mm-4239 === Subject: Several Math Books For Sale I'm getting rid of a number of math texts that I acquired during my studies. If you're interested in any of them or have any questions, email me at mathbooksale@gmail.com. I'll accept any reasonable offer. All books are in very good condition. I'm located in London, Ontario. I'm willing to ship out the books at the buyer's expense. Otherwise, you can arrange to come pick them up. I need to get rid of these ASAP - whichever books haven't been sold after a month will be donated to a local library. Mathematical Logic Joseph R. Shoenfield The Elements of Real Analysis Robert G. Bartle Introduction to Real Variable Theory S. C. Saxena & S. M. Shah Theory of Relativity W. Pauli Probability Theory: An Introductory Course Yakov G. Sinai Calculus Robert T. Smith & Roland B. Minton Single Variable Calculus: Concepts, Applications, Theory Stanley O. Kochman College Geometry Leslie H. Miller Elementary General Topology Theral O. Moore An Introduction To Harmonic Analysis Yitzhak Katznelson Principles of Mathematical Analysis Walter Rudin Complex Variables Herb Silverman === Subject: Re: Several Math Books For Sale >I'm getting rid of a number of math texts that I acquired during my >studies. If you're interested in any of them or have any questions, >email me at mathbooksale@gmail.com. I'll accept any reasonable offer. >All books are in very good condition. I'm located in London, Ontario. >I'm willing to ship out the books at the buyer's expense. Otherwise, >you can arrange to come pick them up. I need to get rid of these ASAP >- whichever books haven't been sold after a month will be donated to a >local library. [...] Elementary General Topology >Theral O. Moore > Moore is actually a quite good, gentle introduction to general topology at the undergraduate level. I don't think it is widely available. If there is anyone out there who doesn't know topolgy yet but would like to learn it, I recommend snapping this one up. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: ? matrix decomp > ... > If A has a complete set of eigenvectors u_n, let S be a matrix whose > columns > are those eigenvectors. Then S^(-1) A S is a diagonal matrix D whose > diagonal entries are the eigenvalues. Now > A = S D S^(-1) = sum_{n=1}^N lambda_n u_n v_n^T > where v_n^T is the n'th row of S^(-1). Note that v_i^T u_j = 1 for i=j, > 0 otherwise. > ... > Do you mean, given such a set of vectors q_n, you want to write every > (real symmetric?) A in that way? > You only have N degrees of freedom in the right side. > How many degrees of freedom on the left side? > Some of my colleagues said that a N-by-N matrix has N^2 degree of > freedom, > but some said there are only N. I am confused at this point. If there are no other requirements, each entry of the N by N matrix > is one degree of freedom, so N^2 is correct. If A must be symmetric, > there are N(N+1)/2. > > So looks like what confused me is that the degree-of-freedom and > > the dimension are two different concept? > > More specifically, a N-dim matrix A can be decomposed using N linearly > dependnet > > rank-one matrix, but A still has N^2 degree-of-freedom, if A is non > symmetric. > > It seems to me that, the degree-of-freedom is the apparent dimension and > > the true dimension is no greater than the degree-of-freedom. In other > words, > > when A is not symmetric, it looks like we need to decompose it into each > > degree-of-freedom, but actually there are some linear dependence among > those > > N^2 degree-of-freedom or variables, so it turns out we only have N linear > > independent direction? No, we really have N^2 linearly independent directions. You may be able to write A as a linear combination of N rank-one matrices, but the rank-one matrices will depend on 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: ? matrix decomp > message > ... > If A has a complete set of eigenvectors u_n, let S be a matrix whose > columns > are those eigenvectors. Then S^(-1) A S is a diagonal matrix D whose > diagonal entries are the eigenvalues. Now > A = S D S^(-1) = sum_{n=1}^N lambda_n u_n v_n^T > where v_n^T is the n'th row of S^(-1). Note that v_i^T u_j = 1 for > i=j, > 0 otherwise. > ... > Do you mean, given such a set of vectors q_n, you want to write > every > (real symmetric?) A in that way? > You only have N degrees of freedom in the right side. > How many degrees of freedom on the left side? > Some of my colleagues said that a N-by-N matrix has N^2 degree of > freedom, > but some said there are only N. I am confused at this point. > If there are no other requirements, each entry of the N by N matrix > is one degree of freedom, so N^2 is correct. If A must be symmetric, > there are N(N+1)/2. > So looks like what confused me is that the degree-of-freedom and > the dimension are two different concept? > More specifically, a N-dim matrix A can be decomposed using N linearly > dependnet > rank-one matrix, but A still has N^2 degree-of-freedom, if A is non > symmetric. > It seems to me that, the degree-of-freedom is the apparent dimension > and > the true dimension is no greater than the degree-of-freedom. In other > words, > when A is not symmetric, it looks like we need to decompose it into > each > degree-of-freedom, but actually there are some linear dependence among > those > N^2 degree-of-freedom or variables, so it turns out we only have N linear > independent direction? No, we really have N^2 linearly independent directions. > You may be able to write A as a linear combination of N rank-one matrices, > but the rank-one matrices will depend on A. on A, you meant that one cannot ARBITRARY choose a set of N linearly independent rank-one matrices to decompose A? For example, any N-dim matrix A has its SVD, so one can decompose A by those N rank-one matrices. However, those rank-one matrices are quite different between two N-dim matrices A and B. One cannot use those rank-one matrices obtained from SVD of A to express B? Also, it means that dimension and degree-of-freedom are two different concepts. smaller/ simpler subproblem, it seems that dimension is more important? by Cheng Cosine May/18/2k7 NC === Subject: Re: ? matrix decomp > > message > ... > If A has a complete set of eigenvectors u_n, let S be a matrix whose > columns > are those eigenvectors. Then S^(-1) A S is a diagonal matrix D whose > diagonal entries are the eigenvalues. Now > A = S D S^(-1) = sum_{n=1}^N lambda_n u_n v_n^T > where v_n^T is the n'th row of S^(-1). Note that v_i^T u_j = 1 for > i=j, > 0 otherwise. > ... > Do you mean, given such a set of vectors q_n, you want to write > every > (real symmetric?) A in that way? > You only have N degrees of freedom in the right side. > How many degrees of freedom on the left side? > Some of my colleagues said that a N-by-N matrix has N^2 degree of > freedom, > but some said there are only N. I am confused at this point. > If there are no other requirements, each entry of the N by N matrix > is one degree of freedom, so N^2 is correct. If A must be symmetric, > there are N(N+1)/2. > So looks like what confused me is that the degree-of-freedom and > the dimension are two different concept? > More specifically, a N-dim matrix A can be decomposed using N linearly > dependnet > rank-one matrix, but A still has N^2 degree-of-freedom, if A is non > symmetric. > It seems to me that, the degree-of-freedom is the apparent dimension > and > the true dimension is no greater than the degree-of-freedom. In other > words, > when A is not symmetric, it looks like we need to decompose it into > each > degree-of-freedom, but actually there are some linear dependence among > those > N^2 degree-of-freedom or variables, so it turns out we only have N > linear > independent direction? No, we really have N^2 linearly independent directions. > You may be able to write A as a linear combination of N rank-one > matrices, > but the rank-one matrices will depend on A. > > > on A, you meant that one cannot ARBITRARY choose a set of N linearly > > independent rank-one matrices to decompose A? For example, any N-dim > > matrix A has its SVD, so one can decompose A by those N rank-one matrices. > > However, those rank-one matrices are quite different between two N-dim > matrices > > A and B. One cannot use those rank-one matrices obtained from SVD of A to > > express B? Right. That's what I meant. -- Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Galathea's Manifesto Many of those papers on emergent gravity actually do not do what I do. I am the only one getting tetrads and spin connections FROM the vacuum coherence - as far as I know. Einstein's GR is simply tetrads and spin connections. Once you have them you have it. Also none of the others realize that tetrads are renormalizable spin 1 with a dimensionless coupling. Spin 2 is secondary from entangled pairs of spin 1 along with spin 1 and spin 0 i.e. 1 + 1 -> 2 + 1 + 0 0r 3x3 = 5 + 3 + 1 dim of the IRREPS of O(3) === Subject: My truly original discovery is my M-Matrix of non-closed 1 forms from the coherent vacuum Goldstone phases of the post-inflation field from which the warped Einstein-Cartan tetrad 1-forms and the spin-connection 1-forms simply emerge in much the same way as does the superfluid velocity field v = (h/m)Grad(Phase). A^a = M^a^a tetrad 1-forms S^a^b = M^[a,b] = - S^b^a spin-connection 1-forms M^a^b = Theta^a/dPhi^b - dTheta^a/Phi^b {Theta^a} & {Phi^b} are 2 Lorentz group 4-vectors of coherent vacuum Goldstone phase 0-forms from 9 real Higgs scalar fields. e^a = I^a + @A^a @ = (Lp^2/zpf)^1/3 Lp^2 = hG/c^3 ds^2 = guvdx^udx^v = e^aea T^a = de^a + S^ac/e^c = torsion field 2-form R^a^b = dS^a^b + S^ac/S^cb = curvature field 2-form Also, my apparently original recognition that quantum gravity is renormalizable at the square root tetrad field level where it is essentially a spin 1 Yang-Mills quantum field theory! Spin 2 is a composite structure from entangled pairs of spin 1 tetrad quanta. There are also spin 1 and spin 0 quantum corrections absent in the classical (i.e. macro-quantum ODLRO c-number) part of the theory. === Subject: Continuity Function Hello! We have f(x,y), f ist affine Furthermore is f continuous on a compact (even convex) set M M:= C1 x C2 , x in C1, y in C2 The mapping x |-> f(x,y) ist affine Let Psi(x) : = max_y f(x,y) (The maximum of f(x,y) taken upon y) How do I show that Psi(x) is continuous on M? Andrew === Subject: Re: Continuity Function > Hello! We have f(x,y), f ist affine Furthermore is f continuous on a compact (even convex) set M M:= C1 x C2 , x in C1, y in C2 The mapping x |-> f(x,y) ist affine Let Psi(x) : = max_y f(x,y) (The maximum of f(x,y) taken upon y) How do I show that Psi(x) is continuous on M? You mean that Psi is continuous on C1, I suppose. I would use the fact that since M is compact, f is uniformly continuous on M. Suppose x_1 is near x_0, and that max_y f(x_i, y) is achieved at y_i, i=0,1. x_1 cannot be much less than x_0 because f(x_1, y_0) is not much different from f(x_0, y_0) = Psi(x_0). But by symmetry it cannot be much greater either. I'll leave you to cast this into epsilon-delta form. === Subject: Re: Continuity Function Hello! We have f(x,y), f ist affine Furthermore is f continuous on a compact (even convex) set M M:= C1 x C2 , x in C1, y in C2 The mapping x |-> f(x,y) ist affine Let Psi(x) : = max_y f(x,y) (The maximum of f(x,y) taken upon y) How do I show that Psi(x) is continuous on M? You mean that Psi is continuous on C1, I suppose. I would use the fact that since M is compact, f is uniformly > continuous on M. Suppose x_1 is near x_0, and that max_y f(x_i, y) is > achieved at y_i, i=0,1. x_1 cannot be much less than x_0 because > f(x_1, y_0) is not much different from f(x_0, y_0) = Psi(x_0). But by > symmetry it cannot be much greater either. I'll leave you to cast > this into epsilon-delta form. Hello! I'll repost the problem (because it was a little bit chaotic stated in the first post): If f(x,y) is a continuous function in (x,y) and is defined on M=C_{1}times C_{2}, where C_{1} and C_{2} are compact sets in the euclidean spaces E^{m} and E^{n}, then Psi(x) is a continuous function in x, where Psi (x): = mathop{max }_{y in C_{2}}f(x,y) My idea is the following (it seems that linearity and convexity is not needed): For every epsilon>0 and x' in C_{1} we have to show that we can find a delta>0 so that left|{Psi (x)-Psi (x')}right| < varepsilon ,forall x in C_{1} with left|{x-x'}right| < delta that: begin{gathered}forall y,y' in C_{2}: hfill left|{f(x',y)- f(x,y)}right| < varepsilon hfill end{gathered} AND left|{f(x',y')-f(x,y')}right| < varepsilon if left|{x-x'}right| < delta for delta>0 Choose y and y' so that: f(x',y') = Psi(x') and f(x,y)=Psi(x) for some x with left|{x-x'}right| < delta for delta>0 Okay, now comes the problem: If I can show that: f(x',y) leqslant f(x',y') and f(x,y') leqslant f(x,y) I would be basically finished. How do I show this, or, why is it true? Andrew === Subject: Re: Euclidean Connection def. Please help ...Maybe you were thinking of the bracket of Could you please help me figure out this one too, i.e, the meaning of the bracket: [X,Y]=XY-YX ? I have a headache from trying to figure out even what type of object I am dealing with when I have, e.g, a function multiplying a connection, etc. === Subject: Re: Euclidean Connection def. Please help <6926289.1179532029889.JavaMail.jakarta@nitrogen.mathforum.org > ...Maybe you were thinking of the bracket of > Could you please help me figure out this one too, > i.e, the meaning of the bracket: [X,Y]=XY-YX ? I have a headache from trying to figure out even what type of object I am dealing with when I have, e.g, a function multiplying a connection, etc. > Remember,del^i (partial derivative with respect to the ith direction)evaluated at some fixed point is a first order operator which is now being called a tangent vector replacing the classical vector e^i (0..1..0) ,all zeros except 1 in the ith spot. X (your formula with Einstein summation) is a tangent vector replacing the classical vector X^ie^i = .The relation is that the operator on some f at a fixed point gives the directional dirivative of f with respect to the classical vector ,the gradient of f dot the classical vector as your formula said. ok now XYf is what you said it is and gives the sum of alot of terms X^idel^i(Y^jdel^j) (2nd partial derivatives d^2/dx^idx^j is the classical notation) it is a second order operator .Similarly for YXf .But if you subtract them and use that taking 2nd partials with respect to x^i then x^j or visa versa gives the same result (ie del^idel^j=del^jdel^i) and use standard summation rules then the second order partial derivatives terms cancel. Remember del^i((Y^j)del^jf )has to be computed by the product rule and one term is (del^i(Y^j))(del^jf) (a first order term) the other is Y^j times del^idel^jf (second order term)this latter term is the one to cancel out and gives that the bracket is a first order operator .Calculate the coeficeints carefully ,they are important. Good luck.smn PS Lee's book has very good material but I hear it is sometimes hard to read for lack of detail.Another reference which I like which might help if you get stuck is by W Boothby (probably in Lee's bibliography (intro to diff manifolds and diff geometry I think) === Subject: Re: Linear Least Squares. Linear Alg perspective > I am working on basic topics relating to the Linear Least Squares > method with a Linear Algebra and Numerical Analysis perspective. > > I need to read a text that shows the matrices, and the theorems with > proofs. > > I need to prove that min||Ax-b||^2 is a quadratic, semidefinite, > positive quantity (or problem). Can anybody help me find a webpage or > a textbook on that material? H. M. Edwards' textbook Linear Algebra has a chapter on that subject. Jose Carlos Santos === Subject: Re: Linear Least Squares. Linear Alg perspective Am 19.05.2007 02:02 schrieb Jos.8e Carlos Santos: > > I am working on basic topics relating to the Linear Least Squares > method with a Linear Algebra and Numerical Analysis perspective. > I need to read a text that shows the matrices, and the theorems with > proofs. > I need to prove that min||Ax-b||^2 is a quadratic, semidefinite, > positive quantity (or problem). Can anybody help me find a webpage or > a textbook on that material? > > H. M. Edwards' textbook Linear Algebra has a chapter on that subject. > > > Jose Carlos Santos Another special good and extensive one with this topic is S.Mulaik, The foundations of factor-analysis Gottfried Helms === Subject: Re: Ultraclasses with some details: > > Hi all, > I have presented a theory which has classes that have proper classes > > as members, I called them 'ultraclasses' > Here is the theory in summary: > Ultraclass set theory: is the set of all sentences entailed ( by first > > order logic with identity) from the following non-logical axioms. > Definitions: > > x is a set <-> Ey ( xey & ~yey ). > > x is a proper class <->( ~xex & Ay( xey -> yey) ) > > x is an ultraclass <-> ( xex & Ay( xey -> yey ) ). > Axioms: > > 1)Extensionality: Two classes are identical iff they have the same > > members. > > 2)Regularity: Every non empty class of sets has a member that is > > disjoint from it. > > 3)Comprehension (Two schemas): > 3.A) Class Comprehension schema : if F is a formula in which x is not > > free then all closures of > > ExAy(yex<->( y is a set & F )) > > are axioms. > 3.B) Ultraclass Comprhension schema : if F is a formula in which x is > > not free then all closures of > > Ey(F & ~y is a set) -> ExAy(yex<->( F v y=x )) > > are axioms. > 4)Pairing: The unordered pair of sets is a set, and the unordered pair > > that has a class that is not a set , is an ultraclass. > > 5)Empty Set: The empty class is a set. > > 6)Union: The union of a set is a set, and the union of a class that is > > not a set is not a set. > > 7)Infinity:Ex(0ex & Ay(yex -> yu{y}ex)). > > 8)Power: The power class of a set is a set, and the power class of > > a class that is not a set is an ultraclass. > > y subclass_of x <-> Az(zey ->zex). > > m=P(n) <-> Ax(xem <-> x subclass_of n). > > 9)Size Limitation: x is a set <-> x is subnumerous to V > > were V is the class of all sets. > > and x is subnumerous to y <->( Ef(f:x->y, f is injective) & > > ~Eg(g:y->x , g is injective)). > > 10) Self inclusion: Ax( x is an ultraclass <-> xex ) > / > Theory definition finished: > Versions: > We can have many versions of this theory by changing the axiom of > > Size limitation to weaker versions, also we can have versions without > > regularity, and in these versions we'll use the known > > paradoxes( Russell's, Burali-Forti, and Cantor's ) in order to prove > > ~VeV and ~DeD, etc... ( see Triclass set theory: Exposition). > The other versions of Size limitation are: > xeV<->Ey( yeV & y is supernumerous to x). > With this weak version of limitation of size, this theory can prove > > ZF, but not ZFC. > Why don't you have a go at writing out a detailed proof that ZF can be > interpreted in your theory? > it is the same proof here, see topic: Triclass set theory:Exposition. > If you want equivalence of the weak version of this theory (the > version with the weak Size limitation ) with ZF, then the prove is : > All what we need to prove here is replacement, since the other axioms > of ZF are clearly theorems in this theory. > Replacement is proved easily with this version, since we already have > AfAxAyAz((f:x->y & z=Range(f)) -> ~( z is supernumerous to x)). > ( Review topic: Triclass set theory:Exposition for details ). > Then it is clear that replacement here is a theorum i.e: > AfAxAyAz((f:x->y & x is a set & z=Range(f))-> z is a set). > which is replacement in ZF. > Zuhair > One concern I have is that in your statement of the axiom of infinity > you never specified that x had to be a set. > Oops, how I forgot that? > Yes axiom of infinity should be: > Ex (x is a set & 0ex & Ay(yex -> yu{y}ex)). > Zuhair > - Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text - > Okay, well how about you repost that giving detailed proofs of all the > alleged theorems, > But why they are essentially the same proofs( they only need some > superficial modification) you can see them > in topic : Triclass set theory:Exposition. > As soon as I got to the first theorem, I saw that it was stated > without proof, and I stopped there. > What first theorem and where, you mean in the Exposition of Triclass > set theory, The first theorem stated in the exposition. > do you mean the Russell paradox theorem that is stated > before the axioms, this is a theorem from first order logic Yes, I know, that's fine, I know the proof of that theorem. > and not > from Triclass set theory, that's why it was put before the axioms, > this theorem is proved from first order logic. > It's not my job to re-invent your > arguments for you. You present the argument and I'll check it for > you. > You don't need to re-invent them, they are essentially the same > arguments. The same arguments as what? State your claims and argue them. Or show > me where I can find the arguments. I don't think I've seen a single > argument yet for any claim you've made in either of the two threads. > and also explain the relation between the theory in > that post and the theory you're discussing in this thread? > You mean the detailed relationship,oh this is too hard, I can't do > that, one thing I know is that this theory is stronger than the other > one, and I think it can prove the consistency of Triclass set theory. > Well, *argue* your claims and I'll check the arguments. I'm not going > to bother thinking about your claims if you haven't got to the stage > of writing up an argument which you think is rigorous. > what are you talking about? no need to argue anything, to me > everything is clear. If it is not clear to you, then that is not my > problem, neither yours. Quite often, when something is clear to you, you have in fact made a > mistake. This happens to everyone sometimes, and very often to you > because you are inexperienced. When you make a mathematical claim, it > is your responsibility to argue for it in a way that convinces every > qualified person. That's part of doing mathematics. If you want > experienced mathematicians to help you with your investigations, > you've got to take the trouble to formulate arguments for your claims. > It's not our responsibility to work out whether your claims are true > or false without any hints from you as to the reasons why you thought > they were true in the first place. If you're not interested in > formulating claims for your arguments, then bugger off and work by > yourself. I didn't ask anybody here about weather my claims are true > or false. > Then what's the point? Do you want us to help you or don't you? If I'm going to help you with your investigations, I have to work out which parts of what you say are correct. You're expecting me to do this without any hints from you as to how you arrived at your conclusions, and you don't mind laughing when I make a mistake. It's rude. If you want people with more experience than you to help you in your work, then be polite and do your fair share of the work. If not, then go away and work by yourself. > The points that I asked help with are not the points you've > raised, I asked if one has any information > about set theories that has proper classes as elements of > other classes like Ackermann's set theory for example. You will not help me by discussing points I am already > sure of. You're foolish to rely on what you're sure of when you haven't even bothered to write out the proofs properly. Everyone needs to write out proofs in full detail and have them checked by others, you far more than most. > I know for sure that this theory proves ZF,and actually > the strong vesion of it proves ZFC, not only that > it also proves all versions NBG and MK. > this is clear, there is no need to discuss that subject. > If you say you know for sure and you can't show me a proof, then it's all just hot air. People make mistakes all the time. For example, I pointed out a mistake in your formulation of the axiom of infinity. Did you know for sure then? So why do you know for sure now? You might be right. You don't know for sure. > Zuhair > Zuhair > Zuhair > - Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Ultraclasses with some details: zuhair, I hope you take what Rupert says in his post here seriously. He's very informed, very helpful, and can help you prove a lot of different things. It's really a shame that his willingness to help you (even if in a critical way sometimes) gets mired with you in personal antagonisms. > Then what's the point? Do you want us to help you or don't you? If I'm > going to help you with your investigations, I have to work out which > parts of what you say are correct. You're expecting me to do this > without any hints from you as to how you arrived at your conclusions, > and you don't mind laughing when I make a mistake. It's rude. If you want people with more experience than you to help you in your > work, then be polite and do your fair share of the work. If not, then > go away and work by yourself. The points that I asked help with are not the points you've > raised, I asked if one has any information > about set theories that has proper classes as elements of > other classes like Ackermann's set theory for example. You will not help me by discussing points I am already > sure of. You're foolish to rely on what you're sure of when you haven't even > bothered to write out the proofs properly. Everyone needs to write out > proofs in full detail and have them checked by others, you far more > than most. I know for sure that this theory proves ZF,and actually > the strong vesion of it proves ZFC, not only that > it also proves all versions NBG and MK. > this is clear, there is no need to discuss that subject. If you say you know for sure and you can't show me a proof, then > it's all just hot air. People make mistakes all the time. For example, > I pointed out a mistake in your formulation of the axiom of infinity. > Did you know for sure then? So why do you know for sure now? You > might be right. You don't know for sure. === Subject: Re: Ultraclasses with some details: Just to illustrate this theory again after the necessary corrections made. Ultraclass set theory: is the set of all sentences entailed ( by first order logic with identity) from the following non-logical axioms. Definitions: x is a set <-> Ey ( xey & ~yey ). x is a proper class <->( ~xex & Ay( xey -> yey) ) x is an ultraclass <-> ( xex & Ay( xey -> yey ) ). Axioms: 1)Extensionality: Two classes are identical iff they have the same members. 2)Regularity: Every non empty class of sets has a member that is disjoint from it. 3)Comprehension (Two schemas): 3.A) Class Comprehension schema : if F is a formula in which x is not free then all closures of ExAy(yex<->( y is a set & F )) are axioms. 3.B) Ultraclass Comprhension schema : if F is a formula in which x is not free then all closures of Ey(F & ~(y is a set)) -> E!xAy(yex<->( F v y=x )) are axioms. 4)Pairing: The unordered pair of sets is a set, and the unordered pair that has a class that is not a set , is an ultraclass. 5)Empty Set: The empty class is a set. 6)Union: The union of a set is a set, and the union of a class that is not a set is not a set. 7)Infinity:Ex(x is a set & 0ex & Ay(yex -> yu{y}ex)). 8)Power: The power class of a set is a set, and the power class of a class that is not a set is an ultraclass. y subclass_of x <-> Az(zey ->zex). m=P(n) <-> Ax(xem <-> x subclass_of n). 9)Size Limitation: x is a set <-> x is subnumerous to V were V is the class of all sets. and x is subnumerous to y <->( Ef(f:x->y, f is injective) & ~Eg(g:y->x , g is injective)). 10) Self inclusion: Ax( x is an ultraclass <-> xex ) / Theory definition finished: Versions: We can have many versions of this theory by changing the axiom of Size limitation to weaker versions, also we can have versions without regularity, and in these versions we'll use the known paradoxes( Russell's, Burali-Forti, and Cantor's ) in order to prove ~VeV and ~DeD, etc... ( D is the proper class of all ordinals that are sets). The other versions of Size limitation are: xeV<->Ey( yeV & y is supernumerous to x). With this weak version of limitation of size, this theory can prove ZF, but not ZFC. Actually we can even have a version of this theory that has antichoice in it by adding on top of the theory that has the weak version of size limitation , the axiom of antichoice which is Antichoice Axiom: D is subnumerous to V. Zuhair === Subject: Re: Ultraclasses with some details: zuhair, I hope you take what Rupert says in his post here seriously. > He's very informed, very helpful, and can help you prove a lot of > different things. It's really a shame that his willingness to help you > (even if in a critical way sometimes) gets mired with you in personal > antagonisms. Then what's the point? Do you want us to help you or don't you? If I'm > going to help you with your investigations, I have to work out which > parts of what you say are correct. You're expecting me to do this > without any hints from you as to how you arrived at your conclusions, > and you don't mind laughing when I make a mistake. It's rude. If you want people with more experience than you to help you in your > work, then be polite and do your fair share of the work. If not, then > go away and work by yourself. > The points that I asked help with are not the points you've > raised, I asked if one has any information > about set theories that has proper classes as elements of > other classes like Ackermann's set theory for example. > You will not help me by discussing points I am already > sure of. You're foolish to rely on what you're sure of when you haven't even > bothered to write out the proofs properly. Everyone needs to write out > proofs in full detail and have them checked by others, you far more > than most. > I know for sure that this theory proves ZF,and actually > the strong vesion of it proves ZFC, not only that > it also proves all versions NBG and MK. > this is clear, there is no need to discuss that subject. If you say you know for sure and you can't show me a proof, then > it's all just hot air. People make mistakes all the time. For example, > I pointed out a mistake in your formulation of the axiom of infinity. > Did you know for sure then? So why do you know for sure now? You > might be right. You don't know for sure.- Hide quoted text - - Show quoted text - No Moe you and Rupert didn't understand what I wanted. I asked a specific question and I wanted an answer to it, That question was the following: Do anybody know anything about theories which contain proper classes as members of other classes like Ackermann's set theory for example. That was my question. I didn't ask for weather my claims of this theory proving ZF or NBG are true or not. I know that these claims are true because people much more expert than Rupert confirmed this to me. I cannot be blamed for refusing to illustrate things that I didn't ask help with. I respect Rubert's willingness to help me, and I thank him for that, but he is trying to discuss something that I didn't ask for help with. Anyhow you and Rupert misunderstood me. Zuhair === Subject: Re: Ultraclasses with some details: > zuhair, I hope you take what Rupert says in his post here seriously. > He's very informed, very helpful, and can help you prove a lot of > different things. It's really a shame that his willingness to help you > (even if in a critical way sometimes) gets mired with you in personal > antagonisms. > Then what's the point? Do you want us to help you or don't you? If I'm > going to help you with your investigations, I have to work out which > parts of what you say are correct. You're expecting me to do this > without any hints from you as to how you arrived at your conclusions, > and you don't mind laughing when I make a mistake. It's rude. > If you want people with more experience than you to help you in your > work, then be polite and do your fair share of the work. If not, then > go away and work by yourself. > The points that I asked help with are not the points you've > raised, I asked if one has any information > about set theories that has proper classes as elements of > other classes like Ackermann's set theory for example. > You will not help me by discussing points I am already > sure of. > You're foolish to rely on what you're sure of when you haven't even > bothered to write out the proofs properly. Everyone needs to write out > proofs in full detail and have them checked by others, you far more > than most. > I know for sure that this theory proves ZF,and actually > the strong vesion of it proves ZFC, not only that > it also proves all versions NBG and MK. > this is clear, there is no need to discuss that subject. > If you say you know for sure and you can't show me a proof, then > it's all just hot air. People make mistakes all the time. For example, > I pointed out a mistake in your formulation of the axiom of infinity. > Did you know for sure then? So why do you know for sure now? You > might be right. You don't know for sure.- Hide quoted text - - Show quoted text - No Moe you and Rupert didn't understand what I wanted. > I asked a specific question and I wanted an answer to it, > That question was the following: Do anybody know anything about > theories which contain proper classes as members of other classes like > Ackermann's set theory for example. > That was my question. I didn't ask for weather my claims of this > theory proving ZF or NBG are true or not. I know that these claims are > true because people much more expert than Rupert confirmed this to > me. I cannot be blamed for refusing to illustrate things that I didn't ask > help with. I respect Rubert's willingness to help me, and I thank him for that, > but he is trying to discuss something that I didn't ask for help with. Anyhow you and Rupert misunderstood me. Zuhair- Hide quoted text - - Show quoted text - Okay, so some qualified people have confirmed one of your claims for you, well that's great, but I would make two points: (1) Apparently you have been having trouble giving a correct statement of the claim, I have already pointed out one problem with your initial definition of the theory, so it would be worthwhile getting to the point where you are sure you have the correct statement of the claim (2) Even then, it would still be a worthwhile exercise for you to write out your own complete proof of this claim in a form which will convince any qualified mathematician. I don't know much about Ackermann set theory, but there's a definition and a brief discussion here: http://en.wikipedia.org/wiki/Ackermann_set_theory If you want me to help you explore the relations between your theory and Ackermann set theory, then I would ask you to make absolutely sure you have given the correct definition of your theory and write up complete proofs of all your claims about it. Then I will check those proofs for you and help you investigate what relations there are between your theory and Ackermann set theory. === Subject: Re: Ultraclasses with some details: > zuhair, I hope you take what Rupert says in his post here seriously. > He's very informed, very helpful, and can help you prove a lot of > different things. It's really a shame that his willingness to help you > (even if in a critical way sometimes) gets mired with you in personal > antagonisms. > Then what's the point? Do you want us to help you or don't you? If I'm > going to help you with your investigations, I have to work out which > parts of what you say are correct. You're expecting me to do this > without any hints from you as to how you arrived at your conclusions, > and you don't mind laughing when I make a mistake. It's rude. > If you want people with more experience than you to help you in your > work, then be polite and do your fair share of the work. If not, then > go away and work by yourself. > The points that I asked help with are not the points you've > raised, I asked if one has any information > about set theories that has proper classes as elements of > other classes like Ackermann's set theory for example. > You will not help me by discussing points I am already > sure of. > You're foolish to rely on what you're sure of when you haven't even > bothered to write out the proofs properly. Everyone needs to write out > proofs in full detail and have them checked by others, you far more > than most. > I know for sure that this theory proves ZF,and actually > the strong vesion of it proves ZFC, not only that > it also proves all versions NBG and MK. > this is clear, there is no need to discuss that subject. > If you say you know for sure and you can't show me a proof, then > it's all just hot air. People make mistakes all the time. For example, > I pointed out a mistake in your formulation of the axiom of infinity. > Did you know for sure then? So why do you know for sure now? You > might be right. You don't know for sure.- Hide quoted text - > - Show quoted text - No Moe you and Rupert didn't understand what I wanted. > I asked a specific question and I wanted an answer to it, > That question was the following: Do anybody know anything about > theories which contain proper classes as members of other classes like > Ackermann's set theory for example. > That was my question. I didn't ask for weather my claims of this > theory proving ZF or NBG are true or not. I know that these claims are > true because people much more expert than Rupert confirmed this to > me. I cannot be blamed for refusing to illustrate things that I didn't ask > help with. I respect Rubert's willingness to help me, and I thank him for that, > but he is trying to discuss something that I didn't ask for help with. Anyhow you and Rupert misunderstood me. Zuhair- Hide quoted text - - Show quoted text - Okay, so some qualified people have confirmed one of your claims for > you, well that's great, but I would make two points: (1) Apparently you have been having trouble giving a correct statement > of the claim, I have already pointed out one problem with your initial > definition of the theory, so it would be worthwhile getting to the > point where you are sure you have the correct statement of the claim > (2) Even then, it would still be a worthwhile exercise for you to > write out your own complete proof of this claim in a form which will > convince any qualified mathematician. I don't know much about Ackermann set theory, but there's a definition > and a brief discussion here: http://en.wikipedia.org/wiki/Ackermann_set_theory If you want me to help you explore the relations between your theory > and Ackermann set theory, then I would ask you to make absolutely sure > you have given the correct definition of your theory and write up > complete proofs of all your claims about it. Then I will check those > proofs for you and help you investigate what relations there are > between your theory and Ackermann set theory. mentioned above later. But I have presented the definition of the theory yesterday after the necessary corrections made. You can see easily that MK is a subset of this theory, so it is clear why ZF,ZFC is proved in this theory. Anyhow working out my claims and proving them one after the other is an important stop no doubt,I will work that later. Zuhair - Hide quoted text - - Show quoted text - === Subject: Re: Debate with Waldyr Rodrigues on Foundations of Einstein's 1915 Theory o === Subject: Re: Debate with Waldyr Rodrigues on Foundations of Einstein's 1915 Theory o > Look you jerk. Waldyr Rodrigues Jr is a fan of my book Space > Time and Beyond and in 1984 paid for my airfare > from San Francisco to Brazil where I was a visiting > several months. Waldyr worked on UFO problems > for the Brazilian Air Force and also believes Uri Geller > has real PK powers. Waldyr introduced me to a > Brazilian Army General tracking Uri's and other's PK > powers as well as UFOs. Fred Alan Wolf was with me > in Brazil at these meetings. Security check! If you are the real Jack Sarfatti, put this text dkue mnqo pscn on your blog at http://destinymatrix.blogspot.com/ Maybe I'm in Jack's killfile, because he has not responded to two previous ID requests like this one. === Subject: Re: Debate with Waldyr Rodrigues on Foundations of Einstein's 1915 Theory o Bytes: 1869 > message > athforum.org... > Look you jerk. Waldyr Rodrigues Jr is a fan of my > book Space > Time and Beyond and in 1984 paid for my airfare > from San Francisco to Brazil where I was a visiting > several months. Waldyr worked on UFO problems > for the Brazilian Air Force and also believes Uri > Geller > has real PK powers. Waldyr introduced me to a > Brazilian Army General tracking Uri's and other's > PK > powers as well as UFOs. Fred Alan Wolf was with me > in Brazil at these meetings. > > Security check! > If you are the real Jack Sarfatti, put this text > dkue mnqo pscn > on your blog at > http://destinymatrix.blogspot.com/ > > Maybe I'm in Jack's killfile, because he has not > responded to two previous > ID requests like this one. > > The respondent isn't Jack Sarfatti. The real one may waste his talents, but I've never known him to waste his time. Tom === Subject: Re: Debate with Waldyr Rodrigues on Foundations of Einstein's 1915 Theory o > message > athforum.org... > Look you jerk. Waldyr Rodrigues Jr is a fan of my > book Space > Time and Beyond and in 1984 paid for my airfare > from San Francisco to Brazil where I was a visiting > several months. Waldyr worked on UFO problems > for the Brazilian Air Force and also believes Uri > Geller > has real PK powers. Waldyr introduced me to a > Brazilian Army General tracking Uri's and other's > PK > powers as well as UFOs. Fred Alan Wolf was with me > in Brazil at these meetings. > Security check! > If you are the real Jack Sarfatti, put this text > dkue mnqo pscn > on your blog at > http://destinymatrix.blogspot.com/ > Maybe I'm in Jack's killfile, because he has not > responded to two previous > ID requests like this one. > > The respondent isn't Jack Sarfatti. The real one > may waste his talents, but I've never known him to > waste his time. > Then what the hell is going on? === Subject: Re: Debate with Waldyr Rodrigues on Foundations of Einstein's 1915 Theory o === Subject: Warning to Trolls: Create an e-annoyance, go to jail This applies to all flaming Trolls who hide their real ID if ever they are found out. You are free to say what you want on the internet as long as you truthfully ID who you are so that an offended party can take legal action. FYI Uri Geller BTW has taken such legal actions and he has the money to do it. JC has made slanderous remarks about Geller - not a good idea. Annoying someone via the Internet is now a federal crime. It's no joke. Last Thursday, President Bush signed into law a prohibition on posting annoying Web messages or sending annoying e-mail messages without disclosing your true identity. In other words, it's OK to flame someone on a mailing list or in a blog as long as you do it under your real name. http://news.com.com/Create+an+e-annoyance,+go+to+jail/2010-1028_3-6022491.ht ml === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail <19676758.1179612583754.JavaMail.jakarta@nitrogen.mathforum.org This applies to all flaming Trolls who hide their real ID if ever they are found out. You are free to say what you want on the internet as long as you truthfully ID who you are so that an offended party can take legal action. FYI Uri Geller BTW has taken such legal actions and he has the money to do it. JC has made slanderous remarks about Geller - not a good idea. Annoying someone via the Internet is now a federal crime. It's no joke. Last Thursday, President Bush signed into law a prohibition on posting annoying Web messages or sending annoying e-mail messages without disclosing your true identity. In other words, it's OK to flame someone on a mailing list or in a blog as long as you do it under your real name. http://news.com.com/Create+an+e-annoyance,+go+to+jail/2010-1028_3-602... Oh, BTW, Jack. I want to go on the record as saying that I personally find the unsolicited e-mails you send to me several times a week annoying. They just go into the spam folder anyways (where they belong AFAIC). How many other people get this junk mail from Jack? (Guess this is my punishment for having to pointed out to him he was wrong, eh?). Matt === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail junoexpress > Oh, BTW, Jack. I want to go on the record as saying that I personally > find the unsolicited e-mails you send to me several times a week > annoying. They just go into the spam folder anyways (where they > belong AFAIC). How many other people get this junk mail from Jack? I had three in my inbox today. But none previously. LH === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail I have no idea who juno express is. Not in my address book. He may be on some open list I reply to. Probably Gordon Novel's or Gary Ford's. If juno sends me a copy of what he is talking about I will try to correct it. As to Hammick, I simply sent you a request to stop harassing me on this forum. === Subject: Juno Express There is no such person Juno Express on any of the lists I reply to. As I thought this is another phony false allegation from a Troll. If there is such a person let him ID his precise e-mail so that I can check it to see what list this person is allegedly on. This kind of smear from anonymous sources is evil. === Subject: Re: Juno Express Also, why did this person simply not contact me earlier about his alleged problem? More lies. === Subject: MTBrenneman@gmail.com I have not sent you any e-mails at all knowingly. I may have unintentionally sent you a few meant for MThorne. Obvious error. All you had to do then was to have informed me of that and I would have taken steps to correct it. Simply bounce back any such e-mails sent to you by error for me to correct. Very simple. I find your e-mails and personal attacks on me in this forum also very annoying. You started this not me remember. I had no idea of your existence until your personal attacks. In any case I am not able to use the Federal Law against you because you are, it seems at least, unlike the other vicious Trolls, hiding your true ID. Also, I am not hiding my true ID and to repeat any e-mails you got from me were unintended meant for MThorne on Gary Ford's open list and not to you. I may have sent you one or two personal e-mails directly responding to your attacks on me in this forum - I don't recall at the moment. === Subject: Re: MTBrenneman@gmail.com typo change In any case I am not able to use the Federal Law against you because you are, it seems at least, unlike the other vicious Trolls, hiding your true ID. to In any case I am not able to use the Federal Law against you because you are NOT, it seems at least, unlike the other vicious Trolls, hiding your true ID. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail > This applies to all flaming Trolls who hide their real ID if ever they are > found out. You are free to say what you want on the internet as long as > you truthfully ID who you are so that an offended party can take legal > action. FYI Uri Geller BTW has taken such legal actions and he has the > money to do it. JC has made slanderous remarks about Geller - not a good > idea. Annoying someone via the Internet is now a federal crime. It's no joke. Last Thursday, President Bush signed into law a prohibition > on posting annoying Web messages or sending annoying e-mail messages > without disclosing your true identity. In other words, it's OK to flame someone on a mailing list or in a blog as > long as you do it under your real name. http://news.com.com/Create+an+e-annoyance,+go+to+jail/2010-1028_3-6022491.ht m l That's interesting. So if you are, for example, Australian, this does not apply? Or does the law come into effect if either party is American? Just curious. -- Glen === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Annoying someone via the Internet is now a federal crime. Sorry Jack, it's not just people in America who want to annoy you. -- Richard -- Consideration shall be given to the need for as many as 32 characters in some alphabets - X3.4, 1963. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail <19676758.1179612583754.JavaMail.jakarta@nitrogen.mathforum.org>, > This applies to all flaming Trolls who hide their real ID if ever they are > found out. You are free to say what you want on the internet as long as you > truthfully ID who you are so that an offended party can take legal action. > FYI Uri Geller BTW has taken such legal actions and he has the money to do > it. JC has made slanderous remarks about Geller - not a good idea. > > Annoying someone via the Internet is now a federal crime. > > It's no joke. Last Thursday, President Bush signed into law a prohibition on > posting annoying Web messages or sending annoying e-mail messages without > disclosing your true identity. > > In other words, it's OK to flame someone on a mailing list or in a blog as > long as you do it under your real name. > > http://news.com.com/Create+an+e-annoyance,+go+to+jail/2010-1028_3-6022491.ht m l Not a very good day in the history of the republic. In 1776, Thomas Paine anonymously published his pamphlet Common Sense, urging independence of the colonies from Britain. I assume that annoyed the hell out of King George. Anonymous -- and annoying -- speech has a long and honorable history in this country. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail <19676758.1179612583754.JavaMail.jakarta@nitrogen.mathforum.org > [..] Not a very good day in the history of the republic. In 1776, Thomas > Paine anonymously published his pamphlet Common Sense, urging > independence of the colonies from Britain. I assume that annoyed the > hell out of King George. Anonymous -- and annoying -- speech has a long > and honorable history in this country Old Farmer George himself wasn't above posting anonymously on forums, for example to disseminate his ideas on improving agriculture. See: http://www.lib.udel.edu/ud/spec/exhibits/sfc/ag.htm Arthur Young (1741-1820). Annals of Agriculture and Other Useful Arts. London: Arthur Young, 1784-1800. 34 volumes. Arthur Young was an agricultural experimenter, a staunch advocate of agricultural reform, and a prolific writer on agricultural subjects. Young began this monthly journal in 1784. It was published continuously to 1809, with two later appearances in 1812 and 1815. In addition to Young, contributors to the Annals included King George III, Joseph Priestley, and Thomas William Coke. John R Ramsden === Subject: Old Farmer George My Gawd you guys are weak-minded when it comes to logic. No one said anonymous speech was illegal. What is illegal is slanderous smears and false allegations of criminal activity by people who hide their true ID on the Internet. In more general terms anything written anonymously on the Internet that gives good cause for the intended target to be annoyed is illegal and criminal. Good cause would be determined by a jury and/or judge. JC accused me of taking illegal drugs. That is false. JC has no way of knowing what I do of course. Any jury and judge would find JC's false allegation good cause for feeling annoyed under the law. Until that law is overturned it is on the books. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail Anonymous speech for purposes of slander has never been condoned. That's nonsense and it is an abuse of the words freedom of speech. Indeed, the Trolls here have invaded my privacy with completely false allegations that I use illegal drugs. They are evil and to do so is not only immoral but criminal. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail <18372649.1179641176768.JavaMail.jakarta@nitrogen.mathforum.org Anonymous speech for purposes of slander has never been condoned. That's nonsense and it is an abuse of the words freedom of speech. Indeed, the Trolls here have invaded my privacy with completely false allegations that I use illegal drugs. They are evil and to do so is not only immoral but criminal. word annoy not slander. There are laws for slander in place (actually, the word you wanted to use was libel, not slander, just so you know the correct word next time), so there was no need for this bill. The point OP is making is that the bill is yet another restriction on our freedoms by Bush. I'm just surprised that someone like Spector, who for the last couple of years seemed to have woken up and become a decent human being, would pass such legislation. Guess Bush is getting tired of all those annoying e-mails calling for his impeachment. The good news is that the language is so vague and the law so stupid, that if it ever were tested in a court of law, I'd bet money it wouldn't hold water legally. M === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail Anonymous speech with intent of harassment on the internet is legally the same as anonymous threatening phone calls. It is only done by cowardly creeps, what the cops call perps. People who do that sort of thing are mentally ill and are prone to commit other crimes as well. === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail Is this the real JS too? http://www.law.duke.edu/slsa/archive/?/digest+2007-04-18+%23001.txt Quote from what I believe is the real JS: Warning to Trolls: Create an e-annoyance, go to jail === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail http://www3.telus.net/ldh/me.jpg === Subject: Re: Warning to Trolls: Create an e-annoyance, go to jail http://www3.telus.net/ldh/me.jpg is a photo of Larry Hammick who lives in Vancouver, Canada. Clearly same guy who has no credentials as an academic mathematician strictly amateur whose only accomplishment seems to be in playing Bridge. Larry stick to games whose rules you understand and stop harassing me. === Subject: Message from Kyoto This is Kyoto-Bastard-Poet-Report Vol.95 [Dr.Yukawa's Taoistic-Japanese-Inn] Dr.Yukawa said, Every matter comes from somewhere. They stay in the room for a while & they travel again. [A Secret Aestheticism of Dr.Yukawa] Dr.Yukawa suddenly confessed [Tanizaki-like-Message](to a famous editorial director, Seigo Matsuoka) [Being a scientist is required something like licking fingers of woman's leg.] (from Novelist Tanizaki) He meant something 'observing & living here & now is totally different thing'. The important thing is 'how we recognize[what it is] [by what]'. This [what it is] & [by what] is usually continuous so far. But he insisted that it does not matter they are separate thing. The identification between [what A] & [what B] would be a real theoretical physics. Three thousand kaleidscopic world inside a light snow the inside is also a light snow (Ryokan) picture: the smallest shrine in Kyoto. === Subject: Re: Message from Kyoto What did Professor Yukawa mean? Well he may have meant that the electrons, (also quarks ...) are Bohm hidden variables something like tiny Kerr black holes with strong short-range Abdus Salam gravity G* > G(Newton). That is renormalization group flow where gravity gets stronger on shorter scales ~ 1 fermi. The effective size of the electron C(p) is something like (in simplest toy model neglecting charge and spin) C(p) ~ (e^2/mc^2)(1 - (e^2/mc^2)(p/h))^1/2 p is the 3-momentum transfer between probe and target electron. That is, the event horizon is at e^2/mc^2 ~ 1 fermi ratio of outer observed circumference C(p) to the fixed physical radius e^2/mc^2 is as above. When the momentum transfer is ~ (1 fermi)^-1 the electron appears as a point. For quarks, their event horizon inside the hadron dark matter positive ZPF pressure bag may be 10^-16 cm. Note e^2 = G*m^2 Blackett relation Therefore, e^2/mc^2 ~ G*m/c^2 Black holes and strings are complementary descriptions. Begin forwarded message: Author: Jack Sarfatti === Subject: Message from Kyoto This is Kyoto-Bastard-Poet-Report Vol.95 [Dr.Yukawa's Taoistic-Japanese-Inn] Dr.Yukawa said, Every matter comes from somewhere. They stay in the room for a while & they travel again. [A Secret Aestheticism of Dr.Yukawa] Dr.Yukawa suddenly confessed [Tanizaki-like-Message](to a famous editorial director, Seigo Matsuoka) [Being a scientist is required something like licking fingers of woman's leg.] (from Novelist Tanizaki) He meant something 'observing & living here & now is totally different thing'. The important thing is 'how we recognize [what it is] [by what]'. This [what it is] & [by what] is usually continuous so far. But he insisted that it does not matter they are separate thing. The identification between [what A] & [what B] would be a real theoretical physics. Three thousand kaleidscopic world inside a light snow the inside is also a light snow (Ryokan) picture: the smallest shrine in Kyoto. === Subject: Frank Tipler sounds the alarm! I fear that in the very near future, education in physics will have to be obtained from some source other than a university. It is becoming increasingly clear that this corruption of education is probably universal across all disciplines. If so, then all advanced education will have to be obtained outside of the university. And if that is the case, then why should universities exist at all? Does this apply to string theorists also? What about loop quantum gravity theorists? ;-) Begin forwarded message: === Subject: General Relativity and The Standard Model not being taught in Universities Postmodern Physics Colleges Fail to Teach Basics - Even in Physics! By Frank Tipler May 16, 2007 EditorOs Note: Guest writer Frank Tipler is a professor of physics at Tulane University. A recent study shows that Shakespeare is no longer a required course for English majors at the overwhelming majority of American elite universities. This is not a surprise: most people are well aware that students are no longer taught the basics in the humanities departments. Unfortunately, the situation is just as bad in physics departments. At the overwhelming majority of physics departments at American universities, even the most elite, key elements of basic physics are no longer taught. For example, I am aware of no American university that requires, for an undergraduate degree in physics, a course in general relativity, which is Albert EinsteinOs theory of gravity. At the overwhelming majority of American universities, including Harvard, M.I.T. and Cal Tech, one is not even required to take a course in general relativity to get a Ph.D. in physics! As a consequence, most American Ph.D.Os in physics do not understand general relativity. If a problem arises that requires knowledge of EinsteinOs theory of gravity, almost all American physicists can only look blank. This is in spite of the fact that general relativity has been known to be the correct theory of gravity for almost a century. The Standard Model has been experimentally confirmed, and some dozen and more Nobel Prizes in physics have been awarded for the discovery and experimental confirmation of the Standard Model. Yet I am aware of no physics department in the United States that requires a course in the Standard Model for an undergraduate degree in physics. Very few, if any, require a course in the Standard Model even for a Ph.D. in physics. ItOs as if law schools stopped requiring students to take courses in crucial subjects like contracts and property law. So one can get an undergraduate degree in physics and even a Ph.D., without knowing anything at all about the fundamental forces that control the universe at the most basic level. Since our entire civilization requires at least somebody knows basic physics, requires that at least people who have Ph.D.Os in physics know basic physics, this is a disaster. If very few physicists know the Standard Model, it is unlikely that anyone will attempt to develop the new source of energy which the Standard Model shows is possible in principle. The basic reasons why modern physics is not covered in required courses are identical to the basic reasons why Shakespeare is not covered: (1) the faculty in both cases want to teach their narrow specialty rather than the basic courses in their field, (2) the faculty members in both cases no longer understand the basic material in their own field, (3) the faculty no longer believe there are fundamental truths in their own disciplines. I'm sure that many members of typical universityOs English faculty no longer have a basic understanding of Shakespeare. How could they, if they themselves have never taken a course on Shakespeare? A degree in English is no longer a guarantee that the degree holder has a basic knowledge of Shakespeare or other great writers. Similarly, a degree in physics from an American university is no guarantee that the student with this degree understands basic physics. The physics facultyOs increasing ignorance of basic physics is starting to show up in their research, as I describe at length in my recent book, The Physics of Christianity (Doubleday, 2007). I show that, across all disciplines, a collapse of belief in Christianity over the past several decades among university faculty has been accompanied by a collapse in the belief that there is fundamental truth which should be imparted to students. Every undergraduate majoring in physics, or at the very least, every graduate student in physics, should be required to take a two-semester sequence: one semester on general relativity, and one semester on the Standard Model. Both courses have been taught for decades to physics students as an elective, but no physics department will require them. Once, on my own initiative, I forced a required course on the Standard Model at the graduate level, since I firmly believe that knowledge of the Standard Model should be required for all Ph.D.Os in physics. I achieved this by changing a required two-semester graduate course in electromagnetism into a one-semester course in electromagnetism, and a one-semester course on the Standard Model. I used an undergraduate textbook for the Standard Model course. The students violently objected. They didnOt see any reason to learn the Standard Model. They saw no reason why they should know any basic physics beyond what was standard 50 years ago. The other faculty backed them up. This occurred more than 10 years ago, and since then not one Ph.D. student at Tulane has been taught the Standard Model. The reason the physics faculty backed the graduate students up N supported them in their desire to remain ignorant of the central fundamental theory of physics N is that they themselves were never taught the Standard Model when they were graduate students, and thus they saw no reason to require their own students to learn it. I wasnOt taught the Standard Model either when I was a graduate student N it was in the process of being discovered when I was a graduate student N but it was obviously something every physicist should know, so I taught myself the theory. These same physics faculty were never taught general relativity either (I was; and in fact my Ph.D. thesis was on a problem in general relativity), so they see no reason why physics Ph.D.Os should be taught general relativity. I fear that in the very near future, education in physics will have to be obtained from some source other than a university. It is becoming increasingly clear that this corruption of education is probably universal across all disciplines. If so, then all advanced education will have to be obtained outside of the university. And if that is the case, then why should universities exist at all? TV dinner still cooling? Check out Tonight's Picks on Yahoo! TV. Jack Sarfatti sarfatti@pacbell.net If we knew what it was we were doing, it would not be called research, would it? - Albert Einstein http://www.authorhouse.com/BookStore/ItemDetail.aspx?bookid=23999 http://lifeboat.com/ex/bios.jack.sarfatti http://qedcorp.com/APS/Dec122006.ppt http://www.flickr.com/photos/lub/sets/72157594439814784 === Subject: Re: Frank Tipler sounds the alarm! Commentary 1 This is a very useful little book by Lochlainn O' Raifaeartaigh in Dublin published by Princeton 1997. It has seminal papers by Weyl, Klein, Fock, Schrodinger, London & Pauli in English from the original German. The two great battle-tested principles of basic theoretical physics are 1. The local gauge principle, i.e. relativity in the most general sense of no action without reaction, no passive absolute Newtonian arenas, Leibniz's relationism of dialogue not monologue. 