if you aren't claiming that cardinality is number of elements, then we have no disagreement. You can use cardinality all you like. > I argue -- for reasons that have > > > never been addressed -- that if you take every second element from the > > set of all naturals you get the set of even naturals. This implies > > that the set of all even naturals should have exactly half the > > elements of the set of all natural numbers. > > That presumes that every set must have a \number\ of elememnts, but \ does > not explain what sort of thing that \number\ is supposed to be. Nor is it intended to, so this is hardly a striking criticism. > > > I can see no reason why > > that wouldn't be the case. So the reply was that if you multiply > > every element in the set of natural numbers by 2 it would BECOME the > > set of even naturals. But this, to me, only holds true if there is an > > element in the set of even naturals for every element in the set of > > naturals. > > In the sense of bijection, there is. Okay, but since the fact that we would take every second element OUT of the set of naturals to get the even naturals seems to belie that. So which is more reasonable or accurate or correct? Or are none of them correct, and it's just whatever the person finds it convenient to believe? > > > And the claim is that this is what the bijection > > demonstrates. But this seems to contradict the reasoning given > > above. > > Only because you want to assume things for which there is no > justification, such as that a set cannot biject with a proper subset of > itself. Oh, no, no, no. I don't assume that at all. I am PERFECTLY willing to accept that ... but just not then willing to assume that bijection works as a notion of \number of elements\ for infinite sets. Basically, it seems to be only because I take the \take half the elements\ case seriously. > > > And that leads me to question that bijection works for > > infinite sets, in that it really is the case that if there is a > > bijection between two infinite sets they have the same number of > > elements. > > What do YOU mean by infinite sets \having the same number of elements\? It would be similar to what could be done by counting on finite sets, and thus the \remove and add\ must be maintained. I don't define it based on bijection, but note that bijection CERTAINLY works for all finite sets. We are crawling towards a \infinite sets work differently, and now we have to choose what differences from finite sets we want to take\ conclusion, which is where I left the thread the last time. === Subject: Re: PARADISE LOST: Debunking Cantor's theory >> >> Why must every set have a \number\ of elements? It is certainly not \ the >> case that the set of natural numbers has a natural number of elements. > Well, Stephen was the one that claimed it had to, not me. And if you Will you stop lying. I never claimed any such thing. You simply do not understand what you are talking about, and resort to making false statements in a vain attempt to bolster your lame arguments. Stephen === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > Implying that you can never finish actually doing it. See, by all \ the > > > > > notions of \do to all\ that we've had with sets, it implies that \ if > > > > > you do something to all elements in a set you do it to the last \ one as > > > > > well. But for those infinite sets, that can't be done; the \ element, > > > > > by your own reasoning, doesn't exist. So what does it mean to do > > > > > something to every element in an infinite set? > > > > > > It means to do it for every element in that infinite set. > > > > > And you can't see why that's completely unhelpful? > > > > To you perhaps. But being helpful to you, and others as ignorant as \ you, > > is not my goal in life. > > You mean people who actually know logic and reasoning? Ignorance is the natural state, which, by means of sufficient tuition from others and and study by oneself, can sometimes be overcome, at least locally. > > Read the section again. I said \What does it mean to do something to > every element in an infinite set?\ You replied: \It means to do it > for every element in that infinite set.\ Your reply merely restates > the question, and thus is not an answer, useful or otherwise. > > It MIGHT be useful if you said \That's a primitive; it can't be > defined by appealing to any other words\, but I'm afraid that that > doesn't seem to be what you meant. Fear not. > > > > > > > > It follows that N=1 and 2*n are defined (by induction) for all n in \ N. > > > > > Okay, so let's accept the inductive argument, and apply it to all the > > > elements in the SET of all natural numbers. What's the number of > > > elements of that set? > > > > Why must every set have a \number\ of elements? It is certainly not \ the > > case that the set of natural numbers has a natural number of elements. > > Well, Stephen was the one that claimed it had to, not me. And if you > want to say that \number of elements\ doesn't really apply to infinite > sets, I'd disagree with you, but it would be a consistent and > reasonable position for you to hold, and would end the debate for my > part; if you aren't claiming that cardinality is number of elements, > then we have no disagreement. You can use cardinality all you like. There are all sorts of different contexts in which \number\ has different meanings. What I said was that that set of natural numbers does not have a NATURAL number of elements. That does not prohibit the use of \number\ for cardinality or ordinality, or even in quaternions. > > > > I argue -- for reasons that have > > > > > never been addressed -- that if you take every second element from \ the > > > set of all naturals you get the set of even naturals. This implies > > > that the set of all even naturals should have exactly half the > > > elements of the set of all natural numbers. > > > > That presumes that every set must have a \number\ of elememnts, but \ does > > not explain what sort of thing that \number\ is supposed to be. > > Nor is it intended to, so this is hardly a striking criticism. I can grok half of a finite set (if its cardinality is even), but \half\ of an infinite set could be, with equal justification, any subset for which the complimentary subset is of equal cardinality. You apparently have a stricter standard. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > Well, true enough, but it would hardly be a stretch of the existing > > axioms to say that the only result in a natural number if that number > > would be less than 10 digits. > > You would forbid 9,999,999,999 + 1? Why the USA national debt is liable > to exceed that before Bush is out. Um, the line you quoted below answers that question for you (or, if you can't figure it out, the answer is \No\). > > > > > Note that I'm not trying to say that we SHOULD define it that way, > > just that Brian's insistence that the axioms insist that there can't > > be a last natural number is not justified by his statement. > > The axioms that most people accept for the naturals, the Peano > properties, for one example: > http://en.wikipedia.org/wiki/Peano_postulates#Peano.27s_axioms > Peano's nine axioms, rephrased in contemporary notation, are: > 1 1 is a natural number. > 2 Every natural number is equal to itself (equality is reflexive). > 3 For all natural numbers a and b, a = b if and only if b = a > (equality is symmetric). > 4 For all natural numbers a, b, and c, if a = b and b = c then a = c > (equality is transitive). > 5 If a = b and b is a natural number then a is a natural number. > 6 If a is a natural number then Sa is a natural number. > 7 If a and b are natural numbers then a = b if and only if Sa = Sb. > 8 If a is a natural number then Sa is not equal to 1. > 9 For every set K, if 1 is in K, and Sx is in K for every natural > number x in K, then every natural number is in K. (It makes no > difference here whether all elements of K are natural numbers.) > Axioms 2, 3, 4, and 5 are now considered basic properties of equality > and taken for granted in most contexts. Thus it is axioms 1, 6, 7, 8, > and 9 which describe the structure of the natural numbers. > > These make it quite clear that a \last\ natural would be most \ unnatural. \successor\. Successor, unrestricted, would indeed claim that there is no last natural. I believe that it could be changed fairly easily to be less stipulative and more descriptive, but I'll concede the point: the current axioms DO imply directly that there is no last natural number, and that then inventing one might indeed be problematic. [snipped] > > > Actually, it doesn't matter if it's done singly or simultaneously. If > > you do it all at the same time I should be able to access any > > individual result ... which should include something in the second > > half of the set (which we can't know about) and potentially the, say, > > largest element that we did it to. > > What \second half\ of what set? > > Take the set of natural powers of two and multiply each of its members > by 2, and you don't have a 'second half' unless {1} is the first half. Um ... this isn't relevant. I can do things to the second half of the set ... it's just that what happens to them isn't the same thing as happens on the sequential set of naturals. > > > > > Basically, it's just a question, and it becomes really important when > > we want to look at a small question like \how large is the set > > produced by multiplying every element in the set of all naturals by > > 2? > > The same size as the original set by cardinality. And that's the premise that's under debate here, so obviously this will NOT settle the point. > > > Dead wrong. If I take any large but finite set of consecutive powers \ of > > > 2, and multiply each of them by 2, I will get at most one number \ outside > > > that set. > > > Oh, come on, you KNOW that isn't what I meant by \consecutive natural > > numbers\, so this is just a bald evasion. > > So that you are complaining because doing something tho change the > value of a natural changes the sets of naturals that the result can > belong to? No, I am complaining in THIS response because you are dishonestly trying to shift the discussion to another set without actually making it APPEAR as if you are. In short, you are being dishonest in applying the results of something that I did not attend to talk about as a proof that what I talked about is wrong. > > > > > > > Now, you can insist that this is different for infinite > > > > sets, but to do so you have to rely on there being no last element \ ... > > > > which makes one wonder what it means to multiply every element in \ that > > > > set by 2. > > > > Try 2*x = x + x. > > > Same question applies: what does it mean to add every element in the > > set of all naturals by itself? > > You can't add it by itself, but it is easy to add it to itself. And you can't see that this is a nitpick? And yes, I did just recently do something similar to MoeBlee, but history shows that I have good reasons for doing so [grin]. > > And what it means follows from the definition of addition for naturals, > which covers the addition of any two naturals to form a natural. But not in the right way: it doesn't tell us what the size of the set ends up as. [snipped] > > Since I think that the axioms and definitions lead to a contradictory > > conclusion, again why do you think that that would be more important > > than people like you showing why those axioms and definitions should > > be maintained even in the light of odd conclusions that at least some > > people are trying to draw. > > Because we who have worked through all those steps that you have not, > and see how well they work, are not willing to give them up for those > who won't do the work necessary to understand. Define \how well they work\. Note that most people seem to be harping on \it works for all infinite sets\ as the main benefit ... but I argue that all sorts of definitions that would work for all infinite sets aren't, in fact, right (I think you'd have to concede that as well; can there BE a wrong set theory?) and find that the initial consequences seem so odd and contradictory to me that I suspect that even if I worked through all of it I'd STILL have my same objections. But part of this is BECAUSE I'm not a mathematician and so don't care about a clean and useful mathematics (and working for all infinite sets is certainly more useful than something that doesn't). But you'll note that I'M not insisting that you have to give them up. I've said repeatedly that you can say that set size is just cardinality and ignore all of these. Now, let me make a point about how all this started, at least for me; I got involved originally (long, long ago) because someone was insisting that BECAUSE cardinality was set size was number of elements people HAD to accept the even/natural size equation. Since they don't have any justification other than \that's what cardinality means\, I protested that that was not a valid move to make. As long as you don't try to convince me of something MERELY on this debated notion of set size, I will have no problem with you maintaining it (well, other than the problem I noted in my latest summary [grin]). > > > > > > > Well, first, I don't particularly care about that theory if it \ leads > > > > to contradictory conclusions > > We don't either, so have eliminated as many of them as we can. But new > people keep coming up with the ones we have thrown out. But then you should be able to easily explain to them why they should be thrown out. But \read a book\ and \it just is that way\ are not going to convince people that you've thrown them out validly. [snipped. You missed the point, but I don't care about that point anyway]. === Subject: Re: PARADISE LOST: Debunking Cantor's theory ... > > > Actually, it doesn't matter if it's done singly or simultaneously. \ If > > > you do it all at the same time I should be able to access any > > > individual result ... which should include something in the second > > > half of the set (which we can't know about) and potentially the, say, > > > largest element that we did it to. > > > > What \second half\ of what set? > > > > Take the set of natural powers of two and multiply each of its members > > by 2, and you don't have a 'second half' unless {1} is the first half. > > Um ... this isn't relevant. I can do things to the second half of the > set ... it's just that what happens to them isn't the same thing as > happens on the sequential set of naturals. What I am objecting to is \half\ of a set when that \half\ has the same \ cardinality as the whole. How does a \half\ differ from a \third\, or a \ quarter, in such cases? > > > > > > > > > > Basically, it's just a question, and it becomes really important when > > > we want to look at a small question like \how large is the set > > > produced by multiplying every element in the set of all naturals by > > > 2? > > > > The same size as the original set by cardinality. > > And that's the premise that's under debate here, so obviously this > will NOT settle the point. I hope that \same cardinality\ is not under debate. > > > > > Dead wrong. If I take any large but finite set of consecutive powers \ of > > > > 2, and multiply each of them by 2, I will get at most one number \ outside > > > > that set. > > > > > Oh, come on, you KNOW that isn't what I meant by \consecutive \ natural > > > numbers\, so this is just a bald evasion. > > > > So that you are complaining because doing something tho change the > > value of a natural changes the sets of naturals that the result can > > belong to? > > No, I am complaining in THIS response because you are dishonestly > trying to shift the discussion to another set without actually making > it APPEAR as if you are. In short, you are being dishonest in > applying the results of something that I did not attend to talk about > as a proof that what I talked about is wrong. > > > > > > > > > > > > Now, you can insist that this is different for infinite > > > > > sets, but to do so you have to rely on there being no last element \ ... > > > > > which makes one wonder what it means to multiply every element in \ that > > > > > set by 2. > > > > > > Try 2*x = x + x. > > > > > Same question applies: what does it mean to add every element in the > > > set of all naturals by itself? > > > > You can't add it by itself, but it is easy to add it to itself. > > And you can't see that this is a nitpick? It is a legitimate nit. > > And yes, I did just recently do something similar to MoeBlee, but > history shows that I have good reasons for doing so [grin]. > > > > > And what it means follows from the definition of addition for naturals, > > which covers the addition of any two naturals to form a natural. > > But not in the right way: it doesn't tell us what the size of the set > ends up as. > > [snipped] > > > > Since I think that the axioms and definitions lead to a contradictory > > > conclusion, again why do you think that that would be more important > > > than people like you showing why those axioms and definitions should > > > be maintained even in the light of odd conclusions that at least some > > > people are trying to draw. > > > > Because we who have worked through all those steps that you have not, > > and see how well they work, are not willing to give them up for those > > who won't do the work necessary to understand. > > Define \how well they work\. Peano arithmetic proves that the commutativity , associativity, distributivity, and other expected properties of natural number arithmetic follow, by induction, from simple definitions, in any set theory that allows the Peano postulates to be theorems. > > > > > Well, first, I don't particularly care about that theory if it \ leads > > > > > to contradictory conclusions > > > > We don't either, so have eliminated as many of them as we can. But new > > people keep coming up with the ones we have thrown out. > > But then you should be able to easily explain to them why they should > be thrown out. But \read a book\ and \it just is that way\ are not > going to convince people that you've thrown them out validly. If the demonstration of something is sufficiently long and technical, no one is going to repeat it over and over, one does it by writing it once and publishing it \as a book\. You are asking for the content of whole books to be adequately summarized in a couple of paragraphs, and it cannot be done. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <8RJti.35955$fJ5.15717@pd7urf1no> > > > > >> Spare yourself the sputtering: Just read a book. > > > > Well, \Just read a book\ alone is not a good advise in many \ instances: > > > > a) One has to understand the main points, as well as the limitations \ of > > > the book. > > > I'd like to add to a): \, before one gives advise to others to \just \ read a book\. > > (1) Even if one has limitations in understanding, the suggestion to > another to at least consult a basic textbook is still a good > suggestion. (2) I don't know what particular point of the suggestor's > understanding or limitations of certain textbooks you think are in > question here. I think you miss the point. I think his point is that before you can tell someone to \just go read a book\, you first have to understand what the main point that person is trying to make is. Otherwise, they may find that after they spend all that time reading the book it says nothing about what point they were actually trying to make. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > On Aug 6, 6:48 am, Allan C Cybulskie > > I looked up what an ordinal number was. The comment was to the extent > > that an ordinal number was basically defined as the set of ordinal > > numbers leading up to it. Which you yourself seem to accept later. > > We PROVE that every ordinal is a set of ordinals, but that is not the > DEFINITION of 'is an ordinal' nor must we define every particular > ordinal in that manner. Sigh. And does this matter? Is it correct to say what I said, or not? Putting aside the \defined\ word that you have such issues with ... > > > > > 1) You can only do a bijection between two sets, not between a set \ and > > > > a \number\. > > > > I don't care very much about the WORD 'number'. We call certain sets > > > 'ordinal numbers' and 'cardinal numbers'. But the word 'number' in > > > those defintions has no part in the deductions we make. We could call > > > them 'ordinal rebmuns' and 'cardinal rebmuns' and our mathematics > > > would not be affected anymore than if you call an elevator an > > > 'elevator' or a 'lift'. > > > And this is relevant because ... ? > > Because you made a point about numbers, especially emphasizing with > scare quotes around the word 'number'. Um, scare quotes aren't used to emphasize in the way you insist I was emphasizing it. Actually, linguistically speaking your translation of my use of \number\ to 'number' is invalid; they mean different things in linguistics, as my professor strongly insisted upon. > > > It has to be between two sets is > > all I said. > > No, you said, even as you are requoted in your own post: > > \You can only do a bijection between two sets, not between a set and a > \number\.\ > > And my point is that we do not depend on the word 'number'. And I never accused you of doing that. Did you somehow miss reading the first part where I flat out said that you can only do it between two sets? And even if that second part of the sentence is supposed to be considered hugely important to my point, I never made a claim about the word 'number'. I said that it had to be two sets. If a \number\ was defined as being a set (like the ordinals are, in at least some way) then it would be valid. I fail to see why you would think anything otherwise. But if a \number\ is not a set, then a bijection cannot be done between a set and that \number\. That first premise was basically just to show that even without looking up what an ordinal number is, I can assume that it has to be a set, or else saying that an ordinal number could be used as a set size would mean nothing to the discussion. In short, you nitpicked over one word instead of looking at what the sentence actually said. > > > > > 2) Therefore, an ordinal number must be directly related to or \ defined > > > > by a set. > > > > Yes, every ordinal number IS a set. > > > Okay, you're doing fine so far ... > > knows virtually nothing about the subject. Um, since this comment is about your understanding of what _I_ was trying to say, I think it is safe to assume that I know more about that than you do ... > > > > > 3) If I do a bijection from a set S to a set that defines or > > > > determines the ordinal number, I prove that those two sets are \ equal > > > > in size. > > > > \Set that defines or determines the ordinal number\. I don't know \ what > > > that is supposed to mean. Every ordinal number IS a set. > > > ... but here you seem to develop stupidity. Basically, what I said > > here is just the same phrasing as I used in 2), which you translated > > as \it IS a set\. > > I didn't translate anything there. I merely QUOTED you then pointed > out that every ordinal is a set. In fact, you said: \> > > > 2) Therefore, an ordinal number must be directly related to or defined > > > > by a set. > > > > Yes, every ordinal number IS a set.\ The \Yes\ implies that my statement is at least in some sense correct, followed by what seems to be a precise statement of the relation/ definition in question. Thus it is safe to conclude that you translate my claim of \directly related to or defined by a set\ to the conclusion \It IS a set\. If that isn't what you meant, then you need to clarify what you meant above. If that is, then your later protestations that you cannot understand what that phrase means would, at best, show that you had no clue what you meant to say you made the above statement. Look, MoeBlee, you're probably a very intelligent person, and I will definitely claim that you know more about mathematics than I do. But when you try to play rhetorical games, you end up painting yourself into a corner, because in general people who AREN'T hugely technical or mathematical are better at rhetoric than those that are, because it plays a bigger role in what they do. And so when you nitpick over words or develop some sort of selective amnesia, you'll get picked apart by anyone who actually cares. But if you stuck to the mathematics, and tried to use your knowledge to explain and translate the questions of the questioners, you'd probably fare far better than those you reply to. > > > But here you can't take that and even doing some > > slight thinking figure out that that's STILL what I'm talking about. > > And since this infects the entire rest of the point, go make that > > substitution and then you can see what I meant. > > Spare yourself the sputtering: Just read a book. So you WON'T do the substitution to be consistent with what YOU said then? And instead will shrug it off with a \Read a book\? Is there any reason why ANYONE should take you seriously when you can't even live up to what you yourself have said in even a small attempt to understand someone else's point? [snipped] > > > You keep arguing on the basis of some notion of \the set that \ defines > > > the ordinal\. Perhaps if you were familiar with the subject you'd \ know > > > how to say whatever it is you mean by that. > > > As I said above, you got it the first time I used that phrasing, but > > seem to have developed memory loss by the time you got to the next > > point ... > > There's a bunch of information for you in all those quotes to which > you did not respond. Then all you can muster is something about some > supposed memorry loss of mine. Basically, all you said was \I don't know what you're talking about\, which is belied by the first statement you made where you clearly do and seem to clearly think you do. Hence the comment about memory loss. > > > I'm just telling you about set theory. I'm not too interested in what > > > you and Stephen may or may not have demanded of each other. > > Then your response is pointless; you were challenging me and yet are > > proving the same thing _I_ was saying; that you only get a number from > > a bijection if you already know it for one of the two sets in the > > bijection. > > Nope. You understand virtually zilch about this subject. A rational > response by you would be to get a book on the subject and find out > something about it. But, alas, you are one of those folks who feels it > makes sense to spew on a subject of which you know virtually nothing. I fail to see how anything you said refutes the comment. One would expect that if you really DID know the mathematics that well, you could explain why I'm wrong by saying that you can only get a number of elements for a set as an actual number if you have an actual number for the number of elements for one of the two sets if you are using bijection. That's the point under discussion, and all you've said agrees with me. If you made a claim about dualism (to turn this into a comp.ai.philosophy example) and you were right, I expect that I'd be versed enough in the theory to at least be able to figure out that you and I were saying the same thing. I also expect that I'd at least try. I have seen no evidence of that from you. > I wouldn't go on the Internet spewing about, e.g., the geology of New > Zealand, since I don't know anything about it. So I find it > fascinating that there are people who choose to spew on sundry > subjects of which they know virtually nothing. Except that I'm not going on the Internet and \spewing about\ mathematics. I'm simply responding to the insistence of some people in sci.math that if you form a set by taking half the elements of another set it is obvious that it has the same number of elements (or set size, but that's trickier since it can depend on some sort of \definition\). And if it's only obvious if I study AND ACCEPT all of Cantorian set theory ... well, you can see why that's an utterly invalid line of argumentation. It's like saying that it is obvious that Judaism is the one true religion ... but only to those who ARE Jews. > > > > ZFC (even just ZF with Scott's method) provides that every set has a > > > cardinality and that the cardinality of a set satisfies certain > > > properties vis-a-vis the set. If you ever study the subject, then you > > > can tell us what alternative theory you endorse that also meets that > > > demand. > > > The debate is over whether or not the cardinality just IS the number > > of elements. Since it seems to do odd things for infinite sets, the > > question is open. > > Definitions don't \do\ odd things to objects. Mathematical definitions don't have consequences? Hardly seems worth having them, then. Note the translation you made from \for\ to \to\, BTW. > > > If you spent more time thinking and less time ranting about non- > > mathematicians who don't know the secret codes, you might get > > somewhere. > > There are no secret codes. You can refer to any number of basic > undergraduate textbooks that are publicly available. When I used the term \secret codes\, it was not to mean that it wasn't available, but that you couldn't understand anything not expressed in them. But you can be profoundly literal ... when it suits you. [snipped] > As to my thinking about the subject, I do my own thinking as I compare > my own intuitions and ideas with an ongoing study of a wide range of > differening mathematical systems and philosophies about mathematics. > Meanwhile, your spewing is based on a virtual ignorance and abysmal > misunderstandings of the even basics. And again, you miss the point that the thinking I asked you to do was to take someone who isn't as well-versed in mathematics as you are and try to map it to the mathematical knowledge you have, instead of nitpicking over words that are used incorrectly. > > But you need to know precisely the \certain ordinal\. My point was > > that if you didn't know that, but just had the mapping, you wouldn't > > know the \number of elements\. Even in a finite case, to make it > > simpler. > > If we know the mapping, then we know the ordinal. As to the instances > in which we don't know the mapping, I discussed that. Set theory (in > classical logic) is not constructive. We know that. And this relates to my point ... how? Let me translate the discussion: Me: Only if you know the precise number of the number of elements (or cardinality or whatever) of one of the two sets in the bijection can you get a number of elements for the OTHER set (or either of them, really) using a bijection. You: But we ALWAYS know the ordinal number and thus the number of elements or cardinality of that set. Well, fine and dandy ... but if you didn't, you wouldn't get what the cardinality of S is by bijecting with the ordinal number. The fact that you just happen to always be able to do that is utterly irrelevant to the discussion. > > > We DON'T define the size of N as N. We PROVE that the cardinality of \ N > > > is N. Your objection is again ill-premised. > > > How did you prove that it was N before defining what N was? > > We don't. You know, if you get a good textbook, you'll see it all > developed step by step. I can't do all that step by step work for you > in just a few posts. Ah, but you don't need to ... all you have to do is point me -- even online -- to the definition of the number \N\ does not include it being the size of the natural numbers. I'm open to that being the case; my first search indicated that that was not the case. > > > Whatever he is demanding of you, you have not given a definition of > > > \number of\ that meets the demands meet by the method of set \ theoretic > > > cardinality - that every set has a cardinality and that the > > > cardinality of a set satisfies certain properties [we could list them > > > here] vis-a-vis the set. > > > And yet I find myself utterly unconcerned about matching what > > cardinality can do. Odd that > > And I am pretty much unconcerned as to what you are utterly > unconcerned with, excpet when you're spewing as ignorantly as you are. Well, except, of course, when you demand that a definition of \number of\ has to meet \the demands met by the method of set theoretic cardinality\, to which my reply is that I really don't give a damn about what that satisfies, nor do I feel bound to give any definition that meets those demands. You may insist that you won't accept any such definition until I do, at which point I will simply counter insist that I will never accept a definition that implies that the set of even naturals has the same \number of elements\ as the full set of naturals, at which point we return to a pointless and unresolvable debate. Which is what I SAID this was when I first returned to it. === Subject: Re: PARADISE LOST: Debunking Cantor's theory On 19 Aug, 14:27, Allan C Cybulskie > > On Aug 6, 6:48 am, Allan C Cybulskie > > > > ZFC (even just ZF with Scott's method) provides that every set has \ a > > > > cardinality and that the cardinality of a set satisfies certain > > > > properties vis-a-vis the set. If you ever study the subject, then \ you > > > > can tell us what alternative theory you endorse that also meets \ that > > > > demand. > > > > The debate is over whether or not the cardinality just IS the number > > > of elements. Since it seems to do odd things for infinite sets, the > > > question is open. > > > Definitions don't \do\ odd things to objects. > > Mathematical definitions don't have consequences? That's right. Mathematical definitions are just abbreviations that ultimately revert to primitives. I'm sure you've been told this a few times already.. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > On 19 Aug, 14:27, Allan C Cybulskie > > > On Aug 6, 6:48 am, Allan C Cybulskie > > > > > ZFC (even just ZF with Scott's method) provides that every set has \ a > > > > > cardinality and that the cardinality of a set satisfies certain > > > > > properties vis-a-vis the set. If you ever study the subject, then \ you > > > > > can tell us what alternative theory you endorse that also meets \ that > > > > > demand. > > > > > > The debate is over whether or not the cardinality just IS the \ number > > > > of elements. Since it seems to do odd things for infinite sets, \ the > > > > question is open. > > > > > Definitions don't \do\ odd things to objects. > > > > Mathematical definitions don't have consequences? > > That's right. Mathematical definitions are just abbreviations that > ultimately revert to primitives. I'm sure you've been told this a few > times already.. One's choice of a definition can have consequences. A good choice can aid understanding and a poor one inhibit it. But it is quite correct that all definitions are merely abbreviations, and could, at least in theory, be eliminated. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46bbdf58@news2.lightlink.com> <46c79699@news2.lightlink.com> > > ... > > > > I do say it because you use too many undefined things. You say the \ same > > > > because you do not *like* ordinal arithmetic. > > > > I say it because it makes no sense to me. You add more, you move to \ the > > > right, whether you're infinitely far from the origin or not. You \ remove, > > > you go left, no matter where you are. > > > But it is proven using ordinary logic and well-given definitions, and > > you do not *like* that result. > > Definitions? Are those axiomatic statements of existence, or theorems > asserting existence based on those axioms preceding it? Neither. It is somewhat disapointing that you still don't know what a mathematical definition is. > I do not like statements that contradict basic assumptions I've adopted. > > So, no, while I have little (but not none) issue with predicate logic, I > do not accept conclusions borne of axioms that are not, IMHO, justified. \ :) The truth of the axioms of a system does not affect what the theorems of that system are. It is somewhat disapointing that you still do not grok the concept of _axiomatic_ mathematics. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> > > >>>> There is no largest natural. Nor is there any natural x such > >>>> that 2*x is not a natural number. These are simple conseqeuences > >>>> of the definition of \natural number\. > >>> Ah. So you avoid the whole problem by simply defining that any time > >>> you multiple a natural number by \2\ it MUST result in a natural > >>> number. Which is, of course, valid. But then it belies considering > >>> us to have a set of \natural numbers\ at all, since if you actually \ do > >>> the multiplication of EVERY element in the set by 2 you should end \ up, > >>> eventually, with an element that isn't in the set anymore. > >> Adding one or multiplying by 2 sequentially only causes problems if \ one > >> insists that the sets being acted on are finite. There is no problem > >> with such infinite processes in infinite sets. > > > Implying that you can never finish actually doing it. See, by all the > > notions of \do to all\ that we've had with sets, it implies that if > > you do something to all elements in a set you do it to the last one as > > well. But for those infinite sets, that can't be done; the element, > > by your own reasoning, doesn't exist. So what does it mean to do > > something to every element in an infinite set? > > > This wouldn't be an issue if we all agreed on the outcome of doing it > > in this case. But we don't, because the argument of \multiply every > > element by 2\ is being used to justify claiming that a set that you > > can form by taking every second element from another set (the set of > > even naturals) does NOT have less elements than the set you formed the > > set of even naturals from (the set of all naturals). And I, at least, > > find that not only odd, but a flat-out paradox. So we need to agree > > on what it means to do something to every element in an infinite set > > before we can settle this question. > > > Note the use of \less elements\ above, and not size. If you claim > > that the set of even naturals and the set of all naturals have the > > same SIZE, where you define size as being \cardinality\, which \ happens > > to map to number of elements for finite sets but not for infinite > > sets, but since number of elements is problematic for you on infinite > > sets so you prefer cardinality, then this whole discussion should go > > away. Tony can go off and work on Elementarian Size, you can maintain > > Cantorian Size, and all the rancor should be eliminated. Since I've > > already flat-out suggested this, I suspect that some people will not > > accept this sort of compromise ... > > Hi Allan - > > I do appreciate your participation in this discussion, as you seem to > have extremely similar objections as I do to standard transfinitology. I > am afraid, however, that you misinterpret the purpose of the debate. :) > > You see, it's rather difficult to formulate a new theory in mathematics, > especially when you're question what's considered the very foundation of > the discipline [snipped] But here's the issue with this: there is no reason why your theory has to be a competitor ... yet. So far, it looks like most of your disagreement is over set size. There is no reason in mathematics why two competing theories cannot exist simultaneously, as long as we are careful not to insist that the differences prove things in the other theory. Look, for example, at the various geometries to see how that can work. So you don't need to compete with them yet, and so it seems problematic to insist on doing so. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> On Aug 19, 4:04 am, Allan C Cybulskie > I'm back, after having FAR better things to do than reply to this > thread for the past couple of weeks [grin]. > > So what I want to do is start with ANOTHER summary post, since things > are drifting away from the key points again in the other posts. I'll > reply to the other replies later. > > Okay, first, the key point: Bijection as a notion of \two sets have > the same size\. How is this derived, and how is it justified? I > claimed in my return to this discussion that bijection get its > justification as a METHOD for determining if two sets have the same > \number of elements\ (where number of elements is loosely defined as > what we'd get if we counted the number of elements on finite sets, and > is up in the air what it would mean for infinite sets for now) because > it happens to work on finite sets. But it is not simply part of the > definition of \number of elements\ or even \set size\; more > stipulations need to be made to get it that way. This is because: > > Take two finite sets, S and T. They were counted and we know that S > has n elements, T has m elements, and n != m. And what, exactly, does \they were counted\ mean (we will restrict this question to finite sets for now)? I claim that \counting\ is, itself, a bijection: counting the elements by saying \1\, \2\, \3\, etc. as I move the elements from one pile to another is one way of / demonstrating/ a one-to-one matching between the elements of the set with those of a certain canonical set. > Someone (as I said > elsewhere, it will never happen; the bijection always does work for > finite sets) proves that there is a bijection between S and T. What > would we conclude? > Obviously, someone screwed up! > First, it is unlikely that in that case we would conclude that because > there is a bijection between those two sets that they would have the > same number of elements. I don't agree that this conclusion is obvious from your \counterfactual\. We might equally reject the notion that n and m respectively actually correspond to a /correct/ \counting\ of the sets in question. Suppose I am watching a group of men and women dance. Every man is dancing with a woman, and every woman is dancing with a man. If you then tell me that you \count\ n men and m women and n != m, I'd naturally say that you can't possibly have \counted\ /correctly/: the set of men and in women are in bijection, so there /must/ be the same \count\ of both men and women. I wouldn't need to actually \count\ the men and women to \intuit\ this. > We'd instead claim that bijection does not > always work as a way to determine if two sets have the same number of > elements. If set size is not simply defined to be cardinality and is > defined to relate in any way to counting the elements, we'd therefore > conclude that bijection is not always a way to determine if two sets > have the same size. > I bring my sheep into the pen. As each sheep enters, I place a rock in a bag, one rock for each sheep. The next morning, as the sheep leave the pen, I discard a rock from the bag for each sheep. If there are rocks left over, I know I've lost a sheep overnight: I don't need to \count\ the rocks or the sheep to know this. Counting /relies/ on bijection: it is transitive. If it weren't transitive, there'd be no point in \counting\ anything to start with. > Second, this shows that \number of elements\ does NOT simply mean that > there is a bijection between the two sizes. Of course not. Intuitively, a bijection between two sets simply means that they have the /same/ size; by itself, it doesn't say anything about /what/ that size is. It so happens that there is a /canonical/ collection of sets that we use when we count things; represented by symbols like \175\. But using symbols like \175\ to denote set sizes would be pointless without the idea that a bijection between the elements between \the sheep on that hill\ and \175\ is what we /mean/ by \there are 175 sheep on that hill\. > If it did, the question I > asked above would be nonsensical; there is no way in which the number > of elements could POSSIBLY be different by counting it. And yet it > seems a perfectly reasonable analysis, and moreover we seem to be able > to conclude that in that case we could -- and would, even -- claim > that bijection fails as a method for determining \number of > elements\ (and hence set size, for me and Tony, at any rate). > It's equally reasonable to claim that the statement \n is the count of elements the first set\ is a false statement (e.g., the way in which the \counting\ took place was flawed: perhaps some elements were \counted\ twice, or were never \counted\ at all). === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> I'm back, after having FAR better things to do than reply to this thread for the past couple of weeks [grin]. So what I want to do is start with ANOTHER summary post, since things are drifting away from the key points again in the other posts. I'll reply to the other replies later. Okay, first, the key point: Bijection as a notion of \two sets have the same size\. How is this derived, and how is it justified? I claimed in my return to this discussion that bijection get its justification as a METHOD for determining if two sets have the same \number of elements\ (where number of elements is loosely defined as what we'd get if we counted the number of elements on finite sets, and is up in the air what it would mean for infinite sets for now) because it happens to work on finite sets. But it is not simply part of the definition of \number of elements\ or even \set size\; more stipulations need to be made to get it that way. This is because: Take two finite sets, S and T. They were counted and we know that S has n elements, T has m elements, and n != m. Someone (as I said elsewhere, it will never happen; the bijection always does work for finite sets) proves that there is a bijection between S and T. What would we conclude? First, it is unlikely that in that case we would conclude that because there is a bijection between those two sets that they would have the same number of elements. We'd instead claim that bijection does not always work as a way to determine if two sets have the same number of elements. If set size is not simply defined to be cardinality and is defined to relate in any way to counting the elements, we'd therefore conclude that bijection is not always a way to determine if two sets have the same size. Second, this shows that \number of elements\ does NOT simply mean that there is a bijection between the two sizes. If it did, the question I asked above would be nonsensical; there is no way in which the number of elements could POSSIBLY be different by counting it. And yet it seems a perfectly reasonable analysis, and moreover we seem to be able to conclude that in that case we could -- and would, even -- claim that bijection fails as a method for determining \number of elements\ (and hence set size, for me and Tony, at any rate). Now, what Tony and I are going on about is that this situation that we cannot have in finite sets happens with infinite sets. It isn't about pigeonhole principles or proper subsets with us, but is about the idea that you can add an element to a set, or remove an element from a set, or form a set by taking half the elements from another set and yet maintain the same \set size\. This claim is mainly justified -- we can ignore the peculiarities of the \infinite numbers\ and addition on them for now; I, at least, insist that we can talk about this without having any actual number at all -- by the fact that there is a bijection between those two infinite sets. But the above analysis on finite sets means that that isn't, itself, conclusive; bijection can be wrong (even if it never is). So we need more of a justification than that. The \add 1 to an infinite number\ argument tends to be the next one advanced. Other than the claim that you don't need a number at all -- I insist that I can know that if I add an element to a set, it has MORE elements even if I don't know precisely how many elements that set has -- this seems to be a mathematical trick of the sort that my old high school math teacher used to put up on the walls, where we could prove things like 2 = 1. But no one took that to seriously prove that 2 = 1, and yet some people here seem to be taking the infinite number case as proof of something quite extraordinary. In addition, Stephen seems to think that if I defined a new number system for infinite number that happened to work out the way I wanted, that would be legitimate. But is the size of a set a property of a set, or a property of the numbers used to express it? Shouldn't it work out the same no matter what number I'm using, or at least in some sense? At any rate, there is a way out of this whole mess. Cantorians can say that in all cases, set size just IS cardinality, and cardinality just IS bijection-based. Fine. But note the consequences of doing that: in the finite set case I described, you'd have to insist that elements is different, because the bijection occurred. If you do so, then you have to leave behind ANY justification of this based on counting elements for finite sets: while it happens to work, it isn't definitive; it just happens to work. So to translate that, you cannot say that it happens to map to \counting\ and so it works for finite sets, since if counting differed from bijection for finite sets you'd reject counting. So you need another justification for claiming that this is a good notion of set size to use. So far, that justification seems to be that you can universally apply it to infinite sets and that it leads to interesting results. But since I don't find that compelling, I'm free to reject that and say that \number of elements\ is a sufficient notion for set size -- even if it isn't fully defined out yet for infinite sets. But this wouldn't cause any great clash, unless we tried to insist that our notion of set size applied to the other \group's\ notion of sets and proved things about it. I am unlikely to do that any time soon [grin]. Second, there does seem to be a problem between bijection and adding of an element to a set. If you said that you added an element to a set, how would you prove to someone that you had actually done that? So, say that we have a set S, and we form a set T by adding one element E to S. How would we prove that T is S with E added? It isn't sufficient to show that E is in T and not in S, since I would insist that you merely replaced an element in S with E, and didn't actually add E at all. So what you have to show is that every element that is in S is in T, and then there is an additional element E in T. Now, bijection is summarized as being \there is a mapping that maps every element in a set onto one and only one element in the other set\. So what does the above examination of adding an element show? Well, at the base level, your obvious mapping would be to map every element in S onto the corresponding element in T; so you'd map the elements in S onto the elements from S that are in T. By the definition above, that would leave E unmapped; all of the elements in S must map directly onto each other in T, and we claimed that if that's done there's one left over ... E. Now, when I first brought this up, the original poster (and if I've missed replying to that reply, or missed the reply to my reply, I apologize; it's hard to keep up sometimes when you only reply once a week [grin]) said that you can change the mapping to include E. I believe the example he used was a set of: {0, 1, 2, 3, ...} {-1, 0, 1, 2, ...} The problem is that this doesn't seem to help. You have to map sizeOf(S) elements onto T, but by the only way to prove that an element was added there are sizeOf(S) elements in T BEFORE considering E; even if E is deliberately included in the mapping, a DIFFERENT element must be excluded. So this is kind of where I stand on this. Discuss if you wish, but before you do, please read. I tire of correcting people who claim to have read and understood my summaries and yet ask questions that I answered in them. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> On 19 Aug, 12:04, Allan C Cybulskie > I'm back, after having FAR better things to do than reply to this > thread for the past couple of weeks [grin]. > > So what I want to do is start with ANOTHER summary post, since things > are drifting away from the key points again in the other posts. I'll > reply to the other replies later. > > Okay, first, the key point: Bijection as a notion of \two sets have > the same size\. How is this derived, and how is it justified? Intuition. > I claimed in my return to this discussion that bijection get its > justification as a METHOD for determining if two sets have the same > \number of elements\ (where number of elements is loosely defined as > what we'd get if we counted the number of elements on finite sets, and > is up in the air what it would mean for infinite sets for now) because > it happens to work on finite sets. But it is not simply part of the > definition of \number of elements\ or even \set size\; more > stipulations need to be made to get it that way. This is because: > > Take two finite sets, S and T. They were counted and we know that S > has n elements, T has m elements, and n != m. Someone (as I said > elsewhere, it will never happen; the bijection always does work for > finite sets) proves that there is a bijection between S and T. What > would we conclude? That their proof is incorrect, or that the theory they are working in is inconsistent. > First, it is unlikely that in that case we would conclude that because > there is a bijection between those two sets that they would have the > same number of elements. We'd instead claim that bijection does not > always work as a way to determine if two sets have the same number of > elements. If set size is not simply defined to be cardinality and is > defined to relate in any way to counting the elements, we'd therefore > conclude that bijection is not always a way to determine if two sets > have the same size. > > Second, this shows that \number of elements\ does NOT simply mean that > there is a bijection between the two sizes. If it did, the question I > asked above would be nonsensical; there is no way in which the number > of elements could POSSIBLY be different by counting it. Indeed, this seems to be the case. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> > On 19 Aug, 12:04, Allan C Cybulskie > > > I'm back, after having FAR better things to do than reply to this > > thread for the past couple of weeks [grin]. > > > So what I want to do is start with ANOTHER summary post, since things > > are drifting away from the key points again in the other posts. I'll > > reply to the other replies later. > > > Okay, first, the key point: Bijection as a notion of \two sets have > > the same size\. How is this derived, and how is it justified? > > Intuition. Excellent. Mine conflicts because of the whole \adding and removing\ thing with infinite sets. But since no one person's intuition is necessarily better than another's, we would end up at a stalemate ... which is where I said this always ends up ... > > > I claimed in my return to this discussion that bijection get its > > justification as a METHOD for determining if two sets have the same > > \number of elements\ (where number of elements is loosely defined as > > what we'd get if we counted the number of elements on finite sets, and > > is up in the air what it would mean for infinite sets for now) because > > it happens to work on finite sets. But it is not simply part of the > > definition of \number of elements\ or even \set size\; more > > stipulations need to be made to get it that way. This is because: > > > Take two finite sets, S and T. They were counted and we know that S > > has n elements, T has m elements, and n != m. Someone (as I said > > elsewhere, it will never happen; the bijection always does work for > > finite sets) proves that there is a bijection between S and T. What > > would we conclude? > > That their proof is incorrect, or that the theory they are working in > is inconsistent. Assume that the proof is correct, and no theory other than what exists now works. What would you conclude? This is a counterfactual ... I am not claiming that this would ever occur. > > > First, it is unlikely that in that case we would conclude that because > > there is a bijection between those two sets that they would have the > > same number of elements. We'd instead claim that bijection does not > > always work as a way to determine if two sets have the same number of > > elements. If set size is not simply defined to be cardinality and is > > defined to relate in any way to counting the elements, we'd therefore > > conclude that bijection is not always a way to determine if two sets > > have the same size. > > > Second, this shows that \number of elements\ does NOT simply mean \ that > > there is a bijection between the two sizes. If it did, the question I > > asked above would be nonsensical; there is no way in which the number > > of elements could POSSIBLY be different by counting it. > > Indeed, this seems to be the case. Except the question above is not nonsensical ... it clearly makes sense, even if it will never actually happen. