mm-1056 > .... Is there a brain > lesion/freezing study that shows where modus ponens lies?. > Given the number of people who think that > If P then Q, Q, therefore P > is a valid inference, probably not. > I used to think that logic, being (as I thought) necessary for survival > in the world, might have evolved. But given the number of stupid people > who seem to do very well, I now have my doubts. Natural selection does not operate in civilized society because civilization is not natural. Simpletons and moral cowards are easier to exploit for profit because they are more easily intimidated, hence they prosper (though without civil rights). I have no doubt anymore that some managers consciously prefer them and seek them out. === Subject: Re: logic is innate? Discussion, linux) > Natural selection does not operate in civilized society because > civilization is not natural. Oh? How did it arise then? What is the distinction between natural and artificial? -- Jesse F. Hughes You see 300 of something, anything, and you go `[Man], thatÕs a lot of stuff.-- Jim Bigler, quoted in the Pittsburgh Post-Gazette. === Subject: Re: logic is innate? > I used to think that logic, being (as I thought) necessary for survival > in the world, might have evolved. But given the number of stupid people > who seem to do very well, I now have my doubts. I think that logic is a structure that is imposed on language to make it understandable and useful. Thinking in general, seems to be associative with some sort of Bayesian component reducing the need for precise thought. (e.g. A person may be adverse to eating a specific food if they get the stomach ßu after eating it.) I think that it pays to look at a list of logical fallacies and see how difficult it is to explain the error to a person making the logical fallacy. (e.g. You are either for the war on terrorism or against it.) -- Ron === Subject: Re: Dumbing down, dumbing up? >>So whatÕs going on here? There are constant barrages of anecdotes >>about how stupid kids are these days. ArenÕt more and more highschool >>kids taking the Calculus AP exam or college level classes? >> http://www.maa.org/features/dualenrollment.html > More are, but fewer are taking real mathematics courses. > The old-fashioned Euclidean geometry course, emphasizing > proofs, was mathematics. So was induction in the old > college algebra course. There was little additional > mathematics in the old program until rigorous analysis > or algebra was introduced. On this, nothing has changed. just to get the terminology down correctly, mathematics means mathematical subject matter with proofs, and not theory means not real mathematics means pattern matching, plugging in formulas, learning to apply symbolic algorithms? > All of the computation courses, given without any theory > foundation, are not real mathematics courses. Honors > students may still be taught axiom-theorem-proof mathematics, > and computational procedures based on that theoretical > background, but the proportion of college entrants who > have had some rigorous mathematics has dropped from almost > all to very few. IsnÕt this also a trend in entry level undergrad calculus courses? That is, more emphasis on doing calculus (computational) and less on the reason why (proof based)? Is the trend also a reason for the introduction of bridge courses (a proof methods class taken as pre-req for upper level undergrad math classes) -- Mitch Harris (remove q to reply) === Subject: Re: Dumbing down, dumbing up? >So whatÕs going on here? There are constant barrages of anecdotes >about how stupid kids are these days. ArenÕt more and more highschool >kids taking the Calculus AP exam or college level classes? > http://www.maa.org/features/dualenrollment.html >> More are, but fewer are taking real mathematics courses. >> The old-fashioned Euclidean geometry course, emphasizing >> proofs, was mathematics. So was induction in the old >> college algebra course. There was little additional >> mathematics in the old program until rigorous analysis >> or algebra was introduced. On this, nothing has changed. >just to get the terminology down correctly, mathematics means >mathematical subject matter with proofs, and not theory means not >real mathematics means pattern matching, plugging in formulas, >learning to apply symbolic algorithms? Plugging into formulas is like using a hammer or a typewriter; it is essentially training, rather than education. There is a huge difference here. The same holds for directly applying symbolic algorithms. Especially for the non-mathematician, the important parts are concepts and language. In my opinion, the use of variables is what freed post-Renaissance mathematics from the restraints which kept the Greeks from doing much more; one variable goes back to Diophantus (about 300), the systematic use of more than one for numbers to Viete, around 1600, functions to Euler, and the full modern use to those looking at the foundations around 1900 and later. But if this is extended to full linguistics, it becomes easy. The other point is concepts. This is much harder than it looks, but is very important. If one has the concepts and the language, one can translate between real world problems and formal problems, and the the full power of mathematics can be applied to the formal problems. A major stumbling block is that the same mathematical objects can have different interpretations. For the ordinary integers, the cardinal and ordinal concepts are quite different, and I believe that one of the problems with the new math was that the apparently easier, but really harder to work with, cardinal concept was chosen, rather than the apparently more complicated, but easier to develop, ordinal concept. There are other concepts which involve extensions to rational numbers, signed numbers, real and complex numbers, which involve introducing additional somewhat related concepts. So when I am asked what 1 means, how many answers should I give? It is quite possible to get an intuitive understanding of concepts, but learning mechanics does not seem to help. It is extremely doubtful if learning to add and multiply by rote helps at all in learning the underlying concepts, while the other does help, although it may leave the person asking why? The non-mathematician needs to understand what proofs are, and the need for them. The mathematician needs to be more familiar, and able to produce them sometimes. Learning to apply algorithms needs to be divided into its parts; formulate the problem as a mathematical problem, see if there are appropriate methods available, and then use them. The first part is language; for the second part, the formulation may be standard, in which case the method can be looked up, or it might have to be derived, or some other trick found. Sometimes the tricks can be found, but often it will take a good mathematician to see them. As for pattern matching, this is a research method, and cannot be taught. It can be demonstrated, and one can hope the student picks up some of it. Research consists in seeing the obvious, and research ability is essentially independent of background, alas. We need more good researchers, and they are not going to be produced by wasting their time going from specific to general. We do not have the time. >> All of the computation courses, given without any theory >> foundation, are not real mathematics courses. Honors >> students may still be taught axiom-theorem-proof mathematics, >> and computational procedures based on that theoretical >> background, but the proportion of college entrants who >> have had some rigorous mathematics has dropped from almost >> all to very few. >IsnÕt this also a trend in entry level undergrad calculus courses? >That is, more emphasis on doing calculus (computational) and less on >the reason why (proof based)? If we are going to do anything else, we will have to assume that the entering college students need to learn basics. >Is the trend also a reason for the introduction of bridge courses >(a proof methods class taken as pre-req for upper level undergrad math >classes) Understanding proofs belongs in elementary school. It has been taught to the upper half of fifth graders, and I think a less pedantic approach would even work for a larger set of third graders. The present proof methods courses are far too weak; a direct abstract algebra course might well work much better, as the students would not have too many examples available. Proof methods in analysis are much less effective. And it would not be a bad idea to teach the number systems by a modification of Landau, starting in first grade. First graders can handle those proofs, but it seems teachers cannot. -- 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: Set inclusion and membership In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, Academic Science, Den Haag, 1976, I find the following sentence, when translated into English: Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. Therefore: a e A is not the same as a c A . Now suppose that I am kind of stubborn, insisting that the 2 concepts should denote the same thing. With other words: I find that a = { a } everywhere. Since I am a physicist by education, this is not a stupid idea. Because, in physics, a member of a set cannot be distinguished from a subset of a set, which contains only that element as a member. Question. Would such an assumption be rather harmless & scratch only the surface of set theory? Or would it have profound consequences? How does it affect the Zermelo Fraenkel system (ZFC), for example? Just a thought. Curious about your answers. Han de Bruijn === Subject: Re: Set inclusion and membership >In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, >Academic Science, Den Haag, 1976, I find the following sentence, when >translated into English: >Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. >Therefore: a e A is not the same as a c A . >Now suppose that I am kind of stubborn, insisting that the 2 concepts >should denote the same thing. With other words: I find that a = { a } >everywhere. Since I am a physicist by education, this is not a stupid >idea. Because, in physics, a member of a set cannot be distinguished >from a subset of a set, which contains only that element as a member. Suppose on has a random set A. Then it may well be that P(A is the empty set) is not zero. However, the probability of the empty set of outcomes is zero. This can even occur in physics; having no photons is a non-trivial event. >Question. Would such an assumption be rather harmless & scratch only >the surface of set theory? Or would it have profound consequences? It would very quickly lead to paradoxes. A 2-element set X is not the same as the set whose only element is X. >How does it affect the Zermelo Fraenkel system (ZFC), for example? As you can see from the above, greatly. -- 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: Set inclusion and membership > Suppose on has a random set A. Then it may well be that > P(A is the empty set) is not zero. However, the probability > of the empty set of outcomes is zero. This can even occur > in physics; having no photons is a non-trivial event. Since IÕm a disaster in statistics and probability, would you mind to elaborate on this a little bit? Especially, I donÕt understand what you mean by a random set. And how that relates to a = {a} . Han de Bruijn === Subject: Re: Set inclusion and membership > In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, > Academic Science, Den Haag, 1976, I find the following sentence, when > translated into English: > Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. > Therefore: a e A is not the same as a c A . > Now suppose that I am kind of stubborn, insisting that the 2 concepts > should denote the same thing. With other words: I find that a = { a } You then get a in a which contradicts the axiom of foundations or regularity. === Subject: Re: Set inclusion and membership > You then get a in a which contradicts the axiom of foundations or > regularity. Confirmative. (I discovered that myself, to be honest.) But I couldnÕt find any other axiom to be violated. Am I wrong? Not that I would like to have more axioms killed, but ... Han de Bruijn === Subject: Re: Set inclusion and membership Discussion, linux) >> In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, >> Academic Science, Den Haag, 1976, I find the following sentence, when >> translated into English: >> Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. >> Therefore: a e A is not the same as a c A . >> Now suppose that I am kind of stubborn, insisting that the 2 concepts >> should denote the same thing. With other words: I find that a = { a } > You then get a in a which contradicts the axiom of foundations or > regularity. This isnÕt any particular problem. Set theory without foundation (or even with an anti-foundation axiom) is perfectly workable. If this was the only issue, then de BruijnÕs suggestion wouldnÕt be so bad. But of course it isnÕt the only issue. On his suggestion, the empty set {} is equal to the set {{}}. Thus: {} = {{}} Now, by definition (A x)(NOT x in {}). Since {} = {{}}, we conclude that (A x)(NOT x in {{}}) and hence (NOT {} in {{}}). Of course, by definition we know that {} *is* in {{}}. Thus we have an inconsistency. Hence, de BruijnÕs suggestion has rather more problems than violating foundations. I didnÕt use much of ZF in the above reasoning. I used the axiom of the empty set (or the axioms that some set exists together with separation, if you prefer), the definition of {a} (derived from axiom of pairing) and a property of equality that most people consider part of basic logic. I didnÕt even use extensionality. You can show the contradiction easily enough without the empty set. Just pick any set S such that (S not in S) or (E y)(y in S and NOT y = S). The same reasoning will yield a contradiction. Thus, to avoid the contradiction, de BruijnÕs theory must consist of *nothing* aside from sets of the form a = {a}. No pairing, no union, no separation, no empty set, etc. Not the most exciting universe of sets. -- A recruitment consultant I know thinks the most important quality in a winner is to be lucky. To avoid wasting his time with unlucky applicants, he takes half the resumes piled on his desk and throws them straight in the bin. -- John Ramsden === Subject: Re: Set inclusion and membership > But of course it isnÕt the only issue. On his suggestion, the empty > set {} is equal to the set {{}}. Thus: > {} = {{}} > Now, by definition (A x)(NOT x in {}). Since {} = {{}}, we conclude > that (A x)(NOT x in {{}}) and hence (NOT {} in {{}}). Of course, by > definition we know that {} *is* in {{}}. Thus we have an > inconsistency. Thus, indeed, we have an inconsistency with the _empty set_ {} . > You can show the contradiction easily enough without the empty set. But _that_ is not true! Suppose that S is _not empty_ in what follows. > Just pick any set S such that (S not in S) or > (E y)(y in S and NOT y = S). The same reasoning will yield a > contradiction. [ .. ] It doesnÕt. It cannot be that (S not in S) because S = {S} implies that (S is in S), always, unconditionally. With other words, JesseÕs would-be (E y)(y in S and NOT y = S) is a set that simply does not exist. I found confirmation of the above on the web: http://plato.stanford.edu/entries/mereology/ Less philosophical / verbose and more mathematical: http://people.imise.uni-leipzig.de/alexander.heussner/files/ Mereology.pdf Mereology is the science about parts. (DidnÕt know that such science could possibly exist.) The gist of the argument presented in the latter a Boolean Algebra, BUT _without the empty set_. The latter is important. It all means, in conclusion, that my would-be axiom is NOT such a stupid idea in the first place; as several people here have pointed out, too. But it definitely has some drawbacks. The most important being that an empty set is NOT compatible with (c) = (e). Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> Discussion, linux) >> But of course it isnÕt the only issue. On his suggestion, the empty >> set {} is equal to the set {{}}. Thus: >> {} = {{}} >> Now, by definition (A x)(NOT x in {}). Since {} = {{}}, we conclude >> that (A x)(NOT x in {{}}) and hence (NOT {} in {{}}). Of course, by >> definition we know that {} *is* in {{}}. Thus we have an >> inconsistency. > Thus, indeed, we have an inconsistency with the _empty set_ {} . >> You can show the contradiction easily enough without the empty set. > But _that_ is not true! Suppose that S is _not empty_ in what follows. >> Just pick any set S such that (S not in S) or >> (E y)(y in S and NOT y = S). The same reasoning will yield a >> contradiction. [ .. ] > It doesnÕt. It cannot be that (S not in S) because S = {S} implies that > (S is in S), always, unconditionally. With other words, JesseÕs would-be > (E y)(y in S and NOT y = S) is a set that simply does not exist. So your only sets satisfy the following two conditions. (1) x is in x (2) x is the only element of x. This is a set theory, is it? What constructions do you expect? Pairing? No. Union? No. Powerset? It is to laugh. > I found confirmation of the above on the web: > http://plato.stanford.edu/entries/mereology/ > Less philosophical / verbose and more mathematical: http://people.imise.uni-leipzig.de/alexander.heussner/files/ Mereology.pdf Where is the analogue to your axiom a = {a}? Let us try to translate this to mereological terms. This axiom seems to say two things. (1) a is a part of a. (2) a is the only part of a. That sounds bloody useless and I donÕt see it occurring in your reference. -- Mathematicians are rather important in the infrastructures of many organizations that protect civilization. IÕve determined that they are a consistent security risk, and seem to have other agendas, other loyalties beyond loyalty to their respective nations. -- James Harris === Subject: Re: Set inclusion and membership > So your only sets satisfy the following two conditions. > (1) x is in x > (2) x is the only element of x. I never said this. My only _singletons_ satisfy these two conditions. But there exist many sets which are not singletons, of course, even in my theory :-) Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> Discussion, linux) >> So your only sets satisfy the following two conditions. >> (1) x is in x >> (2) x is the only element of x. > I never said this. My only _singletons_ satisfy these two conditions. > But there exist many sets which are not singletons, of course, even in > my theory :-) For any set x, {x} is a singleton. You said before that a={a} is true everywhere. Whatever did you mean? ,---- | It doesnÕt. It cannot be that (S not in S) because S = {S} implies that | (S is in S), always, unconditionally. With other words, JesseÕs would-be | (E y)(y in S and NOT y = S) is a set that simply does not exist. `---- This says that there are no sets aside from sets satisfying x={x}. Thus (1) every set is a singleton and (2) every set is an element of itself. Therefore, the is-element relation is equivalent to equality. If this isnÕt what you mean, then what *do* you mean? IÕm sure I canÕt guess. -- Jesse F. Hughes I talk with bigger fish who are playing different games. -- James S Harris has a way with the metaphor. === Subject: Re: Set inclusion and membership [ .. nitpicking as usual .. ] Two points, in case you didnÕt notice: 1. IÕm done with you 2. This thread has ended Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <871xe9ylka.fsf@phiwumbda.org> Discussion, linux) > [ .. nitpicking as usual .. ] > Two points, in case you didnÕt notice: > 1. IÕm done with you > 2. This thread has ended Yes, pointing out that the solution to your question has very little to do with your question is nitpicking. You are a tedious and stupid person. -- Jesse F. Hughes Well, I donÕt claim to be an expert, in fact I am a fry cook with a national burger chain, but I have solved many differential and partial differential equations numerically. --C. Bond === Subject: Re: Set inclusion and membership > You are a tedious and stupid person. Logged. Are you proud of yourself? Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <871xe9ylka.fsf@phiwumbda.org> <87acsvbqu0.fsf@phiwumbda.org> Discussion, linux) >> You are a tedious and stupid person. > Logged. Are you proud of yourself? I thought that the earlier vulgar language prompted you to ignore my posts. Feeling suddenly less sensitive, are we? -- Jesse F. Hughes You shouldnÕt hate Mother Mathematics. -- James S. Harris === Subject: Re: Set inclusion and membership > This says that there are no sets aside from sets satisfying x={x}. > Thus (1) every set is a singleton and (2) every set is an element of > itself. Therefore, the is-element relation is equivalent to equality. > If this isnÕt what you mean, then what *do* you mean? IÕm sure I > canÕt guess. What he means is that he is clueless :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> Discussion, linux) >> It doesnÕt. It cannot be that (S not in S) because S = {S} implies that >> (S is in S), always, unconditionally. With other words, JesseÕs would-be >> (E y)(y in S and NOT y = S) is a set that simply does not exist. > So your only sets satisfy the following two conditions. > (1) x is in x > (2) x is the only element of x. Put differently, a consequence of this brilliant suggestion that x = {x} is The element relation is the same as equality. You havenÕt stumbled across your own mereology. YouÕve take first order logic with equality, added a binary relation and interpreted that relation as equality. -- Reality has a fascinating ability to check us when we get a little too big for our britches... Make no mistake. There isnÕt a mathematician alive today that I canÕt now touch, and not a mathematical career on the planet that I canÕt now affect. --James Harris, render of worlds === Subject: Re: Set inclusion and membership > Put differently, a consequence of this brilliant suggestion that > x = {x} is > The element relation is the same as equality. No, that is only valid for a singleton. Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <87fz2qov5z.fsf@phiwumbda.org> Discussion, linux) >> Put differently, a consequence of this brilliant suggestion that >> x = {x} is >> The element relation is the same as equality. > No, that is only valid for a singleton. For every set x, {x} is a singleton. Therefore, for every set x, x = {x}. If you have something else in mind, youÕll have to express it more ,---- | Now suppose that I am kind of stubborn, insisting that the 2 concepts | should denote the same thing. With other words: I find that a = { a } | everywhere. `---- What did that *mean*? Everywhere sounds fairly sweeping to me. -- I am a force of Nature. Time is a friend of mine, and We talk about things, here and there. And sometimes We muse a bit [...] and then We watch them go... in the meantime, Time and I, We play with some of them, at least for a little while. --- JSH and His pal, Time. === Subject: Re: Set inclusion and membership > If you have something else in mind, youÕll have to express it more > clearly. [ ... ] There are _a lot_ of things I would have to express more clearly. But, as far as I am concerned, the rest of this thread will be nitpicking. I have found my Precious One. ItÕs called Mereology. Which is equivalent to a Boolean Algebra without the empty set. I find this a satisfactory end-result. IÕve also learned that ZFC without the axiom of Foundation is a possibility. Han de Bruijn Note: Precious One as quoted from Lord of the Rings by Tolkien. No argument. Just a pun. Since I have to explain everything to some. === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> <87llciow8j.fsf@phiwumbda.org> <87fz2qov5z.fsf@phiwumbda.org> <874qj5ylom.fsf@phiwumbda.org> Discussion, linux) >> If you have something else in mind, youÕll have to express it more >> clearly. [ ... ] > There are _a lot_ of things I would have to express more clearly. > But, as far as I am concerned, the rest of this thread will be > nitpicking. I have found my Precious One. ItÕs called Mereology. > Which is equivalent to a Boolean Algebra without the empty set. > I find this a satisfactory end-result. Good, but it does not satisfy your desideratum that x = {x}. Neither does anti-well-founded set theory. In both cases, there are certain xÕs for which that is true: discrete xÕs for mereology and the unique x satisfying x = {x} for AczelÕs anti-well-founded set theory. But it is certainly not satisfied everywhere, which is what started this thread. -- I am one of the more important discoverers in mathematical history, but future students will have the luxury of knowing that, and may be puzzled by your behavior now. -- James Harris (At least I have the foresight to quote his pearls of wisdom.) === Subject: Re: Set inclusion and membership > Now, by definition (A x)(NOT x in {}). Since {} = {{}}, we conclude > that (A x)(NOT x in {{}}) and hence (NOT {} in {{}}). Of course, by > definition we know that {} *is* in {{}}. Thus we have an > inconsistency. Hey! That is an ingenious argument. Ingenious enough to be convincing? Anyway, ingenious enough to memorize! > Not the most exciting universe of sets. :-( Han de Bruijn === Subject: Re: Set inclusion and membership <87wtw33f6x.fsf@phiwumbda.org> Discussion, linux) >> Now, by definition (A x)(NOT x in {}). Since {} = {{}}, we conclude >> that (A x)(NOT x in {{}}) and hence (NOT {} in {{}}). Of course, by >> definition we know that {} *is* in {{}}. Thus we have an >> inconsistency. > Hey! That is an ingenious argument. Ingenious enough to be > convincing? I wouldnÕt call it ingenious, but if it doesnÕt convince you that identifying x with {x} is a remarkably bad idea, then I donÕt know what will. -- Jesse F. Hughes And IÕm one of my own biggest skeptics as I had *YEARS* of wrong ideas, and attempts that failed. Worse, for some of them it took *MONTHS* before I figured out where I screwed up. -- James Harris === Subject: Re: Set inclusion and membership > In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, > Academic Science, Den Haag, 1976, I find the following sentence, when > translated into English: > Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. > Therefore: a e A is not the same as a c A . > Now suppose that I am kind of stubborn, insisting that the 2 concepts > should denote the same thing. With other words: I find that a = { a } > everywhere. Since I am a physicist by education, this is not a stupid > idea. Because, in physics, a member of a set cannot be distinguished > from a subset of a set, which contains only that element as a member. > Question. Would such an assumption be rather harmless & scratch only > the surface of set theory? Or would it have profound consequences? > How does it affect the Zermelo Fraenkel system (ZFC), for example? > Just a thought. Curious about your answers. > Han de Bruijn How does adding for all a, aea affect ZFC? Well if you have the foundation axiom, then it creates huge problems. But otherwise IÕm not sure it changes much, if the only unfounded element allowed in a set was itself, then youÕd have an ur-element in the role of the empty set, but we could still denote it {}, and then the set that contains only {} besides itself we could denote {{}}, IÕm not sure what advantage is gained from doing this though. As for the broader question, what if for all a, for all A (aeA <=> acA) were added to ZFC, then again youÕd have problems with foundation, but ignoring that for now, you basically have that every set was itÕs own power set. So to keep the c=e axiom, separation would have to be weakened because while one could make A smaller, one canÕt do it in arbitrary ways. Imagine the smallest set Z that contains the integers. It contains all the integers, so it must also contain every subset of the integers. It contains all the subsets of integers, so it must also contain every set that is a subset of the set of all subsets of the integers. So at least IÕm seeing that B=U{N,P(N),P(P(N)),P(P(P(N))), ...} is a subset of Z, but that means that BeZ, so any set like {B,b} where beB is in Z too. And C=U{B,P(B),P(P(B)),...} is a subset of Z, so therefore in Z. So Z is very very large in the traditional sense. But I havenÕt studied ZFC without foundation much, so maybe itÕs not as large as it looks, but breaking two axioms (separation and foundation) is a pretty good step for getting something not recognizable as ZFC. === Subject: Re: Set inclusion and membership > How does adding for all a, aea affect ZFC? Well if you have the > foundation axiom, then it creates huge problems. But otherwise IÕm > not sure it changes much, if the only unfounded element allowed in a > set was itself, then youÕd have an ur-element in the role of the empty > set, but we could still denote it {}, and then the set that contains > only {} besides itself we could denote {{}}, IÕm not sure what > advantage is gained from doing this though. Advantage or not, I only want to know what the consequences would be if you try. And I think you are right: the axiom of foundation is affected most. You cannot longer define the _transfinite ordinals_, for example. Not that I would find such a disadvantage ... > As for the broader question, what if for all a, for all A (aeA <= acA) were added to ZFC, then again youÕd have problems [ ... ] It occurred to me, yes, but I wouldnÕt go as far as that. [ pretty good stuff deleted ] > breaking two axioms (separation and foundation) is a pretty good step > for getting something not recognizable as ZFC. Han de Bruijn === Subject: Re: Set inclusion and membership > In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, > Academic Science, Den Haag, 1976, I find the following sentence, when > translated into English: > Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. > Therefore: a e A is not the same as a c A . > Now suppose that I am kind of stubborn, insisting that the 2 concepts > should denote the same thing. With other words: I find that a = { a } > everywhere. Since I am a physicist by education, this is not a stupid > idea. Because, in physics, a member of a set cannot be distinguished > from a subset of a set, which contains only that element as a member. > Question. Would such an assumption be rather harmless & scratch only > the surface of set theory? Or would it have profound consequences? > How does it affect the Zermelo Fraenkel system (ZFC), for example? > Just a thought. Curious about your answers. The set of all elements of N is just N itself. The set of all subsets of N (to so-called power set of N) is quite something else, donÕt you agree? Dan Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Set inclusion and membership > The set of all elements of N is just N itself. The set of all subsets of N > (to so-called power set of N) is quite something else, donÕt you agree? The power set of {a,b} is {{},{a},{b},{a,b}} = {,a,b,{a,b}} <> {a,b} . DonÕt you agree? Han de Bruijn === Subject: Set inclusion and membership {a,b} . > DonÕt you agree? No. What I agree to is The power set of {a,b} is { {},{a},{b},{a,b} } <> { a,b }. === Subject: Re: Set inclusion and membership >>The power set of {a,b} is {{},{a},{b},{a,b}} = {,a,b,{a,b}} <> {a,b} . >>DonÕt you agree? > No. What I agree to is > The power set of {a,b} is { {},{a},{b},{a,b} } <> { a,b }. Yes, that would be according to the books. How about the following. If a, b are sets and aÕ,bare the complements of these sets, then we can define A = {a,b} and the following set: P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. If the number of elements in the original set is N (= 2 here) then the number of elements in the above construct is 2^N (= 4 here). Thus the above set is resemblant to the power set in some respects. How far can we go with that? Is there a name for the above construct? IÕve seen it with an explanation of Karnaugh diagrams. Han de Bruijn === Subject: Re: Set inclusion and membership Discussion, linux) >The power set of {a,b} is {{},{a},{b},{a,b}} = {,a,b,{a,b}} <> {a,b} . >DonÕt you agree? >> No. What I agree to is >> The power set of {a,b} is { {},{a},{b},{a,b} } <> { a,b }. > Yes, that would be according to the books. > How about the following. If a, b are sets and aÕ,bare the complements > of these sets, then we can define A = {a,b} and the following set: > P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes > intersection. This is nonsense, as is usual. What is asupposed to mean? WeÕre taking the complement of a with respect to *what*? Is it A a? Why would *that* be relevant? Or are you still on your confusion that a should be equal to {a}, despite the fact that this leads to triviality or inconsistency? LetÕs try our best to make your suggestion coherent, shall we? I see two alternatives. (1) a= (a u b) a = b a (and similarly b= a b). LetÕs see if this makes any particular sense. Let a be the set {0,1} and b the set {1,2}. Then youÕve constructed the set { {0,2}, {2}, {0}, {1} } Now how is that related to the set of all subsets of { {0,1}, {1,2} }? The empty set isnÕt in it. The set A isnÕt in it. The sets {a} and {b} arenÕt in it. Congratulations. { aÕ.b, aÕ.b , a.b, a.b } is disjoint from the set P(A). (2) Replace awith A {a} and a with {a}, so your proposal is { {a}Õ.{b}, {a}Õ.{b} , {a}.{b}, {a}.{b} } Then your proposal comes to { {}, {b}, {a}, {} }. Closer, but still not P(A). It doesnÕt include the set A. What the heck are you trying to do? -- After years of arguing I realize that your intellects are too limited to fully grasp my work. [...] Still, no matter how child-like your minds are, [...] since you have language, [...] thereÕs a chance that IÕll be able to find something that your minds can handle. --JSH === Subject: Re: Set inclusion and membership > What the heck are you trying to do? IÕm trying to do something else. Han de Bruijn === Subject: Re: Set inclusion and membership <87zn0yp0ke.fsf@phiwumbda.org> Discussion, linux) >> What the heck are you trying to do? > IÕm trying to do something else. Good idea. -- I am a force of Nature. Time is a friend of mine, and We talk about things, here and there. And sometimes We muse a bit [...] and then We watch them go... in the meantime, Time and I, We play with some of them, at least for a little while. --- JSH and His pal, Time. === Subject: Re: Set inclusion and membership { a,b }. > How about the following. If a, b are sets and aÕ,bare the complements > of these sets, then we can define A = {a,b} and the following set: > P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. No. P(A) = P({ a,b }) = as before. > If the number of elements in the original set is N (= 2 here) then the > number of elements in the above construct is 2^N (= 4 here). Thus the > above set is resemblant to the power set in some respects. How far can > we go with that? That resemblance is equinumerous. Cantor took equinumerous to the zenith. You just as well may have claimed P(A) = { Bush, Iraq, Oil, Money } or P(A) = { sex, boy, priest, confession } === Subject: Re: Set inclusion and membership >>P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. > No. P(A) = P({ a,b }) = as before. OK, good guy. Define Q(a) = { aÕ.b, aÕ.b , a.b, a.b } . > That resemblance is equinumerous. Cantor took equinumerous to the zenith. > You just as well may have claimed P(A) = { Bush, Iraq, Oil, Money } > or P(A) = { sex, boy, priest, confession } You donÕt get the point, do you? Han de Bruijn === Subject: Re: Set inclusion and membership P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. > No. P(A) = P({ a,b }) = as before. > OK, good guy. Define Q(a) = { aÕ.b, aÕ.b , a.b, a.b } . > That resemblance is equinumerous. Cantor took equinumerous to the zenith. > You just as well may have claimed P(A) = { Bush, Iraq, Oil, Money } > or P(A) = { sex, boy, priest, confession } > You donÕt get the point, do you? Pointfullessness Are you appealing, to some invocation That is revealing, a small revocation For my repealing, your incantation? Now changing the point, to make my point which is the point I deserve for pointing out the most pointedly pointed point ever pointed, may I point out to you, the point I wish you to appoint me without disappointment at our next appointment? === Subject: Re: Set inclusion and membership > Pointfullessness > Are you appealing, to some invocation > That is revealing, a small revocation > For my repealing, your incantation? > Now changing the point, to make my point which is the point I deserve for > pointing out the most pointedly pointed point ever pointed, may I point > out to you, the point I wish you to appoint me without disappointment at > our next appointment? Hmm, your English seems to be as good as my Dutch! :-) Indeed. LetÕs finish, please, this part of the thread: > P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. I apologize for launching this add-on, which has caused nothing else but confusion. I shouldnÕt have opened this can of worms, at all. OK? Han de Bruijn === Subject: Re: Set inclusion and membership Are you appealing, to some invocation > That is revealing, a small revocation > For my repealing, your incantation? > Now changing the point, to make my point which is the point I deserve for > pointing out the most pointedly pointed point ever pointed, may I point > out to you, the point I wish you to appoint me without disappointment at > our next appointment? > P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. > I apologize for launching this add-on, which has caused nothing else > but confusion. I shouldnÕt have opened this can of worms, at all. OK? Depends. If you like the fish you catch with those worms, then eat Ōem and if you donÕt then toss them back, dump the worms into soft soil and pack up your fishing rod. === Subject: Re: Set inclusion and membership Discussion, linux) >P(A) = { aÕ.b, aÕ.b , a.b, a.b } where (.) denotes intersection. >> No. P(A) = P({ a,b }) = as before. > OK, good guy. Define Q(a) = { aÕ.b, aÕ.b , a.b, a.b } . >> That resemblance is equinumerous. Cantor took equinumerous to the zenith. >> You just as well may have claimed P(A) = { Bush, Iraq, Oil, Money } >> or P(A) = { sex, boy, priest, confession } > You donÕt get the point, do you? I donÕt. You are utterly incoherent. You use complementation without specifying complement with respect to what set. Your claim that | Q(A) | = 4 is given without proof. How do we know that aÕ.bis distinct from a.b? What the hell does the intersection of a and b have to do with subsets of {a,b}? -- And the logical extension of free and open-source software in the realm of sex would certainly include