2. The spontaneous breaking (or hiding) of continuous symmetries in the ground state of real on mass shell quanta and also in the vacuum of virtual zero point quanta. AKA More is different (P.W. Anderson) Emergent complexity. ODLRO (Onsager & Oliver Penrose), Goldstone theorem Macroquantum states, Glauber coherent & squeezed states... There remained however the electromagnetic and nuclear forces, and the geometrization of gravity raised the question as to whether these other fundamental forces were 'true' forces operating in curved space of gravitational theory or whether they also were part of the geometry. This question has still not been fully answered. ... these forces and gravitation have a common geometrical structure. This is ... the gauge structure. to be continued === Subject: Re: Frank Tipler sounds the alarm! The book is The Dawning of Gauge Theory - I meant to change the subject line to that and will do so in #2. === Subject: To Troll Larry Hammick I will not do anything you ask me to. Go become a Wickepedio File Jerk. You have the required hack-mindedness (John Nash). You will see yourself as a second hander in Ayn Rand's The Fountainhead and in George Orwell's Animal Farm. You are part of the Digital Maoist cult, the Pol Pot of Cyberspace. You are small-minded and arrogant about it. You should go to the border of Pakistan with Afghanistan and join with your brothers in the Taliban. You have the same approach to superior knowledge and creative thought. You are dead but you don't know it. Same for all The Trolls - Demons from Hell. === Subject: Re: To Troll Larry Hammick >...You should go to the border of Pakistan with Afghanistan and join with >your brothers in the Taliban. ... If anybody's wondering where that came from, Jack Sarfatti has looked me up at Wiki. http://en.wikipedia.org/wiki/Abdul_Hadi_al_Iraqi http://en.wikipedia.org/wiki/Anas_al-Liby http://en.wikipedia.org/wiki/Mohammad_Hasan_Khalil_al-Hakim etc. === Subject: Re: To Troll Larry Hammick Hammick is really stupid. If you go to my Blog you will find there many of my comments made to this forum - I mean the positive ones with Galathea - not my replies to the Trolls may they rot soon. === Subject: Re: To Troll Larry Hammick And yes, Hammick I did not allow your stupid comment on my Blog. === Subject: Re: Why is it called a Divisor? Originator: jdolan@math.UUCP (James Dolan) |magidin@math.berkeley.edu (Arturo Magidin) said: | | [...] | |>algebraic number theory in an attempt to find an equivalent theory |>for unique factorization when the ring is not a UFD. | |Careful. Dedekind did indeed place the emphasis on a theory |equivalent to unique factorization, and his ideas have dominated |algebraic number theory (and gave birth to Ring Theory later |on). When Hilbert and Weyl presented Kronecker's theory of divisors, |they viewed it as an equivalent alternative formulation to Dedekind, |and presented it that way. Dedekind was a gifted writer and |expositor; his writings from over 125 years ago could be used, with |only minor adjustments in wording, to teach the introduction to |algebraic number theory today. Kronecker, on the other hand, was late |to publish, hard to read, and often confusing, and so Kronecker's |works are known mostly through their rephrasing by Hilbert and Weyl, |to parallel Dedekind's. | |According to Edwards, however, who has made a very thorough and |interesting study of the works of both Dedekind and Kronecker, |Kronecker did NOT give unique factorization that much importance. For |Kronecker, the primary objective and main use of divisors was the |ability to define greatest common divisors in the ->absence<- of |unique factorization. The advantage is that his definition is such |that it is not ambient-dependant. That is, his definitions |effectively defines a greatest common divisor of two algebraic |integers in the ring of ALL algebraic integers, so that if you extend |the field, the greatest common divisor does not change at all. By |contrast, the factorization of ideals into prime ideals ->does<- |change when you move fields. (Granted, this is actually very |interesting, and lets Galois Theory come into play, leading to the |theory of tame and wild extensions and the like, all very |fruitful). Ideals themselves change when you change the field. | |Whereas Dedekind obtains greatest common divisors (both in the ring |of all algebraic integers, and in intermediate rings of integers; and |likewise in the function field case/coordinate ring case) as a |consequence once he has substituted unique factorization into unique |factorization of ideals, Kronecker first develops the entire theory |of greatest common divisors, and only later considers the problem of |factoring into prime divisors in intermediate fields. | |According to Edwards (Preface to Divisor Theory): | | The kroneckerian theory of divisors has at least three clear | advantages over the Dedekindian theory of ideals: | | (1) It follows from the single, natural premise that the | content of the product of two polynomials is the product of | the contents. | | (2) It entails an algorithmic test for divisibility, which, in | | Dedekindian terms, gives a specific computation for | deciding whether a given element is in the ideal generated | by a finite set of other elements. (For ideological reasons | that are explained in [Edwards, Harold M. Dedekind's | Invention of Ideals, Bull. Lond. Math. Soc. 15 (1983) | 8-17. (Studies in the History of Mathematics, Esther | R. Phillips, ed., MAA 1987, 8-20)], Dedekind made a | ->virtue<- [emphasis in the original] of the lack of such a | test in his theory, whereas Kronecker was of the opposite | opinion). | | (3) It is independent of the ambient field, so that, unlike the | Dedekindian theory, all statements remain true without | modification when the ambient field is extended. (The very | ->definition<- [emphasis in the original] of an ideal as a | certain kind of subset of the field depends heavily on the | ambient field. | | Another important advantage of the theory of divisors is its | applicability to integral and nonintegral divisors alike; the | theory of ideals involves both integral and nonintegral ideals, | but the integral ideals are far more natural and are usually the | only ones introduced in the early stages of the theory. i don't think that i agree with edwards about very much of this. (it's a bit hard for me to tell though, since i'm not sure if my understanding of this stuff was ever really that good, and having always been a bit shaky, it may have deteriorated somewhat while i've been focusing on other areas recently.) i think that the differences that edwards emphasizes here between dedekind's approach and kronecker's approach seem mostly rather superficial, and miss the point of the real significant distinction between the approaches. (on the other hand, i have the impression that weyl himself tended to over-emphasize the seemingly superficial differences, and since it's generally not that easy to out-guess weyl i may keep an open mind about whether weyl and edwards may have been on to something here.) ideals themselves change when you change the field. (of course presumably we're talking here about ideals of the ring of integers of the field, since fields themselves don't have very interesting ideals.) they don't really change that much, though. they mostly just split (or do a couple of other things which for present purposes are not that different from splitting), which is a pretty mild thing to happen, and which has the effect that pushing ideals forward along a ring homomorphism not only preserves the greatest common factor operation but is in fact a homomorphism of distribitive lattices. (and in the perhaps more revealing geometric language where all of the arrows are turned around, this distributive lattice homomorphism is interpreted as pulling subvarieties back along a map between algebraic varieties.) here again conceptual clarity seems to favor dedekind's approach over kronecker's approach, in connection with the very ideas that edwards seems to be trying to explain. in contrast, i think that the real advantage of kronecker's approach over dedekind's approach, even if just a stylistic advantage, is the way that kronecker emphasized the concept of ideal _number_, before dedekind came along and chopped off the number aspect of it. i think that thinking in terms of ideal numbers in contrast to mere ideals really helps in understanding kronecker's jugendtraum ideas about explicit calculation of ideal numbers as values of certain geometrically interesting transcendental functions at certain geometrically interesting points. unfortunately i have so far only vague suspicions rather than actual knowledge about how kronecker's jugendtraum ideas relate to interesting modern developments. -- jdolan@math.ucr.edu === Subject: Re: Why is it called a Divisor? Originator: jdolan@math.UUCP (James Dolan) |in contrast, i think that the real advantage of kronecker's approach |over dedekind's approach, even if just a stylistic advantage, is |the way that kronecker emphasized the concept of ideal _number_, |before dedekind came along and chopped off the number aspect of it. |i think that thinking in terms of ideal numbers in contrast to mere |ideals really helps in understanding kronecker's jugendtraum |ideas about explicit calculation of ideal numbers as values of |certain geometrically interesting transcendental functions at certain |geometrically interesting points. unfortunately i have so far only |vague suspicions rather than actual knowledge about how kronecker's |jugendtraum ideas relate to interesting modern developments. i should rephrase this slightly since it shows traces of my tendency to conflate together all mathematicians whose name begins with k, kummer and kronecker in particular. i think that ideal number was originally kummer's concept rather than kronecker's, but i still think that it's true that of kummer's successors, kronecker pursued the ideal number concept more faithfully with the jugendtraum program than dedekind did with the theory of ideals. the picture here is complicated however by further questions about substantive versus stylistic differences between the work of kronecker and that of dedekind; thus kronecker who pursued the ideal numbers at a very concrete level was also the one who seemed unwilling to grant such ideal objects a concrete status as part of mathematical reality. -- jdolan@math.ucr.edu === Subject: albert shimoon eom === Subject: ubasic factor ise.. icafe.. please 2^52.. Cc: mcdonewt@yahoo.co.nz UBASIC supports bignums, fractions, complex numbers, polynomials and integer factorisation. It runs under MS-DOS and is written in assembly language. ??? i wish to factor integers up to 2E9 or 4E15 at internet cafe. (2^31, 2^52.) how can i do it. is there on on-line factoriser, please. or excel..? i used to do these easily on Acorn A5000 Archimedes risc-os in 200x. bbc basic 64. sericalc4s. myprog etc.. e.g.. (planet sedna?) convert base 27. will the icafe let me download software and run it. can i get a printout of all the calculation steps + results i did. === Subject: Re: ubasic factor ise.. icafe.. please 2^52.. > UBASIC supports bignums, fractions, complex numbers, polynomials and > integer factorisation. It runs under MS-DOS and is written in assembly > language. > ??? i wish to factor integers up to 2E9 or 4E15 > at internet cafe. E(2^31, E2^52.) how can i do it. > is there on on-line factoriser, please. or excel..? Forget it. i used to do these easily on Acorn A5000 Archimedes risc-os in 200x. > bbc basic 64. > sericalc4s. myprog etc.. e.g.. > (planet sedna?) Econvert base 27. > will the icafe let me download software and run it. > can i get a printout of all the calculation steps + results i did. === Subject: Union of AANR (to Zbigniew Karno) Is it true that X_1 union X_2 is also an AANR? any idea on how to prove it? or reference === Subject: Re: Union of AANR (to Zbigniew Karno) Is it true that X_1 union X_2 is also an AANR? any idea on how to > prove it? or reference > Surely some mistake? === Subject: Re: Union of AANR (to Zbigniew Karno) > Is it true that X_1 union X_2 is also an AANR? any idea on how to > prove it? or reference > Surely some mistake? === Subject: Re: range conversion <7wRNRl5L7CdNS=wI5rywal=I706I@4ax.com >Hi I have a sequence of numbers within a range[0...1] which needs to be >plotted on a vertical scale ranging [0...4] because of another >sequence is plotted on the bigger range. how can I convert the smaller >range sequence to show nicely on the bigger range scale? i.e. expand its range without distorting the rate of change of the >sequence's elements? > Your question is a bit ambiguous. If you have a bunch on numbers in > the interval [0,1] and you want to stretch them to [0,4] you can > simply multiply each number by 4. But I wonder if you are talking about a 2D plot since you mention > sequences. Could you be a bit more specific? --Lynn Is this any help. I had 2 series. Road accidents and deaths. hopefully deaths are less than injuries. So, I devised my own bbc basicV progms to plot 2 series on one plot with a pth-root transformation. this uniformly shrinks or expands one range to allow logarithmic comparison etc. plus polynomial root can often be calc even when value goes to zero. e.g. y^ (1/2.3) etc. square root or cube root or any root 10-th root.. or any power. negative power.. I did some great work on the axes. calculating preferred numbers and placing these at correct distance. because i have not been satisfied with share graphs.. plot a window x min max,, y min max. don.mcdonald nz.. 19/5/07. === Subject: Re: #7 published examples of mistaken proofs by various authors; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof On May 17, 10:31 pm, sttscitr...@tesco.net First of all, VERY FUNNY AND VERY IMMATURE! For those of you who are interested, SOMEONE seems to have said or > done something which resulted in my posting privileges being > (temporarily) revoked. Evidently, someone (whom I will call Aristotle > Plutonium, since his proofs are more Aristotle-like than Archimedes- > like: heavier objects are more attracted to the earth; therefore they > fall faster) didn't understand that I had made a joke post back in > February ... But more about that later. My name popped up in this thread, and I found something even more > important that AP seems to have overlooked. > (#2) --- quoting THE BOOK OF NUMBERS John H. Conway and Richard K. Guy, > Springer-Verlag, 1996, pages 133-135 --- > There Are Always New Primes! > It's not obvious that the sequence of primes continues indefinitely. > There might come a stage, perhaps, when the sieving process stops > because all the numbers have been struck out. However, Euclid also > proved that the primes do indeed continue forever. > Imagine that all the primes you know are > 2, 3, 5, 7, 11, 13. > Then we'll show that there must be another one. Multiply your > primes together and add 1 to get the larger number > 2x3x5x7x11x13 + 1 = 30031. > This number is certainly bigger than 1. What is the smallest number, > bigger than 1, that divides it exactly? This must be a prime, > otherwise one of its factors would be a smaller candidate. But it > can't be one of the old primes 2, 3, 5, 7, 11, or 13, since > each of these leaves remainder 1 when divided into 30031. So > we've found a new prime. > Sometimes the big number here is already prime, but sometimes, > as in the previous example, it isn't: > 1 + 1 = 2 is prime > 2 + 1 = 3 is prime > 2x3 + 1 = 7 is prime > 2x3x5 + 1 = 31 is prime > 2x3x5x7 + 1 = 211 is prime > 2x3x5x7x11 + 1 = 2311 is prime > but > 2x3x5x7x11x13 + 1 = 30031 = 59x509 > and > 2x3x5x7x11x13x17 + 1 = 510511 = 19x97x277, > while > 2x3x5x7x11x13x17x19 + 1 = 9699691 = 347x27953. > The next few prime numbers of the form 2x3x5x...xp + 1 are for p = > 31, 379, 1019, 1021, and 2657. > Although we've known since Euclid that the primes get as large > as you like, it was quite a long time before mathematicians could > explicitly point to some very big ones. > --- end quoting THE BOOK OF NUMBERS John H. Conway and Richard K. Guy, > Springer-Verlag, 1996, pages 133-135 --- > Conway and Guy have a mixed up version of both direct and indirect. > Trouble is > that Conway and Guy are more wrapped up in showing number examples > than in actually giving a proof. Do they mean by Imagine as a > substitute > for Suppose as in a reductio ad absurdum? But if they did mean a > indirect method proof then their 30031 is necessarily a new prime > under the reductio ad absurdum. So theirs is a mixed up proof that > ends as invalid. [...] Conway and Guy do NOT claim that 30031 is prime; AP has clearly not > read this argument carefully, since they clearly state (quoted from > above): > Sometimes the big number here is already prime, but sometimes, > as in the previous example, it isn't. And then further state that > What is the smallest number, > bigger than 1, that divides it exactly? This must be a prime, > otherwise one of its factors would be a smaller candidate. But it > can't be one of the old primes 2, 3, 5, 7, 11, or 13, since > each of these leaves remainder 1 when divided into 30031. So > we've found a new prime. This prime is the smallest divisor of the product of the (finite) > list of primes, + 1, which is greater than 1. And since it's not in > the finite set of primes, we get a contradiction. Unless you are have proven that every n >1 has at least > one prime factor a contradiction can't be found. And this was error in _Aristotle Plutonium's_ original proof. (He doesn't mention that he failed his own test by omitting a result like this from it.) But Guy and Conway have actually shown this, when they say: > What is the smallest number, > bigger than 1, that divides it exactly? This must be a prime, > otherwise one of its factors would be a smaller candidate. This is only used in the IP proof, but also applies to any n > 1. A more thorough proof would be: Let m be the smallest factor of n which is greater than 1. If m is composite, then (according to the definition of composite) we can write m = a*b, where a > 1, b > 1, and a and b are integers. But b >= 2 means a < 2*a <= b*a = m, and clearly if m divides n, then a also divides into n. Hence, m is not the smallest factor of n which is also greater than 1 is. Contradiction; therefore, m is prime. (Note this is true whether or not n is a prime. This is the sort of cleverness that AP can't stand for.) -- Christopher Heckman === Subject: Re: #7 published examples of mistaken proofs by various authors; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof > On May 17, 10:31 pm, sttscitr...@tesco.net This prime is the smallest divisor of the product of the (finite) > list of primes, + 1, which is greater than 1. And since it's not in > the finite set of primes, we get a contradiction. Unless you are have proven that every n >1 has at least > one prime factor a contradiction can't be found. And this was error in _Aristotle Plutonium's_ original proof. (He > doesn't mention that he failed his own test by omitting a result like > this from it.) But Guy and Conway have actually shown this, when they > say: > What is the smallest number, > bigger than 1, that divides it exactly? This must be a prime, > otherwise one of its factors would be a smaller candidate. This is only used in the IP proof, but also applies to any n > 1. A > more thorough proof would be: Let m be the smallest factor of n which is greater than 1. If m is > composite, Here you are assuming what you want to prove. Unless you have a theorem that tells you every n> 1 is in fact divisible by at least one prime, then you can't conclude that any factor of n is composite, i.e. a product of two or more primes. n could be a unit or product of units. As it is a theorem that any n>1 has at least one prime divisor, you can say The product of any n primes +1 is not divisible by any of these n primes and so by the theorem there must be an n+1th prime, so you are constructing a subset of primes which is infinite which implies the set of primes is infinite or If there are only n primes, the product of these n primes +1 is not divsible by any prime, which contradicts the theorem. === Subject: Re: #7 published examples of mistaken proofs by various authors; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof On May 19, 4:58 am, sttscitr...@tesco.net On May 17, 10:31 pm, sttscitr...@tesco.net This prime is the smallest divisor of the product of the (finite) > list of primes, + 1, which is greater than 1. And since it's not in > the finite set of primes, we get a contradiction. > Unless you are have proven that every n >1 has at least > one prime factor a contradiction can't be found. And this was error in _Aristotle Plutonium's_ original proof. (He > doesn't mention that he failed his own test by omitting a result like > this from it.) But Guy and Conway have actually shown this, when they > say: > What is the smallest number, > bigger than 1, that divides it exactly? This must be a prime, > otherwise one of its factors would be a smaller candidate. This is only used in the IP proof, but also applies to any n > 1. A > more thorough proof would be: Let m be the smallest factor of n which is greater than 1. If m is > composite, Here you are assuming what you want to prove. No, I'm not; I'm assuming the opposite. I want to prove that m is prime, i.e. not composite. Then I obtain a contradiction, which shows that my last assumption (that m is composite) is wrong. This is a standard proof method called contradiction (or reducio ad absurdum). > Unless you have a theorem that tells you every n> 1 > is in fact divisible by at least one prime, then you > can't conclude that any factor of n is composite, i.e. a > product of two or more primes. n could be a unit > or product of units. That is what I'm proving here. --- Christopher Heckman === Subject: Re: #7 published examples of mistaken proofs by various authors; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof > On May 19, 4:58 am, sttscitr...@tesco.net prime, i.e. not composite. Then I obtain a contradiction, which shows > that my last assumption (that m is composite) is wrong. This is a > standard proof method called contradiction (or reducio ad absurdum). Unless you have a theorem that tells you every n> 1 > is in fact divisible by at least one prime, then you > can't conclude that any factor of n is composite, i.e. a > product of two or more primes. n could be a unit > or product of units. That is what I'm proving here. Yes, I see what you mean. You are saying that every n>1 is either irreducible or reducible. If the smallest integer m that divides n is reducible this implies there is a smaller m' <>1 which divides n, a contradiction. So the smallest number >1 that divides n must be irreducible. I'm sure I mentioned something similar to you some time ago hence the confusion. === Subject: TI-84 Plus calculator help.... howdy, How do I store a random integer between 0 and 4 in a variable? (TI-84 Plus programming) I tried: randInt(0,4,1)->A But that didn't work.....any help? === Subject: Spread the word about Saving Children Spread the word about Saving Children Easy Recipes in our Free Cookbook This downloadable cookbook is used to promote our free devotional newsletter in which a large percentage of the profits from advertisers goes to the Save the Children Charity. ***There is no obligation and no charge. *****Spread the word. Please Forward this message to everyone you know. Download Our Easy Recipes in our Free Cookbook Here - http://tinyurl.com/yvvxr2 === Subject: Re: Spread the word about Saving Children Spread the word about Saving Children Easy Recipes in our Free Cookbook [..] Why is this posted in sci.math of all places? In any case, if you're suggesting the surplus of children can be cured by killing and cooking them, the idea is not original. Jonathan Swift suggested it a couple of hundred years ago in his satirical tract A Modest Proposal. See http://www.findagrave.com/cgi-bin/fg.cgi?page=gr&GRid=1944 John R Ramsden === Subject: Re: Taking derivative with respect to another function You are right! In fact, I encountered this issue when dealing with a probability problem. This problem states that, given the joint pdf (probability density function) of two random variables X and Y, how can one determine the pdf of their sum? In other words: Given f(x,y), and Z=X+Y, what is g(z)? The solution first obtains the cdf (cumulative distribution function) of z, and then computes its derivative to obtain the pdf of z: G(z) = P{ X+Y<=z } = int(int(f(x,y), x=-inf..z-y), y=-inf..inf) Where by int(m(x), x=a..b) I denoted the definite integral of function m(x) from a to b; and inf stands for infinity. To obtain g(x), one should take derivative of both sides with respect to z, as follows (note that due to ASCII limitations, I denoted rounded derivatives with the same symbol as ordinary derivatives): g(z) = dG(z)/dz = d int(int(f(x,y), x=-inf..z-y), y=-inf..inf) / dz = int( d int(f(x,y), x=-inf..z-y) / dz, y=-inf..inf) So, all we need to do is to compute: H(z,y) = d int(f(x,y), x=-inf..z-y) / dz This is done using Leibniz formula for derivating a bivariate function under integral: H(z,y) = f(z-y,y) + int(df(x,y)/dz, x=-inf..z-y) There comes the problem. I do not know how to compute the 2nd term; i.e. int(df(x,y)/dz, x=-inf..z-y) = ??? The solution states that the above integral is ZERO. I can't figure out why. I think I have a counterexample for this: If x and y are independent exponential random variables, both with parameter L, then: f(x,y)=f(x)f(y)=L^2*exp(-L*(x+y)) and we know that z=x+y. Hence the above integral will not be zero: df(x,y)/dz = d [L^2*exp(-L*z)] / dz != 0 (Where != denotes is not equal to). I'm sure I were WRONG somewhere, because this results in contradiction. But could you please tell me what's wrong? === Subject: Re: Taking derivative with respect to another function > You are right! In fact, I encountered this issue when dealing with a > probability problem. This problem states that, given the joint pdf > (probability density function) of two random variables X and Y, how > can one determine the pdf of their sum? In other words: Given f(x,y), and Z=X+Y, what is g(z)? The solution first obtains the cdf (cumulative distribution function) > of z, and then computes its derivative to obtain the pdf of z: G(z) = P{ X+Y<=z } = int(int(f(x,y), x=-inf..z-y), y=-inf..inf) Where by int(m(x), x=a..b) I denoted the definite integral of function > m(x) from a to b; and inf stands for infinity. To obtain g(x), one should take derivative of both sides with respect > to z, as follows (note that due to ASCII limitations, I denoted > rounded derivatives with the same symbol as ordinary derivatives): g(z) = dG(z)/dz > = d int(int(f(x,y), x=-inf..z-y), y=-inf..inf) / dz > = int( d int(f(x,y), x=-inf..z-y) / dz, y=-inf..inf) So, all we need to do is to compute: H(z,y) = d int(f(x,y), x=-inf..z-y) / dz This is done using Leibniz formula for derivating a bivariate > function under integral: H(z,y) = f(z-y,y) + int(df(x,y)/dz, x=-inf..z-y) There comes the problem. I do not know how to compute the 2nd term; > i.e. int(df(x,y)/dz, x=-inf..z-y) = ??? The solution states that the above integral is ZERO. I can't figure > out why. I think I have a counterexample for this: If x and y are independent exponential random variables, both with > parameter L, then: f(x,y)=f(x)f(y)=L^2*exp(-L*(x+y)) and we know that z=x+y. Hence the above integral will not be zero: df(x,y)/dz = d [L^2*exp(-L*z)] / dz != 0 (Where != denotes is not equal to). I'm sure I were WRONG somewhere, because this results in > contradiction. But could you please tell me what's wrong? Let g(z) be the density of X+Y. It is given by g(z) = int(f(x,z-x),x=-infinity..infinity) = int(f(z-y,y), y=- infinity..infinity). This is easy to get by looking at P{z < X+Y < z +dz} directly, or by looking at a discrete version of the problem and then passing to a continuous limit. You can also get it from your method, if care is used. g(z) = dG(z)/dz = (d/dz) int(int(f(x,y), x=-inf..z-y), y=-inf..inf) = int( d int(f(x,y), x=-inf..z-y) / dz, y=-inf..inf). When you take (d/dz) inside the y-integration, you end up with the However, your next step is incorrect: z appears ONLY in the upper limit of integration, so you should get ONLY f(z-y,y). There is no uncomputable extra term arising from df(x,y)/dz. Look at it this way: we eventually integrate over y, but for any of the y-values, y is fixed inside the x-integral. Therefore, you just have a one-variable problem involving x alone, so you just want to compute something of the form d/dz int(h(x), x=-infinity .. z-y). This is easy, just from Calculus 101: it equals h(z-y). R.G. Vickson === Subject: Re: Please help me correct this typo > I am aware of a few more typos in this volume, but this one is especially > troubling. On page 131 in the paragraph beginning But now there is a > difficulty, there is an expression beginning with a_0.a_1a_2a_3... which > is clearly mathematical gibberish. Years ago I checked the original > German > and found the correct form of the expression. Unfortunately, I do not > have > that available to me any longer. Does anybody have an idea what might be > intended here: > > http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 .djvu If I recall correctly, the expression in the German original the '=' was before the Sigma, and there was a '-' where the '=' appears in the English translation. I believe what follows the '-' should be a_0.a_1a_2a_3...a_k-1, (a_k,a_k+1...=0)??? Note that this term is a number in base g 'decimal' notation. I do clearly recall that the transcription was horribly incorrect. I believe there was an '=' somewhere to the left of the Sigma as well. It may be that the entire right-hand side of the equation should be identical to the expression for d given in the subsequent block of text. Clearly, for the context, we are dealing with the problem of how to interpret expressions of the form (in base 10) a.b...999(repeated). I am livid about not being able to access this book at the University of Maryland. If anybody reading this has access to the resources of a real university, and it is not terribly inconvenient, I would be extremely grateful if you would help me resolve this by looking at the original text. I am truly dismayed that these volumes have not received more attention by the mathematics community. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Is It True About Math >I was in the math building today and overheard some math graduate > students saying modern math is all done with smoke and mirrors. This > is probably some sort of inside joke, but does anyone know what they > meant? > vast majority of the details are commonplace, mechanical, and most importantly standard; thus are omitted. It does give an air of magic to the results however, and someone who is not an expert (such as a grad student, or indeed an interested professor reading in other fields) will have this feeling. -- Glen === Subject: Re: Is It True About Math > I was in the math building today and overheard some math graduate > students saying modern math is all done withsmokeandmirrors. This > is probably some sort of inside joke, but does anyone know what they > meant? A lot of graduate students, when out of earshot of their professors, discuss the unreality of much of modern mathematics. The basic idea is that the real world is finite (or at least, we can only take in a finite amount of information), and so only the mathematics of the finite really matters. The infinite is a useful abstraction as long as it helps us reason about the finite world, but modern mathematics studies the infinite for its own sake without concern over whether it helps us understand the finite world. That might be what they were talking about. === Subject: Re: inequality anyone? Darren > Need help with proving this inequality (to do with proving that a > power-series is analytic): |frac{z^{k+1} - b^{k+1}}{z-b} - (k+1)b^k | le frac{(|b| + |z-b|)^{k+1} - |b|^{k+1}}{|z-b|} - (k+1)|b|^k le (k+1)[(|b| + |z-b|)^k - |b|^k] > I start with |z| le |b| + |z-b|, and get the first term of the second > line. But why subtract the term (k+1)|b|^k ? Wouldn't you add for a > triangle inequality? And, how do you go from the 2nd line to the 3rd line? The 3rd line can be re-written: (k+1)[(|b| + |z-b|)^k - |b|^k] = k(|b| + |z-b|)^k + (|b| + |z-b|)^k - > (k+1)|b|^k The 2nd and 3rd term on the r.h.s can easily be gotten, but I am > having trouble getting k(|b| + |z-b|)^k. > I can get the second inequality. First let's get rid of all those absolute value signs, by writing c = |b| d = |z-b| so we want > frac{(|b| + |z-b|)^{k+1} - |b|^{k+1}}{|z-b|} - (k+1)|b|^k > le (k+1)[(|b| + |z-b|)^k - |b|^k] (c + d)^{k+1} - c^{k+1} - (k+1) d c^k <= (k+1) d [ (c+d)^k - c^k ] i.e. (c + d)^{k+1} - c^{k+1} <= (k+1) d (c+d)^k where c and d are nonnegative. Just write (c + d)^{k+1} and (c+d)^k as sums involving binomial coefficients, and then compare the two above polynomials termwise. It comes down to showing (k+1) choose m <= (k+1)(k choose m) which is easy. LH === Subject: Re: Solution manual ( free ) Solution manual of textbooks in ebook format! Get it in hours! > I have the complete electronic SOLUTION MANUAL in PDF format > containing ALL (odd and even) solutions for the books listed below. > These are not paper books, they are ebooks. The price is $8 each. I accept payment through paypal only. If > interested, just send me an email to books...@gmail.com mentioning > which solution manual you want. I will promptly reply you with where > you can send the paypal payment. As soon as the payment is received, a > download link for the solution manual will be sent to your email > address within a few hours. Fast service! Books for which I have electronic solution manual: > 1. Calculus: Early Transcendentals (5th Ed) by James Stewart [ISBN: > 0534393217] > 2. Digital Communications (4th Ed) by Proakis, John [ISBN: > 0072321113] > 3. Electric Circuits (7th Ed) by Nilsson, James W.; Riedel, Susan > [ISBN:0131465929] > 4. Computer Networks (4th Ed) by Tanenbaum, Andrew S. [ISBN: > 0130661023] > 5. Electric Machinery Fundamentals (4th Ed) by Stephen J. Chapman > [ISBN: 0072465239] > 6. Discrete-Time Signal Processing (2nd Ed) by Oppenheim, Alan V > [ISBN: 0137549202] > 7. Engineering Mechanics , Dynamics (5th Ed) by J. L. Meriam, L. G. > Kraige [ISBN: 0471406457] > 8. Engineering Mechanics - Statics (10th Ed) by Russell C. Hibbeler > [ISBN: 0131411675] > 9. Engineering Mechanics - Dynamics (10th Ed) by Russell C. Hibbeler > [ISBN: 0131416782] > 10. Control Systems Engineering (4th Ed) by Nise, Norman S. [ISBN: > 0471445770] > 11. Adaptive Filter Theory (4th Ed) by Haykin, Simon [ISBN: > 0130901261] > 12. Introduction to Electrodynamics (3rd Ed) by Griffiths, David > [ISBN: 013805326X] > 13. First Course in Probability, A (7th Ed) by Ross, Sheldon [ISBN: > 0131856626] > 14. Vector Mechanics for Engineers, Statics (7th Ed) by Beer, > Ferdinand P.; Johnston Jr., E [ISBN: 0072930780] > 15. Microwave and Rf Design of Wireless Systems (1st Ediiton) by > Pozar, David M. [ISBN: 0471322822] > 16. Communications Systems (4th Ed) by Haykin, Simon [ISBN: > 0471178691] > 17. Automatic Control Systems (8th Ed) by Kuo, Benjamin C.; > Golnaraghi, Farid [ISBN: 0471134767] > 18. Microwave Engineering (3rd Ed) by Pozar, David M. [ISBN: > 0471448788] > 19. Fundamentals of Electric Circuits (2nd Ed ) by Charles Alexander, > Matthew Sadiku [ISBN: 0073048356] > 20. Wireless Communications: Principles and Practice (2nd Ed) by > Theodore Rappaport [ISBN: 0130422320] > 21. Fundamentals of Electromagnetics with Engineering Applications by > Stuart M. Wentworth [ISBN: 0471263559] > 22. Communication Systems Engineering (2nd Ed) by Proakis, John G > [ISBN: 0130617938] > 23. Digital Communications: Fundamentals and Applications (2nd Ed) by > Bernard Sklar [ISBN: 0130847887] > 24. Principles of Communication: Systems, Modulation and Noise (5th > Ed) by R. E. Ziemer, W. H. Tranter [ISBN: 0471392537] > 25. Computer Networks: A Systems Approach (3rd Ed) by Larry L. > Peterson, Bruce S. Davie [ISBN: 155860832X] > 26. Device Electronics for Integrated Circuits (3rd Ed) by Richard > Muller, Theodore Kamins, Mansun Chan [ISBN: 0471593982] > 27. Linear Circuit Analysis: Time Domain, Phasor, and Laplace > Transform Approaches (2nd Ed) by Raymond DeCarlo, Pen-Min Lin [ISBN: > 0195136667] > 28. Introduction to Solid State Physics (8th Ed) by Charles Kittel > [ISBN: 047141526X] > 29. Introduction to Quantum Mechanics (2nd Edition) by David J. > Griffiths [ISBN: 0131118927] > 30. Digital Signal Processing by Thomas J. Cavicchi [ISBN: > 0471124729] > 31. Engineering Electromagnetics (6th Ed) by William H. Hayt, John A. > Buck [ISBN: 0072551666] > 32. Physics for Scientists and Engineers (6th Ed) by Raymond A. > Serway, John W. Jewett [ISBN: 0534408427] > 33. Antennas for All Applications (3rd Ed) by John D. Kraus, Ronald J. > Marhefka [ISBN: 0072321032] > 34. Advanced Modern Engineering Mathematics (3rd Ed) by Glyn James > [ISBN: 0130454257] > 35. Digital Image Processing (2nd Ed) by Rafael Gonzalez, Richard > Woods [ISBN: 0201180758] > 36. Physical Chemistry (7th Ed) by Julio Paula, Peter Atkins [ISBN: > 0716735393] > 37. Semiconductor Device Fundamentals (1st Ed) by Robert F. Pierret, > [ISBN: 0201543931] > 38. RF Circuit Design: Theory and Applications by Reinhold Ludwig, > Pavel Bretchko [ISBN: 0130953237] > 39. Fundamentals of Physics (7th Ed) by David Halliday, Robert > Resnick, Jearl Walker [ISBN: 0471216437] > 40. Fundamentals of Thermodynamics (5th Ed) by Richard E. Sonntag, > Claus Borgnakke, Gordon J. Van Wylen [ISBN: 047118361X] > 41. Signal Processing and Linear Systems by B. P. LAthi [ISBN: > 0195219171] > 42. Fundamentals of Digital Logic with Verilog Design (1st Ed) by > Stephen Brown, Zvonko Vranesic [ISBN: 0072838787] > 43. Feedback Control of Dynamic Systems (4th Ed) by Gene Franklin, > David Powell, Abbas Naeini [ISBN: 0130323934] > 44. Design of Analog CMOS Integrated Circuits (1st Ed) by Behzad > Razavi [ISBN: 0072380322] > 45. Signals and Systems (2nd Ed) by Alan V. Oppenheim, Alan S. Willsky > [ISBN: 0138147574] > 46. Electronic Circuit Analysis (2nd Ed) by Donald Neamen [ISBN: > 0072451947] > 47. Applied Numerical Analysis (7th Ed) by Curtis F. Gerald, Patrick > O. Wheatley [ISBN: 0321133048] > 48. Engineering Circuit Analysis (6th Ed) by William H. Hayt, Jack > Kemmerly, Steven M. Durbin [ISBN: 0072853204] > 49. Introduction to Electric Circuits (6th Ed) by Richard C. Dorf, > James A. Svoboda [ISBN: 0471447951] > 50. Semiconductor Physics And Devices (3rd Ed) by Donald Neamen [ISBN: > 0072321075] > 51. Digital Signal Processing (4th Ed) by John G. Proakis, Dimitris K > Manolakis [ISBN: 0131873741] > 52. Vector Mechanics for Engineers: Dynamics (7th Ed) by Beer, > Ferdinand P.; Johnston Jr., E [ISBN: 0073209260] > Wasn't that supposed to be free??? AA' i am interested in the followin 2 books: > 8. Engineering Mechanics - Statics (10th Ed) by Russell C. Hibbeler > [ISBN: 0131411675] > 9. Engineering Mechanics - Dynamics (10th Ed) by Russell C. 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(<464e8928$0$18538$834e42db@reader.greatnowhere.com>) === Subject: * Sizzling Boobies all new pics and Vidz http://nudepicks.blogspot.com/ - Download them often and get them all! === Subject: Re: * Sizzling Boobies all new pics and Vidz > http://nudepicks.blogspot.com/ - Download them often and get them all! And after doing that we need the helpdesk? -- FREEDOM OF SPEECH, USE IT OR LOSE IT http://communities.zeelandnet.nl/data/novamente/index.php?page=13&showpage=5 2538 ----------------- www.Newsgroup-Binaries.com - *Completion*Retention*Speed* Access your favorite newsgroups from home or on the road ----------------- === Subject: cancel: * Sizzling Boobies all new pics and Vidz Excessive crossposting. === Subject: Re: Skolem Paradox Question > Nam D. Nguyen says... > > I might have *specified* all of what you've said including that > structure of the language, according to Shoenfield. But to him, *after* > constructing a language structure M with all the mappings mentioned above, > you *still have to interpret a formula F in M*: We want to define a formula > F to be valid in M *_if_ all the meanings of F are true in M*. > > I don't have Schoenfield, but I think you are misunderstanding him. > Some authors distinguish between a model and an interpretation > in the way they treat *free variables*. > > A model gives an interpretation to each constant symbol, function > symbol, and relation symbol. The model does *not* give an interpretation > to *variables*. So an open sentence such as > > x > y > > is neither true nor false in a model. Iirc, in my original post the formula in question is a < b where a,b a constants of the language L(a,b,<). So I don't think variables really much affect what I intend to say here. (This is a minor point imho). > > An *interpretation*, on the other hand, assigns a value to > each variable, as well. I've been using model and interpretation > interchangably, because I didn't think that this distinction was > important for this discussion. But if that's the distinction you > mean, then we can certainly talk about it. > > When one distinguishes between interpretations and models, > then one typically says that a formula Phi is *valid* in > model M if Phi is true in every interpretation extending > M. It really boils down to this crucial question: does FOL allow for individual subjectivity, in saying a formula F is true in a language structure M? I think it does allow, albeit rather implicitly at least in Shoenfield's book. Shoenfield used set of (object) n-tuples in talking about n-ary relations (functions, predicates) in a structure M. Where set, object, n-tuple are intuitive concepts and not something formally syntactical such as ZF sets, ZF n-order tuples , etc... But since, eg, the 2-tuple in this example is an intuitive concept, one is free to interpret it as *a is the 1st element*, or alternatively *b is*. The question is why we could be sure of this subjectivity? The short answer is if this subjectivity were not allowed, then language structures/models would become *another fixed syntactical form of reasoning*, rivaling the syntactical inference or provability, which is not an intention the founders of FOL would have in mind, imho. The longer answer is GIT1 is a *stopgap proof* when Godel failed to syntactically prove that PM, or any T as strong as arithmetic, be incomplete. The core part of Hilbert's formalism is complete detachment of intuition/meaning from reasoning: formulae and provability are sheer games of syntactical manipulation. In such syntactical formalism, it seems that Godel couldn't advance the proof of GIT1 (and others?). Hence Godel invented a work around: - Firstly, if syntactically a T is defined as incomplete iff T proves neither F nor ~F (for some F), then Godel invented model (and model truth) in such a way that T's syntactical incompleteness would be equivalent to there exist 2 models of T, in which F be true in one and false in the other. - Secondly, Godel *bypassed* the syntactical definition of arithmetic and *assumed* that there be an arithmetic model (the naturals) already. He needed this bypass so when encoding T using the language of arithmetic, he could equate the model's truth (or falsehood) of T with those of arithmetic. But by allowing a formula be subjectively being true or false in some models, Godel allowed *subjective intuition* to have an equal, and *dangerously rival*, status as that of formal syntactical provability! In the first place, when a structure of a language is finite, subjectivity of interpretation still can render a formula F to be interpreted differently (as true, or false). In the second place, when a language structure is infinite and as complex as the concept of arithmetic, there could be formulae in which it's *really beyond* human intuition to grasp the arithmetic truth of the formulae. (And I don't mean the arithmetic proof of such formula is a big finite proof: I mean if there were such proof it would be infinite! which is of course outside the scope of FOL). === Subject: Re: Skolem Paradox Question <3QAXh.125177$aG1.42708@pd7urf3no> <6SIXh.125961$6m4.58236@pd7urf1no> <28WXh.129711$DE1.51920@pd7urf2no> <6kVYh.141970$DE1.50543@pd7urf2no> <2Ib2i.184845$6m4.177219@pd7urf1no> individual subjectivity, in saying a formula F is true in a language > structure M? I think it does allow, albeit rather implicitly at least > in Shoenfield's book. Shoenfield used set of (object) n-tuples in talking > about n-ary relations (functions, predicates) in a structure M. Where set, > object, n-tuple are intuitive concepts and not something formally > syntactical such as ZF sets, ZF n-order tuples , etc... But since, eg, > the 2-tuple in this example is an intuitive concept, one is free > to interpret it as *a is the 1st element*, or alternatively *b is*. Model theory can be formalized in, e.g., a formal Z set theory. So there is no need for your obfuscation about subjectivity. > The question is why we could be sure of this subjectivity? The short answer is if this subjectivity were not allowed, then language > structures/models would become *another fixed syntactical form of reasoning*, > rivaling the syntactical inference or provability, which is not an intention > the founders of FOL would have in mind, imho. Who are the founders of FOL that you have in mind? And what is the basis for your opinion about what they had in mind or would have had in mind about model theory itself being formalized? Anyway, whatever they might have thought about such issues does not determine that we should not formalize model theory in a formal theory or at least recognize that model theory is in principle formalizable. > The longer answer is GIT1 is a *stopgap proof* when Godel failed to > syntactically prove that PM, or any T as strong as arithmetic, be incomplete. Without my remarking on Godel himself, you do realize that we don't need to invoke any semantical notions to prove that there is no complete, consistent, recursively axiomatizable extension of PA, right? > The core part of Hilbert's formalism is complete detachment of > intuition/meaning from reasoning: formulae and provability are sheer games of > syntactical manipulation. In such syntactical formalism, it seems that Godel > couldn't advance the proof of GIT1 (and others?). Hence Godel invented a work > around: What is the basis for your claim that Godel could not prove incompleteness in a syntactical regard or whatever it is you are claiming? > - Firstly, if syntactically a T is defined as incomplete iff T proves neither > F nor ~F (for some F), then Godel invented model (and model truth) in such > a way that T's syntactical incompleteness would be equivalent to there exist > 2 models of T, in which F be true in one and false in the other. What specific Godel paper on incompleteness do you have in mind? I admit my recollection of his own proof is not fresh, so you could do me a favor by pointing out where in his incompleteness proof there is a dependence on such models as you mention. > - Secondly, Godel *bypassed* the syntactical definition of arithmetic What syntactical definition is that? > and > *assumed* that there be an arithmetic model (the naturals) already. He assumed a MODEL? Or he just assumed some basic mathematics about natural numbers? > He needed > this bypass so when encoding T using the language of arithmetic, he could > equate the model's truth (or falsehood) of T with those of arithmetic. Again, where is this part about models? And what do you mean by truth or falsehood of a model?. What would a true model or a false model be? Especially, where did Godel say anything about the truth of falsehood of a model (whatever 'true model' and 'false model' might mean)? > But by allowing a formula be subjectively being true or false in some models, > Godel allowed *subjective intuition* to have an equal, and *dangerously rival*, > status as that of formal syntactical provability! Again, Where did he do that? > In the first place, when a > structure of a language is finite, subjectivity of interpretation still can > render a formula F to be interpreted differently (as true, or false). No, for a given model, there is no subjectivity that can affect whether a sentence is true or whether it is false in that model. > In the > second place, when a language structure is infinite and as complex as the > concept of arithmetic, there could be formulae in which it's *really beyond* > human intuition to grasp the arithmetic truth of the formulae. Okay, with certain models with infinite universes it is not computable whether any given sentence is true or whether it is false in that model. We all know that. But I don't know how you think that fact supports whatever thesis it is that you're trying to convey. > (And I don't > mean the arithmetic proof of such formula is a big finite proof: I mean if > there were such proof it would be infinite! which is of course outside > the scope of FOL In a given system (a meta-system), which can itself be a first order theory, there may or may not be a proof (which is always finite) that a certain sentence is true in a given model (of an object level theory). That doesn't entail that there is no system in which it can be proven that said sentence is true in said model. MoeBlee === Subject: Re: Skolem Paradox Question <3QAXh.125177$aG1.42708@pd7urf3no> <6SIXh.125961$6m4.58236@pd7urf1no> <28WXh.129711$DE1.51920@pd7urf2no> <6kVYh.141970$DE1.50543@pd7urf2no> <2Ib2i.184845$6m4.177219@pd7urf1no> basis for your opinion about what they had in mind or would have had > in mind about model theory itself being formalized? Anyway, whatever > they might have thought about such issues does not determine that we > should not formalize model theory in a formal theory or at least > recognize that model theory is in principle formalizable. P.S. I don't mean to put words in your mouth that you are claiming that certain pioneers of first order logic did or would (had model theory been contemporaneous with them) disapprove of the formalization of model theory. But at least that seems to me to be close to, even if not exactly, the upshot of your claim. > In a given system (a meta-system), which can itself be a first order > theory, That should be [...], which can itself be a first order system. MoeBlee === Subject: Re: Skolem Paradox Question Nam D. Nguyen says... >But by allowing a formula be subjectively being true or false in some models, >Godel allowed *subjective intuition* to have an equal, and *dangerously rival*, >status as that of formal syntactical provability! No, there is nothing subjective about truth in a model. >In the first place, when a >structure of a language is finite, subjectivity of interpretation still can >render a formula F to be interpreted differently (as true, or false). No, that's not correct. If you specify the structure, then you've specified the interpretation. -- Daryl McCullough Ithaca, NY === Subject: Re: Skolem Paradox Question > Nam D. Nguyen says... > > But by allowing a formula be subjectively being true or false in some models, > Godel allowed *subjective intuition* to have an equal, and *dangerously rival*, > status as that of formal syntactical provability! > > No, there is nothing subjective about truth in a model. > > In the first place, when a > structure of a language is finite, subjectivity of interpretation still can > render a formula F to be interpreted differently (as true, or false). > > No, that's not correct. If you specify the structure, then > you've specified the interpretation. > > -- > Daryl McCullough > Ithaca, NY > Just keep saying No doesn't advance your argument, imho. === Subject: Re: Skolem Paradox Question Nam D. Nguyen says... >It really boils down to this crucial question: does FOL allow for >individual subjectivity, in saying a formula F is true in a language >structure M? And the answer is clearly no. A formula can have multiple interpretations, but once you've specified a structure, you've *specified* an interpretation. Of course, when you specify an interpretation, you specify it in some *language*. So that language itself is open to multiple interpretations. >I think it does allow, albeit rather implicitly at least >in Shoenfield's book. Shoenfield used set of (object) n-tuples in talking >about n-ary relations (functions, predicates) in a structure M. Where set, >object, n-tuple are intuitive concepts and not something formally >syntactical such as ZF sets, ZF n-order tuples , etc... But since, eg, >the 2-tuple in this example is an intuitive concept, one is free >to interpret it as *a is the 1st element*, or alternatively *b is*. No, I don't think so. I don't think that's what Schoenfield was saying. -- Daryl McCullough Ithaca, NY === Subject: cancel: Re: NUDE PICS! FREE DOWNLOAD Excessive crossposting. === Subject: Re: Challenge <59bq43ddtj4uib05d1odt412qcr2ve16p0@4ax.com> <3ndr435ge1i3bat65n04q2q972bigoeij7@4ax.com> Hi all, > how can one find a multiplier m for any given integer N > such that the product m*N can be represented > by m*N=16*x^4+16*x^3+(4+16*i)*x^2+8*i*x+4*(i+1)^2-1 > the unknowns m,x and i being integers. > (i do not have a solution and believe it cannot be solved) > Gerry > What are the restrictions on x, m, i? Or is the question: for which > x,i is there an m such that . . .? Or is it: for which m is there an > x,i...? > At first glance, the right-hand side is always odd, so if N is even > there cannot be a solution. > At first glance, the right-hand side is always odd, so if N is even > there cannot be a solution. > Yes good remark . > Here are some examples : > N=208083;m=1;i=18;x=10; > N=2080831;m=925;i=2036;x=-100; > N=20808311;m=105;i=7261;x=-90; > N=1217381787111313131;m=117861;i=35367772818;x=277513; > Gerry- Hide quoted text - > - Show quoted text - >Hi >i forgot to mention that only coprime values for N of interest so >again: > What do you mean by coprime values for N? > Coprime to what? > quasi- Hide quoted text - > - Show quoted text - >Hi Quasi > thank you for your question >What do you mean by _coprime values for_ N? > you