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b75784@news2.lightlink.com> On 19 Aug, 16:33, Allan C Cybulskie > > > On 19 Aug, 12:04, Allan C Cybulskie > > > > I'm back, after having FAR better things to do than reply to this > > > thread for the past couple of weeks [grin]. > > > > So what I want to do is start with ANOTHER summary post, since things > > > are drifting away from the key points again in the other posts. I'll > > > reply to the other replies later. > > > > Okay, first, the key point: Bijection as a notion of \two sets have > > > the same size\. How is this derived, and how is it justified? > > > Intuition. > > Excellent. Mine conflicts because of the whole \adding and removing\ > thing with infinite sets. But since no one person's intuition is > necessarily better than another's, we would end up at a stalemate ... > which is where I said this always ends up ... That's your key point? Is anyone disagreeing? > > > I claimed in my return to this discussion that bijection get its > > > justification as a METHOD for determining if two sets have the same > > > \number of elements\ (where number of elements is loosely defined \ as > > > what we'd get if we counted the number of elements on finite sets, \ and > > > is up in the air what it would mean for infinite sets for now) \ because > > > it happens to work on finite sets. But it is not simply part of the > > > definition of \number of elements\ or even \set size\; more > > > stipulations need to be made to get it that way. This is because: > > > > Take two finite sets, S and T. They were counted and we know that S > > > has n elements, T has m elements, and n != m. Someone (as I said > > > elsewhere, it will never happen; the bijection always does work for > > > finite sets) proves that there is a bijection between S and T. What > > > would we conclude? > > > That their proof is incorrect, or that the theory they are working in > > is inconsistent. > > Assume that the proof is correct, and no theory other than what exists > now works. What would you conclude? That the theory we are working in has P and ~P as theorems and as such is inconsistent. > > > First, it is unlikely that in that case we would conclude that \ because > > > there is a bijection between those two sets that they would have the > > > same number of elements. We'd instead claim that bijection does not > > > always work as a way to determine if two sets have the same number of > > > elements. If set size is not simply defined to be cardinality and is > > > defined to relate in any way to counting the elements, we'd therefore > > > conclude that bijection is not always a way to determine if two sets > > > have the same size. > > > > Second, this shows that \number of elements\ does NOT simply mean \ that > > > there is a bijection between the two sizes. If it did, the question \ I > > > asked above would be nonsensical; there is no way in which the number > > > of elements could POSSIBLY be different by counting it. > > > Indeed, this seems to be the case. > > Except the question above is not nonsensical ... it clearly makes > sense, even if it will never actually happen. OK, it makes sense. It's just uninteresting (because it is counterfactual). === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > On 19 Aug, 12:04, Allan C Cybulskie > > > > > I'm back, after having FAR better things to do than reply to this > > > thread for the past couple of weeks [grin]. > > > > > So what I want to do is start with ANOTHER summary post, since things > > > are drifting away from the key points again in the other posts. I'll > > > reply to the other replies later. > > > > > Okay, first, the key point: Bijection as a notion of \two sets have > > > the same size\. How is this derived, and how is it justified? > > > > Intuition. > > Excellent. Mine conflicts because of the whole \adding and removing\ > thing with infinite sets. But since no one person's intuition is > necessarily better than another's, we would end up at a stalemate ... > which is where I said this always ends up ... We have a partial ordering of sets by cardinality which satisfies the rules for partial orderings and allows comparison of any two sets (at least in ZZFC or NBG). Until you can come up with a PO which allows comparison of any two sets, it is not a stalemate. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > > The first column has omega elements, but it does > > > > > not have an initial seqment of length omega that has an end. > > > > > The first column has only one initial segment of length omega. > > > > > This initial segment has no end and is not in the bijection > > > > > If the complete first column is not in the bijection with the set \ of > > > > natural numbers (finite sequences of 1's in the lines), then only \ the > > > > finite initial segments of the first column number are in the > > > > bijection with the natural numbers (sequences of 1's in the lines). > > > > Yes. > > > > > This means: > > > > > Either > > > > 1) there are less than omega natural numbers, > > > > No. There are omega finite > > > initial segments of the first column. An infinite set > > > can have all elements finite. > > > So the complete first column is not required to have omega? > > Not if one has in its place the complete set of its finite initial > segments. > > > > > > > or > > > > 2) omega is not represented by the complete first column. > > > > No. Omega is represented by the initial segment that > > > does not have an end. > > > That is the complete first column. So the complete first column is > > required to have omega? > > Not if one has in its place the complete set of its finite initial > segments. The complete set of finite initial segments is absorbed by the set finite initial segments of 1's in the lines. There is no line which corresponds to the infinite segment omega. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > > > > > > > > > > The first column has omega elements, but it does > > > > > > not have an initial seqment of length omega that has an end. > > > > > > The first column has only one initial segment of length omega. > > > > > > This initial segment has no end and is not in the bijection > > > > > > > If the complete first column is not in the bijection with the set \ of > > > > > natural numbers (finite sequences of 1's in the lines), then only \ the > > > > > finite initial segments of the first column number are in the > > > > > bijection with the natural numbers (sequences of 1's in the \ lines). > > > > > > Yes. > > > > > > > This means: > > > > > > > Either > > > > > 1) there are less than omega natural numbers, > > > > > > No. There are omega finite > > > > initial segments of the first column. An infinite set > > > > can have all elements finite. > > > > > So the complete first column is not required to have omega? > > > > Not if one has in its place the complete set of its finite initial > > segments. > > > > > > > > > > > or > > > > > 2) omega is not represented by the complete first column. > > > > > > No. Omega is represented by the initial segment that > > > > does not have an end. > > > > > That is the complete first column. So the complete first column is > > > required to have omega? > > > > Not if one has in its place the complete set of its finite initial > > segments. > > The complete set of finite initial segments is absorbed by the set > finite initial segments of 1's in the lines. Is this supposed to mean something? > There is no line which > corresponds to the infinite segment omega. Quite right, but there is no need of one either. Each line has a first element, and the set of those indexed first elements IS the first column. Nothing more is needed despite WM's delusions to that effect. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > The first column has omega elements, but it does > > > not have an initial seqment of length omega that has an end. > > > The first column has only one initial segment of length omega. > > > This initial segment has no end and is not in the bijection > > > If the complete first column is not in the bijection with the set of > > natural numbers (finite sequences of 1's in the lines), then only the > > finite initial segments of the first column number are in the > > bijection with the natural numbers (sequences of 1's in the lines). > > Each element in the first column corresponds uniquely with the line > which contains it, so the \whole first column\ does biject with the set > of natural numbers as represented by that set of lines. Nevertheless we know that for a complete bijection a line with omega 1's is required. > > Each element of the first column also corresponds uniquely with the > initial segment of that column ending with that element, so that, by > composition, both bijections exist. No. There is nothing in the lines which would correspond to omega. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > The first column has omega elements, but it does > > > > not have an initial seqment of length omega that has an end. > > > > The first column has only one initial segment of length omega. > > > > This initial segment has no end and is not in the bijection > > > > > If the complete first column is not in the bijection with the set of > > > natural numbers (finite sequences of 1's in the lines), then only the > > > finite initial segments of the first column number are in the > > > bijection with the natural numbers (sequences of 1's in the lines). > > > > Each element in the first column corresponds uniquely with the line > > which contains it, so the \whole first column\ does biject with the \ set > > of natural numbers as represented by that set of lines. > > Nevertheless we know that for a complete bijection a line with omega > 1's is required. So that WM is saying that there must be an infinite natural number? Since there is an obvious bijection between finite naturals and lines, and the set of lines must match the set of naturals and must also match the set of row indices, WM is claiming an infinite natural. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b9eade@news2.lightlink.com> <46ba17e4@news2.lightlink.com> > > > > > > > > > > > Can't you read? I am not doing constructive mathematics! > > > > It does not matter. > > > > B is not constructable. This is true in ordinary and > > > in constructive mathematics. > > > > In constructive mathematics B does not exist. > > > > In ordinary mathematics, B exists and is > > > countable, however the > > > diagonal of B is not constructable. > > > This is because there is no list constructable.This is why there is > > no bijection with N. > > No. The fact that you cannot construct a list does not > mean that there is no list [WM: Can't you read? > I am not doing constructive mathematics!]. > There is a list. There is no constructable list. Ok. You say that there is a list, so let me know please where I can find it. Or should it be a ghost list? The bijection between nodes and paths in the binary tree is of same reality. We know that there cannot be more paths than nodes (which are countable) but we cannot construct the complete bijection. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > > > > > > > > > Can't you read? I am not doing constructive mathematics! > > > > > > It does not matter. > > > > > > B is not constructable. This is true in ordinary and > > > > in constructive mathematics. > > > > > > In constructive mathematics B does not exist. > > > > > > In ordinary mathematics, B exists and is > > > > countable, however the > > > > diagonal of B is not constructable. > > > > > This is because there is no list constructable.This is why there is > > > no bijection with N. > > > > No. The fact that you cannot construct a list does not > > mean that there is no list [WM: Can't you read? > > I am not doing constructive mathematics!]. > > There is a list. There is no constructable list. > > Ok. You say that there is a list, so let me know please where I can > find it. Or should it be a ghost list? The bijection between nodes and > paths in the binary tree is of same reality. For finite trees (of more than one node), no such bijection exists. For complete infinite binary trees, no such bijection exists. One wonders what sort of trees WM has in mind for which such bijections do exist. > We know that there cannot > be more paths than nodes (which are countable) but we cannot construct > the complete bijection. What WM claims to know, being unproven, and often disprovable, is not evidence. If one has a complete infinite binary tree at all, then one must have uncountably many paths. WM's only out is to go back to his claim of no infinite sets at all, so that all trees would have to be finite, and the problem of uncountability does not arise. === Subject: Re: PARADISE LOST: Debunking Cantor's theory <46b9eade@news2.lightlink.com> <46ba17e4@news2.lightlink.com> > > > > > > > > > > > Can't you read? I am not doing constructive mathematics! > > > > > It does not matter. > > > > > B is not constructable. This is true in ordinary and > > > > in constructive mathematics. > > > > > In constructive mathematics B does not exist. > > > > > In ordinary mathematics, B exists and is > > > > countable, however the > > > > diagonal of B is not constructable. > > > > This is because there is no list constructable.This is why there is > > > no bijection with N. > > > No. The fact that you cannot construct a list does not > > mean that there is no list [WM: Can't you read? > > I am not doing constructive mathematics!]. > > There is a list. There is no constructable list. > > Ok. You say that there is a list, so let me know please where I can > find it. Take the finite strings. Remove anything that does not define a real number. What is left is the desired list. - William Hughes === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > > > The first column has omega elements, But there are not omega natural numbers, i.e., finite sequences of 1's in the lines. Or do you see a bijection of finite sequences of 1's with omega, i.e., the complete first column? > but it does > > > > > not have an initial seqment of length omega that has an end. No, but one that has no end. > > > > > The first column has only one initial segment of length omega. > > > > > This initial segment has no end and is not in the bijection > > > > > If the complete first column is not in the bijection with the set \ of > > > > natural numbers (finite sequences of 1's in the lines), then only \ the > > > > finite initial segments of the first column number are in the > > > > bijection with the natural numbers (sequences of 1's in the lines). > > > > Yes. > > > > > This means: > > > > > Either > > > > 1) there are less than omega natural numbers, > > > > No. There are omega finite > > > initial segments of the first column. An infinite set > > > can have all elements finite. > > > So the complete first column is not required to have omega? > > The first column does not have an element omega. > Omega is represented by a set of elements of the first > column, the initial segment that does not have an end. If there were omega natural numbers, then the lines were in bijection with the complete first column with omega 1's. The problem gets most transparent by investigating it with the help of set geometry. Geometrical aspects can best be understood by means of vectors. In the matrix M' 11111... 11 111 1111 ... (zeros omitted) the first column C and the first line L can be interpreted as vectors with omega components. C = lim_{n --> oo} (n, 0) L = lim_{n --> oo} (0, n) The diagonal D is then the vectorial sum D = C + L. D = lim_{n --> oo} (n, n) This seems to be a very natural approach. Now replace M' by the matrix M 1 11 111 1111 ... (zeros omitted) which does not contain any line with omega 1's (a strictly increasing sequence cannot assume its limit). If the difference between presence and absence of the line with omega 1's makes any effect (do you think so? or is set theory a bit blurry here?), then the diagonal cannot exist in M because not all of its second (horizontal) indices are available. The first (vertical) indices however must all exist because we explicitly assume the existence of actual infinity in M', and the first column has not been changed when switching from M' to M. So we arrive at the conclusion that the complete infinite set of indices exists in the vertical component of the diagonal but does not exist in the horizontal component of the diagonal. Can that be the case for one and the same diagonal? A possible solution of this problem would be to admit that the actually infinite (which yields the order type omega + 1 after extending it by one element after the non- present end) does not exist in any column either, and hence does not exist anywhere. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > The problem gets most transparent by investigating it with the help of > set geometry. Geometrical aspects can best be understood by means of > vectors. In the matrix M' > > 11111... > 11 > 111 > 1111 > ... > > (zeros omitted) the first column C and the first line L can be > interpreted as vectors with omega components. > C = lim_{n --> oo} (n, 0) > L = lim_{n --> oo} (0, n) > The diagonal D is then the vectorial sum > D = C + L. > D = lim_{n --> oo} (n, n) > > This seems to be a very natural approach. Now replace M' by the matrix > M > > 1 > 11 > 111 > 1111 > ... > > (zeros omitted) which does not contain any line with omega 1's (a > strictly increasing sequence cannot assume its limit). If the > difference between presence and absence of the line with omega 1's > makes any effect (do you think so? or is set theory a bit blurry > here?), then the diagonal cannot exist in M because not all of its > second (horizontal) indices are available. In M' there is one second index for every natural number and no other second indices. In M there is one second index for every natural number and no other second indices. Looks the same to me. - William Hughes === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > The problem gets most transparent by investigating it with the help of > > set geometry. Geometrical aspects can best be understood by means of > > vectors. In the matrix M' > > > 11111... > > 11 > > 111 > > 1111 > > ... > > > (zeros omitted) the first column C and the first line L can be > > interpreted as vectors with omega components. > > C = lim_{n --> oo} (n, 0) > > L = lim_{n --> oo} (0, n) > > The diagonal D is then the vectorial sum > > D = C + L. > > D = lim_{n --> oo} (n, n) > > > This seems to be a very natural approach. Now replace M' by the matrix > > M > > > 1 > > 11 > > 111 > > 1111 > > ... > > > (zeros omitted) which does not contain any line with omega 1's (a > > strictly increasing sequence cannot assume its limit). If the > > difference between presence and absence of the line with omega 1's > > makes any effect (do you think so? or is set theory a bit blurry > > here?), then the diagonal cannot exist in M because not all of its > > second (horizontal) indices are available. > > In M' there is one second index for every natural number > and no other second indices. > > In M there is one second index for every natural number > and no other second indices. > > Looks the same to me. Hence the removal of the line with omega 1's has no influence. This kind of influence is normally only practised by non-existing entities. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > The problem gets most transparent by investigating it with the help \ of > > > set geometry. Geometrical aspects can best be understood by means of > > > vectors. In the matrix M' > > > > 11111... > > > 11 > > > 111 > > > 1111 > > > ... > > > > (zeros omitted) the first column C and the first line L can be > > > interpreted as vectors with omega components. > > > C = lim_{n --> oo} (n, 0) > > > L = lim_{n --> oo} (0, n) > > > The diagonal D is then the vectorial sum > > > D = C + L. > > > D = lim_{n --> oo} (n, n) > > > > This seems to be a very natural approach. Now replace M' by the \ matrix > > > M > > > > 1 > > > 11 > > > 111 > > > 1111 > > > ... > > > > (zeros omitted) which does not contain any line with omega 1's (a > > > strictly increasing sequence cannot assume its limit). If the > > > difference between presence and absence of the line with omega 1's > > > makes any effect (do you think so? or is set theory a bit blurry > > > here?), then the diagonal cannot exist in M because not all of its > > > second (horizontal) indices are available. > > > In M' there is one second index for every natural number > > and no other second indices. > > > In M there is one second index for every natural number > > and no other second indices. > > > Looks the same to me. > > Hence the removal of the line with omega 1's has no influence. Not as far as the existence or not of the diagonal. As the diagonal exists in M', and changing the first line from length omega to length one does not change the indices available, the diagonal exists in M. Alternately, we can note that the only difference between M and M' is that the latter has one's in the first row, where the former has zero's. Howevever, the position, (1,1) which is the only position from the first row that contributes to the diagonal, is 1 in both M and M'. - William Hughes === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > > > > > > > > > > > > The first column has omega elements, > > But there are not omega natural numbers, i.e., finite sequences of 1's > in the lines. There must be omega natural numbers (finite sequences of 1's) if there are to be omega positions in a column. > Or do you see a bijection of finite sequences of 1's > with omega, i.e., the complete first column? Of course we do, the bijection where every finite sequence of 1's matches its own position in that column. > > > but it does > > > > > > not have an initial seqment of length omega that has an end. > > No, but one that has no end. Nor does the sequence of finite lines have an end. > > > > > So the complete first column is not required to have omega? > > > > The first column does not have an element omega. > > Omega is represented by a set of elements of the first > > column, the initial segment that does not have an end. > > > If there were omega natural numbers, then the lines were in bijection > with the complete first column with omega 1's. > The problem gets most transparent by investigating it with the help of > set geometry. Geometrical aspects can best be understood by means of > vectors. In the matrix M' > > 11111... > 11 > 111 > 1111 > ... > > (zeros omitted) the first column C and the first line L can be > interpreted as vectors with omega components. And if one does not omit all those 0's, EVERY line has omega components. > C = lim_{n --> oo} (n, 0) Since that limit has only two members, WM is claiming that a column has only two rows. > L = lim_{n --> oo} (0, n) And a row has only two columns. > The diagonal D is then the vectorial sum WM is trying to add covaraint vectors to contrvariant vectors, which is a no-no. > D = C + L. > D = lim_{n --> oo} (n, n) > > This seems to be a very natural approach. Only for those whose knowledge of vectors and matrices is as deficient as WM's. Now replace M' by the matrix > M > > 1 > 11 > 111 > 1111 > ... > > (zeros omitted) which does not contain any line with omega 1's (a > strictly increasing sequence cannot assume its limit). If the > difference between presence and absence of the line with omega 1's > makes any effect (do you think so? or is set theory a bit blurry > here?), WM is certainly quite blurry here. > then the diagonal cannot exist in M because not all of its > second (horizontal) indices are available. For every first (vertical) index, there is a second (horizontal) index corresponding to the last 1 in that row and the first 1 in that column. So the only row indices that can possibly be unavaliable correespond to column indices that are also not available. > The first (vertical) > indices however must all exist because we explicitly assume the > existence of actual infinity in M', and the first column has not been > changed when switching from M' to M. Those who choose to be blind, like WM, cannot be forced to see. > > So we arrive at the conclusion that the complete infinite set of > indices exists in the vertical component of the diagonal but does not > exist in the horizontal component of the diagonal. WM somehow arrives at conclusions which are directly contradicted by his own evidence. The so called diagonal of WM's last example needs no more than the last element of each row. So unless WM can show that there are missing rows, the diagonal is as well-defined as WM's \matrix\. === Subject: Re: PARADISE LOST: Debunking Cantor's theory > > > > > > > > > > > The first column has omega elements, > > But there are not omega natural numbers, i.e., finite sequences of 1's > in the lines. Yes there are. There are omega initial sequences that have an end. These are exactly the finite sequences of ones in the lines. > Or do you see a bijection of finite sequences of 1's > with omega, i.e., the complete first column? No. But this is irrelevent. The bijection is with the omega initial sequences that have ends. None of theses is the complete first column. (Note there are omega sequences with ends. None of these sequences is of length omega) - William Hughes === Subject: Re: PARADISE LOST: Debunking Cantor's theory >>> Nam D. Nguyen a \.8ecrit : >> >>>> Do you know that FOL can't be used to formalize SR, without an >>>> ambiguity? >>> >>> No, I didn't knew that. I can bet my salary you just made that up >>> (Petti Lounestoo, if still alive, would torn you to shreds). Anyway, >>> it is a grotesque affirmation (you are troling again, right?). >> >> I simply asked a very short and straight forward question. >> >>> You dont even \know\ what means a formalsation of a physical theory, >>> in mathematical terms. >> >> Briefly, state axioms, written in a suitable 1st order languages, >> that you think would express the physical SR's \theory\ (set of >> physical phenomena), then try to see if any theorem of that axiom set >> would not be faithfully modeled by the abstraction of any of the >> phenomena. >> >> Did you know such, before making your assertion above? >> >>> The simplest explanation of difficulties in formalisations could be >>> the wrongness of SR. >> >> It's not true that every physical phenomenon could be expressed, say, >> by 1st order formalization. For instance, could you find a model of >> a 1st order formal theory T, for SR, in which \ Simultaneity(event1,event2) >> be *both* true and false? >> >>> But in fact, SR is one of the best formalized thory. >> >> Precisely, what do you mean by \best formalized\ theory, when the >> formalization and its having models don't seem to rhyme? >> >>> Are you sure you are not confused with quantum gravity? >> >> Are you sure why you asked this question? >> > > I admit I don't fully understand your questions, but they appear to have > sharp points, so it's good not to have you on the \other\ side, Nam. :) What other side? I mean why would you think there are only 2 sides in \ mathematical reasoning? > > Tony === Subject: Windows XP optimization tips Optimize your system http://windowsxpsp2pro.blogspot.com/ === Subject: Re: Windows XP optimization tips format=flowed; reply-type=original > Optimize your system > http:// windowsxpsp2pro. blogspot. com/ > ego at his \look at me, look at me\ blog site. Bozo can't even pick a === Subject: Re: JSH: Factoring integers, more analysis <87eji2x5lz.fsf@phiwumbda.org> > > I am not saying that James is a bot, but even for a bot it would be > > quite an achievment not many AI's would be able to solve problems like > > that. > > Yes, that *is* something! A problem that modern AI cannot easily > solve! Imagine! > > If a bot can't do it, what chance have we? > > -- > \Looking at their behavior I see them endangering not only their own > futures, but that of their families, and now, considering my latest > result, the future of people all over the world.\ -- James S. Harris, > on the shortsightedness of his mathematical critics None? Modern man had three centurys to do it but failed. Seem it took a man configured as a bot to do it, and a bot configured as a man to confirm it. === Subject: PAK Host Online(PHO): Web Hosting, Free Domain Registration \ /Transfer, Website Builder & Web Marketing in Pakistan, Karachi, Lahore, \ Islamabad & Rawalpindi at http://www.pakhostonline.com/ PAKHostOnline.com offered Pakistan No.1 Web Hosting, DOMAIN Registration / Transfer, 99% UP-Time, 24/7 Technical Support at http://www.pakhostonline.com/ === Subject: Re: predicate calculus question On Aug 18, 11:56 pm, \narutocan...@gmail.com\ > hi > > I'm studying proof systems, I'm at the point where the book > \Hirst and Hirst's A Primer for Logic and Proof\ are about to finish > with predicate calculus and entering into first order logic theories. > I like to clear some qeustions before moving on. I've also looked at > several web pages about the refutation and resolution questions on > \logically valid\ formulas but still can't positively get the > conclusions nailed down. (I'm definitely not talking about model > specific formulas but only \logically valid\ formulas). > > questions: > 1. I know the resolution question for \logically valid\ formulas is > answered in the positive (from several sources), but is it positive > \regardless of the number of already proved formulas\ and \regardless > of the resolution strategy\ and \only take finite number of steps\ ? > 2. I know at least one source claimed that the refutation might take > forever (this is really vague). Is it forever (or impossible) due to > the lack of already proved forumula or forever (or impossible) due to > resolution strategy? > Are you asking about formulas of propositional (Boolean) calculus, or of the predicate calculus (also including quantifiers)? There is a decision procedure for propositional logic, e.g. truth tables, while there is none for the logic of the predicate calculus. Goedel proved the completeness of the predicate calculus, namely that the theorems of the predicate calculus are precisely those statements which are valid in every model (of the prescribed formal language). So, as Dr. Ullrich points out, if you check all possible proofs in the predicate calculus for a proof of P, then when P is logically valid, one will (eventually, without effective bound on the number of steps required) find a proof of P. This corresponds to resolution or a system of natural deduction, in which one begins by assuming ~P and continues to infer deductions. If P is logically valid, then ~P is logically invalid and eventually an inconsistency/contradiction will appear in the stream of deductions. On the other hand if P is not logically (universally) valid as a formula of the predicate calculus, Goedel's completeness theorem tells us that there is a model in which P is not valid. Unfortunately such a model may be infinite. In this sense a refutation (showing a model in which the formula is not valid) \might take forever\. Here's an illustration. Let P be the statement that if \<\ is a partial order, then there exists an element maximal with respect to \<\. The statement is valid in every finite model. Therefore a refutation of P would require an infinite model. === Subject: Re: predicate calculus question On Sat, 18 Aug 2007 20:56:04 -0700, \narutocanada@gmail.com\ >hi > >I'm studying proof systems, I'm at the point where the book >\Hirst and Hirst's A Primer for Logic and Proof\ are about to finish >with predicate calculus and entering into first order logic theories. >I like to clear some qeustions before moving on. I've also looked at >several web pages about the refutation and resolution questions on >\logically valid\ formulas but still can't positively get the >conclusions nailed down. (I'm definitely not talking about model >specific formulas but only \logically valid\ formulas). > >questions: >1. I know the resolution question for \logically valid\ formulas is >answered in the positive (from several sources), but is it positive >\regardless of the number of already proved formulas\ and \regardless >of the resolution strategy\ and \only take finite number of steps\ ? >2. I know at least one source claimed that the refutation might take >forever (this is really vague). Is it forever (or impossible) due to >the lack of already proved forumula or forever (or impossible) due to >resolution strategy? Godel proved that there is no algorithm which will decide for every formula P whether or not P is valid. Here an \algorithm\ is by definition a procedure that's guaranteed to terminate in finitely many steps. \Look for a proof of P by enumerating all proofs in the system; if you find a proof of P then P is valid, if you don't find a proof of P then P is not valid\ works but it can take infinitely many steps: If P is not valid you will never find a proof of P. You may never find a proof of \not P\ either... ************************ David C. Ullrich === Subject: Re: predicate calculus question > On Sat, 18 Aug 2007 20:56:04 -0700, \narutocanada@gmail.com\ > > >hi > > > >I'm studying proof systems, I'm at the point where the book > >\Hirst and Hirst's A Primer for Logic and Proof\ are about to finish > >with predicate calculus and entering into first order logic theories. > >I like to clear some qeustions before moving on. I've also looked at > >several web pages about the refutation and resolution questions on > >\logically valid\ formulas but still can't positively get the > >conclusions nailed down. (I'm definitely not talking about model > >specific formulas but only \logically valid\ formulas). > > > >questions: > >1. I know the resolution question for \logically valid\ formulas is > >answered in the positive (from several sources), but is it positive > >\regardless of the number of already proved formulas\ and \regardless > >of the resolution strategy\ and \only take finite number of steps\ ? > >2. I know at least one source claimed that the refutation might take > >forever (this is really vague). Is it forever (or impossible) due to > >the lack of already proved forumula or forever (or impossible) due to > >resolution strategy? > > Godel proved that there is no algorithm which will decide for > every formula P whether or not P is valid. Here an \algorithm\ > is by definition a procedure that's guaranteed to terminate > in finitely many steps. > > \Look for a proof of P by enumerating all proofs in the system; > if you find a proof of P then P is valid, if you don't find > a proof of P then P is not valid\ works but it can take > infinitely many steps: If P is not valid you will never find > a proof of P. You may never find a proof of \not P\ either... There is always some more to prove and more code to write. > > > > ************************ > > David C. Ullrich === Subject: MI5 Persecution: Fitted up 26/4/96 (791) === Subject: Re: MI5? Please can someone explain what's going on here? Distribution: : >The (remote) possibility remains that 'Mike Corley' is either : >not schizophrenic (but is 'pretending' to be so) or 'he' is : >a product of a number of persons (?psychology students). : Given other ways in which I have seen people exploit some of The \ Internet's : capabilities to disrupt or indulge in sophistry, or to exploit a medium : that resembles speech without the non-verbal and intonation cues, etc : as a means of denigrating others, I question your use, albeit in quotes, : of the word \remote\. I'm not saying it isn't remote and therefore it \ is : great, I'm just saying that I don't think we can easily classify it as : remote, moderate, or great. I think you can build up quite a good picture based on what someone says and on their posting patterns. I don't think \The Internet\ (capitals, no less) is as opaque a medium as you make it out to be. : It is not easy to determine the validity of all information on The : Internet without making use of extra supplementary information. : We do have the problem, pointed out by someone else, of the possibly : \too perfect\ textbook characteristics of what is being posted. I explained that one, but I don't mind explaining it again (you don't mind having it explained again to you, do you now?). The reason my \symptoms\ are such a perfect fit to the textbook is because the people causing the campaign \fitted me up\ in such a way that what they did would resemble the symptoms of schizophrenia. Hence TV, radio, other media, people in the streets etc. By a fortunate coincidence (for them) these mthods of harassment are the ones which offer easiest channels of access (for them). It's really quite neat. All it takes is for people to start believing that the \symptoms\ aren't symptoms but reality, though, and the house of cards collapses in a heap. And there are _lots_ of people now who knoiw full well what has gone on. : If harrassment by email, etc, has happened by someone out of the country, : can a complaint be made that results in arrest or whatever upon that : person's entry into the country? An interesting point which Mike may be : able to inform us about, as he's said he will be in the UK in a few weeks : time. Picture the scene at the airport; \I arrest you for being Mike Corley and mailbombing people\ \But my name isn't Corley. Who he? Mailbombing isn't illegal is it? You'd have to lock up a lot of people if sending annoying email was a crime\ \Er.....\ : -- : David Stretch: Greenwood Institute of Child Health, Univ. of Leicester, \ UK. : dds@leicester.ac.uk Phone:+44 (0)116-254-6100 Fax:+44 \ (0)116-254-4127 : assume they are meant for you. Mike, this is called paranoia. But that's the way real abuse works, too. People interject words and phrases into what they say which they know will have meaning for the \ listener. And sometimes, they make it obvious. The very first evening of my job in Oxford, we went for a drink with the technical director, and a couple of other employees. The TD said in an \as-if\ aside to one of the others, \Is this the bloke who's been on TV?\ (he said it directly in front of me, and obviously meant mke to hear him saying it). The other person replied, \Yes, I think so\. I think the subtext of what the TD said was \Why are they bothering with him? He's so insignificant, why would they possibly want to spend the resources going after him and putting all that expensive technology in his home, when there must be much better targets?\. The Technical Director was given to sometimes disrespecting people, you see, and in my case he couldn't see the point of anyone expending money on harassing me. === Subject: Re: Treatment of Schizophrenia Distribution: : Probably 'cos you come across as reasoned & articulate, it's a pity : about the other stuff :) Veracity is so unreasonable. : >>pps. You should still see a doc again Mike. : > : >Doing so. Trouble is, all this mental-illness stuff provides camouflage : >for the harassment, which is real. It alows people who otherwise would : >consider the harassment seriously to disregard it. It makes \ conversations : >with a lawyer or police brief when otherwise it would merit discussion. : The point is that there are two possibilities happening here- : 1. There's a large conspiracy of people out to get you, for no : other reason than that they have the means to do so, and that : it involves a lot of the Media & a proportion of the public : 2. You (who admit to having some headspace problems) are suffering : from acute paranoid schizophrenia. : Possibility #1 is _possible_, but would be unprecendented (OTOH, : how would we know?), unfeasible, and many other things beginning : with _un_ which I can't think of at the moment. Besides, if there : was something going on, chances are some of us here would know : about it, and I'm convinced that nobody does. \Unprecedented\ hits the nail on the head. It _is_ unprecedented, but we have only just reached the technical stage at which it is feasible, and we know video-spying is done to other people (NB the Diana-Hewitt episode) and is a routine tool of security agencies. Perhaps what is unprecedented is not the technical side, but the social manipulation of many people by a concealed element in what other countries would be called the secret police. The most disturbing element is the degree to which people allow themselves to be unquestioningly manipulated by an evil element within the state. 791 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: equality axioms hi Why the equality axioms only use implication? (Reflexivity of equality) for-all x (x=x) (Substitutivity of equality) x=y => A(x,x) => A(x,y) why not: x=y <=> A(x,x) <=> A(x,y) === Subject: Re: equality axioms On Sun, 19 Aug 2007 10:24:02 -0700, \narutocanada@gmail.com\ > > Why the equality axioms only use implication? > Why not? (Hint: In most logical systems implication is more primitive than equivalence. Actually usually we have: A <-> B =df A -> B & B -> A.) > > (Reflexivity of equality) for-all x (x=x) > (Substitutivity of equality) for-all x for-all y [x=y => (A(x,x) => \ A(x,y))] > > why not: > for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))] > (*) (**) (**) Would be possible (though not necessary), but see comment from above. (*) would be \wrong\. For example, from your formula we would get for x := 1, y := 2, and A(x) := x > 0 1 = 2 <=> (1 > 0 <=> 2 > 0). Now 1 > 0 holds and 2 > 0 holds (in a system comprising arithmetic). Hence 1 > 0 <=> 2 > 0 would hold. This means we would get 1 = 2. Not good. F. -- E-mail: infosimple-linede === Subject: Re: equality axioms > On Sun, 19 Aug 2007 10:24:02 -0700, \narutocanada@gmail.com\ > > > > > Why the equality axioms only use implication? > > > Why not? > > (Hint: In most logical systems implication is more primitive than > equivalence. Actually usually we have: A <-> B =df A -> B & B -> A.) > > > > > (Reflexivity of equality) for-all x (x=x) > > (Substitutivity of equality) for-all x for-all y [x=y => (A(x,x) => \ A(x,y))] > > > > why not: > > for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))] > > (*) (**) > > (**) Would be possible (though not necessary), but see comment from > above. > > (*) would be \wrong\. > > For example, from your formula we would get for x := 1, y := 2, and > A(x) := x > 0 > > 1 = 2 <=> (1 > 0 <=> 2 > 0). > > Now 1 > 0 holds and 2 > 0 holds (in a system comprising arithmetic). > Hence 1 > 0 <=> 2 > 0 would hold. This means we would get > > 1 = 2. > > Not good. Did you make a mistake? \for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))]\ if A(x,y) = x>y and x=1 y=2 we have 1=2 <=> (1>1 <=> 1>2) false <=> (false <=> false) so the logic still holds. I can't think of a case where implication would be more useful then equivalence. > > > F. > > -- > > E-mail: infosimple-linede === Subject: Re: equality axioms On Sun, 19 Aug 2007 14:58:48 -0700, \narutocanada@gmail.com\ >>> >>> Why the equality axioms only use implication? >>> >> Why not? >> Note: In most logical systems implication is \more primitive\ than equivalence. Actually usually we have: A <-> B =df A -> B & B -> A.) >>> >>> (Reflexivity of equality) for-all x (x=x) >>> (Substitutivity of equality) for-all x for-all y [x=y => (A(x,x) => \ A(x,y))] >>> >>> why not: >>> for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))] >>> (*) (**) >> >> (**) Would be possible (though not necessary), but see comment from >> above. >> >> (*) would be \wrong\. >> >> For example, from your formula we would get for x := 1, y := 2, and >> A(x) := x > 0 >> >> 1 = 2 <=> (1 > 0 <=> 2 > 0). >> >> Now 1 > 0 holds and 2 > 0 holds (in a system comprising arithmetic). >> Hence 1 > 0 <=> 2 > 0 would hold. This means we would get >> >> 1 = 2. >> >> Not good. > > Did you make a mistake? > I hope not! > > \for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))]\ > Here A refers to a wff of FOPL. Maybe it's helpful to write the axiom schema in the following way: AxAy(x = y <-> (A[x] <-> A[y])). Now we take \z > 0\ for A[z]. Hence we get (as an special axiom): AxAy(x = y <-> (x > 0 <-> y > 0)). Now specialization (x := 1 and y := 2) gives: 1 = 2 <-> (1 > 0 <-> 2 > 0) as \desired\. Now \1 > 0 <-> 2 > 0\ is true (since \1 > 0\ is true and \2 > 0\ is true), but \1 = 2\ is false (at least in standard arithmetic). Hence \1 = 2 <-> (1 > 0 <-> 2 > 0)\ is false. > > I can't think of a case where implication would be more useful then > equivalence. > That's not of relevance from a logical point of view. :-) Actually, implication may be considered a fundamental logical \relation\. Very often (especially in math) we have A -> B but NOT A <-> B. So how would _you_ express A -> B with <-> and, say, negation? F. -- E-mail: infosimple-linede === Subject: Re: equality axioms >> >> I can't think of a case where implication would be more useful then >> equivalence. >> > That's not of relevance from a logical point of view. ;-) > > Actually, implication may be considered a fundamental logical > \relation\. > For example, we may define <-> by using -> (and &) the following way: A <-> B =df A -> B & B -> A. Of course /equivalence/ (as such) is an important notion in mathematics and logic. Incidentally, by adopting <-> we may formulate a _single_ axiom schema for identity theory: For all y: A[y] <-> Ex(x = y & A[x]). (Mentioned in one of Quine's books.) Still the usual axioms (i.e. one axiom and one axiom schema) for identity theory (in the context of FOPL) are: A1 Ax(x = x) A2 AxAy(x = y -> (A[x] -> A[y])). F. -- E-mail: infosimple-linede === Subject: Re: equality axioms <4kfhc31o0a2pl47kre1telaa4cmuetsm32@4ax.com> > On Sun, 19 Aug 2007 14:58:48 -0700, \narutocanada@gmail.com\ > > >>> > >>> Why the equality axioms only use implication? > >>> > >> Why not? > >> > > Note: In most logical systems implication is \more primitive\ than > equivalence. Actually usually we have: A <-> B =df A -> B & B -> A.) > > >>> > >>> (Reflexivity of equality) for-all x (x=x) > >>> (Substitutivity of equality) for-all x for-all y [x=y => (A(x,x) => \ A(x,y))] > >>> > >>> why not: > >>> for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))] > >>> (*) (**) > >> > >> (**) Would be possible (though not necessary), but see comment from > >> above. > >> > >> (*) would be \wrong\. > >> > >> For example, from your formula we would get for x := 1, y := 2, and > >> A(x) := x > 0 > >> > >> 1 = 2 <=> (1 > 0 <=> 2 > 0). > >> > >> Now 1 > 0 holds and 2 > 0 holds (in a system comprising arithmetic). > >> Hence 1 > 0 <=> 2 > 0 would hold. This means we would get > >> > >> 1 = 2. > >> > >> Not good. > > > > Did you make a mistake? > > > I hope not! > > > > > \for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))]\ > > > Here A refers to a wff of FOPL. Maybe it's helpful to write the axiom > schema in the following way: > > AxAy(x = y <-> (A[x] <-> A[y])). > > Now we take \z > 0\ for A[z]. Hence we get (as an special axiom): > > AxAy(x = y <-> (x > 0 <-> y > 0)). > > Now specialization (x := 1 and y := 2) gives: > > 1 = 2 <-> (1 > 0 <-> 2 > 0) > > as \desired\. > > Now \1 > 0 <-> 2 > 0\ is true (since \1 > 0\ is true and \2 > 0\ is > true), but \1 = 2\ is false (at least in standard arithmetic). Hence > \1 = 2 <-> (1 > 0 <-> 2 > 0)\ is false. > > > > > I can't think of a case where implication would be more useful then > > equivalence. > > > That's not of relevance from a logical point of view. :-) > > Actually, implication may be considered a fundamental logical > \relation\. > > Very often (especially in math) we have > > A -> B > > but NOT > > A <-> B. > > So how would _you_ express A -> B with <-> and, say, negation? You are right it should be \for-all x for-all y [x=y (only implies) => (A(x,x) <=> A(x,y))]\ 1. \for-all x for-all y [x=y (only implies) => (A(x,x) <=> A(x,y))]\ 2. \for-all x for-all y [x=y (only implies) => (A(x,x) => A(x,y))]\ But I can still make a case why I might want 1 instead of 2. For example, for refutation purpose? (to detect wrong statements right away) If you use 2, instead of 1, wouldn't the proof just \sound right\ but not \tight\? Maybe there are cases where both might be useful. I just can't think of one for case 2, right now. > > > F. > > -- > > E-mail: infosimple-linede === Subject: Re: equality axioms On Sun, 19 Aug 2007 15:56:42 -0700, \narutocanada@gmail.com\ > > You are right it should be > \for-all x for-all y [x=y => (A(x,x) <=> A(x,y))]\ > Right. > > 1. \for-all x for-all y [x=y => (A(x,x) <=> A(x,y))]\ > 2. \for-all x for-all y [x=y => (A(x,x) => A(x,y))]\ > > But I can still make a case why I might want 1 instead of 2. > Well, you certainly MAY adopt 1 instead of 2. (It's just not the axiom schema usually adopted for \identity theory\.) Note that a possible variant of 2 is AxAy(x = y & A[x] -> A[y]). (Seen in one of Quine's books.) x = y A[x] ------------- (Subst. of identity) A[y] Maybe this might help you to see the \logical relevance\ of the \standard formulation\ of the axiom (schema). > > For example, for refutation purpose? (To detect wrong statements right > away). > > If you use 2, instead of 1, wouldn't the proof just \sound right\ but > not \tight\? > Example?! > > Maybe there are cases where both might be useful. > I just can't think of one for case 2, right now. > ??? Case 2 is the \standard formulation\ of the axiom (schema), i.e. it suffices for \identity theory\ (together with Ax(x = x)). F. -- E-mail: infosimple-linede === Subject: Re: equality axioms <4kfhc31o0a2pl47kre1telaa4cmuetsm32@4ax.com> > > > On Sun, 19 Aug 2007 14:58:48 -0700, \narutocanada@gmail.com\ > > > > >>> > > >>> Why the equality axioms only use implication? > > >>> > > >> Why not? > > >> > > > > Note: In most logical systems implication is \more primitive\ than > > equivalence. Actually usually we have: A <-> B =df A -> B & B -> A.) > > > > >>> > > >>> (Reflexivity of equality) for-all x (x=x) > > >>> (Substitutivity of equality) for-all x for-all y [x=y => (A(x,x) => \ A(x,y))] > > >>> > > >>> why not: > > >>> for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))] > > >>> (*) (**) > > >> > > >> (**) Would be possible (though not necessary), but see comment from > > >> above. > > >> > > >> (*) would be \wrong\. > > >> > > >> For example, from your formula we would get for x := 1, y := 2, and > > >> A(x) := x > 0 > > >> > > >> 1 = 2 <=> (1 > 0 <=> 2 > 0). > > >> > > >> Now 1 > 0 holds and 2 > 0 holds (in a system comprising arithmetic). > > >> Hence 1 > 0 <=> 2 > 0 would hold. This means we would get > > >> > > >> 1 = 2. > > >> > > >> Not good. > > > > > > Did you make a mistake? > > > > > I hope not! > > > > > > > > \for-all x for-all y [x=y <=> (A(x,x) <=> A(x,y))]\ > > > > > Here A refers to a wff of FOPL. Maybe it's helpful to write the axiom > > schema in the following way: > > > > AxAy(x = y <-> (A[x] <-> A[y])). > > > > Now we take \z > 0\ for A[z]. Hence we get (as an special axiom): > > > > AxAy(x = y <-> (x > 0 <-> y > 0)). > > > > Now specialization (x := 1 and y := 2) gives: > > > > 1 = 2 <-> (1 > 0 <-> 2 > 0) > > > > as \desired\. > > > > Now \1 > 0 <-> 2 > 0\ is true (since \1 > 0\ is true and \2 > 0\ \ is > > true), but \1 = 2\ is false (at least in standard arithmetic). Hence > > \1 = 2 <-> (1 > 0 <-> 2 > 0)\ is false. > > > > > > > > I can't think of a case where implication would be more useful then > > > equivalence. > > > > > That's not of relevance from a logical point of view. :-) > > > > Actually, implication may be considered a fundamental logical > > \relation\. > > > > Very often (especially in math) we have > > > > A -> B > > > > but NOT > > > > A <-> B. > > > > So how would _you_ express A -> B with <-> and, say, negation? > > You are right it should be > \for-all x for-all y [x=y (only implies) => (A(x,x) <=> A(x,y))]\ > > 1. \for-all x for-all y [x=y (only implies) => (A(x,x) <=> A(x,y))]\ > 2. \for-all x for-all y [x=y (only implies) => (A(x,x) => A(x,y))]\ > > But I can still make a case why I might want 1 instead of 2. > For example, for refutation purpose? (to detect wrong statements right > away) > > If you use 2, instead of 1, wouldn't the proof just \sound right\ but > not \tight\? > Maybe there are cases where both might be useful. > I just can't think of one for case 2, right now. Well, since I can swap x and y, the case 1 and 2 actaully means the same thing. stupid me. I also find a rule: (model specific axiom => predicate calculus axiom) (predicate calculus axiom never => model specific axiom) > > > > > > > F. > > > > -- > > > > E-mail: infosimple-linede === Subject: Re: equality axioms On Sun, 19 Aug 2007 16:16:35 -0700, \narutocanada@gmail.com\ >> >> 1. \for-all x for-all y [x=y => (A(x,x) <=> A(x,y))]\ >> 2. \for-all x for-all y [x=y => (A(x,x) => A(x,y))]\ >> >> [...] >> > Well, since I can swap x and y, the case 1 and 2 actually means the > same thing. > Right (\in effect\ they mean the same thing). You may take formula 1 or 2 as an axiom schema for your identity theory. But -as already mentioned- the second one is usually preferred (for certain reasons). > > I also find a rule [...] > Great. ;-) F. -- E-mail: infosimple-linede === Subject: Latest models of Gibson guitars Reviews of latest models of best guitars, fender, gibson, yamaha, and many more, with pictures and prices. http://pro-guitars.blogspot.com/ And if you want to win a free guitar go here http://freeguitars.blogspot.com/ === Subject: Car Air Conditioners! Everything you need to know about car air conditioners... http://car-air-conditioning.blogspot.com/ === Subject: MI5 Persecution: Fitted up 26/4/96 (791) === Subject: Re: MI5? Please can someone explain what's going on here? Distribution: : >The (remote) possibility remains that 'Mike Corley' is either : >not schizophrenic (but is 'pretending' to be so) or 'he' is : >a product of a number of persons (?psychology students). : Given other ways in which I have seen people exploit some of The \ Internet's : capabilities to disrupt or indulge in sophistry, or to exploit a medium : that resembles speech without the non-verbal and intonation cues, etc : as a means of denigrating others, I question your use, albeit in quotes, : of the word \remote\. I'm not saying it isn't remote and therefore it \ is : great, I'm just saying that I don't think we can easily classify it as : remote, moderate, or great. I think you can build up quite a good picture based on what someone says and on their posting patterns. I don't think \The Internet\ (capitals, no less) is as opaque a medium as you make it out to be. : It is not easy to determine the validity of all information on The : Internet without making use of extra supplementary information. : We do have the problem, pointed out by someone else, of the possibly : \too perfect\ textbook characteristics of what is being posted. I explained that one, but I don't mind explaining it again (you don't mind having it explained again to you, do you now?). The reason my \symptoms\ are such a perfect fit to the textbook is because the people causing the campaign \fitted me up\ in such a way that what they did would resemble the symptoms of schizophrenia. Hence TV, radio, other media, people in the streets etc. By a fortunate coincidence (for them) these mthods of harassment are the ones which offer easiest channels of access (for them). It's really quite neat. All it takes is for people to start believing that the \symptoms\ aren't symptoms but reality, though, and the house of cards collapses in a heap. And there are _lots_ of people now who knoiw full well what has gone on. : If harrassment by email, etc, has happened by someone out of the country, : can a complaint be made that results in arrest or whatever upon that : person's entry into the country? An interesting point which Mike may be : able to inform us about, as he's said he will be in the UK in a few weeks : time. Picture the scene at the airport; \I arrest you for being Mike Corley and mailbombing people\ \But my name isn't Corley. Who he? Mailbombing isn't illegal is it? You'd have to lock up a lot of people if sending annoying email was a crime\ \Er.....\ : -- : David Stretch: Greenwood Institute of Child Health, Univ. of Leicester, \ UK. : dds@leicester.ac.uk Phone:+44 (0)116-254-6100 Fax:+44 \ (0)116-254-4127 : assume they are meant for you. Mike, this is called paranoia. But that's the way real abuse works, too. People interject words and phrases into what they say which they know will have meaning for the \ listener. And sometimes, they make it obvious. The very first evening of my job in Oxford, we went for a drink with the technical director, and a couple of other employees. The TD said in an \as-if\ aside to one of the others, \Is this the bloke who's been on TV?\ (he said it directly in front of me, and obviously meant mke to hear him saying it). The other person replied, \Yes, I think so\. I think the subtext of what the TD said was \Why are they bothering with him? He's so insignificant, why would they possibly want to spend the resources going after him and putting all that expensive technology in his home, when there must be much better targets?\. The Technical Director was given to sometimes disrespecting people, you see, and in my case he couldn't see the point of anyone expending money on harassing me. === Subject: Re: Treatment of Schizophrenia Distribution: : Probably 'cos you come across as reasoned & articulate, it's a pity : about the other stuff :) Veracity is so unreasonable. : >>pps. You should still see a doc again Mike. : > : >Doing so. Trouble is, all this mental-illness stuff provides camouflage : >for the harassment, which is real. It alows people who otherwise would : >consider the harassment seriously to disregard it. It makes \ conversations : >with a lawyer or police brief when otherwise it would merit discussion. : The point is that there are two possibilities happening here- : 1. There's a large conspiracy of people out to get you, for no : other reason than that they have the means to do so, and that : it involves a lot of the Media & a proportion of the public : 2. You (who admit to having some headspace problems) are suffering : from acute paranoid schizophrenia. : Possibility #1 is _possible_, but would be unprecendented (OTOH, : how would we know?), unfeasible, and many other things beginning : with _un_ which I can't think of at the moment. Besides, if there : was something going on, chances are some of us here would know : about it, and I'm convinced that nobody does. \Unprecedented\ hits the nail on the head. It _is_ unprecedented, but we have only just reached the technical stage at which it is feasible, and we know video-spying is done to other people (NB the Diana-Hewitt episode) and is a routine tool of security agencies. Perhaps what is unprecedented is not the technical side, but the social manipulation of many people by a concealed element in what other countries would be called the secret police. The most disturbing element is the degree to which people allow themselves to be unquestioningly manipulated by an evil element within the state. 