publicly shared sex at a sex party,... queer sexuality and... non-proprietary sexual affection. Annalee Newitz writing in Salon.com === Subject: Re: Set inclusion and membership posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs Where do you believe set theory comes up in physics? === Subject: Re: Set inclusion and membership > Where do you believe set theory comes up in physics? If you believe that music is physics ... I have an application where set theory is employed with the transition of chords. With the chords Am and C, for example, only the notes in the symmetric difference Am<>C need to be changed (to become Off or On respectively). Han de Bruijn === Subject: Re: Set inclusion and membership posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs > Where do you believe set theory comes up in physics? > If you believe that music is physics ... I have an application where set > theory is employed with the transition of chords. With the chords Am and > C, for example, only the notes in the symmetric difference Am<>C need to > be changed (to become Off or On respectively). Not an answer to the question. You said this in your original post: Because, in physics, a member of a set cannot be distinguished from a subset of a set, which contains only that element as a member. So you presumably have something in mind in conventional physics, not your ideas about how physics should be. So where, aside from your ideas, is set theory used in physics? What prompted the above quote? Where in physics is the above concept used? By the way, I donÕt think your music idea is valid beyond your trivial example. Show me how it tells me the right chord progression to go from, oh, C Maj to F# Maj to G minor for instance. - Randy === Subject: Re: Set inclusion and membership > Not an answer to the question. You said this in your original post: > Because, in physics, a member of a set cannot be distinguished > from a subset of a set, which contains only that element as a member. > So you presumably have something in mind in conventional physics, > not your ideas about how physics should be. > So where, aside from your ideas, is set theory used in physics? > What prompted the above quote? Where in physics is the > above concept used? Only at the lowest level. When counting pencils in a box, for example. Can you distinguish a single pencil from a collection that contains only that pencil? I mean, physically? Then devise some experiment to determine what the difference may be. Or do you mean that sets have no counterpart in physics, at all? ThatÕs the reason behind the music application. I find that there *are* counterparts of sets in physics and technology. And they *are* quite useful once in a while. > By the way, I donÕt think your music idea is valid beyond your > trivial example. Show me how it tells me the right chord > progression to go from, oh, C Maj to F# Maj to G minor for > instance. Heh, heh. I gave that trivial example as just an example. My midi-music applications go far beyond that, of course: http://huizen.dto.tudelft.nl/deBruijn/muziek/delphi/index.htm The theory behind this is only available in Dutch, alas. But I would be surprised if nobody else but me has found that set-theoretic approach quite useful. Search for Verzamelingenleer in: http://www.tomaatnet.nl/~handebruijn/muziek/fig_sw.htm Oh well. C Maj = {C,E,G,B} , F# Maj = {F#,A#,C#,F} , G minor = {G,A#,D} Hence F# Maj C Maj = {F#,A#,C#,F} On , C Maj F# Maj = {C,E,G,B} Off. (YouÕve defined disjoint sets.) Now wait and listen. Then: G minor F# Maj = {G,D} On , F# Maj G minor = {F#,C#,F} Off. What you see here are the common (asymmetric) set theoretic differences, denoted by . (Hope that I made no mistakes here.) Han de Bruijn === Subject: Re: Set inclusion and membership >Only at the lowest level. When counting pencils in a box, for example. >Can you distinguish a single pencil from a collection that contains >only that pencil? I mean, physically? Physically, can you distinguish a pencil from the box that contains one pencil? -- IÕm not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Set inclusion and membership > Physically, can you distinguish a pencil from the box that contains > one pencil? I canÕt. Because the box is not a part of the set. Han de Bruijn === Subject: Re: Set inclusion and membership > In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, > Academic Science, Den Haag, 1976, I find the following sentence, when > translated into English: > Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. > Therefore: a e A is not the same as a c A . > Now suppose that I am kind of stubborn, insisting that the 2 concepts > should denote the same thing. With other words: I find that a = { a } > everywhere. Since I am a physicist by education, this is not a stupid > idea. Because, in physics, a member of a set cannot be distinguished > from a subset of a set, which contains only that element as a member. > Question. Would such an assumption be rather harmless & scratch only > the surface of set theory? Or would it have profound consequences? > How does it affect the Zermelo Fraenkel system (ZFC), for example? > Just a thought. Curious about your answers. There is an old subject called ŌmereologyÕ. Whitehead tried to do set theory with only the part-of relation. And itÕs possible, indeed, without too much trouble. There are some modern approaches, too. Mostly associated with logic and ai (spatial reasoning, &c). -- Herman Jurjus === Subject: Re: Set inclusion and membership > There is an old subject called ŌmereologyÕ. Whitehead tried to do set > theory with only the part-of relation. And itÕs possible, indeed, > without too much trouble. There are some modern approaches, too. Mostly > associated with logic and ai (spatial reasoning, &c). I can find a whole lot of references now, with Google. Han de Bruijn === Subject: Re: Set inclusion and membership >In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, >Academic Science, Den Haag, 1976, I find the following sentence, when >translated into English: >Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. >Therefore: a e A is not the same as a c A . >Now suppose that I am kind of stubborn, insisting that the 2 concepts >should denote the same thing. With other words: I find that a = { a } >everywhere. stubborn would not be the right (hypothetical) word here. >Since I am a physicist by education, this is not a stupid >idea. Because, in physics, a member of a set cannot be distinguished >from a subset of a set, which contains only that element as a member. This is nonsense. >Question. Would such an assumption be rather harmless & scratch only >the surface of set theory? Or would it have profound consequences? >How does it affect the Zermelo Fraenkel system (ZFC), for example? >Just a thought. Curious about your answers. >Han de Bruijn ************************ David C. Ullrich === Subject: Re: Set inclusion and membership [ ... ] >>Since I am a physicist by education, this is not a stupid >>idea. Because, in physics, a member of a set cannot be distinguished >>from a subset of a set, which contains only that element as a member. > This is nonsense. Are you a physicist? Han de Bruijn BTW. Gozer: Are you a god? Dr. Raymond Stantz: No. Gozer: Then die. (as quotet from Ghostbusters 1) === Subject: Re: Set inclusion and membership >[ ... ] >Since I am a physicist by education, this is not a stupid >idea. Because, in physics, a member of a set cannot be distinguished >from a subset of a set, which contains only that element as a member. >> This is nonsense. >Are you a physicist? Uh, no. Why? There are no sets in physics per se. There are sets in the mathematics used in physics - thereÕs nothing in mathematical physics that says a = {a}. >Han de Bruijn >BTW. Gozer: Are you a god? Dr. Raymond Stantz: No. Gozer: Then die. >(as quotet from Ghostbusters 1) _ThatÕs_ pretty convincing. A quote from a standard reference on physics explaining that a = {a} might be even better, but of course you canÕt give anything like that. ************************ David C. Ullrich === Subject: Re: Set inclusion and membership > _ThatÕs_ pretty convincing. A quote from a standard reference > on physics explaining that a = {a} might be even better, but > of course you canÕt give anything like that. Look who is saying this! Just go to Google and search for the phrase David C. Ullrich and OFF. Then youÕll know _his_ meaning of being pretty convincing. (And being civilized as well.) Han de Bruijn === Subject: Re: Set inclusion and membership > _ThatÕs_ pretty convincing. A quote from a standard reference > on physics explaining that a = {a} might be even better, but > of course you canÕt give anything like that. > Look who is saying this! Just go to Google and search for the phrase > David C. Ullrich and OFF. Then youÕll know _his_ meaning of > being pretty convincing. (And being civilized as well.) David C. Ullrich containing that text, in 36 it was in quotes from at least one of them is a falsification. And that in 4 1/2 years. So? -- 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: Set inclusion and membership > Look who is saying this! Just go to Google and search for the phrase > David C. Ullrich and OFF. Then youÕll know _his_ meaning of > being pretty convincing. (And being civilized as well.) > David C. Ullrich containing that text, in 36 it was in quotes from > at least one of them is a falsification. And that in 4 1/2 years. So? So what? Substitute idiot, fool, stupid, dishonest, ignorant, youÕre crazy for OFF, to name only a few of his favorite words. Not to mention all of his sneering comments which cannot be tracked by keyword search at all. But what can you expect from an inhabitant of a country, where calling someone an asshole is done by a candidate for presidency. > Back to the present. I take it then that > ŌBTW. Gozer: Are you a god? Dr. Raymond Stantz: No. Gozer: Then die. > (as quotet from Ghostbusters 1)Õ > is in fact the best justification you can come up with for > the curious notion that a = {a} in physics? Ullrich doesnÕt even notice the difference between a would-be argument and a would-be signature. Han de Bruijn Reporter: What do you think of Western Civilization, sir? Ghandi: Well, I think that would be a mighty good idea! === Subject: Re: Set inclusion and membership Discussion, linux) > But what can you expect from an inhabitant of a country, where > calling someone an asshole is done by a candidate for presidency. Speaking of assholes and idiots... off. -- Naomi Klein reports that Microsoft is helping develop e-government in Iraq, Ōwhich Dees admits is a little ahead of the curve, since there is no g-government in Iraq--not to mention functioning phones lines.Õ -- as reported in The Register, http://www.theregister.co.uk === Subject: Re: Set inclusion and membership > Speaking of assholes and idiots... > off. Then you are the next one I will not answer anymore. And thatÕs final. Civilization comes before the math. Han de Bruijn === Subject: Re: Set inclusion and membership Discussion, linux) >> _ThatÕs_ pretty convincing. A quote from a standard reference >> on physics explaining that a = {a} might be even better, but >> of course you canÕt give anything like that. > Look who is saying this! Just go to Google and search for the phrase > David C. Ullrich and OFF. Then youÕll know _his_ meaning of > being pretty convincing. (And being civilized as well.) Answer the question. -- Jesse F. Hughes My experience indicates that the people who post on this newsgroup are about at the level of a 10 year old in the year 2060. -- More wisdom from James Harris, time traveler === Subject: Re: Set inclusion and membership > Answer the question. I decide that myself. Han de Bruijn === Subject: Re: Set inclusion and membership >> _ThatÕs_ pretty convincing. A quote from a standard reference >> on physics explaining that a = {a} might be even better, but >> of course you canÕt give anything like that. >Look who is saying this! Just go to Google and search for the phrase >David C. Ullrich and OFF. Then youÕll know _his_ meaning of >being pretty convincing. (And being civilized as well.) Giggle. You need to read a little more carefully, noting whatÕs a _quote_ as opposed to whatÕs something I said. In the posts I suspect youÕre referring to, my _quoting_ that phrase was indeed a pretty conclusive way to make the point I was making (namely that the person I was quoting has no business complaining about how people are so mean to poor innocent little him.) Back to the present. I take it then that ŌBTW. Gozer: Are you a god? Dr. Raymond Stantz: No. Gozer: Then die. (as quotet from Ghostbusters 1)Õ is in fact the best justification you can come up with for the curious notion that a = {a} in physics? >Han de Bruijn ************************ David C. Ullrich === Subject: Re: Set inclusion and membership >>In the book by S.T.M Ackermans / J.H. van Lint, Algebra en Analyse, >>Academic Science, Den Haag, 1976, I find the following sentence, when >>translated into English: >>Never mix up Ōbeing a member(e) or Ōbeing a subset(c) of a set. >>Therefore: a e A is not the same as a c A . >>Now suppose that I am kind of stubborn, insisting that the 2 concepts >>should denote the same thing. With other words: I find that a = { a } >>everywhere. > stubborn would not be the right (hypothetical) word here. >>Since I am a physicist by education, this is not a stupid >>idea. Because, in physics, a member of a set cannot be distinguished >>from a subset of a set, which contains only that element as a member. > This is nonsense. You are too generous, David. >>Question. Would such an assumption be rather harmless & scratch only >>the surface of set theory? Or would it have profound consequences? In de Bruijn set theory, where {a} = a, then taking a to be the two-element set {1,2} we conclude that {{1,2}} = {1,2}. On the left there is a 1-element set and on the right a 2-element set. Hence 1 = 2. So for every 1 thing, de Bruijn can conjure up another by magic (presumably this works with Euros). Conclusion: de Bruijn is a mathematical creationist. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Set inclusion and membership > Conclusion: de Bruijn is a mathematical creationist. OK. Talking about mathematical creationism ... With the axiom of foundation, *you* create the following sets: 0 := {}, 1 := {{}}, 2 := {{},{{}}}, 3 := {{},{{}},{{},{{}}}}, ... Copy and paste, huh? And you think you have defined the natural numbers with that. But, with de BruijnÕs creationism: 0 := {}, 1 := {{}} = {} = 0, 2 := {{},{{}}} = {0,{}} = {0,0} = {0} = 0 Do I have to continue? No. Because the lesson is: that you canÕt create Something from Nothing. Conservation of mass and energy. A sound physical principle. Han de Bruijn === Subject: Re: Set inclusion and membership Discussion, linux) >> Conclusion: de Bruijn is a mathematical creationist. > OK. Talking about mathematical creationism ... > With the axiom of foundation, *you* create the following sets: > 0 := {}, 1 := {{}}, 2 := {{},{{}}}, 3 := {{},{{}},{{},{{}}}}, ... What the hell has foundation have to do with those sets? Try the axiom of pairing. > Copy and paste, huh? > And you think you have defined the natural numbers with that. > But, with de BruijnÕs creationism: > 0 := {}, 1 := {{}} = {} = 0, 2 := {{},{{}}} = {0,{}} = {0,0} = {0} = 0 > Do I have to continue? No. > Because the lesson is: that you canÕt create Something from Nothing. Wait... Your theory[1] entails that for all x, x = {}. But you think this is a *feature*? Anyway, itÕs a lot worse than that. Your reasoning above entails that {} is an element of {}. This obviously contradicts the definition of the empty set. Is that a feature, too? > Conservation of mass and energy. A sound physical principle. A sound (?) physical principle that has nothing to do with set theory. Footnotes: [1] IÕm being generous. No theory has been given here. Just a hint of an axiom: for all x, x= {x}. -- At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom that evening, Robert Dees urges us Ōto network on behalf of the people of Iraq,Õ -- Naomi Klein reports on MicrosoftÕs efforts to further democracy. === Subject: Re: Set inclusion and membership > Footnotes: > [1] IÕm being generous. No theory has been given here. Just a > hint of an axiom: for all x, x= {x}. You are not so generous. My intention is exactly the same. I just want to figure out how far I can go with that hint of an axiom. Yes, IÕm playing a game. But IÕm not fooling you, because I intend it to be a _serious_ game. And IÕve got some worthwile responses already. Han de Bruijn === Subject: Re: Set inclusion and membership > [ ... ] (presumably this works with Euros). This suggests that you have actually skimmed over my web pages. IÕm ßattered :-) Han de Bruijn === Subject: Re: Set inclusion and membership >>Since I am a physicist by education, this is not a stupid >>idea. Because, in physics, a member of a set cannot be distinguished >>from a subset of a set, which contains only that element as a member. This is nonsense. > You are too generous, David. IÕve studied physics, and I have no recollection of having to feign ignorance of the distinction between a and {a} to solve any physics problem or make any prediction. >>Question. Would such an assumption be rather harmless & scratch only >>the surface of set theory? Or would it have profound consequences? > In de Bruijn set theory, where {a} = a, then taking a to be > the two-element set {1,2} we conclude that {{1,2}} = {1,2}. > On the left there is a 1-element set and on the right a 2-element > set. Hence 1 = 2. So for every 1 thing, de Bruijn can conjure up > another by magic (presumably this works with Euros). > Conclusion: de Bruijn is a mathematical creationist. Actual conclusion: de Bruijn set theory has 1 = 2 for all 1 and 2 such that 1 e a and 2 e A. Robin Chapman made an assumption that a={1,2}, when clearly the equality a={a} implies that a is an ur-element as it is a singleton that contains only itself. If you assume a singleton has two elements, then of course youÕll get weird results. If you had finished the work to conclude 1=2=a then you would have gotten back to a={a}, so the Chapman analysis seems to have no content. === Subject: Re: Set inclusion and membership > IÕve studied physics, and I have no recollection of having to feign > ignorance of the distinction between a and {a} to solve any physics > problem or make any prediction. The fact that a <> {a} is in the mathematical machinery which as such is applied to physics. Are you really sure that this _can do no harm_? Suppose that a = {a} has been established instead by experimentation: an apple cannot be distinguished from an {apple}. Then what? > [ ... ] so the Chapman analysis seems to have no content. Glad the conclusion is yours. :-) Han de Bruijn === Subject: Re: Set inclusion and membership > With other words: I find that a = { a } > everywhere. What do you imagine to be the relevance of these findings of yours to anything in mathematics? === Subject: Re: Set inclusion and membership >>With other words: I find that a = { a } >>everywhere. > What do you imagine to be the relevance of these findings of yours > to anything in mathematics? Other people, unlike you, have come up with quite sensible answers. Han de Bruijn === Subject: Re: A New Abstract for Special Relativity > Is it possible to derive the Lorentz transformation from the Galilean > transformation? In the way you use those words, it is possible to derive anything from 1+1=2 -- when you introduce enough additional postulates to derive the desired result, where you start becomes irrelevant. Tom Roberts tjroberts@lucent.com === Subject: Re: A New Abstract for Special Relativity I have never seen you so terribly confused. I suppose that you never recovered from that severe beating that I gave you. http://www.everythingimportant.org/viewtopic.php?p=3437#3437 http://www.everythingimportant.org/viewtopic.php?p=3441#3441 http://www.everythingimportant.org/viewtopic.php?p=3451#3451 http://www.everythingimportant.org/viewtopic.php?p=3452#3452 http://www.everythingimportant.org/viewtopic.php?p=3466#3466 http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity > I have never seen you so terribly confused. I suppose that you never > recovered from that severe beating that I gave you. > http://www.everythingimportant.org/viewtopic.php?p=3437#3437 > http://www.everythingimportant.org/viewtopic.php?p=3441#3441 > http://www.everythingimportant.org/viewtopic.php?p=3451#3451 > http://www.everythingimportant.org/viewtopic.php?p=3452#3452 > http://www.everythingimportant.org/viewtopic.php?p=3466#3466 > http://www.everythingimportant.org/relativity/special.pdf You couldnÕt beat up a cripple except in your own mind. John Anderson === Subject: Re: A New Abstract for Special Relativity >> I have never seen you so terribly confused. I suppose that you >> never recovered from that severe beating that I gave you. >> http://www.everythingimportant.org/viewtopic.php?p=3437#3437 >> http://www.everythingimportant.org/viewtopic.php?p=3441#3441 >> http://www.everythingimportant.org/viewtopic.php?p=3451#3451 >> http://www.everythingimportant.org/viewtopic.php?p=3452#3452 >> http://www.everythingimportant.org/viewtopic.php?p=3466#3466 >> http://www.everythingimportant.org/relativity/special.pdf > You couldnÕt beat up a cripple except in your own mind. Do you agree with the mathematicians who answered this question or do you agree with Tom Roberts? Eugene Shubert === Subject: Re: A New Abstract for Special Relativity >> Is it possible to derive the Lorentz transformation from the Galilean >> transformation? > In the way you use those words, it is possible to derive anything > from 1+1=2 -- when you introduce enough additional postulates to > derive the desired result, where you start becomes irrelevant. > Tom Roberts tjroberts@lucent.com Robertsdefinition of irrelevant verbiage: a) But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v... b) It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain V = (c+w)/(1+w/c) = c. Androcles. === Subject: Re: A New Abstract for Special Relativity >> Is it possible to derive the Lorentz transformation from the Galilean >> transformation? > In the way you use those words, it is possible to derive anything > from 1+1=2 -- when you introduce enough additional postulates to > derive the desired result, where you start becomes irrelevant. > Tom Roberts tjroberts@lucent.com > Robertsdefinition of irrelevant verbiage: > a) But the ray moves relatively to the initial point of k, when > measured > in the stationary system, with the velocity c-v... > b) It follows, further, that the velocity of light c cannot be altered > by composition with a velocity less than that of light. > For this case we obtain V = (c+w)/(1+w/c) = c. Androclesdefinition of listening: *Plonk*, now I can pretend I canÕt hear you Dirk Vdm === Subject: Re: A New Abstract for Special Relativity >> Is it possible to derive the Lorentz transformation from the >> Galilean transformation? > In the way you use those words, it is possible to derive anything > from 1+1=2 -- when you introduce enough additional postulates > to derive the desired result, where you start becomes irrelevant. Tom, Your comment doesnÕt answer my question. Can I begin with a Galilean transformation in a Galilean universe, pick a constant c out of thin air, reset clocks, define a parameter v as a function of proper velocity and then arrive at the Lorentz transformation? http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity >Is it possible to derive the Lorentz transformation from the >Galilean transformation? >>In the way you use those words, it is possible to derive anything >>from 1+1=2 -- when you introduce enough additional postulates >>to derive the desired result, where you start becomes irrelevant. > Your comment doesnÕt answer my question. Just look up there at the question you asked and the answer I gave. If you cannot apply that answer to the question you posed, then you need to learn how to think. Or to read. IÕll come back to this at the end.... > Can I begin with a > Galilean transformation in a Galilean universe, pick a constant c > out of thin air, reset clocks, define a parameter v as a function > of proper velocity and then arrive at the Lorentz transformation? That is a COMPLETELY different question from the one you asked earlier (quoted above). BTW in standard usage, proper velocity would always be 0, and that phrase is never used. For a Lorentz transform, v must be the relative velocity of the two frames. If by Galilean Universe you mean a model universe in which the Galilean transform is exactly satisfied in all situations, then what you ask is not possible because there are valid velocities greater than whatever value of c you picked out of thin air. Moreover, if you happen to find coordinates on a pair of inertial frames such that a LT applies between them, in at least one of the frames the coordinates will not correspond to standard rulers and clocks in this universe. In general you will have to construct different clocks and rulers for each frame in order for them to correspond to the coordinates you use, and the rulers will need to vary depending upon their orientation (!). Stated differently: You can use whatever coordinates you choose. The coordinates you choose in different frames will determine the coordinate transform between those coordinates in those frames; you are free to select coordinates so the LTs hold between frames[#]. But this has NOTHING WHATSOEVER to do with any physics or geometry in this universe. Specifically: in general the (spatial) metric components will not be diag(1,1,1) when projected onto your spatial coordinates[*]; nor will time intervals correspond to differences in your time coordinate[*]. [#} I make no claim that one set of coordinates on each frame will do this for all frames. But you can certainly do this for any given pair of frames (and a suitable value of c). [*] This is just another way of saying you need to construct special clocks and orientation-dependent rulers in each frame. We will now leave this Galilean universe and discuss another silly statement you made: > IÕm not claiming that thereÕs an isomorphism > between those two mathematical groups [the Lorentz and Galileo groups]. Sure you are! The claim to derive the LT from the GT says precisely that -- you just didnÕt complete the proof (which fails for reasons IÕll give in the second paragraph below). In the usual vocabulary, a mathematical derivation establishes an implication from the axioms to the theorem (i.e. the set of axioms implies the theorem). But in such a derivation every step is reversible, so if you complete the set of theorems, the set of theorems implies the axioms. As you have never selected any specific frames, your derivation applies to every possible pair of frames, thus completing the necessary set of theorems. Had you actually done this, deriving the LT from the GT, you would have established a group homomorphism between the Galileo and Lorentz groups, which is an isomorphism between those two groups. But you did NOT do that. You added a bunch of additional postulates (without mentioning them, and apparently without even noticing that you were using them), and in fact those additions are sufficient to derive the Lorentz transforms without any reference to your starting point (the Galilean transforms). [Yes, this is just a long-winded restatement of what I said earlier. Your technique can derive the LT from 1+1=2. Nobody else would call what you did a derivation from the GT, it is rather a derivation from the GT plus a bunch of other axioms, and the GT itself is not actually needed.] Tom Roberts tjroberts@lucent.com === Subject: Re: A New Abstract for Special Relativity >>Is it possible to derive the Lorentz transformation from the >>Galilean transformation? >In the way you use those words, it is possible to derive anything >from 1+1=2 -- when you introduce enough additional postulates >to derive the desired result, where you start becomes irrelevant. >> Your comment doesnÕt answer my question. > Just look up there at the question you asked and the answer I gave. If > you cannot apply that answer to the question you posed, then you need > to learn how to think. Or to read. IÕll come back to this at the > end.... Yeah, my question was which variant of the PoR (EinsteinÕs version) do we use? Is it a) But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v... or b) It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain V = (c+w)/(1+w/c) = c. And your answer, Roberts, was Is it [... irrelevant verbiage] You didnÕt answer my question, Roberts. If you cannot give an answer to the question I posed, then you need to learn how to think. Or to read. DONÕT YOU? Androcles === Subject: Roberts considers Einstein paper to be irrelevant verbiage. > Is it possible to derive the Lorentz transformation from the > Galilean transformation? >> In the way you use those words, it is possible to derive anything >> from 1+1=2 -- when you introduce enough additional postulates >> to derive the desired result, where you start becomes irrelevant. > Tom, > Your comment doesnÕt answer my question. Of course he canÕt answer your question. He snipped mine as irrelevant verbiage too. Roberts: Standard and well-known derivations of the Lorentz transform are based on the following assumptions/postulates/techniques: 1. The Principle of Relativity (EinsteinÕs version) Androcles: Which one do we use? Is it a) But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v... or b) It follows, further, that the velocity of light c cannot be altered by composition with a velocity less than that of light. For this case we obtain V = (c+w)/(1+w/c) = c. Roberts: Is it [... irrelevant verbiage] This is what Roberts was faced with and was compelled to snip in defense of his religion, burying his head in the sand. Now, IÕve used the words a) and or b) to make a legible sentence, and the rest of the irrelevant verbiage is all EinsteinÕs. We can safely conclude that Roberts considers EinsteinÕs words to be irrelevant verbiage. How say you now, Roberts? I say you are a charlatan. IÕm calling you that because the evidence is quite clear. You are a rogue, a scoundrel, a bogus physicist, as low and debase as your predecessor, Einstein the huckster. You are deliberately perpetrating his hoax to your own profit, under the guise of pretending to be a physicist. You are incompetent, as are your colleagues, and engaged in falsehood for personal gain. You deserve prosecution for fraud. You contribute nothing to the society in which you live, nothing to the scientific community, nothing to humanity. You spread the lies of Einstein, hiding behind them, and do so for your own evil purpose. I despise you, I spit on you. You are . Androcles. === Subject: Re: A New Abstract for Special Relativity <41a8d0af@sys13.hou.wt.net> <41ab47e1@sys13.hou.wt.net> Is it possible to derive the Lorentz transformation from the >> Galilean transformation? > In the way you use those words, it is possible to derive anything > from 1+1=2 -- when you introduce enough additional postulates > to derive the desired result, where you start becomes irrelevant. > Tom, > Your comment doesnÕt answer my question. Can I begin with a > Galilean transformation in a Galilean universe, pick a constant c > out of thin air, reset clocks, define a parameter v as a function > of proper velocity and then arrive at the Lorentz transformation? > http://www.everythingimportant.org/relativity/special.pdf The actual form of the transformation depends on the coordinates in use. It doesnÕt matter if you can find coordinates under which the gallilean group transforms look like the lorentz group transforms in minkowski space, if youÕre starting with the assumption of a galilean universe you must end with a galilean universe or else you are attempting to prove that the galilean transformation group and the lorentz group are isomorphic, which is a logical contradiction. Modern QFT starts with the Lorentz group as a symmetry of spacetime. NoetherÕs theorem says that youÕre going to get different conserved charges if you are in a galilean universe. Your approach is not equivalent to standard special relativity. === Subject: Re: A New Abstract for Special Relativity > Is it possible to derive the Lorentz transformation from the > Galilean transformation? >> In the way you use those words, it is possible to derive >> anything from 1+1=2 -- when you introduce enough >> additional postulates to derive the desired result, where >> you start becomes irrelevant. >> Tom, >> Your comment doesnÕt answer my question. Can I begin with a >> Galilean transformation in a Galilean universe, pick a constant c >> out of thin air, reset clocks, define a parameter v as a function >> of proper velocity and then arrive at the Lorentz transformation? >> http://www.everythingimportant.org/relativity/special.pdf > The actual form of the transformation depends on the coordinates > in use. It doesnÕt matter if you can find coordinates under which > the gallilean group transforms look like the lorentz group > transforms in minkowski space, If it doesnÕt matter, then why canÕt Tom Roberts admit the validity of my approach to SR: This paper demonstrates that for any two inertial frames of reference in either a Galilean or Einsteinian universe, clocks can be reset thereby modifying the usual transformation equations into a Lorentz transformation with a new arbitrary constant c. This trick is exploited to derive the Lorentz transformation from the Galilean transformation. > if youÕre starting with the assumption of a galilean universe you > must end with a galilean universe or else you are attempting to > prove that the galilean transformation group and the lorentz > group are isomorphic, which is a logical contradiction. Obviously! But IÕm not claiming that thereÕs an isomorphism between those two mathematical groups. > Your approach is not equivalent to standard special relativity. Your objection is based on misunderstanding what IÕve written. Why donÕt you object to an actual equation or statement in my derivation? http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity Eugene Shubert a .8ecrit : > If it doesnÕt matter, then why canÕt Tom Roberts admit the validity > of my approach to SR: > This paper demonstrates that for any two inertial frames of > reference in either a Galilean or Einsteinian universe, clocks > can be reset thereby modifying the usual transformation > equations into a Lorentz transformation with a new > arbitrary constant c. This trick is exploited to derive the > Lorentz transformation from the Galilean transformation. What you donÕt understand because of a likely lackness in understanding of what algebra is about (not to say about physics) is that you got the form, the appearance of LTs, not their content. What you are doing is like pretending that x->e^x is linear because its graph is a line on a logarithmic scale. === Subject: Re: A New Abstract for Special Relativity > Eugene Shubert a .8ecrit : >> If it doesnÕt matter, then why canÕt Tom Roberts admit the validity >> of my approach to SR: >> This paper demonstrates that for any two inertial frames of >> reference in either a Galilean or Einsteinian universe, clocks >> can be reset thereby modifying the usual transformation >> equations into a Lorentz transformation with a new >> arbitrary constant c. This trick is exploited to derive the >> Lorentz transformation from the Galilean transformation. > You got the form, the appearance of LTs, not their content. Now that would be real magic. http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity <41a8d0af@sys13.hou.wt.net> <41ab47e1@sys13.hou.wt.net> <41ab5d9c@sys13.hou.wt.net Is it possible to derive the Lorentz transformation from the > Galilean transformation? >> In the way you use those words, it is possible to derive >> anything from 1+1=2 -- when you introduce enough >> additional postulates to derive the desired result, where >> you start becomes irrelevant. >> Tom, >> Your comment doesnÕt answer my question. Can I begin with a >> Galilean transformation in a Galilean universe, pick a constant c >> out of thin air, reset clocks, define a parameter v as a function >> of proper velocity and then arrive at the Lorentz transformation? >> http://www.everythingimportant.org/relativity/special.pdf > The actual form of the transformation depends on the coordinates > in use. It doesnÕt matter if you can find coordinates under which > the gallilean group transforms look like the lorentz group > transforms in minkowski space, > If it doesnÕt matter, then why canÕt Tom Roberts admit the validity > of my approach to SR: > This paper demonstrates that for any two inertial frames of > reference in either a Galilean or Einsteinian universe, clocks > can be reset thereby modifying the usual transformation > equations into a Lorentz transformation with a new > arbitrary constant c. This trick is exploited to derive the > Lorentz transformation from the Galilean transformation. > if youÕre starting with the assumption of a galilean universe you > must end with a galilean universe or else you are attempting to > prove that the galilean transformation group and the lorentz > group are isomorphic, which is a logical contradiction. > Obviously! But IÕm not claiming that thereÕs an isomorphism > between those two mathematical groups. > Your approach is not equivalent to standard special relativity. > Your objection is based on misunderstanding what IÕve written. > Why donÕt you object to an actual equation or statement in my > derivation? You snipped my objection. Essentially there are two possibilities when you are claiming to derive the lorentz transformation from the gallilean group: either you are assuming a gallilean univerise and are claiming to derive SR as a consequence from it, which would not be true since a theory based in a galilean universe will have different conserved quantities and spin representations than a minkowski one, i.e. the dirac equation will no longer work OR you are not claiming to derive special relativity as a consequence of a galilean universe, in which case you are playing with coordinates which is not very interesting to anyone. === Subject: Re: A New Abstract for Special Relativity > you are playing with coordinates which is not very interesting > to anyone. The issue isnÕt what is interesting. The issue is physicists who speak out of ignorance while not even being able to imagine how different clock resynchronization procedures alter the Galilean and Lorentzian transformation equations. http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity > ...OR you are not claiming to derive special relativity as a > consequence of a galilean universe, in which case you are > playing with coordinates which is not very interesting to > anyone. So you are admitting the possibility of deriving the Lorentz transformation from the Galilean transformation by resetting clocks? http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity > Essentially there are two possibilities when you are claiming to > derive the lorentz transformation from the gallilean group: Are you implying that the American Journal of Physics Editor, Jan Tobochnik, is a complete idiot for not seeing any remarkable claim in my derivation? http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Re: A New Abstract for Special Relativity >> Essentially there are two possibilities when you are claiming to >> derive the lorentz transformation from the gallilean group: > Are you implying that the American Journal of Physics Editor, Jan > Tobochnik, is a complete idiot for not seeing any remarkable > claim in my derivation? > http://www.everythingimportant.org/relativity/special.pdf > Eugene Shubert IÕm fairly certain that 1) He rejected your paper. 