are right of course i should have mentioned GCD(N,m)=1 > so for the examples only the first three are good solutions > the last one does not count > Would this be better? > Challenge: > How can one find a multiplier m for any given COPRIME N > such that gcd(N,m)=1 and the product m*N can be represented > by m*N=16*x^4+16*x^3+(4+16*i)*x^2+8*i*x+4*(i+1)^2-1 > the unknowns m,x and i being integers > and x has to be different from zero. >OK N=208083, m=1 GCD(N,m)=1 >OK N=2080831, m=925 GCD(N,m)=1 >OK N=20808311, m=105 GCD(N,m)=1 >NOK N=1217381787111313131, m=117861 GCD(N,m)=3 > Gerry > Provided N is odd, such integers m,x,i can always be found. > Choose m such that > mN =3 (mod 8), > mN not equal to 3, > gcd(m,N)=1 > Since n is odd, such an m is easily found. In fact, there are > infinitely many such values of m. > Let x=(mN-3)/8 > and let i=x-2x^2 or i=-2*x^2-3x-2, it doesn't matter which. > The required equation is then identically satisfied. > quasi- Hide quoted text - > - Show quoted text - Hi Quasi, your solutions are correct thank you, >but can you also find solutions like the following example: N=20808311, m=105 GCD(N,m)=1 mN=2184872655 >N1=2184872655 ,i=7261 ,x=-90 >(with N1=16*x^4+16*x^3+(4+16*i)*x^2+8*i*x+4*(i+1)^2-1) By using your method i found: N=20808311, m=5 GCD(N,m)=1 mN=104041555 >N1=104041555 ,i=-338270128950078 ,x=13005194 The reason is the following: By adding one to x for my example you can factor N because >GCD(N(x,i),N(x+1,i))>1 and N1=2118634875 ,i=7261 ,x=-89 > N1=2118634875 , GCD(N,N1)=9277 This doesn't work for your solutions or could it? N=20808311, m=5 GCD(N,m)=1 mN=104041555 N1=21649291061926935 , > i=- 338270128950078 ,x=13005195 > N1=21649291061926935 , GCD(N,N1)=20808311 Is it possible to restrict the variables such that the solutions >always allow to factor N? Sure, the solutions can yield a factorization of n, but there's no way > to find those solutions unless you already know the factors of N. I can sense that you're looking for a trap door -- a way to factor N > easily by some simple algebraic identity. Forget it, you're looking > for fool's gold. At some point, if you play with identities long enough, you'll gain > the realization that the power of identities is mostly illusory. In > other words, if an identity can be proved easily using only elementary > algebra, don't expect it solve any famous unsolved problems. quasi- Hide quoted text - - Show quoted text - Hi Quasi, yes the posted quartic f(x,i)=16(x^4+x^3)+(4+16i)x^2+8ix+4(i+1)^2-1 allows to factor N for some combinations x,i but apparently like you pointed out not always. It surprised me that for any x,i GCD( f(x,i),f(x+1,i))>1 and GCD( f(x-1,i),f(x),i)>1, and there seem to be more quartics which have this property. Gerry === Subject: Re: Sheaves and the empty set particular open covering as you have chosen with the product set with > the empty indexing set. So you should take the maximum covering which > would include the emptyset, hence the covering would be nonempty. > If I understand correctly by what you mean by the maximum covering, you see this only works when your covering cannot be the empty covering. If your coverings cannot be the empty covering there is a canonical map from the product of those section sheaves of one covering to a finer covering i.e. if {V_i}_I is a finer covering than {U_j}_J you have the following canonical maps F(U) --> prod_J F(U_j) ---> prod_I F(V_i) and then you have your exact sequence because the maps F(U) --> prod_I F(V_i) is the composition of the above But if the empty covering can cover your open set, then the composition is not the same (the composition in rings becomes the zero map, whilst the canonical map F(emptyset)--> prod_I F(V_i) ... V_i=emptyset becomes the identity map, or just the products of them. Jose Capco > -- -kira === Subject: Re: Sheaves and the empty set procedure should keep for all U > F(U) = Z. > Then it should be fine even if there are disjoint open sets. I do realize the problem when trying to speak of sheaves and when one knows of the naive definition of sheaves of rings. Therefore some authors define sheaves requiring F(emptyset)=0 the zero ring/group. Some even define this at once on presheaves. But if you know the more rigorous definition using products and so on, you won't face this problem. An ellaborate explanation on this using the naive definition but with more details on the covering (i.e. the author tries to define coverings very precisely) can be found in the Right sorry, its fine if there are -nonempty- disjoint open sets, but when you deal with the empty set you get the problem. To try to mend this problem when I say a family of f_i in F(U_i) have the gluing property over a covering {U_i}_iin I of U with U_isubset U, I mean that f_i|U_ij = f_j| U_ij for all i,j in I with U_ij=U_i/ U_j. And when I say that these families can be glued together I would mean that there exists a *unique* (which need not be said, because it follows from the first sheaf axiom in the naive definition) f in F(U) such that f|U_i = f_i for all i in I. Now I can take U_i=emptyset , and I=emptyset, I still have a covering of the emptset. And trivially *any* family of f in F(emptyset) has the gluing property remember f_i|U_ij=f_j|U_ij for all ***i,j in I **** the above condition is fulfilled for any family of f in F(emptyset) because there is no i,j in I , the above becomes an empty condition. Now what happens is that there exists indeed an f in F(emptyset) such that f|U_i = f_i for all *** i in I *** (1) but the first sheaf axiom in the naive definition is now violated.. the first sheaf axiom says that f|U_i = g|U_i for all ***i in I*** THEN f=g (2) clearly any f satisfies (1) because we have an empty covering, but then (2) is violated, because this f is not unique anymore. I think the explanation above is the best one can do when one has this naive definition. An easier explanation is when you have the equalizer (or the exact sequence as you call it).. If you have the empty covering the product becomes empty, and empty product is the zero ring. Well, I hope this elaborate explanation has helped you a bit.. but now for the sheafification... F in your definition is a presheaf there is no question to that (unless one defines presheaf with F(emptyset)=0 ). But you are realizing the sheafification incorrectly. The sheafification of F is the sheaf induced by the family of rings F'(U), where F'(U):={f:U --> Z : f locally constant} at emptyset F'(emptyset) consists trivially of one point. Note that the presheaf morphism between a presheaf and the sheaf need not be monomorphism at each section (i.e. there are U by which F(U)->F'(U) is not a monomorphism). > But the definition calls for any *arbitrary* open covering, not just a > particular open covering as you have chosen with the product set with > the empty indexing set. So you should take the maximum covering which > would include the emptyset, hence the covering would be nonempty. > Yes, but when you want to prove that something is NOT a sheaf, then you choose *some* open covering and family of f_i's by which you cannot glue. > -- -kira Jose Capco === Subject: Re: Sheaves and the empty set Please be patient with me, but I still have some questions about your line of thoughts. On 2007-05-19 03:38:34 -0400, Jose Capco said: > > To try to mend this problem when I say a family of f_i in F(U_i) have > the gluing property over a covering {U_i}_iin I of U with U_isubset > U, I mean that f_i|U_ij = f_j| U_ij for all i,j in I with U_ij=U_i/ > U_j. And when I say that these families can be glued together I would > mean that there exists a *unique* (which need not be said, because it > follows from the first sheaf axiom in the naive definition) f in F(U) > such that f|U_i = f_i for all i in I. Now I can take U_i=emptyset , > and I=emptyset, I still have a covering of the emptset. And trivially > *any* family of f in F(emptyset) has the gluing property > > remember > > f_i|U_ij=f_j|U_ij for all ***i,j in I **** > > the above condition is fulfilled for any family of f in F(emptyset) > because there is no i,j in I , the above becomes an empty condition. But if the indexing set I is empty, would not the family of f, which is indexed by I, necessarily be empty too? On the other hand, if you begin with a *nonempty* family of f, for which you want to check if it has the gluing property, then would not the indexing set I for that family of f be nonempty also, in which case the product prod_{i in I} F(U_i) be nonempty? So, if I is nonempty, then there is no problem with having any nonempty family of arbitrary set of f satisfying the gluing property. But if I is empty, there is no member in the empty family of f to glue to begin with. > Now what happens is that there exists indeed an f in F(emptyset) such > that > > f|U_i = f_i for all *** i in I *** (1) > > but the first sheaf axiom in the naive definition is now violated.. > the first sheaf axiom says that > > f|U_i = g|U_i for all ***i in I*** THEN f=g (2) > > clearly any f satisfies (1) because we have an empty covering, > but then (2) is violated, because this f is not unique anymore. > > > I think the explanation above is the best one can do when one has this > naive definition. > An easier explanation is when you have the equalizer (or the exact > sequence as you call it).. If you have the empty covering the product > becomes empty, and empty product is the zero ring. > > Well, I hope this elaborate explanation has helped you a bit.. -- -kira === Subject: Re: Sheaves and the empty set its ok, you can ask as much as you want, I am trying to give a crash course on sheaves and its to my benefit too that I can explain it all very well :) > But if the indexing set I is empty, would not the family of f, which is > indexed by I, necessarily be empty too? Yes you are definitely right, I made a mistake there you have an empty family..Sorry! But still you have to check for sheaveness using the two axioms (I won't call it the naive definition anymore, I think it deserves a better name). The first axiom cannot be satsified, the first axiom is the identity axiom, i.e. if f and g agrees locally for any open covering, then they agree throghout. Formally: Identity axiom of sheaf of rings: If /_{iin I} U_i = U with U_i open and if for f,g in F(U), f|U_i = g|U_i for all i in I then f=g in F(U) > On the other hand, if you > begin with a *nonempty* family of f, for which you want to check if it > has the gluing property, then would not the indexing set I for that > family of f be nonempty also, in which case the product prod_{i in I} > F(U_i) be nonempty? Yes, but I want it to be an empty family.. Actually its my mistake, I kept on saying gluing there and gluing here .. which is the first thing a person tries to think of when he tries to check sheaf.. but it's actually the identity axiom that is not fulfilled. Sorry, my bad. the first condition will not be fulfilled, even if the family is empty.. Let us fix ourselves with your example F(U)=Z for all open U and let us try to prove this in detail. Let f,g be ANY element in F(emptyset)=Z say f=1 and g=0. Let {U_i}_i=emptyset, i.e. the empty covering (thus there are no U_i's!) be the covering of U=emptyset ... we can do this because in set theory /_i=emptyset U_i = emptyset f|U_i = g|U_i for any i in I=emptyset is true for any f,g in your constant presheaf F ... (because we choose I=emptyset) but although f=1 and g=0 fullfills the above, by the identity axiom of sheaves we must have 1=0 which is obviously a contradiction. So, if I is nonempty, then there is no problem with having any nonempty > family of arbitrary set of f satisfying the gluing property. But if I > is empty, there is no member in the empty family of f to glue to begin > with. If I is nonempty then we can find a contradiction, the only vital point is when I is empty. If I were empty, there are no member to glue to begin with. So you are right, the second axiom is fulfilled trivially (second axiom is the gluing axiom). But the first axiom is not fulfilled. Some people write the two sheaf axiom into ONE axiom Definition of sheaf using only one axiom: A sheaf is a presheaf such that ... For any U and any covering U_i (including the empty ones) if f_i's have the gluing property (if they were empty then they have it obviously) then there is a UNIQUE f in U such that (if now f_i's were empty this uniqueness will be violated) f|U_i = f_i Feel free to ask some more if there is anything that you find unclear =) Jose Capco === Subject: Re: Sheaves and the empty set On 2007-05-19 07:04:14 -0400, Jose Capco said: > > Identity axiom of sheaf of rings: > If /_{iin I} U_i = U with U_i open > and if for f,g in F(U), f|U_i = g|U_i for all i in I > then f=g in F(U) > > the first condition will not be fulfilled, even if the family is > empty.. Let us fix ourselves with your example F(U)=Z for all open U > and let us try to prove this in detail. > > Let f,g be ANY element in F(emptyset)=Z say f=1 and g=0. Let > {U_i}_i=emptyset, i.e. the empty covering (thus there are no U_i's!) > be the covering of U=emptyset ... we can do this because in set theory > /_i=emptyset U_i = emptyset > > f|U_i = g|U_i for any i in I=emptyset is true for any f,g in your > constant presheaf F ... (because we choose I=emptyset) > > but although f=1 and g=0 fullfills the above, by the identity axiom of > sheaves we must have 1=0 which is obviously a contradiction. Ok. I see. Since an empty covering does cover the emptyset, the identity axiom for a sheaf is *vacuously* true for any pair of arbitrary elements in F(emptyset). Hence any two member of F(emptyset) are equal => F(emptyset) = 0 the zero ring. It seems that the key argument above rests on the idea that the covering set can be empty. So I wonder, what if we change the definition to require that any covering be *nonempty*? In that case, there seems to be no other reason why F(emptyset) must then be the zero ring. So sheaves now can have non-zero rings assigned to its emptyset. But do we still have a sound theory for sheaves in this case? And if we do, do we gain or lose anything by it? Perhaps other parts of sheaf theory would go wrong if we require coverings be always nonempty (or equivalently that F(emptyset) can be a nonzero ring)? Is there some desirable reasons why we want F(emptyset) to be a terminal object in the category of rings? Is it just because the emptyset is an initial object in the category of open sets of X? -- -kira === Subject: Re: Sheaves and the empty set covering set can be empty. So I wonder, what if we change the > definition to require that any covering be *nonempty*? In that case, > there seems to be no other reason why F(emptyset) must then be the zero > ring. So sheaves now can have non-zero rings assigned to its emptyset. But do we still have a sound theory for sheaves in this case? And if > we do, do we gain or lose anything by it? To your set of questions, I cannot gaurantee that my reasoning is all that great. But I think if I were to define sheaves from scratch (without having known originally what sheaves were as they are now) I would still go on and let the empty set to have the zero ring (for sheaves of rings) or in general the terminal object. I'll explain as we go along. Actually I do believe that originally the idea of sheaves was developed as an inspiration to the continuous functions. We regard the restriction maps in a sheaf just the way we regard the continuous functions being restricted. What is any possible logical way of restricting a continuous function from a nonempty open set to the empty set (which is also an open set)? The empty set is part of our topology and it is foolish to disregard it imo. So we have to have a definition for the continuous functions over the empty set.. I can't think of any other logical thing other than the 0 ring (consider for instance the continuous functions from R to R, R=reals) Perhaps other parts of sheaf theory would go wrong if we require > coverings be always nonempty (or equivalently that F(emptyset) can be a > nonzero ring)? > There is another definition of sheaves, which I haven't yet gone deeply into. It's with using the so called .8etale maps (local homeomorphisms) and sheaf spaces. You see, the section rings of a sheaf of rings ,ie. those F(U)'s, can always be regarded as some sort of continuous functions from U to some topological space, specifically it is shown that any sheaf is isomorphic to the continuous sections of a suitable .8etale map. For further readings (and I think this is a good book also for a beginner who has already touched the basic definition of sheaves as rings) I recommend you the book of Verschoren Relative Invariants of Sheaves. > Is there some desirable reasons why we want F(emptyset) to be a > terminal object in the category of rings? Is it just because the > emptyset is an initial object in the category of open sets of X? Yeah well.. the empty covering imo is just as important as the emptyset is and it may come handy at sometimes, remember we also deduced F(emptyset) to be the zero ring using this exact map and having an empty product, and empty product is sometimes desirable. Just as much as empty products and empty sums in numbers are desirable. But I guess this isn't really a great reason I'm giving, I'm just saying that I think disregarding the empty covering may not be all that good as you may think. Actually, it depends whether you regard the emptyset as an initial object or a terminal object. You can regard your presheaf either as a contravariant functor from X (category induced by your topological space, with injections as morphisms and open sets as objects) to your working category or as a covariant functor (which is my personal taste) from X^op (the dual/opposite of the the category I just described for X) to your category. If you consider it as X^op then the empty set also becomes a terminal object. So you send a terminal object to a terminal object in your category using your covariant functor which doesn't sound very irrational :) .. If I were to choose the action of F I would choose it that way probably :) .. it does less good if you send initial object to initial objects.. the initial object in X^op is X (the whole space) and the initial object in say Crings (commutative unitary rings with unitary homomorphisms) is Z.. then you don't have the continuous maps over X anymore, you just have Z as your F(X), but doing this ignores the original philosophy of sheaves. > -kira Jose Capco === Subject: Re: Sheaves and the empty set > To your set of questions, I cannot gaurantee that my reasoning is all > that great. But I think if I were to define sheaves from scratch > (without having known originally what sheaves were as they are now) I > would still go on and let the empty set to have the zero ring (for > sheaves of rings) or in general the terminal object. I'll explain as > we go along. I haven't followed this discussion, but does it actually matter how F(emptyset) is defined? It doesn't seem to me to come into the definition of a sheaf. Wouldn't it be easier just to say that F(U) is defined for non-empty sets? Incidentally, I find your use of the term zero ring slightly confusing. What exactly is this? Presumably the terminal object depends on the precise category you are talking about; in this case I would have thought the most common case was commutative rings with 1. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Sheaves and the empty set On May 20, 2:06 pm, Timothy Murphy but does it actually matter how F(emptyset) is defined? > It doesn't seem to me to come into the definition of a sheaf. > Wouldn't it be easier just to say that F(U) is defined for non-empty sets? You might want to browse into the older posts of this thread where it is explained that although F(emptyset) is not defined, it can be computed to be the terminal object of your category Incidentally, I find your use of the term zero ring slightly confusing. > What exactly is this? Oh sorry, it is indeed confusing. By that I mean the terminal object of the category of commutative unitary rings. In other words, the trivial commutative unitary ring or simply the ring with only one element (where 0=1). Its debatable whether this is in our category or not, but most category theorist would want it to be in our category of commutative unitary rings (because we want this category to be complete, ie. we want it to have limits for all small diagrams or equivalently we want it to have products and equalizers) > Presumably the terminal object depends on the precise category > you are talking about; > in this case I would have thought the most common case > was commutative rings with 1. > Yes it is. But we can talk about other categories. The main point here is that F(emptyset)=terminal object, no matter what kind of complete category we are working with. Jose Capco === Subject: Re: Sheaves and the empty set On 2007-05-20 01:16:17 -0400, Jose Capco said: > It seems that the key argument above rests on the idea that the > covering set can be empty. So I wonder, what if we change the > definition to require that any covering be *nonempty*? In that case, > there seems to be no other reason why F(emptyset) must then be the zero > ring. So sheaves now can have non-zero rings assigned to its emptyset. > > But do we still have a sound theory for sheaves in this case? And if > we do, do we gain or lose anything by it? > > To your set of questions, I cannot gaurantee that my reasoning is all > that great. But I think if I were to define sheaves from scratch > (without having known originally what sheaves were as they are now) I > would still go on and let the empty set to have the zero ring (for > sheaves of rings) or in general the terminal object. I'll explain as > we go along. > > Actually I do believe that originally the idea of sheaves was > developed as an inspiration to the continuous functions. We regard the > restriction maps in a sheaf just the way we regard the continuous > functions being restricted. What is any possible logical way of > restricting a continuous function from a nonempty open set to the > empty set (which is also an open set)? The empty set is part of our > topology and it is foolish to disregard it imo. So we have to have a > definition for the continuous functions over the empty set.. I can't > think of any other logical thing other than the 0 ring (consider for > instance the continuous functions from R to R, R=reals) > > > Perhaps other parts of sheaf theory would go wrong if we require > coverings be always nonempty (or equivalently that F(emptyset) can be a > nonzero ring)? > > > There is another definition of sheaves, which I haven't yet gone > deeply into. It's with using the so called .8etale maps (local > homeomorphisms) and sheaf spaces. You see, the section rings of a > sheaf of rings ,ie. those F(U)'s, can always be regarded as some sort > of continuous functions from U to some topological space, specifically > it is shown that any sheaf is isomorphic to the continuous sections of > a suitable .8etale map. For further readings (and I think this is a good > book also for a beginner who has already touched the basic definition > of sheaves as rings) I recommend you the book of Verschoren Relative > Invariants of Sheaves. > > > Is there some desirable reasons why we want F(emptyset) to be a > terminal object in the category of rings? Is it just because the > emptyset is an initial object in the category of open sets of X? > > Yeah well.. the empty covering imo is just as important as the > emptyset is and it may come handy at sometimes, remember we also > deduced F(emptyset) to be the zero ring using this exact map and > having an empty product, and empty product is sometimes desirable. > Just as much as empty products and empty sums in numbers are > desirable. But I guess this isn't really a great reason I'm giving, > I'm just saying that I think disregarding the empty covering may not > be all that good as you may think. > > Actually, it depends whether you regard the emptyset as an initial > object or a terminal object. You can regard your presheaf either as a > contravariant functor from X (category induced by your topological > space, with injections as morphisms and open sets as objects) to your > working category or as a covariant functor (which is my personal > taste) from X^op (the dual/opposite of the the category I just > described for X) to your category. If you consider it as X^op then the > empty set also becomes a terminal object. So you send a terminal > object to a terminal object in your category using your covariant > functor which doesn't sound very irrational :) .. If I were to choose > the action of F I would choose it that way probably :) .. it does less > good if you send initial object to initial objects.. the initial > object in X^op is X (the whole space) and the initial object in say > Crings (commutative unitary rings with unitary homomorphisms) is Z.. learned a lot about the motivations of this theory and made corrections on my understanding. (I had previously thought that F(null) = 0 was an arbitrary choice). I will take upon your book suggestions and try to get a deeper definition for a sheaf. -- -kira === Subject: Re: The distinction between multiplication and addition. > What's the difference between multiplication and > addition ? > > Order, disorder, and that's about all. > > The fact that any real number can be expressed > alternatively as a product or > as a sum of other reals is further evidence of the > fact that order and > disorder are existentially indeterminate. > > If you say that 10 = 2+2+2+2+2, and all of the 2's > are identical in > magnitude so that you might as well multiply, > basically that 10 = 2+2+2+2+2 > = 5 * 2, then what you are really saying is that > there is order, structure, > and regularity in the summands, i.e. thay are all of > identical magnitude, > all equal to 2. > > But, 10 = 1 + 9 = 2 + 8 = 3 + 7, etc etc etc. > > So, if all you have is the number 10, there is no way > to determine if it's a > sum or a product. I may be regarded as both, but If > you were being forced > to choose, then the distinction is primarily one or > ORDER, DISORDER and > STRUCTURE. And, because the distinction is arbitrary, > we must regard this > order, disorder and structure as being existentially > indeterminate. > > whatnot that act like > WAVES, because THEY ARE waves and also PARTICLES. I think that all what you express is related to time. Our inability to completely understand the primes is our inability to properly understand the relationship between addition and multiplication and this inability is due as Sylvester describes to an incomplete understanding experience of time: I have sometimes thought that the profound mystery which envelops our conceptions relative to prime numbers depends upon the limitations of our faculties in regard to time, which like space may be in essence poly- dimensional and that this and other such sort of truths would become self-evident to a being whose mode of perception is according to superficially as opposed to our own limitation to linearly extended time. For that reason the Goldbach Conjecture can not been proved in Peano Arithmetic due to the fact that the Peano Axioms are static and there exists a characterization of the mentioned conjecture in an infinite set of even numbers that depends on the time. If we associate the natural numbers to temporal states we can create a dynamical process that allows to capture all the information for knowing if a natural number is prime or not and in some sense this information is due to consider time identified with deformed hyperbolas i.e. bi dimensional time. Fernando. === Subject: Re: The distinction between multiplication and addition. What's the difference between multiplication and > addition ? Order, disorder, and that's about all. The fact that any real number can be expressed > alternatively as a product or > as a sum of other reals is further evidence of the > fact that order and > disorder are existentially indeterminate. If you say that 10 = 2+2+2+2+2, and all of the 2's > are identical in > magnitude so that you might as well multiply, > basically that 10 = 2+2+2+2+2 > = 5 * 2, then what you are really saying is that > there is order, structure, > and regularity in the summands, i.e. thay are all of > identical magnitude, > all equal to 2. But, 10 = 1 + 9 = 2 + 8 = 3 + 7, etc etc etc. So, if all you have is the number 10, there is no way > to determine if it's a > sum or a product. I may be regarded as both, but If > you were being forced > to choose, then the distinction is primarily one or > ORDER, DISORDER and > STRUCTURE. And, because the distinction is arbitrary, > we must regard this > order, disorder and structure as being existentially > indeterminate. whatnot that act like > WAVES, because THEY ARE waves and also PARTICLES. I think that all what you express is related to time. In a way, indeed it is. I can only describe my opinion of time because I dont have any neat identities. I think that time, on the quantum scale behaves probabilistically, it's no different than length, in fact it _is_ length, it's the 4th length, the 4th dimension, and when you start playing around with the number system (a + ~b), that little ~ can do some things which look strange but are not. If time is probabilistic, just like length must be, then you have an opportunity to explain the slits in time experiment. > Our inability to completely understand the primes is our > inability to properly understand the relationship > between addition and multiplication and this inability > is due as Sylvester describes to an incomplete > understanding experience of time: The vexing thing about primes, in my view, is that the distribution of primes is fixed. It does not change. It is what it is, and it will be the same in 10million years from now. The distribution itself is fixed. I've got some very bizarre views of order and disorder. I know that alot of work has been done trying to figure out that distribution, and I dont have any answers except to say that the primes seem to be a particular subset of N, from a whole class of such subsets of N, and that you might get further by studying that class instead of focusing just on this particular subset, the primes. I've tried to think of things like this through the lens of my own non-mainstream views and made no progress. I'll also say that I did very poorly in number theory and I know practically _nothing_ about this area except a few things that I was able to retain somehow, so my opinion is probably not so reliable. I have sometimes thought that the profound mystery > which envelops our conceptions relative to prime numbers > depends upon the limitations of our faculties in regard > to time, which like space may be in essence poly- > dimensional and that this and other such sort of truths > would become self-evident to a being whose mode of > perception is according to superficially as opposed to > our own limitation to linearly extended time. For that reason the Goldbach Conjecture can not been > proved in Peano Arithmetic due to the fact that the > Peano Axioms are static and there exists a > characterization of the mentioned conjecture in an > infinite set of even numbers that depends on the time. If we associate the natural numbers to temporal states > we can create a dynamical process that allows to capture > all the information for knowing if a natural number is > prime or not and in some sense this information is due > to consider time identified with deformed hyperbolas > i.e. bi dimensional time. > Fernando. === Subject: Re: The distinction between multiplication and addition. > once you use the terms multiplication and addition, you have put > yourself on the side of order, disorder is more along the lines of > twos and fives can constuct 10 or 2+2=10, whereas 5 x 2 = 5 > mathematics is a system you can use, if it is convenient, it generally > works. there are others... > Maybe order and disorder are nto the optimal words to use to describe what I'm trying to get at. Maybe symmetry is more accurate. I restarted the idea with a new post because the other one didnt generate any replies. Anyhow, I think that most professionals and academicians probably tend to go as light as possible on the philosophical end of things, and focus on algebraic structures, manipulations, rigor, etc. But philosophy can be interesting. I can think of 3 different kinds of things which are kind of interesting, but not really considered mainstream math. [1] Addition This is basically just putting things together regardless of their size. You have a collection of objects and there are no rules whatsoever governing the magnitudes of the items being added. [2] Multiplication Of course there's no law that you have to think of it this way, but it is obvious that multiplication can be regarded as shorthand for adding a collection of objects of all equal magnitude. [3] Functions, infinite sums, infinite products, etc The exact same line of reasoning can be adapted which borrows functional relationships to establish other things of interest. As an example of this, the well known infinite sum __ 2 ^ ( - n ) = 2 / /------/ n=0 to infinity so, the number two can be considered to be a sum of elements which have a very strange structure, a structure which when added converges nicely, but the structure itself is quite distinct from the other two cases [1] or [2]. The elements os this sequence are not structureless, but neither are they perfectly symmetric as in [2]. It seems that you can add things with no symmetry whatsoever [1], you can add things together with absolute symmetry [2], and then there's everything in between [3]. Seems like this is an area which probably gets a lot of informal contemplation, but how can you really fashion that into a usable tool ? Seems almost impossible to even approach this unless you use something like existential ideas. Can the amount of symmetry be taken as being somehow variable in some sense ? Well, I was thinking about a mathematical gedanken experiment. Instead of an atom smasher, you have a number smasher. It's just a device which smashes any given number into a collection of fragments which must be structured like [1], [2], or [3] in the above. The crazy thing is that any number, say 2, can be smashed by method [1], the number 2 can be smashed by method [2], and can also be smashed by method [3]. This is true of all reals. If you put a number in the smasher you have no way of knowing what will come out. The outcome is topologically indeterminate, unless you set things up intentionally so that the smasher returns a result which is either [1], [2], or [3].. I think that this is what's happening in the double slit. Of course I realize that everything I've done thus far amounts to little more than mathematical bull because it's pure philosophy. But the reason I'm stuck on this is beacuse it seems to mesh perfectly with all of the other nonsense I've been spewing for so long. === Subject: Re: The distinction between multiplication and addition. [1] Addition > This is basically just putting things together regardless of their size. > You have a collection of objects and there are no rules whatsoever governing > the magnitudes of the items being added. You write addition where you seem to mean counting. They are not the same and you are in emminent company, Hume and Russell, in confusing them. Peano characterized addition within pure mathematics. Counting is applied mathematics. Be clear about the distinction then you will be in a good position to continue working on the philosophy of physics. === Subject: Re: The distinction between multiplication and addition. > [1] Addition > This is basically just putting things together regardless of their size. > You have a collection of objects and there are no rules whatsoever governing > the magnitudes of the items being added. You write addition where you seem to mean counting. They are not the same and you are in emminent company, Hume and Russell, in confusing them. Peano characterized addition within pure mathematics. Counting is applied mathematics. Be clear about the distinction then you will be in a good position to continue working on the philosophy of physics. > addition as something very nebulous which resembles the reverse of cell division. A collection of blobs are assembled into the same place. Then, once you have performed the addition, those little blobls have all merged into a single larger blob. Those blobs could be anything, area, length, functions, variables, whatever. But in physics they would be little chunks of length. If you wished maybe you could elaborate a little on that distinction, counting and addition, I'd be most grateful. === Subject: Re: The distinction between multiplication and addition. On May 19, 9:29 am, Tomoko Kanazawa > [1] Addition > This is basically just putting things together regardless of their > size. > You have a collection of objects and there are no rules whatsoever > governing > the magnitudes of the items being added. > You write addition where you seem to mean counting. They are not the same and you are in emminent company, Hume and Russell, in confusing them. > Peano characterized addition within pure mathematics. Counting is applied > mathematics. Be clear about the distinction then you will be in a good > position to continue working on the philosophy of physics. addition as something very nebulous which resembles the reverse of cell > division. A collection of blobs are assembled into the same place. Then, once you have > performed the addition, those little blobls have all merged into a single > larger blob. Those blobs could be anything, area, length, functions, variables, whatever. > But in physics they would be little chunks of length. If you wished maybe you could elaborate a little on that distinction, > counting and addition, I'd be most grateful. I trouble over the arithmetic product some also. Whereas a scalar magnitudinal product can be fashioned into raw addition the signed product does not take this simplistic relation. The assymetry of the signed product in the reals or complex numbers seems to have little physical correspondence: (-1)(-2) = + 2 (+1)(+2) = + 2 Symmetry is broken by the arithmetical product with regard to sign and the physical correspondence of such a simple behavior is puzzling. In terms of spatial behavior no such product makes sense. Yet in stepping up to complex numbers we see rotation forms naturally as a result. I see this as an open problem. Superposition is a refined word for addition which carries more physical correspondence and its accuracy seems beyond refutation. We observe a continuum thinly populated with singularities. Resolving the behavior of the singularities in terms of simplistic product relationships would be desirable. Then the arithmetic product would carry more weight. The product relations of physics tend to yield a new domain rather than a result in the domain of the sources. The fundamental forces expose the second derivative as important. In terms of charge we start to see that the product relationship of the reals takes meaning, but the naked arithmetic product is still burdened. The units puzzle might best be described representationally. If we sum five rocks with three rocks we can see representationally and physically eight rocks. But if we multiply these representationally then magically appear fifteen rocks. Even within the simple construction five rocks we see an implicit representational product of five times rock so our language embeds the scalar product. Still the physical correspondence of five rocks times three rocks is nonexistent. So I do see your musing as appropriate. -Tim === Subject: Re: The distinction between multiplication and addition. On May 19, 6:29 am, Tomoko Kanazawa > [1] Addition > This is basically just putting things together regardless of their > size. > You have a collection of objects and there are no rules whatsoever > governing > the magnitudes of the items being added. > You write addition where you seem to mean counting. They are not the same and you are in emminent company, Hume and Russell, in confusing them. > Peano characterized addition within pure mathematics. Counting is applied > mathematics. Be clear about the distinction then you will be in a good > position to continue working on the philosophy of physics. If you wished maybe you could elaborate a little on that distinction, > counting and addition, I'd be most grateful. Do not be misled by the ambiguities in ordinary language. If you are putting fruit into a bowl you can physically add apples and oranges but you cannot add them mathematically. What you can do is count them and add the numbers. Counting a set of things gives each one an ordinal number. You get a first, a second, a third and so on. when you have counted all you have the number of items. === Subject: Re: The distinction between multiplication and addition. > On May 19, 6:29 am, Tomoko Kanazawa > [1] Addition > This is basically just putting things together regardless of their > size. > You have a collection of objects and there are no rules whatsoever > governing > the magnitudes of the items being added. > You write addition where you seem to mean counting. They are not the same and you are in emminent company, Hume and Russell, in confusing them. > Peano characterized addition within pure mathematics. Counting is applied > mathematics. Be clear about the distinction then you will be in a good > position to continue working on the philosophy of physics. If you wished maybe you could elaborate a little on that distinction, > counting and addition, I'd be most grateful. Do not be misled by the ambiguities in ordinary language. If you are > putting fruit into a bowl you can physically add apples and oranges > but you cannot add them mathematically. What you can do is count them > and add the numbers. Counting a set of things gives each one an > ordinal number. You get a first, a second, a third and so on. when you > have counted all you have the number of items. > exactly what you mean now. I had thought about this quite a bit a while back. The conclusion that I came to was pretty much that : Abstract addition is different from physically trying to add things, you're right, that is actually counting. And I think that it has to do with uniqueness of physical objects. I think that every physical object posseses the property of uniqueness, you cant really add things which are all slightly different. This leads to trivials, etc etc . === Subject: Re: Wikipedia editors cannot tell the difference between an label and a historical fact Re: definition of nickname for encyclopedic use Check out http://en.wikipedia.org/wiki/Rihab_Rashid_Taha for example. There are several others. -- Richard Okay it said nickname Dr. Germ by UN weapons inspectors It is okay if Wikipedia defined nickname and distinguished it from referal name, appellation name, alias name and nickname. Because the Dr. Germ is what I defined as an appellation name, just as Babe Ruth called the Sultan of Swat is not a nickname but an appellation name. A nickname by a person is a shortened name so that guests or company can easily talk to you. A referal name is what other people would call you behind your back. So Wikipedia cannot be blanket charging people with so called nicknames without defining these names. And encyclopedias have no right to be forcing a name on someone, just because they do not have a definition of nickname. I, and everyone else governs and controls what other people can call me as a nickname. What they call me as a Appellation name, such as Bush called the Great Divider, well, the person cannot control that. So Wikipedia is wrong again above in that Dr. Germ was not her nickname, it was her Appellation name or a Society Moniker name. So unless Wikipedia defines these assortment of names and has a fair policy on using names, then it is slander otherwise. 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: Wikipedia editors cannot tell the difference between an label and a historical fact Re: definition of nickname for encyclopedic use > > [...] > You're not the only scientist with a nickname in Wikipedia. > Check out > http://en.wikipedia.org/wiki/Rihab_Rashid_Taha > for example. There are several others. > -- Richard > > Okay it said nickname Dr. Germ by UN weapons inspectors > > It is okay if Wikipedia defined nickname and distinguished it from > referal name, appellation name, alias name > and nickname. Because the Dr. Germ is what I defined as an > appellation name, just as Babe Ruth called > the Sultan of Swat is not a nickname but an appellation name. I trust you understand that these terms are idiosyncratic to you. No one else uses referral name, appellation name, or alias name. In fact, since appellation means name, and alias means a substitute name, both these terms are redundant. Wikipedia doesn't define nickname. The word already has a meaning. > A nickname by a person is a shortened name so that guests or company > can easily talk to you. A referal name is what other people would call > you behind your back. A nickname is just another name. > So Wikipedia cannot be blanket charging people with so called > nicknames without defining these names. Wikipedia is pretty much free to do what it wants. As, no doubt, you've figured out by now. > And encyclopedias have no right to be forcing a name on someone, just > because they do not have a definition of nickname. Encyclopedias report. They don't force anyone to use a name. > I, and everyone else governs and controls what other people can call > me as a nickname. No, you govern what name you repsond to. You do *not* control what other people call you. Refer to some responses to this thread. > What they call me > as a Appellation name, such as Bush called the Great Divider, well, > the person cannot control that. > > So Wikipedia is wrong again above in that Dr. Germ was not her > nickname, it was her Appellation name or a > Society Moniker name. > > So unless Wikipedia defines these assortment of names and has a fair > policy on using names, then it is slander otherwise. First of all, it wouldn't be slander. That's spoken. You're thinking of libel. Remember our last discussion? The words must be false and damaging. Ironically, by making such a big deal of this issue, you've ensured that people will call you Arky, if only to annoy you. Thus making the nickname a reality. Why did you do that? > Archimedes Plutonium === Subject: Re: Wikipedia editors cannot tell the difference between an label and a historical fact Re: definition of nickname for encyclopedic use > [...] You're not the only scientist with a nickname in Wikipedia. > Check out http://en.wikipedia.org/wiki/Rihab_Rashid_Taha for example. There are several others. -- Richard > > Okay it said nickname Dr. Germ by UN weapons inspectors > > It is okay if Wikipedia defined nickname and distinguished it from > referal name, appellation name, alias name > and nickname. Because the Dr. Germ is what I defined as an > appellation name, just as Babe Ruth called > the Sultan of Swat is not a nickname but an appellation name. > > A nickname by a person is a shortened name so that guests or company > can easily talk to you. A referal > name is what other people would call you behind your back. > > So Wikipedia cannot be blanket charging people with so called > nicknames without defining these names. > > And encyclopedias have no right to be forcing a name on someone, just > because they do not have a definition > of nickname. > > I, and everyone else governs and controls what other people can call > me as a nickname. What they call me > as a Appellation name, such as Bush called the Great Divider, well, > the person cannot control that. > > So Wikipedia is wrong again above in that Dr. Germ was not her > nickname, it was her Appellation name or a > Society Moniker name. > > So unless Wikipedia defines these assortment of names and has a fair > policy on using names, then it is > slander otherwise. > > Archimedes Plutonium > www.iw.net/~a_plutonium > whole entire Universe is just one big atom > where dots of the electron-dot-cloud are galaxies Dude, get over wikipedia. Really, move on. ` === Subject: Wikipedia editors in trouble with the law over their arbitrariness in assigning nicknames to people I posted this to the discussion page in Wikipedia entry of Archimedes Plutonium. I do not believe I am the only person who would be irate to see a encyclopedia shoving down your throat a nickname you never heard of before and which you detested and which you had no say as to what your nickname was. --- discussion page --- Exactly. And besides, it is pretty obvious that Arky is a phonetic derivation of Archimedes, unless AP pronounces the ch in his name differently than the usual hard k sound. - Loadmaster 21:40, 11 May Arky, me old plate, here, for your lawyer's pleasure, is the first clever, cute, derogatory, or otherwise substitute name for a person or thing's proper name (for example, Bob, Rob, Robbie, Robin, and Bert are possible nicknames for Robert). I draw your attention to the first and fourth adjectives in that definition. Not that there is anything especially derogatory about Arky, it's just that you don't That definition of nickname does not distinguish between other names such as referal name or appellation name or alias name. Dr. Germ or the Great Divider, or Sultan of Swat are not nicknames according to my definition but an appellation name. And unless Wikipedia comes up with a definition of nickname that is uniformly used and not just when the editors want to demonize a person, then you should not use nicknames at all. The definition has to state how a nickname is created and how much a person has a say in what their nickname is. Arthur Rubin and Greglocock are confused between a nickname and a Appellation name. Arky is neither one of these, but a referal name. A encyclopedia editor would have to ask Francis how many people called me Arky, who they were and had any of those people actually met me to see how I would react when called that name. Because this gets into the issue of encyclopedia's as not recorders of history but of imposing what they like. Dr. Germ had the follow-up that the UN inspectors gave her that name. Where is the follow-up of Arky, was it one drunk Dartmouth student or was it Francis singularly fabricating a name. When you get the attorney letter that will be a new source for the crew there at Wikipedia. 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: x^2 - Ay^2 =1 Supersedes: (supersedes ) > > On 17 May, 11:46, Vincenzo Librandi > Yes, but you are not solving for A in > x^2-Ay^2 =1 > > The difficult question is if A = 23332322391 > > The fundamental solution of this one has 4648 digits. > (found in 1s by Dario Alpern's solver) > > or 109 > > The one given by Fermat to Frenicle in 1657, hence this one can be > solved by hand calculation... > > Yes, but it is the fact that there exists an algorithm > that solves x^2-Ay^2 =1 that is important. > > I suppose you could use the Vincenzo method > on a polynomial identity version of Pell's Equation > > The f,g,h etc are polynimials with integer coefficients > > f^2-gh^2 =1 > (f+1)(f-1) = gh^2 > As GCD(f+1,f-1) =2, do you think so ? even in integers it is false : f = 4, GCD(3,5) = 2 ??? GCD is 1 or 2. > f+1 and f-1 can't share any > polynomial factors that aren't units. OK > f-1 = pq^2 > f+1 = pq^2 + 2 = rs^2 ? > > So > [pq^2+1]^2 - ((pq)^2 + 2p)q^2 =1 > > So this is a solution for A= (pq)^2 +2p > This suffers the same flaw that Vincenzo's : knowing in advance a solution you get an equation whose solutions are given, but this is of no help in solving Pell's equation, that is knowing A find the solutions. > Is there a solution in non-trivial polynomials with > integer coefficients for rs^2-pq^2 = 2 ? > Then > > [pq^2+1]^2 - (pr)(qs)^2 =1 Not too hard to find some example of non trivial solutions in integers of a*x^2 - b*y^2 = 2 for instance a=6, b=13 X0=3, Y0=2 and all others from (53, 78 ; 36, 53)^n * (X, Y) that is (315, 214), (33387, 22682), (3538707, 2404078)... The problem with polynomial Z[X] is worse as you you can't define an euclidian division : you need Q[X], that is polynomial with _rational_ coefficients. Hence many properties required to solve a Pell's like equation are no more available. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: x^2 - Ay^2 =1 <24041933.1179398844762.JavaMail.jakarta@nitrogen.mathforum.org> ) > On 17 May, 11:46, Vincenzo Librandi > Yes, but you are not solving for A in > x^2-Ay^2 =1 > The difficult question is if A = 23332322391 > The fundamental solution of this one has 4648 digits. > (found in 1s by Dario Alpern's solver) > or 109 > The one given by Fermat to Frenicle in 1657, hence this one can be > solved by hand calculation... Yes, but it is the fact that there exists an algorithm > that solves x^2-Ay^2 =1 that is important. I suppose you could use the Vincenzo method > on a polynomial identity version of Pell's Equation The f,g,h etc are polynimials with integer coefficients f^2-gh^2 =1 > (f+1)(f-1) = gh^2 > As GCD(f+1,f-1) =2, do you think so ? > even in integers it is false : f = 4, GCD(3,5) = 2 ??? > GCD is 1 or 2. What I should have said was If d is a polynomial that divides f+1 and f-1 then it must divide the difference (f+1)-(f-1) =2 so GCD is -2,-1,1, 2. > f+1 and f-1 can't share any > polynomial factors that aren't units. OK f-1 = pq^2 > f+1 = pq^2 + 2 = rs^2 ? So > [pq^2+1]^2 - ((pq)^2 + 2p)q^2 =1 So this is a solution for A= (pq)^2 +2p This suffers the same flaw that Vincenzo's : knowing in advance a > solution you get an equation whose solutions are given, but this is of > no help in solving Pell's equation, that is knowing A find the > solutions. Yes, I agree. It doesn't help you find solutions in integers. > The problem with polynomial Z[X] is worse as you you can't define an > euclidian division : There does not have to be a Euclidean algorithm on Z[sqrt(d)], but the continued fraction method still works. > you need Q[X], that is polynomial with _rational_ coefficients. > Hence many properties required to solve a Pell's like equation are no more available. Yes, you would think the cyclic method would work on polynomoal Pell equations === Subject: Re: x^2 - Ay^2 =1 > ... > I suppose you could use the Vincenzo method > on a polynomial identity version of Pell's Equation > The f,g,h etc are polynimials with integer coefficients > f^2-gh^2 =1 > ... > The problem with polynomial Z[X] is worse as you can't define an > euclidian division : > you need Q[X], that is polynomial with _rational_ coefficients. There does not have to be a Euclidean algorithm > on Z[sqrt(d)], but the continued fraction method > still works. No, I speak about Z[X] that is the set of all polynomials in one unknown X, with coefficients in integers Z. [quote] > The f,g,h etc are polynimials with integer coefficients [/quote] You need to have some Euclidean _division_ to be able to get the floor(sqrt(g)), that is find a polynomial which is the approximate square root of polynomial g. Inside the algorithm, you need also to use the Euclidean division to get a floor( some p/q ). There just can't be any method of defining some Euclidean division of polynomials in Z[X]. That is given a and b in Z[X] to find q and r in Z[X] with a = b*q + r and 0 <= f(r) < f(b) with some indicator function f(p) for any p in Z[X] -> f(p) in N, and some additional properties of f(), like f(u*v) >= f(u) for any u,v != 0 The proof for that is quite easy. Not imagine and try all possible functions f() ! but : Suppose there were an Euclidean _division_, then it could be used to define an Euclidean _algorithm_, which, extended, would lead to a Bezout theorem : GCD(a,b) = a*u + b*v Let a = X and b = 2, they are coprime that is GCD = 1, and it is impossible to find u and v with X*u + 2*v = 1, just because the constant term of X*u + 2*v is even, hence contradiction. In Q[X], that is the set of polynomials with coeficients in Q (rationals), such an Euclidean division exists, and f(p) is just the degree of p. For Euclidean division in integers, f() ist just the absolute value, giving the usual 0 <= r < |b| > Hence many properties required to solve a Pell's like equation are no more > available. Yes, you would think the cyclic method would work > on polynomoal Pell equations How do you solve for p in a*p^2 + b*p + c = 0 with a,b,c,p /in Z[X] ??? -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: x^2 - Ay^2 =1 <24041933.1179398844762.JavaMail.jakarta@nitrogen.mathforum.org> ... > That is given a and b in Z[X] to find q and r in Z[X] with > a = b*q + r and 0 <= f(r) < f(b) with some indicator function f(p) for > any p in Z[X] -> f(p) in N, and some additional properties of f(), like > f(u*v) >= f(u) for any u,v != 0 I see you mean division with remainders. The existence of a division algorithm. I thought you meant the Euclidean algorithm > The proof for that is quite easy. > Not imagine and try all possible functions f() ! but : > Suppose there were an Euclidean _division_, then it could be used to > define an Euclidean _algorithm_, which, extended, would lead to a > Bezout theorem : GCD(a,b) = a*u + b*v Let a = X and b = 2, they are coprime that is GCD = 1, and it is > impossible to find u and v with X*u + 2*v = 1, just because the > constant term of X*u + 2*v is even, hence contradiction. But you are not really using a division algorithm for continued fractions sqrt(10) = 3 +d, 0 Hence many properties required to solve a Pell's like equation are no more > available. Yes, you would think the cyclic method would work > on polynomoal Pell equations How do you solve for p in a*p^2 + b*p + c = 0 with > a,b,c,p /in Z[X] ??? There;s no difficulty with p^2 -3p +2 =0 p= 1 or p =2. p^2 -(2n+1)p +(n^2+n) = 0 p= n or p = n+1 === Subject: Re: x^2 - Ay^2 =1 >This suffers the same flaw that Vincenzo's : knowing >in advance a solution you get an equation whose >solutions are given, but this is of no help in solving >Pell's equation, that is knowing A find the solutions > Repeat: For Pell's equation x^2-Ay^2=1 if is knowed A and only A. In this change, how in any my other's case, A is gived how form polynomial: A=9n^2-n Y=216n-12 X=648n^2-72n+1 then for this polynomial , with n>1 A=(8,34,78,140,220,318,...,) Y=(204,420,636,852,1068,1284,...,) X=(577,2449,5617,10081,15841,22897,...,) For Pell's equation x^2-Ay^2=1, when A=9n^2+n and Y=216n+12 X=648n^2+72n+1 (with n>1) A=(10,38,84,148,230,330,...,) Y=(228,444,660,876,1092,1308,...) X=(721,2737,6049,10657,16561,23761,...,) Trover.98 tutte le forme polimoniali di A; anche quelle che legano i fatidici A=109, or A=421, o anche ad altri A che sono numeri primi e che hanno soluzioni di X anche con centinaia di migliaia di cifre. Per ci.98 che attiene le primitive vi ho fatto vedere come si ottengono da una semplice scomposizione. Excuse me for language. A risentirci Vincenzo Librandi vincenzo.librand === Subject: Re: x^2 - Ay^2 =1 > > This suffers the same flaw that Vincenzo's : knowing in advance a > solution you get an equation whose solutions are given, but this > is of no help in solving Pell's equation, that is knowing A find > the solutions > Repeat: > For Pell's equation x^2-Ay^2=1 if is knowed A and only A. No, not given A, but given a specific form of polynomial P(n), you get from n both A=P(n) and a solution X(n), Y(n). And your polynomials have been choosen such as it works, not to give some value of A given in advance. We are waiting for your polynomials giving A = 61, 109 or 421... > In this change, how in any my other's case, A is gived how form > polynomial: A=9n^2-n with n>1 > Y=216n-12 > X=648n^2-72n+1 > A=9n^2+n > Y=216n+12 > X=648n^2+72n+1 (with n>1) This is again false, they don't give the _fundamental_ solutions (hence they are useless). I'm too lazzy to correct, the CF expansion period seems to be 2 and 4, hence calculations should be not too hard. Trovero tutte le forme polimoniali di A; > anche quelle che legano i fatidici A=109, > or A=421, o anche ad altri A che sono numeri > primi e che hanno soluzioni di X anche con > centinaia di migliaia di cifre. Per cio che attiene le primitive vi ho > fatto vedere come si ottengono da una semplice scomposizione. Excuse me for language. I tried to translate this through Google... then a little bit arranged, as direct translation was a mess. > > I will find all the polynomial forms of A. > Good luck, endless task... > > Also those that are related with the fatidic A=109, or A=421, > I understand that you intend to find the solutions for A = n^2 + n + 1 for instance ? Greeat... All the polynomial for A you have given, up to now, are very specific ones, for which you know the solutions in advance. And you never prooved they give the fundamental solutions for that A. (and even several examples where they don't, see two more above) > > or also to other A that are prime numbers > or that give solutions for X with even hundred of [thousands ?] > of digits. > And sure there are endless specific cases. That's why we have a _method_ that works perfectly well to solve any given Pell's equation (Lagrange's algorithm)... a method, not an endless list of cases. And A being prime or not is of little concern here, as you just can't deduce the solutions of x^2 - (p*q)*y^2 = 1 from those of x^2 - p*y^2 = 1 and x^2 - q*y^2 = 1. > > In what concerns the primitive, I have to say > they are obtained from a simple decomposition. > No they are not. Also I'm not quite sure you don't mix up 'prime' and 'primitive', or may be it's Google... Solving a Pell's equation is finding it's fundamental solution, that is the one from which all others result, for this value of A. That is also the smallest >0 solution for this A. This can't be obtained from any 'simple decomposition' of whatever. I give you a solution, which is may be or not the fundamental one, of this equation : x^2 - 21*y^2 = 1 X = 665335, Y = 145188 Could you say if this solution is fundamental or not ? Could you find from this the fundamental solution ? show me your 'simple decomposition'... Another example x^2 - 85*y^2 = 1 X = 285769, Y = 30996 same question. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: x^2 - Ay^2 =1 >We are waiting for your polynomials giving > A = 61, 109 or 421... 1766319049^2-61*226153980^2=1 is tied 29718^2-61*3805^2=-1 find the tie. For others: tied x^2-109*Y^2=-1 or x^2-421y^1=-1, and I fiend what ask. >I give you a solution, which is may be or not the >fundamental one,of this equation : >x^2 - 21*y^2 = 1 >X = 665335, Y = 145188 >Could you say if this solution is fundamental or not ? >Could you find from this the fundamental solution ? >show me your 'simple decomposition'... Another example >x^2 - 85*y^2 = 1 >X = 285769, Y = 30996 >same question. Of course 285769 = (11.83,313) isn't square and 665335=(5,11,12097) isn't square so 55=(5,11)isn't square; and 55^2-21*12^2=1. I find the metod ! Vincenzo Librandi vincenzo.librandweoz@alice.it === Subject: Re: x^2 - Ay^2 =1 > > We are waiting for your polynomials giving > A = 61, 109 or 421... 1766319049^2-61*226153980^2=1 > is tied I don't understand what you mean by 'tied' tied to what ? related with the =-1 equation ? Also, yes you have a solution to that specific equation, but not with your method of polynomial representation for A ! (61 is not in the list of your allready given polynomial forms) > 29718^2-61*3805^2=-1 > find the tie. what tie ??? if X, Y is a solution of any generalized Pell's equation x^2 - A*y^2 = -1, then X^2 + A*Y^2, 2X*Y is a solution of x^2 - A*y^2 = +1 This is true for all A for which x^2 - A*y^2 = -1 has a solution. And generalized into : If X,Y is solution of x^2 - A*y^2 = p and U,V solution of x^2 - A*y^2 = q, then U*X + A*V*Y, U*Y + V*X is solution of x^2 - A*y^2 = p*q > For others: tied x^2-109*Y^2=-1 > or x^2-421y^1=-1, and I fiend what ask. Again, I don't understand your 'tied'. None of 109 and 421 is in your list of polynomials. What do you mean by 'fiend' ? And as above, if we know a solution of x^2 - A*y^2 = -1, then we get immediately the solution for x^2 - A*y^2 = +1. So *if* x^2-109*Y^2=-1 or x^2-421y^1=-1 have solutions (and they do) we can deduce from these solutions the solution of the +1 equations. But if the equation x^2 - A*y^2 = -1 has no solutions, and many of them don't, we can deduce nothing about solution of the corresponding x^2 - A*y^2 = +1 Also the problem is not in just solving these equations, but to apply _your_ method in solving them. That is find polynomial forms for A, X, Y which fit to this equation. I have reordered your answers, the two equations are independent. > I give you a solution, which is may be or not the fundamental one, > of this equation : x^2 - 21*y^2 = 1 > X = 665335, Y = 145188 > Could you say if this solution is fundamental or not ? > Could you find from this the fundamental solution ? > show me your 'simple decomposition'... Of course [...] > 665335=(5,11,12097) isn't square > so 55=(5,11)isn't square; and 55^2-21*12^2=1. > Usually just write 665335 = 5*11*12097 for prime decompositions, writing (a,b,c...) is confusing. > I find the metod ! What method ? You fell into a lucky coincidence for the 55. I would have given X = 73180801, Y = 15969360 instead of the 665335, what would your method have given ? 73180801 = 17*31*138863, yes it is not a square, neither are 17*31 = 527 nor 17*138863 = 2360671 nor 31*138863 = 4304753 neither of these three being solutions of x^2 - 21y^2 = 1 although 73180801 _is_ a solution, and is not the fundamental solution. What do you deduce ??? The same if I had given X = 6049 = 23*263. Yes 55 is the fundamental solution, that you didn't proove, although with y = 12 a proof could be just try and check all values y = 1, 2... 11, or just solve the Pell's equation using classical methods, what you reject. And all solutions of this equation (A=21) are, in increasing order : x = 1, y = 0 (the trivial solution of all Pell's equations) x = 55, y = 12 (the fundamental solution of this equation) x = 6049, y = 1320 x = 665335, y = 145188 x = 73180801, y = 15969360 x = 8049222775, y = 1756484412 x = 885341324449, y = 193197315960 etc... (from allready discussed recurrence relations) and there are _no_ others, just because we start from the fundamental one, apart changing x into -x and/or y into -y. [reordered, this is the second equation, unrelated with the first] > > Another example > x^2 - 85*y^2 = 1 > X = 285769, Y = 30996 > same question. > > Of course [... and] 285769 = (11,83,313) isn't square Do you mean that 913 = 11*83 should be a solution of x^2 - 85*y^2 = 1 ? or just 11, or 83 ... ? Did you proove that 285769 is or not the fundamental solution for this equation ? A _method_ is supposed to work in all cases, or should allways be able to distinguish between cases where it applies and cases where it doesn't ! -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: Quark Force, NASA Pioneer Anomaly, Flat Stellar Rotation Curves - all same explanation The probes drift in opposite directions, one with drag against the sun, the other with drag awayst the sun. === Subject: Residue Classes In extending the natural numbers to the integers, Pickert and G.9arke introduce composite symbols of the form a-a' as solutions for x in x+a'=a and assert that (a,a')->a-a' is a mapping of all pairs of natural numbers onto the extended domain. After developing an equivalence relation over the extended domain, they represent the pairs (a,a'), (b,b'), etc., as A, B, etc., respectively. Then they state that we could, if we wished, simply use the A* in place of the a-a'. They say that the A* are called residue classes with respect to the equivalence relation. I'm looking at Pages 108 and 109: http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 .djvu Does this mean that, for example, if a=7 and a'=4 then 3 would be a member of A*? Or would being a member of A* require something of the form (4,1), (5,2), etc? Is there a distinction between a residue class and an equivalence class? I'm a bit unsure as to whether, at this point in the development, the natural numbers are even considered to be members of the extended domain. Specifically, I mean that I am not sure whether they mean the extended domain to consist exclusively of mappings of pairs of natural numbers. For myself, I'm almost inclined to consider the natural numbers and the integers to be disjoint collections which just happen to have some members which look and act a whole lot alike. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Residue Classes Hatto von Aquitanien > In extending the natural numbers to the integers, Pickert and G.9arke > introduce composite symbols of the form a-a' as solutions for x in x+a'=a > and assert that (a,a')->a-a' is a mapping of all pairs of natural numbers > onto the extended domain. After developing an equivalence relation over > the extended domain, they represent the pairs (a,a'), (b,b'), etc., as A, > B, etc., respectively. Then they state that we could, if we wished, > simply > use the A* in place of the a-a'. They say that the A* are called residue > classes with respect to the equivalence relation. I'm looking at Pages 108 and 109: > http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 . djvu Does this mean that, for example, if a=7 and a'=4 then 3 would be a member > of A*? Or would being a member of A* require something of the form (4,1), > (5,2), etc? > Is there a distinction between a residue class and an > equivalence class? Although I can't read the djvu file, there's no doubt he is defining integers as _equivalence classes_ of pairs of naturals. Two pairs (a,b) and (c,d) are equivalent whenever a+d = b+c. I can think of two reasons why he should be using the term equivalence class rather than residue class. > I'm a bit unsure as to whether, at this point in the development, the > natural numbers are even considered to be members of the extended domain. They aren't, exactly, but in the end one may identify a natural number e with the equivalence class of (e,0), a process which amounts to the usual embedding of the natural numbers into the integers. LH === Subject: Re: Residue Classes > Hatto von Aquitanien > In extending the natural numbers to the integers, Pickert and G.9arke > introduce composite symbols of the form a-a' as solutions for x in x+a'=a > and assert that (a,a')->a-a' is a mapping of all pairs of natural numbers > onto the extended domain. After developing an equivalence relation over > the extended domain, they represent the pairs (a,a'), (b,b'), etc., as A, > B, etc., respectively. Then they state that we could, if we wished, > simply > use the A* in place of the a-a'. They say that the A* are called residue > classes with respect to the equivalence relation. > I'm looking at Pages 108 and 109: > http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 .djvu > Does this mean that, for example, if a=7 and a'=4 then 3 would be a > member of A*? Or would being a member of A* require something of the form > (4,1), (5,2), etc? > Is there a distinction between a residue class and an > equivalence class? > Although I can't read the djvu file, I believe Firefox has the plugin bundled with the distribution. That's what I was told. I don't use Windows very often, so I'm not really sure what's available. I am certain that there is freely available software of a usable quality which will allow you to read djvu. > there's no doubt he is defining > integers as _equivalence classes_ of pairs of naturals. Two pairs (a,b) > and (c,d) are equivalent whenever > a+d = b+c. The authors use the term residue class. In a footnote they state that the term comes from the theory of divisibility and give references to subsequent chapters. The term equivalence class shows up only once in the index, and that refers to a section on the theory of relations. I've looked at a few other sources, and they all seem to give a slightly different explanation for these terms. > I can think of two reasons why he should be using the term equivalence > class rather than residue class. What would those be? > I'm a bit unsure as to whether, at this point in the development, the > natural numbers are even considered to be members of the extended domain. > > They aren't, exactly, but in the end one may identify a natural number e > with the equivalence class of (e,0), a process which amounts to the usual > embedding of the natural numbers into the integers. They did things a bit differently. They simply defined equivalence between integers and natural numbers, and then showed that their definition satisfies the laws of equality. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Residue Classes Hatto von Aquitanien > me: > I can think of two reasons why he should be using the term equivalence > class rather than residue class. What would those be? One is that residue class is a long-established bit of jargon in number theory. Residue is just a synonym for remainder, on division. But division is not involved in the current construction. The second reason is that this formal construction is analogous to various others around math, in which cases residues are irrelevant; examples are -- the definition of a rational number a/b as the equivalence class of the element (a,b) in ZxZ, for the equivalence relation wz = yx between two elements (w,x) and (y,z). -- the definition of a complex number as an equivalence class of polynomials over R, modulo the polynomial x^2+1. There are many others. Anyhow, a residue class is a special case of an equivalence class, so that author's jargon is not outright wrong. LH === Subject: Re: Residue Classes > Hatto von Aquitanien > me: > I can think of two reasons why he should be using the term equivalence > class rather than residue class. What would those be? > One is that residue class is a long-established bit of jargon in number > theory. Residue is just a synonym for remainder, on division. But > division is not involved in the current construction. > The second reason is that this formal construction is analogous to various > others around math, in which cases residues are irrelevant; examples are > -- the definition of a rational number a/b as the equivalence class of the > element (a,b) in ZxZ, Minor nit: isn't it Zx(Z{0}), for Z = the set of integers and 0 its additive identity? > for the equivalence relation > wz = yx > between two elements (w,x) and (y,z). > -- the definition of a complex number as an equivalence class of polynomials > over R, modulo the polynomial x^2+1. > There are many others. > Anyhow, a residue class is a special case of an equivalence class, so that > author's jargon is not outright wrong. > LH === Subject: Re: Residue Classes Virgil > Minor nit: isn't it Zx(Z{0}), for Z = the set of integers and 0 its > additive identity? Yes. And I did notice that, honest ;) === Subject: Re: Residue Classes > > Hatto von Aquitanien > me: > I can think of two reasons why he should be using the term equivalence > class rather than residue class. > What would those be? > One is that residue class is a long-established bit of jargon in number > theory. Residue is just a synonym for remainder, on division. But > division is not involved in the current construction. > The second reason is that this formal construction is analogous to various > others around math, in which cases residues are irrelevant; examples are > -- the definition of a rational number a/b as the equivalence class of the > element (a,b) in ZxZ, > > Minor nit: isn't it Zx(Z{0}), for Z = the set of integers and 0 its > additive identity? > > for the equivalence relation > wz = yx > between two elements (w,x) and (y,z). > -- the definition of a complex number as an equivalence class of polynomials > over R, modulo the polynomial x^2+1. > There are many others. > Anyhow, a residue class is a special case of an equivalence class, so that > author's jargon is not outright wrong. > LH I have not checked into all the references given on that page 109, may be you do if you are really interested. It is just a wording not used today (and not sure what's due to translation to English). What they do is constructing a group from a half group (or so), IN x IN /~ = Z, the integers. What they call residue class w.r.t. equivalence relation is a representant, but no clear definition is given (as over the whole text - Math often was written that way in older times). Thinking about the according footnote I think they have the following in mind: The construction modulo a relation is like ring modulo ideal or classical Z / mZ. A German word is Restklassen-Ring and an element is called Restklasse, which at that time is not the same as residue (not sure how it was used before). The reason is: given some number k you dived by n. Which may not give a natural number, something reminds - a Rest, at school children/students learn(ed?) that as Teilen mit Rest. Probably you want to check the German original text. === Subject: Re: Residue Classes > > Hatto von Aquitanien > me: > I can think of two reasons why he should be using the term equivalence > class rather than residue class. > What would those be? > One is that residue class is a long-established bit of jargon in number > theory. Residue is just a synonym for remainder, on division. But > division is not involved in the current construction. > The second reason is that this formal construction is analogous to various > others around math, in which cases residues are irrelevant; examples are > -- the definition of a rational number a/b as the equivalence class of the > element (a,b) in ZxZ, > > Minor nit: isn't it Zx(Z{0}), for Z = the set of integers and 0 its > additive identity? > > > I have not checked into all the references given on that page > 109, may be you do if you are really interested. I was not objecting to your discussion in general, but merely pointing out that defining your equivalence relation, as you indicated it, on the whole of ZxZ produces fractions with zero denominators. If one constructs the integers from the naturals, then the rationals from the integers, one must specifically exclude zero as a denominator. An alternate approach, which nicely sidesteps that problem of division by zero, is to create the positive rationals from the (positive) naturals first, then the signed rationals, including the zero rational, from the positive rationals. N = set of (positive) naturals. P= the set of positive rationals, NxN modulo the relation that == iff a*d = b*c in N, with the usual arithmetic + == and * == with no zero element at all, but unit == 1' Then Q, the set of ALL rationals, is PxP modulo the relation that === Subject: Re: Residue Classes > ... > > I have not checked into all the references given on that page > 109, may be you do if you are really interested. > > I was not objecting to your discussion in general, but merely pointing > out that defining your equivalence relation, as you indicated it, on the > whole of ZxZ produces fractions with zero denominators. ... Sorry, my answer was thought for the original poster (so you meant him, not you ...) === Subject: Re: Residue Classes > > ... > > I have not checked into all the references given on that page > 109, may be you do if you are really interested. > > I was not objecting to your discussion in general, but merely pointing > out that defining your equivalence relation, as you indicated it, on the > whole of ZxZ produces fractions with zero denominators. > ... > > Sorry, my answer was thought for the original poster > (so you meant him, not you ...) O.K. === Subject: free mobile sms in all over the world www.entertainmentvenues.org venues for live entertainment, mobile venues, free web2sms, mobile themes, free nokia themes, Mobile 3GP Videos, Mobile Games, Mobile secrets, Online Games, free wallpapers, Msn Stuff and many more. === Subject: Re: Where is the sensation about the discovery? <464c3167$0$21145$7a628cd7@news.club-internet.fr Porton, what is post-axiomatic? What is the theory with no axioms? I hope that I will be able to construct theories using only definitions (that kind of definitions which I call analogies and probably other kinds of definitions), with no axioms. It is currently just a hope, not yet a formal theory. Some very rough drafts about that (so called 21 Century Math Method) are listed at the end of http://www.mathematics21.org/post-axiomatic-mathematics-concept.html Here is the programme: http://www.mathematics21.org/21century.html http://www.mathematics21.org/method.html These drafts are VERY too rough. This theory is not yet developed. But it is great if it will succeed. -- Victor Porton - http://www.mathematics21.org === Subject: cancel: Re: * Sizzling Boobies all new pics and Vidz Control: cancel <464ef033$0$4840$8d2e0cab@news.newsgroup-binaries.com> Excessive crossposting. (<464ef033$0$4840$8d2e0cab@news.newsgroup-binaries.com>) === Subject: Module vs. Abelian Group I recall posting about this a long time ago, and someone objected to the use of the term module as a synonym for Abelian group. Does anybody disagree with the text and footnote on page 111 which state that a set with an operation defined in it is called a module or an Abelian group if the operation is commutative and associative, there is a neutral element and every element has an inverse? http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 .djvu -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Module vs. Abelian Group On Sat, 19 May 2007 10:28:24 -0400, Hatto von Aquitanien >I recall posting about this a long time ago, and someone objected to the use >of the term module as a synonym for Abelian group. Does anybody >disagree with the text and footnote on page 111 which state that a set with >an operation defined in it is called a module or an Abelian group if the >operation is commutative and associative, there is a neutral element and >every element has an inverse? That's the definition of Abelian group. Module is much more general - in particular an abelian group is the same thing as (ie is naturally isomorphic to) a Z-module (ie a module over Z'.) http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol- 1.djvu ************************ David C. Ullrich === Subject: Re: Module vs. Abelian Group > On Sat, 19 May 2007 10:28:24 -0400, Hatto von Aquitanien > >I recall posting about this a long time ago, and someone objected to the >use >of the term module as a synonym for Abelian group. Does anybody >disagree with the text and footnote on page 111 which state that a set >with an operation defined in it is called a module or an Abelian group if >the operation is commutative and associative, there is a neutral element >and every element has an inverse? > > That's the definition of Abelian group. Module is much > more general - in particular an abelian group is the same > thing as (ie is naturally isomorphic to) a Z-module > (ie a module over Z'.) What are the axioms of an Abelian group? What are the axioms of a module? >http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol -1.djvu > > > ************************ > > David C. Ullrich -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Module vs. Abelian Group On Sun, 20 May 2007 12:23:45 -0400, Hatto von Aquitanien > On Sat, 19 May 2007 10:28:24 -0400, Hatto von Aquitanien > >I recall posting about this a long time ago, and someone objected to the >use >of the term module as a synonym for Abelian group. Does anybody >disagree with the text and footnote on page 111 which state that a set >with an operation defined in it is called a module or an Abelian group if >the operation is commutative and associative, there is a neutral element >and every element has an inverse? > > That's the definition of Abelian group. Module is much > more general - in particular an abelian group is the same > thing as (ie is naturally isomorphic to) a Z-module > (ie a module over Z'.) What are the axioms of an Abelian group? >What are the axioms of a module? Why are you asking? You just _gave_ the definition of abelian group, and you just quoted the definition of R-module in another post. If what you're really curious about is how an abelian group is a Z-module then you should ask about that... >http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vo l-1.djvu > > > ************************ > > David C. Ullrich ************************ David C. Ullrich === Subject: Re: Module vs. Abelian Group > On Sun, 20 May 2007 12:23:45 -0400, Hatto von Aquitanien > > That's the definition of Abelian group. Module is much > more general - in particular an abelian group is the same > thing as (ie is naturally isomorphic to) a Z-module > (ie a module over Z'.) >What are the axioms of an Abelian group? >What are the axioms of a module? > > Why are you asking? You just _gave_ the definition of > abelian group, and you just quoted the definition of > R-module in another post. There is no contradiction between the definition given for R-module, and the assertion that module and Abelian group are synonyms. > If what you're really curious about is how an abelian > group is a Z-module then you should ask about that... I am not asking about an R-module, nor a Z-module. I am asking what a module is. If I ask what a bluebird is, and someone tells me it is a bird which is blue, that does not tell me anything more about what a bird is than that some of them are blue. The definition (axioms) for module given on page 111 of Behnke, et al., is (are) identical to that (those) which is (are) typically given for an Abelian group.[*] The reason I asked what the axioms for a module are is be cause I am being told that module is not a synonym for Abelian group. http://baldur.globalsymmetry.com/open-source/org/sth/math/behnke-et-al/vol-1 .djvu [*]Strictly speaking, however, I believe the axioms are not orthogonal. The existence of a neutral element follows from invertability. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Module vs. Abelian Group >Does anybody > disagree with the text and footnote on page 111 which state that a set with > an operation defined in it is called a module or an Abelian group if the > operation is commutative and associative, there is a neutral element and > every element has an inverse? I disagree. These are different notions. Module is a generalization of Abelian group: an Abelian group is a module over the ring of integers. Otherwise (just to make things confusing ;) any module has an underlying Abelian group. Module is also a generalization of vector space: a vector space is a module over a field. For definitions, see e.g. http://en.wikipedia.org/wiki/Module_(mathematics) === Subject: Re: Module vs. Abelian Group > >Does anybody > disagree with the text and footnote on page 111 which state that a set > with > an operation defined in it is called a module or an Abelian group if the > operation is commutative and associative, there is a neutral element and > every element has an inverse? > > I disagree. These are different notions. > > Module is a generalization of Abelian group: an Abelian group is a module > over the ring of integers. Weyl defines an Abelian group as a group in which the defining operation is commutative. > Otherwise (just to make things confusing ;) any module has an underlying > Abelian group. It appears that the term module has been used in different ways by different authors. Most references I have found talk about R-modules, Z-modules, left modules, etc. I don't know if those are somehow derived from the usage found in Behnke, et al., or not. I noticed that in chapter 5 Gr.9abner and Lesky reference the section I am referring to when they discuss the module property of an ideal. See page 338. I don't know what usage has precedence, but I suspect the usage I am referring to was in the original 1958 edition. > Module is also a generalization of vector space: a vector space is a > module over a field. That appears to be what Farleigh calls an R-module. > For definitions, see e.g. > > http://en.wikipedia.org/wiki/Module_(mathematics) I don't take Wikipedia as authoritative. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Module vs. Abelian Group On Sat, 19 May 2007 22:53:52 -0400, Hatto von Aquitanien > >Does anybody > disagree with the text and footnote on page 111 which state that a set > with > an operation defined in it is called a module or an Abelian group if the > operation is commutative and associative, there is a neutral element and > every element has an inverse? > > I disagree. These are different notions. > > Module is a generalization of Abelian group: an Abelian group is a module > over the ring of integers. Weyl defines an Abelian group as a group in which the defining operation is >commutative. > Otherwise (just to make things confusing ;) any module has an underlying > Abelian group. It appears that the term module has been used in different ways by different >authors. Most references I have found talk about R-modules, Z-modules, >left modules, etc. I don't know if those are somehow derived from the >usage found in Behnke, et al., or not. I noticed that in chapter 5 Gr.9abner >and Lesky reference the section I am referring to when they discuss the >module property of an ideal. See page 338. I don't know what usage has >precedence, but I suspect the usage I am referring to was in the original >1958 edition. > Module is also a generalization of vector space: a vector space is a > module over a field. That appears to be what Farleigh calls an R-module. Assuming that R means the real numbers: An R-module is a _real_ vector space. In general a vector space over a field K is a K-module. > For definitions, see e.g. > > http://en.wikipedia.org/wiki/Module_(mathematics) I don't take Wikipedia as authoritative. ************************ David C. Ullrich === Subject: Re: Module vs. Abelian Group > On Sat, 19 May 2007 22:53:52 -0400, Hatto von Aquitanien > > >Does anybody > disagree with the text and footnote on page 111 which state that a set > with > an operation defined in it is called a module or an Abelian group if > the operation is commutative and associative, there is a neutral > element and every element has an inverse? > > I disagree. These are different notions. > > Module is a generalization of Abelian group: an Abelian group is a > module over the ring of integers. >Weyl defines an Abelian group as a group in which the defining operation >is commutative. > Otherwise (just to make things confusing ;) any module has an underlying > Abelian group. >It appears that the term module has been used in different ways by >different >authors. Most references I have found talk about R-modules, Z-modules, >left modules, etc. I don't know if those are somehow derived from the >usage found in Behnke, et al., or not. I noticed that in chapter 5 >Gr.9abner and Lesky reference the section I am referring to when they >discuss the >module property of an ideal. See page 338. I don't know what usage has >precedence, but I suspect the usage I am referring to was in the original >1958 edition. > Module is also a generalization of vector space: a vector space is a > module over a field. >That appears to be what Farleigh calls an R-module. > > Assuming that R means the real numbers: An R-module is > a _real_ vector space. In general a vector space over > a field K is a K-module. R is a ring. That is R, and not RR(double-struck). Let R be a ring. A (left) R-module consists of an abelian group M together with an operation of external multiplication of each element of M by each element of R on the left such that for all alpha, beta elements of M and r, s elements of R, the following conditions are satisfied: 1. (r alpha) element of M. 2. r (alpha + beta) = r alpha + r beta. 3. (r + s) alpha = r alpha + s alpha. 4. (r s) alpha = r (s alpha). We shall somewhat incorrectly speak of the R-module M. As for Z; Fraleigh uses both Z and ZZ in the context of discussing modules. His list of symbols at the back of the book indicates that Z(G) is the center of G, which means nothing to me. Z does not appear elsewhere in that list. There does, however, appear to be some connection implied between Z and ZZ. Perhaps the analogous relationship is present between R and RR? I'm not sure what external multiplication means, but I assume it means that the operation is not an injective mapping to R. Since Farleigh's use of the term module is always qualified, it appears there is no contradiction between his usage, and that in Behnke, et al. BTW, I'm not sure that it is strictly proper to assert that, for example, QQ is a ring without further qualification. One must, in order to be fully correct, also state the operations under which it is a ring. QQ is merely a set, a collection of elements. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Module vs. Abelian Group It appears that the term module has been used in different ways by different > authors. Like most words in most languages. http://en.wikipedia.org/wiki/Module_(mathematics) I don't take Wikipedia as authoritative. What you take as authoritative is of little importance. The fact is that both Alex S's posting and this Wikipedia page reflect the common usage of the term module in contemporary mathematics more accurately that the 50-year old textbook you cite. (I suspect also this this book reflects a more Germanic usage of the word module (or Modul)). Victor Meldrew === Subject: Re: Module vs. Abelian Group > It appears that the term module has been used in different ways by > different authors. > > Like most words in most languages. > >http://en.wikipedia.org/wiki/Module_(mathematics) > I don't take Wikipedia as authoritative. > > What you take as authoritative is of little importance. Perhaps to you. To me it is rather essential. > The fact is that both Alex S's posting and this Wikipedia page > reflect the common usage of the term module in contemporary > mathematics more accurately that the 50-year old textbook you > cite. Or, perhaps a failure to understand the abstraction correctly. > (I suspect also this this book reflects a more Germanic > usage of the word module (or Modul)). That would be German, or - more correctly - in this case German language. Germanic is is an adjective denoting an ethno-linguistic complex consisting of people identified by the Romans as Germani, as well as denoting ethno-linguistic descendents of the Germani. I suspect the word appearing in the original German is module. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: about zero vector- > hi,friends > the proposition: the zero vector perpendicular to any vector. is correct? > in chinese high school textbook,it avoid the 0 vector when refer to 2 perpendicular vectors. > can you give me a detail expatiation? > > it means the null vector, not the zero vector, and perpendicular is > probably a mistranslation of orthoganal, perpendicular only applies to > a two dimensional space. > > derek. > perpendiculum = plumb line; perpendiculator = stone-cutter, using a plumb line when and where necessary. My much-coveted fantasy/hypothesis: The perpendiculator in mediaeval times was a free-lancer who took care that tall buildings (in particular churches and church towers) got precisely vertical. His tools were plumb lines of all sizes and weights, one of his main skills was climbing scaffoldings of all heights. Needless to say he must not suffer from fear of heights. Linguistic: Google >perpendiculator yields 19 hits, among which three entries to online dictionaries: Latin-Hungarian: http://latin.oszk.hu/cgi-bin3/index.cgi?function=doc&filename=41769&position =24323403-460 Latin-Russian: http://antic.slimhost.info/list.php?letter=P Explanatory dictionary - Latin only - as if Latin still were the international scientific language: http://www.uni-mannheim.de/mateo/camenaref/gesner/gesner1/v3/books/Gesner_th esaurus_P_1.html The bottom line is that perpendicular is ==not== a wrong translation of orthogonal. Orthogonal means right-angled; perpendiculum literally means something that hangs down. === Subject: Re: Sum of squares- > If a^2 + b^2 = x^2 + y^2 (and a, b, x, y all not zero) > > then I am 99% sure either a = x, b = y or a = y, b = x. But what > makes this true, because it is clearly not true if we remove the > squares: > > 2 + 5 = 1 + 6 > > Geometrically, of course, a^2 + b^2 = r is the equation of a circle. > So if x^2 + y^2 = r as well, both equations represent the same circle. > > But I would like a symbolic proof without appealing to geometry. > The subject of Sums of Squares is a mathematical universe in its own right. Let me just recall just the two theorems that are most relevant to your observation. Both are applicable to positive integers. (A) A prime number p = 4n+1 is the sum of two squares in an essentially unique way. (B) A number which is the product of N different prime numbers of the form 4n+1 is the sum of two squares in essentially 2^(N-1) different ways. Therefore your conjecture is true (even 100% instead of 99%) only in the case that a^2 + b^2 is a prime number. Literature: Hardy and Wright: An introduction to the theory of numbers. Oxford, 1938; reprinted at least four times... and many more books! But this one is among the best, perhaps the best introduction to number theory. Happy studies: Johan E. Mebius === Subject: Re: Sum of squares > If a^2 + b^2 = x^2 + y^2 (and a, b, x, y all not zero) then I am 99% sure either a = x, b = y or a = y, b = x. It is 100% false (at least if you allow a,b,x and y to be real numbers rather than integers). The points (a,b) and (x,y) lie on the same circle, but are otherwise unrelated. R.G.Vickson > But what > makes this true, because it is clearly not true if we remove the > squares: 2 + 5 = 1 + 6 Geometrically, of course, a^2 + b^2 = r is the equation of a circle. > So if x^2 + y^2 = r as well, both equations represent the same circle. But I would like a symbolic proof without appealing to geometry. === Subject: Re: Sum of squares > If a^2 + b^2 = x^2 + y^2 (and a, b, x, y all not zero) > > then I am 99% sure either a = x, b = y or a = y, b = x. It isn't true. > Geometrically, of course, a^2 + b^2 = r is the equation of a circle. > So if x^2 + y^2 = r as well, both equations represent the same circle. And there are lots of different points on a circle. There may even be lots of integer points on a circle. > But I would like a symbolic proof without appealing to geometry. a^2 - x^2 = y^2 - b^2 (a + x) (a - x) = (y + b) (y - b) To find an example in positive integers, take any positive integer that can be factored in two different ways as the product of two distinct positive integers that are both odd or both even. For example, 15 * 1 = 5 * 3 a + x = 15, a - x = 1 => a = 8, x = 7 y + b = 5, y + b = 3 => y = 4, b = 1 8^2 + 1^2 = 7^2 + 4^2 -- 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 squares Hmm okay, I see the counter example. But my book has the following: a^2 + b^2 = 1 It then uses cos^2(t) + sin^2(t) = 1 and says there is some angle t such that a^2 = cos^2(t) b^2 = sin^2(t) How is this justified? === Subject: Re: Sum of squares > Hmm okay, I see the counter example. But my book has the following: > > a^2 + b^2 = 1 > > It then uses cos^2(t) + sin^2(t) = 1 > > and says there is some angle t such that > > a^2 = cos^2(t) > > b^2 = sin^2(t) > > How is this justified? Since a^2 + b^2 = 1, a^2 <= 1 and so -1 <= a <= 1. On the other hand, cosine is continuous, cos(0) = 1, and cos(pi) = -1, and therefore there is some _t_ in [0,pi] such that cos(t) = a. Then sin^2(t) = 1 - cos^2(t) = 1 - a^2 = b^2. Jose Carlos Santos === Subject: Re: Sum of squares > If a^2 + b^2 = x^2 + y^2 (and a, b, x, y all not zero) > > then I am 99% sure either a = x, b = y or a = y, b = x. But what > makes this true, because it is clearly not true if we remove the > squares: > > 2 + 5 = 1 + 6 1^2 + 7^2 = 5^2 + 5^2. Jose Carlos Santos === Subject: SUSPECT NABBED IN PIRANHA ATTACK AT SMITHSONIAN -- Yes, Indeed, It Was Ed Conrad < > BULLETIN < > By Oliver Shagnasty > The Washingtan Pist < WASHINGTON, DC -- City police today apprehended the person they claim had placed piranha in the Smithsonin's indoor swimming pool earlier this week, causing former ousted Secretary Lawrence M. Small his precious manhood and as well as considerable embarrassment. The suspect was identified as one Ed Conrad who was accompanied by his dog, Blue. < http://mysite.verizon.net/edconrad/FOSSILS/EdConrad&Blue.jpg < Conrad, who doesn't know his ass from a hole in the ground, has long thought he was smarter than the Smithsonian experts but even we all know he has a petrified brain made of coal. Where was this freakin' imbecile when real brains were given out? Will somebody pass me the darts? < http://mysite.verizon.net/edconrad/FOSSILS/EdConradDartboard.jpg < > MORE MORE MORE < < > EARLIER EXCLUSIVE < http://mysite.verizon.net/edconrad/FOSSILS/SaveTheWhale.jpg < Lawrence M. Small, former Secretary of the Smithsonian, lost his weener this morning during a swim in the Institution's indoor pool. < Small, who enjoys taking a dip after breakfast, was bitten by one or more piranha which authorities say probably had been put in the pool by one Ed Conrad. < Conrad, respected on the Internet as a leader of conserving Cyberspace, apparently was miffed because the Smithsonian has long denied his discoveries of petrified bones, teeth and soft organs -- some HUMAN -- between anthracite coal veins, which obviously destroys the facetious Theory of man's Evolution.. < The photo was taken by a Paparazzi just seconds before the piranha attack. < Mr. Small, at the hospital preparing for a penis transplant from talk.origins moderator David Iain Greig, a donor, said he may have to find a new line of work and admits he has received several offers to join the carnival and tour as a freak. <. http://www.nurple.com/sideshow/ < HOW TO PROTECT YOUR PENIS http://www.vasectomy-information.com/humor/manual.htm < EXPLAINING THE WEENER (and other good stuff...) http://www.urbandictionary.com/define.php?term=weener < THOSE BASTARD PIRANHA http://en.wikipedia.org/wiki/Piranha < HOW LAWRENCE B. SMALL SAVED TAXPAYERS' MONEY (For Himself) http://www.washingtonpost.com/wp-dyn/content/graphic/2007/03/19/GR2007031900 078.html < < > TRUTH VS. THE MONUMENTAL LIE OF MAN'S EVOLUTION > (Fossils -- SOME HUMAN -- Found Between Coal Veins) < http://mysite.verizon.net/edconrad/FOSSILS/OldestHumanSkull.JPG http://mysite.verizon.net/edconrad/FOSSILS/ManasOldasCoal.jpg http://mysite.verizon.net/edconrad/FOSSILS/Discoveries.jpg http://mysite.verizon.net/edconrad/FOSSILS/MoreFossils.jpg http://mysite.verizon.net/edconrad/FOSSILS/HumanBrain.jpg http://mysite.verizon.net/edconrad/FOSSILS/TestResults.jpg http://mysite.verizon.net/edconrad/FOSSILS/OldestTool.jpg http://mysite.verizon.net/edconrad/FOSSILS/Scorpion.jpg http://mysite.verizon.net/edconrad/FOSSILS/MVC-013F.JPG http://mysite.verizon.net/edconrad/FOSSILS/MVC-022S.JPG < < The anthros, their interests SO vested And Poor Truth, for too long molested Their deceit and deception Near the point of perfection If it were MY call, they'd ALL be arrested < < > PETRIFIED FOSSILS STILL EMBEDDED IN SLATE < http://mysite.verizon.net/edconrad/NewFossils/MVC-002S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-003S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-006S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-007S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-009S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-010S.JPG http://mysite.verizon.net/edconrad/NewFossils/MVC-012S.JPG < < Ed Conrad http://www.edconrad,com Man as Old as Coal < (cc) Sir Isaac Newton Johannes Keppler Ed Conrad Copernicus Galilei Galileo Immanuel Velikovsky Christopher Columbus Plato Socrates Matthew, Mark, Luke and Jack Albert Einstein Thomas Edison Confusius General Electric Time Warner Disney Viacom News Corp. Turner Broadcasting NBC CBS ABS MSNBC Joe Scarborough Fox News Barbara Walters Sean Hannity Ann Coulter (Douche Bag) David Horowitz Michael Savage Rupert Murdoch Bill O'Reilly PBS BBC Lou Dobbs C-span Meet the Press Brian Williams Katie Couric Charles Gibson White House Billy Meier UFO Intelligent Design Keith Olbermann Evolution Charles Darwin Rush Limbaugh Intelligent Design Bill Moyers UFO Penn State News of the World New England Journal of Medicine 20/20 alt.obituaries The View Oprah Winfrey Rosie O'Donnell Meet the Press Nightline Anderson Cooper Jerry Falwell Christianity Hinduism Islam Judaism Buddhism Sikhism American Association for the Advancement of Science L'Osservatore Romano Vatican Pope Benedict Republican Paul Myers PZ Alan Mann Penn Evolutionisists Life After Life Phil Kaufman Lucifer History Channel 60 Minutes Extraterrestrial Cancer Karl Rove President Bush Nature magazine Penthouse Prehistoric University of California-Berkeley Wilton Krogman Raymond Dart Penn State Raymond Rye World-renowned philosopher Clayton Lennon (1900-1996) < === Subject: Re: JSH: What a paper Here is a definition of the Harris property. > Let P(x) be a polynomial of degree n and leading coefficient 1 > and assume all the coefficients are integers. Further assume that > the constant term of P(x) is divisible by a prime, p, but not > divisible by any higher power of p. > A ring R is said to have the Harris Property if the roots of > P(x) are in R and exactly one of the roots is divisible by p > in R. > The ring of algebraic integers provably does not have the Harris > Property. That is its 'flaw'. No contradiction or inconsistency > with Galois theory is implied. Nor does the fact that some > other ring DOES have the Harris Property imply any inconsistency > with Galois theory. > As to why he wants rings to have the 'Harris Property', that > is another story which goes back several years. >Are you sure that's right? It made way too much sense to me. I think Marcus's condition is a bit too strong; there are other > considerations at play which led to James's decision that the roots > should behave the way he wants them to behave, which have to do with > the coefficients for middle terms and other considerations. I don't > know if he even remembers where it all came from in the first place. See The main point being, he wants the roots to behave in a certain way because one step of his proof of FLT requires it. He is refractory to all attempts to explain what is wrong with that step because if it is false, he has nothing. Marcus. > for my latest attempt at a recap. Note in particular the four > coefficient equations, just before the paragraph that starts Now, > if f is not equal to 3, then this means that -- > 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: JSH: What a paper [...] > Here is a definition of the Harris property. > Let P(x) be a polynomial of degree n and leading coefficient 1 > and assume all the coefficients are integers. Further assume that > the constant term of P(x) is divisible by a prime, p, but not > divisible by any higher power of p. > A ring R is said to have the Harris Property if the roots of > P(x) are in R and exactly one of the roots is divisible by p > in R. > The ring of algebraic integers provably does not have the Harris > Property. That is its 'flaw'. No contradiction or inconsistency > with Galois theory is implied. Nor does the fact that some > other ring DOES have the Harris Property imply any inconsistency > with Galois theory. > As to why he wants rings to have the 'Harris Property', that > is another story which goes back several years. >Are you sure that's right? It made way too much sense to me. > I think Marcus's condition is a bit too strong; there are other > considerations at play which led to James's decision that the roots > should behave the way he wants them to behave, which have to do with > the coefficients for middle terms and other considerations. I don't > know if he even remembers where it all came from in the first place. > See > The main point being, he wants the roots to behave in a certain > way because one step of his proof of FLT requires it. He is > refractory to all attempts to explain what is wrong with that step > because if it is false, he has nothing. > Running back to old arguments where you or other sci.math'ers were > successful in confusing people is a very low tactic as the point of my > new paper is to REMOVE the ability to be convincing in objecting to my > research. The arguments in your most recent version of this paper include > old arguments. Then cite the current paper, as if what you are saying is true, why go > back to old discussions? Are you conceding (finally) that the old arguments were wrong? Proof cannot be wrong. I have proof but people like you fight it. > Whether you have proof or not is not a matter of simply asserting that you have proof. In the end, you have to convince other people that you have proof. So far that has not happened. > People like yourself work to confuse people, like you've been doing > lately with your replies in this thread, and readers on these > newsgroups can just look around to see how active you and your people > have been. > What do you mean, your people ? I don't have any people. > AND that is NOT SURPRISING with such a huge result, as people hate > change, and my mathematical research forces massive change in number > theory--an area that is used to glacial change if any. > You have no basis for saying that. Mathematics has progressed at an astonishing rate since the beginning of the 20th century. > And for those who wonder how devastating that paper is to these people > look at how they keep running from its details, now trying to re-visit > the past---back to when they succeeded in convincing a lot of people > that I was wrong. > And even helped kill a math journal. We may never know what killed the journal. It was incompetently > edited. That is clear independent of what happened with your > written > by the editor, who lied to you repeatedly regarding the supposed > peer review of your paper. Sending you a copy of an e-mail of > a note from W. Dale Hall, written AFTER the paper was published, > and claiming that it was one of the peer review reports, is a > transparent > lie and a token of his incompetence and contempt for you. > My current paper covers all the relevant mathematics in extreme > detail. Your current paper says: The question now arises, how are you forced out of the ring > of algebraic integers if an expression that is in that ring is > just being subtracted from an identity? The identity cannot change things. The underlying base > expression is in the ring of algebraic integers, so should you > not remain in the ring? :But you do not. What you are saying here is that the ring of algebraic integers is > not closed under subtraction. Now for the first time you have > stated explicitly what the flaw is in the ring of algebraic > integers. No I'm not. Subtracting from the identity is the first step, while > the paper covers the others that show when you find yourself forced > out of the ring of algebraic integers. > I cannot see why you are denying this. You start with an expression involving algebraic integers. The expression itself is an algebraic integer. Then you subtract from it another algebraic integer. The formulation is essenitially: Y = A - B, where A and B are algebraic integers. Then you say that the result, Y, is NOT in the ring of algebraic integers. If you were correct, you would be saying, the ring of algebraic integers is not closed under subtraction. > The paper is very straightforward, and simply leaves no room for > doubt. > Similarly there is no room for doubt about what you said. You are claiming the ring of algebraic integers is not closed under subtraction. That is the simplest way to say it. > It shows how to generally factor 175x^2 - 15x + 2 = 2(f(x) + 1)* (g(x) + 1) covering the infinity of possible solutions for f(x) and g(x), and in > so doing shows when they can be in the ring of algebraic integers, as > well as when one might think they should since I rely on a method that > involves subtracting from an identity. Remarkably what is subtracted from the identity is kind of complicated > looking: r^2 + rs - (2 + 2xt + tQ(x))st + s^2 = (2x+Q(x))t^2 > Complicated or not, what you are doing is nothing more than subtracting one algebraic integer from another algebraic integer. And you are claiming that the result forces you out of the ring of algebraic integers. So you are just saying that the ring of algebraic integers is not closed under subtraction. See the quotation from the paper that I gave above. > and to me that is rather amazing as well, as who would have guessed > that? > Have you ever seen any of the proofs that the set of algebraic integers IS closed under subtraction? Those proofs do not use Galois theory or any advanced algebraic number theory. Yet you think you have a disproof. > Now if you READ THE PAPER I QUOTED the goddamn paper. What you said in it is obviously equivalent to saying the ring of algebraic integers is not closed under subtraction. Why do you disagree with saying that? > and study it then you can understand exactly > what is happening and I've given the big picture now of what my > methods actually end up doing as they are factorization results. So > the simple answer is I extended from polynomial factorization to > general factorizations of polynomials, so it is a remarkable bit of > number theory in and of itself. Sure maybe you actually believed some of the stuff you and other > sci.math'ers spouted against my earlier work, but nows a chance to get > excited about mathematical discovery! After all, it's not like the truth will change. > For you, the truth seems to change every time you revise the paper. > I'll tell you, I'm really, really happy to have this answer as > questions raised over my research did bug me, and I felt that if it > were really correct, then I should be able to answer them all, but > there were cases where I kept going back to my own answers wondering. > Rightly so. And still so. > It took closing the space to answer even my lingering doubts, as > forced to explain I slowly ended up moving to an argument that covered > ALL factorizations of 175x^2 - 15x + 2 = 2(f(x) + 1)* (g(x) + 1) which took a while, and I credit arguing with quite a few people over > the years with forcing me to finally do that, and connect all the way > back to the identity and what was subtracted from it, which completely > closed down all room for doubt. > Not in the slightest. > So this paper introduces identities, where I call them tautological > spaces, and this paper introduces my newest phrase algebraic residues, > and this paper also includes the object ring as it ties everything > together and I think that makes it a beautiful work. Consider, if humanity has any kind of future that paper may be > considered one of the great works in mathematical history, and I am > telling you that arguing with people fighting me every step of the way > was an integral part of forcing me to do what it took to answer all > objections. Hey, people like you said some nasty stuff, often went over the line, > and repeatedly refused to be convinced by mathematical proof, but you > forced me to simplify, simplify, simplify and in doing so I gained a > much greater understanding myself, and for that I thank you. > Here all I am pointing out is that your current version of the paper claims to prove that the ring of algebraic integers is not closed under subtraction. Really, all I am doing is restating your conclusion. Why do you keep resisting saying this? Marcus. > James Harris === Subject: Re: JSH: What a paper > Here all I am pointing out is that your current version of the > paper claims to prove that the ring of algebraic integers is > not closed under subtraction. Really, all I am doing is > restating your conclusion. Why do you keep resisting saying > this? Let me guess: (a) He doesn't want it to be true. (b) He doesn't understand why it's true. === Subject: Re: JSH: What a paper paper claims to prove that the ring of algebraic integers is > not closed under subtraction. Really, all I am doing is > restating your conclusion. Why do you keep resisting saying > this? Let me guess: > (a) He doesn't want it to be true. > (b) He doesn't understand why it's true. (c) A crank doesn't like to submit to standard notation. --- J K Haugland http://home.no.net/zamunda === Subject: Re: JSH: What a paper Here all I am pointing out is that your current version of the > paper claims to prove that the ring of algebraic integers is > not closed under subtraction. Really, all I am doing is > restating your conclusion. Why do you keep resisting saying > this? Let me guess: > (a) He doesn't want it to be true. Very definitely he doesn't. > (b) He doesn't understand why it's true. > He doesn't have the faintest idea. He doesn't even realize that there it is necessary to prove that the set of algebraic integers is closed under subtraction (and multiplication) before you can call it a ring. He may think: it's obviously a ring because everyone calls it a ring, so it must be closed under subtraction. It has never occurred to him that there has to have been a proof of it. > (c) A crank doesn't like to submit to standard notation. > True also. But I think the real reason is: (d) He doesn't want to say that the set of algebraic integers is not closed under subtraction, because he suspects - rightly - that closure is a well-known accepted fact. He is OK with contradicting Dedekind; he is OK with contradicting Galois Theory; but he knows that contradicting a really basic fact is another independent, and fatal, strike against his proof. He very likely doesn't know how closure is proved - it takes about 3 lines from absolute first principles - a high-school kid could understand it - if he did know the proof, he would also know there is no chance that it is erroneous. He wants to be able to continue saying that when you subtract one particular algebraic integer from another, you do not remain in the ring or you are forced out of the ring, but he cannot stand to have this translated into the statement that the ring is not closed under subtraction - even though this is a totally obviously equivalent statement - because he knows this destroys his claims. So he continues his old pattern of lying, denying, blustering, whining - anything but admitting the obvious. This only makes sense if you assume he is deluded beyond redemption. Marcus > --- > J K Hauglandhttp://home.no.net/zamunda === Subject: Re: JSH: What a paper it takes about 3 lines from absolute first principles - > a high-school kid could understand it - if he did know > the proof, he would also know there is no chance that > it is erroneous. Can you give a reference for this, or the proof itself? I think Arturo Magidin posted a proof that the algebraic integers formed a ring some years back in a JSH thread, but IIRC it was certainly not high-school level (it involved a fair bit of linear algebra). -Rotwang === Subject: Re: JSH: What a paper it takes about 3 lines from absolute first principles - > a high-school kid could understand it - if he did know > the proof, he would also know there is no chance that > it is erroneous. Can you give a reference for this, or the proof itself? I think Arturo > Magidin posted a proof that the algebraic integers formed a ring some > years back in a JSH thread, but IIRC it was certainly not high-school > level (it involved a fair bit of linear algebra). -Rotwang Algebraic Number Theory by Pollard and Diamond (A Carus Monograph, published by the MAA). See the chapter on algebraic integers. Marcus === Subject: Re: JSH: What a paper <5b3mfqF2qcm5bU1@mid.individual.net> <0K63i.2$y6.0@dfw-service2.ext.ray.com> <464d33f2$1@news.auckland.ac.nz> <464dfaba$0$97230$892e7fe2@authen.yellow.readfreenews.net >Wiles in contrast has a paper I can tear apart in multiple ways. >How could that be the case, when you obviously don't have the >necessary background knowledge to evaluate his argument? > Maybe he means he can _literally_ tear it apart in multiple ways. like a really really small phone book, 5 pages or so.