791 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Open Convex Sets do you know a proof of the following quite evident fact? Let A be an open convex subset of R^n. Then (I) if A is bounded, then it is homeomorphic to an open ball of R^n, and its boundary is homemorphic to S^(n-1); (II) if A is not bounded, it is homeomorphic to R^n, and its boundary, if it is not-empty, is homeomorphic to R^(n-1). Maury === Subject: Re: Open Convex Sets <8232966.1187547077087.JavaMail.jakarta@nitrogen.mathforum.org>, > do you know a proof of the following quite evident fact? > Let A be an open convex subset of R^n. Then > > (I) if A is bounded, then it is homeomorphic to an open > ball of R^n, and its boundary is homemorphic to S^(n-1); Take any point p in A. For any q in S^(n-1), the ray p + rq, r in [0,oo), intersects the boundary of A in a unique point that is at distance d(q) from p. Define a map of {|x| <= 1} into cl(A) by setting h(x) = p + d(x/|x|)x for x nonzero, h(0) = p. That will give you the desired homeomorphisms. > (II) if A is not bounded, it is homeomorphic to R^n, > and its boundary, if it is not-empty, is homeomorphic > to R^(n-1). > > Maury === Subject: Re: Open Convex Sets On Sun, 19 Aug 2007 14:10:46 EDT, Maury Barbato >do you know a proof of the following quite evident fact? >Let A be an open convex subset of R^n. Then > >(I) if A is bounded, then it is homeomorphic to an open >ball of R^n, and its boundary is homemorphic to S^(n-1); > >(II) if A is not bounded, it is homeomorphic to R^n, >and its boundary, if it is not-empty, is homeomorphic >to R^(n-1). For case (II), there's another possibility -- the boundary could be homeomorphic to a disjoint union of two copies of R^(n-1). quasi === Subject: Re: Open Convex Sets <0j2hc3l6p0br8akrv57uhc09n0h2cjpavg@4ax.com> > On Sun, 19 Aug 2007 14:10:46 EDT, Maury Barbato > > >do you know a proof of the following quite evident fact? > >Let A be an open convex subset of R^n. Then > > >(I) if A is bounded, then it is homeomorphic to an open > >ball of R^n, and its boundary is homemorphic to S^(n-1); > > >(II) if A is not bounded, it is homeomorphic to R^n, > >and its boundary, if it is not-empty, is homeomorphic > >to R^(n-1). > > For case (II), there's another possibility -- the boundary could be > homeomorphic to a disjoint union of two copies of R^(n-1). Or more generally, S^m x R^k with m+k=n-1. (E.g. A is an infinite cylinder.) Any more possibilities? === Subject: Re: Open Convex Sets > > On Sun, 19 Aug 2007 14:10:46 EDT, Maury Barbato > > > > >do you know a proof of the following quite evident > fact? > > >Let A be an open convex subset of R^n. Then > > > > >(I) if A is bounded, then it is homeomorphic to an > open > > >ball of R^n, and its boundary is homemorphic to > S^(n-1); > > > > >(II) if A is not bounded, it is homeomorphic to > R^n, > > >and its boundary, if it is not-empty, is > homeomorphic > > >to R^(n-1). > > > > For case (II), there's another possibility -- the > boundary could be > > homeomorphic to a disjoint union of two copies of > R^(n-1). > Oh, yes! Besides, I think that if the boundary of A is homeomorphic to a disjoint union of two copies of R^(n-1), then it is exactly a disjoint union of two copies of R^(n-1)!! > Or more generally, S^m x R^k with m+k=n-1. (E.g. A > is an infinite > cylinder.) Any more possibilities? > Ops ... I failed to identify these other cases. They seem to exhaust all the possibilities. === Subject: Re: Truth Among Mathematikers and Empirics >> > Putting assertions in the form >> >of a question is one of his techniques for avoiding that frightening >> >possibility. In this case, it's an extra flourish, since the assertion >> >would make no sense even if it were put in the form of a declarative >> >sentence. >> >> Now, now, Schlock. Be polite. You've already admitted \not\ forms a >> perfectly sensible declarative sentence prefixed by some variation of >> \it is\ as in \Rachel's children were not\. Dissemble thou not! > >Yes, I also \admitted\ that sodium affixed to chlorine makes salt, >and pointed out that chlorine is not salt. A quaint aphorism indeed. I believe Brian is of a similar persuasion only he prefers not to speak whereof he knows not. > As I recall \ you >dismissed that as some sort of \analogical\ reasoning. I dismissed it as too stupid for words. If you require assistance in evasion I suggest you try PD. He isn't quite so stupid as yourself although on a scale of 1 to 10 he's pretty stupid. But you can feel quite at home since it's invariably the monumentally stupid who develop the greatest expertise at evasion. ~v~~ === Subject: Re: Truth Among Mathematikers and Empirics > >> And just how then, pray tell, can >> alternation be defined without the use of negation? > > A v B =df (A -> B) -> B > >You're welcome, dumbass. So I ask about \alternation\ and you prefer to discuss \A v B\? Just for the record I also asked: >\Wherein Schlock the Divine imagines he can divine the meaning of all >logical operations without use of negation. Would Schlock the Divine >care to wax a little less vague and a little more specific?\ In response to which I imagine Schlock the Divine would have preferred to discuss \A /\\= \\/ B <-> MT\ except he couldn't quite bear to subject himself to further ridicule. See, Schlock, the problem isn't that you don't talk a lot. It's that when you do talk you have nothing of probative significance to say. You say something and when asked to demonstrate the truth of what you say you just say something else, which demonstrates the truth of nothing except your ability to talk. Let's face it all dialectics do the same. In trying to demonstrate the truth of what they say dialectics all suffer from acute logorrhea and just run off at the mouth. Look at Bob, Randy, Stephen, Brian, PD, and other assorted mathematikers and empirics, besides you. This is why they want to believe definitions are only abbreviations because they realize they're much too vulnerable when they talk, get tired of hearing all the nonsense anyway, and prefer to shorten what they have to say as much as possible. It's ridiculous I know but who wouldn't? ~v~~ === Subject: Re: Conditional Expectation. Uniform distribution. > > > > > > > I am having a hard time tackling the following problem. > > > Can anybody help me? > > > > Let U be a uniform(0,1) random variable. Suppose that n trials are to > > > be performed and that conditional on U=u these trials will be > > > independent with a common success probability u. Compute the mean and > > > variance of the number of successes that occur in these trials. > > > > Well, the mean should be easy. Why don't you show us what you have > > done so far? > > > The variance is trickier, but can always be done using the original > > definition Var(X) = E(X-EX)^2, or the equivalent Var(X) = E(X^2) - > > (EX)^2. Again, if you show us what you have done so far, we can maybe > > help you over your difficulties. > > > R.G. Vickson > > Let U be a Uniform(0,1) random variable. Suppose that n trials are to > be performed and that conditional on U=u these trials will be > independent with a common success probability u. Compute the mean and > variance of the number of successes that occur in these trials. > > Let Z be the bernoulli variable that is 1 when a trial is a success. > Let T be the number of success. > > Here is what I have for you probawannabe > > P{Z=1|U=u}=u > > P{Z=0|U=u}=1-u > > T|U~Binomial(n,u) > > ET=E{E[T|U]}=E{nu}=n/2 > > VarT=Var{E[T|U]}+E{Var[T|U]}=Var{nu}+E{nu(1-u)} > =n^2/12+n/2-n(n^2/12+1/4) Your answer cannot be right, as it contains a term -n^3/12, which for large n will swamp the other terms and give Var < 0. I think your basic method is OK, but you have made some computational errors. Using Var(T) = E(T^2) - (ET)^2, I get Var(T) = n/6 + n^2/12. R.G. Vickson === Subject: How to test a pseudo random prime number generator? There a a variety of techniques for testing the quality of random number generators (DIEHARD, NIST, Knuth, ect.). How can I test the quality of a \random prime number generator\? Rich === Subject: Re: How to test a pseudo random prime number generator? >There a a variety of techniques for testing the quality of random >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >quality of a \random prime number generator\? First, you have to define what you mean by a random prime. There has to be an underlying distribution, and it has to be _declared_, otherwise there's nothing to test against. Thus, for an ordinary random generator, the distribution is usually declared to be a discrete uniform distribution on a specified set of alternatives. Hence, to test the quality of such a random number generator, one can check to see if it gives results consistent with its declared distribution. quasi === Subject: Re: How to test a pseudo random prime number generator? > >There a a variety of techniques for testing the quality of random > >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the > >quality of a \random prime number generator\? > > First, you have to define what you mean by a random prime. > This part of my problem, I am not sure what I mean by a \random prime\. But suppose we define random_prime(x):=nextprime(random(x)) [Here random(x) is a random number between 0 and x-1 and nextprime(x) is the first prime larger than x.] If the underlying x's are all small, no problem. But what if the x's are large? Rich === Subject: Re: How to test a pseudo random prime number generator? >> >There a a variety of techniques for testing the quality of random >> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >> >quality of a \random prime number generator\? >> >> First, you have to define what you mean by a random prime. >> > >This part of my problem, I am not sure what I mean by a \random >prime\. > >But suppose we define > >random_prime(x):=nextprime(random(x)) > >[Here random(x) is a random number between 0 and x-1 and nextprime(x) >is the first prime larger than x.] If the underlying x's are all >small, no problem. But what if the x's are large? > >Rich For cryptographic purposes the standard definition of \random prime\ is \a prime number in the given range\. The usual method is to generate a series of random number in the range and test each one for primality. After a number of tries you have either found one or you need to use a bigger range. Pseudocode: function PrimeInRange(lo, hi) : integer if not(2 < lo <= hi) then throw exception limit <- 100 * (floor(log2(u)) + 1) repeat limit <- limit - 1 if limit = 0 then throw exception n <- random(lo, hi) until IsPrime(n) return n end function rossum === Subject: Re: How to test a pseudo random prime number generator? Originator: jgamble@ripco.com (John M. Gamble) > >The usual method is to generate a series of random number in the range >and test each one for primality. After a number of tries you have >either found one or you need to use a bigger range. > >Pseudocode: > function PrimeInRange(lo, hi) : integer > if not(2 < lo <= hi) then throw exception > limit <- 100 * (floor(log2(u)) + 1) > repeat > limit <- limit - 1 > if limit = 0 then throw exception > n <- random(lo, hi) > until IsPrime(n) > return n > end function > I'm assuming that u = hi - lo? Hmm, also, unlikely as it may be, the range test should allow 2 <= lo ..., since someone could conceivably ask for PrimeInRange(2,3). Is the 100 factor in the limit calculation something that just works, or is there mathematical reason for it? -- -john February 28 1997: Last day libraries could order catalogue cards from the Library of Congress. === Subject: Re: How to test a pseudo random prime number generator? On Sun, 19 Aug 2007 22:33:07 +0100, rossum > >>> >There a a variety of techniques for testing the quality of random >>> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >>> >quality of a \random prime number generator\? >>> >>> First, you have to define what you mean by a random prime. >>> >> >>This part of my problem, I am not sure what I mean by a \random >>prime\. >> >>But suppose we define >> >>random_prime(x):=nextprime(random(x)) >> >>[Here random(x) is a random number between 0 and x-1 and nextprime(x) >>is the first prime larger than x.] If the underlying x's are all >>small, no problem. But what if the x's are large? >> >>Rich >For cryptographic purposes the standard definition of \random prime\ >is \a prime number in the given range\. > >The usual method is to generate a series of random number in the range >and test each one for primality. After a number of tries you have >either found one or you need to use a bigger range. > >Pseudocode: > function PrimeInRange(lo, hi) : integer > if not(2 < lo <= hi) then throw exception > limit <- 100 * (floor(log2(u)) + 1) > repeat > limit <- limit - 1 > if limit = 0 then throw exception > n <- random(lo, hi) > until IsPrime(n) > return n > end function Yes, but it's not uniformly random on the set of primes in the range lo .. hi. See my recent replies. However if true uniformity is unimportant, then it's fine. quasi === Subject: Re: How to test a pseudo random prime number generator? >> repeat >> n <- random(lo, hi) >> until IsPrime(n) > > Yes, but it's not uniformly random on the set of primes in the range > lo .. hi. See my recent replies. Yes it is. There's no reason for it to prefer any prime in the range over any other. It doesn't use nextprime(). -- Ben === Subject: Re: How to test a pseudo random prime number generator? > >> repeat > >> n <- random(lo, hi) > >> until IsPrime(n) > > > Yes, but it's not uniformly random on the set of primes in the range > > lo .. hi. See my recent replies. > > Yes it is. There's no reason for it to prefer any prime in the range over > any other. It doesn't use nextprime(). > Ok. Using nextprime() is dumb. Now suppose I claimed (which I don't) to have a random prime number generator of equivalent *quality* to the above only faster, say. Can I test empirically the validity of this quality claim w/o converting my purported random sequence of prime numbers {p_n_i} to {n_i}? Rich === Subject: Re: How to test a pseudo random prime number generator? On Sun, 19 Aug 2007 23:05:39 +0100, Ben Rudiak-Gould >>> repeat >>> n <- random(lo, hi) >>> until IsPrime(n) >> >> Yes, but it's not uniformly random on the set of primes in the range >> lo .. hi. See my recent replies. > >Yes it is. There's no reason for it to prefer any prime in the range over >any other. It doesn't use nextprime(). Oops. Yes, you're right, sorry -- I thought you were using nextprime. Your method is definitely uniform. quasi === Subject: Re: How to test a pseudo random prime number generator? >On Sun, 19 Aug 2007 22:33:07 +0100, rossum > >> >>>> >There a a variety of techniques for testing the quality of random >>>> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >>>> >quality of a \random prime number generator\? >>>> >>>> First, you have to define what you mean by a random prime. >>>> >>> >>>This part of my problem, I am not sure what I mean by a \random >>>prime\. >>> >>>But suppose we define >>> >>>random_prime(x):=nextprime(random(x)) >>> >>>[Here random(x) is a random number between 0 and x-1 and nextprime(x) >>>is the first prime larger than x.] If the underlying x's are all >>>small, no problem. But what if the x's are large? >>> >>>Rich >>For cryptographic purposes the standard definition of \random prime\ >>is \a prime number in the given range\. >> >>The usual method is to generate a series of random number in the range >>and test each one for primality. After a number of tries you have >>either found one or you need to use a bigger range. >> >>Pseudocode: >> function PrimeInRange(lo, hi) : integer >> if not(2 < lo <= hi) then throw exception >> limit <- 100 * (floor(log2(u)) + 1) >> repeat >> limit <- limit - 1 >> if limit = 0 then throw exception >> n <- random(lo, hi) >> until IsPrime(n) >> return n >> end function > >Yes, but it's not uniformly random on the set of primes in the range >lo .. hi. See my recent replies. > >However if true uniformity is unimportant, then it's fine. > >quasi As I said, this is for the cryptographic definition, not the mathematical version. Cryptography can get a bit sloppy round the edges mathematically. rossum === Subject: Re: How to test a pseudo random prime number generator? On Sun, 19 Aug 2007 23:02:32 +0100, rossum > >>On Sun, 19 Aug 2007 22:33:07 +0100, rossum >> >>> >>>>> >There a a variety of techniques for testing the quality of random >>>>> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >>>>> >quality of a \random prime number generator\? >>>>> >>>>> First, you have to define what you mean by a random prime. >>>>> >>>> >>>>This part of my problem, I am not sure what I mean by a \random >>>>prime\. >>>> >>>>But suppose we define >>>> >>>>random_prime(x):=nextprime(random(x)) >>>> >>>>[Here random(x) is a random number between 0 and x-1 and nextprime(x) >>>>is the first prime larger than x.] If the underlying x's are all >>>>small, no problem. But what if the x's are large? >>>> >>>>Rich >>>For cryptographic purposes the standard definition of \random prime\ >>>is \a prime number in the given range\. >>> >>>The usual method is to generate a series of random number in the range >>>and test each one for primality. After a number of tries you have >>>either found one or you need to use a bigger range. >>> >>>Pseudocode: >>> function PrimeInRange(lo, hi) : integer >>> if not(2 < lo <= hi) then throw exception >>> limit <- 100 * (floor(log2(u)) + 1) >>> repeat >>> limit <- limit - 1 >>> if limit = 0 then throw exception >>> n <- random(lo, hi) >>> until IsPrime(n) >>> return n >>> end function >> >>Yes, but it's not uniformly random on the set of primes in the range >>lo .. hi. See my recent replies. >> >>However if true uniformity is unimportant, then it's fine. >> >>quasi >As I said, this is for the cryptographic definition, not the >mathematical version. Cryptography can get a bit sloppy round the >edges mathematically. But actually, it is uniform -- I misread the algorithm. I thought you were using nextprime. Sorry. Thus, mathematically, it's not sloppy. (But I was sloppy in my reading of it). quasi === Subject: Re: How to test a pseudo random prime number generator? >On Sun, 19 Aug 2007 22:33:07 +0100, rossum > >> >>>> >There a a variety of techniques for testing the quality of random >>>> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >>>> >quality of a \random prime number generator\? >>>> >>>> First, you have to define what you mean by a random prime. >>>> >>> >>>This part of my problem, I am not sure what I mean by a \random >>>prime\. >>> >>>But suppose we define >>> >>>random_prime(x):=nextprime(random(x)) >>> >>>[Here random(x) is a random number between 0 and x-1 and nextprime(x) >>>is the first prime larger than x.] If the underlying x's are all >>>small, no problem. But what if the x's are large? >>> >>>Rich >>For cryptographic purposes the standard definition of \random prime\ >>is \a prime number in the given range\. >> >>The usual method is to generate a series of random number in the range >>and test each one for primality. After a number of tries you have >>either found one or you need to use a bigger range. >> >>Pseudocode: >> function PrimeInRange(lo, hi) : integer >> if not(2 < lo <= hi) then throw exception >> limit <- 100 * (floor(log2(u)) + 1) >> repeat >> limit <- limit - 1 >> if limit = 0 then throw exception >> n <- random(lo, hi) >> until IsPrime(n) >> return n >> end function > >Yes, but it's not uniformly random on the set of primes in the range >lo .. hi. See my recent replies. > >However if true uniformity is unimportant, then it's fine. To dramatize the issue, write a program to empirically answer the following question ... For 10-digit primes p, what is the expected value of (p - prevprime(p))? If you use your proposed method to generate \random\ primes, you'll get an answer which is not even close to correct. quasi === Subject: Re: How to test a pseudo random prime number generator? >> >There a a variety of techniques for testing the quality of random >> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >> >quality of a \random prime number generator\? >> >> First, you have to define what you mean by a random prime. > >This part of my problem, I am not sure what I mean by a \random >prime\. > >But suppose we define > >random_prime(x):=nextprime(random(x)) > >[Here random(x) is a random number between 0 and x-1 and nextprime(x) >is the first prime larger than x.] If the underlying x's are all >small, no problem. But what if the x's are large? Perhaps better to specify a range, so as to avoid small primes (unless you don't want to avoid them). Thus, suppose you want a random 10-digit prime number. Here's one approach ... Let pi(x) be the number of primes <= x and let p_n denote the n'th prime. Suppose you have an efficient way to calculate pi(x). Let a=pi(10^9) and let b=pi(10^10). Choose a random integer n between a+1 and b inclusive. Then take p_n as the random prime. Note, assuming pi(x) can be efficiently calculated for 10^9 <= x <= 10^10, then, by binary search, p_n can also be efficiently calculated. The method above produces a uniformly random 10-digit prime, however unless pi(x) can be efficiently calculated for 10^9 <= x <= 10^10, the method is impractical. As an alternative, let's try a method similar to the one you outlined, but restricted to producing 10-digit primes. Thus, take a random number x between 10^9 and 10^10-1 inclusive. If x is prime, take x, otherwise take the next prime after x, rejecting it exceeds 10^10-1. This method is efficient but does not produce a uniform distribution of 10-digit primes. The lack of uniformity is due to the fact that the probability that a prime is chosen is proportional to the distance to the previous prime, or to 10^9, whichever distance is less. Thus, if p and p+2 are successive 10-digit primes, and if q and q+4 are also successive 10-digit primes, then q+4 has twice as much chance of being selected as p+2. To correct for that bias, you can first sample 10-digit primes and their next or previous primes, to get an approximate distribution of gaps. Once you have a distribution of gaps, you can then begin choosing \random\ primes as previously described, but selectively reject to maintain the fair distribution. In essence, you restore the correct balance (\affirmative action\ for primes). quasi === Subject: Re: How to test a pseudo random prime number generator? > >>> >There a a variety of techniques for testing the quality of random >>> >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the >>> >quality of a \random prime number generator\? >>> >>> First, you have to define what you mean by a random prime. >> >>This part of my problem, I am not sure what I mean by a \random >>prime\. >> >>But suppose we define >> >>random_prime(x):=nextprime(random(x)) >> >>[Here random(x) is a random number between 0 and x-1 and nextprime(x) >>is the first prime larger than x.] If the underlying x's are all >>small, no problem. But what if the x's are large? > >Perhaps better to specify a range, so as to avoid small primes (unless >you don't want to avoid them). > >Thus, suppose you want a random 10-digit prime number. > >Here's one approach ... > >Let pi(x) be the number of primes <= x and let p_n denote the n'th >prime. > >Suppose you have an efficient way to calculate pi(x). > >Let a=pi(10^9) and let b=pi(10^10). Choose a random integer n between >a+1 and b inclusive. Then take p_n as the random prime. > >Note, assuming pi(x) can be efficiently calculated for 10^9 <= x <= >10^10, then, by binary search, p_n can also be efficiently calculated. > >The method above produces a uniformly random 10-digit prime, however >unless pi(x) can be efficiently calculated for 10^9 <= x <= 10^10, the >method is impractical. > >As an alternative, let's try a method similar to the one you outlined, >but restricted to producing 10-digit primes. > >Thus, take a random number x between 10^9 and 10^10-1 inclusive. If x >is prime, take x, otherwise take the next prime after x, rejecting it >exceeds 10^10-1. > >This method is efficient but does not produce a uniform distribution >of 10-digit primes. The lack of uniformity is due to the fact that the >probability that a prime is chosen is proportional to the distance to >the previous prime, or to 10^9, whichever distance is less. Thus, if p >and p+2 are successive 10-digit primes, and if q and q+4 are also >successive 10-digit primes, then q+4 has twice as much chance of being >selected as p+2. Ignore the proposed method for correcting the non-unformity -- it's not correct. >10-digit primes and their next or previous primes, to get an >approximate distribution of gaps. Once you have a distribution of >gaps, you can then begin choosing \random\ primes as previously >described, but selectively reject to maintain the fair distribution. >In essence, you restore the correct balance (\affirmative action\ for >primes). The distribution of gaps is irrelevant. What is relevant is the actual gaps for the candidate random primes. Here's a corrected version, step by step ... (1) Choose a random number x between 10^9 and 10^10-1 inclusive. (2) If x is prime, let p=x, otherwise let p = nextprime(x), but in the latter case, if p > 10^10-1, go back to step (1). (3) Let d = p - max(prevprime(p),10^9-1) (4) If d=1, then accept p, otherwise accept p with probability 1/d. If p is rejected, go back to step (1). The distribution is now exactly uniform on the set of all 10-digit primes. The main disadvantage is that by this method, for 10-digit primes, only 1 in approximately 22 prime candidates is accepted, on average. Thus, when compared with the method which accepts all candidates, this method is about 22 times slower, but at least it's truly uniform (assuming the underlying RNG is regarded as uniform). quasi === Subject: Re: How to test a pseudo random prime number generator? > > > >There a a variety of techniques for testing the quality of random > > >number generators (DIEHARD, NIST, Knuth, ect.). How can I test the > > >quality of a \random prime number generator\? > > > First, you have to define what you mean by a random prime. > > This part of my problem, I am not sure what I mean by a \random > prime\. > > But suppose we define > > random_prime(x):=nextprime(random(x)) > > [Here random(x) is a random number between 0 and x-1 and nextprime(x) > is the first prime larger than x.] If the underlying x's are all > small, no problem. But what if the x's are large? > > Rich http://mathworld.wolfram.com/PrimalityTest.html === Subject: line sliding in oval Hi Folks, I have one solution but please examine if a general solution is possible. http://i11.tinypic.com/63wfic4.jpg A smooth continuous closed oval pPQq is symmetric with respect to x- axis,which also bisects a pair of radial rays pP and qQ passing through origin.Find its polar equation if pQ = qP = 2c = constant for any such radial ray pair. TIA Narasimham === Subject: Re: line sliding in oval > Hi Folks, I have one solution but please examine if a general solution > is possible. > > http://i11.tinypic.com/63wfic4.jpg > > A smooth continuous closed oval pPQq is symmetric with respect to x- > axis,which also bisects a pair of radial rays pP and qQ passing > through origin.Find its polar equation if pQ = qP = 2c = constant for > any such radial ray pair. TIA For each value of the angle theta in the appropriate interval (-a,a), there are two r values r1 and r2. By the law of cosines, (2c)^2 = r1^2 + r2^2 - 2 r1 r2 cos(2 theta). Write this as (2c)^2 = (r1+r2)^2 - 4 r1 r2 cos(theta)^2 Now r1 and r2 are the roots of the quadratic polynomial X^2 - A X + B where A = r1+r2 and B = r1 r2 so (2c)^2 = A^2 - 4 B cos(theta)^2 i.e. A = sqrt(4 c^2 + 4 B cos(theta)^2). For theta = (+/-) a we want r1 = r2, i.e. A^2 = 4 B, so B(a) = c^2/sin(a)^2 while for |theta| < a, A^2 - 4 B > 0, i.e. B(theta) < c^2/sin(theta)^2. Otherwise, B can be any positive even function of theta. The polar equation is then r^2 - sqrt(4 c^2 + 4 B(theta) cos(theta)^2) r + B(theta) = 0. If you want your \oval\ to be convex, there must be an additional inequality on B and its first and second derivatives, but that looks rather messy to me. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: \ =?utf-8?B?aXNsYW1pYy1qaWhhZCAg2KfZhNis2YfYp9ivINin2YTYpdiz2YTYp9mF2Yo=?= aXNsYW1pYy1qaWhhZCAg2KfZhNis2YfYp9ivINin2YTYpdiz2YTYp9mF2YoKCmh0dHA6Ly9ncm91\ cHMuZ29vZ2xlLmNvbS9ncm91cC9pc2xhbWljLWppaGFkCgoKLi4uLi4uLi4uLi4uLgo= === Subject: Re: islamic-jihad ?????? ???????? > islamic-jihad ?????? ???????? > That \group\ plus the guy who runs http://www.webgroupes.eu/thread-334-0-141066-1304/fr-soc-politique/world-new\ s-network.htm (the name of which is copied from a Sunni Iraqi insurgent/terrorist \ website) plus the guy who ran the defunct plus the current guy nasser1987@gmail.com (see ) are all the same fantasy-jihadi, who seems to be in or from Arabia, probably \ KSA. Next? === Subject: Re: Koos Nolst Trenite - 'The Promised Definition of Friendship' V3.2.1.txt.nfo > > Koos Nolst Trenite ('Cause Trinity') is arguably the most intelligent, > the most caring and loving, the most beautiful and the most truthful > philosopher known. [snip crap] http://www.mazepath.com/uncleal/analysis.jpg -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/lajos.htm#a2 === Subject: May be a short history of JSH? No offense. Greeting, I don't really read sci.math, so I'm not all that familiar with those story going on here. I have noticed JSH for quite a while, and I cannot help but to wonder what makes him such an intersting person, especially which incident had made him hate mathematicians so much. I know the entire history is in the archive, but it would take me hours or even days to read. So if anyone who's familar with the story can just write a paragraph or two (SHORT) to explain the background, it would be nice. I am not judging his work. I am only interesting in the formation of === Subject: Re: May be a short history of JSH? No offense. > Greeting, > > I don't really read sci.math, so I'm not all that familiar with those > story going on here. I have noticed JSH for quite a while, and I > cannot help but to wonder what makes him such an intersting person, > especially which incident had made him hate mathematicians so much. I > know the entire history is in the archive, but it would take me hours > or even days to read. So if anyone who's familar with the story can > just write a paragraph or two (SHORT) to explain the background, it > would be nice. > > I am not judging his work. I am only interesting in the formation of I became interested in doodling at math after it was reported that Andrew Wiles had found a proof of Fermat's Last Theorem, as I wondered if maybe there wasn't still a simpler proof out there. So it was a just a lark for me in many ways but I went at it seriously enough. After bugging mathematicians at universities with some of my ideas for proving FLT--which all failed by the way--I discovered Usenet and a math community online where it seemed I could toss out math ideas in a proper place. That year I was hit with a flood of insults, including being called insane and told to go away. It was a matter of pride for regulars on math newsgroups to be as insulting as possible to \cranks\ and \crackpots\ where it was regulars on the newsgroups who decided who they hated and they would gang up on people insulting them and replying in insulting ways to EVERY post until they'd drive a person off. Then they'd celebrate. The worst ones would ride you with insults until they thought they'd beaten you down enough and then hint or tell you to commit suicide. That is the math world I know. I kept at it partly because I found I liked doodling at math and also because I didn't feel like being intimidated into silence while MOST people are. Think about it, if you when you posted you were insulted, insulted, insulted would you not leave? Finally I got what I thought were solid results that stood up to every attack, but the rain of insults continued so I turned to the math journals, and even got a paper published in a now defunct journal that Members of the sci.math newsgroup erupted in fury when they heard of the publication and some of them conspired ONLINE in posts to mount an email campaign against my paper which convinced the editors of the journal who yanked it out of the electronic edition. They managed one more edition and then quietly shut down. I have mathematical proof and I have experience with a mathematical community that is best described as evil with people who have called me sub-human, continually questioned my sanity, and even turned to race as an issue, who cannot be convinced by mathematical proof. The current math field is corrupt. The people in it are morally bankrupt and they have learned to lie about math. How do you know a mathematical argument called a proof is actually one? You TRUST, right? Well these are people who will call you insane if you disagree with them. Make them mad enough and they'll tell you to kill yourself. Why do you trust them? James Harris === Subject: Re: May be a short history of JSH? No offense. The short background was very nice, just a few more questions: 2) Does JSH acknowledge that he has NPD? 3) Is his writing skill ever a part of what sparks the insult? 4) Is JSH a real person? > > > Greeting, > > > I don't really read sci.math, so I'm not all that familiar with those > > story going on here. I have noticed JSH for quite a while, and I > > cannot help but to wonder what makes him such an intersting person, > > especially which incident had made him hate mathematicians so much. I > > know the entire history is in the archive, but it would take me hours > > or even days to read. So if anyone who's familar with the story can > > just write a paragraph or two (SHORT) to explain the background, it > > would be nice. > > > I am not judging his work. I am only interesting in the formation of > > I became interested in doodling at math after it was reported that > Andrew Wiles had found a proof of Fermat's Last Theorem, as I wondered > if maybe there wasn't still a simpler proof out there. So it was a > just a lark for me in many ways but I went at it seriously enough. > > After bugging mathematicians at universities with some of my ideas for > proving FLT--which all failed by the way--I discovered Usenet and a > math community online where it seemed I could toss out math ideas in a > proper place. > > That year I was hit with a flood of insults, including being called > insane and told to go away. > > It was a matter of pride for regulars on math newsgroups to be as > insulting as possible to \cranks\ and \crackpots\ where it was > regulars on the newsgroups who decided who they hated and they would > gang up on people insulting them and replying in insulting ways to > EVERY post until they'd drive a person off. > > Then they'd celebrate. > > The worst ones would ride you with insults until they thought they'd > beaten you down enough and then hint or tell you to commit suicide. > > That is the math world I know. > > I kept at it partly because I found I liked doodling at math and also > because I didn't feel like being intimidated into silence while MOST > people are. > > Think about it, if you when you posted you were insulted, insulted, > insulted would you not leave? > > Finally I got what I thought were solid results that stood up to every > attack, but the rain of insults continued so I turned to the math > journals, and even got a paper published in a now defunct journal that > > Members of the sci.math newsgroup erupted in fury when they heard of > the publication and some of them conspired ONLINE in posts to mount an > email campaign against my paper which convinced the editors of the > journal who yanked it out of the electronic edition. > > They managed one more edition and then quietly shut down. > > I have mathematical proof and I have experience with a mathematical > community that is best described as evil with people who have called > me sub-human, continually questioned my sanity, and even turned to > race as an issue, who cannot be convinced by mathematical proof. > > The current math field is corrupt. The people in it are morally > bankrupt and they have learned to lie about math. > > How do you know a mathematical argument called a proof is actually > one? > > You TRUST, right? Well these are people who will call you insane if > you disagree with them. > > Make them mad enough and they'll tell you to kill yourself. > > Why do you trust them? > > James Harris === Subject: Re: May be a short history of JSH? No offense. > The short background was very nice, just a few more questions: > > > 2) Does JSH acknowledge that he has NPD? > > 3) Is his writing skill ever a part of what sparks the insult? > > 4) Is JSH a real person? Better you should ask if the person you're replying to is the real JSH. And don't top post, it annoys the regulars. Perhaps you should kill yourself. > > > > > > > > Greeting, > > > > I don't really read sci.math, so I'm not all that familiar with those > > > story going on here. I have noticed JSH for quite a while, and I > > > cannot help but to wonder what makes him such an intersting person, > > > especially which incident had made him hate mathematicians so much. \ I > > > know the entire history is in the archive, but it would take me hours > > > or even days to read. So if anyone who's familar with the story can > > > just write a paragraph or two (SHORT) to explain the background, it > > > would be nice. > > > > I am not judging his work. I am only interesting in the formation of > > > I became interested in doodling at math after it was reported that > > Andrew Wiles had found a proof of Fermat's Last Theorem, as I wondered > > if maybe there wasn't still a simpler proof out there. So it was a > > just a lark for me in many ways but I went at it seriously enough. > > > After bugging mathematicians at universities with some of my ideas for > > proving FLT--which all failed by the way--I discovered Usenet and a > > math community online where it seemed I could toss out math ideas in a > > proper place. > > > That year I was hit with a flood of insults, including being called > > insane and told to go away. > > > It was a matter of pride for regulars on math newsgroups to be as > > insulting as possible to \cranks\ and \crackpots\ where it was > > regulars on the newsgroups who decided who they hated and they would > > gang up on people insulting them and replying in insulting ways to > > EVERY post until they'd drive a person off. > > > Then they'd celebrate. > > > The worst ones would ride you with insults until they thought they'd > > beaten you down enough and then hint or tell you to commit suicide. > > > That is the math world I know. > > > I kept at it partly because I found I liked doodling at math and also > > because I didn't feel like being intimidated into silence while MOST > > people are. > > > Think about it, if you when you posted you were insulted, insulted, > > insulted would you not leave? > > > Finally I got what I thought were solid results that stood up to every > > attack, but the rain of insults continued so I turned to the math > > journals, and even got a paper published in a now defunct journal that > > > Members of the sci.math newsgroup erupted in fury when they heard of > > the publication and some of them conspired ONLINE in posts to mount an > > email campaign against my paper which convinced the editors of the > > journal who yanked it out of the electronic edition. > > > They managed one more edition and then quietly shut down. > > > I have mathematical proof and I have experience with a mathematical > > community that is best described as evil with people who have called > > me sub-human, continually questioned my sanity, and even turned to > > race as an issue, who cannot be convinced by mathematical proof. > > > The current math field is corrupt. The people in it are morally > > bankrupt and they have learned to lie about math. > > > How do you know a mathematical argument called a proof is actually > > one? > > > You TRUST, right? Well these are people who will call you insane if > > you disagree with them. > > > Make them mad enough and they'll tell you to kill yourself. > > > Why do you trust them? > > > James Harris === Subject: Re: May be a short history of JSH? No offense. > The worst ones would ride you with insults until they thought they'd > beaten you down enough and then hint or tell you to commit suicide. > > That is the math world I know. That's not \math world\. It's just your usual Usenet, where you'll find the same happens quite regardless of the subject. It is futile to attempt to draw any conclusions about any group of people based on the behaviour of its purported representatives in the news. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \Wovon man nicht sprechen kann, daruber muss man schweigen\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: May be a short history of JSH? No offense. <87y7g6na06.fsf@huxley.huxley.fi> On Aug 19, 7:46 pm, Aatu Koskensilta > > The worst ones would ride you with insults until they thought they'd > > beaten you down enough and then hint or tell you to commit suicide. > > > That is the math world I know. > > That's not \math world\. It's just your usual Usenet, where you'll > find the same happens quite regardless of the subject. It is futile to > attempt to draw any conclusions about any group of people based on the > behaviour of its purported representatives in the news. > Like I said, that's the math world I know. To me sci.math is representative. James Harris === Subject: Re: May be a short history of JSH? No offense. > Greeting, > > I don't really read sci.math, so I'm not all that familiar with those > story going on here. I have noticed JSH for quite a while, and I > cannot help but to wonder what makes him such an intersting person, > especially which incident had made him hate mathematicians so much. I > know the entire history is in the archive, but it would take me hours > or even days to read. So if anyone who's familar with the story can > just write a paragraph or two (SHORT) to explain the background, it > would be nice. > > I am not judging his work. I am only interesting in the formation of It comes down o a couple of major things. 1) when he was young, people told him - Jimmy - look at how smart you are at adding numbers and taking square roots on your calculator. Somehow, later in life, this turned into being the greatest number theorist of all time. When no one on here would accept anything he called proof - we were all evil because we weren't like family members stroking the young boys ego. I think this has been amplifified by item 2 below. 2) James has Nacissistic Personality Disorder and thinks these egomanical things. One is that he needs any kind of attention - good or bad. He has learned to expertly troll and is a master crank and cheat - just like his entire life. His NPD continues to worsen - now he has to change the tin foil in his tin-foil hat very frequently. He thinks we are all out to get him because that serves his delusional NPD condition. That is really the heart of it. === Subject: Re: May be a short history of JSH? No offense. >> Greeting, >> >> I don't really read sci.math, so I'm not all that familiar with those >> story going on here. I have noticed JSH for quite a while, and I >> cannot help but to wonder what makes him such an intersting person, >> especially which incident had made him hate mathematicians so much. I >> know the entire history is in the archive, but it would take me hours >> or even days to read. So if anyone who's familar with the story can >> just write a paragraph or two (SHORT) to explain the background, it >> would be nice. >> >> I am not judging his work. I am only interesting in the formation of > > It comes down o a couple of major things. > > 1) when he was young, people told him - Jimmy - look at how smart you > are at adding numbers and taking square roots on your calculator. > Somehow, later in life, this turned into being the greatest number > theorist of all time. When no one on here would accept anything he > called proof - we were all evil because we weren't like family members > stroking the young boys ego. I think this has been amplifified by item > 2 below. > > 2) James has Nacissistic Personality Disorder and thinks these > egomanical things. One is that he needs any kind of attention - good > or bad. He has learned to expertly troll and is a master crank and > cheat - just like his entire life. His NPD continues to worsen - now > he has to change the tin foil in his tin-foil hat very frequently. He > thinks we are all out to get him because that serves his delusional > NPD condition. > > That is really the heart of it. > oh, so JSH isn't the greatist mathematician ever?? I think you are the deluded one! === Subject: Re: May be a short history of JSH? No offense. On Sun, 19 Aug 2007 12:08:03 -0700, Gvnaena Pura >Greeting, > >I don't really read sci.math, so I'm not all that familiar with those >story going on here. I have noticed JSH for quite a while, and I >cannot help but to wonder what makes him such an intersting person, >especially which incident had made him hate mathematicians so much. I >know the entire history is in the archive, but it would take me hours >or even days to read. So if anyone who's familar with the story can >just write a paragraph or two (SHORT) to explain the background, it >would be nice. > >I am not judging his work. I am only interesting in the formation of Here's are is a link to a past post by marcus_b which attempts to provide some insights. quasi === Subject: Cheats !!!! Those of you who post replies and accues ones of cheating and who do not know the facts, well mind your own bussness. I got a message from one of you yesterday. I asked for help on retaining a solution manual like a lot of you do. I received rude message yesterday acussing me of cheating, and using improper english in this chat area. Here is the deal. This is not a formal school function or job area. I will use short hand and lol's. As far as the manual I need it becuase I cant get help on the issue I need help on. It isnt math. It is economics. I need to see how to do GPS, which a manual will not give me answers to. So know what you are talking about before replying to someones post. My instructor isnt responding to my emils and hasnt for 2 weeks on this problem and I need to turn my assignment in today. I am appoled at the person who sent me that nasty message. === Subject: Re: Cheats !!!! > Those of you who post replies and accues ones of cheating and who do > not know the facts, well mind your own bussness. I got a message from > one of you yesterday. I asked for help on retaining a solution manual > like a lot of you do. I received rude message yesterday acussing me of > cheating, and using improper english in this chat area. Here is the > deal. This is not a formal school function or job area. I will use > short hand and lol's. As far as the manual I need it becuase I cant > get help on the issue I need help on. It isnt math. It is economics. I > need to see how to do GPS, which a manual will not give me answers to. > So know what you are talking about before replying to someones post. > My instructor isnt responding to my emils and hasnt for 2 weeks on > this problem and I need to turn my assignment in today. >I am appoled > at the person who sent me that nasty message. > I am sincerely \appoled\ to read this post... === Subject: Re: Cheats !!!! ... Funny posture. Who will believe that? You ask in a Math forum unrecognizable about GPS to solve an exercise in Economics? And you are going to honestly deliver that as supported outcome, yes? You got a task. And can not solve it yourself, what is the instructor seems to expect: \I need help on. It isn't math.\. Well, may be you fail the test, but you might learn from that: try to ask (there are magic words which even children do know) in readable mails at appropriate forums or try a library or Google and start not just before deadline. And certainly you will have referred to any help received. === Subject: Re: Cheats !!!! > Those of you who post replies and accues ones of cheating and who do not > know the facts, well mind your own bussness. I got a message from one of > you yesterday. I asked for help on retaining a solution manual like a > lot of you do. I received rude message yesterday acussing me of > cheating, and using improper english in this chat area. Here is the > deal. This is not a formal school function or job area. I will use short > hand and lol's. That's all right. Don't feel offended if readers in this forum think that you are semi-literate though. > As far as the manual I need it becuase I cant get help > on the issue I need help on. It isnt math. It is economics. I need to > see how to do GPS, which a manual will not give me answers to. So know > what you are talking about before replying to someones post. My > instructor isnt responding to my emils and hasnt for 2 weeks on this > problem and I need to turn my assignment in today. I am appoled at the > person who sent me that nasty message. Don't be. You really, really come across as barely literate, which makes us more and more convinced that you are just trying to find a shortcut to hard work. === Subject: Re: Cheats !!!! On Aug 19, 12:16 pm, marydaniels > Those of you who post replies and accues ones of cheating and who do > not know the facts, well mind your own bussness. I got a message from > one of you yesterday. I asked for help on retaining a solution manual > like a lot of you do. I received rude message yesterday acussing me of > cheating, and using improper english in this chat area. Here is the > deal. This is not a formal school function or job area. I will use > short hand and lol's. As far as the manual I need it becuase I cant > get help on the issue I need help on. It isnt math. It is economics. I > need to see how to do GPS, which a manual will not give me answers to. > So know what you are talking about before replying to someones post. > My instructor isnt responding to my emils and hasnt for 2 weeks on > this problem and I need to turn my assignment in today. So, you _are_ a cheater. > I am appoled > at the person who sent me that nasty message. === Subject: Re: Cheats !!!! On Aug 19, 12:16 pm, marydaniels > Those of you who post replies and accues ones of cheating and who do > not know the facts, well mind your own bussness. Actually, many people here are educators and they do consider to to be their business (not bussness) when they feel somebody is trying to get a \free ride\. > I got a message from > one of you yesterday. I asked for help on retaining a solution manual > like a lot of you do. I received rude message yesterday acussing me of > cheating, and using improper english in this chat area. Here is the > deal. This is not a formal school function or job area. I will use > short hand and lol's. The aim is _comminication_. If you don't respect your audience enough to write properly, why should anybody want to help you? > As far as the manual I need it becuase I cant > get help on the issue I need help on. It isnt math. It is economics. I > need to see how to do GPS, which a manual will not give me answers to. > So know what you are talking about before replying to someones post. > My instructor isnt responding to my emils and hasnt for 2 weeks on > this problem and I need to turn my assignment in today. Well, you will just have to get a lower mark on that assignment. After all, you don't know how to do the question! > I am appoled > at the person who sent me that nasty message. I read it, and did not consider it nasty at all (at least, if it was the one that appeared in the newsgroup). Criticism is not nastiness. Criticism is supposed to be FOR YOUR BENEFIT. People are saying: here is what you are doing wrong, and here is how you can improve. We are not born perfect and all-knowing; we learn and improve by having our mistakes pointed out to us, provided, of course, that we WANT to improve. R.G. Vickson Adjunct Professor, University of Waterloo === Subject: Re: Help !!! Where Are the Solution Manuals > > I have sent several of you personal emails that had links to the > > solution manuals i need them today can someone tell me why no one is > > emailing me back. And where can i find the ones for Economics Today: > > The Macro View (13th Ed., Roger LeRoy Miller) And for activiebook > > version 2.0 principles of marketing, Kotler : Armstrong I have to have > > them today my classes end tomarrow i am willing to pay a resonable > > Usually replies are given here in the newsgroup. > > Were you to use 'I' for the first person singular pronoun instead of the > imaginary number i = sqr -1 and write with adequate sentence structure > instead of stream of thought, you would be taken as a serious student. > > As for getting solution books, forget it. You paid money to get \ educated. > If you don't want to learn, go to a diploma mill. In the meantime, I > suggest you take a course in English as a first language. One more thing this whole room does nothing but cheat. Theys send in a question on math and get the answers to it. So you need to look at yourself not me. I have never used this room for math or accounting ever. === Subject: Re: Help !!! Where Are the Solution Manuals > > I have sent several of you personal emails that had links to the > > solution manuals i need them today can someone tell me why no one is > > emailing me back. And where can i find the ones for Economics Today: > > The Macro View (13th Ed., Roger LeRoy Miller) And for activiebook > > version 2.0 principles of marketing, Kotler : Armstrong I have to have > > them today my classes end tomarrow i am willing to pay a resonable > > Usually replies are given here in the newsgroup. > > Were you to use 'I' for the first person singular pronoun instead of the > imaginary number i = sqr -1 and write with adequate sentence structure > instead of stream of thought, you would be taken as a serious student. > > As for getting solution books, forget it. You paid money to get \ educated. > If you don't want to learn, go to a diploma mill. In the meantime, I > suggest you take a course in English as a first language First of all. This isnt a formal student group session. Second The only reason i nead a solution manual is becuase i am stuck on a problem and need to see how to do it. My text does not explain the problem at all. Third I have never nor will i ever cheat. I have paid over 80,000 on my educaion thus far, with no help from a instructor or student and have maintained a B average. So here is my advise back to you but out unless you know the facts. === Subject: Re: Help !!! Where Are the Solution Manuals > > As for getting solution books, forget it. You paid money to get \ educated. > > If you don't want to learn, go to a diploma mill. In the meantime, I > > suggest you take a course in English as a first language > > First of all. This isnt a formal student group session. Second The > only reason i nead a solution manual is becuase i am stuck on a > problem and need to see how to do it. My text does not explain the > problem at all. Third I have never nor will i ever cheat. I have paid > over 80,000 on my educaion thus far, with no help from a instructor or > student and have maintained a B average. So here is my advise back to > you but out unless you know the facts. > Now that you've informed me of the facts about how you're getting ripped off because of the downward trend of education quality and have expressed a need for help, you have some options. You can present the problem and what you have tried for a solution and ask for hints. Many here are willing to give hints. On the other hand, I for one frequently don't read posts which flout the deliberate illiteracy of the poster with blatant disregard of English punctuation and composition. One benefit of this method is that I don't get overloaded answering too many posts, leaving me more time to consider my favor studies or to learn from reading the more difficult and astute threads. Again, my advice to you is to take a course in English as a first language. You may find it of value when you enter the job market. === Subject: Re: One more reason to impreach Bush <74ffc3li1q8pf3b1ain07l72907mr60fd0@4ax.com> Conyer's and Kucinich's move to impeach Cheeny, ignoring the Shrub that will spontaneously combust, when, is a great oppotunity to clean-out the neocons & jihadis in the Admin.; when did Trickier Dick convert to Islam? > But for serious evil, don't forget Cheney. Bush is a willing puppet, > but Cheney's the puppeteer. hus quoth: Only fragments of these remarks have been reported in the West, while in Russia RIA Novosti gave its readers a headline (\Clinton supports placement of BMD in Poland and the Czech Republic\) that was directly opposite to what President Clinton said. http://larouchepub.com/other/2007/3428clinton_at_yalta.html --n~nerfman~n! http://larouchepub.com/pr/2007/070730conyers_impeach_dick.html 14 Italian Senators Call for Cheney Impeachment Aug. 1, 2007 (EIRNS)- The Lyndon LaRouche Political Action Committee (LPAC) issued the following release today. Fourteen members of the Italian Senate have signed a call \to the Members of Congress to support Rep. Kucinich's House Resolution 333 for the Impeachment of Dick Cheney.\ http://larouchepub.com/pr/2007/070801italian_senators_call.html === Subject: Re: One more reason to impreach Bush Reply-to: being_there@bellsouth.net They both need impeaching. 'cause they lied, if for no other reason. That was enough for them to go after Bill Clinton. Unfortunately, they were successful. http://www.cafepress.com/fartotheleft.24612070 http://www.cafepress.com/fartotheleft.24536630 Both are reasons enough. Larry Jay -- Trust your politicians, Do You? If I were you, I'd think again on that one. http://www.cafepress.com/fartotheleft.91990814 === Subject: Re: One more reason to impreach Bush > They both need impeaching. 'cause they lied, if for no other reason. \ That > was enough for them to go after Bill Clinton. Unfortunately, they were > \ successful.http://www.cafepress.com/fartotheleft.24612070http://www.cafepress\ .com/fartotheleft.24536630 > Both are reasons enough. > > Larry Jay > -- > Trust your politicians, Do You? > If I were you, I'd think again on that one. > > http://www.cafepress.com/fartotheleft.91990814 If you impeach one node, another node just takes its place. === Subject: Re: One more reason to impreach Bush > > > They both need impeaching. 