2) He did so in a diplomatic way. 3) He saw nothing remarkable. Androcles. === Subject: Re: A New Abstract for Special Relativity > Essentially there are two possibilities when you are claiming to > derive the lorentz transformation from the gallilean group: I havenÕt claimed to derive anything from a Galilean group structure. http://www.everythingimportant.org/relativity/special.pdf Eugene Shubert === Subject: Some Various Aspects Omnipotence Mathematically Some Various Aspects Omnipotence Mathematically What is Omnipotence? By definition this is a state of unlimited power. Is Omnipotence Presense possible? Well it depends on the state of the universe and that Being. If there is corruption in the universe, Omnipotent presence isnÕt possible. You canÕt make something not corrupted become corrupted. You canÕt mix Heaven With Hell. In the Priority State of Existence and Order of The Universe, where there is No Beginning, The Foundations of Existence in Omnipotence show that, Omnipotence isnÕt to occur but served by a Being or Supreme Being because without this state of a humble servant, 1) It doensÕt allow for the Free Will Choice of othe beings with the respect of other beings in the universe. If one reaches a state of Omnipotence, they must realize immediately that they are to be servants of the Omnipotent State even though they are of a higher level existence or state of being, 2) In the Omnipotence State with respect to presense means absolutely No Movement. If one moves, they have reduced their Omnipotence State of Existence. It takes time to move. Omnipotent Presense takes no time to do anything. So one did exist everywhere at the same time, there would be absolutely no movement and No Words spoken. It takes time to speak. 3) In the Highest States of Existence, it can be observed by revelation that, Free Will Choice ( with the respect of others with Free Will Choice) has always been their in the field structure of the universe and always will be there. The Universe is perfect at these states of existence and they have no beginning. The revealing knowledge of these field structures shows that the State of Good was there from the start and it will always be that way. A violation of these priority states from a state of Omnipotence and no beginning would mean one would force or impose control over or take away from a another being from the beginning. Evil begins in the violation of these priority states of existence. A person might say to you, you are a liar. But if they tell you from the priority states that you are a liar, then it can be seen that those that make a liar are the liars. Those that make criminals are the criminals. Evil is basically a state of existence that disallows your freedom, your will and your choice from the start and tells you what you are to be, before you can even get there to have a say in the matter. As we look at the corruption in the world today, we can see that many try to get wealthy at the expense of others, and they hog all the wealth while the others suffer. But in the beginning states of the universe, all things were to be shared and created equally for, by and with all, with respect for those in these higher existence states. The Foundations of the Universe, in the field structures are set up in a way that All things arenÕt really known by One Being, but it is shared with others beings. The more work, the more power and knowledge. So as one fills in more of the knowledge of the Foundations, the more that person is or the more Powerful that person is and their existence state is to be respected as a higher level. So it is a bunch of bull that someone will say they control you and you have to do what they say or die. All people have the right to live as they fill in the evolving knowledge of the founding structures of the universe freely. Like for example, a mathematician fills in the math knowledge of the universe. A physicist fills in the physics knowledge of the universe and so on. So all people are important and free and have choices and have a chance to move up into higher levels of existence. SmartÕs Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: lattice algorithm Let a subset of ZxZ be given by A = {(x,y) | x,y in Z, min_x <= x <= max_x, min_y <= y <= max_y } Let u1, u2 be a basis for a sublattice of ZxZ, so that every point in the sublattice can be written as (x,y) = u u1 + v u2 for u, v in Z. I am interested in finding an efficient algorithm to visit every point of the sublattice that is also in A, each point being visited once only. I do have an algorithm to solve the problem, but IÕm sure there must be a more efficient one known, something similar to BresenhamÕs algorithm perhaps. Can anyone give me a link or reference to such an algorithm? TIA Chris === Subject: Re: lattice algorithm > Let a subset of ZxZ be given by > A = {(x,y) | x,y in Z, min_x <= x <= max_x, min_y <= y <= max_y } OK, so a rectangle per x,y parameterization. > Let u1, u2 be a basis for a sublattice of ZxZ, so that every point in > the sublattice can be written as (x,y) = u u1 + v u2 for u, v in Z. So a thin-filling (dithering) of the rectangle via a solid filling of a diamond shape per u,v parameterization. The obvious way is to find the bounding rectangle per u,v, and then perform a lexicographic scan of it, being smart about where to start and end to avoid lots of visits outside the diamond. But this isnÕt efficient unless u,v are nearly orthogonal, so I would suggest sliding u against v or vice versa to make the new uÕ,vas nearly orthogonal as possible, and then computing the bounding rectangle in uÕ,vand scanning in uÕ,vÕ. One way to do this is a 2-d version of the usual GCD algorithm: Find which of u or v is smallest in magnitude, and project the other on it to get the component in that direction, then integer-divide to find out how many copies of the small one can be subtracted/added from/to the larger one to make the large one become as small as possible, which will make it as orthogonal to the smaller vector as possible. If the remainder of the large one is now smaller than the former smallest one, reverse roles. Repeat until the integer quotient is zero. > something similar to BresenhamÕs algorithm perhaps. ThatÕs for visiting a single zigzag of lattice points to emulate a single straight line at some arbitrary angle. Since you want a solid diamond in u,v, not a single thin line, I donÕt see how the zigzag-line-drawing algorithm will help you. === Subject: Re: lattice algorithm > Let a subset of ZxZ be given by > A = {(x,y) | x,y in Z, min_x <= x <= max_x, min_y <= y <= max_y } > OK, so a rectangle per x,y parameterization. Indeed, but you canÕt be too careful in sci.math ! > Let u1, u2 be a basis for a sublattice of ZxZ, so that every point in > the sublattice can be written as (x,y) = u u1 + v u2 for u, v in Z. > So a thin-filling (dithering) of the rectangle via a solid filling of a > diamond shape per u,v parameterization. The obvious way is to find the > bounding rectangle per u,v, and then perform a lexicographic scan of > it, What do you mean by a lexicographic scan in this context? > being smart about where to start and end to avoid lots of visits > outside the diamond. Indeed, I want to avoid any visits outside the diamond if possible. Currently I calculate the min and max values for u from the definition of A, and then calculate the min and max values of v for each value of u, but this means a lot of (mainly) double precision integer divisions, which IÕd like to avoid if possible. (A lot because this process is repeated for many thousands of lattices) > But this isnÕt efficient unless u,v are nearly > orthogonal, so I would suggest sliding u against v or vice versa to > make the new uÕ,vas nearly orthogonal as possible, and then computing > the bounding rectangle in uÕ,vand scanning in uÕ,vÕ. One way to do > this is a 2-d version of the usual GCD algorithm: Find which of u or v > is smallest in magnitude, and project the other on it to get the > component in that direction, then integer-divide to find out how many > copies of the small one can be subtracted/added from/to the larger one > to make the large one become as small as possible, which will make it > as orthogonal to the smaller vector as possible. If the remainder of > the large one is now smaller than the former smallest one, reverse > roles. Repeat until the integer quotient is zero. I should have also stated that the basis vectors are already short, having been generated by this kind of lattice reduction technique. I also wasnÕt quite clear that I want to find the (u,v) values rather than the (x,y) values, and that IÕd prefer not to calculate the (x,y) values at all if possible. > something similar to BresenhamÕs algorithm perhaps. > ThatÕs for visiting a single zigzag of lattice points to emulate a > single straight line at some arbitrary angle. Since you want a solid > diamond in u,v, not a single thin line, I donÕt see how the > zigzag-line-drawing algorithm will help you. I was thinking that this problem is similar to the ßood-fill problem in computer graphics, and that could use Bresenham to find the boundary points. Or something like that.. Er.. Um.. Chris === Subject: Re: lattice algorithm Oops, I overlooked one of your questions: > What do you mean by a lexicographic scan in this context? Lexicographic scan is nothing more than the result of nested loops where the outermost loop does the leftmost variable and the next inner loop does the next-to-leftmost variable etc. In this case thereÕs just a nesting two-deep because you have only two variables. for u = ulow to uhigh for v = vlow to vhigh previsit(u,v) (ThatÕs vaguely algolish/pseudocode. I hate C/java syntax, and if I === Subject: Re: lattice algorithm > I want to avoid any visits outside the diamond if possible. Why? Is a visit outside the diamond extremely expensive, so that even one such visit dominates all the compute time of your process? (For example, a visit may mean doing a remote access to a WebPage, where only the WebPages within the diamond are valid, and attempt to connect to anything outside the diamond sits there for several minutes waiting for the host to respond before it finally times out, whereas each visit within the diamond responds within a fraction of a second most of the time so that a hundred inside visits consumes only a small fraction of a minute.) If thatÕs the case, all you have to do to speed it up is do a pre-visit that simply checks whether the point is inside the diamond or not, if so then do a real visit, if not then omit the real visit. Even if you have to use multi-precision exact integer (bignum) arithmetic to make the inside/outside decision, thatÕs just a tiny fraction of a second for each point, much faster to do that calculation than to sit for minutes waiting for a non-existant WebSite or whatever. > this process is repeated for many thousands of lattices In your original query, I assumed you wanted to process rather large diamonds, and not too many of them, so going to extra work to set up the scanning, so that the scan of the large diamond can be efficient, was worth the effort. But if youÕre dealing with a very large number of very tiny diamonds, like where the bounding rectangle is no larger than 3-by-3 most of the time and never much larger than that, and if outside-diamond visits cost no more than inside-diamond visits in fact maybe they actually cost slightly less, the overhead of the set-up is more than the total cost of outside-diamond visits, so thereÕs no need to do anything more than lexicographic scan of the bounding rectangle and test for inside/outside diamond at each point, itÕs best to avoid the overhead of the setup. In fact if the bounding rectangles are always very tiny, it may be best not to bother computing the bounding rectangle at all. Instead, use this algorithm: (1) Find the center of the x,y rectangle, map to u,v coordinates and round x,y to nearest integer each. (2) Start with that point and start a spiral path around it, keeping track on each once-around whether you hit any inside points or not. (3) So long as your last once-around path hit at least one inside point, repeat. When a complete once-around didnÕt hit even one inside point the whole trip, you know youÕre done. That method minimizes the setup to an absolute minimum, and for tiny rectangles/diamonds the total cost of the spiral visiting, even that last once-around that had 0% success rate, isnÕt a lot. > I also wasnÕt quite clear that I want to find the (u,v) values rather > than the (x,y) values, and that IÕd prefer not to calculate the (x,y) > values at all if possible. You really need to try it both ways and see which turns out faster for typical data. One way, compute actual diamond shape in u,v coordinates, which is more work of setup but 50% saving on visiting time. Other way, compute only the u,v bounding rectangle, which is less set-up time but not the 50% saving on visiting time. For large diamonds, extra setup is better, but for small diamonds extra setup costs more than it saves, and for some middle size both methods take the same time, the tradeoff crossover point. Only by actually timing your code both ways on typical data can you know which side of the tradeoff-point your data is. > I was thinking that this problem is similar to the ßood-fill problem > in computer graphics, and that could use Bresenham to find the > boundary points. Yes, if you carefully use FOUR instances of BresenhamÕs method, starting your scan for example from top scanline, hence using top-left and top-right bounds, and carefully noticing when youÕve reached left or right corner and have to switch from top-left to bottom-left or from top-right to bottom-right, and noticing when you reach the bottommost point when the two lines come together, you get a big lot of setup cost but very efficient innermost loop and not too bad middle loop. (That middle loop needs to make the decision when youÕve reached either of the left/right corners, and when youÕve reached bottom corner.) If the corners of the diamond had integer u,v values, all youÕd have to do is find an existing ßood-fill-polygon subroutine where the paint-pixel call was parameterized, and simply replace the paint-pixel call by whatever your visit-point function is, and call the subroutine given a four-point polygon which is your diamond shape. But since u,v values at corners are not integers, you may have to get the source of such a ßood-fill-polygon subroutine and modify it to deal with non-integer u,v values, i.e. initialize BresenhamÕs method for each edge of the polygon in an unsual way, but the rest of the ßood-fill algorithm would be the same. Also, since youÕre modifying the code, given each pair of opposite sides of diamond have the same slope, you can optimize out the four slope calculations to do only two and just copy the result from one side to the opposite site. === Subject: Re: lattice algorithm >> I want to avoid any visits outside the diamond if possible. > Why? Is a visit outside the diamond extremely expensive, so that even > one such visit dominates all the compute time of your process? ItÕs just wasted processing, which IÕd like to avoid, if possible. Also, the outside points canÕt be actually visited, because thereÕs no memory allocated for them, so I need to check whether a point is inside or outside. Ideally IÕd like an algorithm that guaranteed that all points are inside. >> this process is repeated for many thousands of lattices > In your original query, I assumed you wanted to process rather large > diamonds, and not too many of them, so going to extra work to set up > the scanning, so that the scan of the large diamond can be efficient, > was worth the effort. The size of the diamond varies from quite large (several hundred by several hundred) to quite small, so maybe I need to use a different algorithm depending on the size of the diamond. >> I was thinking that this problem is similar to the ßood-fill problem >> in computer graphics, and that could use Bresenham to find the >> boundary points. > Yes IÕve tried using Bresenham to find the lattice points near the edges of the diamond, and then sorting the resulting set of points - this then gives me bounds for the double loop, but I have to adjust the set of points since some are outside the diamond. Currently it doesnÕt give quite the same results as my original algorithm, so I think thereÕs still a bug in there somewhere. Also, IÕm using a generic winding number algorithm to check whether a point is inside the diamond, but I suspect thereÕs a more efficient algorithm for this, since a diamond is such a simple shape. Any pointers? Chris === Subject: dimension of free modules - looking for example somewhere (probably in LangÕs Algebra) I have read that there are rings R and free modules R^n and R^m which are isomorphic although n!=m. Does anybody know of an example? TIA, Tobias -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: dimension of free modules - looking for example > somewhere (probably in LangÕs Algebra) I have read that there are rings R > and free modules R^n and R^m which are isomorphic although n!=m. Does > anybody know of an example? > TIA, > Tobias Let A=Z x Z x Z x ... (countable product of Z) with pointwise addition and multiplication sirix. === Subject: Re: dimension of free modules - looking for example >> somewhere (probably in LangÕs Algebra) I have read that there are rings R >> and free modules R^n and R^m which are isomorphic although n!=m. Does >> anybody know of an example? >> TIA, >> Tobias > Let A=Z x Z x Z x ... (countable product of Z) with pointwise addition > and multiplication This is not an example: as A-modules A^n isomorphic to A^m implies n = m. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: dimension of free modules - looking for example >somewhere (probably in LangÕs Algebra) I have read that there are rings R >and free modules R^n and R^m which are isomorphic although n!