- Hide quoted text - - Show quoted text - It's 200 pages, isn't it? === Subject: functional analysis question Let X_i be an Hilbert space for each i=1,2,... . What is the Hilbert sum of the X_i ? === Subject: Re: functional analysis question > Let X_i be an Hilbert space for each i=1,2,... . > > What is the Hilbert sum of the X_i ? > The set of sequences (x_1, x_2, ...) where x_i in X_i for all i, and sum ||x_i||^2 < infinity. The set can be given the structure of a Hilbert space again. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: functional analysis question <190520071351298717%edgar@math.ohio-state.edu.invalid> On 19 Mag, 19:51, G. A. Edgar Let X_i be an Hilbert space for each i=1,2,... . What is the Hilbert sum of the X_i ? The set of sequences (x_1, x_2, ...) > where x_i in X_i for all i, and sum ||x_i||^2 < infinity. > The set can be given the structure of a Hilbert space again. How do such a set can be given the structure of a Hilbert space? Best, Larry -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: functional analysis question >On 19 Mag, 19:51, G. A. Edgar Let X_i be an Hilbert space for each i=1,2,... . > What is the Hilbert sum of the X_i ? > The set of sequences (x_1, x_2, ...) > where x_i in X_i for all i, and sum ||x_i||^2 < infinity. > The set can be given the structure of a Hilbert space again. How do such a set can be given the structure of a Hilbert space? Did you think about this question at all first? You can make the class of all such sequences into a Hilbert space by defining (*) = sum_j , where of course is the inner product in X_j. Quiz question: Wha question is raised by the supposed definition (*)? Quiz question: What's the _answer_ to that question? >Best, >Larry > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > ************************ David C. Ullrich === Subject: Re: functional analysis question <190520071351298717%edgar@math.ohio-state.edu.invalid> <2lf0539behffb7dhdvs3lvjpt3der4qspg@4ax.com >On 19 Mag, 19:51, G. A. Edgar Let X_i be an Hilbert space for each i=1,2,... . > What is the Hilbert sum of the X_i ? > The set of sequences (x_1, x_2, ...) > where x_i in X_i for all i, and sum ||x_i||^2 < infinity. > The set can be given the structure of a Hilbert space again. How do such a set can be given the structure of a Hilbert space? Did you think about this question at all first? You can make the class of all such sequences into a Hilbert > space by defining (*) = sum_j , where of course is the inner product in X_j. Quiz question: What question is raised by the supposed definition (*)? > Is the norm induced by (*) complete? > Quiz question: What's the _answer_ to that question? Yes, this is assured from the fact that <>_i induces a complete norm for each i and sum ||x_i||^2 < infinity >Best, >Larry > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ************************ David C. Ullrich === Subject: Re: functional analysis question >On 19 Mag, 19:51, G. A. Edgar Let X_i be an Hilbert space for each i=1,2,... . > What is the Hilbert sum of the X_i ? > The set of sequences (x_1, x_2, ...) > where x_i in X_i for all i, and sum ||x_i||^2 < infinity. > The set can be given the structure of a Hilbert space again. >How do such a set can be given the structure of a Hilbert space? > Did you think about this question at all first? > You can make the class of all such sequences into a Hilbert > space by defining > (*) = sum_j , > where of course is the inner product in X_j. > Quiz question: What question is raised by the supposed definition (*)? > >Is the norm induced by (*) complete? I suppose there are many correct answers here, but the _first_ question is why does the series converge. That's obvious or not, depending on how automatic it is for you to apply __ in situations where it's relevant. Then there's the question of why it defines an inner product, which is trivial, and _then_ there's the question of completeness. > Quiz question: What's the _answer_ to that question? Yes, this is assured from the fact that <>_i induces a complete norm >for each i and >sum ||x_i||^2 < infinity It certainly has a lot to do with that, but it seems to me as well that there's a bit more argument required. Hmm. For example: Define Y to be the same as the product space above, except that Y contains only sequences (y_j) such that y_j = 0 for all but finitely many j. It's still true that every X_j is complete, but Y is _not_ complete. >Best, >Larry > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ > ************************ > David C. Ullrich > ************************ David C. Ullrich === Subject: Re: functional analysis question >On 19 Mag, 19:51, G. A. Edgar Let X_i be an Hilbert space for each i=1,2,... . > What is the Hilbert sum of the X_i ? > The set of sequences (x_1, x_2, ...) > where x_i in X_i for all i, and sum ||x_i||^2 < infinity. > The set can be given the structure of a Hilbert space again. >How do such a set can be given the structure of a Hilbert space? Did you think about this question at all first? You can make the class of all such sequences into a Hilbert > space by defining (*) = sum_j , where of course is the inner product in X_j. Quiz question: What question is raised by the supposed definition (*)? > Is the norm induced by (*) complete? My answer, Why does the sum (*) converge? > > > Quiz question: What's the _answer_ to that question? > > Yes, this is assured from the fact that <>_i induces a complete norm > for each i and > sum ||x_i||^2 < infinity I think completeness needs more than this argument. > >Best, >Larry > -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ************************ David C. Ullrich > > -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: the 5 important numbers <464bcfcd$0$97253$892e7fe2@authen.yellow.readfreenews.net> can in terms of digit-sequences be expressed exactly, every > number can be approximated in arbitrary accuracy (e.g., on > the number-line or in the complex-plane). Zero and i can be approximated in ways which can be measured > in terms of accuracy. > Thus zero and i match the above mentioned criterion for > numbers. Thus zero and i might be numbers. (To make sure > you need to check if zero and i match all criteria for > numbers... ) Infinity cannot be approximated in a way which can be measured > in terms of accuracy. > Thus infinity does definitely not match the above mentioned > criterion for numbers, thus it does not match all criteria for > numbers (no matter what these are), thus infinity is definitely > not a number. Ulrich Cantor had much to say on infinity. I believ there are an infinite number of Alephs. === Subject: 3D Sudoku variant Sudoku-Ball(R) is a 3D sudoku variant which has 14 sudokus network around the ball. It's a unique challenge because each individual sudoku is linked to 4 other sudokus. It is a mathematical and 3D challenge as the bottom line is that all sudokus are linked together. The puzzle can be played for free at http://www.sudoku-ball.com Have fun! === Subject: Re: JSH: Suggestions? > I'm not a professional mathematician, so my efforts can be called > recreational. By the same argument, since Fermat was not a professional mathematician (he was a judge), his efforts can be called recreational. Is that what you think? > Years ago I got a paper published, and sci.math'ers went after it, and > got it retracted by the journal editors with some emails. The journal > later died, shutting down after one more edition. Are you under the delusion that there is some connection between those two facts? > Over the years I've worked at explanations of my research that take > away the objections they relied upon so that I can convince that my > original work was valid, and they were wrong. Convince? Who, besides you? > I finally have that paper which I've tested in posts to this newsgroup > and alt.math and alt.math.undergrad as understandably I do not like > the sci.math newsgroup, but was not surprised that sci.math'ers came > over to attack my paper. > > And I have been gratified that their tactics have failed this time as > the paper does cover everything as I need against their kind of > opposition. Isn't that what you said about the previous paper? > But what now? > > Journals are wary of me now. After all, one journal keeled over and > died!!! What has that to do with you? > The paper is available to the world on a Google group of mine, but > that seems like a crap shoot--waiting and hoping that someone will > notice this result. > > And what a result!!! > > I managed to use identities to get some incredible number theory > analysis done. > > And that's so simple of a thing, like people use identities all the > time in mathematics. > > e.g. > > x^2 + 2xy = z^2 > > add y^2 to both sides to get > > x^2 + 2xy + y^2 = y^2 + z^2 > > and you can solve for x: > > x = sqrt(y^2 + z^2) - y What do you mean by sqrt, James? > So I have a remarkable technique that relies on subtracting from > identities, extending mathematics. Growing knowledge. How much mathematics have you extended so far? > So then, what do I do? If mathematicians work to close all the doors > against an inconvenient truth, what options do I have? Keep sucking your thumb. Jose Carlos Santos === Subject: Re: JSH: Suggestions? Jos.8e Carlos Santos > I'm not a professional mathematician, so my efforts can be called > recreational. By the same argument, since Fermat was not a professional mathematician > (he was a judge), his efforts can be called recreational. Is that what > you think? > Years ago I got a paper published, and sci.math'ers went after it, and > got it retracted by the journal editors with some emails. The journal > later died, shutting down after one more edition. Are you under the delusion that there is some connection between those > two facts? Well, we really should give JSH credit for finding a journal that was willing to publish some of his , probably without even reading it. I would have it thought it quite unlikely. And I'm sure he misses the outfit! LH === Subject: measure of continued fractions whose partial quotients sum to n. If we chose a random number,x, in (0,1) x=[0,a(1),...,a(k),...], what is the probability that a(1)+...+a(k)=n. Call this probability p(n). What does p(n) converge to? For instance if x is in (1/2,1) x=[0,1,....], so p(1)=1/2 If x is in (1/3,2/3) x=[0,1,1,...] or x=[0,2,...] so p(2)=1/3 If x is in (1/4,2/5), (3/5, 4/5) then x=[0,3,...],x=[0,1,1,1,..],x=[0,2,1,...] x=[0,1,2,...] so p(3)= 3/10 Define the sequence H(0,0)=0/1, H(0,1)=1/1 H(n,2k)= H(n-1,k) H(n, 2k+1)= med( H(n-1,k),H(n-1,k+1) ) where med(a/b+c/d) =(a+c)/(b+d) then H(n) is a subset of the f(n)th farey sequence, where f(n) is the nth fibonacci number. H(n) also tells you on what intervals x has the desired property. For n>1, p(n)= -H(n,1)+ H(n,3)-H(n,5)+....+H(n,2^n-1) I'll do a couple examples (...) means that a number in this interval has the desired property. 0 ( 1/2 1) 0 (1/3 1/2 2/3 ) 1 0 (1/4 1/3 2/5 ) 1/2 ( 3/5 2/3 3/4) 1 0 (1/5 1/4 2/7 ) 1/3 (3/8 2/5 3/7) 1/2 (4/7 3/5 5/8) 2/3 (5/7 3/4 4/5 ) 1 So this is how far I got so far....what do I do now? === Subject: Re: measure of continued fractions whose partial quotients sum to n. I was just messing around and found that the sequence I defined is almost exactly the same as the stern-Brocot Tree. === Subject: Re: Examples in Category theory >[...] > Although I believe if the coproduct [ in Rng ] exist then it must be > some sort of free ring .. don't even know if coproduct always > exist in this category. Actually it exists. The coproduct of R and S should be R (+) S (+) R(x)S equipped with a suitable multiplication and the obvious injections. The argument runs like this: Let Rng be the category of rings (possibly) without unity and let Ring be the category of unitary rings and unitary homomorphisms. (1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring where on objects F(R) = Z x R with componentwise addition and multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps F(f)(m,r) = (m,fr). (2) Suppose F: X --> A is left adjoint to G: A --> X and assume also, that G has pullbacks. Then for any object z of X, the pair F and G induce another pair of functors F': X/z --> A/F(z) and G': A/F(z) --> X/z such that F' is left adjoint to G'. F' is defined via F'(u: x->z) = F(u): F(x) -> F(z) and for v: a --> Fz, one defines G'v as the pullback of Gv and the original unit X => GF evaluated at z. p ---> Ga | | |G'v |Gv v v z ---> GFz switching via adjoint gives Fp ---> a | | |F'G'v | v v v which exhibits the counit F'G' => A/Fz of the new adjunction. (3) Applying (2) to the (1) with z = 0 (and using Rng/0 = Rng) one calculates ( I write (+) instead of x now ) F(0) = Z, F'(R) = (Z (+) R --> Z) (the projection to Z) G'(u: A --> Z) = ker(u) (4) Both functors F': Rng --> Ring/Z and G': Ring/Z --> Rng are full and faithful. Hence one has an equivalence between Rng and Ring/Z. Therefore the coproduct of R and S in Rng can be calculated by (i) replacing R and S by Z(+)R --> Z and Z(+)S --> Z (ii) calculation the coproduct of the above to objects in Ring/Z p: (Z(+)R) (x) (Z(+)S) --> Z with p( (m+r) (x) (n+s) ) = mn which can be rewritten as p: Z (+) R (+) S (+) R(x)S with p(m + r + s + v) = m (iii) going back to Rng, which gives the kernel of p as the desired coproduct. Marc === Subject: Re: Examples in Category theory Although I believe if the coproduct [ in Rng ] exist then it must be > some sort of free ring .. don't even know if coproduct always > exist in this category. Actually it exists. The coproduct of R and S should be > R (+) S (+) R(x)S equipped with a suitable multiplication > and the obvious injections. > Well I'm confused here alone, what do you mean by (x)? a tensor product?.. how can you make a tensor product of R and S? We don't assume that R and or S are Z-algebras.. if 1 were not say in R, then R can possibly be not a Z-algebra and so I don't know how you do the tensor product. > The argument runs like this: Let Rng be the category of rings (possibly) without unity and > let Ring be the category of unitary rings and unitary homomorphisms. > I suppose first that here you assume Rng has also objects as rings without units. > (1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring > where on objects F(R) = Z x R with componentwise addition and > multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps > F(f)(m,r) = (m,fr). > What gaurantees F(R) to be unitary? If R has no unit.. then ZxR has no unit if its just the componentwise addition/multiplication. Jose Capco === Subject: Re: Examples in Category theory >[...] > Although I believe if the coproduct [ in Rng ] exist then it must be > some sort of free ring .. don't even know if coproduct always > exist in this category. Actually it exists. The coproduct of R and S should be > R (+) S (+) R(x)S equipped with a suitable multiplication > and the obvious injections. > Well I'm confused here alone, what do you mean by (x)? a tensor > product?.. how can you make a tensor product of R and S? We don't > assume that R and or S are Z-algebras.. if 1 were not say in R, then R > can possibly be not a Z-algebra and so I don't know how you do the > tensor product. Actually R and S _are_ Z-algebras --- in fact symmetric two-sided Z-algebras (which is used in the construction of F(R)). Of course they may not be _unitary_ algebras, but that is no problem. R(x)S is just the tensor product of the abelian groups. > > The argument runs like this: Let Rng be the category of rings (possibly) without unity and > let Ring be the category of unitary rings and unitary homomorphisms. > I suppose first that here you assume Rng has also objects as rings > without units. Well, my above unity is not good. I meant Identity. I use Rng as mnemonic for Ring without Identity, so i drop the i from the word Ring. > > (1) The inclusion of Ring into Rng has a left adjoint F: Rng --> Ring > where on objects F(R) = Z x R with componentwise addition and > multiplication (m,r)(n,s) = (mn , nr + ms + rs) and on maps > F(f)(m,r) = (m,fr). > What gaurantees F(R) to be unitary? If R has no unit.. then ZxR has no > unit if its just the componentwise addition/multiplication. _^^^^^^^^^^^^^^ It is important, that the _multiplication_ on Z x R is _not_ componentwise. One has (m,r)(n,s) = (mn , nr + ms + rs) , so the unit is (1,0). One should think of (m,r) = (m,0) + (0,r) as a sum m + r, then it is clear what the product (m+r)(n+s) should be. Also, F(R) can be described as the set of matrices of the form (m r) (0 m) with m in Z and r in R. Then the ordinary addition and multiplication of matrices will give the ring structure of F(R). Marc === Subject: Holomorpic ramified covering Suppose we have f is a holomorphic ramified covering from some domain U -> V. Assume, in addition we know that f is actually unramified. How does it follow then that the degree of f is given by the degree of f, say, at the inner boundary component. (U is supposed to be closed). Seems to be easy, but i don't find appropiate words. Any ideas would be welcome. === Subject: Re: Holomorpic ramified covering > Suppose we have f is a holomorphic ramified covering from some domain > U -> V. Assume, in addition we know that f is actually unramified. How does it > follow then that the degree of f is given by the degree of f, say, at > the inner boundary component. (U is supposed to be closed). Seems to be easy, but i don't find appropiate words. Any ideas would be welcome. > We also know that both U and V are doubly-connected annuli. === Subject: (Q,+) as a Direct Limit I am sorry that probably I am going to be *very* imprecise, but I just want to get the idea of this direct limit construction. Q is a direct limit of an infinity of cyclic groups. I guess that these groups are those of the form <1/r> with r integer >0. It is easy to verify that the groups <1/r> induce a partial order (with respect to the inclusion relation) that is *directed*, i.e. for every u,v there exist w such that u,v < w. For example, G_2=<1/2> and G_3=<1/3> are not comparable, but they are included in <1/3*2>=<1/6>. Obviously, such groups are NOT disjoint, but we can easily think about an infinitude of suitable isomorphic copies that allow us to take their disjoint union. A direct system has *homomorphisms* defined in this way: a(t,t) is the identity G_t -> G_t a(r,s)a(s,t) = a(r,t) whenever r,s < t. Now, my question is: in order to obtain the direct limit, what homomorphisms do we have to consider? Intuitively I would choose a simple inclusion function (is it called an im/embedding?), for example x in <1/2> |--> x in <1/6>. This should work even with compositions: a(2,6) <1/2> ------------> <1/6> | | | | | | | | a(6,30) | | | a(2,30) | | | | V --------------> <1/30> . I know that this could be totally wrong, but I am absolutely a novice with direct limits. Let me know more about this problem, and please point me out what kind of homomorphisms do occur. === Subject: Re: (Q,+) as a Direct Limit > I am sorry that probably I am going to be *very* imprecise, but I just want > to get the idea of this direct limit construction. > Q is a direct limit of an infinity of cyclic groups. I guess that these > groups are those of the form <1/r> with r integer >0. > Now, my question is: in order to obtain the direct limit, what homomorphisms > do we have to consider? Intuitively I would choose a simple inclusion > function (is it called an im/embedding?), for example x in <1/2> |--> x in > <1/6>. You are basically correct. Define the group G_r = (1/r)Z for positive integers r. The partial order on the integers is divisibility and the maps f_{r,s} G_r -> G_s for r|s given by inclusion define a directed system with direct limit isomorphic to Q. Alternatively one can take groups H_s = Z, and consider the maps g_{r,s}: H_r -> H_s given by multiplication by s/r for r|s. Again the direct limit of this directed system is Q. Victor Meldrew === Subject: Function satisfying a Holder condition Hello everybody, I'm trying to do a simulation study. For this, I need an example of a real function on [0,1] which satisfies the Holder condition for one or more index *alpha* such that 1/2 < alpha < 1; but only in this interval. Does anybody knows one? Which one? Rogerio === Subject: fit curve to surround points in the plane I am searching for a way to fit a curve *around* a set of points in the plane. Preferably it should be parametrized, so that the best and ugliest fit involves straight line segments, and successive relaxations yield smoother curves that surround the set of points less tightly. The set of points may be redundant. The ultimate goal is to code this in PostScript. Any pointers to methods? Stijn === Subject: Re: fit curve to surround points in the plane > I am searching for a way to fit a curve *around* a set of points in > the plane. Preferably it should be parametrized, so that the best and > ugliest fit involves straight line segments, and successive > relaxations yield smoother curves that surround the set of points less > tightly. The set of points may be redundant. The ultimate goal is to > code this in PostScript. Any pointers to methods? Check out Zhao, H.-K., Osher, S. and Fedkiw, R., Fast Surface Reconstruction Using the Level Set Method, 1st IEEE Workshop on Variational and Level Set Methods, 8th ICCV, Vancouver, 194-202, 2001, available at http://math.uci.edu/~zhao/publication/mypapers/pdf/surface2.pdf. Might be overkill, but also works in higher dimensions. === Subject: Re: Comprehensive Solution Manual for Textbooks Does anyone have the test banks for : Foundations of Finance - Arthur Keown, William Petty, John Martin, David Scott (5th edition) (ISBN: 0131856057) I need them by tomm, if possible e mail me ASAP. === Subject: Math jobs that wont ask for a transcript. What jobs are there that having a math degree qualifies me for, where they are also unlikely to ask me for an academic transcript? === Subject: Re: Math jobs that wont ask for a transcript. > What jobs are there that having a math degree qualifies me for, where > they are also unlikely to ask me for an academic transcript? In the UK, I have rarely been asked for my degree certificate let alone a detailed transcript of my marks of individual courses - which is what I gather you mean - I had not previously heard of such a thing. There seems to be suggestion that you don't want to have to show that you didn't do any really hard maths courses in your maths degree. Again in the UK early in one's career it is normal to ask which unit one had undertaken. And I would imagine that if you are unlucky they might check with the institution;) Nick === Subject: Re: Math jobs that wont ask for a transcript. they are also unlikely to ask me for an academic transcript? In the UK, I have rarely been asked for my degree certificate let alone a > detailed transcript of my marks of individual courses - which is what I > gather you mean - I had not previously heard of such a thing. There seems to be suggestion that you don't want to have to show that you > didn't do any really hard maths courses in your maths degree. Again in the UK early in one's career it is normal to ask which unit one had > undertaken. And I would imagine that if you are unlucky they might check > with the institution;) Nick In the US I'd be surprised if a job asked to see an academic transcript. Letters of reference are almost a given though. especially in an academic field. === Subject: Re: zeta(1) = (-1)! ?? Am 15.05.2007 15:14 schrieb Gottfried Helms: > Am 15.05.2007 13:30 schrieb [Micro]: > Gottfried Helms a .8ecrit : > in the context of binomials I came across this > substitution as meaningful, or say more precisely > that it would be meaningful to set > zeta(1) / (-1)! = 1 > Could you please tell us how you found this? I guess the Euler Gamma > function is used somewhere. > link tomorrow or so. > version. References should be added, and I'd like helpful comments to correct errors and to improve contents/arguing. You may email or cc any comment to my mailbox. TIA for any consideration - Gottfried Helms === Subject: Presuppositions and Occurrence Relevance Logic Some time ago on sci.crypt I posted some notes about the relationship between the logic of pressupositions, occurrence-relevance logic, liar's paradox and G.9adel's sentence. I have now written an entire paper on the subject. The objective of this paper is to explore the relationship between the logic of presupposition and occurrence relevance logic. In section 1 we recapitulate the logic of presuppositions as proposed by P.F. Strawson. In section 2 we note the similarities between the logic of presuppositions and occurrence -relevance logic proposed by Richard Diaz. In section 3 we recall van Fraassen's solution of Liar's paradox based on the logic of presuppositions. In section 4 we ponder the similarities between Liar's paradox and G.9adel's sentence. Finally in section 5 we justify the view that not every sentence has to have a truth value. http://xnewberry.tripod.com/Presuppositions 2007 05 19.html === Subject: Re: Presuppositions and Occurrence Relevance Logic > Some time ago on sci.crypt I posted some notes about the relationship > between the logic of pressupositions, occurrence-relevance logic, > liar's paradox and G.9adel's sentence. I have now written an entire > paper on the subject. > > The objective of this paper is to explore the relationship between the > logic of presupposition and occurrence relevance logic. In section 1 > we recapitulate the logic of presuppositions as proposed by P.F. > Strawson. In section 2 we note the similarities between the logic of > presuppositions and occurrence -relevance logic proposed by Richard > Diaz. In section 3 we recall van Fraassen's solution of Liar's paradox > based on the logic of presuppositions. In section 4 we ponder the > similarities between Liar's paradox and G.9adel's sentence. Finally in > section 5 we justify the view that not every sentence has to have a > truth value. > > http://xnewberry.tripod.com/Presuppositions_2007_05_19.html > > We have concluded after a long and painful process that This sentence is false is neither true nor false. This seems rather odd. In the course of everyday life when we communicate with others we do not spend hours on each sentence looking for a proof by contradiction that it does not have a truth value. There must be a more direct method. In order to tell whether a picture is true or false we must compare it with reality. [TLP 2.223] Sentences are pictures and therefore to find if This sentence is false is true we have to compare it with reality. What does it say about reality? Nothing. There is nothing to compare it with. It is meaningless. This sentence is false does not have a truth value because it is meaningless. Yes why spend hours with proofs? Why not try a simple truth table: X X<->~X 0 0 1 0 So in a blink of an eye we know everything about reality and have a full picture.... Yawn ... Bye === Subject: Re: Presuppositions and Occurrence Relevance Logic > Some time ago on sci.crypt I posted some notes about the relationship > between the logic of pressupositions, occurrence-relevance logic, > liar's paradox and G.9adel's sentence. I have now written an entire > paper on the subject. > The objective of this paper is to explore the relationship between the > logic of presupposition and occurrence relevance logic. In section 1 > we recapitulate the logic of presuppositions as proposed by P.F. > Strawson. In section 2 we note the similarities between the logic of > presuppositions and occurrence -relevance logic proposed by Richard > Diaz. In section 3 we recall van Fraassen's solution of Liar's paradox > based on the logic of presuppositions. In section 4 we ponder the > similarities between Liar's paradox and G.9adel's sentence. Finally in > section 5 we justify the view that not every sentence has to have a > truth value. > http://xnewberry.tripod.com/Presuppositions 2007 05 19.html > We have concluded after a long and painful process that This > sentence is false is neither true nor false. This seems rather > odd. In the course of everyday life when we communicate with > others we do not spend hours on each sentence looking for a > proof by contradiction that it does not have a truth value. > There must be a more direct method. In order to tell whether a > picture is true or false we must compare it with reality. [TLP > 2.223] Sentences are pictures and therefore to find if This > sentence is false is true we have to compare it with reality. > What does it say about reality? Nothing. There is nothing to > compare it with. It is meaningless. This sentence is false > does not have a truth value because it is meaningless. Yes why spend hours with proofs? Why not > try a simple truth table: X X<->~X > 0 0 > 1 0 So in a blink of an eye we know everything > about reality and have a full picture.... Not that I care for Newberry's theory, but your response seems odd. This sentence is false is paradoxical. X <-> ~X is not paradoxical at all. It is a mere contradiction. Your truth table does not seem so useful. -- These mathematicians are worse than communists, as how do you explain their behavior? I *am* the American Dream, fighting for what should be mine, having to get past weak-minded academics who are fighting to block my success. But I shall prevail!!! -- James S. Harris === Subject: Re: Presuppositions and Occurrence Relevance Logic > Yes why spend hours with proofs? Why not > try a simple truth table: > X X<->~X > 0 0 > 1 0 > So in a blink of an eye we know everything > about reality and have a full picture.... > > Not that I care for Newberry's theory, but your response seems odd. > > This sentence is false is paradoxical. X <-> ~X is not > paradoxical at all. It is a mere contradiction. > > Your truth table does not seem so useful. > This sentence is false is also a mere contradiction. As are most of the paradoxies. Bye === Subject: Re: Presuppositions and Occurrence Relevance Logic > Yes why spend hours with proofs? Why not > try a simple truth table: > X X<->~X > 0 0 > 1 0 > So in a blink of an eye we know everything > about reality and have a full picture.... > Not that I care for Newberry's theory, but your response seems odd. > This sentence is false is paradoxical. X <-> ~X is not > paradoxical at all. It is a mere contradiction. > Your truth table does not seem so useful. > This sentence is false is also a mere contradiction. > As are most of the paradoxies. Wrong. A mere contradiction is a statement that is necessarily false. If This sentence is false, is necessarily false, then it is false. Therefore, This sentence is false asserts a true proposition, namely that it is false. Thus, it is true. No statement is both true and false. Thus, we can confidently claim that This sentence is false is not false and hence it cannot be a contradiction. (Of course, it is not true either, for very plain and obvious reasons.) Your analysis of This sentence is false is remarkably misguided. -- Jesse F. Hughes Disney has now succeeded in preventing anyone from doing to Mickey Mouse what Disney did to Quasimodo. -- Randolph Rackovitz, on Eldred vs. Ashcroft === Subject: Re: Presuppositions and Occurrence Relevance Logic > Wrong. If you look at the truth table you have always a zero on the left hand side. Thus the sentence and the relationship it suggest to itself is not satisfiable and the same time. > A mere contradiction is a statement > that is necessarily false. If Yes, if a sentence is not satisfiable (nicht erf.9fllbar) then it is contradictory (widerspruchsvoll) and vice versa. Content of the first completness theorem from G.9adel for example. Soundness and completness between valuation and proof system. > No statement is both true and false. Thus, we can confidently claim > that This sentence is false is not false and hence it cannot be a > contradiction. Yes, there is no truth value called true and false. But the truth table checks in the first row the value true for X, and in the second row the truth value false for X. > (Of course, it is not true either, for very plain and obvious > reasons.) My saying in the truth table. The result of evaluating the sentence and the relationship it suggesst, for all cases X=true and X=false, yields false. Thus the sentence and the relationship it suggests is not satisfiable. > Your analysis of This sentence is false is remarkably misguided. Why? It is only a picture of what you have said with words. > This sentence is false, is necessarily false, then it is false. > Therefore, This sentence is false asserts a true proposition, namely > that it is false. Thus, it is true. When you are using the modifier necessarily false then ok the sentence needs to be interpreted differently. Then you have something like: This sentence is necessarily false X <-> (diamond (X <-> false)) This needs another analysis then the truth table I presented, yes. But I didn't want to give a picture for the sentence with necessarily in it. My venture was much simpler. === Subject: Re: Presuppositions and Occurrence Relevance Logic On Sun, 20 May 2007 08:50:36 -0400, Jesse F. Hughes > This sentence is false is also a mere contradiction. > As are most of the paradoxies. > Wrong. > Right. As if it were that simple... In my judgment, it would be quite wrong and dangerous from the standpoint of scientific progress to depreciate the importance of this and other antinomies, and to treat them as jokes or sophistries. It is a fact that we are here in the presence of an absurdity, that we have been compelled to assert a false sentence (since (3), as an equivalence between two contradictory sentences, is necessarily false). If we take our work seriously, we cannot be reconciled with this fact. We must discover its cause, that is to say, we must analyze premises upon which the antinomy is based; we must then reject at least one of these premises, and we must investigate the consequences which this has for the whole domain of our research. It should be emphasized that antinomies have played a preeminent role in establishing the foundations of modern deductive sciences. And just as class-theoretical antinomies, and in particular Russell's antinomy (of the class of all classes that are not members of themselves), were the starting point for the successful attempts at a consistent formalization of logic and mathematics, so the antinomy of the liar and other semantic antinomies give rise to the construction of theoretical semantics. Source: Alfred Tarski, The Semantic Conception of Truth and the Foundations of Semantics, 1944. http://www.jfsowa.com/logic/tarski.htm Note the final sentence: ...the antinomy of the liar and other semantic antinomies give rise to the construction of theoretical semantics. --- says who? ;-) > Yawn ... > Very big yawn ... > Very very big yawn ... > Well, I guess, that explains a lot. Your analysis of This sentence is false is remarkably misguided. > Indeed. F. -- E-mail: infosimple-linede === Subject: Re: Presuppositions and Occurrence Relevance Logic On Sun, 20 May 2007 12:03:13 +0200, J. Burse Yes why spend hours with proofs? Why not > try a simple truth table: > X X<->~X > 0 0 > 1 0 > So in a blink of an eye we know everything > about reality and have a full picture.... > > Not that I care for Newberry's theory, but your response seems odd. > > This sentence is false is paradoxical. X <-> ~X is not > paradoxical at all. It is a mere contradiction. > > Your truth table does not seem so useful. > >This sentence is false is also a mere contradiction. Huh? X <-> ~X is in fact a mere contradiction: there's no problem assigning it a truth value, it's a perfectly well-behaved anti-tautology, always false. This sentence is false is certainly not a mere contradiction. Because if it's false it follows that it's true, while if it's true it follow that it's false. >As are most of the paradoxies. Good of you to share this with us. >Bye ************************ David C. Ullrich === Subject: Re: Presuppositions and Occurrence Relevance Logic On Sun, 20 May 2007 12:03:13 +0200, J. Burse > Yes why spend hours with proofs? Why not > try a simple truth table: > X X <-> ~X > 0 0 > 1 0 > So in a blink of an eye we know everything > about reality and have a full picture.... > > Not that I care for Newberry's theory, but your response > seems odd. > Indeed. > This sentence is false is paradoxical. X <-> ~X is > not paradoxical at all. It is a mere contradiction. > > Your truth table does not seem so useful. > Right. > > This sentence is false is also a mere contradiction. > As are most of the paradoxies. > Nonsense. F. -- E-mail: infosimple-linede === Subject: Re: Presuppositions and Occurrence Relevance Logic > This sentence is false is also a mere contradiction. > As are most of the paradoxies. > Nonsense. No not nonsense. But the bare truth. Read for example: http://de.wikipedia.org/wiki/Paradoxon I translate it for you: A paradoxon or paradox or paradoxie is a special *contradiction*. Old greek para=contra, doxa=opinion. 1) Performative contradiction: Contradiction as a result of a negative self referentiality. Example paradox of Eubulides: * This sentence is false. (It is true, if it is false, and it is false, if it is true) Here my penny of analysis of this sentence: This sentence is false. This is .. : X <-> .. .. is false : (.. <-> false) So in summa: X <-> (X <-> false) Which we can also write as: X <-> ~X Yawn ... Very big yawn ... Very very big yawn ... Bye === Subject: Re: Presuppositions and Occurrence Relevance Logic > This sentence is false is also a mere contradiction. > As are most of the paradoxies. > Nonsense. No not nonsense. But the bare truth. > Read for example: http://de.wikipedia.org/wiki/Paradoxon I translate it for you: A paradoxon or paradox or paradoxie is a special > *contradiction*. Old greek para=contra, doxa=opinion. I don't really care how Wikipedia uses the term contradiction. In logic, a contradiction is a statement which is necessarily false. The liars' paradox is neither true nor false and hence it is not a contradiction. There is a related contradiction. The liar's sentence is true iff the liar's sentence is not true. If we use X to denote The liar's sentence is true, then we see that this can be written as X <-> ~X. Now that's a plain contradiction, since that sentence is necessarily false. But that's not the liar's sentence. The liar's sentence is neither true nor false. It is not necessarily false and hence it is not a contradiction. [...] > Here my penny of analysis of this sentence: This sentence is false. This is .. : X <-> .. > .. is false : (.. <-> false) So in summa: X <-> (X <-> false) Which we can also write as: X <-> ~X What is X? It seems to be standing in for the liar's sentence. But if X is the liar's sentence, then the liar's sentence cannot be X <-> ~X, since X <-> ~X is not the same sentence as X. Honestly, Jan, this is just a horrible analysis, contrary to every published discussion on this paradox[1]. How is it that everyone else has come to this mistaken belief that the liar sentence can neither be true nor false when you've been able to show that it is merely false in a few lines? Either you have remarkable insight into a paradox that has been misanalyzed for a couple of thousand years or your correspondents are correct and you are simply on the wrong track. Which do you think is more plausible? > Yawn ... > Very big yawn ... > Very very big yawn ... Hey, don't yawn! This is a great breakthrough! *Everyone* thought that the liar sentence was a paradox, neither true nor false. You've just shown it is no more paradoxical than A & ~A. Golly! Footnotes: -- Jesse F. Hughes Hey look, Captain, next time someone wants to tie us up, let's put up a fight. --Adventures by Morse === Subject: Re: Presuppositions and Occurrence Relevance Logic On Sun, 20 May 2007 09:04:08 -0400, Jesse F. Hughes The liars' paradox is neither true nor false and hence it is not a > contradiction. > Careful with that axe, Eugene. You are talking about the liar _sentence_ here. [...] [...] The liar's sentence ... > See?! :-) ... is true iff the liar's sentence is not true. If we use X to denote > The liar's sentence is true, then we see that this can be written as > X <-> ~X. Now that's a plain contradiction, since that sentence is > necessarily false. But that's not the liar's sentence. > Exactly. The liar's sentence is neither true nor false. > Or maybe it's both... (->Dialetheism) It is not necessarily false and hence it is not a contradiction. > Right. > Here my penny of analysis of this sentence: > [Snipped some crap] > Honestly, Jan, this is just a horrible analysis, contrary to every > [or at least almost all --G.F.] published discussion on this paradox. > Indeed. > Yawn ... > Very big yawn ... > Very very big yawn ... > Hey, don't yawn! This is a great breakthrough! > He really shouldn't! In my judgment, it would be quite wrong and dangerous from the standpoint of scientific progress to depreciate the importance of this and other antinomies, and to treat them as jokes or sophistries. It is a fact that we are here in the presence of an absurdity, that we have been compelled to assert a false sentence (since (3), as an equivalence between two contradictory sentences, is necessarily false). If we take our work seriously, we cannot be reconciled with this fact. We must discover its cause, that is to say, we must analyze premises upon which the antinomy is based; we must then reject at least one of these premises, and we must investigate the consequences which this has for the whole domain of our research. It should be emphasized that antinomies have played a preeminent role in establishing the foundations of modern deductive sciences. And just as class-theoretical antinomies, and in particular Russell's antinomy (of the class of all classes that are not members of themselves), were the starting point for the successful attempts at a consistent formalization of logic and mathematics, so the antinomy of the liar and other semantic antinomies give rise to the construction of theoretical semantics. (Alfred Tarski) F. http://plato.stanford.edu/entries/dialetheism/ Alfred Tarski, The Semantic Conception of Truth and the Foundations of Semantics, 1944. http://www.jfsowa.com/logic/tarski.htm -- E-mail: infosimple-linede === Subject: Re: Presuppositions and Occurrence Relevance Logic > On Sun, 20 May 2007 09:04:08 -0400, Jesse F. Hughes > The liars' paradox is neither true nor false and hence it is not a > contradiction. > Careful with that axe, Eugene. You are talking about the liar > _sentence_ here. [...] Yes, you're right. -- Jesse F. Hughes [Lancelot] sighed, defeated. 'It is as practical to hurry an acorn toward treeness as to urge a damsel when her mind is set.' -- John Steinbeck, /The Acts of King Arthur and His Noble Knights/ === Subject: Re: Presuppositions and Occurrence Relevance Logic On Sun, 20 May 2007 13:12:48 +0200, J. Burse Here my penny of analysis of this sentence: > Your analysis is crap. (Sorry.) The following one is much more reasonable: http://www.jfsowa.com/logic/tarski.htm#sec7 Tarski's conclusion: In my judgment, it would be quite wrong and dangerous from the standpoint of scientific progress to depreciate the importance of this and other antinomies, and to treat them as jokes or sophistries. It is a fact that we are here in the presence of an absurdity, that we have been compelled to assert a false sentence (since (3), as an equivalence between two contradictory sentences, is necessarily false). If we take our work seriously, we cannot be reconciled with this fact. We must discover its cause, that is to say, we must analyze premises upon which the antinomy is based; we must then reject at least one of these premises, and we must investigate the consequences which this has for the whole domain of our research. It should be emphasized that antinomies have played a preeminent role in establishing the foundations of modern deductive sciences. And just as class-theoretical antinomies, and in particular Russell's antinomy (of the class of all classes that are not members of themselves), were the starting point for the successful attempts at a consistent formalization of logic and mathematics, so the antinomy of the liar and other semantic antinomies give rise to the construction of theoretical semantics. Tarski? Who the is Tarski?! F. -- E-mail: infosimple-linede === Subject: Re: Presuppositions and Occurrence Relevance Logic > This sentence is false is also a mere contradiction. > As are most of the paradoxies. > Nonsense. No not nonsense. But the bare truth. > Read for example: http://de.wikipedia.org/wiki/Paradoxon I translate it for you: A paradoxon or paradox or paradoxie is a special > *contradiction*. Old greek para=contra, doxa=opinion. 1) Performative contradiction: Contradiction as > a result of a negative self referentiality. Example > paradox of Eubulides: * This sentence is false. (It is true, if it is > false, and it is false, if it is true) > Here my penny of analysis of this sentence: This sentence is false. This is .. : X <-> .. > .. is false : (.. <-> false) So in summa: X <-> (X <-> false) Which we can also write as: X <-> ~X > You're absolutely, utterly mistaken. Completely missed the point there by substituting .. for sentence. Can you even see that these two statements are different: False. This statement is false. Because you are asserting they are equal. > Yawn ... > Very big yawn ... > Very very big yawn ... > Agreed. -- Glen === Subject: Re: Relationship between RBFs and Implicit Functions > I'm sorry if this question has an obvious answer, but what's the > relationship between Radial Basis Functions and Implicit functions? I now know that implicit surfaces *can* be created from scattered 1. Variational Implicit Surfaces by Grek Turk and James F. O'brien http://graphics.cs.uiuc.edu/~jch/cs497jch/variational-is.pdf Compactly Supported Radial Basis Functions http://www.cs.umbc.edu/~rheingan/pubs/smi2001.pdf Unfortunately, although both papers treat radial basis functions (RBFs) in sufficient depth, neither seems to have taken care to explain how the maths of RBFs produce a(n implicit) function that has the form, F(x) = 0. My primary concerns are that: 1. The RBF interpolation functions (three of them for the 3D interpolation problem) interpolate the height field, which does not do not equate to zero. 2. Only one implicit function is required, the RBF interpolation problem produces 2 or 3 equations. Is the single implicit function produced by making each of the interpolation equations to zero and summing them? i.e: X = F1(x) -> G1(x) = X - F1(x) = 0 Y = F2(y) -> G2(y) = Y - F2(y) = 0 and G(x,y) = X + Y - F1(x) - F2(y) = 0 . === Subject: Re: Somewhat Advanced Set Theory Question *weeps* === Subject: Integral of interest What's up? I am tinkering with the following anti derivative: Integral[dx/ (x^3+1)] I know what the integral is equal to due to the use of th CRC and Mathematica The problem is all the substitutions that I've tried to work through it myself have resulted in stuff that doesn't help me see a way out. I've tried, u:=x^3+1 , x:=(u^3 - 1)^(1/3) , dx/du:=u^2(u^3 - 1)^(-2/3) As well as, w:=1/(x^3+1) , x:=((1/w) - 1)^(1/3) , dx/dw:=(1/ (3w^2))((1/w) - 1)^(-2/3) and some trig stuff that failed to make sense. I've also tried integration by parts. I'm pretty sure i can split that integrand up into a couple of fractions but i cannot remember how to get there at the moment. Am I heading in the right direction with any of this or am I missing the point? By the way, the integrals in the CRC have answers that contain natural log and inverse tangent === Subject: Re: Integral of interest > What's up? I am tinkering with the following anti derivative: Integral[dx/ > (x^3+1)] I know what the integral is equal to due to the use of th CRC and > Mathematica The problem is all the substitutions that I've tried to > work through it myself have resulted in stuff that doesn't help me see > a way out. I've tried, u:=x^3+1 , x:=(u^3 - 1)^(1/3) , dx/du:=u^2(u^3 - > 1)^(-2/3) As well as, w:=1/(x^3+1) , x:=((1/w) - 1)^(1/3) , dx/dw:=(1/ > (3w^2))((1/w) - 1)^(-2/3) and some trig stuff that failed to make sense. I've also tried > integration by parts. 1/(x^3+1) = 1/[(x+1)*(x^2-x+1)]. Expand this in partial fractions. I'm pretty sure i can split that integrand up into a couple of > fractions but i cannot remember how to get there at the moment. Am I heading in the right direction with any of this or am I missing > the point? By the way, the integrals in the CRC have answers that contain natural > log and inverse tangent Maple gets 1/3*ln(x+1)-1/6*ln(x^2-x+1)+1/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2)) R.G. Vickson === Subject: Re: Integral of interest === Subject: Re: Integral of interest On 19 May 2007 16:16:02 -0700, Eratosthenes (x^3+1)] >I'm pretty sure i can split that integrand up into a couple of >fractions but i cannot remember how to get there at the moment. Review partial fraction decomposition. 1 / (x^3 + 1) = 1/3 [ 1/(x + 1) - (x - 2)/(x^2 - x + 1) ] === Subject: Re: Integral of interest >What's up? I am tinkering with the following anti derivative: Integral[dx/ >(x^3+1)] I know what the integral is equal to due to the use of th CRC and >Mathematica The problem is all the substitutions that I've tried to >work through it myself have resulted in stuff that doesn't help me see >a way out. I've tried, u:=x^3+1 , x:=(u^3 - 1)^(1/3) , dx/du:=u^2(u^3 - >1)^(-2/3) As well as, w:=1/(x^3+1) , x:=((1/w) - 1)^(1/3) , dx/dw:=(1/ >(3w^2))((1/w) - 1)^(-2/3) and some trig stuff that failed to make sense. I've also tried >integration by parts. I'm pretty sure i can split that integrand up into a couple of >fractions but i cannot remember how to get there at the moment. > > Yes, you use partial fractions. x^3 + 1 = (x + 1) (x^2 - x + 1). >Am I heading in the right direction with any of this or am I missing >the point? By the way, the integrals in the CRC have answers that contain natural >log and inverse tangent. > -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Complete electronic solution manual in PDF! Get it in hours! I have the complete electronic SOLUTION MANUAL in PDF format containing ALL (odd and even - except where noted) solutions for the books listed below. These are not paper books, they are ebooks. The price is $8 each. I accept payment through paypal only. If interested, just send me an email to deltabooks@gmail.com mentioning which solution manual you want. I will promptly reply you with details of where you can send the paypal payment. As soon as the payment is received, a download link for the solution manual will be sent to your email address within a few hours. Fast service! NB: Don't post replies here. If interested, send me an email to deltabooks@gmail.com Books for which I have electronic solution manual: 1. Calculus: Early Transcendentals (5th Ed) by James Stewart [ISBN: 0534393217] 2. Digital Communications (4th Ed) by Proakis, John [ISBN: 0072321113] 3. Engineering Mechanics - Dynamics (10th Ed) by Russell C. Hibbeler [ISBN: 0131416782] 4. Engineering Mechanics - Statics (10th Ed) by Russell C. Hibbeler [ISBN: 0131411675] 5. Engineering Mechanics , Dynamics (5th Ed) by J. L. Meriam, L. G. Kraige [ISBN: 0471406457] 6. Electric Machinery Fundamentals (4th Ed) by Stephen J. Chapman [ISBN: 0072465239] 7. First Course in Probability, A (7th Ed) by Ross, Sheldon [ISBN: 0131856626] 8. Vector Mechanics for Engineers, Statics (7th Ed) by Beer, Ferdinand P.; Johnston Jr., E [ISBN: 0072930780] 9. Microwave and Rf Design of Wireless Systems (1st Ediiton) by Pozar, David M. [ISBN: 0471322822] 10. Communication Systems Engineering (2nd Ed) by Proakis, John G [ISBN: 0130617938] 11. Introduction to Solid State Physics (8th Ed) by Charles Kittel [ISBN: 047141526X] 12. Semiconductor Physics And Devices (3rd Ed) by Donald Neamen [ISBN: 0072321075] 13. Advanced Modern Engineering Mathematics (3rd Ed) by Glyn James [ISBN: 0130454257] 14. Physical Chemistry (7th Ed) by Julio Paula, Peter Atkins [ISBN: 0716735393] 15. Introduction to Quantum Mechanics (2nd Edition) by David J. Griffiths [ISBN: 0131118927] 16. Digital Signal Processing by Thomas J. Cavicchi [ISBN: 0471124729] 17. Electric Circuits (7th Ed) by Nilsson, James W.; Riedel, Susan [ISBN:0131465929] 18. Computer Networks (4th Ed) by Tanenbaum, Andrew S. [ISBN: 0130661023] 19. Adaptive Filter Theory (4th Ed) by Haykin, Simon [ISBN: 0130901261] 20. Wireless Communications: Principles and Practice (2nd Ed) by Theodore Rappaport [ISBN: 0130422320] 21. Principles of Communication: Systems, Modulation and Noise (5th Ed) by R. E. Ziemer, W. H. Tranter [ISBN: 0471392537] 22. Automatic Control Systems (8th Ed) by Kuo, Benjamin C.; Golnaraghi, Farid [ISBN: 0471134767] 23. Microwave Engineering (3rd Ed) by Pozar, David M. [ISBN: 0471448788] 24. Discrete-Time Signal Processing (2nd Ed) by Oppenheim, Alan V [ISBN: 0137549202] 25. Introduction to Electrodynamics (3rd Ed) by Griffiths, David [ISBN: 013805326X] 26. Communications Systems (4th Ed) by Haykin, Simon [ISBN: 0471178691] 27. Control Systems Engineering (4th Ed) by Nise, Norman S. [ISBN: 0471445770] 28. Design of Analog CMOS Integrated Circuits (1st Ed) by Behzad Razavi [ISBN: 0072380322] 29. Signals and Systems (2nd Ed) by Alan V. Oppenheim, Alan S. Willsky [ISBN: 0138147574] 30. Signal Processing and Linear Systems by B. P. LAthi [ISBN: 0195219171] 31. Engineering Circuit Analysis (6th Ed) by William H. Hayt, Jack Kemmerly, Steven M. Durbin [ISBN: 0072853204] 32. RF Circuit Design: Theory and Applications by Reinhold Ludwig, Pavel Bretchko [ISBN: 0130953237] 33. Engineering Electromagnetics (6th Ed) by William H. Hayt, John A. Buck [ISBN: 0072551666] 34. Physics for Scientists and Engineers (6th Ed) by Raymond A. Serway, John W. Jewett [ISBN: 0534408427] 35. Antennas for All Applications (3rd Ed) by John D. Kraus, Ronald J. Marhefka [ISBN: 0072321032] 36. Digital Communications: Fundamentals and Applications (2nd Ed) by Bernard Sklar [ISBN: 0130847887] 37. Digital Image Processing (2nd Ed) by Rafael Gonzalez, Richard Woods [ISBN: 0201180758] 38. Fundamentals of Digital Logic with Verilog Design (1st Ed) by Stephen Brown, Zvonko Vranesic [ISBN: 0072838787] 39. Electronic Circuit Analysis (2nd Ed) by Donald Neamen [ISBN: 0072451947] 40. Fundamentals of Electric Circuits (2nd Ed ) by Charles Alexander, Matthew Sadiku [ISBN: 0073048356] 41. Device Electronics for Integrated Circuits (3rd Ed) by Richard Muller, Theodore Kamins, Mansun Chan [ISBN: 0471593982] 42. Linear Circuit Analysis: Time Domain, Phasor, and Laplace Transform Approaches (2nd Ed) by Raymond DeCarlo, Pen-Min Lin [ISBN: 0195136667] 43. Computer Networks: A Systems Approach (3rd Ed) by Larry L. Peterson, Bruce S. Davie [ISBN: 155860832X] 44. Fundamentals of Electromagnetics with Engineering Applications by Stuart M. Wentworth [ISBN: 0471263559] 45. Feedback Control of Dynamic Systems (4th Ed) by Gene Franklin, David Powell, Abbas Naeini [ISBN: 0130323934] 46. Fundamentals of Physics (7th Ed) by David Halliday, Robert Resnick, Jearl Walker [ISBN: 0471216437] 47. Introduction to Electric Circuits (6th Ed) by Richard C. Dorf, James A. Svoboda [ISBN: 0471447951] 48. Fundamentals of Thermodynamics (5th Ed) by Richard E. Sonntag, Claus Borgnakke, Gordon J. Van Wylen [ISBN: 047118361X] 49. Semiconductor Device Fundamentals (1st Ed) by Robert F. Pierret, [ISBN: 0201543931] 50. Applied Numerical Analysis (6th Ed) by Curtis F. Gerald, Patrick O. Wheatley [ISBN: 0321133048] 51. Digital Signal Processing (4th Ed) by John G. Proakis, Dimitris K Manolakis [ISBN: 0131873741] 52. Physics for Scientists and Engineers (1st Ed) by Randall D. Knight [ISBN: 0805386858] 53. Physics for Scientists and Engineers, Volume 1 (1st Ed) by Randall D. Knight [ISBN: 0805389644] 54. Physics for Scientists and Engineers, Volume 4 (1st Ed) by Randall D. Knight [ISBN: 0805389733] 55. Vector Mechanics for Engineers: Dynamics (7th Ed) by Beer, Ferdinand P.; Johnston Jr., E [ISBN: 0073209260] 56. Physics: Principles with Applications (6th Ed) by Douglas C. Giancoli [ISBN: 0130606200] 57. Physics for Scientists and Engineers (5th Ed) by Paul A. Tipler, Gene Mosca [ISBN: 0716783398] 58.Materials Science and Engineering: An Introduction (6th Ed) by William D. Jr. Callister [ISBN: 0471135763] 59. Engineering Mechanics - Statics (11th Ed) by Russell C. Hibbeler [ISBN: 0132215004] 60. Elementary Differential Equations and Boundary Value Problems (8th Ed) by William E. Boyce, Richard C. DiPrima [ISBN: 0471433381] (Mostly all solutions, only a few missing) 61. Partial Differential Equations and Boundary Value Problems with Fourier Series (2nd Ed) by Nakhle H. Asmar [ISBN: 0131480960] (Student solution manual) Thornton , Jerry B. Marion [ISBN: 0534408966] 63. Power System Analysis by(1st Ed) John Grainger, Jr., William Stevenson [ISBN: 0070612935] 64. Fundamentals of Applied Electromagnetics (1st Ed) Fawwaz Ulaby [ISBN: 013185089X] 65. Microelectronics (2nd Ed) by Jacob Millman, Arvin Grabel [ISBN: 007100596X] 66. Microeconomic Theory by Andreu Mas-Colell, Michael D. Whinston, Jerry R. Green [ISBN: 0195073401] 67. Econometric Analysis (5th Ed) by William H. Greene [ISBN: 0130661899] 68. Fluid Mechanics (5th Ed) by Frank M. White [ISBN: 0072831804] 69. Mechanical Engineering Design (7th Ed) by Charles Mischke, Joseph Shigley, Richard Budynas [ISBN: 0072921935] 70. Linear Algebra And It's Applications (3rd Ed) by David C. Lay [ISBN: 0321287134] 71. Single Variable Calculus (5th Ed) by James Stewart [ISBN: 0534393306] 72. Modern Physics (4th Ed) by Paul Tipler, Ralph Llewellyn [ISBN: 0716743450] 73. Thermodynamics (5th Ed) by Michael Boles, Yunus Cengel [ISBN: 0073107689] 74. University Physics with Modern Physics (11th Ed) by Hugh Young, Roger Freedman [ISBN: 0805387773] 75. Fundamentals of Thermal-Fluid Sciences (2nd Ed) by Yunus Cengel, Robert Turner [ISBN: 0072976756] 76. Fundamentals of Engineering Thermodynamics (5th Ed) by Michael Moran, Howard Shapiro [ISBN: 0471274712] 77. Principles and Applications of Electrical Engineering (5th Ed) by Giorgio Rizzoni [ISBN: 0073220337] 78. Modern Digital and Analog Communication Systems (3rd Ed) by B. P. Lathi [ISBN: 0195110099] 79. Mechanics of Materials (6th Ed) by Russell C. Hibbeler [ISBN: 