'cause they lied, if for no other reason. \ That > > was enough for them to go after Bill Clinton. Unfortunately, they were > > \ successful.http://www.cafepress.com/fartotheleft.24612070http://www.cafepress\ .co... > > Both are reasons enough. > > > Larry Jay > > -- > > Trust your politicians, Do You? > > If I were you, I'd think again on that one. > > >http://www.cafepress.com/fartotheleft.91990814 > > If you impeach one node, another node just takes its place. \No Matter Who You Vote For, The Government Always Gets In\ --- song by the Bonzo Dog Band === Subject: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES I'M OUTTA THIS FUKKEN PLACE. I'M TIRED OF READING THE BULLSHIT POSTED BY A CROWD OF MIDGET-MINDED DIPSHITS, NOT ONE OF WHOM HAS ENOUGH BRAINS TO BE ABLE TO POUR PISS OUT OF A BOOT OR COUNT TO ELEVEN WITHOUT UNZIPPING HIS PANTS. LITTLE MORE ABOUT MATH. THAT ISN'T POSSIBLE IN THIS PLACE. I DOUBT IF ANYONE HERE IS BRIGHT ENOUGH TO EVEN KNOW WHAT MATH IS. MUCH AGAIN THAT IS POSTED BY A SIMPLE-MINDED QUEERS WHO IDENTIFY THEMSELVES AS MATHEMATICIANS. IT'S HARD TO BELIEVE THAT SCUMBAGS WITH SUCH A SHORTAGE OF BRAINS CAN SURVIVE IN THE WORLD BUT THERE THEY ARE. ADIOS, DIPSHIT. I WISH YOU A FUTURE WITH ONE COCK ALWAYS IN YOUR MOUTHS AND ANOTHER STUCK UP YOUR ASSES. OF COURSE ONE CAN FIND HERE AN ABUNDANCE OF OTHER BRAIN-DEAD PUKES WHOSE ONLY REASON FOR LIVING IS TO BABBLE INCOHERENTLY ABOUT ONE IDIOTIC THING OR ANOTHER. I CAN SEE YOU IN MY MIND. MOST OF THE DAY RECTUM. ONCE EVERY THIRTY MINUTES YOU FOOLS REVERSE THE POSITIONS OF THE THUMBS AND CONTINUE TO MOVE SIMPLE THOUGHTS THROUGH A MIND SO DIM IT CANNOT CONTEMPLATE ANYTHING MORE COMPLEX THAN THE NEXT TIME YOU LOCK YOURSELF IN A DARK CLOSET AND PUMP YOUR POTATO. WHAT A SORRY JOKE THIS PLACE IS AND YOU PEOPLE ARE. TOO BAD YOU SCROATS DON'T HAVE THE BRAINS TO SEE IT. :^( === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES > I'M OUTTA THIS FUKKEN PLACE. I'M TIRED OF READING THE BULLSHIT > POSTED > BY A CROWD OF MIDGET-MINDED DIPSHITS, NOT ONE OF WHOM HAS ENOUGH > BRAINS TO BE ABLE TO POUR PISS OUT OF A BOOT OR COUNT TO ELEVEN > WITHOUT UNZIPPING HIS PANTS. > LITTLE > MORE ABOUT MATH. THAT ISN'T POSSIBLE IN THIS PLACE. I > DOUBT IF ANYONE HERE IS BRIGHT ENOUGH TO EVEN KNOW WHAT MATH IS. MUCH > AGAIN > THAT IS POSTED BY A SIMPLE-MINDED QUEERS WHO IDENTIFY THEMSELVES AS > MATHEMATICIANS. IT'S HARD TO BELIEVE THAT SCUMBAGS WITH SUCH A > SHORTAGE > OF BRAINS CAN SURVIVE IN THE WORLD BUT THERE THEY ARE. ADIOS, DIPSHIT. > I > WISH YOU A FUTURE WITH ONE COCK ALWAYS IN YOUR MOUTHS AND ANOTHER > STUCK > UP YOUR ASSES. > OF COURSE ONE CAN FIND HERE AN ABUNDANCE OF OTHER BRAIN-DEAD > PUKES > WHOSE ONLY REASON FOR LIVING IS TO BABBLE INCOHERENTLY ABOUT ONE > IDIOTIC THING OR ANOTHER. I CAN SEE YOU IN MY MIND. MOST OF THE DAY > RECTUM. ONCE EVERY THIRTY MINUTES YOU FOOLS REVERSE THE POSITIONS OF > THE THUMBS AND CONTINUE TO MOVE SIMPLE THOUGHTS THROUGH A MIND SO DIM > IT CANNOT CONTEMPLATE ANYTHING MORE COMPLEX THAN THE NEXT TIME YOU > LOCK YOURSELF IN A DARK CLOSET AND PUMP YOUR POTATO. WHAT A SORRY > JOKE > THIS PLACE IS AND YOU PEOPLE ARE. TOO BAD YOU SCROATS DON'T HAVE THE > BRAINS TO SEE IT. :^( So do you think my new sock puppet style is ready for responding to AP? --- Christopher Heckman === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES > I'M OUTTA THIS FUKKEN PLACE. C'mon, don't be so shy. Tell us what you really think. --Lynn === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES >> >> I'M OUTTA THIS FUKKEN PLACE. >> >> > C'mon, don't be so shy. Tell us what you really think. > DICK --- UMMM... UMMM... --- AHHH.... UMMM... === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES ( snip agressive sexually frustrated bogus ) 1) its not our fault you havent learned anything 2) this is not an education institute, this is a forum 3) you wouldnt know anything about us since this is your first post ; your a \ newbie 4) and a sexually frustrated one... 5) maybe you want to learn us something then if your so much smarter maybe another FLT short proof :-) or JSH factoring ?? or maybe this is the reason why you left ?? we are not all FLT short provers etc ... mabye math isnt your thing try icehockey , your agressive enough for it ... tommy1729 === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES [Snip rant] The caps lock key is usually situated at the far left hand side of your keyboard near the center row. Hope this helps. === Subject: Re: ADIOS YOU SIMPLE-MINDED SONSOFBITCHES > com... > > [Snip rant] > > The caps lock key is usually situated at the > far left hand side of your keyboard near the > center row. Hope this helps. I have been at least ten minutes laughing. Fernando. === Subject: Re: Shubert's Derivation of the Lorentz Transformations Dude, what about x^2 - ct^2? What about reduction to the Galilean transformation? The first is more fundamental, but the second one is more painful - you got picked apart by a CHEMIST ... (me :-) === Subject: Re: Shubert's Derivation of the Lorentz Transformations Alfred Ziegler gives an overview of representative derivations of the LT found in textbooks. See, The role of the two postulates of special VII. A SURVEY OF TEXTBOOKS A. Linearity The majority of textbooks sees no need to justify the linearity of the transformation5,7,8,9,10,11,12,13,14,15. When given it is derived either from homogeneity of time and space3,16,17, 18,19,20 or from Newton's first law which requires uniform and thus straight motion to remain uniform and straight in another system of reference4,21,22. Resnick3 is the most explicit text in this respect. Actually a lot more \obvious\ statements regarding the possible choices of coordinates have to be demonstrated which among the texts listed here are only given in Rindler17. Wow! According to this paragraph by Ziegler, it seems that the majority of physicists that write for the purpose of explaining the foundations of relativity are satisfied with voodoo physics. And when physicists try to explain the foundations, they argue that linearity follows from the homogeneity of time and space, which is totally false, or they invoke a coordinate dependent version of Newton's first law, which presupposes an unacknowledged law of physics. Is it an intrinsic law of nature that says that somehow the universe favors a linear clock synchronization scheme or is it a law of the universe that clock synchronization schemes are unconstrained and free to be arbitrary? Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Re: Shubert's Derivation of the Lorentz Transformations [...] Physicists are not mathematicians. That linearity follows from the homogeneity of space and time is /good enough/. === Subject: Re: Shubert's Derivation of the Lorentz Transformations > [...] > > Physicists are not mathematicians. That linearity follows from the > homogeneity of space and time is /good enough/. To be specific, they are applied mathematicians where certain reasonable mathematical properties are assumed by insight into what you are applying it \ to. For example, when applying it infinite Markov chains, one usually assumes the limit process in the chain states going to infinity can be interchanged with the one raising it to a power. Interestingly a deeper analysis using renewal theory shoes it is always justified. But that is just in that case - often one makes such assumptions because the infinite case is really an approximation to a finite one - infinite chains can not actually physically occur. Same sort of considerations are made all the time in physics. Bill === Subject: Re: Shubert's Derivation of the Lorentz Transformations <0a7yi.23152$4A1.21755@news-server.bigpond.net.au> > > > > [...] > > > Physicists are not mathematicians. That linearity follows from the > > homogeneity of space and time is /good enough/. > > To be specific, they are applied mathematicians where certain reasonable > mathematical properties are assumed by insight into what you are applying \ it > to. For example, when applying it infinite Markov chains, one usually > assumes the limit process in the chain states going to infinity can be > interchanged with the one raising it to a power. Interestingly a deeper > analysis using renewal theory shoes it is always justified. But that is > just in that case - often one makes such assumptions because the infinite > case is really an approximation to a finite one - infinite chains can not > actually physically occur. Same sort of considerations are made all the > time in physics. I got a good example too. Right now I'm playing around with Einstein-Cartan theory. I'm trying to get the Schwarzschild solution that has spin. I'm toying with the idea that the Kerr solution might be obtainable [not through Schwarzschild - need an axially symmetric metric] as well. But Schwarzschild first. Without making some specific assumptions about the nature of the solution, I wouldn't get very far. If I didn't assume that spin is constant [w.r.t. the spherically symmetric coordinate system...], it would be impossible to solve. If I didn't assume reduction to Minkowski in the m,spin-->0 limit, I'd be clueless. If I didn't assume reduction to the Schwarzschild solution in the spin-->0 limit, I would not have made the cute discovery that the g_tt component splits as A(r)+B(t). Which was really goddamn cool. Ok, that's it because the third assumption *did* reduce the 5 PDE [4 unique] system to 3 [unique] PDEs of substantially lower difficulty. But I'm still stuck. However, if I played physics like Eugene wants to do, I could never do anything because I would be wasting my time going for full generality regardless of physical utility. > > Bill === Subject: Re: Shubert's Derivation of the Lorentz Transformations Alfred Ziegler gives an overview of representative derivations of the LT found in textbooks. See, The role of the two postulates of special VII. A SURVEY OF TEXTBOOKS A. Linearity The majority of textbooks sees no need to justify the linearity of the transformation5,7,8,9,10,11,12,13,14,15. When given it is derived either from homogeneity of time and space3,16,17, 18,19,20 or from Newton's first law which requires uniform and thus straight motion to remain uniform and straight in another system of reference4,21,22. Resnick3 is the most explicit text in this respect. Actually a lot more \obvious\ statements regarding the possible choices of coordinates have to be demonstrated which among the texts listed here are only given in Rindler17. Wow! According to this paragraph by Ziegler, it seems that the majority of physicists that write for the purpose of explaining the foundations of relativity are satisfied with voodoo physics. And when physicists try to explain the foundations, they argue that linearity follows from the homogeneity of time and space, which is totally false, or they invoke a coordinate dependent version of Newton's first law, which presupposes an unacknowledged law of physics. It is an intrinsic law of nature that says that somehow the universe favors a linear clock synchronization scheme or is it a law of the universe that clock synchronization schemes are unconstrained and free be arbitrary? Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Re: Shubert's Derivation of the Lorentz Transformations > Alfred Ziegler gives an overview of representative derivations of the > LT found in textbooks. See, The role of the two postulates of special > > VII. A SURVEY OF TEXTBOOKS > > A. Linearity > > The majority of textbooks sees no need to justify the linearity > of the transformation5,7,8,9,10,11,12,13,14,15. When given it > is derived either from homogeneity of time and space3,16,17, > 18,19,20 or from Newton's first law which requires uniform and > thus straight motion to remain uniform and straight in another > system of reference4,21,22. Resnick3 is the most explicit text > in this respect. Actually a lot more \obvious\ statements > regarding the possible choices of coordinates have to be > demonstrated which among the texts listed here are only given > in Rindler17. > > Wow! According to this paragraph by Ziegler, it seems that the > majority of physicists that write for the purpose of explaining the > foundations of relativity are satisfied with voodoo physics. And when > physicists try to explain the foundations, they argue that linearity > follows from the homogeneity of time and space, which is totally > false, or they invoke a coordinate dependent version of Newton's first > law, which presupposes an unacknowledged law of physics. It is an > intrinsic law of nature that says that somehow the universe favors a > linear clock synchronization scheme or is it a law of the universe > that clock synchronization schemes are unconstrained and free be > arbitrary? > > Shubeehttp://www.everythingimportant.org/relativity/special.pdf You may find also interesting Eugene's discussion on section 10.3.1 on http://www.arxiv.org/pdf/physics/0504062 I find beatiful the part {BLOCKQUOTE There are two common features of these derivations, which we nd troublesome. First, they assume an abstract (i.e., independent on real phys- ical processes and interactions) nature of events occupying space-time points (t, x, y, z). Second, they postulate the isotropy and homogeneity of space around these points. It is true that these assumptions imply linear univer- sal character of Lorentz transformations, and, after some algebra, speci c expressions (J.2) - (J.5) follow. The main problem with these approaches is that in physics we should be interested in transformations of observables cannot make an assumption that transformations of these observables are completely what system. One can reasonably assume that all directions in space are exactly equivalent for participates in interactions. distance from each other. Suppose that we want to derive boost transformations for directions in space are not equivalent: For example, the direction pointing to the 2 is di erent from other directions. So, the assumption of the spatial isotropy cannot be applied here. } Maxwell electrodynamics, QED, and GR are esentially one-body theories. Those theories doe not modell the *full* two-body system. That is the reason that textbooks you cite earlier introduce the one-body Lagrangian and maybe some two-body approximated Lagrangians (e.g. Darwin one) but never the *full* two-body system. http://canonicalscience.blogspot.com/2007/08/relativistic-lagrangian-and-lim\ itations.html Note for readers] Because some past episodes of flamming in sci.physics.relativity, both comments in my blog and my newsgroup e-mail are disabled. Note for 'experts' and 'professionals'] Avoid to reply this message if you are in my blacklist below. The capacity of any human for correcting your endless conceptual nonsenses and foolish mathematical mistakes is, unfortunately, just finite. Also my occupations do not include to teach you to read others, not to teach you dimensional analysis or even pre-university physics. Since you will be sanely ignored here in thereafter you are open to misread, misquote, or misinterpret me in any way you want, specially when that adds some light to your grey existence. You are open to write any triviality; to invent any mistake I did not really did. You can cite discredited, outdated, or wrong references. You can manipulate or misread references. You are also open to address any insult you consider supports your points and you can, of course, extend your insults to any poster, institution, colleague, friend, theories, or journal discrediting you. You can also try to falsify ratings, voting against me dozens or several hundred of times simulating different people. You can use the same dishonest tactic for increasing the rating of your akins. {BLACKLIST { Bilge; Bill Hobba; Dono (once Karandash2); Eric Gisse; Tim Shuba; }} === Subject: Re: Shubert's Derivation of the Lorentz Transformations On Aug 19, 7:13 am, \Juan R.\ > You may find also interesting Eugene's discussion on section 10.3.1 > on http://www.arxiv.org/pdf/physics/0504062 \There is a significant number of publications which claim that Lorentz transformation formulas (J.2) - (J.5) can be derived even without using the Einstein's second postulate (see [127, 128, 129, 130, 131] and references cited there). There are two common features of these derivations, which we find troublesome. First, they assume an abstract (i.e., independent on real physical processes and interactions) nature of events occupying space-time points (t, x, y, z). Second, they postulate the isotropy and homogeneity of space around these points.\ Why should the isotropy and homogeneity of space be considered troublesome? It's just a postulate. Can Eugene prove that this postulate is inconsistent? Eugene might call it a naive postulate but to the best of my understanding and belief, there is, at the moment, no conclusive argument or experimental evidence that contradicts it. Likewise, why should assuming the nature of space and time being independent of real physical processes and interactions be troublesome to anyone? Physicists are free to create any axiom set they wish. If Eugene is smart enough to devise mathematical equations for a new quantum theory where spacetime is changing while matter and energy are undergoing physical processes and interactions, God bless him. But not until Eugene gets there can his book be considered physics. Eugene Stefanovich is merely in the pursuit of physics. The meaning of words is important here. For my definition of axiomatic physics, see section 2 of http://www.everythingimportant.org/relativity/special.pdf mainstream physics is that too many physicists are driven by hotly pursuing grand theories while demonstrating a very poor understanding of the simplest fundamentals. Shubee http://www.everythingimportant.org/relativity/special.pdf > {BLACKLIST > { Bilge; Bill Hobba; Dono (once Karandash2); Eric Gisse; Tim > Shuba; }} === Subject: Re: Shubert's Derivation of the Lorentz Transformations On Aug 19, 7:13 am, \Juan R.\ > You may find also interesting Eugene's discussion on section 10.3.1 > on http://www.arxiv.org/pdf/physics/0504062 \There is a significant number of publications which claim that Lorentz transformation formulas (J.2) - (J.5) can be derived even without using the Einstein's second postulate (see [127, 128, 129, 130, 131] and references cited there). There are two common features of these derivations, which we find troublesome. First, they assume an abstract (i.e., independent on real physical processes and interactions) nature of events occupying space-time points (t, x, y, z). Second, they postulate the isotropy and homogeneity of space around these points.\ Why should the isotropy and homogeneity of space be considered troublesome? It's just a postulate. Can Eugene prove that this postulate is inconsistent? Eugene might call it a naive postulate but to the best of my understanding and belief, there is, at the moment, no conclusive argument or experimental evidence that contradicts it. Likewise, why should assuming the nature of space and time being independent of real physical processes and interactions be troublesome to anyone? Physicists are free to create any axiom set they wish. If Eugene is smart enough to devise mathematical equations for a new quantum theory where spacetime is changing while matter and energy are undergoing physical processes and interactions, God bless him. But not until Eugene gets there can his book be considered physics. Eugene Stefanovich is merely in the pursuit of physics. The meaning of words is important here. For my definition of axiomatic physics, see section 2 of http://www.everythingimportant.org/relativity/special.pdf. mainstream physics is that too many physicists are driven by hotly pursuing grand theories while demonstrating a very poor understanding of the simplest fundamentals. Shubee http://www.everythingimportant.org/relativity/special.pdf > {BLACKLIST > { Bilge; Bill Hobba; Dono (once Karandash2); Eric Gisse; Tim > Shuba; }} === Subject: Re: Shubert's Derivation of the Lorentz Transformations On Aug 19, 7:13 am, \Juan R.\ > You may find also interesting Eugene's discussion on section 10.3.1 on > http://www.arxiv.org/pdf/physics/0504062 \There is a significant number of publications which claim that Lorentz transformation formulas (J.2) - (J.5) can be derived even without using the Einstein's second postulate (see [127, 128, 129, 130, 131] and references cited there). There are two common features of these derivations, which we find troublesome. First, they assume an abstract (i.e., independent on real physical processes and interactions) nature of events occupying space-time points (t, x, y, z). Second, they postulate the isotropy and homogeneity of space around these points.\ Why should the isotropy and homogeneity of space be considered troublesome? It's just a postulate. Can Eugene prove that this postulate is inconsistent? Eugene might call it a naive postulate but to the best of my understanding and belief, there is, at the moment, no conclusive argument or experimental evidence that contradicts it. Likewise, why should assuming the nature of space and time being independent of real physical processes and interactions be troublesome to anyone? Physicists are free to create any axiom set they wish. If Eugene is smart enough to devise mathematical equations for a new quantum theory where spacetime is changing while matter and energy are undergoing physical processes and interactions, God bless him. But not until Eugene gets there can his book be considered physics. Eugene Stefanovich is merely in the pursuit of physics. The meaning of words is important here. For my definition of axiomatic physics, see section 2 of http://www.everythingimportant.org/relativity/special.pdf. mainstream physics is that too many physicists are driven by hotly pursuing grand theories while demonstrating a very poor understanding of the simplest fundamentals. Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Re: Shubert's Derivation of the Lorentz Transformations Alfred Ziegler gives an overview of representative derivations of the LT found in textbooks. See, The role of the two postulates of special http://www.arxiv.org/abs/0708.0988 VII. A SURVEY OF TEXTBOOKS A. Linearity The majority of textbooks sees no need to justify the linearity of the transformation5,7,8,9,10,11,12,13,14,15. When given it is derived either from homogeneity of time and space3,16,17, 18,19,20 or from Newton's first law which requires uniform and thus straight motion to remain uniform and straight in another system of reference4,21,22. Resnick3 is the most explicit text in this respect. Actually a lot more \obvious\ statements regarding the possible choices of coordinates have to be demonstrated which among the texts listed here are only given in Rindler17. Wow! According to this paragraph by Ziegler, it seems that the majority of physicists that write for the purpose of explaining the foundations of relativity are satisfied with voodoo physics. And when physicists try to explain the foundations, they argue that linearity follows from the homogeneity of time and space, which is totally false, or they invoke a coordinate dependent version of Newton's first law, which presupposes an unacknowledged law of physics. It is an intrinsic law of nature that says that somehow the universe favors a linear clock synchronization scheme or is it a law of the universe that clock synchronization schemes are unconstrained and free be arbitrary? Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Re: GOLDBACH CONJECTURE <46c111e5$0$21143$7a628cd7@news.club-internet.fr> On Aug 13, 7:22 pm, Denis Feldmann quasi a ?crit : > > > > > >> Let P i be an odd prime < N. > > >> Let P j + P k <= N & P j + P (k+1) > N. > > >> For a given N, what is the probability that for some j & k; P j + > >> P k = N? > > > It's kind of a silly question. > > > Firstly, since you specified that P j, P k are required to be odd > > primes, if N is odd, the probability is 0. > > > If N is even, N>=6, the probability is 1 iff N is a sum of 2 primes. > > > Thus, if Goldbach's conjecture is true, then for all even N, N>=6, the > > probability is 1. Equivalently, if for some even N, N>=6, the > > probability is less than 1, then Goldbach's conjecture is false. > > > Thus, in effect, you are asking if Goldbach's Conjecture is true. > > > quasi > > As it is formulated, of course, you are right. What the OP probably > means is that, with a distribution of numbers P i following prime > density (ie Prob(P i is prime)=1/log P i ; those are no real > probabilities, of course, but just asymptotic density formulations), > perhaps with correction to allow for N even, what is the \probability\ > that... > > These, and similar questions, have long be studied (Hardy and Littlewood > were probably the first to do it, though I would suspect Gauss had a few > resultzs of his own) and give surpisingly good results (ie close to > experimental ones) ... which dont prove a thing, as possible > discrepancies could well appear for very large N (as the results for > Li(x) and the Skewes number have shown) > > If we suppose those \probabilistic\ estimates and methods are right, > then Goldbach conjecture, twin primes conjecture and many other are > true (in Goldbach case, \extremely\ true) > > But nobody has yet found a way to a proof by that road... The point is moot. The GC cannot be true unless there are an infinite number of primes! Bill J === hexagon result & its generalization via proof\, The Montana Mathematics Enthusiast, June 2007, Vol. 4, No. 2, which can be downloaded at my homepage at \ http://mysite.mweb.co.za/residents/profmd/homepage4.html illustration of the 'discovery' function of proof leading to a generalization, which would be accessible to talented high school students as well as undergraduate students. Also click on the blue link called 'hexagon centroids' to view a short online animation video clip illustrating the generalization for the hexagon case. My homepage also contains the usual bi-monthly mathematical/ mathematics education quote and cartoon at the bottom. === Subject: Set notation - Substitution Two questions: 1. Say I have a set S={a,b,c,d,...}. Now, I want to define a new set R such that R is the same as S, except that say c is replaced by some other element #. So, R={a,b,#,d,...}. Is there a way I could define R in terms of S, without it being it too awkward? For example, I find \R=S\\{a} U {#} \ awkward. Is there a better (and clear) way to express this? 2. Game theorists often define a game G as \G=<...>\, where the \<\ and \>\ look like big parentheses with a sharp angle. Is there any reason to use this notation rather than simply \{\ and \}\? === Subject: Re: Set notation - Substitution > For example, I find \R=S\\{a} U {#} \ awkward. Is there a better (and > clear) way to express this? There is no shorter standard way to write S\\{a} union {#}. If you're writing a paper you could just introduce some ad hoc notation -- e.g. S[a/#]. Be sure it's worth it, though, and does not merely serve to confuse your readers. > Game theorists often define a game G as \G=<...>\, where the \<\ and > \>\ look like big parentheses with a sharp angle. Is there any > reason to use this notation rather than simply \{\ and \}\? Yes. A game G is a tuple, not a set. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \Wovon man nicht sprechen kann, daruber muss man schweigen\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Set notation - Substitution <87bqd3ns35.fsf@huxley.huxley.fi> === Subject: Re: Set notation - Substitution > > 1. Say I have a set S={a,b,c,d,...}. Now, I want to define a new set R > such that R is the same as S, except that say c is replaced by some > other element #. So, R={a,b,#,d,...}. > > Is there a way I could define R in terms of S, without it being it too > awkward? For example, I find \R = S \\ {c} u {#}\ awkward. > Awkward or not, this is the most \natural\ way to get what you want. R = (S \\ {c}) u {#} ---> R is S (except) containing # instead of c. (Using the outdated \+\ symbol for \u\ and \-\ this could be written: R = S - {c} + {#}. With preference from left to right.) > > Is there a better (and clear) way to express this? > I don't think that the following is much clearer: R = {x e S : x =/= c} u {#}. > > 2. Game theorists often define a game G as \G = <...>\, where the \<\ > and \>\ look like big parentheses with a sharp angle. Is there any > reason to use this notation rather than simply \{\ and \}\? > No idea concerning game theory, but in general set theory \<.,.>\ usually denotes an ordered pair. (The concept can be generalized to more than two \elements\). F. -- E-mail: infosimple-linede === Subject: Solutions for Mechanical Design of Machine Elements and Machines: A \ Failure I am interested in the solutions manual for Mechanical Design of Machine Elements and Machines: A Failure Prevention Perspective (Collins). If anyone has this solutions manual let me know? === Subject: Sum of infinite Power Series of Square Matrix Hi world, Can some good soul help me out, i need the closed form of following expression, S^-1 = sum(1 to inf){ (A^-1)*(C)*((A')^-1) } Now A , C & S are squre matrixes ......n X n order as such my target is S (inverse of LHS), But if i can get closed form of RHS even then i will able to get S (quite obviuos isn't it ;) ) if we try scalar case result then the above series converges for |A| <1...... Now how do i interpret in Matrix case. any type of help will be appriciated.................. Jang === Subject: Re: Sum of infinite Power Series of Square Matrix > Hi world, > > Can some good soul help me out, i need the closed form of following > expression, > > S^-1 = sum(1 to inf){ (A^-1)*(C)*((A')^-1) } My guess is that the sum you want is sum_{j=1}^infty A^(-j) C (A')^(-j) (which would converge for ||A^(-1)|| < 1 where ||.|| is any (sub-multiplicative) matrix norm). However, I don't see why this sum should necessarily be invertible. It certainly won't be if C is singular and commutes with A. Please tell us what the actual question is. > Now A , C & S are squre matrixes ......n X n order > > as such my target is S (inverse of LHS), But if i can get closed > form of RHS even then i will able to get S (quite obviuos isn't > it ;) ) > > if we try scalar case result then the above series converges for |A| > <1...... Now how do i interpret in Matrix case. > > any type of help will be appriciated.................. > > Jang > -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Sum of infinite Power Series of Square Matrix On Aug 19, 8:56 pm, Robert Israel > > Hi world, > > > Can some good soul help me out, i need the closed form of following > > expression, > > > S^-1 = sum(1 to inf){ (A^-1)*(C)*((A')^-1) } > > My guess is that the sum you want is sum_{j=1}^infty A^(-j) C (A')^(-j) > (which would converge for ||A^(-1)|| < 1 where ||.|| is any > (sub-multiplicative) matrix norm). However, I don't see why > this sum should necessarily be invertible. It certainly won't be if > C is singular and commutes with A. > > Please tell us what the actual question is. > > > Now A , C & S are squre matrixes ......n X n order > > > as such my target is S (inverse of LHS), But if i can get closed > > form of RHS even then i will able to get S (quite obviuos isn't > > it ;) ) > > > if we try scalar case result then the above series converges for |A| > > <1...... Now how do i interpret in Matrix case. > > > any type of help will be appriciated.................. > > > Jang > > -- > Robert Israel isr...@math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Mr. Robert got it correctly; let me state the problem again, S^-1 = sum (j= 1 to inf) { (A^-j) * (C) * ((A')^-j) }.............. Can a closed form for RHS exit if so under what conditions, Mr. Robert, I understand resulting RHS might not be invertible, I am having different A & C matrixes only invertible result interests me not the condition of existence right now...... actually I have to run simulation for any n order system, and I am unable to do the infinite series sum in Matlab. What I can do is if I know the condition for convergence then I can choose A & C accordingly and with given closed form solution quickly find out the sum. With that only invertible results either S are to be used up ahead in simulations. please let me know ur thoughts. Jang === Subject: Re: Sum of infinite Power Series of Square Matrix > Hi world, > > Can some good soul help me out, i need the closed form of following > expression, > > S^-1 = sum(1 to inf){ (A^-1)*(C)*((A')^-1) } > > Now A , C & S are squre matrixes ......n X n order > > as such my target is S (inverse of LHS), But if i can get closed > form of RHS even then i will able to get S (quite obviuos isn't it ;) > ) > > if we try scalar case result then the above series converges for |A| > <1...... Now how do i interpret in Matrix case. > > any type of help will be appriciated.................. > > Jang > It appears to me that the term being added infinitely many times is constant: that is, it doesn't vary from one term to the next in the infinite sum. If that is correct, you aren't going to get convergence if that (infinitely-repeated) term is nonzero. Dale === Subject: confused about field axioms Hello I have a basic question about field axioms we define in real analysis. There are basic 8 axioms and plus there are axioms about ordering. Using basic 8 axioms we derive that (a)(0)=0 for some real number 'a'. Now the way we define basic multiplication as follows. 2 x 2 = 2 added two times = 2+2 = 4 2 x 1 = 2 added one time = 2 similarly, can't we define 2 x 0 = 2 added no times = nothing = 0 ? Beacause if we can do this, we can make the statement (a)(0) = 0 as one of the axioms and then eliminate few axioms from list of 8 axioms. May be I am wrong here. But I always had trouble with this list of 8 axioms. Because some of the axioms just seem 'obvious'. Take for example following one. For real numbers a,b we have a+b = b+a Now if I take 2 apples in one hand and 3 apples in another, or other way round, take 3 apples in hand and 2 in another, we will end up with 5 apples. Now I understand that I am only dealing with 'whole' numbers here, but cutting apples suitably we can extend same argument for non integral numbers. So I have trouble understanding 'deep truth' in these axioms. We also prove that (-1)(-1)=1 using these axioms. Again like my argument above, why cant we take this statement as axiom and eliminate some in original list of axioms. I am basically from physics background but I have taken introductory course in analysis. and I am really having trouble with these 'axioms'.I love math very much. But some concepts like this in analysis just drive me nuts. Wood === Subject: Re: confused about field axioms Hi Wood === Subject: Re: confused about field axioms > Hello > > I have a basic question about field axioms we define in > real analysis. There are basic 8 axioms and plus there > are axioms about ordering. Using basic 8 axioms we derive > that (a)(0)=0 for some real number 'a'. Now the way we > define basic multiplication as follows. > > 2 x 2 = 2 added two times = 2+2 = 4 > 2 x 1 = 2 added one time = 2 > > similarly, can't we define > > 2 x 0 = 2 added no times = nothing = 0 ? > > Beacause if we can do this, we can make the statement > > (a)(0) = 0 > > as one of the axioms and then eliminate few axioms from > list of 8 axioms. May be I am wrong here. But I always > had trouble with this list of 8 axioms. Because some > of the axioms just seem 'obvious'. Take for example > following one. For real numbers a,b we have > > a+b = b+a > > Now if I take 2 apples in one hand and 3 apples in > another, or other way round, take 3 apples in > hand and 2 in another, we will end up with > 5 apples. Now I understand that I am only > dealing with 'whole' numbers here, but cutting > apples suitably we can extend same argument > for non integral numbers. > So I have trouble understanding 'deep truth' > in these axioms. We also prove that (-1)(-1)=1 > using these axioms. Again like my argument above, > why cant we take this statement as axiom and > eliminate some in original list of axioms. > I am basically from physics background but > I have taken introductory course in analysis. > and I am really having trouble with these > 'axioms'.I love math very much. But some concepts > like this in analysis just drive me nuts. > > Wood These axioms are \obvious\ since they're supposed to be _abstractions_ of your arithmetic (+, -, *, /). You take every \reasonable\ property that +, -, *, / should be expected to have, and just call any operation satisfying those axioms +, -, *, or /. This way you can define an \addition\, \subtraction\, \multiplication\ and \division\ on any set you want, not just real numbers(*). Don't you think that if you have something called \addition\ it should commute, eh? Axioms are assumed as truth from the start, after all isn't that why they're called \axioms\? (*) Actually that is not quite true -- you can only have a field on a set of p or p^n elements where p is a prime number and n is a nonnegative integer, or of infinite elements. === Subject: Re: confused about field axioms On Sun, 19 Aug 2007 18:08:49 EDT, woodland > > But I always had trouble with this list of 8 axioms. > Because some of the axioms just seem 'obvious'. > That's exactly why it is reasonable (in certain cases) to explicitly state them as axioms. After all we use axioms as a starting point for our proofs. > > Take for example following one. For real numbers a,b we have > > a+b = b+a > > Now if I take 2 apples in one hand and 3 apples in > another, or other way round, take 3 apples in > hand and 2 in another, we will end up with > 5 apples. Now I understand that I am only > dealing with 'whole' numbers here, but cutting > apples suitably we can extend same argument > for non integral numbers. > See comment above. (Note though that sometimes axioms are less obvious.) > > So I have trouble understanding 'deep truth' in these axioms. > Axioms usually DON'T formulate 'deep truths'. We have to _derive_ them from our axioms. > > We also prove that (-1)(-1) = 1 using these axioms. > Right. Since usually this statement is not adopted as an axiom we have to _derive_ it from our axioms (i.e. we have to /prove/ it). > > [...] why cant we take this statement as axiom > and eliminate some in original list of axioms. > We MIGHT do that, of course. But you would have to show then that your new system of axioms allows to derive the same theorems as your old one. (Note that an axiom -for technical reasons- also counts as a theorem.) So if you eliminate some axioms from your original list of axioms, say A, B, C, and adopt (-1)(-1) = 1 as a new axiom instead, you would have to show that A, B, C can be derived from the new system. Actually it turns out that \(-1)(-1) = 1\ is not very helpful (used) as an axiom. F. -- E-mail: infosimple-linede === Subject: Re: confused about field axioms <8643824.1187561360302.JavaMail.jakarta@nitrogen.mathforum.org>, > Hello > > I have a basic question about field axioms we define in > real analysis. There are basic 8 axioms and plus there > are axioms about ordering. Using basic 8 axioms we derive > that (a)(0)=0 for some real number 'a'. Now the way we > define basic multiplication as follows. > > 2 x 2 = 2 added two times = 2+2 = 4 > 2 x 1 = 2 added one time = 2 > > similarly, can't we define > > 2 x 0 = 2 added no times = nothing = 0 ? > > Beacause if we can do this, we can make the statement > > (a)(0) = 0 > > as one of the axioms and then eliminate few axioms from > list of 8 axioms. May be I am wrong here. But I always > had trouble with this list of 8 axioms. Because some > of the axioms just seem 'obvious'. Take for example > following one. For real numbers a,b we have > > a+b = b+a > > Now if I take 2 apples in one hand and 3 apples in > another, or other way round, take 3 apples in > hand and 2 in another, we will end up with > 5 apples. Now I understand that I am only > dealing with 'whole' numbers here, but cutting > apples suitably we can extend same argument > for non integral numbers. > So I have trouble understanding 'deep truth' > in these axioms. We also prove that (-1)(-1)=1 > using these axioms. Again like my argument above, > why cant we take this statement as axiom and > eliminate some in original list of axioms. I think you have it backwards. Axioms should be \obvious\; you derive deep truths from them. (-1)(-1)=1 is not that obvious; lots of students have trouble conceptualizing it. Better to derive it than just assume it. > I am basically from physics background but > I have taken introductory course in analysis. > and I am really having trouble with these > 'axioms'.I love math very much. But some concepts > like this in analysis just drive me nuts. > > Wood === Subject: Re: confused about field axioms > Hello > I have a basic question about field axioms we define in > real analysis. There are basic 8 axioms and plus there > are axioms about ordering. Using basic 8 axioms we derive > that (a)(0)=0 for some real number 'a'. Now the way we > define basic multiplication as follows. > 2 x 2 = 2 added two times = 2+2 = 4 > 2 x 1 = 2 added one time = 2 > similarly, can't we define > 2 x 0 = 2 added no times = nothing = 0 ? > Beacause if we can do this, we can make the statement > (a)(0) = 0 Yes, there is some redundancy in the field axioms. If F is a field, then for each a in F we can conclude from the distributive law that a*0 = a*(0+0) = a*0 + a*0 and therefore a*0 = 0. > as one of the axioms and then eliminate few axioms from > list of 8 axioms. May be I am wrong here. But I always > had trouble with this list of 8 axioms. Because some > of the axioms just seem 'obvious'. Take for example > following one. For real numbers a,b we have > a+b = b+a The axioms may seem less obvious if you recall that we are not just talking about the real numbers here. There are lots of fields besides the reals. The axioms provide the definition of a field, so that we can distinguish fields from non-fields. > Now if I take 2 apples in one hand and 3 apples in > another, or other way round, take 3 apples in > hand and 2 in another, we will end up with > 5 apples. Now I understand that I am only > dealing with 'whole' numbers here, but cutting > apples suitably we can extend same argument > for non integral numbers. > So I have trouble understanding 'deep truth' > in these axioms. We also prove that (-1)(-1)=1 > using these axioms. Again like my argument above, > why cant we take this statement as axiom and > eliminate some in original list of axioms. > I am basically from physics background but > I have taken introductory course in analysis. > and I am really having trouble with these > 'axioms'.I love math very much. But some concepts > like this in analysis just drive me nuts. A triangle is defined to be a polygon with three sides. Is there a \deep truth\ in that definition? Not at all. It's nothing more than a way to distinguish between triangles and non-triangles. Not everything is a triangle. Similarly, not everything is a field. For example, the integers are not a field. The ring of 2x2 matrices is not a field. The rationals are a field. The hyperreals are a field. The integers modulo p are a field whenever p is prime, and not otherwise. We need a way to tell the difference between fields and non-fields, so that when we prove theorems about fields, we know when the theorems apply and when they don't. That's what definitions do. -- Dave Seaman Oral Arguments in Mumia Abu-Jamal Case heard May 17 U.S. Court of Appeals, Third Circuit === Subject: Re: confused about field axioms >> as one of the axioms and then eliminate few axioms from >> list of 8 axioms. May be I am wrong here. But I always >> had trouble with this list of 8 axioms. Because some >> of the axioms just seem 'obvious'. Take for example >> following one. For real numbers a,b we have > >> a+b = b+a > > The axioms may seem less obvious if you recall that we are not just > talking about the real numbers here. There are lots of fields besides > the reals. The axioms provide the definition of a field, so that we can > distinguish fields from non-fields. Indeed. But it turns out that this particular axiom is redundant, since a + b = b + a <=> a + b + (-a) + (-b) = 0 <=> a + b + (-1)*a + (-1)*b = 0 <=> 1*(a + b) + (-1)*(a + b) = 0 <=> (1 + (-1))*(a + b) = 0 <=> 0*(a + b) = 0, which is true. Of course, saying that it is 'redundant' doesn't mean that it is 'obvious'. Jose Carlos Santos === Subject: Re: confused about field axioms > > [...] the field axiom [a+b = b+a] is redundant [...] You can find an in-depth discussion of this in my prior post [1]. have commutative addition. Such semirings are simply subsemirings embeds canonically into a commutative group, its group of differences. See [1] for definitions, examples, and references. --Bill Dubuque === Subject: Re: confused about field axioms > You can find an in-depth discussion of this in my prior post [1]. > have commutative addition. Such semirings are simply subsemirings > embeds canonically into a commutative group, its group of differences. > See [1] for definitions, examples, and references. > > --Bill Dubuque > Jose Carlos Santos === Subject: Re: confused about field axioms \ <5is30nF3q7so1U1@mid.individual.net> > > >> For real numbers a,b we have > > > >> a+b = b+a > > Indeed. But it turns out that this particular axiom is redundant, since > > a + b = b + a <=> a + b + (-a) + (-b) = 0 > > <=> a + b + (-1)*a + (-1)*b = 0 > > <=> 1*(a + b) + (-1)*(a + b) = 0 > > <=> (1 + (-1))*(a + b) = 0 > > <=> 0*(a + b) = 0, > > which is true. > By the same reasoning, a + b = b + a is redundant in any ring with identity. However now, is not modern terminology 'unity' to distinguish between additive and multiplicative identities? Also 0a = 0 is not a ring axiom, but a ring theorem. === Subject: Re: confused about field axioms >>>> For real numbers a,b we have >>>> a+b = b+a >> Indeed. But it turns out that this particular axiom is redundant, since >> >> a + b = b + a <=> a + b + (-a) + (-b) = 0 >> >> <=> a + b + (-1)*a + (-1)*b = 0 >> >> <=> 1*(a + b) + (-1)*(a + b) = 0 >> >> <=> (1 + (-1))*(a + b) = 0 >> >> <=> 0*(a + b) = 0, >> >> which is true. >> > By the same reasoning, a + b = b + a is redundant in any ring with > identity. However now, is not modern terminology 'unity' to distinguish > between additive and multiplicative identities? > > Also 0a = 0 is not a ring axiom, but a ring theorem. I know. And so are x = y <=> x + (-y) = 0 or -a = (-1)*a. I never claimed that I was working directly from axioms. Jose Carlos Santos === Subject: Re: JSH: Perfect factoring algorithm [JSH ] > [yet another \perfect factoring algorithm\, quickly > followed by corrections and renewed & intensified > claims of perfection] >> ... [various people try to make sense of it, even try it, >> & get piss-poor results] ... > ... > But I think you made a mistake. That's what you always think. > I'm going to review the proof tomorrow, but it's so trivial that the > check is a formality. That's what you always say. > You can either look over that proof explained by me in two threads > today and find an error, or go back over your code and figure out what > you did wrong. Third choice: he can just ignore it. Saves time for him, and doesn't make \ any difference to you :-) > Sorry if I did make a mistake and just can't see it yet. > > But the proof looks VERY simple to me at this point, so it's hard to > see where there is any failure on my part so I think the failure is > yours. While I bet your posts in this thread will be deleted from Google Groups within a week -- depends on how long it takes you to do some /actual/ thinking, eh? === Subject: Re: JSH: Perfect factoring algorithm > [JSH ] > > > [yet another \perfect factoring algorithm\, quickly > > followed by corrections and renewed & intensified > > claims of perfection] > >> ... [various people try to make sense of it, even try it, > >> & get piss-poor results] ... > > ... > > But I think you made a mistake. > > That's what you always think. > > > I'm going to review the proof tomorrow, but it's so trivial that the > > check is a formality. > > That's what you always say. > > > You can either look over that proof explained by me in two threads > > today and find an error, or go back over your code and figure out what > > you did wrong. > > Third choice: he can just ignore it. Saves time for him, and doesn't \ make > any difference to you :-) In fact I went for option four: remain unaware of James' reply until you quoted it, since he deleted it from Google before I had had a chance to read it. Having since done so, though, I feel that it's a pity for James that he did delete it since his reply is a rare example of him acting in a civilised manner. For a start, when he was unclear about what he intended in the specification of his algorithm and I asked him a couple of questions, he actually gave unambiguous answers. Also, despite the fact that I was claiming results that were at odds with his predictions, he didn't once call me a liar. He even went as far as apologising in advance if he turned out to be mistaken! He deserved credit for this, though of course he has since gone back to his usual \sci.math posters are villains\ routine. === Subject: Re: JSH: Perfect factoring algorithm > While I bet your posts in this thread will be deleted from Google Groups > within a week Not really going out on a limb. Especially since you posted this after James deleted his posts. - William Hughes === Subject: Re: JSH: Perfect factoring algorithm ... [Tim Peters, goes out on a limb] >> While I bet your posts in this thread will be deleted from Google Groups >> within a week [William Hughes] > Not really going out on a limb. Especially since you posted this > after James deleted his posts. So I did! I was replying in the order messages showed up in my newsreader, \ and at the time I didn't realize James had /already/ tried to erase history \ on this one. Or that's just another sci.math lie from just another of the habitual sci.math liars. It always clarifies seeming mysteries to see things from JSH's POV, eh, \William Hughes\? === Subject: Re: JSH: Perfect factoring algorithm > > > While I bet your posts in this thread will be deleted from Google \ Groups > > within a week > > Not really going out on a limb. Especially since you posted this > after James deleted his posts. > > - William Hughes While I think Tim posted after James deleted his posts from Google Groups, it probably wasn't immediately apparent to him because Tim was posting from another newsgroup service in which the text, at least of James' most recent antecedent post, was still visible (and quotable). But James does seem to have made a big deal about his courage to fail in a spectacular way in another sci.math thread. As he rationalizes there, Newton and Gauss never needed to apologize for repeated public errors because they lived in \simpler times\. Note: Newton's survival of the Great Plague of 1665 and Gauss's survival of the Napoleonic Wars must seem like child's play to the instigator of the Math Wars!! === Subject: Re: JSH: Perfect factoring algorithm Analysis on which this was based was flawed. I'm deleting everything out of Google Groups. James Harris > Algorithm for guaranteeing the factorization of a target composite T > of any size. > > Use (x+k)^2 = y^2 + 2k^2 + n*T. > > 1. Pick k=2 and an n that gives you an absolute value of 2k^2 + n*T > above the minimums, where you can choose > > abs(2k^2 + n*T)> 2x_res*T where > > x_res = k*(2)^{-1} mod T > > and solve for x and y by factoring 2k^2 + n*T, iterating through > integer combinations to check with all possible values for integer x > and y. > > 2. If your first n does not factor non-trivially increment its > absolute value by 1 and try again, and do this for a maximum of 3 > increments and you are guaranteed to factor T without regard to the > size of T. > > And that is the perfect factoring algorithm. > > James Harris === Subject: Re: JSH: Perfect factoring algorithm > Analysis on which this was based was flawed. I'm deleting everything > out of Google Groups. > > James Harris > Concession accepted. === Subject: Re: JSH: Perfect factoring algorithm > Analysis on which this was based was flawed. I'm deleting everything > out of Google Groups. > > James Harris cunt === Subject: Re: JSH: Perfect factoring algorithm > > >>Analysis on which this was based was flawed. I'm deleting everything >>out of Google Groups. >> >>James Harris > > > cunt > Ummm, I think calling James an \ASS\ would be way more appropriate... after all, you are what you eat (and James has certainly never tasted the sweet stuff)! Bob Terwilliger === Subject: Re: JSH: Perfect factoring algorithm <13chbd158npqub7@corp.supernews.com> On Aug 19, 2:50 pm, Bob Terwilliger > > >>Analysis on which this was based was flawed. I'm deleting everything > >>out of Google Groups. > > >>James Harris > > > cunt > > Ummm, I think calling James an \ASS\ would be way more appropriate... > after all, you are what you eat (and James has certainly never tasted > the sweet stuff)! > > Bob Terwilliger I'm just curious: He conceded he was wrong, so what's the problem? === Subject: Re: JSH: Perfect factoring algorithm > On Aug 19, 2:50 pm, Bob Terwilliger >>>> Analysis on which this was based was flawed. I'm deleting everything >>>> out of Google Groups. >>>> James Harris >>> cunt >> Ummm, I think calling James an \ASS\ would be way more appropriate... >> after all, you are what you eat (and James has certainly never tasted >> the sweet stuff)! >> >> Bob Terwilliger > > I'm just curious: He conceded he was wrong, > so what's the problem? > At a minimum, the fact that he has not apologized (and will not apologize) to all those he called liars who were indeed speaking the truth. === Subject: Re: JSH: Perfect factoring algorithm > I'm going to review the proof tomorrow, but it's so trivial that the > check is a formality. > > You can either look over that proof explained by me in two threads > today and find an error, or go back over your code and figure out what > you did wrong. > > Sorry if I did make a mistake and just can't see it yet. > > But the proof looks VERY simple to me at this point, so it's hard to > see where there is any failure on my part so I think the failure is > yours. Your proof is correct. We have already written a simple Java implementation of the algorithm that can factor a 120-digit number in less than an hour. We should soon be able to speed this up greatly. MIT/Math === Subject: Re: JSH: Perfect factoring algorithm >Algorithm for guaranteeing the factorization of a target composite T >of any size. > >Use (x+k)^2 = y^2 + 2k^2 + n*T. > >1. Pick k=2 and an n that gives you an absolute value of 2k^2 + n*T >above the minimums, where you can choose > >abs(2k^2 + n*T)> 2x_res*T where > >x_res = k*(2)^{-1} mod T > >and solve for x and y by factoring 2k^2 + n*T, iterating through >integer combinations to check with all possible values for integer x >and y. > >2. If your first n does not factor non-trivially increment its >absolute value by 1 and try again, and do this for a maximum of 3 >increments and you are guaranteed to factor T without regard to the >size of T. > >And that is the perfect factoring algorithm. > > >James Harris As I have done previously, I tested James' latest version of his surrogate method on 500 random composite odd numbers that are multiples of two different primes, each in the range 500 to 1000. The results are compared to Fermat's method, trial factorisation (both forward and reverse) and random picking: Fermat average = 7.43 probes. JSH average = 1942.92 probes. Probe ratio = 1 : 261.567 Trial average = 120.76 probes. Reverse average = 12.08 probes. Random average = 783.65 probes. 