=m. Does >anybody know of an example? >TIA, >Tobias >>Let A=Z x Z x Z x ... (countable product of Z) with pointwise addition >>and multiplication > This is not an example: > as A-modules A^n isomorphic to A^m implies n = m. Sorry, I thought Tobias wanted to have ring isomorphism... didnÕt read carefully enought. sirix. === Subject: Re: dimension of free modules - looking for example > somewhere (probably in LangÕs Algebra) I have read that there are rings R > and free modules R^n and R^m which are isomorphic although n!=m. Does > anybody know of an example? R = End_K(V) where V is an infinite-dimensional vector space over the field K. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: dimension of free modules - looking for example >> somewhere (probably in LangÕs Algebra) I have read that there are rings R >> and free modules R^n and R^m which are isomorphic although n!=m. Does >> anybody know of an example? > R = End_K(V) > where V is an infinite-dimensional vector space over the field K. Hm, a decomposition V = V_1 oplus V_2 with dim(V_1) = dim(V_2) = dim(V) induces linear maps R^2 --> R --> R^2 so that their composition is the identity. This gives rise to a split exact sequence so that R^2 is a direct summand of R. But are they isomorphic? Or is the direct summand free? -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Re: dimension of free modules - looking for example >I have read that there are rings R >and free modules R^n and R^m which are isomorphic although n!=m. Does >anybody know of an example? >>R = End_K(V) >>where V is an infinite-dimensional vector space over the field K. > But are they isomorphic? HereÕs my try - IÕve also seen it somewhere before but IÕm not sure whether solution is ok... Let V=V_1 oplus V_2, V_1~V_2~V (~ is isomorphism of vector spaces) Every fin R may be written as a direct sum of f_1in End(V_1,V_1),..., f_4in End(V_2, V_2). So we have (thatÕs for sure) homomorphism of abelian groups R -> R^4, which is (thatÕs hopefully) R-linear: isnÕt it true that f(a+b+c+d) = fa+fb+fc+fd (+ is direct sum)? Sorry if itÕs stupid. sirix. === Subject: Re: dimension of free modules - looking for example > Let V=V_1 oplus V_2, V_1~V_2~V (~ is isomorphism of vector spaces) > Every fin R may be written as a direct sum of f_1in End(V_1,V_1),..., > f_4in End(V_2, V_2). So we have (thatÕs for sure) homomorphism of > abelian groups R -> R^4, which is (thatÕs hopefully) R-linear: isnÕt it > true that f(a+b+c+d) = fa+fb+fc+fd (+ is direct sum)? Sorry if itÕs > stupid. Right, I didnÕt think about using f_2 in Hom(V_1,V_2) and f_3 in Hom(V_2,V_1). -- everyone who casts a shadow seems to stand in the sun reverse my forename for mail! - saibot === Subject: Rational sine For what all x values are Sin[x] and Cos[x] together rational? Is the solution set finite? Do only the ratios between Pythagorean triples as (m^2-n^2)/(m^2+n^2)and 2 m n /(m^2+n^2) from integers m,n determine rational Sin[x],Cos[x]? If not, how else? Am in doubt,TIA for clarifications. === Subject: Re: Rational sine A rational projective line pair transform is: integers n,m x=n/m-->{ 2*x/(1+x^2),(1-x^2)/(1+x^2)} Since the projective line is the real line to the circle and the rationals are a subset of the reals, it is simple to prove that the n^2+m^2=l^2 Pythagorean triples are projected by a sine and cosine as you suggest. A related topic is the Gauss map of z=x+i*y {xÕ,yÕ,zÕ}={2*x/(1+Abs[z]),2*y/ (1+Abs[z]),(1+Abs[z])/(1-Abs[z ])} If x=n/m and y=p/q, that projects four integers to a sphere in a similar manner. >For what all x values are Sin[x] and Cos[x] together rational? Is the >solution set finite? Do only the ratios between Pythagorean triples as >(m^2-n^2)/(m^2+n^2)and 2 m n /(m^2+n^2) from integers m,n determine >rational Sin[x],Cos[x]? If not, how else? Am in doubt,TIA for >clarifications. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: Rational sine > For what all x values are Sin[x] and Cos[x] together rational? The standard notations are sin(x) and cos(x). > Is the > solution set finite? No. Take any rational t and consider 2t/(1+t^2) and (1-t^2)/(1+t^2)/ > Do only the ratios between Pythagorean triples as > (m^2-n^2)/(m^2+n^2)and 2 m n /(m^2+n^2) from integers m,n determine > rational Sin[x],Cos[x]? If I understand your question aright, the answer is yes. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: The origin of quaternions IÕve read about how quaternions were invented by Hamilton. What did he find unsatisfactory about 3 dimensional numbers? Was it specifically that they canÕt be used for representing rotations in 3d space (IÕm not clear as to whether Hamilton had this purpose in mind at the time?) Or is there some more fundamental ßaw making it impossible to sensibly define +, * operators? G.A. === Subject: Re: The origin of quaternions > IÕve read about how quaternions were invented by Hamilton. What did he find > unsatisfactory about 3 dimensional numbers? > Was it specifically that they canÕt be used for representing rotations in 3d > space (IÕm not clear as to whether Hamilton had this purpose in mind at the > time?) Or is there some more fundamental ßaw making it impossible to > sensibly define +, * operators? > G.A. Plus is easy for three numbers (a,b,c) + (d,e,f,) = (a+d,b+e,c+f) is an example. Multiplication is harder depending on what you want it to do, for instance (a,b,c) *(d,e,f) = (0,0,0) is an example of a multiplication, but itÕs kind of useless. Another one is (a,b,c)*(d,e,f)=(a*b,b*e,c*f), but while it seems more useful, you are basically just doing addition and multiplication in three unrelated spaces, there is no rhyme or reason for making a combined object. So since there are many ways to define + and *, Hamilton had to decide what kind of properties he wanted. Properties similar to rotations are nice properties, but yes then you need four numbers to get nice formulas. === Subject: Re: The origin of quaternions > IÕve read about how quaternions were invented by Hamilton. What did he find > unsatisfactory about 3 dimensional numbers? > Was it specifically that they canÕt be used for representing rotations in 3d > space (IÕm not clear as to whether Hamilton had this purpose in mind at the > time?) Or is there some more fundamental ßaw making it impossible to > sensibly define +, * operators? .... Yes, there is. numbers on a plane. Mathematicians like generalizing things, so a natural question then was What about numbers represented in 3-dimensional space? This turned out to be difficult, and in particular Hamilton struggled with it for a long time before he reached the insight that 4 dimensions would work better than 3. Suppose you want to start with complex numbers x + iy and extend them to 3-dimensional numbers x + iy + jz (where x, y and z are real). Then the product ij must itself be some such number, say ij = a + ib + jc where a, b, c are some real numbers. Then i(ij) = i(a + ib + jc) = ia + (i^2)b + (ij)c = ia - b + (a + ib + jc)c = (ac - b) + i(a + bc) + j(c^2). But i(ij) = (i^2)j by the associative law = - j so (ac - b) + i(a + bc) + j(c^2) is the same number as - j. Equating the coefficients gives ac - b = 0 and a + bc = 0 and also c^2 = - 1, although c is supposed to be a real number. That may answer your question ... is there some more fundamental ßaw ...? Ken Pledger. === Subject: Re: The origin of quaternions >numbers on a plane. Mathematicians like generalizing things, so a >natural question then was What about numbers represented in >3-dimensional space? This turned out to be difficult, and in >particular Hamilton struggled with it for a long time before he reached >the insight that 4 dimensions would work better than 3. > Suppose you want to start with complex numbers x + iy and extend >them to 3-dimensional numbers x + iy + jz -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com A: Maybe because some people are too annoyed by top-posting. Q: Why do I not get an answer to my question(s)? A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? === Subject: Re: The origin of quaternions My understanding was that he was studying vector analysis and looking at A divid B for two vectors. ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). The consideration of A divid B , not as a vector but as an operation carrying a representative segment B into a coterminous representative segment of A, led Sir William Hamilton to the study of quaternions. The use of quaternions in a rotation of spherical triangle problem considerable simplified the math involved. >IÕve read about how quaternions were invented by Hamilton. What did he find >unsatisfactory about 3 dimensional numbers? >Was it specifically that they canÕt be used for representing rotations in 3d >space (IÕm not clear as to whether Hamilton had this purpose in mind at the >time?) Or is there some more fundamental ßaw making it impossible to >sensibly define +, * operators? >G.A. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: The origin of quaternions > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. > ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). > The consideration of A divid B , not as a vector but as an operation > carrying a representative segment B into > a coterminous representative segment of A, led Sir William Hamilton to > the study of quaternions..... This may be a misunderstanding. I remember that one of HamiltonÕs text-books does begin with the idea of dividing one vector by another, but that was written long after the original discovery/invention. Ken Pledger. === Subject: Re: The origin of quaternions - HamiltonÕs 1844 paper Perhaps the following is a relevant contribution to the newsgroup thread on the origin of quaternions. W.R.HamiltonÕs second paper on quaternions presented to the Royal Irish Academy: W.R.Hamilton: On quaternions, or on a new system of imaginaries in algebra; with some geometrical illustrations. Communicated November 11th 1844 to the Royal Irish Academy, Proc. Roy. Irish Acad. vol. 3 (1847), 1-16. If I remember well, it is in this paper that Hamilton tries to set up an algebra of proportionalities of spatial directions. In the horizontal plane one can say: North is to East as South is to West. In the meridional plane on can likewise say that Zenith is to South as Nadir is to North. These are geometrical rephrasings of properties of the complex number system. Hamilton did not succeed in obtaining a consistent algebra of spatial directions. Instead he found that when one more dimension to the three dimensions North-South, East-West, Zenith-Nadir, one indeed obtains a nice closed and consistent algebra of directions and their quotients. All this is actually a geometrical rephrasing of HamiltonÕs discovery (or invention?) of the quaternion algebra. H. knew that this fourth dimension can in no way be a spatial dimension. He imagined this dimension as laid out on a scale and coined the term scalar for this kind of numbers. H. was predominantly a phycisist and so identified this fourth dimension with time. To no avail, as would become apparent over the decades to come. In hindsight one may guess that HamiltonÕs motives for his quest were (1) to generalise the algebra of complex numbers to 3D, and to maintain in the process: (2) the concepts of quotient of directions and of quotient of vectors, and (3) the law of moduli of complex numbers, and of course (4) the often repeated questions at breakfast by his sons, aged six and eight in the autumn of 1843: Well, Papa, can you multiply triplets? (*) All this still does not answer the questions of exactly how H. got his brainwave to add a fourth dimension, and of why he got it then and there. My cherished speculation is that HamiltonÕs walk on October 16th 1843 from Dunsink to the city of Dublin helped to clear up his mind and open it up for a brainwave (a ßow of endorphin set in motion by enjoying a 5-mile walk in the cool autumn air). And then the brainwave will of course involve the subject closest at hand. (*) Se.87n OÕDonnell: William Rowan Hamilton - Portrait of a Prodigy. Boole Press Dublin, 1983, ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) Johan E. Mebius >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> > This may be a misunderstanding. I remember that one of HamiltonÕs >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. > Ken Pledger. === Subject: Re: The origin of quaternions - HamiltonÕs 1844 paper big snip In hindsight one may guess that HamiltonÕs motives for his quest were > (1) to generalise the algebra of complex numbers to 3D, and to maintain > in the process: (2) the concepts of quotient of directions and of > quotient of vectors, and (3) the law of moduli of complex numbers, and > of course (4) the often repeated questions at breakfast by his sons, > aged six and eight in the autumn of 1843: Well, Papa, can you multiply > triplets? (*) Well, I donÕt claim to be on the order of Hamilton but I do have a construction that extends complex properties to three dimensions and beyond. Please see polysigned numbers in sci.math. The solution is not orthogonal. It comes as a result the additive identity in higher signs. Complex numbers are equivalent to three-signed numbers. In four-signed numbers you can interpret them graphically as rays extending from the center of a tetrahedron to each corner. Labeling them -,+,*,# and enforcing the additive identity: - a + a * a # a = 0, where a is either an unsigned scalar or a four-signed value. Thence points in 3D space can be defined as sums in the four signs. The algebraic product rules follows: | - + * # ----------------- - | + * # - | + | * # - + | * | # - + * | # | - + * # This is simply a process of sign counting as the product of any two components are taken, wrapping the count at the highest sign. This obviously yields the rotational nature of the arithmetic. And so a general 3D product like: ( - a + b * c # d )( - e + f * g # h ) yields: + ae * af # ag - ah * be # bf - bg + bh # ce - cf + cg * ch - de + df * dg # dh This procedure in three-signed math yields the complex numbers. This procedure in two-signed math yields the real numbers. All of these can be looked at as number-line style algebra where each sign has a ray emanating from an origin. Taking the step of graphical interpretation or dimensional analysis shows that an n-signed system has dimension n - 1. This is easily seen right from the additive identity. All the commutative and associative properties that apply to the reals and complex numbers also work in higher signs, or higher dimensions. This approach is much simpler than queternions. -Tim > All this still does not answer the questions of exactly how H. got his > brainwave to add a fourth dimension, and of why he got it then and > there. My cherished speculation is that HamiltonÕs walk on October 16th > 1843 from Dunsink to the city of Dublin helped to clear up his mind and > open it up for a brainwave (a ßow of endorphin set in motion by > enjoying a 5-mile walk in the cool autumn air). And then the brainwave > will of course involve the subject closest at hand. > (*) Se.87n OÕDonnell: William Rowan Hamilton - Portrait of a Prodigy. > Boole Press Dublin, 1983, > ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) > Johan E. Mebius > >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> > This may be a misunderstanding. I remember that one of HamiltonÕs >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. > Ken Pledger. > === Subject: Re: The origin of quaternions - HamiltonÕs 1844 paper > big snip In hindsight one may guess that HamiltonÕs motives for his quest were > (1) to generalise the algebra of complex numbers to 3D, and to maintain > in the process: (2) the concepts of quotient of directions and of > quotient of vectors, and (3) the law of moduli of complex numbers, and > of course (4) the often repeated questions at breakfast by his sons, > aged six and eight in the autumn of 1843: Well, Papa, can you multiply > triplets? (*) > Well, I donÕt claim to be on the order of Hamilton but I do have a > construction that extends complex properties to three dimensions and > beyond. Please see polysigned numbers in sci.math. > The solution is not orthogonal. It comes as a result the additive > identity in higher signs. > Complex numbers are equivalent to three-signed numbers. > In four-signed numbers you can interpret them graphically as rays > extending from the center of a tetrahedron to each corner. Labeling > them -,+,*,# and enforcing the additive identity: > - a + a * a # a = 0, where a is either an unsigned scalar or a > four-signed value. > Thence points in 3D space can be defined as sums in the four signs. > The algebraic product rules follows: Funny, When I view this table in the original it is all screwed up, but now it looks good again. Put some spaces on the end just in case. > | - + * # > ----------------- > - | + * # - > | > + | * # - + > | > * | # - + * > | > # | - + * # > This is simply a process of sign counting as the product of any two > components are taken, wrapping the count at the highest sign. This > obviously yields the rotational nature of the arithmetic. > And so a general 3D product like: > ( - a + b * c # d )( - e + f * g # h ) > yields: > + ae * af # ag - ah * be # bf - bg + bh # ce - cf + cg * ch - de + df > * dg # dh > This procedure in three-signed math yields the complex numbers. > This procedure in two-signed math yields the real numbers. > All of these can be looked at as number-line style algebra where each > sign has a ray emanating from an origin. Taking the step of graphical > interpretation or dimensional analysis shows that an n-signed system > has dimension n - 1. This is easily seen right from the additive > identity. All the commutative and associative properties that apply to > the reals and complex numbers also work in higher signs, or higher > dimensions. > This approach is much simpler than queternions. > -Tim > All this still does not answer the questions of exactly how H. got his > brainwave to add a fourth dimension, and of why he got it then and > there. My cherished speculation is that HamiltonÕs walk on October 16th > 1843 from Dunsink to the city of Dublin helped to clear up his mind and > open it up for a brainwave (a ßow of endorphin set in motion by > enjoying a 5-mile walk in the cool autumn air). And then the brainwave > will of course involve the subject closest at hand. (*) Se.87n OÕDonnell: William Rowan Hamilton - Portrait of a Prodigy. > Boole Press Dublin, 1983, > ISBN 0-906783-06-2 (hc) and 0-906783-15-1 (pbk) Johan E. Mebius > >My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions..... >> > This may be a misunderstanding. I remember that one of HamiltonÕs >text-books does begin with the idea of dividing one vector by another, >but that was written long after the original discovery/invention. Ken Pledger. > > === Subject: Re: The origin of quaternions > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. > ( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). > The consideration of A divid B , not as a vector but as an operation > carrying a representative segment B into > a coterminous representative segment of A, led Sir William Hamilton to > the study of quaternions. > The use of quaternions in a rotation of spherical triangle problem > considerable simplified the math involved. Interesting, Roger. So perhaps Hamilton was aware that he needed a system with a