013191345X] 80. Introduction to Fluid Mechanics (6th Ed) by Robert Fox, Alan McDonald, Philip Pritchard [ISBN: 0471202312] 81. Fundamentals of Heat and Mass Transfer (5th Ed) by Frank Incropera, David DeWitt [ISBN: 0471386502] 82. Engineering Mechanics - Statics (5th Ed) by J. L. Meriam (Author), L. G. Kraige [ISBN: 0471406465] 83. Materials Science and Engineering: An Introduction (7th Ed) by William D. Jr. Callister [ISBN: 0471736961] 84. Calculus (5th Ed) by James Stewart [ISBN: 053439339X] 85. Probability, Random Variables and Stochastic Processes (4th Ed) by Athanasios Papoulis , S. Unnikrishna Pillai [ISBN: 0072817259] 86. Field and Wave Electromagnetics (2And Ed) by David K. Cheng [ISBN: 0201128195] 87. Optical Fiber Communications (3rd Ed) by Gerd E. Keiser [ISBN: 0071164685] 88. Analysis and Design of Analog Integrated Circuits (4th Ed) by Paul R. Gray, Paul J. Hurst, Stephen H. Lewis, Robert G. Meyer [ISBN: 0471321680] 89. Design with Operational Amplifiers and Analog Integrated Circuits (3rd Ed) by Sergio Franco [ISBN: 0072320842] 90. Communication Systems (4th Ed) by A. Bruce Carlson, Paul B. Crilly, Janet Rutledge [ISBN: 0070111278] 91. Electric Machinery (6th Ed) by A. E. Fitzgerald, Jr., Charles Kingsley, Stephen D. Umans, Stephen Umans [ISBN: 0073660094] 92. Computer System Architecture (3rd Ed) by M. Morris Mano [ISBN: 0131755633] 93. An Introduction to Numerical Analysis by Endre S.9fli , David F. Mayers [ISBN: 0521007941] 94. Digital Signal Processing (3rd Ed) by Sanjit K Mitra [ISBN: 0073048372] 95. Modern Operating Systems (2nd Ed) by Andrew Tanenbaum [ISBN: 0130313580] (up to chapter 12) 96. Operating Systems Design and Implementation (3rd Ed) by Andrew S Tanenbaum, Albert S Woodhull [ISBN: 0131429388] 97. Basic Engineering Circuit Analysis (8th Ed) by J. David Irwin, R. Mark Nelms [ISBN: 0471487287] === Subject: Re: Racing cars on a track Good point. At high speed, a car's power is probably insufficient to spin the wheels, so the limit on acceleration becomes an issue depending on whether the car is gunning the engine or simply going into a turn. Dan in Philly === Subject: Re: Racing cars on a track > Consider this curve: The outside edge has a begin-of-curve with N,E coordinates of 1000,1000 . > The curve has tangent intersections with N,E coordinates of 1400,1000 . > The curve has an end-of-curve with coordinates of 1400,1400 . And the > curve itself has a radius of 400. Then the inside edge has a begin-of-curve with N,E coordinates of > 1000,1030 . The inside edge has tangent intersections at 1370,1030 . The > inside edge has an end-of-curve at 1370,1400 . And the inside curve has a > radius of 370 . Now the radius point has N,E coordinates of 1000,1400 and the central > angle is 90 degrees. Oh, the inside curve has a midpoint on an azimuth of 315 degrees 370 feet > from the radius point. So the midpoint of the inside curve has N,E > coordinates of 1261.63 , 1138.37 . Now run a motorcycle through the points with N,E coordinates of 1000,1000; > 1261.63,1138.37; and 1400,1400 . What is the radius of that curve and is > it a circular or elliptical ? Graphically speaking it appears to be elliptical ? > Okay, very neat... A circular curve that begins on the fixed tangent at N,E coordinates of 930,1000 and that ends on the other fixed tangent at N,E coordinates of 1400,1470...approximately hits the low point at 1261.63 , 1138.37 . And the 470 tangent distances along with the 90 degree central angle makes for a 470 radius. So the centerline radius of 385 is increased to about 470 by entering wide and then arcing down low... That's by graphical trial and error. Of course it could be calculated exactly by trial and error. In fact the distance from the new radius point at 930,1470 to the curve low point at 1261.63 , 1138.37 is 469 so I missed it by one unit... === Subject: Re: Racing cars on a track > Consider this curve: > The outside edge has a begin-of-curve with N,E coordinates of 1000,1000 . > The curve has tangent intersections with N,E coordinates of 1400,1000 . > The curve has an end-of-curve with coordinates of 1400,1400 . And the > curve itself has a radius of 400. > Then the inside edge has a begin-of-curve with N,E coordinates of > 1000,1030 . The inside edge has tangent intersections at 1370,1030 . The > inside edge has an end-of-curve at 1370,1400 . And the inside curve has a > radius of 370 . > Now the radius point has N,E coordinates of 1000,1400 and the central > angle is 90 degrees. > Oh, the inside curve has a midpoint on an azimuth of 315 degrees 370 feet > from the radius point. So the midpoint of the inside curve has N,E > coordinates of 1261.63 , 1138.37 . > Now run a motorcycle through the points with N,E coordinates of > 1000,1000; 1261.63,1138.37; and 1400,1400 . What is the radius of that > curve and is it a circular or elliptical ? > Graphically speaking it appears to be elliptical ? > > Okay, very neat... A circular curve that begins on the fixed tangent at N,E coordinates of > 930,1000 and that ends on the other fixed tangent at N,E coordinates of > 1400,1470...approximately hits the low point at 1261.63 , 1138.37 . And > the 470 tangent distances along with the 90 degree central angle makes for > a 470 radius. So the centerline radius of 385 is increased to about 470 by > entering wide and then arcing down low... That's by graphical trial and error. Of course it could be calculated > exactly by trial and error. In fact the distance from the new radius point > at 930,1470 to the curve low point at 1261.63 , 1138.37 is 469 so I missed > it by one unit... > Oh, one unit on the azimuth of 315 is 0.707 units on the North tangent... So the begin of the curve is at N,E coordinates of 929.29 , 1000 and the end of the curve is at N,E coordinates of 1400 , 1470.707 ... And the new radius is at 929.29 , 1470.707 ... And the distance to the low point is 470 ! A two step calculation... === Subject: Re: Racing cars on a track > Consider this curve: > The outside edge has a begin-of-curve with N,E coordinates of 1000,1000 > . The curve has tangent intersections with N,E coordinates of 1400,1000 > . The curve has an end-of-curve with coordinates of 1400,1400 . And the > curve itself has a radius of 400. > Then the inside edge has a begin-of-curve with N,E coordinates of > 1000,1030 . The inside edge has tangent intersections at 1370,1030 . The > inside edge has an end-of-curve at 1370,1400 . And the inside curve has > a radius of 370 . > Now the radius point has N,E coordinates of 1000,1400 and the central > angle is 90 degrees. > Oh, the inside curve has a midpoint on an azimuth of 315 degrees 370 > feet from the radius point. So the midpoint of the inside curve has N,E > coordinates of 1261.63 , 1138.37 . > Now run a motorcycle through the points with N,E coordinates of > 1000,1000; 1261.63,1138.37; and 1400,1400 . What is the radius of that > curve and is it a circular or elliptical ? > Graphically speaking it appears to be elliptical ? > Okay, very neat... > A circular curve that begins on the fixed tangent at N,E coordinates of > 930,1000 and that ends on the other fixed tangent at N,E coordinates of > 1400,1470...approximately hits the low point at 1261.63 , 1138.37 . And > the 470 tangent distances along with the 90 degree central angle makes > for a 470 radius. So the centerline radius of 385 is increased to about > 470 by entering wide and then arcing down low... > That's by graphical trial and error. Of course it could be calculated > exactly by trial and error. In fact the distance from the new radius > point at 930,1470 to the curve low point at 1261.63 , 1138.37 is 469 so I > missed it by one unit... > > Oh, one unit on the azimuth of 315 is 0.707 units on the North tangent... So the begin of the curve is at N,E coordinates of 929.29 , 1000 and the > end of the curve is at N,E coordinates of 1400 , 1470.707 ... And the new > radius is at 929.29 , 1470.707 ... And the distance to the low point is 470 ! A two step calculation... > No, the new radius is now 470.71 and I have the distance from the radius point to the low point at 470... === Subject: Re: Racing cars on a track > Consider this curve: > The outside edge has a begin-of-curve with N,E coordinates of 1000,1000 > . The curve has tangent intersections with N,E coordinates of 1400,1000 > . The curve has an end-of-curve with coordinates of 1400,1400 . And the > curve itself has a radius of 400. > Then the inside edge has a begin-of-curve with N,E coordinates of > 1000,1030 . The inside edge has tangent intersections at 1370,1030 . > The inside edge has an end-of-curve at 1370,1400 . And the inside curve > has a radius of 370 . > Now the radius point has N,E coordinates of 1000,1400 and the central > angle is 90 degrees. > Oh, the inside curve has a midpoint on an azimuth of 315 degrees 370 > feet from the radius point. So the midpoint of the inside curve has N,E > coordinates of 1261.63 , 1138.37 . > Now run a motorcycle through the points with N,E coordinates of > 1000,1000; 1261.63,1138.37; and 1400,1400 . What is the radius of that > curve and is it a circular or elliptical ? > Graphically speaking it appears to be elliptical ? > Okay, very neat... > A circular curve that begins on the fixed tangent at N,E coordinates of > 930,1000 and that ends on the other fixed tangent at N,E coordinates of > 1400,1470...approximately hits the low point at 1261.63 , 1138.37 . And > the 470 tangent distances along with the 90 degree central angle makes > for a 470 radius. So the centerline radius of 385 is increased to about > 470 by entering wide and then arcing down low... > That's by graphical trial and error. Of course it could be calculated > exactly by trial and error. In fact the distance from the new radius > point at 930,1470 to the curve low point at 1261.63 , 1138.37 is 469 so > I missed it by one unit... > Oh, one unit on the azimuth of 315 is 0.707 units on the North tangent... > So the begin of the curve is at N,E coordinates of 929.29 , 1000 and the > end of the curve is at N,E coordinates of 1400 , 1470.707 ... And the new > radius is at 929.29 , 1470.707 ... > And the distance to the low point is 470 ! A two step calculation... > > No, the new radius is now 470.71 and I have the distance from the radius > point to the low point at 470... Calculated by trial and error... The motorcycle turns off the entry tangent at N,E coordinates of 927.57 , 1000 . The motorcycle hits the road width low point of 1261.63 , 1138.37 . And the motorcycle hits the exit tangent at 1400 , 1472.43 . The radius of the motorcycle path is 472.43 ... while the road centerline radius is 385 ... and that's to calculate some radius of some definition...whether or not the actual path is circular or elliptical. === Subject: Re: Racing cars on a track > Consider this curve: > The outside edge has a begin-of-curve with N,E coordinates of > 1000,1000 . The curve has tangent intersections with N,E coordinates > of 1400,1000 . The curve has an end-of-curve with coordinates of > 1400,1400 . And the curve itself has a radius of 400. > Then the inside edge has a begin-of-curve with N,E coordinates of > 1000,1030 . The inside edge has tangent intersections at 1370,1030 . > The inside edge has an end-of-curve at 1370,1400 . And the inside > curve has a radius of 370 . > Now the radius point has N,E coordinates of 1000,1400 and the central > angle is 90 degrees. > Oh, the inside curve has a midpoint on an azimuth of 315 degrees 370 > feet from the radius point. So the midpoint of the inside curve has > N,E coordinates of 1261.63 , 1138.37 . > Now run a motorcycle through the points with N,E coordinates of > 1000,1000; 1261.63,1138.37; and 1400,1400 . What is the radius of that > curve and is it a circular or elliptical ? > Graphically speaking it appears to be elliptical ? > Okay, very neat... > A circular curve that begins on the fixed tangent at N,E coordinates of > 930,1000 and that ends on the other fixed tangent at N,E coordinates of > 1400,1470...approximately hits the low point at 1261.63 , 1138.37 . And > the 470 tangent distances along with the 90 degree central angle makes > for a 470 radius. So the centerline radius of 385 is increased to about > 470 by entering wide and then arcing down low... > That's by graphical trial and error. Of course it could be calculated > exactly by trial and error. In fact the distance from the new radius > point at 930,1470 to the curve low point at 1261.63 , 1138.37 is 469 so > I missed it by one unit... > Oh, one unit on the azimuth of 315 is 0.707 units on the North > tangent... > So the begin of the curve is at N,E coordinates of 929.29 , 1000 and the > end of the curve is at N,E coordinates of 1400 , 1470.707 ... And the > new radius is at 929.29 , 1470.707 ... > And the distance to the low point is 470 ! A two step calculation... > No, the new radius is now 470.71 and I have the distance from the radius > point to the low point at 470... Calculated by trial and error... The motorcycle turns off the entry tangent at N,E coordinates of 927.57 , > 1000 . The motorcycle hits the road width low point of 1261.63 , 1138.37 . > And the motorcycle hits the exit tangent at 1400 , 1472.43 . The radius of the motorcycle path is 472.43 ... while the road centerline > radius is 385 ... and that's to calculate some radius of some > definition...whether or not the actual path is circular or elliptical. Oh, from the graphical approximation to the result: Okay the tangent on the North direction is 470 while the distance to the low point from the radius is 469 on the 315 azimuth tangent. L is distance to add to radius P is distance to add to tangent r is the starting radius R is the required radius t is the starting tangent distance T is the required tangent distance t as (t) is known r as (r) is known 0.7071 * L = P L + (r) = R P + (t) = T T = R L + (r) = R P/.7071 + (r) = R P + (t) = T T = R P/.7071 + (r) = P + (t) 1.414227 P - P = (t) - (r) 0.414227 P = (t) - (r) (r) = 469 (t) = 470 0.414227 * P = 1 P = 2.41 T = 470 + 2.41 T = 472.41 === Subject: Re: Racing cars on a track > Oh, from the graphical approximation to the result: Okay the tangent on the North direction is 470 while the distance to > the low point from the radius is 469 on the 315 azimuth. L is distance to add to radius > P is distance to add to tangent > r is the starting radius > R is the required radius > t is the starting tangent distance > T is the required tangent distance > t as (t) is known > r as (r) is known 0.7071 * L = P > L + (r) = R > P + (t) = T > T = R L + (r) = R > P/.7071 + (r) = R > P + (t) = T > T = R > P/.7071 + (r) = P + (t) > 1.414227 P - P = (t) - (r) > 0.414227 P = (t) - (r) > (r) = 469 > (t) = 470 > 0.414227 * P = 1 > P = 2.41 T = 470 + 2.41 > T = 472.41 === Subject: Re: Racing cars on a track > Oh, from the graphical approximation to the result: > Okay the tangent on the North direction is 470 while the distance to > the low point from the radius is 469 on the 315 azimuth. > L is distance to add to radius > P is distance to add to tangent > r is the starting radius > R is the required radius > t is the starting tangent distance > T is the required tangent distance > t as (t) is known > r as (r) is known > 0.7071 * L = P > L + (r) = R > P + (t) = T > T = R > L + (r) = R > P/.7071 + (r) = R > P + (t) = T > T = R > P/.7071 + (r) = P + (t) > 1.414227 P - P = (t) - (r) > 0.414227 P = (t) - (r) > (r) = 469 > (t) = 470 > 0.414227 * P = 1 > P = 2.41 > T = 470 + 2.41 > T = 472.41 > Now forget the graphical approximation: t is the tangent distance for the outside curve r is the radius distance to the midpoint of the inside curve t = 400 r = 370 1.414214 P - P = (t) - (r) 0.414214 P = 30 P = 72.43 T = 400 + 72.43 T = 472.43 Now some terminology explaination: The curve has a central angle of 90 degrees. The tangent is the straight into the curve and also the straight out of the curve. But the tangent-distance is the distance from the begin-of-curve to the intersection of the tangents. And the tangent-distance is also the distance from the intersection of the tangents to the end-of-curve. === Subject: Re: Racing cars on a track > Oh, from the graphical approximation to the result: > Okay the tangent on the North direction is 470 while the distance to > the low point from the radius is 469 on the 315 azimuth. > L is distance to add to radius > P is distance to add to tangent > r is the starting radius > R is the required radius > t is the starting tangent distance > T is the required tangent distance > t as (t) is known > r as (r) is known > 0.7071 * L = P > L + (r) = R > P + (t) = T > T = R > L + (r) = R > P/.7071 + (r) = R > P + (t) = T > T = R > P/.7071 + (r) = P + (t) > 1.414227 P - P = (t) - (r) > 0.414227 P = (t) - (r) > (r) = 469 > (t) = 470 > 0.414227 * P = 1 > P = 2.41 > T = 470 + 2.41 > T = 472.41 > > Now forget the graphical approximation: t is the tangent distance for the outside curve > r is the radius distance to the midpoint of the inside curve t = 400 > r = 370 1.414214 P - P = (t) - (r) > 0.414214 P = 30 > P = 72.43 T = 400 + 72.43 > T = 472.43 Now some terminology explaination: The curve has a central angle of 90 degrees. The tangent is the straight into the curve and also the straight out of > the curve. But the tangent-distance is the distance from the > begin-of-curve to the intersection of the tangents. And the > tangent-distance is also the distance from the intersection of the > tangents to the end-of-curve. > So I suppose this is a Circular Path Curve...within a given circular curve width... And the calculation should easily develop for central angles other than 90 degrees... === Subject: Re: Racing cars on a track > So I suppose this is a Circular Path Curve...within a given circular > curve width... And the calculation should easily develop for central angles other than 90 > degrees... Central Angle = 45 degrees Tangent length outside curve = 211.49 Radius inside curve = 482.8426 and also as r Radius outside curve = 512.8426 and also as t AngleA = (180 - CentralAngle) / 2 P/Sin(A) + (r) = P + (t) * T/R P/Sin(A) - P = (t) - (r) * T/R P/0.923880 - P = (t) - (r) * T/R 1.082392 P - 1 P = (t) - (r) * T/R 0.082392 P = 30 * T/R P = 364.11 * T/R T = 211.49 + (364.11 * T/R) T = 361.64 === Subject: Re: Racing cars on a track > So I suppose this is a Circular Path Curve...within a given circular > curve width... > And the calculation should easily develop for central angles other than > 90 degrees... Central Angle = 45 degrees > Tangent length outside curve = 211.49 > Radius inside curve = 482.8426 and also as r > Radius outside curve = 512.8426 and also as t > AngleA = (180 - CentralAngle) / 2 > P/Sin(A) + (r) = P + (t) * T/R > P/Sin(A) - P = (t) - (r) * T/R > P/0.923880 - P = (t) - (r) * T/R > 1.082392 P - 1 P = (t) - (r) * T/R > 0.082392 P = 30 * T/R > P = 364.11 * T/R T = 211.49 + (364.11 * T/R) > T = 361.64 No, that result is just approximate... === Subject: Re: Racing cars on a track > So I suppose this is a Circular Path Curve...within a given circular > curve width... > And the calculation should easily develop for central angles other than > 90 degrees... > Central Angle = 45 degrees Tangent length outside curve = 211.49 Radius inside curve = 482.8426 and also as r Radius outside curve = 512.8426 and also as t AngleA = (180 - CentralAngle) / 2 P/Sin(A) + (r) = P + (t) * TangentDist/OutsideRadius P/Sin(A) - P = (t) - (r) * 0.412388 P/0.923880 - P = (t) - (r) * 0.412388 1.082392 P - 1 P = (t) - (r) * 0.412388 0.082392 P = 30 * 0.412388 P = 364.11 * 0.412388 T = 211.49 + (364.11 * 0.412388) T = 361.64 === Subject: Re: Racing cars on a track > So I suppose this is a Circular Path Curve...within a given circular > curve width... Make that a Circular Curve Path...with a given circular curve width... > And the calculation should easily develop for central angles other than > 90 degrees... > > Central Angle = 45 degrees > Tangent length outside curve = 211.49 > Radius inside curve = 482.8426 and also as r > Radius outside curve = 512.8426 and also as t > AngleA = (180 - CentralAngle) / 2 > P/Sin(A) + (r) = P + (t) * TangentDist/OutsideRadius > P/Sin(A) - P = (t) - (r) * 0.412388 > P/0.923880 - P = (t) - (r) * 0.412388 > 1.082392 P - 1 P = (t) - (r) * 0.412388 > 0.082392 P = 30 * 0.412388 > P = 364.11 * 0.412388 T = 211.49 + (364.11 * 0.412388) > T = 361.64 No, that result is just approximate... === Subject: Question on retractions Let X_1 and X_2 be metrizable spaces and I the unit interval. Let int denote intersection of sets and {0} the singleton which contains the element 0. Let U denote union of sets. Question: Is it true that (X_1 x {0}) U ( (X_1 int X_2) x I ) is a retract of X_1 x I ? I'm not sure if this is correct, but first we must show that the LHS is contained in the RHS. Well X_1 int X_2 is a subset of X_1 and hence (X_1 int X_2) x I is a subset of X_1 x I. Also X_1 x {0} is a subset of X_1 x I and therefore (X_1 x {0} ) U ( (X_1 int X_2) x I ) is a subset of X_1 x I. Now define a retraction r: (X_1 x I) -> (X_1 x {0}) U ( (X_1 int X_2) x I ) By the identity map, i.e r(x,t) = (x,t). Then we must check that for all elements z of X_1 x {0}) U ( (X_1 int X_2) x I ) , r(z) = z. So first case, z is of the form (x,0). Then r(z) = r(x,0) = (x,0). Second case, z is of the form (x,0) where x is in the intersection of X_1 and X_2, and thus in particular is in X_1. Hence r(z) = r(x,t) = (x,t) thus r is a retraction. I think there is something wrong here, can someone check please? === Subject: Re: Question on retractions > Now define a retraction r: (X_1 x I) -> (X_1 x {0}) U ( (X_1 int X_2) x I ) By the identity map, i.e r(x,t) = (x,t). This doesn't make sense. What will you do with points of X_1 x I that are not in (X_1 x {0}) U ( (X_1 int X_2) x I ) ? For instance, (x,1) where x in X_1 X_2. (Of course if X_1 is a subset of X_2 everything is trivial.) === Subject: Cyclotomes When T, S, A, B, and n are all positive integers (all >= 1, n is odd, A and B prime to each other), T and S are called cyclotomes when they are represented in either of the following forms; T^n = A^(n-1) + A^(n-2)*B + A^(n-3)*B^2 ~~+ B^(n-1) S^n = A^(n-1) - A^(n-2)*B + A^(n-3)*B^2 ~~- B^(n-1). When n >= 5, then there does not exist cyclotomes S or T for any positive integers A and B. These claims have something to do with my proof on the FLT problem. Any comments will be appreciated. === Subject: Re: Dependant Choice. >Hi all, >What is dependant choice? Can one brief me about its basic idea? and >what it's formula is exactly as writtin in first order logic? >In a previous post I have suggested the following sentence as an axiom >instead of axiom of infinity ( since it implies it, but it is not >equivalent to it ) , I thought of the following axiom as the logical >backround of axiom of infinity. >AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). >were V={x|x is a set}. i.e V is the proper class of all sets. >It is clear that this axiom implies axiom of infinity. >I was told that this is dependant choice. I need some clarification >about this subject? No, this is not; it is true with no form of the axiom of choice. Dependent choice comes in set and class forms. In the set form, if Q is a relation on a non-empty set w such that (Ax)((xew) -> (Ey)(yew & xQy)), then (Am)[mew -> (Ef)(f < Q & meD(f) & (x)(xeR(f) -> xeD(f)))], where D means domain R range, and < subset. It should be clear what the class form is. >Zuhair -- 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: Dependant Choice. Hi all, >What is dependant choice? Can one brief me about its basic idea? and >what it's formula is exactly as writtin in first order logic? >In a previous post I have suggested the following sentence as an axiom >instead of axiom of infinity ( since it implies it, but it is not >equivalent to it ) , I thought of the following axiom as the logical >backround of axiom of infinity. >AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). >were V={x|x is a set}. i.e V is the proper class of all sets. >It is clear that this axiom implies axiom of infinity. >I was told that this is dependant choice. I need some clarification >about this subject? No, this is not; it is true with no form of the axiom > of choice. This is a vague statement. Let me try get that right: You mean that: we can replace both pairing and infinity in MK by the following axiom. AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). without changing Mk to another theory. And by then in this theory both pairing and infinity can be proved without the need for any form of choice. Is that correct? If this correct then this is what I wanted to say. Zuhair Dependent choice comes in set and class forms. In the > set form, if Q is a relation on a non-empty set w such > that (Ax)((xew) -> (Ey)(yew & xQy)), then > (Am)[mew -> (Ef)(f < Q & meD(f) & (x)(xeR(f) -> xeD(f)))], > where D means domain R range, and < subset. It should be clear what the class form is. Zuhair -- > 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 > hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Dependant Choice. > Hi all, > What is dependant choice? Can one brief me about its basic idea? and > what it's formula is exactly as writtin in first order logic? > One formulation of it is Given a set S and a function from S to P(S), > if f(x) is nonempty for all x, then there exists an infinite sequence > (x_1, x_2, ...) such that x_2 is a member of f(x_1), x_3 is a member > of f(x_2), and so on. There are other neater ways to formulate it. In > the presence of the axioms of ZF, it is stronger than countable > choice, but weaker than the full axiom of choice. > In a previous post I have suggested the following sentence as an axiom > instead of axiom of infinity ( since it implies it, but it is not > equivalent to it ) , I thought of the following axiom as the logical > backround of axiom of infinity. > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > were V={x|x is a set}. i.e V is the proper class of all sets. > It is clear that this axiom implies axiom of infinity. > Given what other axioms? > The axioms of any theory that has proper classes and especially V and > that is equi-interpretable with ZF or even Z( other than infinity of > course ). Why don't you have a go at precisely stating this claim and proving > it? > Example the version of NBG that is equi-interpretable with > ZF. Well, let's assume you mean NBG without the axiom of infinity, to make > it interesting. You do realize that NBG has two different kinds of > variables, don't you? So clarify which variables in your sentence are > of which kind. Sorry it is not NBG. I meant MK wihtout infinity . The claim is that: in MK without infinity IF we add the following sentence as an axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > MK also has a two-sorted language. I will assume that you want f to be a class variable and m, x, and y to be set variables. I think you're probably right that your statement is equivalent to the axiom of infinity in the presence of the other axioms, in fact I think that might even be true with NBG, but it depends what we mean by the other axioms, that is, it depends what axiomatization of MK or NBG we are working with. Why don't you have a go at making it clear what you mean by MK without infinity. > Then we will have the following sentence as a theorem in this thoery. Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. Also we have E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Extensionality. Zuhair This was not my problem since it is trivial. > My question was is the sentence AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). equivalent to axiom of dependant choice. > No. > Also I have another question. How can I write this sentence in Z. Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). I think this is equivalent to the above sentence. > Well, Z hasn't got any class variables, and don't you need f to be a class variable, if it denotes a function whose domain is the universe? > Now if I am right about this equivalence then my question can also be > changed to. IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. > Well, as I say, I don't think this can work in a theory whose language lacks class variables. But, more to the point, basically what you're doing is trying to find an alternative formulation of the axiom of infinity. Lots of such formulations are known. Why is yours particularly interesting? > Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. Zuhair- Hide quoted text - - Show quoted text - === Subject: Re: Dependant Choice. > Hi all, > What is dependant choice? Can one brief me about its basic idea? and > what it's formula is exactly as writtin in first order logic? > One formulation of it is Given a set S and a function from S to P(S), > if f(x) is nonempty for all x, then there exists an infinite sequence > (x_1, x_2, ...) such that x_2 is a member of f(x_1), x_3 is a member > of f(x_2), and so on. There are other neater ways to formulate it. In > the presence of the axioms of ZF, it is stronger than countable > choice, but weaker than the full axiom of choice. > In a previous post I have suggested the following sentence as an axiom > instead of axiom of infinity ( since it implies it, but it is not > equivalent to it ) , I thought of the following axiom as the logical > backround of axiom of infinity. > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > were V={x|x is a set}. i.e V is the proper class of all sets. > It is clear that this axiom implies axiom of infinity. > Given what other axioms? > The axioms of any theory that has proper classes and especially V and > that is equi-interpretable with ZF or even Z( other than infinity of > course ). > Why don't you have a go at precisely stating this claim and proving > it? > Example the version of NBG that is equi-interpretable with > ZF. > Well, let's assume you mean NBG without the axiom of infinity, to make > it interesting. You do realize that NBG has two different kinds of > variables, don't you? So clarify which variables in your sentence are > of which kind. Sorry it is not NBG. I meant MK wihtout infinity . The claim is that: in MK without infinity IF we add the following sentence as an axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). MK also has a two-sorted language. I will assume that you want f to be > a class variable and m, x, and y to be set variables. I think you're probably right that your statement is equivalent to the > axiom of infinity in the presence of the other axioms, in fact I think > that might even be true with NBG, but it depends what we mean by the > other axioms, that is, it depends what axiomatization of MK or NBG we > are working with. Why don't you have a go at making it clear what you > mean by MK without infinity. Then we will have the following sentence as a theorem in this thoery. Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. Also we have E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Extensionality. Zuhair This was not my problem since it is trivial. > My question was is the sentence AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). equivalent to axiom of dependant choice. No. Also I have another question. How can I write this sentence in Z. Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). I think this is equivalent to the above sentence. Well, Z hasn't got any class variables, and don't you need f to be a > class variable, if it denotes a function whose domain is the universe? Now if I am right about this equivalence then my question can also be > changed to. IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. Well, as I say, I don't think this can work in a theory whose language > lacks class variables. So you think that the second formula which is intended to replace infinity in Z. cannot do the job because this would be a theory whose language lacks class variables. I disagree with you about that I think that the following formula Af(AyE!f(y)-> Am(Ex(mex & Ay(yex->f(y)ex))). I think this formula is equivalent to the above one. About the distinction between class and set vairables I think this is not necessary in MK. In MK all what we have are classes. Some classes are sets Other classes are proper classes I think what you mean by a class variable something that can be a proper class or a set, but Rupert in MK all variables(unless otherwise specified) are class variables, you are confusing MK with NBG. To make it short for you using your language all variables in the formula I have mentioned are class variables. The formula is AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). so all vairables are class vairables , ALL of them Rupert f,m,x,y,V,f(y). Remember I am speaking in MK, were we can quantify over proper classes. But, more to the point, basically what you're doing is trying to find > an alternative formulation of the axiom of infinity. Lots of such > formulations are known. Why is yours particularly interesting? NO, that is not basically what I am doing. The formula that I have presented is NOT equivalent to axiom of infinity, IT implicates axiom of infinity in the presence of other axioms, but it is not equivalent to axiom of infinity. Axiom of infinity is a specific axiom, it states the existence of the set of all natural numbers, anlthoug when it acts as an axioms it rule is limited to permitting infinite sets to exists. While the formula above do not only implicate infinity, it is by far much more than that. It tells you for example that y={x,P(x),P(P(x)),P(P(P(x))),......} is a set This formula is much stronger than what you think. This formula tells you that for every x , {x} is a set. you see {x} is not infinite set. Proof: Let f:V->V,f(x)=x, then for every x we have according to this axiom {x} as a set. Also this axiom proves pairing of any two sets A and B. Let f:V->V, f(A)=B and Az(~z=A->f(z)=B) Now Let m=A then from the above axiom we have {A,B}. So actually this axiom replaces pairing as well. And I think that with choice among the other axioms it might replace finite(binary)union. and perhaps even other axioms, I don't know, sometimes I think that this formula with choice can replace all other axioms except Extensionlity and Regularity ,this is just an impression and I think it is false. Perhaps I might be wrong about pairing and union, but definitly this axiom is more than infinity. Infinity is only one aspect of this axiom. Zuhair Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. Zuhair- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Dependant Choice. > Hi all, > What is dependant choice? Can one brief me about its basic idea? and > what it's formula is exactly as writtin in first order logic? > One formulation of it is Given a set S and a function from S to P(S), > if f(x) is nonempty for all x, then there exists an infinite sequence > (x_1, x_2, ...) such that x_2 is a member of f(x_1), x_3 is a member > of f(x_2), and so on. There are other neater ways to formulate it. In > the presence of the axioms of ZF, it is stronger than countable > choice, but weaker than the full axiom of choice. > In a previous post I have suggested the following sentence as an axiom > instead of axiom of infinity ( since it implies it, but it is not > equivalent to it ) , I thought of the following axiom as the logical > backround of axiom of infinity. > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > were V={x|x is a set}. i.e V is the proper class of all sets. > It is clear that this axiom implies axiom of infinity. > Given what other axioms? > The axioms of any theory that has proper classes and especially V and > that is equi-interpretable with ZF or even Z( other than infinity of > course ). > Why don't you have a go at precisely stating this claim and proving > it? > Example the version of NBG that is equi-interpretable with > ZF. > Well, let's assume you mean NBG without the axiom of infinity, to make > it interesting. You do realize that NBG has two different kinds of > variables, don't you? So clarify which variables in your sentence are > of which kind. > Sorry it is not NBG. > I meant MK wihtout infinity . > The claim is that: in MK without infinity > IF we add the following sentence as an axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). MK also has a two-sorted language. I will assume that you want f to be > a class variable and m, x, and y to be set variables. I think you're probably right that your statement is equivalent to the > axiom of infinity in the presence of the other axioms, in fact I think > that might even be true with NBG, but it depends what we mean by the > other axioms, that is, it depends what axiomatization of MK or NBG we > are working with. Why don't you have a go at making it clear what you > mean by MK without infinity. > Then we will have the following sentence as a theorem in this thoery. > Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. > Also we have > E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Extensionality. > Zuhair > This was not my problem since it is trivial. > My question was is the sentence > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > equivalent to axiom of dependant choice. No. > Also I have another question. > How can I write this sentence in Z. > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this is equivalent to the above sentence. Well, Z hasn't got any class variables, and don't you need f to be a > class variable, if it denotes a function whose domain is the universe? > Now if I am right about this equivalence then my question can also be > changed to. > IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. Well, as I say, I don't think this can work in a theory whose language > lacks class variables. So you think that the second formula which is intended to replace > infinity in Z. cannot do the job because this would be a theory whose > language lacks class variables. > I disagree with you about that I think that the following formula Af(AyE!f(y)-> Am(Ex(mex & Ay(yex->f(y)ex))). I think this formula is equivalent to the above one. > If we had a function whose domain was the universe, then, assuming the members of the function were Kuratowski ordered pairs, if we applied the union operator to the function twice we would get the universe. But the universe is not a set. So the hypothesis of your conditional can be proved false in Z-infinity. That means your formula can be proved in Z-infinity, so it certainly doesn't entail the axiom of infinity. > About the distinction between class and set vairables > I think this is not necessary in MK. > In MK all what we have are classes. > Some classes are sets > Other classes are proper classes I always thought the language of MK was two-sorted. What's your source here? > I think what you mean by a class variable something > that can be a proper class or a set, but Rupert in MK > all variables(unless otherwise specified) are class variables, you are > confusing MK with NBG. > I always thought that both MK and NBG had two-sorted languages. However, I think they are probably both equi-interpretable with theories in one-sorted languages, so it doesn't matter very much. > To make it short for you using your language all variables in the > formula I have mentioned are class variables. > The formula is AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). so all vairables are class vairables , ALL of them Rupert > f,m,x,y,V,f(y). Remember I am speaking in MK, were we can quantify > over proper classes. > I have no problem with you doing this in the language of MK or NBG. It's the language of Z that I'm worried about. > But, more to the point, basically what you're doing is trying to find > an alternative formulation of the axiom of infinity. Lots of such > formulations are known. Why is yours particularly interesting? NO, that is not basically what I am doing. The formula that I have > presented is NOT equivalent to axiom of infinity, IT implicates axiom > of infinity in the presence of other axioms, but it is not equivalent > to axiom of infinity. > Proof? > Axiom of infinity is a specific axiom, it states the existence of the > set of all natural numbers, anlthoug when it acts as an axioms it rule > is limited to permitting infinite sets to exists. While the formula above do not only implicate infinity, it is by far > much more than that. > It tells you for example that y={x,P(x),P(P(x)),P(P(P(x))),......} is > a set Well, that's easy to prove in MK or NBG. It would be news in Z, yes, but I don't think you've given a coherent way to add your axiom to Z. > This formula is much stronger than what you think. This formula tells you that for every x , {x} is a set. Huh? That follows from the axiom of pairing. I can see that your axiom proves lots of stuff, yes, but I don't think that replacing the axiom of infinity with your axiom is going to increase the strength of the theory. > you see {x} is not infinite set. Proof: Let f:V->V,f(x)=x, then for every x we have according to this > axiom {x} as a set. Also this axiom proves pairing of any two sets A and B. Let f:V->V, f(A)=B and Az(~z=A->f(z)=B) > Now Let m=A then from the above axiom we have {A,B}. So actually this axiom replaces pairing as well. > And I think that with choice among the other axioms > it might replace finite(binary)union. and perhaps even > other axioms, I don't know, sometimes I think that this formula with > choice can replace all other axioms except Extensionlity and > Regularity ,this is just an impression and I think it is false. > I agree. I very much doubt that the axiom can replace union either. > Perhaps I might be wrong about pairing and union, but definitly this > axiom is more than infinity. Infinity is only one aspect of this > axiom. > Well, what you're saying now is that you may be able to replace *several* axioms with this axiom and get the same theory. I grant you that might be slightly more interesting, but still, essentially what you're doing is coming up with an alternative axiomatization. Also, I think you overestimate how much work this axiom will be able to do. But my main point is that the axiom only makes sense in the language of MK or NBG. If you try to translate it into the language of Z you get a vacuous statement. > Zuhair > Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. > Zuhair- Hide quoted text - > - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: Dependant Choice. > Hi all, > What is dependant choice? Can one brief me about its basic idea? and > what it's formula is exactly as writtin in first order logic? > One formulation of it is Given a set S and a function from S to P(S), > if f(x) is nonempty for all x, then there exists an infinite sequence > (x_1, x_2, ...) such that x_2 is a member of f(x_1), x_3 is a member > of f(x_2), and so on. There are other neater ways to formulate it. In > the presence of the axioms of ZF, it is stronger than countable > choice, but weaker than the full axiom of choice. > In a previous post I have suggested the following sentence as an axiom > instead of axiom of infinity ( since it implies it, but it is not > equivalent to it ) , I thought of the following axiom as the logical > backround of axiom of infinity. > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > were V={x|x is a set}. i.e V is the proper class of all sets. > It is clear that this axiom implies axiom of infinity. > Given what other axioms? > The axioms of any theory that has proper classes and especially V and > that is equi-interpretable with ZF or even Z( other than infinity of > course ). > Why don't you have a go at precisely stating this claim and proving > it? > Example the version of NBG that is equi-interpretable with > ZF. > Well, let's assume you mean NBG without the axiom of infinity, to make > it interesting. You do realize that NBG has two different kinds of > variables, don't you? So clarify which variables in your sentence are > of which kind. > Sorry it is not NBG. > I meant MK wihtout infinity . > The claim is that: in MK without infinity > IF we add the following sentence as an axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > MK also has a two-sorted language. I will assume that you want f to be > a class variable and m, x, and y to be set variables. > I think you're probably right that your statement is equivalent to the > axiom of infinity in the presence of the other axioms, in fact I think > that might even be true with NBG, but it depends what we mean by the > other axioms, that is, it depends what axiomatization of MK or NBG we > are working with. Why don't you have a go at making it clear what you > mean by MK without infinity. > Then we will have the following sentence as a theorem in this thoery. > Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. > Also we have > E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Extensionality. > Zuhair > This was not my problem since it is trivial. > My question was is the sentence > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > equivalent to axiom of dependant choice. > No. > Also I have another question. > How can I write this sentence in Z. > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this is equivalent to the above sentence. > Well, Z hasn't got any class variables, and don't you need f to be a > class variable, if it denotes a function whose domain is the universe? > Now if I am right about this equivalence then my question can also be > changed to. > IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. > Well, as I say, I don't think this can work in a theory whose language > lacks class variables. So you think that the second formula which is intended to replace > infinity in Z. cannot do the job because this would be a theory whose > language lacks class variables. > I disagree with you about that I think that the following formula Af(AyE!f(y)-> Am(Ex(mex & Ay(yex->f(y)ex))). I think this formula is equivalent to the above one. If we had a function whose domain was the universe, then, assuming the > members of the function were Kuratowski ordered pairs, if we applied > the union operator to the function twice we would get the universe. > But the universe is not a set. So the hypothesis of your conditional > can be proved false in Z-infinity. That means your formula can be > proved in Z-infinity, so it certainly doesn't entail the axiom of > infinity. I think you might have committed an error, but I am not sure. So I if we apply the union operator to the function twice. In 'Z-I+this axiom'theory (which is a theory from first order logic) no union operator is defined on f. Remember f is not provable to be a set in Z-I+this axiom ,nor in Z-I . Of course I am speaking of f that is present in the formula Af(AyE!f(y)-> Am(Ex(mex & Ay(yex->f(y)ex))). f is a function symbole, certainly it is not provable to be a set in Z- I+this axiom ,nor in Z-I, so you cannot apply the union operator to it. This actually was a question that I wanted to ask you about: can we really quantify over f in Z-I+this axiom ? My question is are we allowed to quantify over f in a first order language theory that doesn't have proper classes in its universe of discourse? Since If we cannot then perhaps I should use the metatheoratical language , I mean a schema. Perhaps I should write it as: if F is a formula in which x is not free then all closures of AyE!zF(y,z) -> Am(Ex(mex & Ay(yex->Az(F(y,z)->zex)))). is an axiom. Perhaps that's the way this axiom should be writtin in a set theory that do no permit proper classes in its universe of discourse. Zuhair === Subject: Re: Dependant Choice. > > Hi all, > What is dependant choice? Can one brief me about its basic idea? and > > what it's formula is exactly as writtin in first order logic? > One formulation of it is Given a set S and a function from S to P(S), > if f(x) is nonempty for all x, then there exists an infinite sequence > (x_1, x_2, ...) such that x_2 is a member of f(x_1), x_3 is a member > of f(x_2), and so on. There are other neater ways to formulate it. In > the presence of the axioms of ZF, it is stronger than countable > choice, but weaker than the full axiom of choice. > In a previous post I have suggested the following sentence as an axiom > > instead of axiom of infinity ( since it implies it, but it is not > > equivalent to it ) , I thought of the following axiom as the logical > > backround of axiom of infinity. > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > were V={x|x is a set}. i.e V is the proper class of all sets. > It is clear that this axiom implies axiom of infinity. > Given what other axioms? > The axioms of any theory that has proper classes and especially V and > that is equi-interpretable with ZF or even Z( other than infinity of > course ). > Why don't you have a go at precisely stating this claim and proving > it? > Example the version of NBG that is equi-interpretable with > ZF. > Well, let's assume you mean NBG without the axiom of infinity, to make > it interesting. You do realize that NBG has two different kinds of > variables, don't you? So clarify which variables in your sentence are > of which kind. > Sorry it is not NBG. > I meant MK wihtout infinity . > The claim is that: in MK without infinity > IF we add the following sentence as an axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > MK also has a two-sorted language. I will assume that you want f to be > a class variable and m, x, and y to be set variables. > I think you're probably right that your statement is equivalent to the > axiom of infinity in the presence of the other axioms, in fact I think > that might even be true with NBG, but it depends what we mean by the > other axioms, that is, it depends what axiomatization of MK or NBG we > are working with. Why don't you have a go at making it clear what you > mean by MK without infinity. > Then we will have the following sentence as a theorem in this thoery. > Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. > Also we have > E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Extensionality. > Zuhair > This was not my problem since it is trivial. > My question was is the sentence > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > equivalent to axiom of dependant choice. > No. > Also I have another question. > How can I write this sentence in Z. > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this is equivalent to the above sentence. > Well, Z hasn't got any class variables, and don't you need f to be a > class variable, if it denotes a function whose domain is the universe? > Now if I am right about this equivalence then my question can also be > changed to. > IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. > Well, as I say, I don't think this can work in a theory whose language > lacks class variables. > So you think that the second formula which is intended to replace > infinity in Z. cannot do the job because this would be a theory whose > language lacks class variables. > I disagree with you about that > I think that the following formula > Af(AyE!f(y)-> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this formula is equivalent to the above one. If we had a function whose domain was the universe, then, assuming the > members of the function were Kuratowski ordered pairs, if we applied > the union operator to the function twice we would get the universe. > But the universe is not a set. So the hypothesis of your conditional > can be proved false in Z-infinity. That means your formula can be > proved in Z-infinity, so it certainly doesn't entail the axiom of > infinity. I think you might have committed an error, but I am not sure. So I > if we apply the union operator to the function twice. > In 'Z-I+this axiom'theory (which is a theory from first order logic) > no union operator is defined on f. Remember f is not provable to be a > set in Z-I+this axiom ,nor in Z-I . Well, that's precisely the problem. In the language of Z-I we can only talk about sets. You can't quantify over class variables. I've been trying to get this across to you for quite a while now. === Subject: Re: Dependant Choice. Well, that's precisely the problem. In the language of Z-I we can only > talk about sets. You can't quantify over class variables. I've been > trying to get this across to you for quite a while now.- Hide quoted text - - Show quoted text - But you didn't say anything about the ulternative schema that I have suggested and I will write it again below: Axiom of Countability:if F is a formula in which x is not free then all closures of AyE!zF(y,z) -> Am(Ex(mex & Ay(yex->Az(F(y,z)->zex)))). are axioms. I think that does the job perfectly. And I do think that the following theory is Z-R itself. 1) Extensionality 2) Separation 3) Countability 4) Union 5) Power Also I think that If I add Size Comprehension and Regularity as below, I think we can get ZF. 6) Size Comprehension: If F is a formula in which x is not free then all closures of Ex( Ay(yex<->F)<->Ez(z supernumerous to x ) ). are axioms. 7)Regularity. Axiom of Size comprehension parallels the idea in NBG that a set should have another set that it should be subnumerous to. Size Comprehension avoids Russell's paradox : Let F<->~yey Then ~Ez( z supernumerous to x) Proof: Since from Regularity Ax ~xex then Ay(yex<->~yey)<->x={y|y=y}, since AzAy(yez->y=y) then z subset_of x And no subset is supernumerous to its set Thus ~Ez( z supernumerous to x) we have ~Ay(yex<->~yey) as a theorem from FOL. Then Size comprehension would be Ex(false<->false) which is trivially true. Regarding the case when F<->yey we have from regularity Ax ~xex Then it is clear that Ay(yex<->yey)<->x={} and from Separation and Countability we have {{}} as a set , which is supernumerous to { } , So size comprehension is true. I don't know if I should add another axiom stating that 8)Size limitation: AxEy( y supernumerous to x ). I think with these 8 axioms we will have a theory that is equivalent to ZF. It is easy to prove that This theory ->ZF Since Replacement can be proved easily in this theory. And Infinity and Pairing follows from Countability. But I don't know how to prove that ZF -> This theory. So what I am sure of is that theory proves ZF, and what I am not sure of is weather ZF prove this theory or not. Zuhair === Subject: Re: Dependant Choice. > And I think that with choice among the other axioms > it might replace finite(binary)union. and perhaps even > other axioms, I don't know, sometimes I think that this formula with > choice can replace all other axioms except Extensionlity and > Regularity ,this is just an impression and I think it is false. Yes this was incorrect. What actually this axiom can do is that it can Replace pairing and infinity in NBG and MK, and an equivalent version of this axiom in Z, can also replace these two axioms in Z. However it is stronger than both of pairing and infinity together. This axiom is very strong regarding building countable sets, that's why I prefer to name it as Axiom of Countablility, since I don't think it is dependent choice. Zuhair === Subject: Re: Dependant Choice. Sorry it is not NBG. I meant MK wihtout infinity . The claim is that: in MK without infinity IF we add the following sentence as an axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). Then we will have the following sentence as a theorem in this thoery. Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. Also we have E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Extensionality. Zuhair This was not my problem since it is trivial. > My question was is the sentence AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). equivalent to axiom of dependant choice. Also I have another question. How can I write this sentence in Z. Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). I think this is equivalent to the above sentence. Now if I am right about this equivalence then my question can also be > changed to. IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. Zuhair- Hide quoted text - - Show quoted text - Based on the above I recommend the following: 1) Axiom of infinity in Z,ZF,ZFC be replaced by the following Axiom: Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). 2) Axiom of infinity in NBG,MK be replaced by the following Axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). Zuhair === Subject: Re: Dependant Choice. > Sorry it is not NBG. I meant MK wihtout infinity . The claim is that: in MK without infinity IF we add the following sentence as an axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). Then we will have the following sentence as a theorem in this thoery. Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. Also we have E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). Proof: Extensionality. Zuhair This was not my problem since it is trivial. > My question was is the sentence AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). equivalent to axiom of dependant choice. Also I have another question. How can I write this sentence in Z. Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). I think this is equivalent to the above sentence. Now if I am right about this equivalence then my question can also be > changed to. IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. Zuhair- Hide quoted text - - Show quoted text - Based on the above I recommend the following: 1) Axiom of infinity in Z,ZF,ZFC be replaced by the following Axiom: Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). 2) Axiom of infinity in NBG,MK be replaced by the following Axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). Zuhair- Hide quoted text - - Show quoted text - First of all, I don't think your suggestion can work in the first case, where we only have set variables. Second, why is this decision about how we present the theory of the least importance? === Subject: Re: Dependant Choice. > Sorry it is not NBG. > I meant MK wihtout infinity . > The claim is that: in MK without infinity > IF we add the following sentence as an axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > Then we will have the following sentence as a theorem in this thoery. > Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. > Also we have > E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Extensionality. > Zuhair > This was not my problem since it is trivial. > My question was is the sentence > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > equivalent to axiom of dependant choice. > Also I have another question. > How can I write this sentence in Z. > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this is equivalent to the above sentence. > Now if I am right about this equivalence then my question can also be > changed to. > IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. > Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. > Zuhair- Hide quoted text - > - Show quoted text - Based on the above I recommend the following: 1) Axiom of infinity in Z,ZF,ZFC be replaced by the following Axiom: Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). 