500 trials, 0 misfactors found. For Fermat a \probe\ is the extraction of a square root. For JSH a probe is taking a GCD. For trial factorisation, reverse trial factorisation and random picking a probe is taking a modulus. As seems usual with James' methods it will factorise the target number, but more slowly than existing methods. rossum === Subject: solutions I was just wondering if anyone here have the solution for Materials Science And Engineering 6th Edition by William D. Callister,Jr. and Fundamentals of Engineering Thermodynamics 6th edition(2007) by Moran Please let me kno ASAP === Subject: Re: Help with tensor question [..] > Even if you ignore the second solution and just look at the original > equation, how can the sum of those n terms be 1 when i==j? Shouldn't > it be n? As Tommy eluded to, you are misinterpreting (Dx'/Dx) to be the inverse of (Dx/Dx') which is not the case as the former is such that y and z are held constant and the latter such that x' and z' are held constant and note that x' != x and y' != y. === Subject: Re: Help with tensor question > are held constant and the latter such that x' and z' are held constant CORRECTION: \such that y' and z' are held constant\ === Subject: Re: Help with tensor question > > > > > Hi > > > I am having trouble interpreting a tensor question from Shaum's > > Outlines Tensor Calculus. > > > On Page 22, question 2.42 a) reads: > > > Show that for independent functions x'_i = x'_i(x_1, x_2, ..., x_n) > > this relation holds > > > (Dx'_i / Dx_r) (Dx_r / Dx'_j) = K_ij > > > where D denotes partial diff, K the kronecka delta tensor and Einstein > > notation is used. > > > My first solution is to parameterize x_i with x'_i > > > x_i=x_i(x'_1, ..., x'_n) > > > then apply the chain rule to x'_i > > > (Dx'_i/Dx'_j) = (Dx'_i/Dx_r) (Dx_r/Dx'_j) > > > which reduces the LHS to the kronecka delta function since (Dy_p/Dy_q) > > = K_pq. > > > Is that a valid solution? > > > Now, my second solution is to swap the denominators and simplfy > > > (Dx'_i / Dx_r) (Dx_r / Dx'_j) = (Dx'_i / Dx'_j) (Dx_r / Dx_r ) = K_ij > > * 1 > > > But how can that be correct? Suppose i=j=1 and we are summing over 3 > > indices then expanding the sums (and renaming components intuitively) > > we get > > > 1 = (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') > > > If we swap denominators for each term, shouldn't the sum be equal to > > 3? > > > Even if you ignore the second solution and just look at the original > > equation, how can the sum of those n terms be 1 when i==j? Shouldn't > > it be n? > > > I think I am not interpreting something correctly or have made an > > error. Can someone point it out to me please? > > 1 = (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') > > > If we swap denominators for each term, shouldn't the sum be equal to > > 3? > > > Even if you ignore the second solution and just look at the original > > equation, how can the sum of those n terms be 1 when i==j? Shouldn't > > it be n? > > > I think I am not interpreting something correctly or have made an > > error. Can someone point it out to me please? > > Hey > You must look at this question from the algebraic point of view... not > in terms of differentials. > x'_i = x'_i(x_1, x_2, x_3) > x_i = x_i(x'_1, x'_2, x'_3) > represent transformation equations from one set of co-ordinates x to > another co-ordinate system x' > define function r(x_1, x_2, x_3) , which can be interpreted as the > intersection of the 3 surfaces x_1, x_2 , x_3 (because its not really > defined anywhere else except in the intersection) > notice that (Dx'_i/Dx_1) (Dx_1/Dx'_j) + (Dx'_i/Dx_2) (Dx_2/Dx'_j) + > (Dx'_i/Dx_3) (Dx_3/Dx'_j) is grad(x'_i)(dotproduct)Dr/Dx'_j > > now you can represent E^i = grad(x'_i) and E_i = Dr/Dx'_j ans your > covariant and contravariant basis respectivley > > so then it is clear to see..... E^i(dotproduct)E_j = K_ij by > definition convention and expanding the sum. If we expand the below equation for when i=j K_ij = (Dx'_i/Dx_r) (Dx_r/Dx'_j) we get (r being the dummy index) 1 = (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') It seems to me the sum should equal 3, not 1? Why isn't 3? === Subject: Re: Help with tensor question > > 1 = (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') the answer was in my original post.. (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') = grad(x') (dotproduct)Dr/Dx' = 1 by def. of covariant and contravariant basis. then adding 1 + 1 + 1 = 3.. which you cant really do.. because x' = x'(x,y,z) and x = x(x',y',z') are transformation equations for the x co-ordinate, so they are partial derivatives. (Dx'/ Dx) (Dx/Dx') is like saying the derivative of the x' transformation equation w.r.t x multiplied by the partial deriv. of the x transformation equation w.r.t x' , look at the transformation from cartesian to polar co-ordinates as an example. its a bit confuzzling, but im sure u'll be allright === Subject: Re: Help with tensor question hey > > 1 = (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') the answer was in my original post.. (Dx'/Dx) (Dx/Dx') + (Dx'/Dy) (Dy/Dx') + (Dx'/Dz) (Dz/Dx') = grad(x') (dotproduct)Dr/Dx' = 1 by def. of covariant and contravariant basis. then adding 1 + 1 + 1 = 3.. which you cant really do.. because x' = x'(x,y,z) and x = x(x',y',z') are transformation equations for the x co-ordinate, so they are partial derivatives. (Dx'/ Dx) (Dx/Dx') is like saying the derivative of the x' transformation equation w.r.t x multiplied by the partial deriv. of the x transformation equation w.r.t x' , look at the transformation from cartesian to polar co-ordinates as an example. its a bit confuzzling, but im sure u'll be allright === Subject: Log Linear Models Why would you use log linear models in a study? === Subject: Re: Log Linear Models >Why would you use log linear models in a study? If your theory suggested that they are appropriate. Why would you use linear models? -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Why is it so difficult? <1186767883.483331@athprx03> <1187214872.333199@athprx03> <1187226156.666042@athprx04> <1187256998.555858@athprx04> <1187294996.327177@athprx04> <1187310648.32445@athprx03> > > > > Well I read the page, and no, I won't discuss it publicly. > > I'm curious though as to what it is that makes it unsuitable > > for public discussion. > > Oh, nothing profound. I guess I just don't like to needlessly flaunt \ personal > information about what I know in this forum. Many people know me here over \ many > years and several professional high caliber mathematicians in this forum \ (who > have helped even withtetration) know roughly 10^^10 times more math than I \ do, > so me going into endless tirades about my own achievements will likely \ make them > puke. > Heh. I don't think it's good for _anyone_, no matter how much math they know, to endlessly flaunt their stuff like crazy as then it is no longer a humble pursuit of knowledge, but an ego game (hardball.). > The characteristic of true wisdom (in anything) is silence and modesty. \ Most > serious work throughout the ages has been done in silence and with \ modesty. And > I like that. > Mmmhm. > > Just curiousity, and I'll respect > > your wishes. Perhaps you could send me an email instead? > > (which is private?) > > No problem. Feel free to contact me off list, if you wish. My email is \ listed on > the \About\ section, but be warned in advance that I can construct and \ send > dangerous tetration functions which may cause you to fall off your chair \ from > laughing :-) :) > -- > I.N. Galidakis === Subject: Re: JSH: Math Wars, reconsidered <5fs5qeF3dh44iU1@mid.individual.net> > And he himself has serious character weaknesses which have led > him to his present sad state of arrogance, laziness, self-delusion > and paranoid bitterness. > So if he wasn't lazy, arrogant, etc. he probably would have done a lot better in school and maybe actually had a REAL argument for posting here. Also, and this has nothing to do with James, would you consider not insulting anybody, even the most disagreeable one, a weakness of character? === Subject: JSH: Our lying world Reality is that people just do lie. And they lie across the board so why wouldn't they lie about math? Mathematicians check mathematicians so if you get paid by presenting stuff in a mathematical way that looks complicated and abstruse so you can just berate anyone who questions you on it--claim they just don't understand--why can't you just lie? Posters reply like that's silly as if mathematicians don't get anything for what they do, and act as if to be worth it a con needs to make someone rich. But if you can make $50,000 per year doing nothing but babbling math- ese that isn't even right, why wouldn't you? And if there is a critical mass of people like you who can laugh and joke about the stupidity of the world trusting you, why couldn't you? The world was so shocked when the tapes of Enron people laughing about loss of power in California came out, as if, as if in today's age we are really so gullible. And Catholic priests molesting children? How dare you even suggest such a thing, right? But it was true. People lie about everything. There are no real checks for most mathematicians except some other mathematician. I say, they lie. They lie repeatedly. And they lie knowing full well why they lie, and they rely on complex stuff that few people can even follow so that they can say things that are not true, so that if you could follow the argument you'd find it lead nowhere. We have a lying world. People start wars over lies. Keep wars going over lies. Burnt bodies and mutilated corpses over lies. Why would any sane human being think that mathematicians are somehow above being human? James Harris === Subject: Re: Our lying world > Reality is that people just do lie. You lie. All the time. And your getting ready to tell some woppers! > > And they lie across the board so why wouldn't they lie about math? You lie about math almost 100% of the time. > > Mathematicians check mathematicians so if you get paid by presenting > stuff in a mathematical way that looks complicated and abstruse so you > can just berate anyone who questions you on it--claim they just don't > understand--why can't you just lie? JSH, you simply are *not smart enough* to understand math. > Posters reply like that's silly as if mathematicians don't get > anything for what they do, and act as if to be worth it a con needs to > make someone rich. You are the con job. > > But if you can make $50,000 per year doing nothing but babbling math- > ese that isn't even right, why wouldn't you? But it is always right. > > And if there is a critical mass of people like you who can laugh and > joke about the stupidity of the world trusting you, why couldn't you? Stinking Troll. > > The world was so shocked when the tapes of Enron people laughing about > loss of power in California came out, as if, as if in today's age we > are really so gullible. You are troll. > > And Catholic priests molesting children? How dare you even suggest > such a thing, right? But it was true. No wrong math today, Troll? > > People lie about everything. You lie about everything. > > There are no real checks for most mathematicians except some other > mathematician. what is a real check? you don not even speak the language of Math, booger off. > > I say, they lie. They lie repeatedly. And they lie knowing full well > why they lie, and they rely on complex stuff that few people can even > follow so that they can say things that are not true, so that if you > could follow the argument you'd find it lead nowhere. You lie all the time. > > We have a lying world. That is your world. > > People start wars over lies. Keep wars going over lies. Burnt bodies > and mutilated corpses over lies. Your mutilated math corpse is stinking up sci.math, go crawl into some other \ newsgroup and stink up and lie there. > > Why would any sane human being think that mathematicians are somehow > above being human? Because they are *Smarter* than you are. You are a simplistic dummy with an \ IQ twice your shoe size. > > > James Harris > === Subject: Re: Our lying world <46c9bb73$0$97238$892e7fe2@authen.yellow.readfreenews.net> > Because they are *Smarter* than you are. You are a simplistic dummy with \ an > IQ twice your shoe size. Prince song \Kiss\?). James Harris is/was a member of the high-IQ society called the Mega Foundation (snicker). James Harris and his \friend\ (if you know what I mean) Quinn Tyler Jackson are in a \closed list\ (ha!), an exclusive priivate group of \thinkers\ (guffaw). I bet the Mega Foundation let James Harris in as a member because the Mega Foundation needed the dues. You can hear the membership commitee chair saying, \Need the dues, man!\ And so it came to pass: James Harris, member of the Mega Foundation hahahahahahahahahahahahahahahahahahahahahahahaha === Subject: Re: JSH: Our lying world > Reality is that people just do lie. > > And they lie across the board so why wouldn't they lie about math? > > Mathematicians check mathematicians so if you get paid by presenting > stuff in a mathematical way that looks complicated and abstruse so you > can just berate anyone who questions you on it--claim they just don't > understand--why can't you just lie? > > Posters reply like that's silly as if mathematicians don't get > anything for what they do, and act as if to be worth it a con needs to > make someone rich. > > But if you can make $50,000 per year doing nothing but babbling math- > ese that isn't even right, why wouldn't you? > > And if there is a critical mass of people like you who can laugh and > joke about the stupidity of the world trusting you, why couldn't you? > > The world was so shocked when the tapes of Enron people laughing about > loss of power in California came out, as if, as if in today's age we > are really so gullible. > > And Catholic priests molesting children? How dare you even suggest > such a thing, right? But it was true. > > People lie about everything. > > There are no real checks for most mathematicians except some other > mathematician. > > I say, they lie. They lie repeatedly. And they lie knowing full well > why they lie, and they rely on complex stuff that few people can even > follow so that they can say things that are not true, so that if you > could follow the argument you'd find it lead nowhere. > > We have a lying world. > > People start wars over lies. Keep wars going over lies. Burnt bodies > and mutilated corpses over lies. > > Why would any sane human being think that mathematicians are somehow > above being human? > > > James Harris > Typical babble from someone who doesn't have a clue what he's doing. My parents don't know anything whatsoever about my main area of interest (differential equations) so if they don't understand what I'm talking about, does that mean I'm lying? Dave === Subject: Re: JSH: Our lying world >> >> Mathematicians check mathematicians so if you get paid by presenting >> stuff in a mathematical way that looks complicated and abstruse so you >> can just berate anyone who questions you on it--claim they just don't >> understand--why can't you just lie? In your heart of hearts don't you think that some of that abstruse and complicated stuff might be interesting for a clever person like you to learn about? I bet you don't really think that it's all a con: that's just sour grapes, because you think the grapes are out of reach. But what if the grapes are not out of your reach? >> There are no real checks for most mathematicians except >> some other mathematician. (I take it you don't mean paychecks!) And those checks are not perfect. For instance, a solution by Dulac in 1923 of part of Hilbert's sixteenth problem, a solution which was extended by Petrovsky and Landis in 1957, was only found to be incorrect in 1980, when a counterexample turned up, and then a gap in the proof was identified in 1981. Source: Jeffrey L. Nunemacher, book review, Jeremy J. Gray, The Hilbert Challenge, Amer. Math. Monthly 110, examples of uncertainty about the validity of proofs at the highest levels of mathematics in Mark Ronan's brilliant popular book on mathematics \Symmetry and the Monster\. In some cases, world experts found examples of structures they had proved could not exist, and they could not find the errors in the proofs, so that just decided not to look (presumably in the hope that easier and more verifiable proofs would turn up, in the course of the vast task of revising the whole project of classifying the finite simple groups). >> Why would any sane human being think that mathematicians >> are somehow above being human? >> >> James Harris The awful fact is, you're human too. On the good side, that might mean that as a human being you might be able to participate intelligently (but not necessarily without conflict, anger or rivalry) in the human activity that those other human beings, mathematicians, do. Yes, you're unique. But so are they. (\I'm not!\ - voice from crowd in \The Life of Brian\.) (I won't keep this up. I seem to remember trying to talk to James in 2005 and getting nowhere. Hence the killfile. I just couldn't resist a peek at this thread, for some no doubt unworthy reason.) -- Angus Rodgers Contains mild peril === Subject: Re: JSH: Our lying world >Reality is that people just do lie. > >And they lie across the board so why wouldn't they lie about math? > >Mathematicians check mathematicians so if you get paid by presenting >stuff in a mathematical way that looks complicated and abstruse so you >can just berate anyone who questions you on it--claim they just don't >understand--why can't you just lie? I don't think that anyone has ever said that mathematicians never lie. Otoh you'd have a more interesting point if you could give some examples of lies they've told about your work. You state quite regularly that things we say here about your work are lies, but it always seems to happen that you eventually retract your claims. Happens over and over: You: I've proved P! Us: The proof is wrong because... You: Liar, liar, pants on fire [many repetitions of last two steps] You: Oops, my proof was wrong. But you never include an apology for calling people liars when they were telling the simple truth about what you turn out to be non-lies you never drop this business about how we're all liars. Very curious, to put it, um, mildly. >Posters reply like that's silly as if mathematicians don't get >anything for what they do, and act as if to be worth it a con needs to >make someone rich. > >But if you can make $50,000 per year doing nothing but babbling math- >ese that isn't even right, why wouldn't you? > >And if there is a critical mass of people like you who can laugh and >joke about the stupidity of the world trusting you, why couldn't you? > >The world was so shocked when the tapes of Enron people laughing about >loss of power in California came out, as if, as if in today's age we >are really so gullible. > >And Catholic priests molesting children? How dare you even suggest >such a thing, right? But it was true. > >People lie about everything. > >There are no real checks for most mathematicians except some other >mathematician. > >I say, they lie. They lie repeatedly. And they lie knowing full well >why they lie, and they rely on complex stuff that few people can even >follow so that they can say things that are not true, so that if you >could follow the argument you'd find it lead nowhere. > >We have a lying world. > >People start wars over lies. Keep wars going over lies. Burnt bodies >and mutilated corpses over lies. > >Why would any sane human being think that mathematicians are somehow >above being human? > > >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Our lying world >Reality is that people just do lie. > >And they lie across the board so why wouldn't they lie about math? > >Mathematicians check mathematicians so if you get paid by presenting >stuff in a mathematical way that looks complicated and abstruse so you >can just berate anyone who questions you on it--claim they just don't >understand--why can't you just lie? Oh dear. I do feel sorry for you James. For a moment, when you saw that your claims for surrogate factoring were wrong, there was a chance that you would get yourself onto a more positive track. Now it appears that you are going back into your old ways. That is not good. Did your old ways bring you contentment last time? Why do you think that things will be different this time? Everyone makes mistakes, but to carry on repeating the same mistakes in the expectation of a different result is not sensible. You cannot blame others for the latest failure of surrogate factoring. You were the one who did the work not them. You were the one who made the mistakes, not them. rossum > >James Harris === Subject: Re: Our lying world > Mathematicians check mathematicians so if you get paid by presenting > stuff in a mathematical way that looks complicated and abstruse so > you Yeah... that may be... But suppose somebody attacks you with a redcurrant? -- Clive Tooth http://www.shutterstock.com/cat.mhtml?gallery_id=61771 === Subject: A simple group problem Here's a simple group problem from my old exams. I know it is simple because I remember that I solved this problem in class back then. But now I forgot the solution. [problem] G is a finite solvable group and N is properly normal in G such that no subgroup of N is normal in G. We want to prove that N is abelian. [end problem] [attempt] If N is simple, then solvability of G forces N to be abelian. If N is not simple then there exist a (maximal?) subgroup H of N. Then gHg^-1 < gNg^-1 = N. It is tempting to claim that since H is normal in N, gHg^-1 = H, and thus H is normal in G, but I think advance for any hint . === Subject: Re: A simple group problem > > Here's a simple group problem from my old exams. I know it is simple > because I remember that I solved this problem in class back then. But > now I forgot the solution. > > [problem] G is a finite solvable group and N is properly normal in G > such that no subgroup of N is normal in G. We want to prove that N is > abelian. [end problem] > > [attempt] If N is simple, then solvability of G forces N to be > abelian. If N is not simple then there exist a (maximal?) subgroup H > of N. Then gHg^-1 < gNg^-1 = N. It is tempting to claim that since H > is normal in N, gHg^-1 = H, and thus H is normal in G, but I think > advance for any hint . *********************************************************** If G is solvable then N is solvable ==> the commutator sbgp. N' of N is not N, and thus N' is a characteristic sbgp. of N and thus normal in G itself ==> N has to be 1 <==> N is abelian. Tonio === Subject: Re: A simple group problem I was able to search for small groups of such properties by brutal force, and it happens that all such N (that I got so far) are p- > > > > > > Here's a simple group problem from my old exams. I know it is simple > > because I remember that I solved this problem in class back then. But > > now I forgot the solution. > > > [problem] G is a finite solvable group and N is properly normal in G > > such that no subgroup of N is normal in G. We want to prove that N is > > abelian. [end problem] > > > [attempt] If N is simple, then solvability of G forces N to be > > abelian. If N is not simple then there exist a (maximal?) subgroup H > > of N. Then gHg^-1 < gNg^-1 = N. It is tempting to claim that since H > > is normal in N, gHg^-1 = H, and thus H is normal in G, but I think > > advance for any hint . > > *********************************************************** > If G is solvable then N is solvable ==> the commutator sbgp. N' of N > is not N, and thus N' is a characteristic sbgp. of N and thus normal > in G itself ==> N has to be 1 <==> N is abelian. > Tonio === Subject: Re: A simple group problem > > I was able to search for small groups of such properties by brutal > force, and it happens that all such N (that I got so far) are p- > > > > G is a finite solvable group and N is properly normal in G > > > such that no subgroup of N is normal in G. Yes, all such N are p-groups. Since N is abelian, the Sylow p- subgroup P of N is precisely the set of p-elements of N, and so is a characteristic subgroup of N. Hence P is normal in G and contained in N, so either P=1 or P=N. Taking some p that divides |N|, one must have P=N is a p-group. You might also look at the subgroup K = { g^p : g in N }. K should be normal in G, but you should be able to show K cannot be all of N. This should tell you even more about the subgroup N. If you want to read more about such N, even when G is not solvable, then you might look for the phrases \minimal normal subgroups\ \socle\ and \characteristically simple group\. For solvable groups, quite a bit is learned by studying the N as you are doing. If you want to continue in this way, also look for \chief factor\. === Subject: Re: A simple group problem days. My association with the Department is that of an alumnus. Cc: Please don't top-post. http://www.xs4all.nl/~hanb/documents/quotingguide.html http://www.caliburn.nl/topposting.html http://www.html-faq.com/etiquette/?toppost Edited to remove top-posting. >> > [problem] G is a finite solvable group and N is properly normal in G >> > such that no subgroup of N is normal in G. We want to prove that N is >> > abelian. [end problem] >> >> > [attempt] If N is simple, then solvability of G forces N to be >> > abelian. If N is not simple then there exist a (maximal?) subgroup H >> > of N. Then gHg^-1 < gNg^-1 = N. It is tempting to claim that since H >> > is normal in N, gHg^-1 = H, and thus H is normal in G, but I think >> > advance for any hint . Typo fixed below: >> *********************************************************** >> If G is solvable then N is solvable ==> the commutator sbgp. N' of N >> is not N, and thus N' is a characteristic sbgp. of N and thus normal >> in G itself ==> N' has to be 1 <==> N is abelian. > >I was able to search for small groups of such properties by brutal >force, and it happens that all such N (that I got so far) are p- I don't follow your question... You are correct that \normal in N\ does not imply \normal in G.\ However, the point is that this ->does<- hold for a special kind of normal subgroups of N, those that are called \characteristic subgroups.\ The commutator subgroup ->is<- such a subgroup. Namely: THEOREM. Let G be a group, let N be normal subgroup of G. Then [N,N] is normal in G. (try proving it; it is very easy). So: you start with G which is solvable, and a proper subgroup N that is normal in G and has the property that no (proper nontrivial) normal subgroup of N is normal in G. You want to prove that N is abelian. As Tonico suggests, look at [N,N]. This is a subgroup of N, which is necessarily normal in G. By assumption, it cannot be a proper nontrivial subgroup of N, so that leaves you two possibilities: either [N,N]={1}, or else [N,N]=N. However, because G is solvable, you also know that N is solvable. And since N is solvable, [N,N]=N cannot hold. So we conclude that [N,N]={1}. And this tells you that N is abelian. So you are done. Why are you doing brute force searches? -- \It's not denial. I'm just very selective about what I accept as reality.\ --- Calvin (\Calvin and Hobbes\ by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: A simple group problem > Please don't top-post. Sorry about that, after using Outlook for 2 month, bad habit just stays (bad excuse). > THEOREM. Let G be a group, let N be normal subgroup of G. Then [N,N] > is normal in G. This follows almost directly from the invariance of [N,N] under conjugation and the normality of N. > (try proving it; it is very easy). > > So: you start with G which is solvable, and a proper subgroup N that > is normal in G and has the property that no (proper nontrivial) normal > subgroup of N is normal in G. You want to prove that N is abelian. > > As Tonico suggests, look at [N,N]. This is a subgroup of N, which is > necessarily normal in G. By assumption, it cannot be a proper > nontrivial subgroup of N, so that leaves you two possibilities: either > [N,N]={1}, or else [N,N]=N. > > However, because G is solvable, you also know that N is solvable. And > since N is solvable, [N,N]=N cannot hold. > > So we conclude that [N,N]={1}. And this tells you that N is > abelian. So you are done. Yes, I understood that, and the original problem was done. > Why are you doing brute force searches? In addition to the original question, I'm also want to see if there are some other properties in such groups. So I did a brutal force search, and one thing pops out is that all such N has p-groups such that all elements has order p for some prime p. So now I want to see if this property can be proved. I think the answer is yes and the proof should be fairly simple. === Subject: Re: A simple group problem days. My association with the Department is that of an alumnus. Typo: >> [problem] G is a finite solvable group and N is properly normal in G >> such that no subgroup of N is normal in G. We want to prove that N is >> abelian. [end problem] >*********************************************************** >If G is solvable then N is solvable ==> the commutator sbgp. N' of N >is not N, and thus N' is a characteristic sbgp. of N and thus normal >in G itself ==> N has to be 1 <==> N is abelian. Last clause was obviously: N' has to be 1 <==> N is abelian. -- \It's not denial. I'm just very selective about what I accept as reality.\ --- Calvin (\Calvin and Hobbes\ by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: Langlands and FLT > I don't know whether a flat representation has to be irreducible. Sorry, I'm being stupid. Wiles already states that flat representations are absolutely irreducible, right after defining them. So, ordinary representations are reducible and flat representations are absolutely irreducible. Another detail I have slighted completely is that rho0 is actually a representation of Gal(Q_Sigma/Q) and we have been laboring over its restriction to the decomposition group Dp of Q_Sigma at p. In other words, we're not just studying representations of Gal(Qpbar/Qp) that happen to have the properties we have been calling ordinary and flat; the representations of Gal(Qpbar/Qp) are required to extend to \ Gal(Q_Sigma/Q). I printed out the published version of Wiles' paper from JSTOR and started reading it from the beginning. It's visually more pleasing to read than the preprint and has some slight differences. Even though I don't really understand ordinary and flat representations as well as I would like, I think I do understand them as stated. I read a little further and a now looking at Wiles' descriptions of the deformations he is interested in. To try to appreciate that, I'm digging out the papers I have on deformations. For some reason, even though I still don't understand them, I feel less bewildered than I did when I first looked at them. I complained earlier about the notion of rho0 appearing by magic from a finite flat group scheme. Actually, it does appear by magic. Let G be a finite flat k-vector group over Zp. Then the Qpbar-valued points of G form a vector space over k and, by composition, Gal(Qpbar/Qp) acts on that vector space. If that isn't magic, I don't know what is. Anyway, now I'm more comfortable with Wiles' talking about a flat representation as arising from a finite flat group scheme. Even though I'm moving on in Wiles' paper, I still think it will be educational to play with the Zp algebra of all Gal(Qpbar/Qp)-equivariant functions from rho0 to Zpbar. To avoid circumlocuation and obscure \ notation, I'm going to refer to this algebra as \the rat\ of rho0, since I am using it for a lab rat. Accordingly, I'll denote it LeRat(rho0) or, if I want to emphasize p, LeRat_p(rho0). If we can somehow find an excuse to call \ something a \horse\ and something else a \barrel\, that would fit very well with \ one of the stories in Fantasia Mathematica. One other comment: you have been trying very hard to explain some of the relevant concepts by referring to elliptic curves and abelian varieties, while I have tended to dismiss such attempts as just examples of the \ relevant concepts, not the concepts themselves. If every finite flat commutative \ grouo scheme arises as the kernel of an isogeny of abelian schemes, then that seems to promote your efforts from mere examples to necessary theory. I \ think that some of the papers I've been trying to read to understand better when finite group schemes over Qp extend to Zp say that this is the case, but I don't really know yet. Also, I don't know yet what Honda systems are but my vague impression at this point is that they are based on such a result in the case of commutative finite group schemes of order a power of p. -- Ignorantly, Allan Adler * Disclaimer: I am a guest and *not* a member of the MIT CSAIL. My actions \ and * comments do not reflect in any way on MIT. Also, I am nowhere near \ Boston. === Subject: Is the complexity class P equal to {0, 1}* See the following proof: L1= A language over {0,1}*, containing all odd numbers. L2= A language over {0,1}*, containing all even numbers. Clearly, L2 = {0,1}* \\ L1. It's also obvious that both L1 and L2 belong to the complexity class P. Can we deduce that the \union\ of L1 and L2 belongs to P? Does it mean that P={0,1}* ? (I'm almost sure that this is not true, otherwise P=NP!) PS: There's another point here that I don't understand. In the set theory, we cannot have two \complement\ sets belonging to another set (unless the latter set is the Universal set). However, in the complexity theory, such situations may arise, as above. === Subject: Re: Is the complexity class P equal to {0, 1}* > See the following proof: > > L1= A language over {0,1}*, containing all odd numbers. > L2= A language over {0,1}*, containing all even numbers. > > Clearly, L2 = {0,1}* \\ L1. > > It's also obvious that both L1 and L2 belong to the complexity class > P. > > Can we deduce that the \union\ of L1 and L2 belongs to P? Yes. There is a polynomial-time algorithm which accepts L1 union L2; namely the following: f(x) return 1; > Does it mean that P={0,1}* ? No. {0,1}* is a set of strings. P is a set of sets (which themselves contain strings). {0,1}* is an element of P, though. (Think of the difference between {1} and {{1}}.) > (I'm almost sure that this is not true, otherwise P=NP!) > > PS: There's another point here that I don't understand. In the set > theory, we cannot have two \complement\ sets belonging to another set > (unless the latter set is the Universal set). However, in the > complexity theory, such situations may arise, as above. Not sure what you mean here. It certainly is possible for a set S to contain a set A and the complement of A. --- Christopher Heckman === Subject: Re: Is the complexity class P equal to {0, 1}* > > PS: There's another point here that I don't understand. In the set > > theory, we cannot have two \complement\ sets belonging to another set > > (unless the latter set is the Universal set). However, in the > > complexity theory, such situations may arise, as above. > > Not sure what you mean here. It certainly is possible for a set S to > contain a set A and the complement of A. I mean this: Suppose that B denotes the complement of A, and S is a set. Then: \A is a subset of S\ and \B is a subset of S\ => \S=The universal \ set\ Let me re-state the problem: How can a set like NP, and its complement coNP, share the same subset (P)? We know that the intersection of complementary sets must be empty. === Subject: Re: Is the complexity class P equal to {0, 1}* > \A is a subset of S\ and \B is a subset of S\ => \S=The universal \ set\ But L1 = {0,1}*\\L2, so you are taking complement respect to {0,1}*. This just shows that {0,1}* is in NP (at best). > Let me re-state the problem: How can a set like NP, and its complement > coNP, share the same subset (P)? We know that the intersection of > complementary sets must be empty. coNP is not the complement of NP. It's just problems who's complement (counterexamples) is in NP (right?). So P is clearly in both of them. === Subject: Re: Is the complexity class P equal to {0, 1}* > > Let me re-state the problem: How can a set like NP, and its complement > coNP, coNP is not the complement of NP > share the same subset (P)? We know that the intersection of > complementary sets must be empty. > === Subject: Calculating a resultant phase angle G'day G'day Folks, Electrical technicians often have to work out the resultant when combining two phasors, say two AC currents with various amplitudes and angles of lead or lag. The standard method for doing this is resolve the phasors into x and y components then to sum the components, take their absolute values and use Pythagoras to find the magnitude of the resultant. Inverse tan of the sum of the y components divided by the sum of the x components is used to get the phase angle of the resultant. Careful students like to have a method of checking their calculations. A variant of the cosine rule provides a simple means of checking the magnitude of the resultant if there are only two phasors. If we let the two phasors be A and B and their resultant C then, C^2= A^2 + B^2 + 2 x A x B x Cos(Angle between A and B) What I'm wondering is, is there an independent method of checking the phase angle of resultant? If so could someone please illustrate it. Best wishes and thank you, -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / \\ /\\ \... and the blind dog was leading.\ http://homepages.paradise.net.nz/quentin === Subject: Re: Calculating a resultant phase angle On Mon, 20 Aug 2007 17:57:17 +1200, Quentin Grady >G'day G'day Folks, > >Electrical technicians often have to work out the resultant when >combining two phasors, say two AC currents with various amplitudes and >angles of lead or lag. The standard method for doing this is resolve >the phasors into x and y components then to sum the components, take >their absolute values and use Pythagoras to find the magnitude of the >resultant. Inverse tan of the sum of the y components divided by the >sum of the x components is used to get the phase angle of the >resultant. > >Careful students like to have a method of checking their calculations. > >A variant of the cosine rule provides a simple means of checking the >magnitude of the resultant if there are only two phasors. > >If we let the two phasors be A and B and their resultant C then, > > C^2= A^2 + B^2 + 2 x A x B x Cos(Angle between A and B) > >What I'm wondering is, is there an independent method of checking the >phase angle of resultant? > >If so could someone please illustrate it. See this page: Line (21) corresponds to the relation above for C^2. Line (22) is, I think, what you asked for. === Subject: Re: Calculating a resultant phase angle This post not CC'd by email On Mon, 20 Aug 2007 04:11:42 -0400, Passerby >On Mon, 20 Aug 2007 17:57:17 +1200, Quentin Grady > >>G'day G'day Folks, >> >>Electrical technicians often have to work out the resultant when >>combining two phasors, say two AC currents with various amplitudes and >>angles of lead or lag. The standard method for doing this is resolve >>the phasors into x and y components then to sum the components, take >>their absolute values and use Pythagoras to find the magnitude of the >>resultant. Inverse tan of the sum of the y components divided by the >>sum of the x components is used to get the phase angle of the >>resultant. >> >>Careful students like to have a method of checking their calculations. >> >>A variant of the cosine rule provides a simple means of checking the >>magnitude of the resultant if there are only two phasors. >> >>If we let the two phasors be A and B and their resultant C then, >> >> C^2= A^2 + B^2 + 2 x A x B x Cos(Angle between A and B) >> >>What I'm wondering is, is there an independent method of checking the >>phase angle of resultant? >> >>If so could someone please illustrate it. > >See this page: > > >Line (21) corresponds to the relation above for C^2. > >Line (22) is, I think, what you asked for. Line (21) does indeed correspond to the relation for c^2 Line (22) summarises the method that the students have already used to find the phase angle of the resultant phasor. Therefore, unfortunately, it won't provide an alternative method to check their calculations. -- Quentin Grady ^ ^ / New Zealand, >#,#< [ / \\ /\\ \... and the blind dog was leading.\ http://homepages.paradise.net.nz/quentin === Subject: Re: Calculating a resultant phase angle On Mon, 20 Aug 2007 20:35:36 +1200, Quentin Grady >This post not CC'd by email > On Mon, 20 Aug 2007 04:11:42 -0400, Passerby > >>On Mon, 20 Aug 2007 17:57:17 +1200, Quentin Grady >> >>>G'day G'day Folks, >>> >>>Electrical technicians often have to work out the resultant when >>>combining two phasors, say two AC currents with various amplitudes and >>>angles of lead or lag. The standard method for doing this is resolve >>>the phasors into x and y components then to sum the components, take >>>their absolute values and use Pythagoras to find the magnitude of the >>>resultant. Inverse tan of the sum of the y components divided by the >>>sum of the x components is used to get the phase angle of the >>>resultant. >>> >>>Careful students like to have a method of checking their calculations. >>> >>>A variant of the cosine rule provides a simple means of checking the >>>magnitude of the resultant if there are only two phasors. >>> >>>If we let the two phasors be A and B and their resultant C then, >>> >>> C^2= A^2 + B^2 + 2 x A x B x Cos(Angle between A and B) >>> >>>What I'm wondering is, is there an independent method of checking the >>>phase angle of resultant? >>> >>>If so could someone please illustrate it. >> >>See this page: >> >> >>Line (21) corresponds to the relation above for C^2. >> >>Line (22) is, I think, what you asked for. > > > Line (21) does indeed correspond to the relation for c^2 > >Line (22) summarises the method that the students have already used to >find the phase angle of the resultant phasor. Therefore, >unfortunately, it won't provide an alternative method to check their >calculations. > Oops, sorry, needed to read more carefully. 8-( OK, how about this ... Let's do this example (all angles are in degrees). 5 sin(wt + 30) + 4 sin(wt + 140) By the usual method, find the approx. solution 5.2268 sin(wt + 75.9832) Check the solution ------------------ Using your suggested method above, find C^2 = 5^2 + 4^2 + 2 * 5 * 4 * cos(110) C ~ 5.2268 which checks, so say we feel safe using it below Now consider the triangle with sides a = 4, b = 5.2268, c = 5 and use the law of cosines in this form cos(A) = (b^2 + c^2 - a^2) / (2bc) cos(A) = (5.2268^2 + 5^2 - 4^2) / (2 * 5.2268 * 5) cos(A) ~ 0.69487 A ~ 45.9832 A + 30 ~ 75.9832 which is the angle found by the usual method HTH === Subject: Exploring 'Phi Time' Relationships o/than Gravity/Light,Energy Hello Math Forum: What I mean by 'Phi Time' relationships is the mathematical and otherwise \ observable behavior of mass/gravity, Light and Energy. I propose that there are other such 'Phi Time' Relationships, Mathematical \ and Observable. One such relationship could be said to be the following: Apple + Orange + Pear = N*Fruit I think this could be a similiar relationship to the mass/gravity, and \ Energy mathematical relationships. What this means to Physics and Math., I am not sure. Does anyone out there agree with this possibility? Zim Olson Creative Mathematics http://www.zimmathematics.com === Subject: Exploring 'Phi Time' Relationships o/than Gravity/Light,Energy Hello Math Forum: What I mean by 'Phi Time' relationships is the mathematical and otherwise \ observable behavior of mass/gravity, Light and Energy. I propose that there are other such 'Phi Time' Relationships, Mathematical \ and Observable. One such relationship could be said to be the following: Apple + Orange + Pear = N*Fruit I think this could be a similiar relationship to the mass/gravity, and \ Energy mathematical relationships. What this means to Physics and Math., I am not sure. Does anyone out there agree with this possibility? Zim Olson Creative Mathematics http://www.zimmathematics.com === Subject: Topology question... I'm not much of a mathematician right now....but would like to be.... Anyway, my question is - Would it be possible to cut off a small portion of a football bladder (thus, making a hole in it) and stretch the rest of it on a plane without having to cause inconsistencies in the bladder's thickness? .................[or] Would it be possible to take a plane paper and wrap it on a sphere without having to cause folds on the paper? -- rAgAv === Subject: Re: Topology question... > Would it be possible to cut off a small portion of a football bladder > (thus, making a hole in it) and stretch the rest of it on a plane > without having to cause inconsistencies in the bladder's thickness? > Thickness is not a topological property. A sphere with a point or a small disk removed, is homeomorphic to the real plane. > Would it be possible to take a plane paper and wrap it on a sphere > without having to cause folds on the paper? > No, the real plane is not homeomorphic to a sphere. === Subject: Re: Calculus I: What to expect? > > I bought a TI-83+ [...] I have not had any need for a higher calculator, > I don't use Mathematica or Maple or [or Macsyma] or any of those things. That seems seriously underpowered for someone who often professes [1] fondness for swatting flies with cannons. Just think of all the bugs you could squash by aiming one of those high-powered computer algebra cannons at all those pesky bugs buzzing around Vladimir Bondarenko. Now that is one cannon I would not hesitate to help you fire! --Bill Dubuque === Subject: Re: Calculus I: What to expect? On 20 Aug 2007 02:46:32 -0400, Bill Dubuque >> >> I bought a TI-83+ [...] I have not had any need for a higher calculator, \ >> I don't use Mathematica or Maple or [or Macsyma] or any of those things. \ > >That seems seriously underpowered for someone who often professes [1] >fondness for swatting flies with cannons. Just think of all the bugs >you could squash by aiming one of those high-powered computer algebra >cannons at all those pesky bugs buzzing around Vladimir Bondarenko. >Now that is one cannon I would not hesitate to help you fire! > >--Bill Dubuque > I'm with you there. When i was in the business world, they didn't seem to mind me using whatever means at my disposal to solve problems. There was never, ever as situation where I was not permitted to use any technological means available to solve a problem! Brian === Subject: f(2*x) = f(x) +1 or f ^ [-1](x) /2 = f ^[ -1](x -1)? With bijective functions {f, g, h} equations f(h(x)) = g(f(x)) may be rewritten f ^ [-1](g ^ [-1](x)) = h ^ [-1](f ^[ -1](x)) , (conditions, intervals must be precised). So, the right answer to the question could be : which equation is simpler to solve or which equation has got, for the solver, a known solution. Let us suppose he already knows a solution to f ^ [-1](x) = 2* f ^ [-1](x -1) , (finite difference) f ^ [-1](x) = c*2^x , c a non null constant. Direct inversion from c*2^f(x) gives : f(x) =ln(x/c)/ln(2) Do you know interesting cases , perhaps more general ones (i.e more variables) , for which this method works. Alain === Subject: Re: f(2*x) = f(x) +1 or f ^ [-1](x) /2 = f ^[ -1](x -1)? > With bijective functions {f, g, h} equations f(h(x)) = g(f(x)) > may be rewritten f ^ [-1](g ^ [-1](x)) = h ^ [-1](f ^[ -1](x)) , > (conditions, intervals must be precised). > > So, the right answer to the question could be : > which equation is simpler to solve or which equation > has got, for the solver, a known solution. > > Let us suppose he already knows a solution to > f ^ [-1](x) = 2* f ^ [-1](x -1) , (finite difference) > f ^ [-1](x) = c*2^x , c a non null constant. > Direct inversion from c*2^f(x) gives : > f(x) =ln(x/c)/ln(2) > So what? Is there something you'd like to point out or conclude about what it gave? Do you mean to state that it's a solution for f(2x) = f(x) + 1 ? f(2x) = (log x/c + log 2/c)/log 2 = f(x) + (log 2 - log c)/log2 = f(x) + 1 - (log c)/log 2 Usually it isn't. > Do you know interesting cases , perhaps more general > ones (i.e more variables) , for which this method works. > Here's more general f(x) = a.log x / log p I'll leave to you, as you do, to make a complete statement out of it. Let's suppose the question is to solve f(2x) = f(x) + 1 What happens if x = 0? Can you show for the domain of f, that f is injection? What is the domain of f? What method are you proposing? Would you be less rambling in your presentation? f(2x) = f(x) + 1, how do you get to f^-1(x) = 2f^-1(x - 1)? f^-1f(2x) = f^-1(f(x) + 1) 2x = f^-1(f(x) + 1) Set x = f^-1(x) 2f^-1(x) = f^-1(f(f^-1(x) + 1) 2f^-1(x) = f^-1(x + 1) Set x = x - 1. Ok, that's how. === Subject: Re: Vectors >> >> >I have another one: >> >> >> >a and b are vectors. >> >> >> >(3a - b) * (a + 4b) = 0 >> >> >(2a + 5b) * (3a - 4b) = 0 >> >> >> >I need to find angle between a and b. >> >> >> >here is what I did so far: >> >> >> >I know that cos(angle a and b) = (a*b) / |a| * |b| >> >> >> >I calculated that a*b = (-4/5)*(b^2) but that >still >dont >leads me \ to >> >> >solution . I need help >> >> >> So you get |b|^2 = -5/4 * a*b. >> >> Now derive |a|^2 = -16/3 * a*b and insert |a| and >|b| >> >> in the angle formula. >> >> >> Best wishes >> >> Torsten. >> >> >I dont think you can derive that way. Anyway if I do >>that the result >> >is 53.13 degrees. I got in my book solution for this >>problem and it >> >doesnt match. >> >> I get cos(alpha) = +/- sqrt(0.15) and so >> alpha = 67.21 degrees or alpha = 112.79 degrees. >> >> Best wishes >> Torsten. .. and since a*b is negative, the unique solution is alpha = 112.79 degrees. Best wishes Torsten. === Subject: Why Can't Two Cubes Become Another Cube? It appears to be an inclusive set relationship, where the larger cubic geometry contains the first cube and the intermediate numerical sequencing cannot in itself, be a cube. [u][u][u] [u][u][u] [u][u][u] [o][o][o][o][o] [o][u][u][u][o] [o][u][u][u][o] [o][u][u][u][o] [o][o][o][o][o] The inner u cube units consist of 27 = 3^3 units The total number of o cube units is 50 + 48 = 98 the total o cube units plus total u cube units equals 5^3 = 125 Two layers of 25 o cube units plus three layers of 16 peripheral o cube units include the 27 u cube units within. 1^3 + 26 = 3^3 [o][o][o] [o][u][o] [o][o][o] 3^3 + 98 = 5^3 5^3 + 218 = 7^3 5^3 + 604 = 9^3 5^3 + 1206 = 11^3 5^3 + 2072 = 13^3 === Subject: Re: Why Can't Two Cubes Become Another Cube? > It appears to be an inclusive set relationship, where the larger cubic > geometry contains the first cube and the intermediate numerical > sequencing cannot in itself, be a cube. > > [u][u][u] > [u][u][u] > [u][u][u] > > [o][o][o][o][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][o][o][o][o] > > The inner u cube units consist of 27 = 3^3 units > > The total number of o cube units is 50 + 48 = 98 > > the total o cube units plus total u cube units equals 5^3 = 125 > > Two layers of 25 o cube units plus three layers of 16 peripheral o > cube units include the 27 u cube units within. > > 1^3 + 26 = 3^3 > > [o][o][o] > [o][u][o] > [o][o][o] > > 3^3 + 98 = 5^3 > > 5^3 + 218 = 7^3 > > 5^3 + 604 = 9^3 > > 5^3 + 1206 = 11^3 > > 5^3 + 2072 = 13^3 Au contraire, two cubes CAN become another cube. There's 0^3+0^3=0^3 also 0^3+1^3=1^3 and 1^3+(-1)^3=0^3 Oh, wait, did you mean non-trivial cubes. But then there's (1.1)^3 +(1.2)^3=(1.451643004818508114. ......)^3 and plenty others involving irrational and complex numbers. Oh wait, did you mean only rational cubes and/or cubes of integers? Well, there's an old old proof that there's no non-trivial rational and/or integer cubes which add to another cube. And there is a recent proof of the generalization to all integer powers larger than squares. So, the answer to your original poorly-phrased question of \why\ is \because there are mathematical proofs which say you cannot add non- trivial rational and/or integer cubes to form another cube\. You don't like it? It sounds counter-intuitive? Tough. Perhaps you won't like the squaring-the-circle or the doubling-the- cube or the trisecting-the-angle proofs either. === Subject: Re: Why Can't Two Cubes Become Another Cube? > It appears to be an inclusive set relationship, where the larger cubic > geometry contains the first cube and the intermediate numerical > sequencing cannot in itself, be a cube. Huh? Are you wondering why one cannot find positive integers a,b,c such that a^3 + b^3 = c^3? This is easily proved. In fact a generalization to higher exponents is true as well, but the proof is either awfully complicated or only available in Latin: Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet. > > [u][u][u] > [u][u][u] > [u][u][u] > > [o][o][o][o][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][o][o][o][o] > > The inner u cube units consist of 27 = 3^3 units > > The total number of o cube units is 50 + 48 = 98 > > the total o cube units plus total u cube units equals 5^3 = 125 > > Two layers of 25 o cube units plus three layers of 16 peripheral o > cube units include the 27 u cube units within. > > 1^3 + 26 = 3^3 > > [o][o][o] > [o][u][o] > [o][o][o] > > 3^3 + 98 = 5^3 > > 5^3 + 218 = 7^3 > > 5^3 + 604 = 9^3 > > 5^3 + 1206 = 11^3 > > 5^3 + 2072 = 13^3 === Subject: Re: Why Can't Two Cubes Become Another Cube? > It appears to be an inclusive set relationship, where the larger cubic > geometry contains the first cube and the intermediate numerical > sequencing cannot in itself, be a cube. > Huh? Would you explain what this means? On second thought, don't bother. I fear the explanation would be equally mystifying. > [u][u][u] > [u][u][u] > [u][u][u] > > [o][o][o][o][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][u][u][u][o] > [o][o][o][o][o] > > The inner u cube units consist of 27 = 3^3 units > > The total number of o cube units is 50 + 48 = 98 > > the total o cube units plus total u cube units equals 5^3 = 125 > > Two layers of 25 o cube units plus three layers of 16 peripheral o > cube units include the 27 u cube units within. > > 1^3 + 26 = 3^3 > > [o][o][o] > [o][u][o] > [o][o][o] > > 3^3 + 98 = 5^3 > > 5^3 + 218 = 7^3 > > 5^3 + 604 = 9^3 > > 5^3 + 1206 = 11^3 > > 5^3 + 2072 = 13^3 > > === Subject: Re: Another Inconvenient Truth > >>According to the above thread, ANY finite set (and there are no >>others in \Bit mapped Set Theory\) can be equivalenced with just >>one unique natural number. Thus the set of all naturals would be >>equivalent to the set of all sets. This is Russell's paradox. No? > > No. I still think it's yes. Han de Bruijn === Subject: Re: Another Inconvenient Truth >>On 18 aug, 04:20, \T.H. Ray\ >> >>> >>> >>>>>>>It has certainly been the case in the past >> >>that >> >>>>new measuring devices >>>> >>>>>>>have been invented and that they have measured >>>> >>>>things never before >>>> >>>>>>>measured. I think it quite possible that not >> >>all >> >>>>possible measuring >>>> >>>>>>>devices have yet been invented. In fact I >> >>think it >> >>>>the height of >>>> >>>>>>>arrogance to suppose otherwise. >>> >>> >>>>>>So much arrogance that the majority of >> >>physicists >> >>>>will agree with >>>> >>>>>>me that actual infinity is unmeasurable, even >> >>in >> >>>>principle. >>> >>>>>What about measuring the focal length of lenses >>>>>with neither positive (converging) nor negative >>>> >>>>(diverging) >>>> >>>>>focal planes? Does that count? >>> >>>>Potential infinite counts. Actual infinite counts >>>>not. The focal planes >>>>are at a very large (\potential infinite\) >> >>distance >> >>>>from the lense. >>>>The distance is so large (and the error in it is >> >>so >> >>>>large as well) that >>>>it doesn't matter how large the distance actually >> >>is: >> >>>>it's \infinite\ >>>>by physical, not by mathematical standards. >>> >>>>Han de Bruijn >>> >>>How can an actual physical measurement be >> >>potentially >> >>>infinite? A measurement is self limiting, >> >>sensitively >> >>>dependent on the measuring instrument. That there >> >>can >> >>>be no infinitely accurate measurement does not >> >>imply >> >>>that the physical standard is infinity. The >> >>standard >> >>>is precisely defined by instrument and interval. >> >>Repeated measurements? Many of them? > > Absolutely. How do you think Mandelbrot answered the > question, \How long is the coastline of England?\ > Scale counts. Yes. I've absorbed \The Fractal Geometry of Nature\. (Excellent book!) Han de Bruijn === Subject: Re: Another Inconvenient Truth >>>How can an actual physical measurement be >> >>potentially >> >>>infinite? A measurement is self limiting, >> >>sensitively >> >>>dependent on the measuring instrument. That there >> >>can>> >>>be no infinitely accurate measurement does not >> >>imply >> >>>that the physical standard is infinity. The >> >>standard >> >>>is precisely defined by instrument and interval. >> >>Repeated measurements? Many of them? > > Absolutely. How do you think Mandelbrot answered the > question, \How long is the coastline of England?\ > Scale counts. Yes. I've absorbed \The Fractal Geometry of Nature\. (Excellent book!) Han de Bruijn Well, then, how can you say that an actual physical measurement is potentially infinite? Many scale varieties are available to measure and all are actual, even to infinity. Isn't the measure function self limiting, as the measure object is self organized? Tom === Subject: Re: Another Inconvenient Truth This part of the thread IMHO has become kind of a mess. So let's repeat relevant parts of the current debate in David R Tribble style. > So much arrogance that the majority of physicists will agree with > me that actual infinity is unmeasurable, even in principle. > What about measuring the focal length of lenses > with neither positive (converging) nor negative (diverging) > focal planes? Does that count? > Potential infinite counts. Actual infinite counts not. The focal planes > are at a very large (\potential infinite\) distance from the lense. > The distance is so large (and the error in it is so large as well) that > it doesn't matter how large the distance actually is: it's \infinite\ > by physical, not by mathematical standards. > How can an actual physical measurement be potentially > infinite? A measurement is self limiting, sensitively > dependent on the measuring instrument. That there can > be no infinitely accurate measurement does not imply > that the physical standard is infinity. The standard > is precisely defined by instrument and interval. I didn't say that an actual physical measurement can be potentially infinite. All I've said is that \potential infinite counts\. I don't understand your statement that \the physical standard is infinity\. > Repeated measurements? Many of them? In the case of the focal length of lenses, if the lense is nearly flat, repeated measurements will result in large values with relatively large errors. Thus indicating a \real\ value which is (physically) infinite, which is just another word for \very large\, so large that exactly \how large\ doesn't count anymore. > Absolutely. How do you think Mandelbrot answered the > question, \How long is the coastline of England?\ > Scale counts. > Yes. I've absorbed \The Fractal Geometry of Nature\. (Excellent book!) http://www.physionet.org/tutorials/epn/html/chp2.htm http://mathforum.org/library/drmath/view/54521.html But the coastline problem is quite a bit different from the lenses focal distance problem. > Well, then, how can you say that an actual physical > measurement is potentially infinite? Many scale > varieties are available to measure and all are actual, > even to infinity. Isn't the measure function self > limiting, as the measure object is self organized? I think I now understand what you mean. Fractals are a valid model of actual coastlines. But you argue that these are (actual) infinite in e.g. length. This will result in scaling problems when trying to measure the length of coastlines. But did I deny that? The outcome of a bunch of measurements will be a set of large lengths with large errors in them. Therefore \physical infinity\. And BTW, the coastline fractals are just a (though very good) mathematical model, not the \real\ thing. Fractals can be \smoothed\ (I call this \re-normalized\). After that, they have \ a finite length. But these lengths do _not_ converge to a definite length. Han de Bruijn === Subject: Re: Another Inconvenient Truth > >> >> >>>>Heh, heh. I didn't define my set theory to be about finite sets >>>>per se. But, quite fortunately, it just turns out that infinite >>>>sets are impossible with it. >> >>>So what? It is no great achievement to come up with a set >>>theory that does not contain infinite sets. >> >>>Your claim is that standard set theory gets the >>>wrong answers for \what is the probablity >>>that a number chosen at random from the set of all >>>numbers is even?\ and \how many balls are in the vase >>>at noon?\. >> >>>However, your set theory does not give different answers >>>to these questions, it just declares the questions >>>meaningless. >> >>I have another theory that answers the first question, but >>not the second one. And the first answer is 1/2. See below. >> >>>You are confident that you know what the correct >>>answers are, but you have no theory that leads >>>to these answers. >> >>Not the \Bit mapped Set Theory\ but the \Naturals Construction >>Set\ theory: >> > > So bit map set theory cannot answer either question, > > Natural constructions set theory can answer the first question > > You do not have a theory that will answer the second question. Indeed. I don't have a theory that will answer this nonsense question. > Despite this you are confident you know the correct answer > to both questions. The flaw in the \Balls in a Vase\ question is that \noon\ doesn't \ exist. Han de Bruijn === Subject: Re: Another Inconvenient Truth <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <11cf$46c94c04$82a1e228$15235@news1.tudelft.nl> > > > > >>>>Heh, heh. I didn't define my set theory to be about finite sets > >>>>per se. But, quite fortunately, it just turns out that infinite > >>>>sets are impossible with it. > > >>>So what? It is no great achievement to come up with a set > >>>theory that does not contain infinite sets. > > >>>Your claim is that standard set theory gets the > >>>wrong answers for \what is the probablity > >>>that a number chosen at random from the set of all > >>>numbers is even?