non-commutative multiplication from the outset? ShouldÕve narrowed his search a bit. I understand youÕre an aficionado of fractals. Quaternion fractals are very nice, but I feel that a dimensionÕs being wasted. Did you ever wonder if you could define a 3d number system (I assume you only need +, * and magnitude operators) for drawing Julia sets etc? I donÕt have any idea if the impossibility of making a division ring implies that you couldnÕt create a system that makes nice Julia sets. cheers, G.A. === Subject: Re: The origin of quaternions Terry Gintz is the expert on quaternion and octonion iterative functions, but IÕve done chaos differential equation solutions based on quaternions. The Euler rotation is what you are looking for ( based on so(3)/ SO(3)). It leaves out the fourth part. >>My understanding was that he was studying vector analysis >> and looking at A divid B for two vectors. >>( ref. page 15 Theoretical Mexhanics, Ames and Murnaghan, Dover,1957). >> The consideration of A divid B , not as a vector but as an operation >>carrying a representative segment B into >>a coterminous representative segment of A, led Sir William Hamilton to >>the study of quaternions. >>The use of quaternions in a rotation of spherical triangle problem >>considerable simplified the math involved. >> >Interesting, Roger. So perhaps Hamilton was aware that he needed a system >with a non-commutative multiplication from the outset? ShouldÕve narrowed >his search a bit. >I understand youÕre an aficionado of fractals. Quaternion fractals are very >nice, but I feel that a dimensionÕs being wasted. Did you ever wonder if >you could define a 3d number system (I assume you only need +, * and >magnitude operators) for drawing Julia sets etc? I donÕt have any idea if >the impossibility of making a division ring implies that you couldnÕt create >a system that makes nice Julia sets. >cheers, >G.A. -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: The origin of quaternions itÕs not clear to me that there really was 3d numbers, since a sufficient notation didnÕt exist for its arithmetic, other than *maybe* simple addition of vectors.... that may be why some folks get all het-up on grassmannians. if you look at GibbÕs cross & inner products, they are directly adapted from HamiltonÕs terms; quaterions solve both siimultaneously. maybe this is part of the reason that Fermat went a further with linear algebra than did Descartes, but especially in the plane. you can see that most purely directional operations are done on the unit sphere, using quaternions with real part being zero; every two rotations can be achieved with one rotation; with two purely imaginary quaternions ... the product is, or is not neccesarily purely imaginary? > My understanding was that he was studying vector analysis > and looking at A divid B for two vectors. >Was it specifically that they canÕt be used for representing rotations in 3d >space (IÕm not clear as to whether Hamilton had this purpose in mind at the >time?) Or is there some more fundamental ßaw making it impossible to >sensibly define +, * operators? --Advice 0.05; Free, if wrong, again! http://tarpley.net/bush6.htm === Subject: Re: The origin of quaternions > IÕve read about how quaternions were invented by Hamilton. What did he find > unsatisfactory about 3 dimensional numbers? If you meant to ask Why did he not define a product in R^3 that, together with the usual sum, would turn it into a division ring?, then the answer is because thatÕs not possible. Jose Carlos Santos === Subject: Re: The origin of quaternions > IÕve read about how quaternions were invented by Hamilton. What did he > find unsatisfactory about 3 dimensional numbers? What are these 3 dimensional numbers of which you speak? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: CantorÕs diagonal proof wrong? at 01:34 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >What does distinguish the limit from the diagonal? What distinguishes a banana from a townhouse? TheyÕre not remotely similar. The diagonal is a sequence of digits, not a number. In the case of the Cantor diagonal argument youÕre referring to, you take the sequence of digits as coefficients in a series and it is trivial to prove that the series converges. >Do we need different words here? We need words that are applicable, and we need to ensure that we have a common understanding of what they refer to. In particular, the term limit has a precise meaning. >However, how, then, can Cantor change all the digits of this >limit? He doesnÕt change anything. He defines a new number in terms of a sequence of representations of numbers. >And why canÕt we consider my proof in the limit? What proof? >So donÕt stop, but consider the limit of my proof. What limit? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? > at 01:34 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >What does distinguish the limit from the diagonal? > What distinguishes a banana from a townhouse? TheyÕre not remotely > similar. > The diagonal is a sequence of digits, not a number. In the case of the > Cantor diagonal argument youÕre referring to, you take the sequence of > digits as coefficients in a series and it is trivial to prove that the > series converges. Ok, the abbreviation diagonal should expand to diagonal number. >Do we need different words here? > We need words that are applicable, and we need to ensure that we have > a common understanding of what they refer to. In particular, the term > limit has a precise meaning. >However, how, then, can Cantor change all the digits of this >limit? > He doesnÕt change anything. He defines a new number in terms of a > sequence of representations of numbers. He defines it by changing the digits a_nn of the diagonal number D to aÕ_nn of the new number DÕ, because he must make sure that in any case aÕ_nn /= a_nn. >And why canÕt we consider my proof in the limit? > What proof? I have definined a Cantor-list, which always contains the diagonal number D_n constructed up to line n in line Z(n+1) by construction. I found this very same list also appearing in this thread: 0.000... 0.100... 0.11000... 0.111000... ... Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). We see that either of the two statements: A) Cantors changed diagonal number differs from every real in a line not A) Cantors diagonal number does not differ from every real in a line can be taken for granted. There is no logical priority in favour of A or not A, as long as all lines are enumerated by natural, hence finite numbers. === Subject: Re: CantorÕs diagonal proof wrong? > 0.000... > 0.100... > 0.11000... > 0.111000... > ... > Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). > We see that either of the two statements: > A) Cantors changed diagonal number differs from every real in a line > not A) Cantors diagonal number does not differ from every real in a > line > can be taken for granted. > If can be taken for granted is supposed to mean has to be true, > and line is supposed to mean the number indicated by a line. > There is no logical priority in favour of A or not A, as long as all > lines are enumerated by natural, hence finite numbers. > Uh, there is a proof for A. ThatÕs quite a logical priority. It does > not get better than that. This proof is necessarily wrong, if A is assumed. Hence, there is no logical priority. === Subject: Re: CantorÕs diagonal proof wrong? <41aa5d5a$14$fuzhry+tra$mr2ice@news.patriot.net> at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >I have definined a Cantor-list, which always contains the diagonal >number D_n constructed up to line n None of those is the number that Cantor defines. >We see that either of the two statements: >A) Cantors changed diagonal number differs from every real in a line >not A) Cantors diagonal number does not differ from every real in a >line can be taken for granted. What does that have to do with CantorÕs proofs? The issue is not what appears in a line but rather what appears in a list. >There is no logical priority in favour of A or not A, as long as all >lines are enumerated by natural, hence finite numbers. Which is, of course, impossible, as Cantor proved. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? > at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >I have definined a Cantor-list, which always contains the diagonal >number D_n constructed up to line n > None of those is the number that Cantor defines. Then he must include numbers which cannot be represented by n. ThatÕs not what he pretended to do. >We see that either of the two statements: >A) Cantors changed diagonal number differs from every real in a line >not A) Cantors diagonal number does not differ from every real in a >line can be taken for granted. > What does that have to do with CantorÕs proofs? The issue is not what > appears in a line but rather what appears in a list enumerated by solely natural numbers. >There is no logical priority in favour of A or not A, as long as all >lines are enumerated by natural, hence finite numbers. > Which is, of course, impossible, as Cantor proved. Pardon, Cantor was talking about positive finite (today we say: natural) numers. If you think, that then his list would be impossible, we do agree. === Subject: Re: CantorÕs diagonal proof wrong? <41aa5d5a$14$fuzhry+tra$mr2ice@news.patriot.net> <41ae4af6$6$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> at 10:24 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >>I have definined a Cantor-list, which always contains the diagonal >>number D_n constructed up to line n >> None of those is the number that Cantor defines. > Then he must include numbers which cannot be represented by > n. ThatÕs not what he pretended to do. Wrong. D is not part of the list, it is merely derived from it. And it is not equal to any D_n. Because for any given n, you will find a non-zero difference |D-D_n|. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? > at 01:34 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: > >What does distinguish the limit from the diagonal? What distinguishes a banana from a townhouse? TheyÕre not remotely > similar. The diagonal is a sequence of digits, not a number. In the case of the > Cantor diagonal argument youÕre referring to, you take the sequence of > digits as coefficients in a series and it is trivial to prove that the > series converges. > Ok, the abbreviation diagonal should expand to diagonal number. > >Do we need different words here? We need words that are applicable, and we need to ensure that we have > a common understanding of what they refer to. In particular, the term > limit has a precise meaning. > >However, how, then, can Cantor change all the digits of this >limit? He doesnÕt change anything. He defines a new number in terms of a > sequence of representations of numbers. > He defines it by changing the digits a_nn of the diagonal number D to > aÕ_nn of the new number DÕ, because he must make sure that in any case > aÕ_nn /= a_nn. >And why canÕt we consider my proof in the limit? What proof? > I have definined a Cantor-list, which always contains the diagonal > number D_n constructed up to line n in line Z(n+1) by construction. I > found this very same list also appearing in this thread: > 0.000... > 0.100... > 0.11000... > 0.111000... > ... > Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). > We see that either of the two statements: > A) Cantors changed diagonal number differs from every real in a line > not A) Cantors diagonal number does not differ from every real in a > line > can be taken for granted. There is no logical priority in favour of A > or not A, as long as all lines are enumerated by natural, hence finite > numbers. Why is I) D_k =/= 0 for every k II) for every Z_n: not ( Z_n,k =/= 0 for every k ) or, denoting E the property of a sequence to have *all* of its elements =/= 0 I) D has the property E II) no Z_n has the property E *not* a logical priority for ( D differs from every Z_n ) ? Just *because* all lines are enumerated by natural numbers from which follows that there is no other candidate for equality than *numbered* lines, and because a sequence can only be equal to the sequence D if it *has* got the property E. -- Horst === Subject: Re: CantorÕs diagonal proof wrong? > at 01:34 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: > >What does distinguish the limit from the diagonal? What distinguishes a banana from a townhouse? TheyÕre not remotely > similar. The diagonal is a sequence of digits, not a number. In the case of the > Cantor diagonal argument youÕre referring to, you take the sequence of > digits as coefficients in a series and it is trivial to prove that the > series converges. Ok, the abbreviation diagonal should expand to diagonal number. > >Do we need different words here? We need words that are applicable, and we need to ensure that we have > a common understanding of what they refer to. In particular, the term > limit has a precise meaning. > >However, how, then, can Cantor change all the digits of this >limit? He doesnÕt change anything. He defines a new number in terms of a > sequence of representations of numbers. He defines it by changing the digits a_nn of the diagonal number D to > aÕ_nn of the new number DÕ, because he must make sure that in any case > aÕ_nn /= a_nn. > >And why canÕt we consider my proof in the limit? What proof? I have definined a Cantor-list, which always contains the diagonal > number D_n constructed up to line n in line Z(n+1) by construction. I > found this very same list also appearing in this thread: 0.000... > 0.100... > 0.11000... > 0.111000... > ... Changing the diagonal elements 0 -> 1, we have D_n = Z(n+1). We see that either of the two statements: > A) Cantors changed diagonal number differs from every real in a line > not A) Cantors diagonal number does not differ from every real in a > line > can be taken for granted. There is no logical priority in favour of A > or not A, as long as all lines are enumerated by natural, hence finite > numbers. > Why is > I) D_k =/= 0 for every k > II) for every Z_n: > not ( Z_n,k =/= 0 for every k ) > or, denoting E the property of a sequence to have *all* of its > elements =/= 0 > I) D has the property E > II) no Z_n has the property E > *not* a logical priority for ( D differs from every Z_n ) ? Just > *because* all lines are enumerated by natural numbers from which > follows that there is no other candidate for equality than *numbered* > lines, and because a sequence can only be equal to the sequence D if > it *has* got the property E. Your proof shows, that D does not completely consist of elements stemming of lines which are enumerated by natural numbers, because any D_n consisting of elements a_nn (with finite n) does contain only a finite number of elements equal 1 (and following zeros). But Cantor assumed his list to consist only of lines enumerated by finite numbers n. === Subject: Re: CantorÕs diagonal proof wrong? at 01:26 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >Your ideas are perfectly reasonable, if you add the remark that an >infinite sequence of digits (which would not form a natural number) >could only emerge from a list with at least one line enumerated by >omega or infinity. No. A sequence is a mapping from N to a set, and N has order type Omega, not Omega plus 1. >In particular actual infinity, the same as finished infinity, is >obviously a contradiction. No. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? > at 01:26 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >Your ideas are perfectly reasonable, if you add the remark that an >infinite sequence of digits (which would not form a natural number) >could only emerge from a list with at least one line enumerated by >omega or infinity. > No. A sequence is a mapping from N to a set, and N has order type > Omega, not Omega plus 1. It is not a matter of order type, but a matter of finity. Each n is finite. Each line-number of Cantors list is finite. Otherwise we could not find and change the diagonal element a_nn. As long as we are capable of doing so, the number of exchanged digits is finite. The infinite set of finite numbers is a contradicito in adjecto. >In particular actual infinity, the same as finished infinity, is >obviously a contradiction. > No. Yes = No to No. === Subject: Re: CantorÕs diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> at 01:26 PM, mueckenh@rz.fh-augsburg.de (W. Mueckenheim) said: >>Your ideas are perfectly reasonable, if you add the remark that an >>infinite sequence of digits (which would not form a natural number) >>could only emerge from a list with at least one line enumerated by >>omega or infinity. >> No. A sequence is a mapping from N to a set, and N has order type >> Omega, not Omega plus 1. > It is not a matter of order type, but a matter of finity. Each n is > finite. Each line-number of Cantors list is finite. Otherwise we could > not find and change the diagonal element a_nn. As long as we are > capable of doing so, the number of exchanged digits is finite. Wrong. There is an infinite number of finite numbers. > The infinite set of finite numbers is a contradicito in adjecto. Your saying so does not make it so. Not even when said in a fancy way. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? >>Your ideas are perfectly reasonable, if you add the remark that an >>infinite sequence of digits (which would not form a natural number) >>could only emerge from a list with at least one line enumerated by >>omega or infinity. >> >> No. A sequence is a mapping from N to a set, and N has order type >> Omega, not Omega plus 1. > It is not a matter of order type, but a matter of finity. Each n is > finite. Each line-number of Cantors list is finite. Otherwise we could > not find and change the diagonal element a_nn. As long as we are > capable of doing so, the number of exchanged digits is finite. > Wrong. There is an infinite number of finite numbers. In terms of potential infinity, of course. But that does not change anything. Potential infinity means, that the sequence of natural numbers does never end, but that each one is finite. With each finite number n a finite sequence 1,2,3,...,n is connected (i.e. a bijection is possible), and the cardinal number of each of such sequences remains finite too. > The infinite set of finite numbers is a contradicito in adjecto. > Your saying so does not make it so. Not even when said in a fancy > way. If you insist on your opinion, then tell me what is wrong with my proof: For n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, and, in particular with its inversion: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a natural number. === Subject: Re: CantorÕs diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> It is not a matter of order type, but a matter of finity. Each n is >> finite. Each line-number of Cantors list is finite. Otherwise we could >> not find and change the diagonal element a_nn. As long as we are >> capable of doing so, the number of exchanged digits is finite. >> Wrong. There is an infinite number of finite numbers. > In terms of potential infinity, of course. No. The Peano axioms donÕt talk about processes or potential. They state properties that make a number belong to the set of naturals. An infinite set is one that can be placed into bijection with a proper subset of it, and the successor operation is a trivial bijection of the set of naturals onto a subset of it. > But that does not change anything. Potential infinity means, that > the sequence of natural numbers does never end, but that each one is > finite. Completely irrelevant. We are not interested in any fuzzy pseudophilosophical word plays. > With each finite number n a finite sequence 1,2,3,...,n is connected > (i.e. a bijection is possible), and the cardinal number of each of > such sequences remains finite too. Sure. But that tells you nothing about whether a union of an infinite number of sets with those properties still has the same properties. Because if you use induction for that proof, it is valid only for forming the union of a finite number of sets. >> The infinite set of finite numbers is a contradicito in adjecto. >> Your saying so does not make it so. Not even when said in a fancy >> way. > If you insist on your opinion, then tell me what is wrong with my > proof: For n of IN there are elements of {2, 4, 6, ..., 2n} > surpassing its cardinal number, No problem with that. > and, in particular with its inversion: If there are no elements of > {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a > natural number. This is a heap of babbling hogwash, since it is not an inversion, but intended to be an equivalent statement. And the equivalent statement would be: There exists no n of IN such that no element of {2, 4, 6 ... 