2) Axiom of infinity in NBG,MK be replaced by the following Axiom: AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). Zuhair- Hide quoted text - - Show quoted text - First of all, I don't think your suggestion can work in the first > case, where we only have set variables. Can you tell me why? Is there something wrong with this formula Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). Can you give some detailed answer, why this formula fails to do the job. Second, why is this decision about how we present the theory of the > least importance? You mean ZF is less important than NBG or MK? Many people might not agree with you here. Zuhair === Subject: Re: Dependant Choice. > Sorry it is not NBG. > I meant MK wihtout infinity . > The claim is that: in MK without infinity > IF we add the following sentence as an axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > Then we will have the following sentence as a theorem in this thoery. > Ex( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Since 0eV > Let f:V->V,f(x)=xU{x} > then this theorem follows directly > from the axiom above. > Also we have > E!x( xeV & 0ex & Ay( yex -> yU{y}ex)). > Proof: Extensionality. > Zuhair > This was not my problem since it is trivial. > My question was is the sentence > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > equivalent to axiom of dependant choice. > Also I have another question. > How can I write this sentence in Z. > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > I think this is equivalent to the above sentence. > Now if I am right about this equivalence then my question can also be > changed to. > IF we add this sentence as an axiom to Z-I, then > we will have Ex(0ex & Ay(yex->yU{y}ex)) as a theorem > in this theory. > Proof: Let f(y)=yU{y} then > we have AyE!f(y) then > Since E0 then Ex(0ex & Ay(yex->yU{y}ex)) > follows directly from the above sentence. > Zuhair- Hide quoted text - > - Show quoted text - > Based on the above I recommend the following: > 1) Axiom of infinity in Z,ZF,ZFC be replaced by the following Axiom: > Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). > 2) Axiom of infinity in NBG,MK be replaced by the following Axiom: > AfAm((f:V->V & meV) -> Ex(xeV & mex & Ay(yex -> f(y)ex))). > Zuhair- Hide quoted text - > - Show quoted text - First of all, I don't think your suggestion can work in the first > case, where we only have set variables. Can you tell me why? Is there something wrong with this formula Af( AyE!f(y) -> Am(Ex(mex & Ay(yex->f(y)ex))). Can you give some detailed answer, why this formula fails to do the > job. > I have actually explained it already. Your hypothesis is that f is a function whose domain is the entire universe. This is impossible if f is a set. > Second, why is this decision about how we present the theory of the > least importance? You mean ZF is less important than NBG or MK? No, of course not. The question of whether we formulate the axiom of infinity in one of the standard ways, or in your way, I don't see as being very important, that was the point I was making. > Many people might not agree with you here. Zuhair- Hide quoted text - - Show quoted text - === Subject: cancel: Re: * Free Pussy and Tit flash Download! Control: cancel <78ev43tk2pfp7eed56jdihdmcndgoplnd1@4ax.com> Excessive crossposting. (<78ev43tk2pfp7eed56jdihdmcndgoplnd1@4ax.com>) === Subject: Expanding 1/f(x) by a (generalized) polynomial? Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized polynomial? By generalized polynomial I mean expansion like g(X)=sum ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where the sum is over k. A generalized polynomial is like a regular polynomial except that exponents pk, qk, ...uk can be any REAL number (even negative), as opposed to natural numbers in regular polynomials. I am interested in the most compact expansion of the given function. Any ideas? H.M. === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? Here's a more general answer ... Let f be a rational function. Suppose f=p where p is a generalized polynomial. The exponents of the monomials of p must all be integers otherwise p will have lots of problems with negative values of the variables. But then multiplying p by some monomial will yield an actual polynomial. This implies f is either a polynomial or else a rational function with a monomial denominator. If so, f actually is a generalized polynomial, otherwise not. quasi === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? I'll take a stab at this. As far as I can see, the answer is no -- f can't be expressed as a generalized polynomial. Suppose f=p where p is a generalized polynomial. Since f is a continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p have nonnegative exponents in the variable x1. On the other hand, at (x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a negative exponent in x1, contradiction. Therefore f is not a generalized polynomial. quasi === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? >Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? I'll take a stab at this. As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. Therefore f is not a generalized polynomial. quasi Actually, I see you used 6 variables, whereas the proof I gave above assumed only 4 variables. However, assuming my proof is correct, the proof strategy works just as well for 6 variables. quasi === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? I'll take a stab at this. As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. Therefore f is not a generalized polynomial. quasi Actually, I see you used 6 variables, whereas the proof I gave above > assumed only 4 variables. However, assuming my proof is correct, the proof strategy works just > as well for 6 variables. quasi- Hide quoted text - - Show quoted text - How about the best generalized polynomial approximation of f? Any idea about that? One can assume x1...x6 are real and POSITIVE if that helps. === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? >Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? >I'll take a stab at this. >As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. >Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. >Therefore f is not a generalized polynomial. >quasi > Actually, I see you used 6 variables, whereas the proof I gave above > assumed only 4 variables. > However, assuming my proof is correct, the proof strategy works just > as well for 6 variables. > quasi- Hide quoted text - > - Show quoted text - How about the best generalized polynomial approximation of f? Any idea >about that? One can assume x1...x6 are real and POSITIVE if that >helps. I have no idea what you mean by an approximation. At a point? Globally? What's the criterion for best? quasi === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? >I'll take a stab at this. >As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. >Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. >Therefore f is not a generalized polynomial. >quasi > Actually, I see you used 6 variables, whereas the proof I gave above > assumed only 4 variables. > However, assuming my proof is correct, the proof strategy works just > as well for 6 variables. > quasi- Hide quoted text - > - Show quoted text - How about the best generalized polynomial approximation of f? Any idea >about that? One can assume x1...x6 are real and POSITIVE if that >helps. I have no idea what you mean by an approximation. At a point? Globally? What's the criterion for best? quasi- Hide quoted text - - Show quoted text - Neither local, nor everywhere... say within the hypercube of [0,1]^6 (the hypercube is for X's only, exponents can be any real number). For critertion, let's say minmizing mean squared error. === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? >Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? >I'll take a stab at this. >As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. >Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. >Therefore f is not a generalized polynomial. >quasi > Actually, I see you used 6 variables, whereas the proof I gave above > assumed only 4 variables. > However, assuming my proof is correct, the proof strategy works just > as well for 6 variables. > quasi- Hide quoted text - > - Show quoted text - >How about the best generalized polynomial approximation of f? Any idea >about that? One can assume x1...x6 are real and POSITIVE if that >helps. > I have no idea what you mean by an approximation. > At a point? Globally? What's the criterion for best? > quasi- Hide quoted text - > - Show quoted text - Neither local, nor everywhere... say within the hypercube of [0,1]^6 >(the hypercube is for X's only, exponents can be any real number). For >critertion, let's say minmizing mean squared error. One problem is that f is unbounded on [0,1]^6, which makes approximating f problematic. Any approximation involving integration will need to worry about whether the improper integral converges. Of course, you can always avoid the issue of unboundedness by appropriately restricting the domain of f, however there is a more fundamental problem ... If you choose a compact domain where f is continuous, then f can already be approximated to any degree of accuracy by polynomials. Such polynomials are not unique -- for any given error tolerance, there will be infinitely many such polynomials. In any case, since polynomials can get arbitrarily close, generalized polynomials certainly can't do any better. quasi === Subject: Re: Expanding 1/f(x) by a (generalized) polynomial? <273153hhdgn6d2akh4gnocoomk5f375rlo@4ax.com >Is it possible to expand f(X)=1/(x1*x2+x3*x4+x5*x6) as a generalized >polynomial? By generalized polynomial I mean expansion like g(X)=sum >ak*(x1)^(pk)*(x2)^(qk)*(x3)^(rk)*(x4)^(sk)*(x5)^(tk)*(x6)^(uk) where >the sum is over k. A generalized polynomial is like a regular >polynomial except that exponents pk, qk, ...uk can be any REAL number >(even negative), as opposed to natural numbers in regular polynomials. >I am interested in the most compact expansion of the given function. >Any ideas? >I'll take a stab at this. >As far as I can see, the answer is no -- f can't be expressed as a >generalized polynomial. >Suppose f=p where p is a generalized polynomial. Since f is a >continuous at (x1,x2,x3,x4)=(0,1,1,1), it follows that all terms of p >have nonnegative exponents in the variable x1. On the other hand, at >(x1,x2,x3,x4)=(1,0,0,0), f(x)=1/x1 so some term of p must have a >negative exponent in x1, contradiction. >Therefore f is not a generalized polynomial. >quasi > Actually, I see you used 6 variables, whereas the proof I gave above > assumed only 4 variables. > However, assuming my proof is correct, the proof strategy works just > as well for 6 variables. > quasi- Hide quoted text - > - Show quoted text - >How about the best generalized polynomial approximation of f? Any idea >about that? One can assume x1...x6 are real and POSITIVE if that >helps. > I have no idea what you mean by an approximation. > At a point? Globally? What's the criterion for best? > quasi- Hide quoted text - > - Show quoted text - Neither local, nor everywhere... say within the hypercube of [0,1]^6 >(the hypercube is for X's only, exponents can be any real number). For >critertion, let's say minmizing mean squared error. One problem is that f is unbounded on [0,1]^6, which makes > approximating f problematic. Any approximation involving integration > will need to worry about whether the improper integral converges. Of course, you can always avoid the issue of unboundedness by > appropriately restricting the domain of f, however there is a more > fundamental problem ... If you choose a compact domain where f is continuous, then f can > already be approximated to any degree of accuracy by polynomials. Such > polynomials are not unique -- for any given error tolerance, there > will be infinitely many such polynomials. In any case, since > polynomials can get arbitrarily close, generalized polynomials > certainly can't do any better. quasi- Hide quoted text - - Show quoted text - === Subject: A worthy collection site A worthy collection site http://www.hotalways.com === Subject: Re: #18 Valid Euclid IP compared to Twinkle Twinkle Little Star ; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof <462F8925.4040103@dtgnet.com> On Apr 30, 6:51 am, sttscitr...@tesco.net I am going to end this book with poetry. Yes, stick to poetry in future you talents are > more literarary than mathematical. [...] I've brought this up in the past: If AP had wished to introduce ideas to minimize the amount of criticism, he would have used the medium of science fiction. (I've even set up a scenario which would allow an astronaut to visit the Atom Totality Universe, which for us would be a parallel universe, but for AP would be this universe.) weren't accepted at the time but would be justified later on); from our point of view, it would be entertainment. --- Christopher Heckman === Subject: Re: #18 Valid Euclid IP compared to Twinkle Twinkle Little Star ; monograph-book: Correcting the Logical Flaws of Euclid's Infinitude of Primes Proof <462F8925.4040103@dtgnet.com> comparison. Compare a mathematics > proof such as Euclid's Infinitude of Primes with a famous melody of > the song Twinkle Twinkle Little Star. Now Wikipedia has a error filled Euclid IP, but has a error free > Twinkle Twinkle song as thus: Twinkle Twinkle Little Star in ABC notation, C major: CCGGAAGFFEEDDCGGFFEEDGGFFEEDCCGGAAGFFEEDDC Now let us say Euclid's IP is this song of Twinkle Twinkle Little > Star. And this is another place where AP goes astray. He claims that there is only ONE possible proof of IP (namely his, which means all others are wrong)(You must actually forgive him for this, since he's at the age where he's developed blinders on the sides of his eyes). However, mathematics is full of theorems with multiple proofs. For instance, the Pythagorean Theorem has over 50 proofs. And in fact, Twinkle, Twinkle, Little Star is not unique, either; there are 12 versions, corresponding to the transpositions into the 11 other keys (C#, D, D#, etc., up to B). And he's forgotten to include note lengths in the melody. Since ABC notation has this facility, he's made yet another mistake. The proper ABC notation for Twinkle, Twinkle, Little Star is CCGGAAG2FFEEDDC2GGFFEED2GGFFEED2CCGGAAG2FFEEDDC2 since the 2 means the previous note is held twice as long as the default value. (See http://abcnotation.org.uk/ , for an explanation of abc notation.) --- Christopher Heckman === Subject: Re: Quo Vadis? <12845178.1179466152269.JavaMail.jakarta@nitrogen.mathforum.org>, > Look you jerk you don't own this forum. Bug off. I am having an intelligent conversation with Galathea. Don't you dare impose your stupidity on me and do not tell me what to do. I do as I please. Go back to Hell. I never told you what to do. It is yourself issuing injunctions. -- Michael Press === Subject: Re: Quo Vadis? I was alluding to the creep you cited I suppose. I do not pay much attention to who is saying what on these stupid hostile personal attacks. Best to stay out of it completely as it is evil. I have more important things to think about like gauge theory and I treat the insults by cowardly Trolls who hide their IDs like annoying mosquitos. === Subject: Scale If I place a can of soup in front of my face ten inches away from my eyes, then begin to move can away from my face at what distance will it appear to be three fourths as tall and at what distance will it appear to be half as tall and will those distances be equal? Ron === Subject: Re: Scale Ucoaaa ooi iUiuia > If I place a can of soup in front of my face ten inches away from my > eyes, then begin to move can away from my face at what distance will > it appear to be three fourths as tall and at what distance will it > appear to be half as tall and will those distances be equal? Hint: Suppose that the height of the can is h inches and at 10 inches appears as 2*theta degrees tall. Then: tan(2*theta)=h/10 (1) For it to show half as tall, you want y, such that: tan(theta)=h/(10+y) (2) (1) => 2*tan(theta)/(1-tan(theta)^2)=h/10 (3) Substitute (2) into (3) and solve for y. I get: y=sqrt(100+h^2). So when the can moves at a distance of 10 + sqrt(100+h^2) inches, it will appear half as tall as it appears at 10 inches, unless I made a typo somewhere. The second question is similar. > Ron -- I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/ ---------------------------------------------------------- There's ALWAYS a mistake somewhere === Subject: Re: Scale On Sun, 20 May 2007 09:28:28 +0300, Ioannis [NonBreakingSpace].8b.96.87.8c .97.99.95 .93fi.92.9b.93.87 > If I place a can of soup in front of my face ten inches away from my > eyes, then begin to move can away from my face at what distance will > it appear to be three fourths as tall and at what distance will it > appear to be half as tall and will those distances be equal? Hint: Suppose that the height of the can is h inches and at 10 inches appears >as 2*theta degrees tall. Then: tan(2*theta)=h/10 (1) For it to show half as tall, you want y, such that: tan(theta)=h/(10+y) (2) (1) = >2*tan(theta)/(1-tan(theta)^2)=h/10 (3) Substitute (2) into (3) and solve for y. I get: y=sqrt(100+h^2). So when the can moves at a distance of 10 + sqrt(100+h^2) inches, it will >appear half as tall as it appears at 10 inches, unless I made a typo >somewhere. The second question is similar. Hmmm ... I got a different answer. It's the height that supposed to appear 3/4 or 1/2 of the original, not the angle. quasi === Subject: Re: Scale > If I place a can of soup in front of my face ten inches away from my > eyes, then begin to move can away from my face at what distance will > it appear to be three fourths as tall and at what distance will it > appear to be half as tall and will those distances be equal? > Hint: Suppose that the height of the can is h inches and at 10 inches appears > as 2*theta degrees tall. Then: > tan(2*theta)=h/10 (1) > For it to show half as tall, you want y, such that: > tan(theta)=h/(10+y) (2) > (1) => 2*tan(theta)/(1-tan(theta)^2)=h/10 (3) > Substitute (2) into (3) and solve for y. I get: > y=sqrt(100+h^2). > So when the can moves at a distance of 10 + sqrt(100+h^2) inches, it will > appear half as tall as it appears at 10 inches, unless I made a typo > somewhere. The second question is similar. > > Hmmm ... > > I got a different answer. > > It's the height that supposed to appear 3/4 or 1/2 of the original, > not the angle. Indeed. But if you look ahead, an object A *looks* twice as tall as an object B precisely when the vertical angle that A takes is the double of the angle that B takes. Jose Carlos Santos === Subject: Re: Scale > If I place a can of soup in front of my face ten inches away from my > eyes, then begin to move can away from my face at what distance will > it appear to be three fourths as tall and at what distance will it > appear to be half as tall and will those distances be equal? > Hint: Suppose that the height of the can is h inches and at 10 inches > appears as 2*theta degrees tall. Then: > ... > So when the can moves at a distance of 10 + sqrt(100+h^2) inches, it will > appear half as tall as it appears at 10 inches, unless I made a typo > somewhere. The second question is similar. > > Hmmm ... > > I got a different answer. > > It's the height that supposed to appear 3/4 or 1/2 of the original, > not the angle. Indeed. But if you look ahead, an object A *looks* twice as tall as an > object B precisely when the vertical angle that A takes is the double of > the angle that B takes. > No, the angle is not what is used when it looks like. This is known from middle age with (re)discovery of perspective laws. Consider a screen AC, at distance OA = 10, and the can AC just behind this screen, the image of the can on the screen is just AC. Then put the same can farther at DE. It looks like means the image of the can on the screen, that is AB. AB = k*AC = k*DE is as Quasi said OA = k*OD (just similar triangles OAB and ODE) . C + + E . /| / | can . / | / | . / | / | . / + B | . / / | | . / / | | . // | | . +-------+-------+ . O eye A D You could also take a picture with a camera, reversing the role of O and screen in the opposite way : . C + + E . /|can / | can . / | / | . / | / | . / +B | . / / | | . film / / | | . H // | | . +-------+-------+-------+- . . | //O A D . | / / lens . | / / . F+ / . | / . | / . |/ . + E with HF = k*HE on the film, multiply everything by the camera factor m AB = m*HF, AC = m*HE you get AB = k*AC and OA = k*OD as above. -- Philippe C., mail : chephip+news@free.fr site : http://chephip.free.fr/ (recreational mathematics) === Subject: Re: Scale No, the angle is not what is used when it looks like. > This is known from middle age with (re)discovery of perspective laws. > It looks like means the image of the can on the screen, that is AB. Not always. It depends on which definition one uses. You and quasi used the definition of linear magnification: http://scienceworld.wolfram.com/physics/Magnification.html I used the definition of angular magnification: http://scienceworld.wolfram.com/physics/AngularMagnification.html Both solutions give roughly the same results if h is small compared to the distance of 10 inches. -- I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/ ---------------------------------------------------------- There's ALWAYS a mistake somewhere === Subject: Re: Scale <1179642512.678193@athprx03> <1suv435531r9iuk7l9fcr0i6j7386hhgjq@4ax.com> <5baaa6F2s3aohU1@mid.individual.net> <1179655803.804740@athprx03 Not always. It depends on which definition one uses. You and quasi used the > definition of linear magnification:http://scienceworld.wolfram.com/physics/Magnification.html I used the definition of angular magnification:http://scienceworld.wolfram.com/physics/AngularMagnification.h t ml I am trying to place an object that will be constructed and rendered in one piece of software into a bitmap of a digital photo, in the perspective of the photo. Does that involve angular magnification or linear magnification? === Subject: Re: Scale > Not always. It depends on which definition one uses. You and quasi used the > definition of linear magnification: > http://scienceworld.wolfram.com/physics/Magnification.html I used the definition of angular magnification: > http://scienceworld.wolfram.com/physics/AngularMagnification.html I am trying to place an object that will be constructed and rendered > in one piece of software into a bitmap of a digital photo, in the > perspective of the photo. Does that involve angular magnification or > linear magnification? Better to use linear. Angular is used for optical instruments usually. -- I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/ ---------------------------------------------------------- There's ALWAYS a mistake somewhere === Subject: Re: Scale <1179642512.678193@athprx03> <1suv435531r9iuk7l9fcr0i6j7386hhgjq@4ax.com> <5baaa6F2s3aohU1@mid.individual.net> <1179655803.804740@athprx03> <1179658472.711572@athprx04 Better to use linear. Angular is used for optical instruments usually. Ok, thank you. And I would like to thank everyone; quasi, loannis Jose Carols Santos and Philippe 92, that helped me understand this problem. Your assistance is very much appreciated. Brgds, Ron === Subject: Re: Scale On 19 May 2007 22:53:52 -0700, ronviers@gmail.com >If I place a can of soup in front of my face ten inches away from my >eyes, then begin to move can away from my face at what distance will >it appear to be three fourths as tall and at what distance will it >appear to be half as tall and will those distances be equal? Let h be the height of the can. Let x be the distance to the can now. Let y be the distance to the can when it is r times as tall as it is now. By similar triangles, (r*h)/x = h/y. Solving for y gives y=x/r. So r=3/4 gives y=(4/3)x and r=1/2 gives y=2x. quasi === Subject: Scale If I place a can of soup in front of my face ten inches away from my eyes, then begin to move can away from my face at what distance will it appear to be three fourths as tall and at what distance will it appear to be half as tall and will those distances be equal? Ron Sorry about the double post but I somehow messed up the body of the last one. === Subject: Dedekind sums, sawtooth cotangent sums > If p and q are coprime positive integers, then > s(p,q) := sum_{i=1 ... q-1} { ((i/q))((p*i/q)) } > where ((x)) = 0 if x is in Z and > ((x)) = x - floor(x) - 1/2 . > So ((.)) is discontinuous at 0, -1/2 < ((x)) < 1/2 and ((1/2)) = 0. > [((.)) is a sawtooth function]. > s(p,q) is known as a Dedekind sum, as described at Mathworld, > http://mathworld.wolfram.com/DedekindSum.html > Dedekind Sums Carus Monograph 16, ISBN: 9781114314917 . > As I understand it, the expression for s(p, q) as a sum > of products of two sawtooth-like terms can be proven to > be equal to a sum of products of two cotangent terms using > Fourier series techniques. > I plan to try Fourier series techniques on expressions such as > sum{k=1 ... 177} { (2k-355)cot(113k*pi/355) } . (***) > I can compute the above with PARI-gp , for example, but > haven't found a way to obtain good upper bounds on > its absolute value that carry over to the general case. > > > I've been experimenting with sawtooth functions. > > I have a numerical result for the 355/113 convergent, but the > one for 22/7 is simpler: > > gp > sum(X=1,21,X*cotan(7*X*Pi/22))/22 > 5.880716623706207807311508174 > > So the above is of the same general type as (***) above... > > Next, numerically I get: > > gp > sum(X=1,21,((19*X/22)-floor(19*X/22)-1/2)*cotan(Pi*X/22)) > 5.880716623706207807311508189 > > (19*X/22)-floor(19*X/22)-1/2 is the same as sawtooth(19*X/22) . > > Note: 19 is a multiplicative inverse of 7 , modulo 22. > > Maybe there is an explanation from the discrete Fourier transform > and the Plancherel theorem, > http://en.wikipedia.org/wiki/Discrete_Fourier_transform . 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I'd bet that's no BS who's videoed. r === Subject: Re: * Free Pussy and Tit flash Download! >http://nudepyyicks.blogspot.com/ - take it easy these girls like it deep >try the pussy flash game for free! That's abuse@telus.com Anybody who does send an abuse report, make sure to mention this person is supplying PORNOGRAPHY TO MINORS. Of course post entire header of message. -- Symantec false positive cripples thousands of Chinese PCs http://tinyurl.com/2s6wow === Subject: Re: * Free Pussy and Tit flash Download! > >http://nudepyyicks.blogspot.com/ - take it easy these girls like it deep >try the pussy flash game for free! > > > That's abuse@telus.com > > Anybody who does send an abuse report, make sure to mention this > person is supplying PORNOGRAPHY TO MINORS. > Done. Top === Subject: cancel: Re: * Free Pussy and Tit flash Download! Control: cancel <465001c5$0$3462$a3f2974a@nnrp1.numericable.fr> Excessive crossposting. (<465001c5$0$3462$a3f2974a@nnrp1.numericable.fr>) === Subject: Exercises on Integral closure of rings We consider commutative unitary rings here.. I think I have a general idea of the proofs of the following, but nevertheless I'd like to share Conjectures: 1. Let I be an index set and A_i be integrally closed subrings of B_i, for all i in I then prod_I A_i is integrally closed in prod_I B_i This is quite easy, the below one I haven't worked out but it could turn out into a generalization of this one: 2. Let I, A_i and B_i be as above but this time I is a directed set then the directed limit lim A_i is integrally closed in the direct limit B_i (note that if A_i is a subring of B_i for all i in I, then by universal property of direct limit, the direct limits of A_i is a subring of the direct limit B_i Jose Capco === Subject: Re: physical origins of matrix algebra? <2203.708T1750T6225183@kltpzyxm.invalid> I wonder how many more people would do well in math if they were > given things to tie it to. A whole plenty lot. I present the fact that MOST of math did > originate from physical problems (allow me to include intangibles > like gambling or annuities, so that I can encompass probability and > statistics?) as defacto evidence that inventions seldom happen > unless they're mothered by necessity Perhaps not as many as you think. One of the essential skills for > advanced math Mathematicians are singularities that are confined in a zoo. Out > here in the real world, there's not much advanced math at work in > the world. There would be, but the engineers and consultants and other so called problem solvers are often too ignorant of mathematics to apply it. My favorite example is the program that was supposed to calculate the optimal ratio between paying out dividends and paying salary to the owners of a privately held business. It was obviously a bog standard LP problem. The software took ages to come up with the answer and the reason was obvious. The solution process consisted of discretizing the two decision variables, then enumerating through all the possible solution pairs to find the minimum. Let's not even get started with bad statistics being done every day. > As far as I know, you can gain a BSEE or a BS Physics > without touching any truly advanced mathematics. I'm not sure if > any advanced math was required to put a man on the moon. You can probably get a BS in math from most places without doing anything remotely advanced. As for space missions, most of it is probably just high precision engineering, but try to compute controls for a spacecraft which needs to have trajectories that remain accurate for years without drifting off course while having a lag of minutes in any control inputs. You'll be spending a fair amount of time working on your numerical methods. > Look. Let's say I could demonstrate that making mathematics > INTERESTING to an average voter, would triple your salary - would you > have a change of heart? Mathematics is a profession that has real worth, whether you acknowledge it or not. Why aren't other professions constantly required to pander to the ignorant masses to make things interesting? === Subject: Solving g(xy)=g(2x)*g(y/3)+g(2x)+g(y/3) , x>0 Is there a known method to solve such an equation with g(x) strictly monotonous? Alain === Subject: Re: Solving g(xy)=g(2x)*g(y/3)+g(2x)+g(y/3) , x>0 On May 20, 11:55 am, alainvergh...@yahoo.fr such an equation with g(x) strictly monotonous? Alain Let y = 3*u, x = v/2 and a = 3/2 => g(a*u*v) = g(u)*g(v) + g(u) + g(v), for v > 0 (1) Is there a general solution for (1)? Theron === Subject: Re: Solving g(xy)=g(2x)*g(y/3)+g(2x)+g(y/3) , x>0 > On May 20, 11:55 am, alainvergh...@yahoo.fr Is there a known method to solve > such an equation with g(x) strictly monotonous? Alain Let y = 3*u, x = v/2 and a = 3/2 = > g(a*u*v) = g(u)*g(v) + g(u) + g(v), for v > 0 (1) Is there a general solution for (1)? Theron *********** YES (1) is equivalent to 1 + g(a*u*v) = (1 + g(u))*(1 + g(v)) in (u,v) homogeneous {a*u*v} Vs {u}*{v} We may choose (1 + g(v)) = (a*v)^b , b constant Alain === Subject: Re: Solving g(xy)=g(2x)*g(y/3)+g(2x)+g(y/3) , x>0 On May 20, 4:55 am, alainvergh...@yahoo.fr such an equation with g(x) strictly monotonous? Alain Starting with some trivial cases might help. For exampls, we have: g(x) = g(2)*g(x/3) + g(2) + g (x/3) = g(2x)*g(1/3) + g(2x) + g(1/3) One thing that comes to mind is: g(2x) = g(2x)*g(2/3) + g(2x) + g(2/3). (since 2x = x * 2) let y = g(2x) and C=g(2/3). This gives us the equation: 0 = Cy + C -1 = y Thus, g(2x) is the constant function -1, which may or may not be a valid solution, depending on whether your definition of (say)monotonoic increasing means x > y implies g(x) > g(y) or if it means x > y implies g(x) >= g(y). If the latter, g(x) = -1 is easily seen to be a valid solution for all x, and therefore definitely on the domain you list. === Subject: denumerable type Let H be an Hilbert space. Then H is said to be of denumerable type if it is ..... ? Separable? Second coutable? Please, let me understand the meaning of denumerable type referred to an Hilbert space. Best, Larry === Subject: Re: denumerable type larry > Let H be an Hilbert space. Then H is said to be of denumerable type if it is ..... ? Separable? > Second coutable? Bourbaki: A metrizable space is said to be of countable type (or seperable) if its topology has a countable base. A few lines later he proves that an equivalent condition is that the space has a countable dense subset. In Dieudonn.8e the usuge is the same except that he is referring only to metric spaces. I hope it's safe to assume that denumerable and countable still mean the same thing :) LH === Subject: Re: Cardinal, Natural, Ordinal, Whole, Counting, Integer? The counting numbers are 1, 2, 3, 4.... The natural numbers are 0, 1, 2, 3, 4... The integers are ... -4, -3, -2, -1, 0, 1, 2, 3, 4... The term whole number is sometimes used to distinguish integers from fractions. Whether or not negative numbers are included would depend on whether the full set of rational numbers, including negative rationals, is being worked with yet. As already noted in this discussion, the terms cardinal and ordinal, as applied to integers, belong to grammar; cardinal numbers being 1, 2, 3..., or one, two, three, and ordinal numbers being 1st, 2nd, 3rd..., or first, second, third. The terms cardinal and ordinal have, however, been given a new meaning which is mathematical, but this belongs to axiomatic set theory, a topic not covered in elementary school. The cardinality of a set is a parameter which distinguishes whether or not the members of two sets, irrespective of order, can be placed in a one-to-one correspondence with each other. Thus, Cantor's diagonal proof showed that the set of real numbers had a higher _cardinality_ than the set of integers. A set with an order-relation which is well-ordered also has an order- type, and ordinal comparisons may be made on the order type. Thus, the set {0, 1, 2, 3... } has order type (lowercase) omega, and the set {1, 2, 3, 4..., 0} with the same members, has order type omega+1, which is greater. There are aleph-1 (a cardinality) distinct order types applicable to sets with cardinality aleph-null. The number of real numbers is known to be greater than the number of integers, but this number, the _cardinality of the continuum_, may be equal to or greater than aleph-1, the next cardinality after aleph-null, the cardinality of the integers; that it is equal is the continuum hypothesis, which remains an open question. John Savard === Subject: Re: Cardinal, Natural, Ordinal, Whole, Counting, Integer? > The counting numbers are > > 1, 2, 3, 4.... > > The natural numbers are > > 0, 1, 2, 3, 4... There is clearly some inconsistency in the definition. Merriam Webster and American Heritage give {1,2,3,...}, as do Behnke, et al. Stoll (Set Theory and Logic,) OTOH, gives {0,1,2,...}; as does Quigley (Axiomatic Set Theory). > The cardinality of a set is a parameter which distinguishes whether or > not the members of two sets, irrespective of order, can be placed in a > one-to-one correspondence with each other. Thus, Cantor's diagonal > proof showed that the set of real numbers had a higher _cardinality_ > than the set of integers. If you accept the proof. I do not. I side with Weyl. Talking about a completed mapping of an infinite set onto, or into another infinite set is to use language and concepts derived from, and applicable to the finite in a domain where they do not properly apply. > A set with an order-relation which is well-ordered also has an order- > type, and ordinal comparisons may be made on the order type. Thus, the > set > > {0, 1, 2, 3... } > > has order type (lowercase) omega, and the set > > {1, 2, 3, 4..., 0} > > with the same members, has order type omega+1, which is greater. There > are aleph-1 (a cardinality) distinct order types applicable to sets > with cardinality aleph-null. The number of real numbers is known to be > greater than the number of integers, but this number, the _cardinality > of the continuum_, may be equal to or greater than aleph-1, the next > cardinality after aleph-null, the cardinality of the integers; that it > is equal is the continuum hypothesis, which remains an open question. > > John Savard -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Cardinal, Natural, Ordinal, Whole, Counting, Integer? > > > > I believe the term 'ordinal number' is used (out side of any specialized > discipline) to designate members of the set of utterances (or concepts > associated with utterances) indicating position in a listing (also > called > an enumeration). That is to say the set of {first, second, third,...}. > > > > You are correct that this is the elementary school definition. Cardinals > are one, two, three, ... and ordinals are first, second, third, ... > > This is distinction without a difference, and it's of little > mathematical importance. > > The true distinction between cardinals and ordinals arises when > mathematicians consider infinite sets. For example consider two ways of > ordering the counting numbers: > > * The usual ordering: 1, 2, 3, 4, 5, ... > > * An alternative ordering where we simply put 1 at the very end: 2, 3, > 4, 5, ..., 1. > > In the usual ordering, there is a smallest element but no largest > element. In our alternative ordering, there is a smallest element and a > largest element. So these two orderings are different. > > Our two orderings represent two distinct ordinal numbers. But since they > are different orderings of the same underlying set, they have the same > cardinality. > > This is the true distinction between cardinals and ordinals. An infinite > set of a given cardinality can be ordered in many different ways. > > ? classical logic was abstracted from the mathematics of finite sets and > their subsets ?. Forgetful of this limited origin, one afterwards mistook > that logic for something above and prior to all mathematics, and finally > applied it, without justification, to the mathematics of infinite sets. > This is the Fall and original sin of [Cantor's] set theory ?. (Weyl, > 1946) After considering this a bit further I think I should add that Cantor's theory of transfinite numbers may represent a logically consistent structure, but I do not believe his theory is a logical consequence of every possible and useful axiomatic foundation for mathematics. I will, however, not concede that one can begin with an axiomatic system which defines mathematical objects as discrete entities and arrive at a completed continuum through construction. -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis === Subject: Re: Wiener's Tauberian Theorem proof On Sun, 13 May 2007 05:24:05 -0500, David C. Ullrich >Writing up some notes on WTT for a student I stumbled >on a proof that seems to me to be infinitely superior >to the proofs I've seen. On the one hand I can't believe >it's not well known, since it's so simple; on the >other hand if it's well known I don't understand why >I've never seen it before. On the third hand, I don't >know the literature on Banach algebras very well, >hence this post - if anyone knows that this is >well known I'd appreciate hearing about it. I see that a few posts appeared on sci.math.research but not here. If anyone's curious, it turns out that no, the argument's not new, for example it appears in the proof of Lemma 25C in Loomis An Introduction to Abstract Harmonic Analysis. >Let's call the following the fundamental theorem >of commutative Banach algebras: ftcba: If A is a commutative Banach algebra with >identity and J is a proper ideal then J lies >in the kernel of some (non-zero) complex homomorphism >of A. And now let L^1 = L^1(R) (actually everything here >works just as well on an arbitrary locally compact >abelian group), and let f^ denote the Fourier transform >of f. WTT: If J is an ideal in L^1 and for every y there >exists g in J with g^(y) <> 0 then J is dense in L^1. >In particular, if f in L^1 and f^ has compact support >then f is in J. Let's say f is in J_E if there exists g in J >with f^ = g^ on E, and f is in J near y if >f is in J_V for some neighborhood V of y. The >usual proof of WTT proceeds in two steps: (i) For every y, f is in J near y. This is proved by an argument that's sort of >inspired by the proof of ftcba, instead of as >a corollary of ftcba itself. Then, using a >partition of unity or something analogous, >we show that (i) implies (ii): (ii) If K is compact then f is in J_K, and we're done. Not that the proof is all that hard, but it >turns out that we can prove (ii) directly, >without going through (i), and more to the >point we can prove (ii) as a simple application >of ftcba itself! Let sigma(A) denote the space of all (non-zero) >complex homorphisms of A. Lemma. Suppose A is a commutative Banach algebra >and I is a closed ideal in A. Then A/I is a >Banach algebra, and sigma(A/I) is equal to >the space of all phi in sigma(A) such that >phi vanishes on I. Proposition. Suppose E is a closed subset of R >and let I be the set of all f in L^1 such that >f^ vanishes on E. Then sigma(L^1/I) = E (ie >if phi is in sigma(L^1/I) then there exists >y in E with phi(f+I) = f^(y)). If E is >compact then L^1/I has an identity. Pf: The first statement follows from the lemma. >Suppose E is compact. If j in L^1 is such that >j^ = 1 on E then j+I is an identity in L^1/I. QED. Pf of WTT: Suppose J is as in the statement of WTT >and f in L^1 such that f^ has compact support K. >Let I be the set of g with g^=0 on K. Let >q : L^1 -> L^1/I be the quotient map q(g) = g + I. Now q(J) is an ideal in L^1/I, and the lemma shows >that g(J) is not annihilated by any complex >homorphism of L^1/I. Hence ftcba shows that >q(J) = L^1/I. In particular there exists j in >J such that j^ = 1 on K; hence f = f*j. QED ************************ David C. Ullrich ************************ David C. Ullrich === Subject: Re: Wiener's Tauberian Theorem proof I receieved an email about this from Julia Kuznetsova. She(?) says she intended to post the same comments to sci.math.research, but she's not certain whether she did it right. Since the post has in any case not appeared yet, and since I'm going to be away for the next week and a half, I thought that I'd post a reply to the email instead: >Hello David, I'd like to reply to your post in the sci.math.research group. >I posted a message there but I'm new to Usenet and not sure if >I've done everything correctly. That's why I decided to send you >a message by mail also. It repeats the post in the internet. Julia Kuznetsova you have any reply to _this_ post I won't be able to reply to it for about 10 days (see above). >David, I think your proof does not differ much from standard >but you omit some statements and therefore it seems shorter. I will use your notations. >The (i) in your text is what's called regularity; and you do use it >in your proof implicitly in your Proposition. To state that sigma(L^1/I) >equals E you need to show that it is no more than E; i.e. for every >t not in E there is f in I with f^(t)<>0; as f is in I, f^(E)=0. >You see it is exactly regularity. The only regularity I know of in this context is the definition that an ideal I in A is regular if A/I has a unit. I don't quite see what (i) has to do with that, exactly - maybe you're using the word in a different sense. But that doesn't matter. The important point, if there is one, is that no, I'm simply _not_ using (i) in my proof! Honest - it seems to me you're misunderstanding something, exactly what I'm not sure. Briefly, the as f is in I, f^(E)=0 is exactly backwards - in fact f is in I _because_ f^(E) = 0. In detail: Let's recall what (i) stated, including context explaining the notation and hypotheses: >WTT: If J is an ideal in L^1 and for every y there >exists g in J with g^(y) <> 0 then J is dense in L^1. >In particular, if f in L^1 and f^ has compact support >then f is in J. >Let's say f is in J_E if there exists g in J >with f^ = g^ on E, and f is in J near y if >f is in J_V for some neighborhood V of y. The >usual proof of WTT proceeds in two steps: >(i) For every y, f is in J near y. Now about the proof of the proposition: >Proposition. Suppose E is a closed subset of R >and let I be the set of all f in L^1 such that >f^ vanishes on E. Then sigma(L^1/I) = E (ie >if phi is in sigma(L^1/I) then there exists >y in E with phi(f+I) = f^(y)). If E is >compact then L^1/I has an identity. (Note that the I in the Proposition is not the same as the J in the statement of WTT...) You're exactly right when you say I need to prove the following: (iii) If t is not in E then there exists f in I with f^(t) <> 0. Here's the proof of (iii): Let phi be a function in C^2_c(R) (twice continuously differentiable, with compact support) such that g(t) <> 0 while g(E) = 0. There exists f in L^1 such that g = f^. QED. Yes, I omitted the proof of (iii) in my post. That was because the proof is entirely trivial; it certainly does not use (i). (Offhand I don't see how it _could_ use (i), since (i) is a statement about J while (iii) is a statement about I. But in any case the trivial proof of (iii) does _not_ use (i), explicitly or implicitly.) In fact it's very easy to give a proof of (iii) that's valid in the context of locally compact abelian groups, using just the Plancherel theorem. Say G is an LCA group with dual group Gamma. Say E is a closed subset of Gamma, let I be the set of f in L^1(G) with f^(E) = 0, and now consider (iii') If t is in Gamma but t is not in E then there exists I in E with f^(t) <> 0. Proof: Let U and V be open sets of finite measure such that t is in U + V but U + V is disjoint from E. Let phi and psi be the characteristic functions of U and V. The Plancherel theorem shows that there exists f in L^1(G) with f^ = phi*psi (the convolution of phi and psi), and it follows that f^(t) <> 0 while f^(E) = 0. QED. >Then, you need to say a few words on why the functions with f^ having >compact supports are dense in L^1 -- you use this also. Yes, I use that fact - that part of the proof is exactly the same as in the standard proof. And that fact is also very trivial. For R it follows using an explicit approximate identity. For G: The Plancherel theorem shows that the set of f in L^2(G) such that f^ has compact support is dense in L^2(G). Say f is in L^1(G). Choose g in L^2(G) with f = gg. Choose g_n in L^2(G) such that g_n^ has compact support ang g_n -> g in L^2. Then (g_n g_n)^ has compact support and g_n g_n -> f in L^1. QED. >And these two facts >together are the usual requirements of the abstract tauberian theorem, >as in Loomis 25D (An introduction to abstract harmonic analysis). I think the proof of WTT itself you give is very close to that given >in Loomis, although he doesn't transfer to quotients. I can't look up that proof for a little while (see above). It sounds to me like it's the same as the standard proof that I know. In any case: Does the proof in Loomis involve an _application_ of what I called the fundamental theorem for commutative Banach algebras? If not then it doesn't seem like the same proof to me - my whole point is that we can get WTT by _applying_ that theorem, instead of by using arguments that are like the proof of the theorem. Or: Look at the proof of (i) in Loomis. It's nowhere near as trivial as the proof of (iii) above, right? >As far as I know, it was long a question whether it's possible to prove WTT >without appealing to regularity. And indeed it is, see J. Esterle, >A complex-variable proof for the Wiener Tauberian theorem, >Ann. Inst. Fourier 30 no. 2 (1980), 91-96. The proof is no short though. ************************ David C. Ullrich === Subject: Jordan form of the BLOCK Frobenius matrix and its inverse 1) -It is well known that Jordan form of the SCALAR Frobenius matrix is Vendermonde matrix (single eigenvalues) or confluent V. matrix for non-single eig. -Inverse of the calssic Vandermonde consist from basic symmetrical polynomials, for confluent Vand. is a little bit more sophysticated, but known. 2) Now let's take into the accound the BLOCK Frobenius matrix. (Frobenius with sub-matrices) Anyone know whether for such a matrix the Jordan form and its inverse is known in the literature ? Is it possible to calculate it in a general analytical form ? Especially the inverse ? John === Subject: msn display pictures www.entertainmentvenues.org venues for live entertainment, mobile venues, free web2sms, mobile themes, free nokia themes, Mobile 3GP Videos, Mobile Games, Mobile secrets, Online Games, free wallpapers, Msn Stuff and many more. === Subject: Re: Topological sum are there topological spaces A and B, such that the disjoint union >(coproduct) >A U A is homeomorphic to B U B, but A is not homeomorphic to B? Florian > According to the abstract available at http://www.ams.org/proc/2006-134-12/S0002-9939-06-08596-0/ there are compact metric spaces A and B having this property. -- David Hartley === Subject: Re: Topological sum are theretopologicalspaces A and B, such that the disjoint union > (coproduct) > A U A is homeomorphic to B U B, but A is not homeomorphic to B? [I thought I posted this yesterday but it has not appeared. Sorry if it's a duplicate.] The answer is no. First consider the case where A and B are discrete, so this is a statement about sets and bijections. If A or B is finite, it's obvious. If A and B infinite, it's a theorem of cardinal arithmetic (using Zorn's lemma) that A U A is in bijective correspondence with A, so the statement is true there as well. Next, suppose A is a disjoint union of homeomorphic components, i.e. A = X x D, X connected, D discrete. Show that B U B is a disjoint union of components homeomorphic to X. A homeomorphism from A U A to B U B gives a bijection on their respective components, and by the previous case there is a bijection between the components of A and B, giving a homeomorphism since those components are homeomorphic. Finally, for general A, classify the components by homeomorphism type. The existence of a homeomorphism will show that the components of B U B admit a similar classification, and then the previous part can be applied to each class. === Subject: Re: Topological sum > are theretopologicalspaces A and B, such that the disjoint union > (coproduct) > A U A is homeomorphic to B U B, but A is not homeomorphic to B? [I thought I posted this yesterday but it has not appeared. Sorry if >it's a duplicate.] The answer is no. First consider the case where A and B are discrete, so this is a >statement about sets and bijections. If A or B is finite, it's >obvious. If A and B infinite, it's a theorem of cardinal arithmetic >(using Zorn's lemma) that A U A is in bijective correspondence with A, >so the statement is true there as well. Next, suppose A is a disjoint union of homeomorphic components, i.e. A >= X x D, X connected, D discrete. Show that B U B is a disjoint union >of components homeomorphic to X. A homeomorphism from A U A to B U B >gives a bijection on their respective components, and by the previous >case there is a bijection between the components of A and B, giving a >homeomorphism since those components are homeomorphic. Finally, for general A, classify the components by homeomorphism >type. The existence of a homeomorphism will show that the components >of B U B admit a similar classification, and then the previous part >can be applied to each class. > In general, a topological space is not the disjoint union of its components. E.g. the rationals with the usual topology. > -- David Hartley === Subject: Re: Topological sum are theretopologicalspaces A and B, such that the disjoint union > (coproduct) > A U A is homeomorphic to B U B, but A is not homeomorphic to B? [I thought I posted this yesterday but it has not appeared. Sorry if >it's a duplicate.] The answer is no. First consider the case where A and B are discrete, so this is a >statement about sets and bijections. If A or B is finite, it's >obvious. If A and B infinite, it's a theorem of cardinal arithmetic >(using Zorn's lemma) that A U A is in bijective correspondence with A, >so the statement is true there as well. Next, suppose A is a disjoint union of homeomorphic components, i.e. A >= X x D, X connected, D discrete. Show that B U B is a disjoint union >of components homeomorphic to X. A homeomorphism from A U A to B U B >gives a bijection on their respective components, and by the previous >case there is a bijection between the components of A and B, giving a >homeomorphism since those components are homeomorphic. Finally, for general A, classify the components by homeomorphism >type. The existence of a homeomorphism will show that the components >of B U B admit a similar classification, and then the previous part >can be applied to each class. In general, a topological space is not the disjoint union of its > components. E.g. the rationals with the usual topology. === Subject: Re: Theoretical vs Actual Is this question difficult? :( === Subject: Re: proper classes in ZF zuhair says... >So what this quotation is telling us, is that all proper classes in >NBG can be reduced in ZFC (or any thoery that doesn't have proper >classes as objects in its universe of discourse) to open formulas. However the source is mentioning that in MK there are proper classes >that cannot be reduced to open formulas in ZFC and the alike theories. I want to know which proper classes in MK that cannot be reduced to >open formulas in ZFC, and why? Let's formulate MK using two different sorts of variables. Upper case variables such as X, Y, Z range over classes, and lower case variables such as x,y,z range over sets. Every set is a class, but not vice-versa. If Phi is a formula that only involves set variables, not class variables, then let's call it a pure formula. The pure formulas of MK correspond to the formulas of ZFC. For any formula Phi(x), MK proves (EX) (Ax) x e X <-> Phi(x) So MK proves that the class { x | Phi(x) } exists for any formula Phi, whether pure or not. (Note that x is lower-case, which means that this is the collection of all *sets* satisfying Phi(x). MK does not allow talking about collections of proper classes) If Phi is *not* pure (that is, if it contains class variables), then the corresponding class { x | Phi(x) } does not correspond to an open formula of ZFC (because ZFC cannot talk about proper classes). Okay, there is actually a proof obligation here. If Phi is not pure, how do you know that there isn't some other formula Phi' that is pure such that Phi(x) <-> Phi'(x)? In other words, how do you know that you get any new classes by allowing impure formulas? Well, here's a way, using Cantor's theorem. Every pure formula Phi(x) can be coded up as a set using Godel coding. So now let's define a function Comp from sets to classes as follows: If z is a code for a formula Phi(x), then Comp(z) = { x | Phi(x) }. If z is not a code for a formula, then Comp(z) = the empty set. Every proper class that is definable by an open formula in ZFC is equal to Comp(x) for some set x. Now, let R = { x | x is not an element of Comp(x) }. Clearly, R cannot be equal to Comp(x) for any x. So R is a proper class that is not definable by a formula of ZFC. (It turns out that R *can* be defined in MK.) -- Daryl McCullough Ithaca, NY === Subject: Re: proper classes in ZF zuhair: Also, did you read Keith Ramsay's post? You should read it and give it some thought too. (Indeed, you should pay attention to any post he makes to you, as he his posts are consistently informative and worth saving to your hard drive.) MoeBlee === Subject: Re: proper classes in ZF > I will quote the following from Planet math. - - - In many set theories there are formally no proper classes; ZFC > is an example of just such a set theory. In these theories one usually > means by a proper class an open formula , possibly with set > parameters . Notice, however, that these do not exhaust all possible > proper classes that should really exist for the universe, as it only > allows us to deal with proper classes that can be defined by means of > an open formula with parameters. The theory NBG formalises this usage: > it's conservative over ZFC (as clearly speaking about open formulae > with parameters must be!). There is a set theory known as Morse-Kelley set theory which allows us > to speak about and to quantify over an extended class of > impredicatively defined proper classes that can't be reduced to simply > speaking about open formulae - - -. Quotation finished. One thing that realy perplex me about this notion, is that if proper > classes in ZF are the open formulas, then we know that MK quantify > over these proper classes, this mean that MK which is a first order > logic set theory is actually quantiying over open formulas in ZF. See, this an example of the kind of problem I find with this kind of loose way of talking. Are ZFC open formulas ACTUALLY proper classes; or rather is there a RELATION BETWEEN ZFC formulas and proper classes? I would be very wary of committing to the propostion that ZFC open formulas are ACTUALLY proper classes. In fact, in my meta-theory, ZFC open formuas are definitely NOT themselves proper classes (in my meta- theory, which is itself a ZFC one step up from object level ZFC, ZFC open formulas are sequences of symbols, and sequences are a certain kind of function, and functions are a certain kind of set, so ZFC formulas are sets, and not proper classes). However, I do understand that we might formalize a certain relation between open formulas in the language of ZFC and objects that fulfill the role of Bernays proper classes. But that's just my own way of looking at it. > This > provok me to ask the following question: can a theory in first order > logic be equi-interpretable with a theory in second order logic? since > it appears to me that MK which is a first order logic thoery is > working like second order ZF, and I think they would be equi- > interpretable. I don't know about equi-interpretability of Morse and ZFC (my guess is that they are not equi-interpretable or at least not proven to be equi- interpretable, since, as I understand, there's not a known relative consistency proof from ZFC to Morse), but, as I understand, there are instances of equi-interpretability between theories in languages of different order. But, dealing with this precisely at least begins with a definition of 'interpretable'. There are definitions in mathematical logic of these things; so we should learn them and be very clear about them if we are to really understand what is involved in such questions as yours. Anyway, as you get further into this you will discover relations among first order theories, first order theories in multi-sorted languages, relativizations of theories, first order set theories, second order theories of arithmetic, hierarchies in set theories, etc. MoeBlee === Subject: Re: proper classes in ZF > I will quote the following from Planet math. - - - In many set theories there are formally no proper classes; ZFC > is an example of just such a set theory. In these theories one usually > means by a proper class an open formula , possibly with set > parameters . Notice, however, that these do not exhaust all possible > proper classes that should really exist for the universe, as it only > allows us to deal with proper classes that can be defined by means of > an open formula with parameters. Just as a matter of personal taste, I kind of don't like speaking of these things in this manner, so I'm somewhat uncomfortable here. But I think the point here is that there are only a denumerable number of formulas, yet Bernays set theory proves the existence of more than just a denumerable number of classes, and since ZFC and Bernays set theory are in certain resepect essentially the same (there are formal ways of saying that, but, for now I demur from a formulation since there are a couple of points regarding it that I feel I haven't worked out properly in my own understanding), then we have to recognize that there are (even in a ZFC context, which again I've not made formal) classes that don't correspond to a ZFC formula. > The theory NBG formalises this usage: > it's conservative over ZFC (as clearly speaking about open formulae > with parameters must be!). There is a set theory known as Morse-Kelley set theory which allows us > to speak about and to quantify over an extended class of > impredicatively defined proper classes that can't be reduced to simply > speaking about open formulae - - -. This probably has to do with the fact that Morse allows the 'P' in the comprehension schema to have general class variables (i.e., not restricted to set variables). In other words, IF I understand all of this correctly: Z (thus ZFC too) is impredicative since the 'P' in the separation schema allows variables whose range includes the very set that is being asserted to exist; then Bernays set theory is impredicative in that way too, but it is not impredicative over proper classes too, since the variables in the 'P' in the comprehension schema are restricted to set variables; but Morse is impredicative over everything (including proper classes) since, for asserting the existence of a proper class, the 'P' in the comprehension schema allows variables whose range includes the very proper class that is being asserted to exist. (If that is not a correct summary, then hopefully someone will correct or sharpen it.) as providing an extended (more?) proper classes even than Bernays set theory. However, I don't quite catch what difference is being reduced to open formulas. > So what this quotation is telling us, is that all proper classes in > NBG can be reduced in ZFC (or any thoery that doesn't have proper > classes as objects in its universe of discourse) to open formulas. That's not my impression (though I could be wrong). I'm not so sure theory proves the existence of more proper classes than can be specified in ZFC by a ZFC formula for each proper class. > However the source is mentioning that in MK there are proper classes > that cannot be reduced to open formulas in ZFC and the alike theories. saying, as it seems to me that you are thinking, that ZFC can catch all the Bernays proper classes but not all of the Morse proper classes (maybe it is saying that, but it's just not clear to me that it is). Anyway, again, the main difference has to do with the differences in what variables are allowed in the respective comprehension schemata. But I'm not clear how that plays out toward what proper classes can be captured by a ZFC formula. > I want to know which proper classes in MK that cannot be reduced to > open formulas in ZFC, and why? I don't know enough about Morse in particular. But at least on general grounds, again, as I mentioned, all of these theories prove the existence of more than can specified by a formula for each class (or even for each set). There are only denumerably many formulas. But there are more than just a denumerable number of sets (and, I would think, in Bernays set theory, even more than just a denumerable number of proper classes; and, I would think, even more proper classes in Morse than in Bernays). P.S. One of the reasons I'm uncomfortable talking about this is that I haven't worked out a problem I discussed with Aatu. See, I'd rather not use a two-sorted language, since, at least for me, that complicates comparing among first order theories. So I started working on a strict formalization of Bernays set theory using relativization rather than two-sorted logic. But what I found out is that when you use relativization, you don't get that Bernays set theory is truly, technically, a conservative extension of ZFC, but rather there is yet another technical qualification that has to be made to get a rigorous formulation of the statement that Bernays is essentially conservative over ZFC. However, that is a concern that is perhaps mainly one of my fussiness as to technicals, while for the working purposes of most people it is enough to understand the basic sense in which Bernays set theory is convservative over ZFC. MoeBlee === Subject: Re: proper classes in ZF > Just as a matter of personal taste, I kind of don't like speaking of > these things in this manner, so I'm somewhat uncomfortable here. But I > think the point here is that there are only a denumerable number of > formulas, yet Bernays set theory proves the existence of more than > just a denumerable number of classes, and since ZFC and Bernays set > theory are in certain resepect essentially the same (there are > formal ways of saying that, but, for now I demur from a formulation > since there are a couple of points regarding it that I feel I haven't > worked out properly in my own understanding), then we have to > recognize that there are (even in a ZFC context, which again I've > not made formal) classes that don't correspond to a ZFC formula. You seem to be disregarding the qualifier with parameters in my text. In NBG the existence of any class definable by a formula with set parameters (and free class variables) is provable. There are certainly more than denumerable number of them -- in fact there are at least as many proper classes as there are sets. probably phrase it differently. There is nothing seriously wrong it, though. > P.S. One of the reasons I'm uncomfortable talking about this is that I > haven't worked out a problem I discussed with Aatu. See, I'd rather > not use a two-sorted language, since, at least for me, that > complicates comparing among first order theories. So I started working > on a strict formalization of Bernays set theory using relativization > rather than two-sorted logic. But what I found out is that when you > use relativization, you don't get that Bernays set theory is truly, > technically, a conservative extension of ZFC, but rather there is yet > another technical qualification that has to be made to get a rigorous > formulation of the statement that Bernays is essentially conservative > over ZFC. However, that is a concern that is perhaps mainly one of my > fussiness as to technicals, while for the working purposes of most > people it is enough to understand the basic sense in which Bernays set > theory is convservative over ZFC. As said, it's a mere technicality. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: proper classes in ZF > there are more than just a denumerable number of sets P.S. Just to be clear, since this comes up sometimes: I'm not denying Lowenheim-Skolem when I say there are more than a denumerable number of sets, since I'm not claiming that there aren't denumerable models of ZFC; and all I am claiming it that there is a theorm of ZFC that we read off from the symbols as There exists an uncountable set. MoeBlee === Subject: Re: jack sarfatti, bio > http://www.law.duke.edu/slsa/archive/?/digest+2007-04-18+%23001.txt > I think your hero is in trouble. > That's a long, dull page. Why not just summarize what you think is relevant? -- Jesse F. Hughes The American people would have been incredibly proud of watching our military folks dispense with basic health care needs to people who needed help. --George W. Bush, March 13, 2007 === Subject: Re: jack sarfatti, bio Another cowardly Troll creep jc with lies hiding their true ID. Note that such annoying slanderous use of the Internet for harassment is actually in violation of a new Federal Law - yet to be tested. === Subject: Re: jack sarfatti, bio <28225140.1179530677939.JavaMail.jakarta@nitrogen.mathforum.org Another cowardly Troll creep jc with lies hiding their true ID. Note that such annoying slanderous use of the Internet for harassment is actually in violation of a new Federal Law - yet to be tested. First of all, there were no lies in my post. second, I think the law would be more harsh on you for pretending to be Sarfatti. -jc === Subject: JC's inability to think clearly & his criminal internet activity Your charges 1. I am some kind of illegal drug addict 2. I never published in respectable journals, books since 1974 are slander. Also you are really stupid since some of the same stuff here on Math Forum is on my blog. You are violating the law by hiding your identity and making slanderous remarks. === Subject: Re: Problem on an nxn grid Cc: jberry@islandnet.com > It has taken me days to understand the equivalent of > See Spot Run in the vocabulary of Graph Theory. > Sparing everyone the litany of that journey... Asinops and Robert Israel, for their help. The problem is a maximum-weight matching with n up to > around 300. Does there exist code (or better, pseudo-code) > to solve this? It could be argued that Chapter 5.3 of > Gibbons's Algorithmic Graph Theory gives a pseudo-code > for the Edmonds algorithm, but it is couched in, er, > Graphic, language. As, for the purposes of this discussion, > a lukewarm mathematician, and a lukewarm coder, it looks > seems to be something approaching pseudo- code for related > problems (such as the Stable Roommate Problem) and code for > parts of the Edmonds algorithm (such as the Hungarian > algorithm), I could not find code or pseudo-code for the > whole kahuna, an algorithm which solves the maximum-weight > matching problem in polynomial time. Maybe I haven't looked > in the right places. But an email from one of you confirms: > no readily- available code. I'm guessing that the reason will be that it is too easy. > Too easy a subject for a degree thesis, that is. Anyway, if somebody can point me to pseudo-code, I'd > appreciate it. In a separate post, I will put forward an idea that might > not reduce the big-O Order of working this through by brute > force, but might reduce the real-world CPU time in cases > where the values (weights) are not overly close to each > other. -- > Jonathan Berry My last post to this thread seems to have gone missing, an all too frequent event with Google Groups. Below I will refer to the weight of edge from node i to node j as A(i,j), consistent with the earlier grid/matrix notation in this thread. All weights are assumed to be nonnegative. We seek a minimum (total) weight perfect matching of the (underlying) complete graph on n nodes. This paper: http://www.dcg.ethz.ch/publications/ctw04.pdf by Wattenhofer and Wattenhofer gives pseudo-code for two approximation algorithms assuming that the weights of the graph satisfy the triangle inequality, i.e. A(i,j) <= A(i,k) + A(k,j) for all nodes i,j,k. This restriction is in one sense inessential as adding max(A) to every weight forces that condition to be true but at the expense of messing up the approximation property (because it now applies to the total inflated by max(A) * n/2). It suggests the question of whether exact solutions for weighted graphs satisfying triangle inequalities can be found just as efficiently, e.g. with nearly quadratic algorithms rather than the improvement of Edmonds original O(n^4) to O(n^3) by Gabow (and in an earlier paper by Lawler). Although most implementation details are omitted, a helpful discussion of the history and tests is here: http://www2.isye.gatech.edu/~wcook/papers/match_ijoc.pdf in a paper by Cook and Rohe. If the weights can be realized as the Euclidean distances between nodes assigned coordinates, then the problem is said to be geometric, and of course the weights must then satisfy the triangle inequality condition. It is known that an exact solution can be obtained in near quadratic time if the coordinates are in a plane: www.cs.brown.edu/cgc/cgc98/final/final21.ps by divide and conquer techniques. === Subject: Re: Cantor Confusion > Which of the following statments is wrong? [...] There are two: (1) The union of all finite paths q =/= p does not cover p. (2) The union of all finite paths q =/= p covers p. (1) ==> There is a node which belongs to p but not to any other path q =/= p. (2) ==> p is nothing but a union of finite paths. (1) is obviously wrong, whereas (2) shows the countability of all paths. Which is wrong? === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > ... > > The set is finitely defined. Not all lucky numbers can get finitely > > defined. > > In general a recursive definition is not considered a finite definition. > > Every definition which ends after finitely many words is a finite > definition. > > Ah, so you disagree with common mathematical terminology. > > No. A finite definition means a definition by a finite number of > words. Every other definition is nonsense. I agree with the common > mathematical definition which implies that there are only finitely > many definitions. Which implication? Again contradicting the axiom of infinity? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion ... > > The set is finitely defined. Not all lucky numbers can get finitely > > defined. > > In general a recursive definition is not considered a finite definition. > > Every definition which ends after finitely many words is a finite > > definition. > Ah, so you disagree with common mathematical terminology. > No. A finite definition means a definition by a finite number of > words. Every other definition is nonsense. I agree with the common > mathematical definition which implies that there are only finitely > many definitions. Which implication? Again contradicting the axiom of infinity? Pardon, I meant countably many definitions. This is implied by the finity of every definition. If there were infinite definitions, then there were uncountably many definitions. === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl > ... > Each of the numbers is finitely definable, because each is a halting > Turing machine, and Turing machines are finitely definable. It is > the set that is not finitely definable. > > If each number is finitely defined, then there are only finitely many > numbers, because there are only finitely many Turing machines. Of > course, this finite set then is finitely defined too. > > Back again to how you started. There are only finitely many natural numbers... > That is not mathematics. > > That is the obvious basis which can be proven by mathematics. It is most ridiculous to state that there are only finitely many Turing machines. Can you prove that, using mathematics? > Now you come up again with a new term. 'Well defined'. What is > 'well defined'? > > A number is well defined if it is defined by a finite number of words > such that, in principle (i.e., given an infinite amount ressources) > the Cauchy epsilon can be made arbitrary small. > > So pi is well defined as the limit of the circumference of the inscribed > n-gon? The sequence of those circumferences *is* a Cauchy sequence... > > According to current mathematics, pi is well defined. Even according > to MatheRealism pi is well defined (as an idea). Your distinction between number and idea is just terminology, and not more than that. > I showed that there are countable sets which cannot be bijected with > the set of natural numbers, for instance the set of finitely definable > numbers. > > But that set *can* be bijected with the set of natural numbers. > > Wrong. Just assertion? > It is > indeed easy to show that there is an injection from that set to the set > of natural numbers. Consider all finite sentences over some alphabet > (let's say the 26 latin letters plus a space). Each such sentence can > be considered as a base-27 number, so we have an injection. > > Claim of injection is correct. Claim of bijection is wrong. (pi for > instance is defined by many different definitions). Do you not know the theorem that if there is an injection of some set S to a countable set T that S is also countable? > An injection is also possible for the set of all paths into the set of > all nodes. (There are two nodes per path.) *Give* that injection. > What node is bijected with the branch-off of 0.101010101010...? > > For an injection you can choose whatever node you want. Wrong. For an injection it is needed that two paths do not map to the same node, so you have to be careful in your mapping. You simply refuse to give an injection because you are not able to give one. > Obviously for > every number represented in the tree there is a node. And there is no > diagonal construction possible in the tree. So pray, *give* an injection. > I claim that there are no more branching-offs than odes and that there > are no more real numbers represented in the tree than are branching- > offs. There is no path ever finished, but it is only branching off > from other paths in infinity. But the number of paths separated from > other paths cannot surpass the number of branching-offs. > > The number of paths is the same from the root node, because every path > starts at the root. Or are you suggesting that there are paths *not* > starting at the root? > > Every path starts at the root node. But in order to count the paths, > they must be distinguishable, i.e. separated. Makes no sense. > One bunch goes in, because the two which come out have not yet been > separated when the paths which they consist of, went in. > > So it is your opinion that bunches that start of the root nevertheless do > not come in at some node but only come out? > > Every bunch starts at the root node. But in order to count the > bunches, they must be distinguishable, i.e. thy must be separated > bunches. The number of separated bunches is doubled at every level. You were talking about bunches going in and out of nodes. What you are doing is counting edges, not bunches, and the number of edges is countable. > Another question about chapter 10. Do you understand what a normal number > is? I think not. Off-hand I do not know whether there are normal numbers > that are normal with respect to all bases (although it is expected that pi > is one). > > Such numbers are called absolutely normal. But to know that is neither > required for the readers of my book in order to understand > MatheRealism nor would it be useful to expand the number of pages and > the price of the book by a large factor. So you prefer to talk nonsense? > But if a number is normal with respect to some base that does > *not* mean that the digits are unpredictable. Nor does unpredictability > of digits mean that a number is normal. > > There are different notions (for instance weakly normal numbers and > absolutely normal numbers). Of course normal numbers can be > constructed, one of the simplest cases is the rational number > 0.12012012... with respect to base 3, That number is not normal to base 3. > but as there must be included > also normally distributed frequencies of 10^100-tuples and larger > tuples most normal numbers cannot be constructed. Do you know about the Chapernowne numbers? But be also aware that the Copeland-Erdos number is normal to base 10. A quote: While Borel proved the normality of almost all numbers with respect to Lebesgue measure, with the exception of a number of special classes of constants, the only numbers known to be normal (in certain bases) are artificially constructed ones such as the Champernowne constant and the Copeland-Erdos constant. So, as I said: if a number is normal with respect to some base that does *not* mean that the digits are unpredictable. Nor does unpredictability of digits mean that a number is normal. But, of course, most normal numbers cannot be constructed, as most numbers cannot be constructed, and most numbers are normal. (Now try to interprete the three occurrences of the word most.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion According to current mathematics, pi is well defined. Even according > to MatheRealism pi is well defined (as an idea). Your distinction between number and idea is just terminology, and not > more than that. Its is more. You cannot answer the question whether the numbers P = [pi*10^10^100] and P' = P with the last digit replaced by 3 nsatisfy P < P'. > I showed that there are countable sets which cannot be bijected with > > the set of natural numbers, for instance the set of finitely definable > > numbers. > But that set *can* be bijected with the set of natural numbers. > Wrong. Just assertion? Define the bijection. > It is > indeed easy to show that there is an injection from that set to the set > of natural numbers. Consider all finite sentences over some alphabet > (let's say the 26 latin letters plus a space). Each such sentence can > be considered as a base-27 number, so we have an injection. > Claim of injection is correct. Claim of bijection is wrong. (pi for > instance is defined by many different definitions). Do you not know the theorem that if there is an injection of some set S to > a countable set T that S is also countable? I use it for the paths and nodes of the tree. But you keep on asking for a bijection. The injection has already been shown. > An injection is also possible for the set of all paths into the set of > all nodes. (There are two nodes per path.) *Give* that injection. Map every node onto the path which leaves it to the left-hand side. > What node is bijected with the branch-off of 0.101010101010...? > For an injection you can choose whatever node you want. Wrong. For an injection it is needed that two paths do not map to the same > node, so you have to be careful in your mapping. You simply refuse to give > an injection because you are not able to give one. Map every node onto the path which leaves it to the left-hand side. > Obviously for > every number represented in the tree there is a node. And there is no > diagonal construction possible in the tree. So pray, *give* an injection. > I claim that there are no more branching-offs than odes and that there > > are no more real numbers represented in the tree than are branching- > > offs. There is no path ever finished, but it is only branching off > > from other paths in infinity. But the number of paths separated from > > other paths cannot surpass the number of branching-offs. > The number of paths is the same from the root node, because every path > starts at the root. Or are you suggesting that there are paths *not* > starting at the root? > Every path starts at the root node. But in order to count the paths, > they must be distinguishable, i.e. separated. Makes no sense. How would you count inseparated paths? > One bunch goes in, because the two which come out have not yet been > > separated when the paths which they consist of, went in. > So it is your opinion that bunches that start of the root nevertheless do > not come in at some node but only come out? > Every bunch starts at the root node. But in order to count the > bunches, they must be distinguishable, i.e. thy must be separated > bunches. The number of separated bunches is doubled at every level. You were talking about bunches going in and out of nodes. What you are > doing is counting edges, not bunches, and the number of edges is countable. The number of paths cannot be larger than the number of edges. === Subject: Re: Cantor Confusion Another question about chapter 10. Do you understand what a normal number > is? I think not. Off-hand I do not know whether there are normal numbers > that are normal with respect to all bases (although it is expected that pi > is one). > Such numbers are called absolutely normal. But to know that is neither > required for the readers of my book in order to understand > MatheRealism nor would it be useful to expand the number of pages and > the price of the book by a large factor. So you prefer to talk nonsense? I consider the difference between absolutely normal and weakly normal not important with respect to the topic of my book, in particular since even such experts as you seem to have no clue about that. > But if a number is normal with respect to some base that does > *not* mean that the digits are unpredictable. Nor does unpredictability > of digits mean that a number is normal. > There are different notions (for instance weakly normal numbers and > absolutely normal numbers). Of course normal numbers can be > constructed, one of the simplest cases is the rational number > 0.12012012... with respect to base 3, That number is not normal to base 3. That number is weakly normal, namely normal to base 3. If you don't know about the definition of normal numbers you should first inform you. Online for instance http://eom.springer.de/N/n067560.htm > but as there must be included > also normally distributed frequencies of 10^100-tuples and larger > tuples most normal numbers cannot be constructed. Do you know about the Chapernowne numbers? But be also aware that the > Copeland-Erdos number is normal to base 10. A quote: > While Borel proved the normality of almost all numbers with respect > to Lebesgue measure, with the exception of a number of special classes > of constants, the only numbers known to be normal (in certain bases) > are artificially constructed ones such as the Champernowne constant > and the Copeland-Erdos constant. Another quote: The weakly-normal number (to base 10) 0.01234567890123456789... is of course rational. http://eom.springer.de/N/n067560.htm === Subject: Re: Cantor Confusion Nntp-Posting-Host: apps.cwi.nl ... > It is not inaccessible to me. > > So you saw my answer? > > No. I say it is not inaccessible to me. Not that I do access it. > > This set F of functions can be bijected with the > set R of reals. The reals form an intercession with the rationals Q. > The rationals form a bijection with the naturals N. > > Sorry, makes no sense without context. > > The context was your due posting. Either you recall it or you access > my group or yu vcannot get an answer on your question which uttered an > unjustified doubt. So you refuse to post an answer in the forum where I asked the question, giving proper references ... I have given you reasons *why* I do not access that group. That you ignore those reasons just shows arrogance. We have seen this same behaviour earlier by James Harris. Are you going to mimick him? I have *no* idea what you mean with this set F of functions. > If you are interested in the discussion which we had about Cantor's > mistake or not mistake concerning his second diagonal proof, you may > enter > > If I look correctly, again, it is not the case that Cantor made an error. > being that Cantor gave an additional prove that there are sets with > cardinality larger than the naturals. But in essence this is quite > irrelevant. Cantor may have erred at times, that is *not* related to the > current ways of set theory. > > It is. Set theory is simply biased. Consider the list > > 0.666... > 0.3666... > 0.33666... > 0.333666... > ... > > If the diagonal number is defined by replace 6 by 3, then we have > two answers none of which can be preferred by logic, but the second of > which is suppressed by convention. But, again, that is *not* the diagonal proof of Cantor. And even with that notation you write nonsense. Replace 6 by 3 yields the sequence 0.33333..., which is not in the list. I have no idea what the second answer would be. > 1) Every entry of the list differs at some place from the diagonal > number. > 2) Every initial segment of the diagonal number is represented by an > entry of the list. Both are true, if you replace (2) by: 2) Every initial segment of the diagonal number is represented by the initial segment of an entry of the list. To wit: 0.333666... does *not* represent 0.333, or you have a very strange interpretation of the word represent. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Cantor Confusion Set theory is simply biased. Consider the list > 0.666... > 0.3666... > 0.33666... > 0.333666... > ... > If the diagonal number is defined by replace 6 by 3, then we have > two answers none of which can be preferred by logic, but the second of > which is suppressed by convention. But, again, that is *not* the diagonal proof of Cantor. The following wm-proof certainly even in your opinion belongs to the diagonal proofs considered by Cantor: 0) mmm... 1) wmmm... 2) wwmmm... 3) wwwmmm... 4) wwwwmmm... ... .......... And if the list can be considered as a completed entity, then there must be all natural numbers in the first column. And there must be a line with all natural indexes mapped on w's, i.e., no w must be missing (as would be the case if one m was present). This argument shows that the list cannot be complete and that replacing the diagonal digit of every number of the list proves literally nothing. === Subject: Re: Cantor Confusion > It is not inaccessible to me. > > So you saw my answer? > No. I say it is not inaccessible to me. Not that I do access it. > > This set F of functions can be bijected with the > > set R of reals. The reals form an intercession with the rationals Q. > > The rationals form a bijection with the naturals N. > Sorry, makes no sense without context. > The context was your due posting. Either you recall it or you access > my group or you cannot get an answer on your question which uttered an > unjustified doubt. So you refuse to post an answer in the forum where I asked the question, > giving proper references ... I have given you reasons *why* I do not access > that group. That you ignore those reasons just shows arrogance. We have > seen this same behaviour earlier by James Harris. Are you going to mimick > him? I have *no* idea what you mean with this set F of functions. You defined the functions: (1) If two functions f and g are not equal, there is a smallest n such that f(n) != g(n). (2) We define f < g if f(n) < g(n), and f > g if f(n) > g(n). This is a complete ordering on that set of functions. > If you are interested in the discussion which we had about Cantor's > > mistake or not mistake concerning his second diagonal proof, you may > > enter > If I look correctly, again, it is not the case that Cantor made an error. > being that Cantor gave an additional prove that there are sets with > cardinality larger than the naturals. But in essence this is quite > irrelevant. Cantor may have erred at times, that is *not* related to the > current ways of set theory. > It is. Set theory is simply biased. Consider the list > 0.666... > 0.3666... > 0.33666... > 0.333666... > ... > If the diagonal number is defined by replace 6 by 3, then we have > two answers none of which can be preferred by logic, but the second of > which is suppressed by convention. But, again, that is *not* the diagonal proof of Cantor. And even with > that notation you write nonsense. Replace 6 by 3 yields the sequence > 0.33333..., which is not in the list. For the entries E(n) of the list we find lim[n-->oo] (E(n) - 0.333...) = 0. It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. > 1) Every entry of the list differs at some place from the diagonal > number. > 2) Every initial segment of the diagonal number is represented by an > entry of the list. Both are true, if you replace (2) by: > 2) Every initial segment of the diagonal number is represented by the > initial segment of an entry of the list. > To wit: > 0.333666... > does *not* represent > 0.333, > or you have a very strange interpretation of the word represent. If very initial segment of the diagonal number is represented by the initial segment of an entry of the list, then the full diagonal number is represented by an entry of the list. === Subject: Re: Cantor Confusion WM says... > It is. Set theory is simply biased. Consider the list > > 0.666... > 0.3666... > 0.33666... > 0.333666... > ... > > If the diagonal number is defined by replace 6 by 3, then we have > two answers none of which can be preferred by logic, but the second of > which is suppressed by convention. The diagonal number is 0.333... which is not on the list. >For the entries E(n) of the list we find >lim[n-->oo] (E(n) - 0.333...) = 0. >It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. That's true. In this particular case, the limit of the sequence is equal to the diagonal of the sequence. So what? >If every initial segment of the diagonal number is represented by the >initial segment of an entry of the list, then the full diagonal number >is represented by an entry of the list. That's false. To say that the number r appears on the list r_0, r_1, ... is to say that there is some natural number j such that r = r_j. If we let D_j = |r - r_j|, then the criterion for r appearing on the list is that exists j such that D_j = 0 In the case we are talking about, that is false. If r_0 = 0.6666...., r_1 = .3666..., etc and r = 0.333..., then |r-r_0| = .333... |r-r_1| = .0333... ... in general, |r - r_j| = 0.333... * 10^{-j} So we have forall j, D_j > 0 So r does not appear on the list. -- Daryl McCullough Ithaca, NY === Subject: Re: Paths <46471798$0$97257$892e7fe2@authen.yellow.readfreenews.net> <46474bce$0$97218$892e7fe2@authen.yellow.readfreenews.net> <46478a24$0$97217$892e7fe2@authen.yellow.readfreenews.net> If no single path of your set suffices, then no number of paths does, > because it can be shown that the effect of any *finite* wrong > set of paths > can be > achieved by one single path. An infinite set is nothing but a finite set which can be increased > infinitely, but always remains a finite set. And the effect of this finite set of paths can be achieved by > one single path. Look! Over there! A pink elephant! It is always the same path. Why should it? It is always ONE path. That is enough. === Subject: Re: Paths <46471798$0$97257$892e7fe2@authen.yellow.readfreenews.net> <46474bce$0$97218$892e7fe2@authen.yellow.readfreenews.net> <46478a24$0$97217$892e7fe2@authen.yellow.readfreenews.net> If no single path of your set suffices, then no number of paths does, > because it can be shown that the effect of any *finite* wrong > set of paths > can be > achieved by one single path. An infinite set is nothing but a finite set which can be increased > infinitely, but always remains a finite set. And the effect of this finite set of paths can be achieved by > one single path. Look! Over there! A pink elephant! It is always the same path. Why should it? It is always ONE path. That is enough. === Subject: Re: Paths <46471798$0$97257$892e7fe2@authen.yellow.readfreenews.net> <46474bce$0$97218$892e7fe2@authen.yellow.readfreenews.net> <46478a24$0$97217$892e7fe2@authen.yellow.readfreenews.net> <46488889$0$97257$892e7fe2@authen.yellow.readfreenews.net> <4648cfc3$0$97246$892e7fe2@authen.yellow.readfreenews.net> <464c3d65$0$97214$892e7fe2@authen.yellow.readfreenews.net> <464cdc0d$0$97246$892e7fe2@authen.yellow.readfreenews.net> On 18 Mai, 00:49, Franziska Neugebauer path leaving each of its nodes left and the path leaving each of its > nodes right? The union of these paths is the finite path which consists only of the root node. > 2. cover A path or set of paths A covers a path or set of paths B, if and only > if the set of nodes belonging to A includes the set of nodes belonging > to B as a (proper or improper) subset. Definition: A path p covers a path q iff A k (k in q -> k in p) Definition: A path p covers a set of paths Q iff A q (q e Q -> p covers q) Definition: A set of paths P covers a path q iff A k (k in q -> E p (p e P & k in p)) Definition: A set of paths P covers a set of paths Q iff A q (q e Q -> P covers q) Though we cannot decide your claim Also the union of all paths q does not cover all nodes of p. since we do not know what union of all paths q means. There are two options (as is often the case in life): (1) The union of all finite paths q =/= p does not cover p. (2) The union of all finite paths q =/= p covers p. (1) ==> There is a node which belongs to p but not to any other path q =/= p. (2) ==> p is nothing but a union of finite paths. (1) is obviously wrong, whereas (2) shows the countability of all paths. === Subject: Re: Paths > On 18 Mai, 00:49, Franziska Neugebauer > What is the union of the paths (2^n) and (2^(n+1) - 1), i.e. the > path leaving each of its nodes left and the path leaving each of its > nodes right? > > The union of these paths is the finite path which consists only of the > root node. Let us apply the set theoretic definitions: p1 := (2^n) = { (0, 1), (1, 2), (2, 4), ... } p2 := (2^(n+1) - 1) = { (0, 1), (1, 3), (2, 7), ... } p3 := p1 Op p2 = { (0, 1) } p3 is the finite path consisting only of the root node. The Operation Op is commonly referred to as _intersection_ and not as /union/. Do you want to rename a well known operation? Until now we have not been discussion finite paths but used paths according to the definition ,----[ <464cdc0d$0$97246$892e7fe2@authen.yellow.readfreenews.net> ] | A path (p_i) is an infinite sequence of nodes whith the properties | | 1. p_0 = 1 (the root node) and | 2. (p_i, p_(i+1)) is edge in CIBT. `---- > 2. cover > A path or set of paths A covers a path or set of paths B, if and > only if the set of nodes belonging to A includes the set of nodes > belonging to B as a (proper or improper) subset. [(U)] > Definition: A path p covers a path q iff > A k (k in q -> k in p) > Definition: A path p covers a set of paths Q iff > A q (q e Q -> p covers q) > Definition: A set of paths P covers a path q iff > A k (k in q -> E p (p e P & k in p)) > Definition: A set of paths P covers a set of paths Q iff > A q (q e Q -> P covers q) > Though we cannot decide your claim > Also the union of all paths q does not cover all nodes of p. > since we do not know what union of all paths q means. > > > There are two options (as is often the case in life): > (1) The union of all finite paths q =/= p does not cover p. > (2) The union of all finite paths q =/= p covers p. 1. Are your still talking about _all_ the paths in the CIBT or only about the finite paths? When you stated (U) we were talking about _all_ the paths (the infinitly long paths). 2. Before deciding (1) or (2) I want to know what union of all paths in (U) means. > (1) ==> There is a node which belongs to p but not to any other path q > =/= p. > (2) ==> p is nothing but a union of finite paths. > > (1) is obviously wrong, whereas (2) shows the countability of all > paths. We are not that quick on the trigger. F. N. -- xyz === Subject: Re: Paths <46471798$0$97257$892e7fe2@authen.yellow.readfreenews.net> <46474bce$0$97218$892e7fe2@authen.yellow.readfreenews.net> <46478a24$0$97217$892e7fe2@authen.yellow.readfreenews.net> infinitely, but always remains a finite set. > Therefore the belief that an infinite path could suffice is wrong. > The belief that an infinite set is finite is a self-contradiction in > logical circles, so that the circles that WM is spinning around in are > illogical ones. The belief that an infinite set is actual, complete, or finished is a self-contradiction but nevertheless being believed in circles which consider themselves logical > In ZF and NBG, and similar set theories, there are actually infinite > sets to which no further elements can be added without forming > different sets. In WM's limited world, CIBT's cannot exist at all, as no such thing as > infinite sets can exist, they must either be not infinite or not sets. The IBT exists, but it is never complete. One can follow a path infinitely long, never reaching the realm of uncountability. === Subject: Re: Paths > > An infinite set is nothing but a finite set which can be increased > infinitely, but always remains a finite set. > Therefore the belief that an infinite path could suffice is wrong. > The belief that an infinite set is finite is a self-contradiction in > logical circles, so that the circles that WM is spinning around in are > illogical ones. > > The belief that an infinite set is actual, complete, or finished is a > self-contradiction but nevertheless being believed in circles which > consider themselves logical There cannot be in any set theory of repute any infinite set which is not actually infinite. Whatever WM insists on as being potentially but not actually infinite, it cannot be a set. Call it a pre-set or a quasi-set of a pseudo-set, but not a set. > > In ZF and NBG, and similar set theories, there are actually infinite > sets to which no further elements can be added without forming > different sets. In WM's limited world, CIBT's cannot exist at all, as no such thing as > infinite sets can exist, they must either be not infinite or not sets. > > The IBT exists, but it is never complete. Then it is not a set, as sets are complete in the sense that their membership is fixed and invariant in ways in which WM's pseudo infinite pseudo-sets are not. > One can follow a path > infinitely long, never reaching the realm of uncountability. But if one ever posits a set of all infinite paths, such a set, if it is to be a set at all, must be uncountable. === Subject: Re: Paths <46471798$0$97257$892e7fe2@authen.yellow.readfreenews.net> <46474bce$0$97218$892e7fe2@authen.yellow.readfreenews.net> <46478a24$0$97217$892e7fe2@authen.yellow.readfreenews.net> <46488889$0$97257$892e7fe2@authen.yellow.readfreenews.net> <4648cfc3$0$97246$892e7fe2@authen.yellow.readfreenews.net> <464c3d65$0$97214$892e7fe2@authen.yellow.readfreenews.net> Paths are isomorphic to initial segments of linear sets (e.g., of > initial segments of the set of natural numbers). Isomorphic as what? What is allegedly being morphed? Nodes of the paths are counted or indexed by natural numbers: 0.000... 0, 1, 2, 3, ... The union of such > segments should be known. Outside of WM's MAthUnRealism, N is not one of them, but is their union. N is the union of all finite initial segments {1,2,3,...,n}. p is the union of all finite paths p' which satisfy, as sets of nodes, p' c p. Therefore p has not more nodes than Up'. Therefore there cannot be more paths p constructed than are paths p' in the tree. > 2. cover A path or set of paths A covers a path or set of paths B, if and only > if the set of nodes belonging to A includes the set of nodes belonging > to B as a (proper or improper) subset. Do you mean the set of nodes belonging to A, or to members of A, > includes the set of nodes belonging to B, or to members of B ? Since a path is a set of nodes, a SET of paths does not, strictly > speaking, contain any nodes as members. But the union of a set of paths is, strictly speaking, a path. Therefore, if we neglect order, U(U{p' | p' c p})) is a set of nodes. === Subject: Re: Paths > > Paths are isomorphic to initial segments of linear sets (e.g., of > initial segments of the set of natural numbers). Isomorphic as what? What is allegedly being morphed? > > Nodes of the paths are counted or indexed by natural numbers: > 0.000... > 0, 1, 2, 3, ... Non-responsive. > The union of such > segments should be known. Outside of WM's MAthUnRealism, N is not one of them, but is their union. > > N is the union of all finite initial segments {1,2,3,...,n}. There is no need to repeat what I have just said. > > p is the union of all finite paths p' which satisfy, as sets of nodes, > p' c p. Therefore p has not more nodes than Up'. Therefore there > cannot be more paths p constructed than are paths p' in the tree. There are as many infinite paths as there can be infinite unions of nested sequences of finite paths, and that allows uncountably many. The set of finite paths of length n (n edges) bijects obviously with the set of finite binary strings of n terms, so in the obvious way , the set of infinite paths bijects with the set of infinite binary strings, which is well known to be uncountable. > 2. cover > A path or set of paths A covers a path or set of paths B, if and only > if the set of nodes belonging to A includes the set of nodes belonging > to B as a (proper or improper) subset. Do you mean the set of nodes belonging to A, or to members of A, > includes the set of nodes belonging to B, or to members of B ? Since a path is a set of nodes, a SET of paths does not, strictly > speaking, contain any nodes as members. > > But the union of a set of paths is, strictly speaking, a path. By whose definition of path? There is nothing in mathematics which makes such an interpretation necessary. If a path were no more than a set of nodes, then every set of nodes would be a path. While there is a modeling of trees in which every path is a suitable set of nodes, there is nothing in the definitions of trees requiring that model to be 'the' model. And even when one uses that model, one finds the CIBT bijects with the set of infinite binary strings, which is well known to be uncountable. === Subject: induction and matrix inversion Let A = (aij) be a q?q positive definite matrix and D = diag(sigma 1, sigma 2, . . . , sigma q) be a diagonal matrix with positive diagonal entries sigma i. Then, inv (D + A) can be represented as a convex combination of the inverses of the principal submatrices of A. (Here, to fit the dimension the inverses are taken and filled with entries 0 according to where the principal submatrices were taken.) Induction: Basis step is allright (for q=1 and q=2 it is shown). For the inductive step assume that it is true for q = k, for a fixed k, say for the matrix B. B is a kxk matrix for which the assumption holds. Then, for q = k + 1 we have the following bordered matrix: C= [B v ] [v' sigma k+1,k+1-r k+1,k+1 ] where v is a kx1 vector, sigma k+1,k+1 is a positive number and r k+1,k +1 is a negative number. We know that for the block matrix B the assertion holds, and yhe aim is to show that the assertion holds for the whole matrix C. === Subject: =?gb2312?B?z+vSqrj2xq/BwbXEsaaxpsLwo6w=?= === Subject: JSH: Rebuttal, newsgroup at faul This newsgroup acted against a paper of mine published in a peer reviewed mathematical journal, and its members who did so have maintained that my paper was in error. I have completed a rebuttal paper, which goes over the entire argument in ways meant to end their ability to easily confuse about the details of my research: Those posters who went after my earlier paper should apologize both to me and the editors of the journal, which is now unfortunately a dead journal. Civilization depends first and foremost on civilized behavior. When angry mobs can trounce the instruments of civilization, like breaking the formal peer review process with false accusations of error in a major paper, then it is our progress as a species that is put in doubt. James Harris === Subject: Re: JSH: Rebuttal, newsgroup at faul >This newsgroup acted against a paper of mine published in a peer >reviewed mathematical journal, and its members who did so have >maintained that my paper was in error. Yes, they've maintained that. Probably the reason is that, to the extent that it made any sense at all, it _was_ in error. Various errors were clearly explained to you long before publication... >I have completed a rebuttal paper, which goes over the entire argument >in ways meant to end their ability to easily confuse about the details >of my research: >Those posters who went after my earlier paper should apologize both to >me and the editors of the journal, which is now unfortunately a dead >journal. Civilization depends first and foremost on civilized behavior. When angry mobs can trounce the instruments of civilization, like >breaking the formal peer review process with false accusations of >error in a major paper, then it is our progress as a species that is >put in doubt. Yeah, that's bad for civilzation all right. What about when people publish papers about things they really don't understand very well, and as a result their publications are total nonsense? >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Rebuttal, newsgroup at faul <7h51539bf976b19dbc3tgkfi8qnhjd7u4a@4ax.com> On May 20, 10:42 am, David C. Ullrich This newsgroup acted against a paper of mine published in a peer >reviewed mathematical journal, and its members who did so have >maintained that my paper was in error. Yes, they've maintained that. Probably the reason is that, > to the extent that it made any sense at all, it _was_ in > error. Various errors were clearly explained to you long > before publication... > Except the method I use involves subtracting from an identity. Now then, can subtracting from an identity change the ring? No. Yet, the primary objection against my paper was that while starting in the ring of algebraic integers provably the conclusion is not valid in that ring, or do you claim otherwise? Now then, if I start in the ring of algebraic integers, and my method involves subtracting from an identity, how does the ring change? >I have completed a rebuttal paper, which goes over the entire argument >in ways meant to end their ability to easily confuse about the details >of my research: >Those posters who went after my earlier paper should apologize both to >me and the editors of the journal, which is now unfortunately a dead >journal. Civilization depends first and foremost on civilized behavior. When angry mobs can trounce the instruments of civilization, like >breaking the formal peer review process with false accusations of >error in a major paper, then it is our progress as a species that is >put in doubt. Yeah, that's bad for civilzation all right. What about when people publish papers about things they > really don't understand very well, and as a result their > publications are total nonsense? > But what if their publication is correct and people claim it's nonsense when it's not? My answer has been to write another paper which deliberately is written to remove areas where posters on sci.math claimed error. That paper is now available with a link leading to it in the post that starts this thread, and it is a rebuttal to your claims and that of other posters with you. Now if you dare to support civilization in actuality versus talking as if you do, then answer my very civilized response of rebutting through a paper. James Harris === Subject: Re: Rebuttal, newsgroup at faul > This newsgroup acted against a paper of mine published in a peer > reviewed mathematical journal, and its members who did so have > maintained that my paper was in error. It has multiple obvious errors, mistakes, and very poor quality, never should have been submitted. I have completed a rebuttal paper, which goes over the entire argument > in ways meant to end their ability to easily confuse about the details > of my research: > You can't be serious, man. What a goof, no pure math at all. This is a funny paper, should be in with the Sunday Comics. > Those posters who went after my earlier paper should apologize both to > me and the editors of the journal, which is now unfortunately a dead > journal. The editor was wise to withdraw your paper, which means rejected, because it had nothing in it. > Civilization depends first and foremost on civilized behavior. And you need to go take a nath course first, your very first one. When angry mobs can trounce the instruments of civilization, like > breaking the formal peer review process with false accusations of > error in a major paper, then it is our progress as a species that is > put in doubt. Your major paper is toilet paper, and I for one, piss on it. > James Harris > === Subject: Re: JSH: Rebuttal, newsgroup at faul > This newsgroup acted against a paper of mine published in a peer > reviewed mathematical journal, and its members who did so have > maintained that my paper was in error. I have completed a rebuttal paper, which goes over the entire argument > in ways meant to end their ability to easily confuse about the details > of my research: > Those posters who went after my earlier paper should apologize both to > me and the editors of the journal, which is now unfortunately a dead > journal. Civilization depends first and foremost on civilized behavior. When angry mobs can trounce the instruments of civilization, like > breaking the formal peer review process with false accusations of > error in a major paper, then it is our progress as a species that is > put in doubt. James Harris Your original paper was in error, and the one you are pushing now is also in error. There seems to be more than a few people finding fault with your new, improved paper. I thought that there could be no room for doubt with it? I am seeing some definite doubts. === Subject: Yet another fancy integral I've been toying with this for the past few months and pursuing some strange avenues of study in the exploration. Integral[sin(cos(x))dx] Now i know that there is no closed form for this particular integral, but according to my topology professor you can represent the integrand as an appropriate infinite hyper geometric series. My question is what hyper geometric series do I use? I know that a hyper geometric distribution is a quotient of a product and sum of gamma functions but I'm not familiar with the series form. thoughts? === Subject: law of sines and proving identity I have the following identity: (a - b)/(a + b) = tan[(alpha - beta) / 2] / tan[(alpha + beta) / 2] I've expanded the right side to get: { sin(alpha - beta) * [1 + cos(alpha + beta)] } / { sin(alpha+ beta) * [1 + cos(alpha - beta)] } Now using the law of sines, I can reduce both sin(alpha - beta) and sin(alpha + beta) to alpha - beta and alpha + beta, respectively. However, I am unsure of how to factor out [1 + cos(alpha + beta)] and [1 + cos(alpha - beta)] -- conrad === Subject: cancel: Re: * Free Pussy and Tit flash Download! Control: cancel Excessive crossposting. () === Subject: Re: Random Numbers <464c0bb0$0$23424$afc38c87@news.optusnet.com.au> On May 17, 1:00 am, Peter Webb > Tell me again why the following number is not a random number! 7548968 I closed my eyes and pressed the number keys haphazardly.. > ---Bill J. There is no such thing as a random number. There are only random > distributions of numbers. === Subject: Re: JSH: Why proof is not enough > [Rupert] > There exist two algebraic integers r that satisfy the equation. The > claim under discussion was that *every* algebraic integer would > satisfy the equation. ... > But that would be just silly! Indeed, but that is how he interpreted your claim. This should be > telling you two things: firstly, that learning how to present your > results in a familiar format is not something people tell you to do > because they favour style over substance, but something that people > tell you to do because at present it is usually impossible to know > what you are actually claiming to have proven. ... Pedantic. The paper is a revolutionary one, and whether or not you or Rupert can > quite understand it or not, is irrelevant. People who can understand it can then proceed to understanding the > short proof of Fermat's Last Theorem which uses the same techniques > and understand as well why Galois Theory is not what a lot of > mathematicians think it is. Now you think you can rest easy on the notion I'm some crank so you > can talk down to me as you don't think any of that is true and you > have your training in whatever and your respect for mathematicians > etc. but that is just about your small-mindedness. > You are a crank. Here is a direct quote from your revolutionary paper: ---------------------------------------------------------------------------- ------------------------------ The question now arises, how are you forced out of the ring of algebraic integers if an expression that is in that ring is just being subtracted from an identity? The identity cannot change things. The underlying base expression is in the ring of algebraic integers, so should you not remain in the ring? But you do not. ---------------------------------------------------------------------------- ------------------------------ These are your words; 100% your words. What you are saying here and in the text of the paper preceding this, is that you can find two numbers A and B in the ring of algebraic integers, such that A - B is not in the ring. That's why you say you are forced out of the ring. That's why you say you do not remain in the ring. This is equivalent to saying the ring of algebraic integers is not closed under subtraction. Anyone can see that equivalence. But you deny it. Why don't you try to resolve this issue before you get in any deeper? Marcus > Reality is you should not have the nerve to talk down to me, and you > should show a lot more respect to mathematical arguments that prove my > case, especially considering the journey where an entire mathematical > in denial about that as well--and the impact of these results on the > world are hard to conceive. What is interesting now to me is how long you'll be able to post like > you do in response to me, and I know the answer: up until the moment > that some mathematicians you respect say that I'm right, and then > you'll shut up like a clam. All they will have to is SAY I am right, and then you'll come to heel > like a well-trained dog. For you it's all a social thing. For others it is about proof. James Harris- Hide quoted text - - Show quoted text - === Subject: Re: JSH: Why proof is not enough > [Rupert] > There exist two algebraic integers r that satisfy the equation. The > claim under discussion was that *every* algebraic integer would > satisfy the equation. ... > But that would be just silly! > Indeed, but that is how he interpreted your claim. This should be > telling you two things: firstly, that learning how to present your > results in a familiar format is not something people tell you to do > because they favour style over substance, but something that people > tell you to do because at present it is usually impossible to know > what you are actually claiming to have proven. ... Pedantic. The paper is a revolutionary one, and whether or not you or Rupert can > quite understand it or not, is irrelevant. People who can understand it can then proceed to understanding the > short proof of Fermat's Last Theorem which uses the same techniques > and understand as well why Galois Theory is not what a lot of > mathematicians think it is. Now you think you can rest easy on the notion I'm some crank so you > can talk down to me as you don't think any of that is true and you > have your training in whatever and your respect for mathematicians > etc. but that is just about your small-mindedness. You are a crank. No, you are just not a mathematician. You seem to be lost on a logical point. > Here is a direct quote from your revolutionary paper: ---------------------------------------------------------------------------- - ----------------------------- > The question now arises, how are you forced out of the ring > of algebraic integers if an expression that is in that ring is > just being subtracted from an identity? The identity cannot change things. The underlying base > expression is in the ring of algebraic integers, so should you > not remain in the ring? But you do not. > ---------------------------------------------------------------------------- - ----------------------------- These are your words; 100% your words. > Yup. If you are forced out of the ring of algebraic integers, and you cannot be if an expression in that ring is JUST being subtracted from an identity, what must be true? Clearly then something more than just subtracting from the identity MUST be happening. Remember, it is not possible for subtracting from an identity to change the ring. Concentrate, go over it a few times and see if you can figure it out. James Harris === Subject: simple residue Hi all! I have a function f(z)= Sum_{k=1} until infinity [ (-1)^k z^(2k-3) ]/ (2k)! and i want to calculate it's residue at point 0, Res(f,0) , i think it will correspond to the coeficient of the potence z^0,but in this never happen. Can someboby help me? Sofia Nunes === Subject: Re: simple residue > Hi all! > > I have a function f(z)= Sum_{k=1} until infinity [ (-1)^k z^(2k-3) ]/ > (2k)! > > and i want to calculate it's residue at point 0, Res(f,0) , i think it > will correspond to the coeficient of the potence z^0, ???? This is a trivial problem if you know the definition of residue. Do you?