\ and \how many balls are in the vase > >>>at noon?\. > > >>>However, your set theory does not give different answers > >>>to these questions, it just declares the questions > >>>meaningless. > > >>I have another theory that answers the first question, but > >>not the second one. And the first answer is 1/2. See below. > > >>>You are confident that you know what the correct > >>>answers are, but you have no theory that leads > >>>to these answers. > > >>Not the \Bit mapped Set Theory\ but the \Naturals Construction > >>Set\ theory: > > > > So bit map set theory cannot answer either question, > > > Natural constructions set theory can answer the first question > > > You do not have a theory that will answer the second question. > > Indeed. I don't have a theory that will answer this nonsense question. However, you are confident that you know the correct answer to this \nonsense question\. - William Hughes === Subject: Re: Another Inconvenient Truth > >> >> >> >> >>>>>>Heh, heh. I didn't define my set theory to be about finite sets >>>>>>per se. But, quite fortunately, it just turns out that infinite >>>>>>sets are impossible with it. >> >>>>>So what? It is no great achievement to come up with a set >>>>>theory that does not contain infinite sets. >> >>>>>Your claim is that standard set theory gets the >>>>>wrong answers for \what is the probablity >>>>>that a number chosen at random from the set of all >>>>>numbers is even?\ and \how many balls are in the vase >>>>>at noon?\. >> >>>>>However, your set theory does not give different answers >>>>>to these questions, it just declares the questions >>>>>meaningless. >> >>>>I have another theory that answers the first question, but >>>>not the second one. And the first answer is 1/2. See below. >> >>>>>You are confident that you know what the correct >>>>>answers are, but you have no theory that leads >>>>>to these answers. >> >>>>Not the \Bit mapped Set Theory\ but the \Naturals Construction >>>>Set\ theory: >> >> >>>So bit map set theory cannot answer either question, >> >>>Natural constructions set theory can answer the first question >> >>>You do not have a theory that will answer the second question. >> >>Indeed. I don't have a theory that will answer this nonsense question. > > However, you are confident that you know the correct answer > to this \nonsense question\. Snipping enough relevance of a poster will give others the impression that you are probably right. I'm no longer interested in this kind of debating \art\. Han de Bruijn === Subject: Re: Another Inconvenient Truth <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <11cf$46c94c04$82a1e228$15235@news1.tudelft.nl> <2b5f$46c99b35$82a1e228$29154@news1.tudelft.nl> > > > > > > >>>>>>Heh, heh. I didn't define my set theory to be about finite sets > >>>>>>per se. But, quite fortunately, it just turns out that infinite > >>>>>>sets are impossible with it. > > >>>>>So what? It is no great achievement to come up with a set > >>>>>theory that does not contain infinite sets. > > >>>>>Your claim is that standard set theory gets the > >>>>>wrong answers for \what is the probablity > >>>>>that a number chosen at random from the set of all > >>>>>numbers is even?\ and \how many balls are in the vase > >>>>>at noon?\. > > >>>>>However, your set theory does not give different answers > >>>>>to these questions, it just declares the questions > >>>>>meaningless. > > >>>>I have another theory that answers the first question, but > >>>>not the second one. And the first answer is 1/2. See below. > > >>>>>You are confident that you know what the correct > >>>>>answers are, but you have no theory that leads > >>>>>to these answers. > > >>>>Not the \Bit mapped Set Theory\ but the \Naturals Construction > >>>>Set\ theory: > > > >>>So bit map set theory cannot answer either question, > > >>>Natural constructions set theory can answer the first question > > >>>You do not have a theory that will answer the second question. > > >>Indeed. I don't have a theory that will answer this nonsense question. > > > However, you are confident that you know the correct answer > > to this \nonsense question\. > > Snipping enough relevance of a poster will give others the impression > that you are probably right. I'm no longer interested in this kind of > debating \art\. > > Han de Bruijn When you can't think of a reply, you change the subject. In this case by accusing me of unfair debating tactics. Interestingly, though you accuse me of snipping a relevant part, you do not indicate what I snipped, nor why you think it was relevant. The only part I snipped was \The flaw in the \Balls in a Vase\ question is that \noon\ doesn't exist.\ While this might explain why you think this is a \nonsense question\, it does not explain why think you know the correct answer. - William Hughes === Subject: Re: Another Inconvenient Truth > >> Snipping enough relevance of a poster will give others the impression >> that you are probably right. I'm no longer interested in this kind of >> debating \art\. > > When you can't think of a reply, > you change the subject. In this case by accusing me > of unfair debating tactics. Interestingly, though you > accuse me of snipping a relevant part, you do not indicate > what I snipped, nor why you think it was relevant. > > The only part I snipped was > > \The flaw in the \Balls in a Vase\ question > is that \noon\ doesn't exist.\ Ah, you know _very well_ what you snipped. > While this might explain why you think this is a \nonsense question\, > it does not explain why think you know the correct answer. I only know that the question is incorrect. I do not know and I can not know the correct answer to an incorrect question. It's simply undefined. (I have only an answer for \time stamps before noon\.) Han de Bruijn === Subject: Re: Another Inconvenient Truth <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <11cf$46c94c04$82a1e228$15235@news1.tudelft.nl> <2b5f$46c99b35$82a1e228$29154@news1.tudelft.nl> <66d48$46c9a983$82a1e228$21619@news2.tudelft.nl> On Aug 20, 10:47 am, Han de Bruijn > > >> Snipping enough relevance of a poster will give others the impression > >> that you are probably right. I'm no longer interested in this kind of > >> debating \art\. > > > When you can't think of a reply, > > you change the subject. In this case by accusing me > > of unfair debating tactics. Interestingly, though you > > accuse me of snipping a relevant part, you do not indicate > > what I snipped, nor why you think it was relevant. > > > The only part I snipped was > > > \The flaw in the \Balls in a Vase\ question > > is that \noon\ doesn't exist.\ > > Ah, you know _very well_ what you snipped. And you still have given no indication why you think it was relevant. > > > While this might explain why you think this is a \nonsense question\, > > it does not explain why think you know the correct answer. > > I only know that the question is incorrect. I do not know and I can not > know the correct answer to an incorrect question. So, your repeated assertions that the answer \the vase is empty at noon\ is wrong because the obvious correct answer is \the vase contains an infinte number of balls at noon\, are based on something you \do not know\ and \can not know\. >It's simply undefined. So your repeated assertion that the answer \the vase is empty at noon\ is wrong is simply noise. > (I have only an answer for \time stamps before noon\.) > So you feel that you can make assertions about what happens at noon, even though you only have answers for \time stamps before noon\. - William Hughes === Subject: Re: Another Inconvenient Truth > >> >>> >>>>I have developed a kind of set theory (called \Bit mapped Set \ Theory\) >>>>in which even the set of all naturals gives rise to Russell's paradox. >>> >>>Please explain why a theory that gives rise to a paradox would be >>>interesting or useful. >> >>If that theory has something to do with truth, then paradoxes in the >>theory are very interesting and useful, because they delineate truth. > > No, the existence of paradoxes shows that the theory is inconsistent. So any theory that admits a proof by contradiction is inconsistent? Euclidian geometry is inconsistent? >>In our \pet\ case, Russell's paradox makes infinite sets \not true\. > > Makes them \not true\ in that (inconsistent) theory. Non sequitur. Han de Bruijn === Subject: Re: Another Inconvenient Truth <2247d$46bb0ddd$82a1e228$29035@news2.tudelft.nl> <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <87zm0pd44o.fsf@phiwumbda.org> On Aug 18, 11:07 pm, Han.deBru...@DTO.TUDelft.NL ............................................................................\ ........ > According to the above thread, ANY finite set (and there are no > others in \Bit mapped Set Theory\) can be equivalenced with just > one unique natural number. Thus the set of all naturals would be > equivalent to the set of all sets. This is Russell's paradox. No? > > Han de Bruijn- ***************************************************************** It's useless. Completely, desperately, hopelessly and even humorously useless: he doesn't even know what Russell's Paradox is!!! The only thing I can do is to write LOL. Some STILL try to debate a crank about set theory, infinite, Cantor and stuff...and he doesn't even know Russell's Paradox!! Good luck Tonio === Subject: Re: Another Inconvenient Truth > On Aug 18, 11:07 pm, Han.deBru...@DTO.TUDelft.NL > \ .............................................................................\ ....... > >>According to the above thread, ANY finite set (and there are no >>others in \Bit mapped Set Theory\) can be equivalenced with just >>one unique natural number. Thus the set of all naturals would be >>equivalent to the set of all sets. This is Russell's paradox. No? > > ***************************************************************** > It's useless. Completely, desperately, hopelessly and even humorously > useless: he doesn't even know what Russell's Paradox is!!! > The only thing I can do is to write LOL. > Some STILL try to debate a crank about set theory, infinite, Cantor > and stuff...and he doesn't even know Russell's Paradox!! > First read the thread: > > > It's quite obvious that this Bit mapped Set theory is _consistent_. > Also note that Bit mapped Set Theory is a _part_ of common set theory. > > Now form _within_ Bit mapped Set Theory the following set: > > A = the set of all sets which are not an element of itself. > > The first few examples with Bit mapped Set Theory show that such sets, > which are not an element of itself, surely exist. More formally: > > A = { x | not ( x in x ) } ( : like in common set theory ) > > Then if (not ( A in A)) it follows that (A in A) > and if (A in A) it follows that (not (A in A)). > This is like in Russell's paradox. > > Thus the assumption that A exists leads to a _contradiction_. But Bit > mapped Set theory is consistent. Therefore the set A does _not exist_ > within Bit mapped Set Theory. > > On the other hand, it is evident that any set in Bit mapped Set Theory > is uniquely characterized by a natural number. In fact: natural = set . > > Thus the set A = { x | not ( x in x ) } would be equivalent to the set > of all natural numbers. But we have just proved that the former set does > not exist. Therefore the set of _all_ naturals also cannot exist within > the realm of Bit mapped Set Theory. This completes the proof. Han de Bruijn === Subject: Re: Another Inconvenient Truth linux) <2247d$46bb0ddd$82a1e228$29035@news2.tudelft.nl> <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <87zm0pd44o.fsf@phiwumbda.org> > According to the above thread, ANY finite set (and there are no > others in \Bit mapped Set Theory\) can be equivalenced with just > one unique natural number. Thus the set of all naturals would be > equivalent to the set of all sets. This is Russell's paradox. No? Not in any form I can understand. But don't you know what Russell's paradox is? Never mind Google, then, since it's not producing the answers you expected. Just give a nice simple proof that the assumption there exists infinite sets leads to a contradiction analogous to Russell's (in your theory). -- Jesse F. Hughes \Part of the problem here, Peter, is that you are an idiot.\ -- Daryl McCullough gives a diagnosis === Subject: Re: Another Inconvenient Truth > >>According to the above thread, ANY finite set (and there are no >>others in \Bit mapped Set Theory\) can be equivalenced with just >>one unique natural number. Thus the set of all naturals would be >>equivalent to the set of all sets. This is Russell's paradox. No? > > Not in any form I can understand. But don't you know what Russell's > paradox is? > > Never mind Google, then, since it's not producing the answers you > expected. Just give a nice simple proof that the assumption there > exists infinite sets leads to a contradiction analogous to Russell's > (in your theory). > First read the thread: It's quite obvious that this Bit mapped Set theory is _consistent_. Also note that Bit mapped Set Theory is a _part_ of common set theory. Now form _within_ Bit mapped Set Theory the following set: A = the set of all sets which are not an element of itself. The first few examples with Bit mapped Set Theory show that such sets, which are not an element of itself, surely exist. More formally: A = { x | not ( x in x ) } ( : like in common set theory ) Then if (not ( A in A)) it follows that (A in A) and if (A in A) it follows that (not (A in A)). This is like in Russell's paradox. Thus the assumption that A exists leads to a _contradiction_. But Bit mapped Set theory is consistent. Therefore the set A does _not exist_ within Bit mapped Set Theory. On the other hand, it is evident that any set in Bit mapped Set Theory is uniquely characterized by a natural number. In fact: natural = set . Thus the set A = { x | not ( x in x ) } would be equivalent to the set of all natural numbers. But we have just proved that the former set does not exist. Therefore the set of _all_ naturals also cannot exist within the realm of Bit mapped Set Theory. This completes the proof. Han de Bruijn === Subject: Re: Another Inconvenient Truth linux) <2247d$46bb0ddd$82a1e228$29035@news2.tudelft.nl> <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <87zm0pd44o.fsf@phiwumbda.org> <87vebbgby6.fsf@phiwumbda.org> <828d4$46c9536d$82a1e228$16932@news1.tudelft.nl> > >> >>>According to the above thread, ANY finite set (and there are no >>>others in \Bit mapped Set Theory\) can be equivalenced with just >>>one unique natural number. Thus the set of all naturals would be >>>equivalent to the set of all sets. This is Russell's paradox. No? >> Not in any form I can understand. But don't you know what Russell's >> paradox is? >> Never mind Google, then, since it's not producing the answers you >> expected. Just give a nice simple proof that the assumption there >> exists infinite sets leads to a contradiction analogous to Russell's >> (in your theory). > > First read the thread: > > > It's quite obvious that this Bit mapped Set theory is _consistent_. > Also note that Bit mapped Set Theory is a _part_ of common set theory. > > Now form _within_ Bit mapped Set Theory the following set: > > A = the set of all sets which are not an element of itself. > > The first few examples with Bit mapped Set Theory show that such sets, > which are not an element of itself, surely exist. More formally: > > A = { x | not ( x in x ) } ( : like in common set theory ) How do you \form\ this set? And so what? If we suppose this set exists, we get a contradiction. How does this prove that there are no infinite sets? Try to recall: You claimed that in your theory there are no infinite sets *and* that the proof of this fact is more or less Russell's paradox. Were you drunk? -- Jesse F. Hughes \I talk with bigger fish who are playing different games.\ -- James S Harris has a way with the metaphor. === Subject: Re: Another Inconvenient Truth > >> >>> >>>>According to the above thread, ANY finite set (and there are no >>>>others in \Bit mapped Set Theory\) can be equivalenced with just >>>>one unique natural number. Thus the set of all naturals would be >>>>equivalent to the set of all sets. This is Russell's paradox. No? >>> >>>Not in any form I can understand. But don't you know what Russell's >>>paradox is? >>>Never mind Google, then, since it's not producing the answers you >>>expected. Just give a nice simple proof that the assumption there >>>exists infinite sets leads to a contradiction analogous to Russell's >>>(in your theory). >> >>First read the thread: >> >> >>It's quite obvious that this Bit mapped Set theory is _consistent_. >>Also note that Bit mapped Set Theory is a _part_ of common set theory. >> >>Now form _within_ Bit mapped Set Theory the following set: >> >>A = the set of all sets which are not an element of itself. >> >>The first few examples with Bit mapped Set Theory show that such sets, >>which are not an element of itself, surely exist. More formally: >> >>A = { x | not ( x in x ) } ( : like in common set theory ) > > How do you \form\ this set? Okay. Replace this by: Now assume that there exists a set A = ... . > And so what? If we suppose this set exists, we get a contradiction. > How does this prove that there are no infinite sets? I assume that you are not so much disabled that you can _read_ the rest of the argument. There are some simple technicalities which distinguish Bit mapped Set Theory from common set theory. You should absorb them in the first place. > Try to recall: You claimed that in your theory there are no infinite > sets *and* that the proof of this fact is more or less Russell's > paradox. Yes. > Were you drunk? Why so offensive? Han de Bruijn === Subject: Re: Another Inconvenient Truth linux) <6c38e$46c164f2$82a1e228$17760@news1.tudelft.nl> <34ba5$46c2fb7e$82a1e228$6105@news2.tudelft.nl> <200708151706.l7FH6CGQ052704@walkabout.empros.com> <87zm0pd44o.fsf@phiwumbda.org> <87vebbgby6.fsf@phiwumbda.org> <828d4$46c9536d$82a1e228$16932@news1.tudelft.nl> <87k5rq4b8p.fsf@phiwumbda.org> > >> >>>It's quite obvious that this Bit mapped Set theory is _consistent_. >>>Also note that Bit mapped Set Theory is a _part_ of common set theory. >>> >>>Now form _within_ Bit mapped Set Theory the following set: >>> >>>A = the set of all sets which are not an element of itself. >>> >>>The first few examples with Bit mapped Set Theory show that such sets, >>>which are not an element of itself, surely exist. More formally: >>> >>>A = { x | not ( x in x ) } ( : like in common set theory ) >> How do you \form\ this set? > > Okay. Replace this by: Now assume that there exists a set A = ... . > >> And so what? If we suppose this set exists, we get a contradiction. >> How does this prove that there are no infinite sets? > > I assume that you are not so much disabled that you can _read_ the rest > of the argument. There are some simple technicalities which distinguish > Bit mapped Set Theory from common set theory. You should absorb them in > the first place. I still don't see that Russell's paradox has anything to do with the obvious fact that one can prove your toy theory has no infinite sets. Can you make the argument clearer? The fact that {x | not x in x} would equal N (if it existed) is quite irrelevant. At best, Russell's paradox shows that one particular infinite set (namely, N) does not exist in your theory. But it only does that *after* you've already shown that every set is finite. So, it is not particularly relevant. > >> Try to recall: You claimed that in your theory there are no infinite >> sets *and* that the proof of this fact is more or less Russell's >> paradox. > > Yes. > >> Were you drunk? > > Why so offensive? Sorry, that was offensive without cause. Just a bad mood, I guess. > > Han de Bruijn > -- Jesse F. Hughes \Depression hits more people than thought.\ --headline in Lexington, KY newspaper, as reported on NPR's Morning Edition === Subject: Re: A CHALLANGE TO CANTORIANS > solve this equation > > 2^x = aleph_0 > Since aleph_0 occurs, this refers to cardinalities and \2^\ and \=\ should be interpreted as: Find a set X such that the power set of X can be bijected to the set of natural numbers. It is easily shown that no such set exists. But this is no problem of infinities, it exists already with finite sets where naive counting works. > hint : 2^1729 is only finite and 2^aleph_0 is aleph_1 Hint: 2^x = 1729 cannot be solved in cardinalities either. Or can you present a set X such that the power set of X has exactly 1729 elements? If you want to come up with x=10.75572... you have switched to a different interpretation of \2^\. > > hahaha > > tommy1729 > > ps if your going to say 2log(aleph_0) , you better simplify that ... > > maybe its aleph_-1 ? aleph_(root(2)-1)? > > or aleph_i ? > > hahahahaha > > cantor \seems\ to be incomplete tommy1729 seems to be completely Cantor challenged. hagman === Subject: Re: A CHALLANGE TO CANTORIANS 1. potential 2. actual === Subject: Re: A CHALLANGE TO CANTORIANS In sci.math, Aatu Koskensilta on 19 Aug 2007 07:31:29 +0300 <87ir7cno0u.fsf@huxley.huxley.fi>: > >> His question is still somewhat interesting, though; what >> precisely is x in the above equation? Or can it be shown >> that no transfinite x can solve that equation? > > Trivially, and without choice, there is no cardinal kappa such that > 2^kappa = aleph-0, since kappa < 2^kappa for all kappa and there are > no infinite cardinals < aleph-0. > Fair enough. :-) -- #191, ewill3@earthlink.net Windows. Because it's not a question of if. It's a question of when. -- === Subject: Re: A CHALLANGE TO CANTORIANS > On Aug 18, 2:52 pm, tommy1729 > > solve this equation > > > > 2^x = aleph_0 > > > > Solve > > 1/x = 0 > > for x. hahaha 1/x = 0 ; so what is x ? aleph_0 or aleph_1 or what ???? WE ALREADY KNOW ZERO IS THE INVERSE OF INFINITY hahaha :-) > > > hint : 2^1729 is only finite and 2^aleph_0 is > aleph_1 > > > ^^^^^^^^^^^^^^^^^^^^ > That is the continuum hypothesis, and it cannot be > proven > or disproven from Zermelo-Fraenkel set theory, even > with > the axiom of choice. It is independent of it -- and > one > can assume it's truth either way and you will get > different > math results. yes it cannot be proven , i know in fact it is part of my critic on cantor many people say : cantor proves this and cantor proved that ... he DID NOT !! even worse wheiter you accept the continuum hypotheses or not , there will \ still be a problem with the equation i gave since if 2^aleph_0 = aleph_0 ( fitting the equation AND not accepting the \ continuum hypotheses ) then how are we going to deal with \uncountable sets\ ?? does aleph_1 still exist then, if it is not 2^aleph_0 ?? and if not ; what is left of cantor ??? we could just as well say infinity is aleph_0 AND THEY ARE JUST SYNONYMS AND THERE ARE NO OTHER \ INFINITIES !!! debunking cantor nice , but perhaps debunking to much since uncountable sets are used in calculus for example ... so either way , we are in trouble and cantor is to blame ... in fact id say math is more complete without cantor therefore cantor is a bad thing... > > > > > hahaha > > > > tommy1729 > > > > ps if your going to say 2log(aleph_0) , you better > simplify that ... > > > > maybe its aleph_-1 ? aleph_(root(2)-1)? > > > > or aleph_i ? > > > > hahahahaha > > > > cantor \seems\ to be incomplete > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > Godel's incompleteness theorem says this is so, in > fact. > But that does not mean Cantor's theory is BAD. > yes i know , but it is still bad what is even WORSE is that people SAY cantor proved this and cantor proved that ... HE DID NOT , not even the continuum hypothesis and he is incomplete and inconsistant too and there is NO WAY to fix that (apart from replacing cantor set theory) funny that 1/x=0 solution :-) tommy1729 === Subject: Re: A CHALLANGE TO CANTORIANS In sci.math, tommy1729 on Sun, 19 Aug 2007 17:13:11 EDT <27024001.1187558021869.JavaMail.jakarta@nitrogen.mathforum.org>: > >> On Aug 18, 2:52 pm, tommy1729 >> > solve this equation >> > >> > 2^x = aleph_0 >> > >> >> Solve >> >> 1/x = 0 >> >> for x. > > > hahaha 1/x = 0 ; so what is x ? aleph_0 or aleph_1 or what ???? > > WE ALREADY KNOW ZERO IS THE INVERSE OF INFINITY > > hahaha > >:-) The extended reals is not a group. >> >> > hint : 2^1729 is only finite and 2^aleph_0 is >> aleph_1 >> >> >> ^^^^^^^^^^^^^^^^^^^^ >> That is the continuum hypothesis, and it cannot be >> proven >> or disproven from Zermelo-Fraenkel set theory, even >> with >> the axiom of choice. It is independent of it -- and >> one >> can assume it's truth either way and you will get >> different >> math results. > > yes it cannot be proven , i know > > in fact it is part of my critic on cantor > > many people say : cantor proves this and cantor proved that ... > > he DID NOT !! > > even worse wheiter you accept the continuum hypotheses or not , there will \ still be a > problem with the equation i gave > > since if 2^aleph_0 = aleph_0 ( fitting the equation AND not accepting the \ continuum hypotheses ) > > then how are we going to deal with \uncountable sets\ ?? There would be no such, in that case. However, a fairly easy proof exists that card(N) < card(2^N). Look up \power set\. > > does aleph_1 still exist then, if it is not 2^aleph_0 ?? No, it does not. Beth_1 = 2^Beth_0; Aleph_1 <= Beth_1. (Aleph_0 = Beth_0). > > and if not ; what is left of cantor ??? Nothing. > > we could just as well say > > infinity is aleph_0 AND THEY ARE JUST SYNONYMS AND THERE ARE NO OTHER \ INFINITIES !!! > > debunking cantor > > nice , but perhaps debunking to much > > since uncountable sets are used in calculus for example ... Actually, they are not, unless you get into Lebesgue theory. Standard Riemannian calculus does not need uncountable sets, merely a continuum. > > so either way , we are in trouble > > and cantor is to blame ... > > in fact id say math is more complete without cantor > > therefore cantor is a bad thing... That's quite a bit of a leap there. >> >> > >> > hahaha >> > >> > tommy1729 >> > >> > ps if your going to say 2log(aleph_0) , you better >> simplify that ... >> > >> > maybe its aleph_-1 ? aleph_(root(2)-1)? >> > >> > or aleph_i ? >> > >> > hahahahaha >> > >> > cantor \seems\ to be incomplete >> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >> Godel's incompleteness theorem says this is so, in >> fact. >> But that does not mean Cantor's theory is BAD. >> > > yes i know , but it is still bad > > what is even WORSE is that people SAY > > cantor proved this and cantor proved that ... > > HE DID NOT , not even the continuum hypothesis > > and he is incomplete and inconsistant too > > and there is NO WAY to fix that > > (apart from replacing cantor set theory) > > > funny that 1/x=0 solution :-) > > tommy1729 -- #191, ewill3@earthlink.net Windows. Because it's not a question of if. It's a question of when. -- === Subject: Re: A CHALLANGE TO CANTORIANS > In sci.math, tommy1729 > > on Sat, 18 Aug 2007 16:52:30 EDT > <29860408.1187470380912.JavaMail.jakarta@nitrogen.math > forum.org>: > > solve this equation > > > > 2^x = aleph_0 > > > > > > hint : 2^1729 is only finite and 2^aleph_0 is > aleph_1 > > > > > > hahaha > > > > tommy1729 > > > > ps if your going to say 2log(aleph_0) , you better > simplify that ... > > > > maybe its aleph_-1 ? aleph_(root(2)-1)? > > > > or aleph_i ? > > > > hahahahaha > > > > cantor \seems\ to be incomplete > > Cantor *is* incomplete; Goedel proved it. yep , and therefore also a reason to remove cantor ... math would be so more complete without cantor ... > > In any event, that equation is similar to the > equation system > > x + 1 = infinity > x != infinity not really ? why ?? ( isnt x = infinity a solution here ?? you did not use aleph notation ...) you just made that up not ?? even so doenst change anything rather shows that cantor is as inconsistant and \ simple\ as x+1=0 x+2=0 > > or > > x^2 = -1 (over R). > > Not all equations have solutions. > > -- > #191, ewill3@earthlink.net > People think that libraries are safe. They're wrong. > They have ideas. > (Also occasionally ectoplasmic slime and cute > librarians.) > > -- > http://www.teranews.com > as you said x^2 = -1 OVER R SO ALL EQUATIONS DO HAVE A SOLUTION !! in this case over R and R is extendable to complex however 'cantor equations' are not solvable NOR extendable and therefore \ BOGUS for all clarity , im not agianst you , im against cantor tommy1729 === Subject: Re: A CHALLANGE TO CANTORIANS <10573065.1187557074305.JavaMail.jakarta@nitrogen.mathforum.org> > > > cantor \seems\ to be incomplete > > > Cantor *is* incomplete; Goedel proved it. > > yep , and therefore also a reason to remove cantor ... > > math would be so more complete without cantor ... Tommy, how can something be \more complete\? It's either complete or not complete in view of underlying postulates. And Godel proved (to paraphrase) that a mathematical system cannot be both complete and consistent. So \math\ would not necessarily be \(more) complete without cantor\ although \math\ may be consistent without Cantor. Would you prefer \math\ to be consistent or complete? You can't have both. === Subject: Re: A CHALLANGE TO CANTORIANS > for all clarity , im not agianst you , im against cantor You're doing a magnificent job at it, too! Go on, expose the rotten core for all to see. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \Wovon man nicht sprechen kann, daruber muss man schweigen\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A CHALLANGE TO CANTORIANS On Sat, 18 Aug 2007 16:52:30 EDT, tommy1729 >solve this equation > >2^x = aleph_0 There is no solution to this equation. So what? >hint : 2^1729 is only finite and 2^aleph_0 is aleph_1 What? Exactly how do you prove that 2^alph_0 = aleph_1? Again, even when you pretend to be \challenging\ us, the things you say are simply oozing ignorance. > >hahaha > >tommy1729 > >ps if your going to say 2log(aleph_0) , you better simplify that ... Nobody's going to say that, because it makes no sense. > >maybe its aleph_-1 ? aleph_(root(2)-1)? > >or aleph_i ? > >hahahahaha > >cantor \seems\ to be incomplete The arithmetic of cardinals is imcomplete, in the sense that not every equation has a solution. So what? Has anyone ever claimed otherwise? ************************ David C. Ullrich === Subject: Re: A CHALLANGE TO CANTORIANS > >hint : 2^1729 is only finite and 2^aleph_0 is aleph_1 > > What? Exactly how do you prove that 2^alph_0 = aleph_1? > > Again, even when you pretend to be \challenging\ us, > the things you say are simply oozing ignorance. The confusion here is that many people, including the OP, are misinformed as to the definition of aleph. They assume that the definition is: aleph_0 = card(N) aleph_(n+1) = 2^card(aleph_n) This has nothing to do with being a crank. I've seen many books and websites (not graduate-level set theory texts, of course, but books geared towards a wider audience) that define aleph this way. Indeed, the following website is devoted to such confusion: http://www.ii.com/math/ch/#confusion Notice that, since such people think that aleph_1 is defined to be c, they state ~CH as being the existence of a cardinal between aleph_0 and aleph_1! So what the OP is really saying is: >> hint : 2^1729 is only finite and 2^aleph_0 is c or even more to the point: >> hint : 2^1729 is only finite and 2^aleph_0 > aleph_0 since his argument is concerning the nonexistence of a kappa such that 2^kappa = aleph_0. Similarly, the OP interprets Prof. Ullrich's reply as: > What? Exactly how do you prove that 2^aleph_0 = c? Notice that the Ghost in the Machine mentioned the beth numbers, which are defined the way many people think the aleph numbers are. Thus one can replace all of the OP's alephs by beths in order to understand what he is really saying. So I don't fault the OP for assuming that aleph_1 is defined as 2^aleph_0, since he mostly likely read the definition in a book. There are other things the OP this isn't one of them. === Subject: Re: A CHALLANGE TO CANTORIANS > > The confusion here is that many people, including the OP, > are misinformed as to the definition of aleph. [bla bla] > > This has nothing to do with being a crank. > Of course it has. \Cranks who contradict some mainstream opinion in some highly technical field, such as mathematics or physics, almost always 1. exhibit a marked lack of technical ability, 2. misunderstand or fail to use standard notation and terminology, 3. ignore fine distinctions which are essential to correctly understanding mainstream belief.\ [In short:] cranks [...] often seem to represent [...] individuals with an exceptional degree of ignorance concerning the subject of their cranky belief.\ http://en.wikipedia.org/wiki/Crank_%28person%29 F. -- E-mail: infosimple-linede === Subject: Re: A CHALLANGE TO CANTORIANS > > > The confusion here is that many people, including the OP, > > are misinformed as to the definition of aleph. [bla bla] > > > This has nothing to do with being a crank. > > Of course it has. Normally, I'd agree, but in the special case of the definition of aleph_1, the error is so widespread that many people who are _not_ trying to refute Cantor give the wrong meaning. In fact, according to Wikipedia, even the Encyclopedia Britannica once defined aleph_1 as 2^aleph_0: http://en.wikipedia.org/wiki/Wikipedia:Errors_in_the_Encyclop%C3%A6dia_Brita\ nnica_that_have_been_corrected_in_Wikipedia#Transfinite_numbers Apparently by now the error in EB has been fixed, but the defines aleph_1 incorrectly: http://jedidiah.stuff.gen.nz/wp/?p=17|The One math book I once read (IIRC it was entitled \Kingdom of Infinite Number\) not only defines aleph_1 as 2^aleph_0, but claims that \aleph_1 = omega\ is equivalent to CH! Not only that, I dug up some old sci.math threads and saw even the mathematicians who are not refuting Cantor have made errors related to CH and the alephs. The following was written in a 2005 Tony Orlow/Ross Finlayson thread, not by TO or RF but the standard mathematicians: \The Continuum Hypothesis does not ask if there is a cardinality between |N| and |P(N)|, but rather asks whether c, 2^|N|, is one of the Alephs or not.\ Whether c is an aleph is a matter of AC, not of CH. One mathematician corrected the error, but then made an error of his own: \c must be an aleph, since the alephs are all the infinite cardinalities, and aleph_omega is known to be bigger than c\ Yet ZFC+\c > aleph_omega\ is consistent if ZFC itself is! Here's one more example from 1999: \NOTATION: Let N* be the cardinality of the natural numbers and R* be the cardinality of the reals. Recall that 2^N* = R*. Let N** be the smallest uncountable cardinal number (i.e. the next largest infinity after N*). The continuum hypothesis (CH) is the statement N** = R*. CH is known to be independent of the ZFC axioms of set theory.\ So the author is using N* to denote aleph_0, N** to denote aleph_1, and R* to denote c. So far, so good. But then: \If CH fails, the situation becomes more complicated. For instance, it is possible for there to be more than N*, more than R*, or even more than 2^R* many distinct cardinal numbers lying between N* and R*, and also between R* and 2^R*. In fact, the number of cardinal numbers lying between N* and R* and the number of cardinal numbers lying between R* and 2^R* can be virtually anything. For instance, it is consistent with the ZFC axioms for there to be 38 cardinals between N* and R* and 19 cardinals between R* and 2^R*; or 559 cardinals between N* and R*, while none lie between R* and 2^R*; or more than 2^R* lying between N* and R* while 12 lie between R* and 2^R*. [However, it is known that there cannot be exactly N* distinct cardinals lying between N* and R*. This is why I used *virtually* in \virtually anything\ above.]\ The last line is intended to be a reference to the cofinality argument that aleph_omega cannot be equal to c. But since N* is intended to be a _cardinality_, what is actually written is that the set of all cardinalities between aleph_0 and c cannot have cardinality aleph_0. Thus the statement also forbids c = aleph_(omega + 1), since the set of all cardinals between aleph_0 and aleph_omega + 1 has cardinality aleph_0 (although it is order isomorphic to omega + 1, via the relation <). In other words, the author has just confused cardinality with ordinality. What's worse is the assertion that there may be more than 2^R* cardinals between N* and R*, i.e., that the set of all cardinals between aleph_0 and c may even have a cardinality even more than 2^c. But this is absurd -- the set of all _ordinals_ between aleph_0 and c clearly has cardinality c, and since every cardinal is an ordinal, there surely can't be strictly more _cardinals_ than _ordinals_ between aleph_0 and c! And the author of all this wasn't TO, RF, or any other \crank,\ but a mathematician who was working on a problem that's much more complicated than something a \crank\ would understand. The author was probably trying to write about set theory without mentioning alephs and whatnot by using the N*, N**, R* notation, but instead inadvertently The point I'm trying to make is that not everyone #3 about cranks failing to notice fine distinctions is correct, but you cannot blame someone who actually read the wrong definition in a book. The OP doesn't even understand why what he's written is wrong. === Subject: Re: A CHALLANGE TO CANTORIANS > > > > > The confusion here is that many people, including the OP, > > > are misinformed as to the definition of aleph. [bla bla] > > > > This has nothing to do with being a crank. > > > Of course it has. > > Normally, I'd agree, but in the special case of the definition > of aleph_1, the error is so widespread that many people who > are _not_ trying to refute Cantor give the wrong meaning. ************************************************************ Aaaah! Yes indeed. We all can be wrong, but the crank's nature drives him into pompous, ostentatious, showy and boastful display of their stupidity, and they even mock and scoff others many times about stuff they don't have much idea about. The crank doesn't ask, the crank doesn't try to learn: he/she *KNOWS* and, of course, he/she is MUCH more intelligent, witty, handsome, rich and sexy than any other person around, specially more than any other one that dares to contradict them. Just read Tommy1979's \HAHAHAHAHAHAHAH\ histerical posts, full with capital letters, in this and other threads;schyzophrenia, apparently. Of course, he's not the only one....fortunately for many of us that can still get some laugh out of their nonsenses from time to time :) Tonio === Subject: Re: A CHALLANGE TO CANTORIANS > Just read Tommy1979's \HAHAHAHAHAHAHAH\ histerical posts, full with > capital letters, in this and other threads;schyzophrenia, apparently. > Of course, he's not the only one....fortunately for many of us that > can still get some laugh out of their nonsenses from time to time :) Much amusement can be derived from the antics of various colourful characters in the news, no doubt. Remote psychiatric diagnosis, such as your attributing Tommy's erratic babbling to \schyzophrenia\, on the other hand is baseless and somewhat silly. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \Wovon man nicht sprechen kann, daruber muss man schweigen\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A CHALLANGE TO CANTORIANS > The confusion here is that many people, including the OP, > are misinformed as to the definition of aleph. They > assume that the definition is: > > aleph_0 = card(N) > aleph_(n+1) = 2^card(aleph_n) > > I've seen many books and websites (not graduate-level set theory > texts, of course, but books geared towards a wider audience) that > define aleph this way. Such books and websites are simply wrong. It is not at all uncommon to find all sorts of mathematical non-sense on the web or in popular and semi-popular books. Some caution is called for, surely, before mouthing off on the horrible contradictions and incompleteness of this or that piece of mathematics. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) \Wovon man nicht sprechen kann, daruber muss man schweigen\ - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: A CHALLANGE TO CANTORIANS ( and is being ignorent once more !! ) > On Sat, 18 Aug 2007 16:52:30 EDT, tommy1729 > > > >solve this equation > > > >2^x = aleph_0 > > There is no solution to this equation. So what? > > >hint : 2^1729 is only finite and 2^aleph_0 is > aleph_1 > > What? Exactly how do you prove that 2^alph_0 = > aleph_1? CMON DAVID . IT IS THE BASIC OF CANTOR SET THEORY AND VISION ON INFINITY SURELY YOU MUST HAVE SEEN THAT BEFORE EVERYBODY ON THE FORUM KNOWS ABOUT THAT it is related to \uncountable set \ \axiom of choice\ \zermelo \ fraenkel\ \continuum hypothesis\ and no it cannot be proven without accepting an axiom but that is the WELLKNOWN viewpoint of cantor JESUS YOU KNOW LESS THAN I EXPECTED !! > > Again, even when you pretend to be \challenging\ us, > the things you say are simply oozing ignorance. NO YOU ARE THE IGNORANT FOOL 2^aleph_0 = aleph_1 exept you !!!!!!!!! > > > > >hahaha hahaha agian for ignorant ullrich and trying to \ project \ his ignorence on me projecting ; wiki for it. > > > >tommy1729 > > > >ps if your going to say 2log(aleph_0) , you better > simplify that ... > > Nobody's going to say that, because it makes no > sense. you might have , since you even never heard of 2^aleph_0 = aleph_1 hahaha your not even good enough to be a good crakpot ... because they know what they are fighting against and know the weaknesses of their opponents theories ... > > > > >maybe its aleph_-1 ? aleph_(root(2)-1)? > > > >or aleph_i ? > > > >hahahahaha > > > >cantor \seems\ to be incomplete > > The arithmetic of cardinals is imcomplete, in the > sense > that not every equation has a solution. So what? Has > anyone ever claimed otherwise? > > ************************ > > David C. Ullrich stop the \ ullrichism \ LADIES AND GENTLEMAN ONE OF THE GREATEST CRITICS OF ME AND DEFENDER OF CANTOR HOWEVER HE DOES NOT EVEN UNDERSTAND ALEPH 0 OR ALEPH 1 HE DOES NOT EVEN KNOW WHAT HE IS DEFENDING !!!! HAHAHAHAHA typical for my critics :-) tommy1729 ps : go whine to your lawyer and ask him to post another ode to you here on \ sci.math and an apology too === Subject: Re: A CHALLANGE TO CANTORIANS On Sun, 19 Aug 2007 17:40:35 EDT, tommy1729 > >( and is being ignorent once more !! ) > >> [...] > >NO YOU ARE THE IGNORANT FOOL > > >2^aleph_0 = aleph_1 > >exept you !!!!!!!!! Ahem! Even I know that what everybody is familiar with is the fact that 2^aleph_0 >= aleph_1, and Cantor's continuum hypothesis (i.e. conjecture) is that this inequality is in fact an equation, i.e. 2^aleph_0 is the smallest cardinal that is strictly greater than aleph_0, i.e. 2^aleph_0 (= c) is the smallest uncountable cardinal. (It's still all a bit theological to me, too, but there is no ambiguity about it.) >> >hahaha > >hahaha agian for ignorant ullrich > >and trying to \ project \ his ignorence on me > >projecting ; wiki for it. Tommy, try looking up \irony\ some time! -- Angus Rodgers Contains mild peril === Subject: Re: A CHALLANGE TO CANTORIANS On Sun, 19 Aug 2007 17:40:35 EDT, tommy1729 >> >> What? Exactly how do you prove that 2^alph_0 = aleph_1? >> > IT IS THE BASIC OF CANTOR SET THEORY AND [bla bla] > Nonsense. We are talking about Cantor's continuum /hypothesis/. \Cantor believed the continuum hypothesis to be true and tried for many years to prove it, in vain. It became the first on David Hilbert's list of important open questions that was presented at the International Mathematical Congress in the year 1900 in Paris.\ Source: http://en.wikipedia.org/wiki/Continuum_hypothesis >> >> Again, even when you pretend to be \challenging\ us, >> the things you say are simply oozing ignorance. >> Or maybe just a sign of some sort of brain damage. Who knows? > > [...] > > tommy1729 > I'd suggest to visit your resident psychiatrist. F. === Subject: Re: A CHALLANGE TO CANTORIANS G. FREGE WROTE > On Sun, 19 Aug 2007 17:40:35 EDT, tommy1729 > > > >> > >> What? Exactly how do you prove that 2^alph_0 = > aleph_1? > >> > > IT IS THE BASIC OF CANTOR SET THEORY AND [bla bla] > > > Nonsense. We are talking about Cantor's continuum > /hypothesis/. > > \Cantor believed the continuum hypothesis to be true > and tried for > many years to prove it, in vain. It became the first > on David > Hilbert's list of important open questions that was > presented at the > International Mathematical Congress in the year 1900 > in Paris.\ > > Source: > http://en.wikipedia.org/wiki/Continuum_hypothesis > > >> > >> Again, even when you pretend to be \challenging\ > us, > >> the things you say are simply oozing ignorance. > >> > Or maybe just a sign of some sort of brain damage. > Who knows? > > > > > [...] > > > > tommy1729 > > > I'd suggest to visit your resident psychiatrist. > > > F. > OH PLEASE I MENTIONED CONTINUUM HYPOTHESE TOO YOU JUST SNIPPED IT !!! GIVING A LINK TO HILBERT DOES NOT PROVE YOUR BETTER THAN ME !!! AND EITHER HOW CANTOR CONTINUUM HYPOTHESIS IS RELATED TO CANTOR SET THEORY \ !! AND EVEN IF NOT DAVID C ULLRICH SHOULD HAVE KNOW ABOUT THE CONTINUUM HYPOTHESIS AND 2^ALEPH_0 = ALEPH_1 HA !!!! SO YOU MIGHT DISCUSS THE REASON WHY THE ABOVE IS TRUE OR WHY DAVID WAS IGNORANT BUT HE REMAINS IGNORENT EITHER HOW !!!! HE CANT EVEN DO AN INTEGRAL i hope your better at math then david !! tommy1729 === Subject: Re: A CHALLANGE TO CANTORIANS > > > solve this equation > > > 2^x = aleph_0 > > > hint : 2^1729 is only finite and 2^aleph_0 is aleph_1 > > > hahaha > > > tommy1729 > > > ps if your going to say 2log(aleph_0) , you better simplify that ... > > This reminds me of one of Tony Orlow's posts. He once referred > to the logarithm of an infinite set in one of his posts. The > context, BTW, was that Orlow thinks of infinite numbers as > infinitely long binary strings, and he considers his unit > infinity, N, to consist of one followed by log_2(N) zeroes. > > Of course, in the hyperreals, one may take logarithms of > infinite hyperreals with ease. Which means there's also a solution in the Surreal Numbers (which was one of my first thoughts when reading tommy's post), because the Surreal Numbers contain the hyperreals (IIRC). --- Christopher Heckman > > maybe its aleph_-1 ? aleph_(root(2)-1)? > > Those numbers do not exist either in ZFC or even in the > hyperreals eithers. > > > cantor \seems\ to be incomplete > > Of course there exists no cardinal number kappa such that > 2^kappa = 3 either. I've seen some intriguing arguments > against Cantor, but this one -- at least on its own -- > is not one of them. > > Notice that in ZF+GCH, 2^kappa = tau has a solution iff > tau is a successor cardinal, of course. In ZF+~GCH > there may not be a solution even if tau is a successor, > and if a solution exists it may not be unique. For some > cardinals, such as aleph_omega, a solution never exists > regardless of CH or GCH (cofinality argument). === Subject: Re: A CHALLANGE TO CANTORIANS > > On Aug 18, 1:52 pm, tommy1729 > > > > > solve this equation > > > > > 2^x = aleph_0 > > > > > hint : 2^1729 is only finite and 2^aleph_0 is > aleph_1 > > > > > hahaha > > > > > tommy1729 > > > > > ps if your going to say 2log(aleph_0) , you > better simplify that ... > > > > This reminds me of one of Tony Orlow's posts. He > once referred > > to the logarithm of an infinite set in one of his > posts. The > > context, BTW, was that Orlow thinks of infinite > numbers as > > infinitely long binary strings, and he considers > his unit > > infinity, N, to consist of one followed by log_2(N) > zeroes. > > > > Of course, in the hyperreals, one may take > logarithms of > > infinite hyperreals with ease. > > Which means there's also a solution in the Surreal > Numbers (which was > one of my first thoughts when reading tommy's post), > because the > Surreal Numbers contain the hyperreals (IIRC). > > --- Christopher Heckman > > > > maybe its aleph_-1 ? aleph_(root(2)-1)? > > > > Those numbers do not exist either in ZFC or even in > the > > hyperreals eithers. > > > > > cantor \seems\ to be incomplete > > > > Of course there exists no cardinal number kappa > such that > > 2^kappa = 3 either. I've seen some intriguing > arguments > > against Cantor, but this one -- at least on its own > -- > > is not one of them. > > > > Notice that in ZF+GCH, 2^kappa = tau has a solution > iff > > tau is a successor cardinal, of course. In ZF+~GCH > > there may not be a solution even if tau is a > successor, > > and if a solution exists it may not be unique. For > some > > cardinals, such as aleph_omega, a solution never > exists > > regardless of CH or GCH (cofinality argument). > > and he is correct ... i have dabated about hyperreals vs cantor before i will avoid a deja-vu feeling , and just say that proginoskes is \ correct... tommy1729 ps: that guy ( lwal..@lausd.net) said he has heard better arguments against \ cantor ... i believe him i tried to keep the arguments as simple as possible but just out of curiousity , what are those better arguments ?? note that i have given several arguments in several threaths against cantor im curious ... plz tell me more tommy1729 === Subject: Re: A CHALLANGE TO CANTORIANS <21643395.1187558950833.JavaMail.jakarta@nitrogen.mathforum.org> > ps: that guy ( lwa...@lausd.net) said he has heard better arguments \ against cantor ... > > i believe him > > i tried to keep the arguments as simple as possible > > but just out of curiousity , what are those better arguments ?? > > note that i have given several arguments in several threaths against \ cantor > > im curious ... > > plz tell me more Well, in one of the other many Cantor threads, there was a discussion about Maddy's MAXIMIZE principle. Now MAXIMIZE is not a formal axiom per se -- indeed, it was argued that MAXIMIZE could imply both CH and ~CH, depending on how one interpreted it. Now this isn't about Maddy's MAXIMIZE principle, but of one of her other principles. This one is mentioned in the FAQ of this very newsgroup: \3. If GCH is true, this implies that aleph_0 has certain unique properties: e.g. that it's that cardinal before which GCH is false and after which it is true. Some would like to believe that the set-theoretic universe is more 'uniform' (homogeneous) than that, without this kind of singular occurrence. Such a 'uniformity' principle tends to imply not GCH.\ In other words, if we let phi be the formula \kappa is a cardinal such that 2^kappa = kappa+\, then in ZF+GCH we have that phi(kappa) is false for arbitrary large finite cardinals, since 2^n is much larger than n+1 (except for n=0 and n=1), but CH implies that phi(aleph_0) is true and by GCH, phi(kappa) is true for all infinite cardinals larger than aleph_0. The fact that 2^kappa is much larger than kappa+ for large finite cardinals, but they suddenly become equal at infinity, is enough for many set theorists to prefer ~GCH to GCH, and this is called Maddy's UNIFORMITY principle. But now, what if we tried applying the UNIFORMITY principle to the formula phi, given by \kappa is a cardinal such that 1+kappa = kappa\ (where + denotes cardinal addition) in ZFC. Then once again, phi(kappa) is false if kappa is finite, but is true if kappa happens to be infinite. And so, to repeat: 3. If the Axiom of Infinity is true, this implies that aleph_0 has certain unique properties: e.g. that it's that cardinal before which \1+kappa = kappa\ is false and after which it is true. Some (so-called \cranks\) would like to believe that the set-theoretic universe is more \uniform\ (homogeneous) than that, without this kind of singular occurrence. Such a \uniformity\ principle tends to imply not (Axiom of Infinity). And indeed, this is what many \cranks\ argue -- that infinite cardinals should work just like finite cardinals, in accord with Maddy's UNIFORMITY principle mentioned above. So what's the difference between a set theorist who uses Maddy's UNIFORMITY principle to argue against GCH and a so-called \crank\ who uses UNIFORMITY to argue against the Axiom of Infinity? === Subject: Re: A CHALLANGE TO CANTORIANS > > So what's the difference between a set theorist [...] > and a [...] \crank\ [...]? > See: http://en.wikipedia.org/wiki/Crank_(person) F. === Subject: spivak,calculus on manifolds 1-30 let f:[a,b]->R be an increasing function. if x1,....xn in [a,b] and distinct, show that o(f,x1)+....o(f,xn) let f:[a,b]->R be an increasing function. > if x1,....xn in [a,b] and distinct, > show that o(f,x1)+....o(f,xn) > o(f,x)=oscillation of f at x. Try considering n+1 points y_i such that a = y_1 < x_1 < y_2 < x_2 < ... < x_n < y_{n+1} = b Now, f(y_2)-f(y_1) + f(y_3)-f(y_2) + ... + f(y_{n+1})-f(y_n) = ? Then? Kiuhnm === Subject: Re: spivak,calculus on manifolds 1-30 <46c965a1$0$4796$4fafbaef@reader4.news.tin.it> > > let f:[a,b]->R be an increasing function. > > if x1,....xn in [a,b] and distinct, > > show that o(f,x1)+....o(f,xn) > > o(f,x)=oscillation of f at x. > > Try considering n+1 points y_i such that > a = y_1 < x_1 < y_2 < x_2 < ... < x_n < y_{n+1} = b > Now, > f(y_2)-f(y_1) + f(y_3)-f(y_2) + ... + f(y_{n+1})-f(y_n) = ? > Then? > > Kiuhnm Moreover,how we know y_1 > > let f:[a,b]->R be an increasing function. > > if x1,....xn in [a,b] and distinct, > > show that o(f,x1)+....o(f,xn) > > o(f,x)=oscillation of f at x. > > Try considering n+1 points y_i such that > a = y_1 < x_1 < y_2 < x_2 < ... < x_n < y_{n+1} = b > Now, > f(y_2)-f(y_1) + f(y_3)-f(y_2) + ... + f(y_{n+1})-f(y_n) = ? > Then? > > Kiuhnm to be sure when adding the f(y_i) I understand that each o(f,x_i)=lim (f(x_i+delta_i)-f(x_i-delta_i) because this is increasing function so conclude o(f,x_i) but why o(f,x_i) f(x) < f(y). Let's suppose we have u < x < v, where u, x, v are in [a,b]. There is a d such that u < x-d < x < x+d < v. Let S be the set {f(y) : |y-x| < d}. It's clear that sup(S) <= f(x+d) < f(v) and that inf(S) >= f(x-d) > f(u). Therefore, sup(S)-inf(S) < f(v)-f(u). Note that o(f,x) = lim_{d->0} M(x,f,d)-m(x,f,d) <= sup(S)-inf(S) and then o(f,x) < f(v)-f(u). (1) Now note that (1) holds even when u <= x < v or u < x <= v. Kiuhnm === Subject: #19 If the Nobel in physics had been fair and true, then there \ would not have been 26% Jew, nor lopsided USA, nor so many omissions like \ Tesla, Edison ; new book: \What Religion is, and how science has started to \ replace it\ I need to focus on finishing another book before the end of August, so I am going to stop on this thread and pick it up some long time later. But before I stop I have to get include some ideas on the tip of my mind such as the Bardeen error. I spoke of the omissions of the Physics Nobel such as Tesla and Edison and Wright brothers. And the awards for trivia such as Dalen or Glashow, Salam, Weinberg, or Mather, Smoot. I spoke of the Nobel physics prize running into a 51% or greater error rate for recent decades and where fake physics is being rewarded with the prize. Such fake physics as Big Bang, black-holes, BCS superconductivity, Standard Model, Quark theory, and encroaching String theories. I did ot speak of the idea that Neutron stars are better explained is dense metallic stars and that neutron stars are more fakery. Anyway, there is a large list of fake physics for which the Nobel prize is harming the science communities by entrenching and rewarded fake physics. I did not address the obvious error of Bardeen where he received the physics Nobel twice. I should mention this story because it indicates how skylarking the Nobel Committee is doing. A rational person would think that a Committee that picks the \best of physics\ would understand that if you give the Nobel twice that the recipient must be a giant of physics. The giants of physics for the 20th century was probably Bohr and Dirac for having lead physics into Quantum Mechanics. So to have given Bardeen the physics Nobel twice is a case study of how poor of a job the Nobel Committee is doing. Even John Bell who showed us with his Bell Inequality that Quantum Mechanics is on the large scale and in later years showed us a concept of Superdeterminism that would be a concept that enters most every other human subject, that John Bell should have received the Nobel physics prize twice, not Bardeen who should be lucky to have received it once. And yet John Bell never got it at all. So the Nobel physics prize is falling into an error rate of greater than 50% for the past decades. That one out of two Nobel physics prizes from about 1979 onwards is a fake physics or trivia physics. If the Nobel prizes continue on this bad track, it will lose its reputation and reach a point where most physicists will reject the prize saying \who wants to be on a list that is a mockery of physics\ The greatest harm of the Nobel physics prize, if it continues on this road, is that it leads people astray as to what is true in physics and where they do their career research into fake physics. The Nobel Committee should always think twice whenever they reward some \theory of physics\ compared to \experimental physics\. Because of the recent huge error rates of the Nobel physics, they flunk and have done a poor and lousy job. They should have investigated why 26% Jew wins and how to fix that bias and dishonesty. They should have reviewed their judging as to why they missed Tesla and Edison and Wright Brothers and how to remedy those omissions. I have offered several possible remedies: (1) perhaps skip some years where no physics is awarded because there was no \brilliant work or discoveries\. The pressure of having to give a prize every year can lead to mistakes. (2) include the idea that geniuses of science popp-up randomly based on population size, so that if you see a race or country with a high number of wins for example 26% Jews, then you know something is deeply wrong in the judging process (3) allow the judging to change and be a flexible system so that it can perfect itself so that the bylaws of the Nobel \will\ can improve with time and not be a dogmatic selection process (4) think about a concept of a Physics or Science King or leader in any given generation (30 years) and let this King form a relationship to the Nobel Committee where the King sort of guides the judging process. Bohr could have filled that position since he led physics into Quantum Mechanics. And whether anyone likes it or not, I am serving that position now. And I am even telling the Nobel Committee their errors of the past and how sloppy and lousy they are as of recent. (5) have the King or Queen of Sweden suspend the Nobel prizes starting now and to review the past performance and to change the bylaws to improve the prizes. In other words, the errors and mistakes are so bad that shut down of the prizes until a reform is in place. A shutdown is warranted because of the entrenched system that gave 26% to Jews is a massive corruption. Science, and especially physics, should not be a neon billboard advertisement to a religion. Physics is above religion, for physics is the highest truth. All of us know how religion tries to draw and paint and picture God, such as Michelangelo's painting of an old bearded man touching the finger of Adam. But physics is now telling us who God is, what God really looks like and telling us the purpose of life. So we have entered the 21st century with a physics prize that is meddled (sorry for the pun) with religion. 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: MI5 Persecution: Fitted up 26/4/96 (791) === Subject: Re: MI5? Please can someone explain what's going on here? Distribution: : >The (remote) possibility remains that 'Mike Corley' is either : >not schizophrenic (but is 'pretending' to be so) or 'he' is : >a product of a number of persons (?psychology students). : Given other ways in which I have seen people exploit some of The \ Internet's : capabilities to disrupt or indulge in sophistry, or to exploit a medium : that resembles speech without the non-verbal and intonation cues, etc : as a means of denigrating others, I question your use, albeit in quotes, : of the word \remote\. I'm not saying it isn't remote and therefore it \ is : great, I'm just saying that I don't think we can easily classify it as : remote, moderate, or great. I think you can build up quite a good picture based on what someone says and on their posting patterns. I don't think \The Internet\ (capitals, no less) is as opaque a medium as you make it out to be. : It is not easy to determine the validity of all information on The : Internet without making use of extra supplementary information. : We do have the problem, pointed out by someone else, of the possibly : \too perfect\ textbook characteristics of what is being posted. I explained that one, but I don't mind explaining it again (you don't mind having it explained again to you, do you now?). The reason my \symptoms\ are such a perfect fit to the textbook is because the people causing the campaign \fitted me up\ in such a way that what they did would resemble the symptoms of schizophrenia. Hence TV, radio, other media, people in the streets etc. By a fortunate coincidence (for them) these mthods of harassment are the ones which offer easiest channels of access (for them). It's really quite neat. All it takes is for people to start believing that the \symptoms\ aren't symptoms but reality, though, and the house of cards collapses in a heap. And there are _lots_ of people now who knoiw full well what has gone on. : If harrassment by email, etc, has happened by someone out of the country, : can a complaint be made that results in arrest or whatever upon that : person's entry into the country? An interesting point which Mike may be : able to inform us about, as he's said he will be in the UK in a few weeks : time. Picture the scene at the airport; \I arrest you for being Mike Corley and mailbombing people\ \But my name isn't Corley. Who he? Mailbombing isn't illegal is it? You'd have to lock up a lot of people if sending annoying email was a crime\ \Er.....\ : -- : David Stretch: Greenwood Institute of Child Health, Univ. of Leicester, \ UK. : dds@leicester.ac.uk Phone:+44 (0)116-254-6100 Fax:+44 \ (0)116-254-4127 : assume they are meant for you. Mike, this is called paranoia. But that's the way real abuse works, too. People interject words and phrases into what they say which they know will have meaning for the \ listener. And sometimes, they make it obvious. The very first evening of my job in Oxford, we went for a drink with the technical director, and a couple of other employees. The TD said in an \as-if\ aside to one of the others, \Is this the bloke who's been on TV?\ (he said it directly in front of me, and obviously meant mke to hear him saying it). The other person replied, \Yes, I think so\. I think the subtext of what the TD said was \Why are they bothering with him? He's so insignificant, why would they possibly want to spend the resources going after him and putting all that expensive technology in his home, when there must be much better targets?\. The Technical Director was given to sometimes disrespecting people, you see, and in my case he couldn't see the point of anyone expending money on harassing me. === Subject: Re: Treatment of Schizophrenia Distribution: : Probably 'cos you come across as reasoned & articulate, it's a pity : about the other stuff :) Veracity is so unreasonable. : >>pps. You should still see a doc again Mike. : > : >Doing so. Trouble is, all this mental-illness stuff provides camouflage : >for the harassment, which is real. It alows people who otherwise would : >consider the harassment seriously to disregard it. It makes \ conversations : >with a lawyer or police brief when otherwise it would merit discussion. : The point is that there are two possibilities happening here- : 1. There's a large conspiracy of people out to get you, for no : other reason than that they have the means to do so, and that : it involves a lot of the Media & a proportion of the public : 2. You (who admit to having some headspace problems) are suffering : from acute paranoid schizophrenia. : Possibility #1 is _possible_, but would be unprecendented (OTOH, : how would we know?), unfeasible, and many other things beginning : with _un_ which I can't think of at the moment. Besides, if there : was something going on, chances are some of us here would know : about it, and I'm convinced that nobody does. \Unprecedented\ hits the nail on the head. It _is_ unprecedented, but we have only just reached the technical stage at which it is feasible, and we know video-spying is done to other people (NB the Diana-Hewitt episode) and is a routine tool of security agencies. Perhaps what is unprecedented is not the technical side, but the social manipulation of many people by a concealed element in what other countries would be called the secret police. The most disturbing element is the degree to which people allow themselves to be unquestioningly manipulated by an evil element within the state. 791 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: New mathematics / physical sciences positions at \ http://jobs.phds.org, Aug 20, 2007 There are new job listings at http://jobs.phds.org -------------------------------------------------------------------- Title: Senior Quantitative Analyst Employer: Pinetum Partners LLC (Executive Search Firm) Location: New York, NY, United States Company Background Pinetum Partners, an executive search firm, has retianed by a subsidiary of a major Wall Street firm that is a in providing institutional investors and asset managers with risk measurement solutions. These... 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Full details: -------------------------------------------------------------------- Title: Sr. Quant, Chief Scientist Employer: Chicago Loop based Trading Group Location: Chicago, IL, United States Requirements: 10+ years experience in the financial industry and a PhD in a math heavy field. Expert in some of: stochastic probability, statistics, and time series analysis. Deep of the modeling of vanilla... Full details: -------------------------------------------------------------------- Title: Quant Employer: Chicago Loop based Trading Group Location: Chicago, IL, United States Requirements: Highly skilled in mathematics and data analysis. and passionate about solving puzzles and a wide variety of Strong software engineering skills. Research experience in a field. Excellent communication skills... 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Full details: -------------------------------------------------------------------- Title: Assistant Professor of Statistics #53883 (2 Positions) Employer: Georgia Southern University Location: Statesboro, GA, United States The Department of Mathematical Science in the Allen E. Paulson of Science and Technology invites nominations and applications for position of Assistant Professor of Statistics #53883 (2 Georgia Southern University, a member... Full details: -------------------------------------------------------------------- Title: Assistant Professor of Mathematics #53882 Employer: Georgia Southern University Location: Statesboro, GA, United States The Department of Mathematical Science in the Allen E. Paulson of Science and Technology invites nominations and applications for position of Assistant Professor of Mathematics #53882. Georgia University, a member institution of... 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Full details: -------------------------------------------------------------------- Title: Director, Analytical Consulting Employer: Benton Search Associates, Inc. Location: Paterson, NJ, United States Interested candidates please e-mail resume and cover letter to: Linda Benton, President Benton Search Associates, Inc. www.BentonSearch.com Full details: -------------------------------------------------------------------- Title: VP, Advanced Analytics Pharma Research Employer: Benton Search Associates, Inc. Location: Philadelphia, PA, United States Vice President, Advanced Analytics $150k - $200k + 20% Small Research Company \.89a near Philadelphia - approx $10M/yr \ Relocation candidate must be US Citizen, PR or Green Card Holder Position The Vice President... Full details: -------------------------------------------------------------------- Title: Marketing Scientist Employer: Benton Search Associates, Inc. Location: Arkansas, United States MARKETING SCIENTIST $100k - $150k For Market Research Supplier AK relocation assistance Must be US Citizen, PR or Green Card Holder POSITION a Responsibilities include all aspects of client study design, business development,... Full details: \ -------------------------------------------------------------------- Title: Positions in investment Bank Financial Engineering (CHINESE Employer: Options Groups Location: New York, NY, United States Quant /IT Developer position available, Top IB Company: Top IB and Fund Location: NYC/Charlotte/Greenwich/Dallas/Tokyo/Hong Kong Quant Developer/Financial Engineer (Salary highly competitive) Team Overview The... Full details: -------------------------------------------------------------------- Post your job (free!): http://jobs.phds.org/jobs/post PhDs.org: Science, Math, and Engineering Career Resources --------------------------------------------------------- * Job Listings: http://jobs.phds.org/ - Job board with hundreds of listings for Ph.D.s - Reach tens of thousands of Ph.D.s each month * Graduate School Rankings: http://graduate-school.phds.org/ - Comprehensive, customizable rankings of graduate programs * Career Resources: http://www.phds.org/ - Pointers to the best resources on the web for: + getting into graduate school + writing your dissertation + jobs for Ph.D.s in academia and industry * Engineering Science Weblog: http://blog.phds.org/ - Building better scientists and engineers === Subject: MI5 Persecution: Bernard Levin 1/6/96 (2014) === Subject: Re: \FANATIC'S FARE FOR THE COMMON MAN\ Distribution: : bu765@torfree.net (Mike Corley) quoted a typical, overblown Bernard Levin : piece, which sounds exactly like any number of other pieces by Levin over : the past thirty years. : He then goes on to draw an anguished parallel with a totally unrelated : incident that had happened between him and a friend a few days earlier. It is in the same style as his other pieces, but it's so close to what actually happened a few days previously, that it's completely reasonable to assume that someone told Levin about the meeting, hammed it up a Look at the parallels; 1. \madman running loose about London\ 2. it's about a meeting which took place recently 3. the police are called (he must be controlled, however much that consumes in resources) 4. he might be violent, approach with caution, yes we're laughing but really we're afraid of him, or is it perhaps our own propaganda that we're letting manipulate us 5. he \bursts into tears, and swears it's all true - and it is\ a very exaggerated impression of the conclusion of that meeting Don't worry if you don't believe me, neither my friend to whom I explained this nor my GP believe me either (until the friend found out otherwise!). If someone just presented this to you as I am presenting it now then I wouldn't find it credible (most likely), so it's not until you actually find out it's true that you can find it credible. And by then it's too late, because they've got you in their trap and explained how \funny\ the abuse is. 2014 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: alt.fan.mike-corley 6/6/96 (3237) === Subject: A Mike Corley newsgroup? > >To be honest, having made the mistake of reading his magnum opus twice > >(like many forms of discomfort, one is prone to repeat the experience to > >but am tempted by some of the sub-threads that develop. > > I made the grave error of showing one of his masterpieces to my > mother, together with some of the (less than respectful) responses > to it. Regrettably, she is now addicted to the Corley saga, is > very much an admirer of him (viewing him as a pugilistic underdog > deserving of great admiration) and I have to waste reams of paper > printing all this bilge out every other day, for her delecation. Given that Mr Corley clearly holds a certain fascination for some people, and that his website, although \informative\, isn't exactly ideal for encouraging discussion of the sort that we all enjoy so much (including Mike himself, by the look of things), would anyone be up for a Mike Corley newsgroup - uk.fan.mike-corley, alt.fan.mike-corley, whatever? (Sorry, I'm not very well up on these things.) This would be a perfect place for Mike to post his writings, and perhaps the occasional \look at the ...mike-corley newsgroup\ posting in uk.misc and so on would be accepted in place of the current cross-posts. -- Richard Fairhurst (assistant editor, Keyboard Review) \I am the vine: and you are the branches\ 3237 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: dynamics systems with multiciphered function multiciphered function? i.e. dinamics sistem (X-?,N,f), where f(x)=(x_1,x_2,....,x_n), where n<= infinity, i.e, for any x\\in X we have (x_1,x_2,....,x_n), but in classicaly theory of dynamic sistem for any x\\in X we have f(x)=y, then we construct orbits of point x ---- {f ^n(x), n\\in N}, in this case the orbit of dynamic sistem is graph.... === Subject: Polynomial Roots (using only multiplication, divsion, addition, subtraction and radicals) for the roots of a polynomial of greater than degree 4 does not exist. Without the restriction of an algebraic solution, what is the highest degree polynomial for which a closed-form solution for the roots has been found? (I did see a reference once for the roots of a quintic using John Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil \ Naval Research Laboratory 4555 Overlook Avenue, SW Washington, DC 20375-5337 === Subject: Re: Polynomial Roots On Mon, 20 Aug 2007 07:03:16 -0400, wood@itd.nrl.navy.mil (J. B. Wood) >(using only multiplication, divsion, addition, subtraction and radicals) >for the roots of a polynomial of greater than degree 4 does not exist. >Without the restriction of an algebraic solution, what is the highest >degree polynomial for which a closed-form solution for the roots has been >found? (I did see a reference once for the roots of a quintic using Have a look at the final few paragraphs here: Also possibly of some interest: > >John Wood (Code 5550) e-mail: wood@itd.nrl.navy.mil \ >Naval Research Laboratory >4555 Overlook Avenue, SW >Washington, DC 20375-5337 === Subject: Re: Polynomial Roots The polynomial prod((x-k),k=1..2117) has roots 1,2,..., 2117. That makes clear that your question is not very \ precise. === Subject: MI5 Persecution: Old_500 5/7/96 (4460) === === Subject: Now we tell the truth as it really is Summary: Keywords: :AND a second posting which appeared in uk.legal on 5th July 1996 (isn't \ Mr :Corley popular ?) > I've just spoken to a man who knows and he has told me you > have never ever been a target. > you made complaints. Come off it \Big Ears\, if you're going to pretend someone else is giving \ me advice then at least get one of your fellow coppers to write it so that it isn't obviously written in the same style as your posts. Also both your post and the one you're pretending isn't yours were posted \ from pipex via news.dircon.co.uk, viz: YOURS; \NOT YOURS\; :NCIS leaks and most policemen regard it as unaccountable and occasionally :insecure. M Howard promises to make NCIS publicly accountable. You know a lot about what policemen think, don't you? 4460 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Solutions manual I would like to help me to sent about solutions manual of Bioprocess Engineering Principles by Doran, 1995 Best Regard === Subject: Re: Solutions manual > I would like to help me to sent about solutions manual of Bioprocess > Engineering Principles by Doran, 1995 Best Regard I can send you a copy. What are you offering in return? === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > The well known Arzela-Ascoli theorem \ http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theoremgives us a criterion \ whether some set of continuous functions is compact in C(X) for compact X. > > Is there an analogue of such a theorem, which would characterize the \ compact subsets of an L^p space ? > Perhaps the Rellich-Kondrachov theorem is also of interest: http://en.wikipedia.org/wiki/Rellich-Kondrachov_theorem === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > The well known Arzela-Ascoli theorem > http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theore > m gives us a criterion whether some set of continuous > functions is compact in C(X) for compact X. > > Is there an analogue of such a theorem, which would > characterize the compact subsets of an L^p space ? > Yes, there is. It is called the Kolmogorov-Riesz theorem. The Kolmogorov-Riesz theorem says that a subset S of L^p([a, b]), where \ 1<=p 0} sup_{x in S} int_a^b |x(s + h) - x(s)|^p ds = 0. Hope this helps Michael === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > > The well known Arzela-Ascoli theorem > > > http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theore > > > m gives us a criterion whether some set of > continuous > > functions is compact in C(X) for compact X. > > > > Is there an analogue of such a theorem, which > would > > characterize the compact subsets of an L^p space ? > > > > Yes, there is. > > It is called the Kolmogorov-Riesz theorem. > > The Kolmogorov-Riesz theorem says that a subset S of > L^p([a, b]), where 1<=p closed, bounded and > > lim_{h --> 0} sup_{x in S} int_a^b |x(s + h) - > x(s)|^p ds = 0. > > Hope this helps > Michael === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > The well known Arzela-Ascoli theorem > http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theore > m gives us a criterion whether some set of continuous > functions is compact in C(X) for compact X. > > Is there an analogue of such a theorem, which would > characterize the compact subsets of an L^p space ? > I believe there are results involving uniform integrable conditions over \ subsets of the measure space. === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > The well known Arzela-Ascoli theorem > http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theorem gives us a > criterion whether some set of continuous functions is compact in C(X) for > compact X. > > Is there an analogue of such a theorem, which would characterize the > compact subsets of an L^p space ? If p =/= 1, then the closed unit ball is compact for the weak topology induced by the conjugate space L^q, where q=p/(p+1) (Banach-Alaglao). This is the usual way compactness appears. For example a compact operator on a Hilbert space is an operator which is continuous from the weak topology to the norm topology. This is probably also true on L^p for 1 < p < oo. Dunford & Schwarz is the place to look. -- rusty === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > The well known Arzela-Ascoli theorem > http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theorem gives us a > criterion whether some set of continuous functions is compact in C(X) for \ > compact X. > > Is there an analogue of such a theorem, which would characterize the > compact subsets of an L^p space ? > I think I should disagree that the A.A. Theorem characterizes the compact subsets of C(X). Of course, strictly speaking, since uniform convergence implies convergence in L^p([a,b]), the AA Theorem still holds for certain subsets of L^p([a,b]). -Bill === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > >> The well known Arzela-Ascoli theorem >> http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theorem gives us a >> criterion whether some set of continuous functions is compact in C(X) for \ >> compact X. >> >> Is there an analogue of such a theorem, which would characterize the >> compact subsets of an L^p space ? >> > > >I think I should disagree that the A.A. Theorem characterizes the compact >subsets of C(X). ??? The theorem says that a subset of C(X) is compact if and only if it is pointwise bounded and (uniformly) equicontinuous. Why do you feel you should not agree that that's a characterization of compactness in C(X)? > Of course, strictly speaking, since uniform convergence >implies convergence in L^p([a,b]), the AA Theorem still holds for certain >subsets of L^p([a,b]). > >-Bill > ************************ David C. Ullrich === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces > > ??? The theorem says that a subset of C(X) is compact if and only if > it is pointwise bounded and (uniformly) equicontinuous. Why do you > feel you should not agree that that's a characterization of > compactness in C(X)? That only characterizes relative compactness, right? -- G. A. Edgar \ http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces On Sun, 19 Aug 2007 18:26:19 -0400, \G. A. Edgar\ >> >> ??? The theorem says that a subset of C(X) is compact if and only if >> it is pointwise bounded and (uniformly) equicontinuous. Why do you >> feel you should not agree that that's a characterization of >> compactness in C(X)? > >That only characterizes relative compactness, right? The theorem says that a subset is relatively compact if and only if what I said. Equivalently, it says that a subset of C(X) is compact if and only if it is closed, pointwise bounded and (uniformly) equicontinuous. ************************ David C. Ullrich === Subject: Re: Analogue of Arzela-Ascoli for L^p spaces >>> The well known Arzela-Ascoli theorem >>> http://en.wikipedia.org/wiki/Arzel%C3%A0-Ascoli_theorem gives us a >>> criterion whether some set of continuous functions is compact in C(X) >>> for >>> compact X. >>> >>> Is there an analogue of such a theorem, which would characterize the >>> compact subsets of an L^p space ? >>> >> >> >>I think I should disagree that the A.A. Theorem characterizes the compact >>subsets of C(X). > > ??? The theorem says that a subset of C(X) is compact if and only if > it is pointwise bounded and (uniformly) equicontinuous. Why do you > feel you should not agree that that's a characterization of My mistake, David. I erroneously thought I might be able to construct a compact subset of C(X) which was not equicontinuous. Now, I see why this is \ incorrect. === Subject: MI5 Persecution: Silly-billy 6/7/96 (5683) === Subject: MC Exposed as a Fraud { snip } Because he has made himself into a martyr and now he finds it impossible to back down. Mike's big secret has now been exposed. He made everything up. Every time Mike makes allegations against the police, MI5, Tom-Dick-and-Harry etc. those organisations have taken people off vitally important work to ascertain whether or not there was any substance in \ Mike's ranting allegations. When I'm told by someone who really does know such things that there was never any plot to get Mike then I trust that person sufficiently enough to accept his word. The problem is Mike has escalated matters to an extreme level and like many silly billys he will find it impossible to give up all his self created \ crap and to live a normal life. After all what can Mike do now this crap has been exposed as untrue ? What new cause can Mike dedicate himself to ? (Some might read that as: who else can Mike now start upsetting ?) So Mike write to John Major, c/o the Private Secretary, 10 Downing Street, London SW1A 2AA and ask the Prime Minister to help you. I'd normally disclose his fax numbers but if I did that you might jam up the lines with abusive postings just like you do here on Usenet. Is anyone interested in joining with me to do a mass e-mailing of protests to Anon Penet and to Toronto Free Net in an attempt to flood their computer systems thus forcing them to seriously consider pulling the plug on Mike ? I'm not sure but is my proposal called a \flame\ ? === Subject: Re: MC Exposed as a Fraud John Youles commented: > I've found that torfree.net don't respond, perhaps if they got a high > enough number of complaints they might take notice. See their web pages > for email addresses. Perhaps we should give Mike 7 days from today (6 July 96) to come to his senses and they start a massive multiple E-mailing campaign to Toronto Free Net. If all us victims send multiple copies of E-mails to the right addresses in Toronto then perhaps the Canadians will begin to get the appropriate message. Copy e-mailed to Mike, just so he knows what is coming if he persists. Please think of the people sleeping in shop doorways every night ------------------------------------------------------------------- === Subject: I'm wasting MI5's time! I should be arrested! Summary: Keywords: It's such a pity that the Security Service has no powers of arrest. How are they supposed to implement a proper secret police state without the ability \ to snatch Joe Citizen off the street whenever they feel like it? >Every time Mike makes allegations against the police, MI5, >Tom-Dick-and-Harry etc. those organisations have taken people off vitally >important work to ascertain whether or not there was any substance in \ Mike's >ranting allegations. I see. So they're not wasting six years of manpower to persecute you. \ They're wasting six years of manpower to prove I'm _not_ being persecuted. I wish I was clever enough to think of something like that. I really, really \ do. >When I'm told by someone who really does know such things that there was >never any plot to get Mike then I trust that person sufficiently enough to >accept his word. Come on \Big Ears\, why don't you admit that you made this posting? This \ is exactly the same line you were feeding me a few months ago. The posts are \ from the same news server, it's even the same newsreading software (Forte Agent .99e/16.227). I'm afraid that if you friend \in the know\ denied the existence of a plot \ then he was being, as they say, economical with the truth. Judging by the \ extreme reaction in London in May, and the panic you're showing now, as well as the replay post yesterday, I would say that things are hotting up again. But \ for you this time, not for me. I am out of harm's way, and there is very little \ you can do as long as I stay out of the UK. 5683 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Hyperbolic distance question Suppose D = {z in C, |Im z| < pi/2 , 0 <= Re(z) <= C }, for some positive constant C, with vertical sides identified. D' = {z in C, |Im z| < pi/2 , 0 <= Re(z) }. Clearly there is a natural projection map p : D' -> D. Then the paper claims that for any e > 0 and z in D, there is z' in \ p^{-1}(z) such that in the hyperbolic distance d_D' of D', such that d_D'(z', z' + C ) < C + e. Does anybody have an ideas how it can be proved. === Subject: distinguishing between slopes of regression lines The problem is like so: I have 9 data points (Y-axis and X-axis) am fitting \ a regression line using the Matlab function \regress\ to the point 1-4 and \ another regression line to the point 2-9 (note that the sizes of each \ regression line are different). I want to ask the following question: are the \ slopes of the two lines are significantly different? I know it should be done using the t-test somehow where I use the b values \ (the regression slopes) but I am not sure exactly how? What formula should I \ use and how do I calculate all of it's variables to get a t-score? And also \ how many degrees of freedom does it have? === Subject: Re: distinguishing between slopes of regression lines > The problem is like so: I have 9 data points (Y-axis and X-axis) am \ fitting a regression line using the Matlab function \regress\ to the point \ 1-4 and another regression line to the point 2-9 (note that the sizes of each \ regression line are different). I want to ask the following question: are the \ slopes of the two lines are significantly different? > I know it should be done using the t-test somehow where I use the b values \ (the regression slopes) but I am not sure exactly how? What formula should I \ use and how do I calculate all of it's variables to get a t-score? And also \ how many degrees of freedom does it have? It's easiest to use dummy variables and an interaction term. St d = 1 if the observation comes from the first sample, 0 otherwise. Also compute d*x for every observation. Then fit the model y = b0 + b1 x + b2 d + b3 d*x + e Fit all the data and observe that you can split the model into the two data sets by substituting for d. One slope is b1, the other is (b1+b3). The two slopes are equal if b3 = 0, which you can test with the t-test. === Subject: Re: distinguishing between slopes of regression lines > > > The problem is like so: I have 9 data points (Y-axis and X-axis) am \ fitting a regression line using the Matlab function \regress\ to the point \ 1-4 and another regression line to the point 2-9 (note that the sizes of each \ regression line are different). I want to ask the following question: are the \ slopes of the two lines are significantly different? > > I know it should be done using the t-test somehow where I use the b \ values (the regression slopes) but I am not sure exactly how? What formula \ should I use and how do I calculate all of it's variables to get a t-score? \ And also how many degrees of freedom does it have? > > It's easiest to use dummy variables and an interaction term. St d = 1 > if the observation comes from the first sample, 0 otherwise. Also > compute d*x for every observation. Then fit the model > > y = b0 + b1 x + b2 d + b3 d*x + e > > Fit all the data and observe that you can split the model into the two > data sets by substituting for d. > One slope is b1, the other is (b1+b3). The two slopes are equal if b3 > = 0, which you can test with the t-test. Bonsoir === Subject: Re: Cartesian product dilemma [...] > Is there any standard way out of this dilemma? Any classic references > would be especially appreciated. There is no \time\ in mathematics. The axioms are all simultaneously true irrespective of the order of their declarations. Thus a cyclic- dependency, as you have, is not really a problem. You can if you want forward declare the existence of a relation before defining what it is. For example 1. Declare the existence of a \relation\ . 2. Define the Cartesian product using the relation 'and'. 3. Define a relation in terms of the Cartesian product. 4. Define 'and'. Parsing that in order should avoid problems. This same problem arises in compiler design. Forward declarations solve it. Alternatively, you can avoid set-theoretic formulations of mathematics and use functional ones like the Lambda Calculus. I like lambda calculus quite a lot but they carry some theoretical problems (but practically they are superior). === Subject: Re: Cartesian product dilemma > [...] > > > Is there any standard way out of this dilemma? Any classic references > > would be especially appreciated. > > There is no \time\ in mathematics. The axioms are all simultaneously > true irrespective of the order of their declarations. Thus a cyclic- > dependency, as you have, is not really a problem. > > You can if you want forward declare the existence of a relation before > defining what it is. > > For example > > 1. Declare the existence of a \relation\ . > > 2. Define the Cartesian product using the relation 'and'. > > 3. Define a relation in terms of the Cartesian product. > > 4. Define 'and'. > > Parsing that in order should avoid problems. This same problem arises > in compiler design. Forward declarations solve it. > > Alternatively, you can avoid set-theoretic formulations of mathematics > and use functional ones like the Lambda Calculus. I like lambda > calculus quite a lot but they carry some theoretical problems (but > practically they are superior). Also, another solution is to embed the definition of a 'relation' and of 'and' together with that of the 'cartesian product' in their n'order logic forms (making sure to avoid cyclic-dependencies). The penalty with this is that you unnecessarily complicate the definition of a cartesian product and you (redundantly) duplicate part of that logic when you define a 'relation' and 'and' separately. === Subject: Re: Cartesian product dilemma > >>Hmm, Spencer Brown's formula formation rules (()) and ()() bear >>great resemblance to the formation rules for my 'Bit mapped Set >>Theory', as exposed in: >> >> >>Quote: >> >>>Formation rules: >>>0. {} is a set (the empty set) >>>1. if A is a set and B are sets then AB are sets >>>2. if A are sets then {A} is a set >>>3. There are no other ways to form (sets or) a set >> >>Coincidence or not? > > Perhaps. Re rule 1: If A is a set and B is a set, what is AB? Then AB are sets (i.e. not a _single_ set). Han de Bruijn === Subject: Re: Cartesian product dilemma > If we take consciousness as the fundamental substrate of the world ... > > \Outside time, without extension\ ... > > There is an awakening, conciousness takes cognizance of itself, > takes cognizance of taking cognizance of itself ... > > Thus is all the world generated?? Yes, in my opinion. Consciousness taking cognizance of itself generates the Naturals from {}. > Do we all re-enact the primal act of consciousness? Depends on your view. In effect, what I (and you, understanding) have done \ is regress back to the time of the first Mind and see the process of Creation. \ The real \Big Bang\ is the creation of the Naturals by Consciousness. The \ material \Big Bang\ is a later consequence of the genesis of the Naturals. > I'm thinking something along these lines. The world as a > manifestation of the Brahman, which remains, nevertheless, outside > the world, outside time, unmanifest. > > Something like that. > > Still, an interesting essay on your part connecting conciousness to > mathematics. What's interesting is having at least one other human who understands it. That's the pinnacle of communication between Brahman and human mind. It's that \little something\ which makes the difference between {} and AN \ ;o) -- I.N. Galidakis === Subject: Re: Cartesian product dilemma > > Still, an interesting essay on your part connecting conciousness to > > mathematics. > > What's interesting is having at least one other human who understands it. Well, \understanding\ may be too strong a term. You might want to have a look at \Laws of Form\ by G. Spencer Brown. For a non-mathematical approach I really like \An Introduction to Zen\ by D.T. Suzuki, with a forward by Carl Jung. Also, \The Field of Zen\ by the same author, if it can be found. Actually, anything by D.T. Suzuki, Professor of Occidental Philosophy at Otani University for many years (now deceased) is worth looking at. -- hz === Subject: quotable quote: \25-standard deviation moves\ The Financial Times quotes Goldman SachsOs chief financial officer (August 13 2007): OWe were seeing things that were 25-standard deviation moves, several days in a row,O said David Viniar, GoldmanOs chief financial officer. OThere \ have been issues in some of the other quantitative spaces. But nothing like what we saw last week.O Source: < http://www.ft.com/cms/s/d2121cb6-49cb-11dc-9ffe-0000779fd2ac.html > David Bernier === Subject: n-root of a complex number? I have this complex number: (4i)^0.5 Normaly a complex number is written as: z = a + ib. Is it correct that a = 0 \ and b = 4 in the above complex number? I can then convert to polarform where modulus: r = (0^2 + 4^2)^(0.5) = 4 and the argument is: cos = 0/4 and sin = 4/4 <=> = pi/2 which gives me the following polarform: z = 4*e^(i*pi/2) using this form I can now get the quadratic root of 4i like: 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) Is the above procedure correct? I know that the polarform is: 2e^(i*pi/4) === Subject: Re: n-root of a complex number? > I have this complex number: > > (4i)^0.5 > > > which gives me the following polar form: > > z = 4*e^(i*pi/2) > > using this form I can now get the quadratic root of 4i like: > > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) > > Is the above procedure correct? That's one of the two complex square roots. You obtain the other one by writing z as z = 4e^{i(\\pi/2 + 2\\pi)}. Can you calculate it now? Can you generalize this for n-th roots of a (non-zero) complex number? Faton Berisha === Subject: Re: n-root of a complex number? <1dKdnZP61cYSBlTbnZ2dneKdnZylnZ2d@giganews.com> > > I have this complex number: > > > (4i)^0.5 > > > > which gives me the following polar form: > > > z = 4*e^(i*pi/2) > > > using this form I can now get the quadratic root of 4i like: > > > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) > > > Is the above procedure correct? > > That's one of the two complex square roots. You obtain the other one by > writing z as > > z = 4e^{i(\\pi/2 + 2\\pi)}. > > Can you calculate it now? > > Can you generalize this for n-th roots of a (non-zero) complex number? > > Faton Berisha- Masquer le texte des messages pr\.8ec\.8edents - > > - Afficher le texte des messages pr\.8ec\.8edents - Bonsoir Berisha === Subject: Re: n-root of a complex number? > which gives me the following polarform: > > z = 4*e^(i*pi/2) > > using this form I can now get the quadratic root of 4i like: > > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) > > Is the above procedure correct? > I know that the polarform is: > > 2e^(i*pi/4) This is _one_ of the complex roots of the number, but you'll always have others. When you square e^(pi*i), you obtain e^(2*pi*i) = 1, so that you should also consider another root, 2e^(i*pi/4) * e^(pi*i) = 2e^(i*(pi/4+pi))=2e^(i*5*pi/4) If you put this back into rectangular form, you'll find that it's not equal to your original root. When you square this, you have (2e^(i*5*pi/4))^2 = 2^2 * (e^(i*pi/4))^2 * (e^(pi*i))^2 = 4 * e^(i*pi/ 2) * 1 = 4i In general, when you wish to take the nth root of re^(i*w), you have to look at (r^(1/n))*e^(i*w/n)*e^(2*pi*i*m/n) where m runs from 0 to n-1, giving you exactly n roots. This is what you should expect since the equation z^n = y should have n solutions, not just one. It's worthwhile trying this for higher n, figuring out how the roots are spaced out on the complex plane. === Subject: Re: n-root of a complex number? > I have this complex number: > > (4i)^0.5 > > Normaly a complex number is written as: z = a + ib. Is it correct that a = \ 0 > and b = 4 in the above complex number? > > I can then convert to polarform where modulus: > > r = (0^2 + 4^2)^(0.5) = 4 > > and the argument is: > > cos = 0/4 and sin = 4/4 <=> = pi/2 > > which gives me the following polarform: > > z = 4*e^(i*pi/2) > > using this form I can now get the quadratic root of 4i like: > > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) ... which is sqrt(2) + i*sqrt(2) > > Is the above procedure correct? Yes, but this omits -sqrt(2) - i*sqrt(2) Note that you can replace pi/2 with pi/2+2pi without changing the number. This corresponds to adding pi to the argument in the result (or multiplying with -1). > I know that the polarform is: > > 2e^(i*pi/4) === Subject: Re: n-root of a complex number? > > > > > > > I have this complex number: > > > (4i)^0.5 > > > Normaly a complex number is written as: z = a + ib. Is it correct that a \ = 0 > > and b = 4 in the above complex number? > > > I can then convert to polarform where modulus: > > > r = (0^2 + 4^2)^(0.5) = 4 > > > and the argument is: > > > cos = 0/4 and sin = 4/4 <=> = pi/2 > > > which gives me the following polarform: > > > z = 4*e^(i*pi/2) > > > using this form I can now get the quadratic root of 4i like: > > > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) > > ... which is > > sqrt(2) + i*sqrt(2) > > > > > Is the above procedure correct? > > Yes, but this omits > -sqrt(2) - i*sqrt(2) > Note that you can replace pi/2 with pi/2+2pi without changing the > number. > This corresponds to adding pi to the argument in the result (or > multiplying with -1). > > > > > I know that the polarform is: > > > 2e^(i*pi/4)- Masquer le texte des messages pr\.8ec\.8edents - > > - Afficher le texte des messages pr\.8ec\.8edents -- Masquer le texte des \ messages pr\.8ec\.8edents - > > - Afficher le texte des messages pr\.8ec\.8edents - Well Hagman gives the good answer. z =(4I)^0.5 = a +I*b z^2 = 4I = a^2-b^2 + 2*I*a*b Identifying a*b =4 and a = b = 2 so (4I)^0.5 = sqrt(2)*(1 + I) ; from I=exp(I*Pi/2)*I , 4*I=(2*exp(I*Pi/4) )^2 (4*I)^(1/2) = 2*exp(I*Pi/4) =... Alain === Subject: Re: n-root of a complex number? > I have this complex number: > (4i)^0.5 > Normaly a complex number is written as: z = a + ib. Is it correct that a = \ 0 > and b = 4 in the above complex number? > I can then convert to polarform where modulus: > r = (0^2 + 4^2)^(0.5) = 4 > and the argument is: > cos = 0/4 and sin = 4/4 <=> = pi/2 > which gives me the following polarform: > z = 4*e^(i*pi/2) Yes, correct. Note that this is the same as 4*e^(i*pi/2 + 2*n*pi) for any integer n. > using this form I can now get the quadratic root of 4i like: > 4^(0.5)*e^((i*pi/2)/2) = 2*e^(i*pi/4) > Is the above procedure correct? > I know that the polarform is: > 2e^(i*pi/4) Every complex number has n n-th roots. In the case n=2, there are two square roots. One is as you said, and the other is 2*e^(5*i*pi/4). -- Dave Seaman Oral Arguments in Mumia Abu-Jamal Case heard May 17 U.S. Court of Appeals, Third Circuit === Subject: Re: Prime Spectrum and Axiom of Choice <0q78c3tnr0rdc7ss203bffpqu5cireborl@4ax.com> > On Thu, 16 Aug 2007 02:05:21 -0700, Jose Capco > > > >Ok.. I just talked with my professor.. > > >Yeah, the condition I provided showed actually that P is not-isolated > >(i.e. P is not clopen) .. we showed this just a while ago =) .. I can > >post the results. > > Perhaps you missed my other reply -- I posted a counterexample. > > quasi I did.. good example, the question is.. whether all rings are actually an example, are there special rings that cant be regarded as a counterexample.. one can see actually that the negation of my condition for P implies P is isolated (ie. clopen).. otherwise suppose P is not clopen. then P is in the closure of X\\{P} if X is dense in Spec R (which is equivalent to I(X)=0) , which implies that /\\_{X\\{P}} Q = (0) This then implies that for all P_0 in Spec R \\{P} /\\_{X\\{P}} Q \\ P_0 = emptyset contradiction our assumption (which was the negation of my Hypothesis about P).. one could ask oneself, whether this is now a characterisation of isolated points.. Jose Capco === Subject: Re: Prime Spectrum and Axiom of Choice On Mon, 20 Aug 2007 06:21:02 -0700, Jose Capco >> On Thu, 16 Aug 2007 02:05:21 -0700, Jose Capco >> >> >> >Ok.. I just talked with my professor.. >> >> >Yeah, the condition I provided showed actually that P is not-isolated >> >(i.e. P is not clopen) .. we showed this just a while ago =) .. I can >> >post the results. >> >> Perhaps you missed my other reply -- I posted a counterexample. >> >> quasi > >I did.. good example, the question is.. whether all rings are actually >an example, are there special rings that cant be regarded as a >counterexample.. one can see actually that the negation of my >condition for P implies P is isolated (ie. clopen).. > >otherwise suppose P is not clopen. then P is in the closure of X\\{P} >if X is dense in Spec R (which is equivalent to I(X)=0) , which >implies that /\\_{X\\{P}} Q = (0) > >This then implies that for all P_0 in Spec R \\{P} > >/\\_{X\\{P}} Q \\ P_0 = emptyset > >contradiction our assumption (which was the negation of my Hypothesis >about P).. one could ask oneself, whether this is now a >characterisation of isolated points.. I think you should restate your new conjecture in full. It's too confusing to try to piece together the various scattered parts to see what you are now claiming. quasi === Subject: Commutativity of addition in rings without unity Another recent thread has addressed the fact that the commutativity of addition can be derived from the other axioms for a ring with unit element. Is the same true if we do not assume a unit element? -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Commutativity of addition in rings without unity On Aug 20, 9:27 am, \Stephen J. Herschkorn\ > Another recent thread has addressed the fact that the commutativity of > addition can be derived from the other axioms for a ring with unit > element. Is the same true if we do not assume a unit element? > > -- > Stephen J. Herschkorn sjhersc...@netscape.net > Math Tutor on the Internet and in Central New Jersey and Manhattan As Arturo mentioned, any nonabelian group can form the additive group if you define the multiplication to be zero. This is associative, and left and right distributive. There are also a few other \trivial\ multiplications (maybe xy=x or something?) but I forget the details. Here is a \nontrivial\ example: There is a small endomorphism near-ring example with nonzero associative multiplication that is left and right distributive over a associative non-commutative addition with additive inverse. The example is a subset of the functions from the symmetric group on three points to itself. Addition is point-wise permutation multiplication, (f+g)(x) = f(x)*g(x), and multiplication is function composition. The example is the smallest subset closed under addition and multiplication that contains the function which takes the first row to the second row in order: [ [ (), (2,3), (1,2), (1,2,3), (1,3,2), (1,3) ], [ (), (2,3), (1,3), (), (), (1,2) ] ] More explicitly, the example contains six functions, which take that first row to the following rows (one row per function): [ (), (), (), (), (), () ] [ (), (), (1,2,3), (), (), (1,3,2) ] [ (), (), (1,3,2), (), (), (1,2,3) ] [ (), (2,3), (2,3), (), (), (2,3) ] [ (), (2,3), (1,2), (), (), (1,3) ] [ (), (2,3), (1,3), (), (), (1,2) ] Michael Slone had asked me about this before, and suggested endomorphism near rings were a good place to look. I just programmed a simple computer search in GAP. === Subject: Re: Commutativity of addition in rings without unity days. My association with the Department is that of an alumnus. >Another recent thread has addressed the fact that the commutativity of >addition can be derived from the other axioms for a ring with unit >element. Is the same true if we do not assume a unit element? No. Take any nonabelian group G, and endow it with a \zero multiplication\, a*b = e for all a,b in G (e being the group identity of G). It satisfies all ring axioms except commutativity of addition and existence of multiplicative unity. -- \It's not denial. I'm just very selective about what I accept as reality.\ --- Calvin (\Calvin and Hobbes\ by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: MI5 Persecution: Silly-billy 6/7/96 (791) === Subject: MC Exposed as a Fraud { snip } Because he has made himself into a martyr and now he finds it impossible to back down. Mike's big secret has now been exposed. He made everything up. Every time Mike makes allegations against the police, MI5, Tom-Dick-and-Harry etc. those organisations have taken people off vitally important work to ascertain whether or not there was any substance in \ Mike's ranting allegations. When I'm told by someone who really does know such things that there was never any plot to get Mike then I trust that person sufficiently enough to accept his word. The problem is Mike has escalated matters to an extreme level and like many silly billys he will find it impossible to give up all his self created \ crap and to live a normal life. After all what can Mike do now this crap has been exposed as untrue ? What new cause can Mike dedicate himself to ? (Some might read that as: who else can Mike now start upsetting ?) So Mike write to John Major, c/o the Private Secretary, 10 Downing Street, London SW1A 2AA and ask the Prime Minister to help you. I'd normally disclose his fax numbers but if I did that you might jam up the lines with abusive postings just like you do here on Usenet. Is anyone interested in joining with me to do a mass e-mailing of protests to Anon Penet and to Toronto Free Net in an attempt to flood their computer systems thus forcing them to seriously consider pulling the plug on Mike ? I'm not sure but is my proposal called a \flame\ ? === Subject: Re: MC Exposed as a Fraud John Youles commented: > I've found that torfree.net don't respond, perhaps if they got a high > enough number of complaints they might take notice. See their web pages > for email addresses. Perhaps we should give Mike 7 days from today (6 July 96) to come to his senses and they start a massive multiple E-mailing campaign to Toronto Free Net. If all us victims send multiple copies of E-mails to the right addresses in Toronto then perhaps the Canadians will begin to get the appropriate message. Copy e-mailed to Mike, just so he knows what is coming if he persists. Please think of the people sleeping in shop doorways every night ------------------------------------------------------------------- === Subject: I'm wasting MI5's time! I should be arrested! Summary: Keywords: It's such a pity that the Security Service has no powers of arrest. How are they supposed to implement a proper secret police state without the ability \ to snatch Joe Citizen off the street whenever they feel like it? >Every time Mike makes allegations against the police, MI5, >Tom-Dick-and-Harry etc. those organisations have taken people off vitally >important work to ascertain whether or not there was any substance in \ Mike's >ranting allegations. I see. So they're not wasting six years of manpower to persecute you. \ They're wasting six years of manpower to prove I'm _not_ being persecuted. I wish I was clever enough to think of something like that. I really, really \ do. >When I'm told by someone who really does know such things that there was >never any plot to get Mike then I trust that person sufficiently enough to >accept his word. Come on \Big Ears\, why don't you admit that you made this posting? This \ is exactly the same line you were feeding me a few months ago. The posts are \ from the same news server, it's even the same newsreading software (Forte Agent .99e/16.227). I'm afraid that if you friend \in the know\ denied the existence of a plot \ then he was being, as they say, economical with the truth. Judging by the \ extreme reaction in London in May, and the panic you're showing now, as well as the replay post yesterday, I would say that things are hotting up again. But \ for you this time, not for me. I am out of harm's way, and there is very little \ you can do as long as I stay out of the UK. 791 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Re: When is mathematical induction taught? <1epc1lxk58bwp$.1ljea7t0zp9x2$.dlg@40tude.net> > Sat, 18 Aug 2007 12:28:16 -0400 from Brian M. Scott > : > > > On Sat, 18 Aug 2007 11:24:27 -0400, \Stephen J. Herschkorn\ > > sci.math,alt.math.undergrad,k12.math.ed: > > > In the U.S., which math course introduces mathematical > > > induction these days? Do high school students ever > > > learn it? > > > Very rarely, unless they do so on their own. > > Wow! Times have certainly changed, and not for the better. > > If memory serves, I learned mathematical induction in ninth grade, at > age 13-14, in the mid-1960s. This was a second-year algebra course. > > I'm pretty sure high schoolers still need two years of algebra to > graduate, in mst states anyway. Is it really true that even high- > school graduates no longer study what was routine for ninth graders > forty years ago? I learned it also in 2nd year algebra, not too long after you (probably 1970). I don't think 2 years of algebra are required in every state. In the DC area (including Maryland and Virginia) there may only be one year required by the state, though individual school systems may require more. I can still remember an editorial by a respected columnist, Colman McCarthy, in the Washington Post, arguing that even one year of algebra shouldn't be required. After all, he argued, he had never seen an advertisement asking for an algebra expert. And in his opinion nobody would ever use the quadratic equation after high school. I can still get mad remembering that editorial all these years later. - Randy === Subject: Re: When is mathematical induction taught? > ... In the U.S., which math course introduces mathematical induction \ these > days? Do high school students ever learn it? Do non-math majors in > college ever learn it? When would a math major first see it? (I > shudder to consider the possibility that even a math major may never > encounter induction.) I first learned induction in abstract algebra, many years ago. \ E At the college where I teach now, math majors encounter it formally in either discrete math or abstract algebra. EAbstract algebra is \ required for our majors, and discrete is required for our el.ed. majors 'concentrating' in mathematics. E --charlie === Subject: Re: When is mathematical induction taught? >In the U.S., which math course introduces mathematical induction these >days? Do high school students ever learn it? Do non-math majors in >college ever learn it? When would a math major first see it? (I >shudder to consider the possibility that even a math major may never >encounter induction.) > Back in the Olden Days (1960s), I took a calculus course from the textbook by Thomas. Mathematical induction was in there. Non-math majors take this course. So the question would be whether the instructor skips that section. -- G. A. Edgar \ http://www.math.ohio-state.edu/~edgar/ === Subject: Re: When is mathematical induction taught? >>In the U.S., which math course introduces mathematical induction these >>days? Do high school students ever learn it? Do non-math majors in >>college ever learn it? When would a math major first see it? (I >>shudder to consider the possibility that even a math major may never >>encounter induction.) >Back in the Olden Days (1960s), I took a calculus course from the >textbook by Thomas. Mathematical induction was in there. Non-math >majors take this course. So the question would be whether the >instructor skips that section. The base case of mathematics was covered in primary school. That was understood. In succeeding cases, if the previous material was understood, the new material was understood. Therefore, by induction, my students already know induction. Therefore, the section can be skipped. Gene Wirchenko Computerese Irregular Verb Conjugation: I have preferences. You have biases. He/She has prejudices. === Subject: Re: When is mathematical induction taught? > >Back in the Olden Days (1960s), I took a calculus course from the > >textbook by Thomas. Mathematical induction was in there. Non-math > >majors take this course. So the question would be whether the > >instructor skips that section. It's still there. At Amazon.com I browsed [Calculus and Analytic Geometry (9th Edition) , George B. Thomas, Ross L. Finney] and found that Appendix A1 is: Mathematical Induction. By the way, I note the book has 1288 pages... -- G. A. Edgar \ http://www.math.ohio-state.edu/~edgar/ === Subject: Re: When is mathematical induction taught? > [Sorry if this is a repeat post. The first has not > come through on my > newsreader.] > > In the U.S., which math course introduces > mathematical induction these > days? Do high school students ever learn it? Do > non-math majors in > college ever learn it? When would a math major first > see it? (I > shudder to consider the possibility that even a math > major may never > encounter induction.) > > I cannot recall in which course I first saw > induction. I do remember > learning it as a teenager on my own from an older > sibling's college > algebra textbook (the 1962 book by Lehmann which I > mention so often). > However, it seems like such textbooks today do not > touch the subject. > > -- > Stephen J. Herschkorn > sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey > and Manhattan > When is proof theory taught? I wonder if it ever was. It's much easier to teach mathematics as if it sprung from the brow of Zeus. Does one recall ever being asked what proof method--let alone induction-- was or could be used to support any theorem? I think it's demonstrably true that we are exposed less and less (I saw two daughters through public school in the 80s and 90s)to the subjects that lead us in fundamental ways to increase our knowledge by self educating. E.g., two years of Latin was an absolute high school graduation requirement in my generation. Art and music these days are often taught by part time and itinerant instructors. Historians of the future may explore how we managed to confuse information with education to such an extent that we substituted one for the other. Tom === Subject: Re: When is mathematical induction taught? >> [Sorry if this is a repeat post. The first has not >> come through on my >> newsreader.] >> In the U.S., which math course introduces >> mathematical induction these >> days? Do high school students ever learn it? Do >> non-math majors in >> college ever learn it? When would a math major first >> see it? (I >> shudder to consider the possibility that even a math >> major may never >> encounter induction.) >> I cannot recall in which course I first saw >> induction. I do remember >> learning it as a teenager on my own from an older >> sibling's college >> algebra textbook (the 1962 book by Lehmann which I >> mention so often). >> However, it seems like such textbooks today do not >> touch the subject. <> -- <> Stephen J. Herschkorn <> sjherschko@netscape.net <> Math Tutor on the Internet and in Central New Jersey <> and Manhattan >When is proof theory taught? I wonder if it ever >was. It's much easier to teach mathematics as if it >sprung from the brow of Zeus. Does one recall ever >being asked what proof method--let alone induction-- >was or could be used to support any theorem? Nonsense. One does not have to teach proof THEORY, but just what a proof is. Induction is another matter. There are more general forms of induction, but a basic form is that for the positive or non-negative integers. It is a key property of those systems, and is not present in other systems, One rarely can make use of it for other subsets of the real or complex numbers. The general notion of proofs works for arbitrary mathematical objects. Induction, as usually stated,' works for positive integers ONLY, or classes like them. >I think it's demonstrably true that we are >exposed less and less (I saw two daughters through >public school in the 80s and 90s)to the subjects that >lead us in fundamental ways to increase our knowledge >by self educating. E.g., two years of Latin was an >absolute high school graduation requirement in my >generation. Art and music these days are often taught >by part time and itinerant instructors. >Historians of the future may explore how we >managed to confuse information with education to such >an extent that we substituted one for the other. >Tom -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: When is mathematical induction taught? When is mathematical induction taught? If it's taught in year X, then it's also taught in year X + 1. === Subject: MI5 Persecution: BBC+ITN=MI5 23/7/96 (2014) === Subject: MI5 Buy into the Media : I was at speakers' corner on Sunday. There was one chap who was bellowing \ : about something or other, I don't know what, but one thing he said to : someone caught my ear: : \BBC, MI5, same thing.\ Can't disagree with that sentiment. Wasn't it documented that MI5 sometimes \bought\ journalists and \ broadcasters? I remember reading a report by some jouralist who had been offered an extra \ tax-free income by MI5 to become their covert mouthpiece, and had refused. Bet you lots of others didn't refuse. Why do you think MI5 have such easy access to the BBC and media? Because they're directly paying them off, \ that's why. Tue, 23 Jul 1996 20:01:15 uk.misc Thread \ 20 of 25 Lines 17 Re: MI5 Buy into the Media No \ responses Iain@cummings.demon.co.uk Iain Cummings \ at Fish! > >Wasn't it documented that MI5 sometimes \bought\ journalists and \ broadcasters? >I remember reading a report by some jouralist who had been offered an \ extra >tax-free income by MI5 to become their covert mouthpiece, and had refused. > > It was Jon Snow of Channel 4. -- Iain C*mmings - iain@cummings.demon.co.uk VISIT THE FEARFUL WORLD OF JIMMY McNULTY - VIOLENT NUTTER! This web-site is not suitable for mature prudes. http://www.geocities.com/SunsetStrip/7433/ > : >mouthpiece, and had refused. > : > : It was Jon Snow of Channel 4. > > Was it reported in any of the papers? It has been reported several times. The most recent was in Private Eye, a few months back. As I recall they also wanted information from him; journalists would be a natural choice for members of the Security Service and the Secret Intelligence Service for information sources. > It might be interesting to see what he had to say regarding their > attempt to recruit him. He was most concerned that many others would have accepted such an offer. However, we can probably make an educated guess as to some of those who accepted: Nigel West (Rupert Allason, MP) and Chapman Pincher would come near to the top of the list. -- \\/ David Boothroyd. Socialist and election analyst. Omne ignotum pro \ magnifico. British Elections and Politics at \ http://www.qmw.ac.uk/~laws/election/home.html I wish I was in North Dakota. Next General Election must be before 22nd May \ '97 The House of Commons now : C 324, Lab 272, L Dem 25, UU 9, PC 4, SDLP 4, SNP \ 4, UDUP 3, Ind 1, Ind UU 1, Spkrs 4. Government majority = 1. Telephone Tate \ 6125. 2014 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: spivak,calculus on manifolds 1-26a Show that every straight line through (0,0) contains an interval around (0,0) which is in R^2-A Help please, I don't understand how to show that === Subject: Re: spivak,calculus on manifolds 1-26a >Show that every straight line through (0,0) contains an interval >around (0,0) which is in R^2-A > >Help please, I don't understand how to show that > If the straight line has equation y=mx, then the interval {(x,mx) | |x| x^2 if m is positive while mx <= 0 if m is zero or negative. -- Daniel Mayost === Subject: Re: spivak,calculus on manifolds 1-26a > Show that every straight line through (0,0) contains an interval > around (0,0) which is in R^2-A > > Help please, I don't understand how to show that > What is A??? Bernard === Subject: Re: spivak,calculus on manifolds 1-26a > > > Show that every straight line through (0,0) contains an interval > > around (0,0) which is in R^2-A > > > Help please, I don't understand how to show that > > What is A??? > > Bernard OOPSi... forgot to mention A={(x,y) in R^2 : x>0 and 0 S. Moreover, A is somewhere dense in S iff there exists an open U < S such that cl(A/\\U) > \ U. I'm also trying to prove that A is somewhere dense in S (in the sense of 2) iff int(cl(A))<>0. It's clear that if A is s.d. then int(cl(A))<>0, but what about the converse? Kiuhnm === Subject: DL over GF(p^k), p small Can someone coment on theese fast DLs over GF(p^k) with p smal? Atached pari programme use p-adic logs, time is less than an minute: (Numbers may wrap) (13:19) gp > \\r fil.gp q=370801^18 ;po1=17 ; po2=314159265358979323846264338327950288419716939937510582097494459230781640\ 6286208998628034825342117109 ; tt=4702816082614016182965692121592448734541768804994600705383687298520660242\ 493249158332154303805784930 ;po1^tt- po2 mod q=0 t2=4604699313934016650709034561513625024599985206475117146466882438732720461\ 76819757800330790823114756571379207136337632502204306103062994072458017797881\ 07829618296187553498768740217764857679385834036527096610226088132909804871500\ 85772768555484868141284896671538170542973039385814425472475475378541658353264\ 10008421423060023496593441240423113905771041361954122450148086307851425110941\ 25656912846715271821518325318857294740147530820821910283761699863431400217626\ 35311850424979567088960104429193012288368633030067209790624107940455199531567\ 6104058601306583435730299863940037603350130072634964196048067365501 po1^t2-po2 mod q=0 (13:20) { \\\\ Tries to find l such that \\\\ po1^l = po2 mod p^k \\\\ Complexity p-1 \\\\ This program is distributed under the terms of the GPL dlani(p,k,po1,po2)= local(i,j,j2,j3,z1,z2,l1,l2,lo,q,kk); q=p^k; kk=k+1; po1=Mod(1,q)*lift(po1); po2=Mod(1,q)*lift(po2); z1=polrootspadic(x-lift(po1),p,kk); z1=z1[1]; z2=polrootspadic(x-lift(po2),p,kk); z2=z2[1]; l1=log(z1); l2=log(z2); lo=lift(Mod(1,p^(k-1))*lift(l2/l1)); \\\\ ;) j=po1^lo; if(j==po2,return(lo)); j3=po1^(p^(k-1)); j2=j3; for(i=1,p-1, if((j*j3)==po2,return(lo+i*p^(k-1))); j3=j3*j2; ); return(0); } p=370801; k=18; q=p^k; ro=znprimroot(q);\\\\XXX this may be slow pre=round(log(q)/log(10))-1; default(realprecision,pre); sec=floor(Pi()*10^pre)+42; po1=Mod(ro,q); po2=sec; tt=dlani(p,k,po1,po2); print(\q=\,p,\^\,k,\ ;po1=\,lift(po1),\ ; po2=\,lift(po2),\ ; tt=\,tt,\ ;po1^tt-po2 mod q=\,lift(po1^tt-po2)); p=2; q=p^k; po1=Mod(3,q);po2=19;\\\\ 7^2, 13^2 t2=dlani(p,k,po1,po2); print(\q=\,p,\^\,k,\ ; po1=\,lift(po1),\ ; po2=\,lift(po2),\ ; t2=\,t2,\ po1^t2-po2 mod q=\,lift(po1^t2-po2)); 2/0; \\\\ Exit nicely === Subject: MI5 Persecution: Latest technology 31/7/96 (3237) === Subject: Surveillance Distribution: world >> ... Normal surveillance >>\mini-cameras\ are quite noticeable and require visible supporting >>circuitry. It seems to me the best place to put a small video \ surveillance >>device would be additional to a piece of electronic equipment such as a \ TV >>or video. > >I would imagine it is quite easy to compleatly hida a camera, even in >a familar enviroment, such as a persons home. You only need a hole the >size of a pin hole, for the camera to see through. This is true. I frequently employ private detectives to spy on people (there's a good reason for this and I'll tell you if you guess correctly) and one such detective showed me the latest technology only last week. A small rucksack. One strap of the rucksack has a tiny hole in it that you wouldn't notice unless someone pointed to it. That's the camera: the wires lead into the rucksack itself where there is a video tape recorder little bigger than a Walkman. I don't know if that's what Roger Cook uses but it certainly explains why it is that the subject can look straight into the lens and not realise (s)he's being filmed. Now, I say \latest technology\ and I think that's the technology that the police use as well as the retired police who become private detectives, but no doubt the security services have far more sophisticated equipment. -- Jon 3237 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: Just too crazy 30/9/96 (4460) === Subject: Re: Is MI5 persecuting Mike Corley? Distribution: : > In summary, for there to be a serious possibility that Mike : >Corley's allegations are true, there needs to be evidence of two : >things. Firstly, there must be evidence of subversive activity on : >his part. Secondly, there needs to be evidence of MI5 taking : >measures to counter this subversive activity, in proportion to the : >magnitude of the threat posed. Upon reading Mike Corley's posts, : Alan, : What about the following scenario? MI5 conducts \experiments\ including : field trials of novel surveillance methods. Say there was a \research : project\ underway to see just how invasive a surveillance of an : individual could be. Perhaps it wasn't research, perhaps it was for : training of novice personnel. Either way, there is a sizeable risk of : detection. In order to minimise the consequences of this, you choose : someone who is out of the public eye, and if possible someone whose : credibility is already damaged - by diagnosis of schizophrenia, for \ example. : This will facilitate the \plausible denial\ if something goes wrong. : I think this would be closer to Mike's theory. : Mike, if you are reading, please don't post the story again. Anyone can : think that I actually *believe* the above. On cost alone, it is extremely : unlikely to be true, and some elements of the story (people in the media : being \in on it\) are just too crazy (though I can see how the paranoia : builds to the extent that you can see \them\ everywhere). My argument is that the fact that it's so totally crazy is what makes it so plausible and so hard to prove. I shall continue to try to find ways to kick down the house of cards. \ Because it is a brittle structure. It only needs one person to corroborate, and \ then it's all over. : P.S. Anyone else ever thought of \what-ifs\ along the same lines? I : remember as a boy daydreaming that maybe I was the focus of national : attention in a \let's follow the life of one person from life to : death\ study - that while I was out or in bed there were programmes on : the television about my life so far, etc. It makes for a good fiction, Perhaps MI5 have become victims of their own paranoia. Their agency has \ been thought of as acting disreputably, so when their decision arrives, they \ don't think twice about breaking the law. Whereas say CSIS (the Canadian \ equivalent) is at pains to point out how it would never ever do anything in \ contravention of the law, and how it is strictly controlled in what it does and how it \ does it. If this matter does ever make it into the public gaze, then it won't just be \ a few individuals in the media or security service who'll get hit. The UK is supposed to be a civilized democratic country (East Germany called \ themselves a democracy and they had the Stasi, but anyway). In a civilized country these things shouldn't be happening - and it is ultimately Parliament and the government who are answerable for not imposing sufficient restraints and accountability on those who are supposed to be ensuring security for \ citizens, not jeopardising it. ..................................................................... 4460 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: Usual targets of such abuse 10/10/96 (5683) === Subject: Re: MI5 says \Kill Yourself\ Distribution: >Indeed. If you've ever had a 'conversation' with someone suffering >from florid schizophrenia, you'll know how difficult it can be to >'argue' with them. I don't have florid symptoms. But I'm in a difficult situation, because \ those people who don't know, aren't going to believe, and those who do, they just \ go along with the crowd. It's never a good idea to go against the grain, and \ the grain here is defined by interests in the establishment and the media. Even people who could say out loud what was happening won't, because then there's \ a risk that they'll be seen as traitors and ostracised. Usually this type of 'hidden abuse' is racial and targetted at a racial minority within a country. You keep the minorities out of the good jobs, \ but you don't admit discrimination exists. It happens everywhere, not just in Britain. The persecution that is going on now is in reality a refined form \ of racism. Instead of \nigger\ it's \nutter\, and abusing the mentally ill \ is still socially acceptable today. In 50 years it might not be, but today \ there isn't any social or legal sanction against it. So really they've refined racial harassment down to a minority of one. The words may be different, but the methods are the same. 5683 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: Excellent web page 19/10/96 (6906) === Subject: Re: MI5 persecution ;; Cost of the Harassment > > :On 4 Oct 1996 13:19:16 +0100, drh92@aber.ac.uk (DANIEL ROBERT > : > :> > :> Oh NO! > :> > :> Not here too. > :> > :> This MI5 persecution thread's been doing the rounds of UK.MISC for \ ages; > :> it concerns some paranoid loony who thinks that MI5 are out to get \ him. > :> > :> They're not; everyone else on that newsgroup IS. > :> > :> Killfile the moron NOW! > : Bit sad really, the guy (must be the same one) has an excellent web page. Definite waste of talent. PS: I am NOT disclosing his URL. > The man's been changing his anon remailer address periodically, too. > > Why on earth he does I don't know; he's in serious danger of creating \ some > real enemies, apart from the phantoms his sadly deranged mind has \ created. > ............................................................................\ ............ === Subject: Re: MI5 persecution // BBC TV and Radio >I expect he will soon be seen in uk.out.to.lunch alt.folklore.urban which I saw and each of those included other groups in the Newsgroups line) I doubt he'll be seen anywhere for long. Using the xs4all remailer doesn't really help if you post the url to a web site which makes it clear who you are. Nuala, who thinks that the sad thing is he used frames much better than many sites done by supposedly 'sane' people. -- Out of the ash I rise/With my red hair/And I eat men like air - Plath You were everything to me/For twenty minutes/ Now I'd rather you would leave - Baby Chaos ............................................................................\ ............ === Subject: Praise Totally excellent page. Although I am a little confused whi is being percecuted? mike.monty@zetnet.co.uk ............................................................................\ ............ 6906 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: WTGROMT 18/11/96 (8129) === Subject: Re: MIKE AND THE DOWNFALL OF UK.LEGAL Distribution: world >Actually, I see uk.misc as a source of occasional interesting information, >and I like jokes. Some people can laugh, you see. > Not much. I thought they might, but the realization has arrived that this particular avenue of exploration is at a dead end. WTGROMT (well that's got rid of me then!) ....................................................................... === Subject: Re: MIKE AND THE DOWNFALL OF UK.LEGAL >>Actually, I see uk.misc as a source of occasional interesting information, \ >>and I like jokes. Some people can laugh, you see. >> > >Not much. I thought they might, but the realization has arrived that this >particular avenue of exploration is at a dead end. Oh please let this be true and not just another wind-up! >WTGROMT (well that's got rid of me then!) one of the colourful \characters\ of uk.misc. Have a custard cream. Dave. ....................................................................... 8129 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: No Justice 20/11/96 (9352) (sent 20/11/96) === Subject: No Justice for those with mental illness Summary: Keywords: Well, the \legal option\ has just foundered on the rock of lawyers \ refusing to deal with me on the grounds that my perception of harassment must be due to \ the disease. So we're back to square one again, the same place we were two years ago. Now perhaps one of our uk.legal participants can clarify this point. To me \ it seems illogical that lawyers should have the final say on whether you are allowed to proceed with a civil case or not. Is there a default mechanism \ or agency for cases such as mine where it is difficult to find a solicitor to represent you? What exactly is the \official Solicitor\? What is it \ possible to do if you can't find a lawyer to represent you? A chance to get a useful response out of the uk newsgroups! perhaps they can \ be .......................................................................... Represent yourself....like the defendants in the McLibel trial. you will need to read some law..but you CAN legally represent yourself I believe. Mike W. .......................................................................... Yes, a lot of people do this nowadays. Saves on legal fees too. Lots of reading up required first, though. -- Paul .......................................................................... (posted 30/11/96 from bu765) === Subject: Re: No Justice for those with mental illness Distribution: >>Represent yourself....like the defendants in the McLibel trial. >>you will need to read some law..but you CAN legally represent yourself I >>believe. >>Mike W. > >Yes, a lot of people do this nowadays. Saves on legal fees too. Lots >of reading up required first, though. What can I do to get competent legal help though? If the case is on the face \ of it so bizarre that no solicitor would represent me or talk to me, then \ surely there must be a default mechanism? What is the \official Solicitor\ that's \ been mentioned? .......................................................................... All you need do is go to a good library & read up about it - that's what the McLibel 2 did. -- Ban Everything or Ban Nothing ! http://www.mahayana.demon.co.uk/ ISO 1386-C compliant .sig All words written in the above posting are my opinions .......................................................................... 9352 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: ti-84 calc question The package claims that my ti-84 plus is 2.5x faster than the ti-83 plus but when i had them do the exact same calculation (one that takes a little time to do) the ti-84 plus goes at the same speet of the 83! could somebody please explain why its not going at its maxinum speed? === Subject: Re: ti-84 calc question On Mon, 20 Aug 2007 09:34:32 -0700, win2tanker >The package claims that my ti-84 plus is 2.5x faster than the ti-83 >plus but when i had them do the exact same calculation (one that takes >a little time to do) the ti-84 plus goes at the same speet of the 83! >could somebody please explain why its not going at its maxinum speed? My guess is the advertisement is a lie, or at least mostly a lie. Probably certain tasks go faster. The marketing types then escalate that restricted truth to a lie, claiming greater overall speed. quasi === Subject: Re: ti-84 calc question > On Mon, 20 Aug 2007 09:34:32 -0700, win2tanker > > >The package claims that my ti-84 plus is 2.5x faster than the ti-83 > >plus but when i had them do the exact same calculation (one that takes > >a little time to do) the ti-84 plus goes at the same speet of the 83! > >could somebody please explain why its not going at its maxinum speed? > > My guess is the advertisement is a lie, or at least mostly a lie. > > Probably certain tasks go faster. The marketing types then escalate > that restricted truth to a lie, claiming greater overall speed. > > quasi tasks go faster\ idk if shells like ION i have on mi calc is causing it to slow down. === Subject: Re: ti-84 calc question I've emailed Ti on the issue this morning but so far no reply. If they don't reply its probably because they don't want to admit that they had some false advertising. === Subject: Re: ti-84 calc question > The package claims that my ti-84 plus is 2.5x faster than the ti-83 > plus but when i had them do the exact same calculation (one that takes > a little time to do) the ti-84 plus goes at the same speed of the 83! > could somebody please explain why its not going at its maxinum speed? Is anybody else having the same isssue? === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I would like to get the solutions to: Modern Control Systems, 9th Ed., by Richard C. Dorf, Robert H Bishop. === Subject: continuity (mistake?) Here is the exercise: \Assume that f:M->N is a function from one metric space to another which satisfies the following condition: if a sequence (p_n) in M converges then the sequence (f(p_n)) in N converges. Prove that f is continuous.\ Well, let's take the following function: f:M->M, where M = {1, 1/2, 1/3, ..., 1/n, ...} U {0}; f(x) = x, x in (0,1); f(0) = 1; f(1) = 0. It satifies the condition enunciated in the problem and yet it isn't continuous. Am I wrong? Kiuhnm === Subject: Re: continuity (mistake?) >Here is the exercise: >\Assume that f:M->N is a function from one metric space to another which >satisfies the following condition: if a sequence (p_n) in M converges >then the sequence (f(p_n)) in N converges. Prove that f is continuous.\ > >Well, let's take the following function: > f:M->M, where M = {1, 1/2, 1/3, ..., 1/n, ...} U {0}; > f(x) = x, x in (0,1); > f(0) = 1; f(1) = 0. >It satifies the condition enunciated in the problem and yet it isn't >continuous. No, it does not satisfy the condition. Consider the sequence (p_n) given by / 1/n if n is odd p_n = { \\ 0 if n is even. This sequence converges to 0 in M. However, the sequence f(p_n) is given by: / 1/n if n is odd f(p_n) = { \\ 1 if n is even and this sequence does ->not<- converge in M. -- \It's not denial. I'm just very selective about what I accept as reality.\ --- Calvin (\Calvin and Hobbes\ by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: Re: continuity (mistake?) > No, it does not satisfy the condition. > > Consider the sequence (p_n) given by > / 1/n if n is odd > p_n = { > \\ 0 if n is even. No... you spoiled the exercise :-( I'm obviously joking, but the fact is that I can use your \trick\ to solve the exercise: (x_n)->x => (x_1,x,x_2,x,x_3,...)->x => (f(x_1),f(x),f(x_2),f(x),...)->y => the \odd\ and the \even\ subsequences -> y => f(x_n)->f(x) and f is continuous. Kiuhnm === Subject: Re: continuity (mistake?) days. My association with the Department is that of an alumnus. >> No, it does not satisfy the condition. >> >> Consider the sequence (p_n) given by >> / 1/n if n is odd >> p_n = { >> \\ 0 if n is even. > >No... you spoiled the exercise :-( > >I'm obviously joking, but the fact is that I can use your \trick\ to >solve the exercise: Indeed; figuring out exactly why your example does not work (i.e., does not satisfy the hypothesis) is all that one needs to solve the problem. Moral: looking at examples is good, and looking at non-examples is also good. They each have something to offer. >(x_n)->x => >(x_1,x,x_2,x,x_3,...)->x => >(f(x_1),f(x),f(x_2),f(x),...)->y => >the \odd\ and the \even\ subsequences -> y => >f(x_n)->f(x) and f is continuous. Right. -- \It's not denial. I'm just very selective about what I accept as reality.\ --- Calvin (\Calvin and Hobbes\ by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: MI5 Persecution: David Hepworth (1) 26/2/97 (10575) === Subject: \Absolute Obscene\ David Hepworth (GLR) Summary: Keywords: Last night (21/Feb/97), I was listening to BBC GLR. You have to understand \ that I was listening by stealth. Back in 1990 I used to have Capital blaring out \ of the speakers all over the garden (\if he listens to Capital then he can't \ be with the volume turned right down. It could not possibly be overheard by \ any listening device, no matter how sensitive, because sound does not carry \ from the headphones. Yet somehow they are still able to tell which station I am listening to on \ the walkman. And last night, I bravely tuned to GLR 94.9FM at around 8.45pm. Everything went well for the first half hour or so. The DJ (today GLR told \ me it was David Hepworth) had a \rock star spelling competition\ (how do you \ spell Shakespear's Sister?), no trouble there. I was recording the show onto tape just in case anything unpleasant happened, as I had reason to think it \ might. Around 9.10pm, it did. Here, precisely, is what Hepworth said after the \ song \Come Around\ by the Muttonbirds (I have this on tape, and at some point \ will be posting the audio on my website); \New album's called 'Envy of Angels', that comes out soon, that's the \ single that's already out, I would imagine, that's 'Come Around', the Muttonbirds, \ er, coming up after this, we got the rock-and-roll A-level, we have, I assume, Brian do we have an embarrassment of prizes in there, we do, don't we, (EMPHASIS) absolute obscene (END-EMPHASIS) amounts of prizes, there will no doubt be a riot at the back door\, and that's, er, A-level coming up after \ this\ The key phrases in what he said are, \EMBARRASSMENT of prizes\, and what \ he himself emphasized verbally, \ABSOLUTE OBSCENE amount of prizes\. It is \ my belief (based on content and tone of voice) that when he spoke these phrases \ he knew I was listening, and that the phrases refer directly to my situation. \ The \EMBARRASSMENT\ is the embarrassment he and other media people would feel \ at having their wrongdoing exposed; the \ABSOLUTE OBSCENE\ (which he \ verbally emphasized) described the disgusting sexual abuse which the harassers have \ been throwing at me. Needless to say, I can't prove this is what he meant, although this has happened enough times that it's hard not to recognise it when you see it. \ But I shall be sending him a copy of this explanation. We'll see what he has to \ say for himself. .................................................................... === Subject: Re: \Absolute Obscene\ David Hepworth (GLR) Distribution: > >> >> >Last night (21/Feb/97), I was listening to BBC GLR. You have to \ understand that >> >I was listening by stealth. Back in 1990 I used to have Capital blaring \ out of >> huh? > >I can't understand it myself. Can anyone else help us ? I will try to explain with greater clarity. I am prone to have ideas of \interactive watching\ by people on TV, and \interactive listening\ by people on the radio; which means as I listen \ to their programme, they are aware that I am listening (because my home is \ spied on, and those doing the spying phone up the radio station and tell them that \ I am listening to them), and they then interject nasty remarks in what they \ say. This is what I believe may have happened on Friday night with GLR, and what \ I It's happened before with other radio stations, most worryingly when I am listening quietly on my walkman as happened on Friday. This time, the interjected words were \embarrassment\ and \absolute obscene\. The \ \absolute obscene\ referred to the abuse which is currently being directed at me. I hope that's a little clearer. .................................................................... Facility thusly: ~ huh? Just because you need to read it twice in order to understand it, is that any reason to inflict it on the rest of us again ? -- '2% of people *do*, 98% of people wish they *had*' - David Fanshawe http://www.mahayana.demon.co.uk/ Read Flatland ! .................................................................... Mike, try playing a cassette in your walkman. There is no way they can reach you then. HTH -- Don Whybrow - Correct email address: don@whybrow.demon.co.uk .................................................................... > ............................ This time, the > interjected words were \embarrassment\ and \absolute obscene\. The \ \absolute > obscene\ referred to the abuse which is currently being directed at me. But it is possible that those two sets of interjected words could have validly referred to something else. > I hope that's a little clearer. .................................................................... >Some people, new to this group, may not realise the Mike is our resident >Paranoid Schizophrenic. This is not a joke, it is serious. He has >already admitted stopping taking prescribed drugs for his condition. Though I have to say I've never heard one as obscure as that, even from Mike. How he can believe a rubbish trail for a crap competition is directed at him is beyond me! He's sent a copy of the post to the jock, too. I hope he doesn't read it out. -- The John Shuttleworth Homepage: It's just an Austin Ambassador of a site! http://www.steviep.demon.co.uk/shuttle.htm .................................................................... 10575 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: Striking out action 10/3/97 (11798) === Subject: \Scandalous, Frivolous or Vexatious\ Summary: Keywords: A couple of weeks ago I issued a summons against the BBC in my local county court, for the tort of private nuisance caused by the spying by their newsreaders on my home. My argument was that their spying had prevented me watching the news at home, and therefore interfered with my normal use of my \ home. The BBC's Litigation Dept at White City have replied not with a defence, \ but with an application for my claim to be struck out because; (a) it discloses no reasonable cause of action; and/or (b) it is scandalous, frivolous or vexatious. Their application will be heard next week. They have not made any affidavit \ in support of their application, nor have they given particulars as to why \ they consider my summons to be unarguable in law, which would be a necessary condition for there to be no reasonable cause of action. I am more worried about point (b). Allegations are scandalous (says Stuart Sime's book) if they impute dishonesty against another party; which my allegations do, against the BBC's newsreaders. As for frivolous or \ vexatious, I think that will be up to me to make a good argument for the effect the \ BBC's spying has had on my life, and up to the district judge's opinion of my \ case. Apparently seeking to have a claim struck out in this way is common \ practice when the plaintiff is a litigant-in-person. Even if it is struck out, there \ is always the opportunity to appeal. I think we could be in for a fight next \ week. .......................................................................... Sun, 02 Mar 1997 20:38:59 uk.legal Thread \ 52 of 54 Lines 13 Re: \Scandalous, Frivolous or Vexatious\ Respno \ 1 of 1 Kate@carterce.demon.co.uk \ KKKKatie > Well, that'll liven up the dear old District Judge. Almost worth taking > the day off to see how Mike fares. almost worth taking a day off to see if he exists Kate -- Just back from the US - you've got to love a country that puts \Vertical Clearance Impeded\ for \Low Bridge\ .......................................................................... === Subject: Re: \Scandalous, Frivolous or Vexatious\ Distribution: >Almost worthwhile taking the day off to go and punch him in the >face for all abuse he's given us on these groups. Which court is it? I'm not telling you which court it is. I don't want to be hounded by irate uk-miscreants! As for raising eyebrows, the member of staff who took the summons form \ didn't change their facial expression at all. Seen it all before, no doubt. .......................................................................... > >-I'm not telling you which court it is. I don't want to be hounded by irate \ >-uk-miscreants! >- >-As for raising eyebrows, the member of staff who took the summons form \ didn't >-change their facial expression at all. Seen it all before, no doubt. > >Take no notice Mike, some people are like that. I hope >you finally get this matter into court. I'm not so fond of the threats of violence against the persistent (yet Corley. Yet, encouraging the fantasies of the mentally ill isn't exactly healthy either. Do you go up to homeless mad people as say things like: \They're coming to get you\, or \Look out behind you?\. Smid .......................................................................... Quite the contary Mike. Your recent posts have been no problem. By friends than you know. I hope that you sort out your problem, sincerely. -- *********************************************************************** I'm Alan Packer and I move in a very mysterious way. If replying to me .......................................................................... > >I think it's good for Mike to vent his anger and frustration on >these two newsgroups. Consider it to be part of your duty to >the comunity in general. Yep. Just not, the, I think 180 posts, one week, when his illness got really bad. Oh yeah, and there was only three real posts, just repeated 60 times. >Mike does a first class job of drawing out the real personality >of the person hiding behind a node name. The way people react >to Mike gives it all away. Erm, explain this rather dubious statement. >As for you. Don't you think it's a trifle condescending to refer >to people as mad. I know they are homeless and without internet >access, but maybe they are just eccentric. Congratulations. This is my first real flame for about a year. I try to control my anger when I come across another uninformed naive idiot on usenet, but sometimes it goes free. We've had Corley have killed a couple of usenet groups I really rather lied. uk.media, to name but one. I gets my goat to read another useless fucker thinking he is a harmless eccentric. Mike is mentally ill. He is unwilling to deal with it. He seems to think that uk.misc is some sort of forum that MI5 reads. And it should be avenged. He's mailbombed a large quantity of people, 1. He thinks MI5 watches him through his television 2. He thinks all references to mad people, refer to him 3. He thinks all people shouting, are shouting at him. 4. He's been diagnosed as mentally ill, just not a paranoid schizophrenic. 5. He gives not a shit about any newsgroups he abuses. 6. He goes through quiet periods, then _very_ nasty periods. 7. All evidence of this great conspiracy is laughable, to say the least. 8. He gives internet/usenet a bad name to the media. Including mailbombing Chris Tarrant and faxing various celebrities. I do not condescend to him. I actually know what he does, and has done in the past, and am frankly not too respectful of him. He can be openly referred to as \mad\ because he is. Let's check my mailbox saves: Repost of when I thought I'd seen the last of loopy mike: >>Actually, I see uk.misc as a source of occasional interesting information, \ >>and I like jokes. Some people can laugh, you see. >> > >Not much. I thought they might, but the realization has arrived that this >particular avenue of exploration is at a dead end. > >WTGROMT (well that's got rid of me then!) Have you actually visited his web page, and read the massive conspiracy that is supposed to be against him? Apparently MI5 watch him, but for no reason. They do it for a laugh. Because they can. It costs them a lot of money, but they still do it. Reality is but a memory for this man. >Anyway I understood the UKMTC members upset Mike Corely , but >I could be wrong. Heaven forbid. Smid .......................................................................... > >> Anyway I understood the UKMTC members upset Mike Corely , but >> I could be wrong. > >There was a pitched battle between MC and \the artist formerly >known as Big Ears\ but that was before the UMTC by which time such outrageous proportions that we had to mount a concerted attempt to get his account(s) closed. The reason for the lack of he can bring an action as a LIP against the BBC (all solicitors having quite rightly and honourably declined to act for him). When this course of action is exhausted without result, I confidently predict that we will once again be the innocent victims of his ire .......................................................................... 11798 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Quasiconcave Functions let U:[0,+inf)^n->R a continuous function such that, if x>=y (that is x_i >= y_i for every i) and x=/=y, then u(x)>u(y). Let us suppose that for every p=(p_1,...,p_n), with p_i > 0 for every i, and for every w>0, the problem max u(x) s. t. p_1*x_1+...+p_n*x_n <= w has only one solution, x*. Then I think that u must be stricly quasiconcave, that is, for every x,y, with u(x)>=u(y), u(tx+(1-t)y)>u(y) for every t in (0,1). This problem originates from economic theory, so I hope someone knows something about the subject. However, I thank you very very much for your attention. Maury === Subject: Re: Guitar Reviews And how do we get these free guitars? And most of all, how is it related to .NET? Do you record your guitar using \ one of your .NET App? ;) > Custom guitar, acoustic guitar, electric guitars, reviews, > specifications, pictures, prices, all here!!!!! > > http://pro-guitars.blogspot.com/ > > Free guitars!!!! > > http://freeguitars.blogspot.com/ > === Subject: JSH: Graded, and Simply Not Smart Enough JSH has languished for YEARS and keeps getting F's on all his math postings \ in sci.math. His cumulative GPA (out of 4.0) is 0.0 for Math, (missing knowledge of complex numbers, doesn't know what a \ proof is,...) 1.0 for Social Skills as mildly amusing, 1.5 for Crackpot/troll 3.5 for lying 3.9 for posting crap JSH was awarded by Google, a single star (1 star) at one time for posting. Therefore, We the Commitiee to Block JSH Works and Save the World, have opened a position in Panjmonya, North Korea for JSH to teach a small class (about 2) of 1st graders Binary Mathematics, on and off, using a small pen flashlight. JSH will be paid by the North Koreans for his efforts, about 1/5 won. === Subject: Re: JSH: Graded, and Simply Not Smart Enough > JSH has languished for YEARS and keeps getting F's on all his math \ postings > in sci.math. > > His cumulative GPA (out of 4.0) is > 0.0 for Math, (missing knowledge of complex numbers, doesn't know what \ a > proof is,...) > 1.0 for Social Skills as mildly amusing, > 1.5 for Crackpot/troll > 3.5 for lying > 3.9 for posting crap > > JSH was awarded by Google, a single star (1 star) at one time for \ posting. > > Therefore, We the Commitiee to Block JSH Works and Save the World, have > opened a position in Panjmonya, North Korea for JSH to teach a small \ class > (about 2) of 1st graders Binary Mathematics, on and off, using a small \ pen > flashlight. > > JSH will be paid by the North Koreans for his efforts, about 1/5 won. That's because JSH doesn't like to learn things. He just likes to wuss around. === Subject: MI5 Persecution: I am being ignored 17/4/97 (13021) === Subject: I am being ignored Summary: Keywords: Something mildly unusual is happening. Nobody has been getting at me for \ the last couple of weeks. Best of all, about three weeks ago I bought a video recorder to try to get some evidence from the News etc, and despite having watched Martyn Lewis, Michael Buerk and the whole lot of them, not one of \ them has said anything to me in the last three weeks. And I have been listening \ to and recording Capital, with no ill effect. All in all, I am being comprehensively ignored. Of course, I should have video-taped the news programmes back in 1990/91/92 (and even 93) when it was still all going on. Anthony Johnsson (casually name-drop to stick the knife in) will tell you how intelligent a person I \ am, yet I didn't have the good sense to record the programmes at the time, and \ it's a bit late now. Today another avenue of exploration closed when my second summons against \ the BBC was struck out (basically through lack of evidence), and an order was \ made by the district judge saying that I could not sue John Birt again without \ the express permission of the court. She explained that this was for my own \ good, to save me the summons fees. Costs were not awarded though (the BBC didn't \ even really seek costs), and I am free to sue anyone else I take a shine to. .................................................................. It is not that you are being ignored.... it is just that as your mental health improves, so your paranoia eases. OR... The dark forces are gathering strength for the final battle with you... Seriously though, it's nice to see you back - cos as long as they are persecuting you, I can relax knowing they haven't got time to bother me! Harry -- Harry Adams harry@smcat.com http://www.smcat.com .................................................................. Great news! Keep taking the tablets. .................................................................. (posted 11/apr/1997, re Neil Long at 8.35pm Capital Radio) === Subject: Re: I am being ignored Distribution: [unreasonable optimism snipped] Well, that didn't last very long, did it? Capital Radio this evening approx 8.35pm, Neil Long's programme a few \ moments after I switched to it (I am listening to it as I type this). The following spouted from his mouth; \want to see Jacko in town? what a gig it's going to be, that guy is \ amazing, and PSYCHOTIC, but amazing on stage, 0171-4200958, on stage he's a MADMAN \ but he's wonderful\ God, I am so tired of this. It's been going on for seven years. When's it \ going to stop? I have the above extract on tape. When I get around to it I will post it, \ and the Hepworth snippet, as audio files on the website. Of course, all the doubting Thomases and Smids can carry on doubting, because the mere fact \ that moments after switching on the DJ starts talking about \psychotic\, that's \ not proof is it? copied to Richard Park, Group Director of Programmes. I suppose they'll \ just lie as usual. Bastards, basically. .................................................................. The solution is surely to not have a TV, video, radio or telephone in the house, Mike. That way they can't be used to watch you. Dave -- dave@ llondel.demon.co.uk Any advice above is worth what I paid for it. .................................................................. === Subject: Re: I am being ignored Distribution: >The solution is surely to not have a TV, video, radio or telephone in >the house, Mike. That way they can't be used to watch you. No. I tried that strategy in 1990/91 when I sold my television and stopped listening to the radio. It didn't work because they got at me through co-workers and eventually the general public. The solution is to gather evidence by recording everything, and then \ complain or sue if possible. I wish I'd recorded it all five or six years ago, then \ I might be more successful in a legal action now. .................................................................. Did anyone else watch the X-Files on Sky last night (Sunday 13th April) and recognise Mike's symptons in the chap with the tattoo? It was quite interesting to see someone responding in precisely the same way, picking up on individual words and phrases in ordinary sentences and applying parts to themselves as though they were intended. Not a pleasant ending though. You take care, Mike, and keep taking the medication. uk.misc just wouldn't be the same without you now. Claire -- ****************************************************************************\ ** * Claire Speed [ENTX] * Network & Operations Unit, Manchester Computing \ * * Dial-up, ISDN, TICTAC * C.Speed@mcc.ac.uk http://www.mcc.ac.uk/Claire/ \ * ****************************************************************************\ ** .................................................................. >moments after switching on the DJ starts talking about \psychotic\, \ that's not >proof is it? No, it isn't, but you knew that anyway. Mike, how is it that you think that this can't be a coincidence because it happened 'moments' after you switched on the radio? Haven't you just told us that you haven't heard a thing for two weeks? Going two weeks watching TV & listening to the radio without a reference to 'mad', 'crazy' or 'loony' is quite an acheivement, especially given the trashy radio you seem to prefer. I've just spent a lot of the weekend trying to keep a manic depressive under control, so that she doesn't have to go back into hospital and lose the last few months' memories to ECT. One thing I've noticed is that manic depressives almost seek to foster their fantasies, they enjoy the control and the high. You seem to be the same - you refuse to answer the points put to you that explain away what you call persecution totally - do you have the courage of your convictions to reply to this? Or do you enjoy the attention your illness brings you? -- Illtud Daniel idaniel@jesus.ox.ac.uk -see Twin Town- -Buy Apollo 440- .................................................................. > Something mildly unusual is happening. Nobody has been getting at me for \ the > last couple of weeks. Do please tell the *truth*. It has been longer than *2* weeks as I'm sure you have noticed, hasn't it ? Well I claim responsibility for the absence of persecution. When I rang John Major's personal private secretary, not the personal political one or any of the others in the same office, and I said you would vote for the Conservative Party at the General Election the secretary said Johnny would be so happy and as a sign of gratitude he rang-up the Head of MI5 and told them to save money for tax cuts by abandoning forthwith their campaign against you. Consequently MI5 and MI6 have been able to redeploy urgently needed resources (i.e. manpower and telecoms interceptions) against an international band of (*****censored*****). On behalf of the law-abiding and decent citizens in the British islands, I would like to sincerely thank you for your invaluable contribution to crime fighting by your instrumental achievement in releasing badly needed and very scarce MI5 and MI6 resources. > Best of all, about three weeks ago I bought a video recorder to try to get \ some > evidence from the News etc, and despite having watched Martyn Lewis, \ Michael > Buerk and the whole lot of them, not one of them has said anything to me \ in the > last three weeks. And I have been listening to and recording Capital, with \ no ill > effect. All in all, I am being comprehensively ignored. Please don't forget it is National Election time and the news broadcasts at 21:00 on BBC1 are preoccupied with British politics. > Of course, I should have video-taped the news programmes back in \ 1990/91/92 > (and even 93) when it was still all going on. Any particular date in mind ? I have a few in the library. You can also purchase copies from the BBC on 0181-743 8000 (central switchboard). > Anthony Johnsson (casually name-drop to stick the knife in) will tell you \ how intelligent > a person I am, yet I didn't have the good sense to record the programmes \ at the time, > and it's a bit late now. Perhaps it is for the best. Now that it is all over, you can return to Canada. Did you manage to get your Canadian citizenship ? What did your contracts at MI5 say about you becoming Canadian ? You never did mention the name of your MI5 liaison officer nor of any of the others. > Today another avenue of exploration closed when my second summons against \ the > BBC was struck out (basically through lack of evidence), and an order was \ made > by the district judge saying that I could not sue John Birt again without \ the > express permission of the court. She explained that this was for my own \ good, > to save me the summons fees. Costs were not awarded though (the BBC didn't \ even > really seek costs), and I am free to sue anyone else I take a shine to. That's true. But as it is now all over there is no need to worry about the past. MI5 said your file has been removed from the Registry and put in a burn-bag so there is nothing left for anyone ever to find. By the way, what was the URL of your famous Corley web site ? .................................................................. === Subject: Re: I am being ignored Distribution: : >Illtud Daniel typed: : > : >> You seem to be the same - you refuse : >> to answer the points put to you that explain away what you call \ persecution : >> totally - do you have the courage of your convictions to reply to \ this? : >> Or do you enjoy the attention your illness brings you? I have to admit that it is actually mildly amusing to have newsreaders \ watching you, although it is tiresome after a bit. I am currently in an interesting situation, because after my lawsuits against them I think the BBC \ newscasters have stopped watching me entirely; I also think that ITN, in particular \ John Suchet, *are* still watching but they're being subtle about it (his blink pattern followed mine). Being called nasty words is quite horrible though, and seems gratuitous; \ what do they gain by it, except to exercise their sadism? : >I think we all ought to remember that schizophrenia is a : >psychosis, and as such, unless you are American and believe : >that *anything* can be dealt with through counselling and/or : >psychotherapy, will not respond to this type of treatment. : Tell me about it. My friend is now unhappily resident in Warneford : mental health unit, or whatever they call psychiatric hospitals \ nowardays. I've never been in a hospital as an in-patient, ever. I think I've been \ very lucky in that regard. : Having said that, one should seek to disabuse even psychotics of their : delusions. The medicines haven't made any difference to my ideas. They obviously make \ a difference in the emotional impact and reaction. .................................................................. >I've never been in a hospital as an in-patient, ever. I think I've been \ very >lucky in that regard. You have. They're not pleasant atmospheres, especially the secure wings, which have more in common with prisons ('airlock' doors, list of 'banned' -- Illtud Daniel idaniel@jesus.ox.ac.uk -see Twin Town- -Buy Apollo 440- .................................................................. 13021 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: Continuous Non-Convex Functions let f:[a,b]->R a continuous function which is not convex. Then I suppose that the following statement holds: there are two distinct point c, d in [a,b] such that for every x in [a,b] we have f(x) >= {[f(c)-f(d)]/(c-d)}*(x-d) + f(d). It's a quite evident fact, but I couldn't prove it ... Maury === Subject: Re: Need a solution manual ppl all ur solution manual problems can be solved by \ http://solutionmanual.net just register(free) and have free access to all solution manuals u can dream of!!!!! Ahmed Sheheryar Operations Head SolutionManual.Net Team === Subject: =?iso-8859-1?q?Re:_An_Introduction_to_G=F6del's_Theorems?= > > > When do you think the paperback edition will be on the shelves of U.S. > > bookstores? > > When do you think it will show up in used bookstores? I coudn't figure out why you made that sardonic comment, but then I thought that perhaps you took my question as implying that I was waiting for some future publication of a paperback edition. But I know that the paperback edition is being published along with the hard back; the point of my question to Peter was just when it's going to be on the shelves in stores. MoeBlee === Subject: =?iso-8859-1?q?Re:_An_Introduction_to_G=F6del's_Theorems?= > It is published immediately in paperback, and Amazon US is reporting > it as in stock, so I assume/hope it is generally available in the USA! > > http://www.amazon.com/Introduction-Theorems-Cambridge-Introductions-P... > > Hurry, hurry while stocks last :-) No luck at the Borders Books in my neighborhood this weekend. I'll try again next weekend. MoeBlee === Subject: Re: An Introduction to =?iso-8859-1?Q?G=F6del's?= Theorems Originator: jgamble@ripco.com (John M. Gamble) >On Sat, 18 Aug 2007 11:57:42 -0700, Peter_Smith said: >>> , which is the main reason I've stuff with >>> publishing my last three books with them. >> >> Or even stuck .... > >I liked \stuff\. > It's not stuffy. -- -john February 28 1997: Last day libraries could order catalogue cards from the Library of Congress. === Subject: FREE SOLUTIONS- END TO ALL TEXT BOOK WORRIES ppl all ur solution manual problems can be solved by \ http://solutionmanual.net just register(free) and have free access to all solution manuals u can dream of!!!!! Ahmed Sheheryar Operations Head SolutionManual.Net Team === Subject: Optimization for Windows XP Optimize your operating system http://windowsxpsp2pro.blogspot.com/ === Subject: MI5 Persecution: Peak Practice 26/4/97 (14244) === Subject: \Peak Practice\ 1/4/97 wanted Summary: Keywords: I am looking for a recording of \Peak Practice\ from yesterday 1.April on \ VHS tape. Willing to pay a reasonable sum for a recording. ................................................................. >I am looking for a recording of \Peak Practice\ from yesterday 1.April on \ VHS >tape. Willing to pay a reasonable sum for a recording. Uh-oh. Peak Practice slagging you off as well now, eh Mike? -- Illtud Daniel idaniel@jesus.ox.ac.uk -see Twin Town- -Buy Apollo 440- ................................................................. === Subject: Re: \Peak Practice\ 1/4/97 wanted Distribution: >Uh-oh. Peak Practice slagging you off as well now, eh Mike? You better believe it. Tuesday's episode had one of its characters being labelled as a \lunatic\ with \something wrong with them\. I changed \ channels rapidly at this point, but in retrospect perhaps I should have carried on watching and taped the program. A few days ago I bought a video recorder (about time too) for evidence gathering purposes. But strangely the TV news (both BBC and ITN) haven't uttered a sound over the last few days. Almost as if they know they're \ being taped... ................................................................. >>Uh-oh. Peak Practice slagging you off as well now, eh Mike? > >You better believe it. Tuesday's episode had one of its characters being >labelled as a \lunatic\ with \something wrong with them\. And? There are many characters in TV programmes who _are_ lunatics with 'something wrong with them'. Why do you think this alludes to you? In the eponymous book, Cervantes' Don Quixote is described as 'crazy' 'mad' and 'lunatic'. Do you think this is a reference to yourself, albeit written by a time-travelling 17th century Spaniard? If you accept that the character of Don Quixote can be described thus without it being a reference to you, what problem do you have with Peak Practice? Or do you enjoy tilting at your own little windmills? A friend's lodger has turned out to be paranoid schizophrenic. Her obsessions are strikingly similar to yours, she is convinced that she is being bugged, and that students at her college are making fun of her for being foreign. She has been looking for the services of a private detective. Maybe Oxford has this unfortunate effect on foreigners? I must say that I'm surviving rather well, or that's what the man under the stairs told me yesterday. -- Illtud Daniel idaniel@jesus.ox.ac.uk -see Twin Town- -Buy Apollo 440- ................................................................. So Mike, you are a woman, married to a doctor,have two children, speak spanish, and living in the Peak District. Is that a fair description of you? BAZZA ................................................................. 14244 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access === Subject: MI5 Persecution: Continuing Silence 9/5/97 (15467) === Subject: The Continuing Silence Summary: Keywords: The silence continues. I am enjoying listening to Radio 4 without any ill effects whatsoever. Admittedly I am recording every moment, but still. Even better, I have fished out a recording from ITN last year which at the time \ I was convinced had been about me, and on re-seeing it I am now convinced it \ is not about me at all. But that is not to disclaim all the many \live incidents\ that have \ happened. Everything written on my website is true. For the newcomers, its URL (as \ seen in the Observer and Private Eye) is ....................................................................... > > The silence continues. I am enjoying listening to Radio 4 without any ill > effects whatsoever. Already the New Labour govt improves the quality of life of the populace! -- Why not drop by my web page sometime? http://www.man.ac.uk/~zlsiida Manchester's cheapest poker game still needs more players - drop me a line if interested, or if you know another game that I might like. ....................................................................... Mike, you're a ray of sunshine on a cold and miserable day. Long may the silence continue. David -- Right that's enough junk mail. If you really want to e-mail me, remove the y from my address ....................................................................... >better, I have fished out a recording from ITN last year which at the time \ I >was convinced had been about me, and on re-seeing it I am now convinced it \ is >not about me at all. You've cheered me up with this news, Mike. Seriously. The last sentence suggests that your health is improving, and you are beginning to become pragmatic, in some way... Take a deep breath, and look to the future. I wish you well. Smid ....................................................................... 15467 -- ------->>>>>>http://www.NewsDemon.com<<<<<<------ Unlimited Access, Anonymous Accounts, Uncensored Broadband Access