2n} surpasses the setÕs cardinal number. And indeed, that is true. But it tells us nothing about the set of all even numbers, since the set of all even numbers is not of the form {2, 4, 6 ... 2n} (with n in N). And since this set is not of the form for which you prove something by induction, your proof does not apply to it. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? <41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net> >It is not a matter of order type, but a matter of finity. Your at least one line enumerated by omega or infinity would seem to rule out finitude. Adding a terminal element to an infinite ordered set creates an ordered set with a different order type. >It is not a matter of order type, but a matter of finity. Each n is >finite. Each line-number of Cantors list is finite. Otherwise we >could not find and change the diagonal element a_nn. The list, however, is infinite. >As long as we are capable of doing so, the number of exchanged > digits is finite. The Cantor proof is not a construction or process. There is no sequencing or temporality in it. It uses a purely functional definition of the antidiagonal number in terms of the numbers on the list. >The infinite set of finite numbers is a contradicito in adjecto. No. There is no contradiction, because there are an infinit number of finite numbers. >Yes = No to No. If there is a contradiction in finished infinity then produce it with a logical argument rather than handwaving. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? >It is not a matter of order type, but a matter of finity. > Your at least one line enumerated by omega or infinity would seem to > rule out finitude. Adding a terminal element to an infinite ordered > set creates an ordered set with a different order type. It need not be a single element, it could be a set of infinitely many omegas. That, however, is irrelevant, because we are talking about the complete set IN of natural numbers n only. >It is not a matter of order type, but a matter of finity. Each n is >finite. Each line-number of Cantors list is finite. Otherwise we >could not find and change the diagonal element a_nn. > The list, however, is infinite. Every natural number is an ordinal number and is simultaneously the cardinal number of the sequence of all its successors including itself: 1,2,3,...,n. The sequence up to a natural number n can never have the cardinal number aleph_0. But IN does not contain other than natural numbers. >As long as we are capable of doing so, the number of exchanged > digits is finite. > The Cantor proof is not a construction or process. There is no > sequencing or temporality in it. It uses a purely functional > definition of the antidiagonal number in terms of the numbers on the > list. There need not be any sequencing or any temporality, but in order to identify the digit a_nn to be exchanged, both must be identified at first, n and a_nn. As long as n is natural, the sequence of lines has a finite cardinal number. But others than lines enumerated by natural numbers could not be identified. >The infinite set of finite numbers is a contradicito in adjecto. > No. There is no contradiction, because there are an infinit number of > finite numbers. I proved that for n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number. The true inversion of this statement is: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, than n is not a natural number. >Yes = No to No. > If there is a contradiction in finished infinity then produce it > with a logical argument rather than handwaving. Infinity is nothing else but the latin translation of unfinished or even unfinishable. Finished unfinished or finished unfinishable is a contradiction. Finished infinity means actual infinity like the infinite set of the finite natural numbers, which is self-contradictory (see above). === Subject: Re: CantorÕs diagonal proof wrong? >An infinite table of reals would seem to have an infinite number of >reals of infinite length. That would be seem to be of size oo * oo >instead of 2^oo like P(N). YouÕre using oo as though it were the name of a number; it isnÕt. The symbol oo only has meaning as part of a phrase that has been defined, e.g., in y= lim 1=0 oo x_i. You have not given a definition for your usage of oo, and your usage does not match any of the conventional definitions. Note that Set Theory has names like Aleph Null and Aleph 1 for distinct cardinal numbers; oo is not a symbol defined in Set Theory. If youÕre going to talk about size oo, oo*oo or 2^oo then itÕs your responsibility to define what those mean. After that you can justify, or fail to, any claims you make about them, but until then itÕs not even wrong. Plus, of course, reals donÕt have lengths, although representations of them do. >But we can easily create a 1 to 1 mapping of a table with the >natural numbers by numbering the rows. And CantorÕs proof tells us >that the table must be missing values. No. >So this seems to imply that a set of size oo*oo >has cardinality of N and not R. A countable union of countable sets is countable, but NxN is not the same as 2^N. >But it is still curious that it happens at 2^oo and not oo^oo. Why? Is it curious that 3*3 is smaller than 3^3, or that 16*16 is smaller than 2^16? -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? In sci.math, Shmuel (Seymour J.) Metz <41a968d3$6$fuzhry+tra$mr2ice@news.patriot.net>: >>An infinite table of reals would seem to have an infinite number of >>reals of infinite length. That would be seem to be of size oo * oo >>instead of 2^oo like P(N). > YouÕre using oo as though it were the name of a number; it isnÕt. A number is an abstract concept. So is oo, although one could try to do cute things like Russelize it (e.g., a number is a level 1 concept -- a, say, bird would be level 0 as one can hold onto a bird -- and infinity is a level 2 concept, as itÕs not a number). Of course, this doesnÕt mean oo * oo is a well-defined concept in its own right -- and one has to make a number of assumptions here to even fit it into the integers (itÕs not clear, for instance, that oo is required to equal aleph_0). > The symbol oo only has meaning as part of a phrase that has been > defined, e.g., in y= lim 1=0 oo x_i. IÕm assuming y = lim (i -> oo) x_i here. Your notation got a little mangled over here... :-) > You have not given a definition > for your usage of oo, and your usage does not match any of the > conventional definitions. Note that Set Theory has names like Aleph > Null and Aleph 1 for distinct cardinal numbers; oo is not a symbol > defined in Set Theory. If youÕre going to talk about size oo, oo*oo or > 2^oo then itÕs your responsibility to define what those mean. After > that you can justify, or fail to, any claims you make about them, but > until then itÕs not even wrong. Plus, of course, reals donÕt have > lengths, although representations of them do. There is a mapping between positive real numbers and a length, given a unit length -- but thatÕs about it. Negative real numbers can be mapped to lengths if one specifies a direction as well. Complex numbers can be mapped to a point on a Euclidean plane. OTOH, this is an externally-defined metric, not one that is inherently, erm, real... ;-) >>But we can easily create a 1 to 1 mapping of a table with the >>natural numbers by numbering the rows. And CantorÕs proof tells us >>that the table must be missing values. > No. This specification is so confused IÕm having trouble understanding it in its original form, but CantorÕs proof is fairly simple (in fact, he has two; IÕm assumimg the second here). Assume a denumerable list of real numbers L (with, naturally, a bijective mapping from N to elements of L), where each number on the list (for simplicity) is between 0 and 1 inclusive. Then construct a number d between 0 and 1 that is guaranteed to not be on that list. In fact, one can construct an uncountable number of such numbers d (assume another list d(L) which has diagonal numbers, then construct a number dthat is not on *that* list) and this extension could go on indefinitely, though one might have to take more than one digit in the diagonal construction in later iterations -- one way to resolve this is to note that the naturals and the rationals have Lebesgue measure 0 over the reals using the standard metric. >>So this seems to imply that a set of size oo*oo >>has cardinality of N and not R. > A countable union of countable sets is countable, but NxN is not the > same as 2^N. I think Cantor proved that, too. :-) >>But it is still curious that it happens at 2^oo and not oo^oo. > Why? Is it curious that 3*3 is smaller than 3^3, or that 16*16 is > smaller than 2^16? I suspect that, if a set S has cardinality of at least card(N), then a bijective mapping can be found between the elements of 2^S and the elements of n^S, where n > 1 is an integer. However, IÕd have to look. -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: Re: CantorÕs diagonal proof wrong? > However, how, then, can Cantor change all the digits of this > limit? > He changes the digits of the sequence, resulting in a different value > as the indicated limit. Sorry, but you are mistaken! Cantor does not (only) intend to show that the exchanged diagonal differs from the indicated limit, i. e. the former value of the diagonal, but he has to show that the exchanged diagonal is in fact different from any real number given in his original Cantor-list. And that proof is valid only, if the exchanged diagonal differs from any real number of the original Cantor-list by at least one digit. Any consideration of an indicated limit is completely irrelevant in this respect. > And why canÕt we consider my proof in the limit? > You are again trying to befuddle people by word games. considering a > proof in the limit is a completely meaningless and nonsensical > expression. > If we try to turn your word games into something making even remotely > sense, the question might have been Why does my proof that is valid > for every element of a sequence say nothing about the limit of the > sequence?. And the answer is, of course, that the limit is not a > member of the sequence. As I tried to express above already, CantorÕs proof has nothing to do with any limit, but only with one element a_nn for each number Z(n) given in line number n. Even if we wanted to consider the limit as > some weird element with index omega or whatever, for whatever crazy > reason, your proof is an induction proof, and induction makes a > statement only for all elements of N, and such a hypothetical omega > could not be a member of N. Of course, but if my proof should break down, it would do so only for some miraculous omega. It remains valid for every natural n and every number Z(n) in line number n. > So whatever one likes to call the limit of the sequence of truncated > decimals of D, your proof does not extend to it. I didnÕt intend that, because Cantor didnÕt too. My intention is to contradict Cantor, and not some miracle created by others. >> That the value specified by the diagonal (which is basically the >> limit of the corresponding series) canÕt be completely specified if >> you stop at any digit. > So donÕt stop, but consider the limit of my proof. > Proofs have no limit. This is nonsensical. They have validity. > Your proof is valid for all numbers corresponding to a truncated > diagonal. The number corresponding to the entire diagonal is not > covered by it. Which number is that? Please give a plain answer. Has Cantor to be satisfied with a truncated diagonal too? Or how can he exchange further elements not covered by my numbers. === Subject: Re: CantorÕs diagonal proof wrong? > Sorry, but you are mistaken! Cantor does not (only) intend to show > that the exchanged diagonal differs from the indicated limit, i. e. > the former value of the diagonal, but he has to show that the > exchanged diagonal is in fact different from any real number given in > his original Cantor-list. And that proof is valid only, if the > exchanged diagonal differs from any real number of the original > Cantor-list by at least one digit. Any consideration of an indicated > limit is completely irrelevant in this respect. > And it does because by definition the exchanged diagonal differs > from *every* real in the list by at least one position. > It differs from real[k] in position k, for every k. because it is > constructed like this. And as there are no reals in the list but those > who have a number k, it differs from *every* real in the list. Your arguing is true only for finite sequences D_k of the diagonal. The number of digits of the complete diagonal D is infinite (aleph_0), hence complete induction and your proof (or construction) fail. If you donÕt accept this reasoning (I wouldnÕt do either), then you must accept the reasoning below. > If you are still assuming that there might be a real in the list from > which it does not differ you are denying that > It differs from every single real in the list > is logically equivalent to > There is no real in the list from which it does not differ > is logically equivalent to > There is no real in the list which is equal to it > Which of the two assumed logical equivalences do you not accept ? All three forms are equipollent, but wrong in case of the following list: 0.0 0.1 0.11 0.111 The diagonal squence D_n constructed by aÕ_nn = 1 + a_nn is always contained in line Z(n+1). This is correct for every finite n by construction of the matrix. This fact does never change. Therefore it is true for the whole diagonal D, unless it contains elements with non-natural numbers as indices. You see, both statements differs by definition and doesnÕt differ by definition can be applied with exactly the same justification. There is not unique resolution. === Subject: Re: CantorÕs diagonal proof wrong? >Your arguing is true only for finite sequences D_k of the diagonal. No. >The number of digits of the complete diagonal D is infinite >(aleph_0), hence complete induction and your proof (or construction) >fail. ThereÕs no induction to fail. >All three forms are equipollent, but wrong in case of the following >list: No. >The diagonal squence D_n The Cantor antidiagonal proof does not construct a sequence of real num bers, it constructsw a single real number. Your D_n has nothing to do with CantorÕs proofs. >This fact does never change. Therefore it is true for the whole >diagonal D, No, because the antidiagonal real is not any of your D_n. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: CantorÕs diagonal proof wrong? >Your arguing is true only for finite sequences D_k of the diagonal. > No. >The number of digits of the complete diagonal D is infinite >(aleph_0), hence complete induction and your proof (or construction) >fail. > ThereÕs no induction to fail. >All three forms are equipollent, but wrong in case of the following >list: > No. >The diagonal squence D_n > The Cantor antidiagonal proof does not construct a sequence of real > num bers, it constructsw a single real number. Your D_n has nothing to > do with CantorÕs proofs. >This fact does never change. Therefore it is true for the whole >diagonal D, > No, because the antidiagonal real is not any of your D_n. If you accept that the whole set of natural numbers is created by the Peano axioms, in particular the second, namely n has a successor n + 1, but if you simultaneously deny that induction is capable of reaching all natural numbers (all lines of the list), then we are talking about mathemagics, but not about mathematics. === Subject: Re: CantorÕs diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Sorry, but you are mistaken! Cantor does not (only) intend to show >> that the exchanged diagonal differs from the indicated limit, i. e. >> the former value of the diagonal, but he has to show that the >> exchanged diagonal is in fact different from any real number given in >> his original Cantor-list. And that proof is valid only, if the >> exchanged diagonal differs from any real number of the original >> Cantor-list by at least one digit. Any consideration of an indicated >> limit is completely irrelevant in this respect. >> And it does because by definition the exchanged diagonal differs >> from *every* real in the list by at least one position. >> It differs from real[k] in position k, for every k. because it is >> constructed like this. And as there are no reals in the list but those >> who have a number k, it differs from *every* real in the list. > Your arguing is true only for finite sequences D_k of the diagonal. Since every digit of the diagonal is part of at least one finite sequence, this is perfectly sufficient. > The number of digits of the complete diagonal D is infinite > (aleph_0), hence complete induction and your proof (or construction) > fail. The number of digits is not used in the proof. The proof is for each position k that is a natural number. The proof does not rely on their being an infinitude of them, it just so happens that there is. > The diagonal squence D_n constructed by aÕ_nn = 1 + a_nn is always > contained in line Z(n+1). This is correct for every finite n by > construction of the matrix. > This fact does never change. Therefore it is true for the whole > diagonal D, Wrong. Therefore it is true for any D_n. D itself is no such D_n. > unless it contains elements with non-natural numbers as indices. No. The diagonal _entries_ contains only natural numbers as indices, numbers that also form the indices of the elements of all the D_n. So if you prove some property holding for all elements of every D_n, then this property will also hold for all elements of D. But that does not make D itself inherit any properties from the D_n. In particular, D is not the image of any Z(n). -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? > The diagonal squence D_n constructed by aÕ_nn = 1 + a_nn is always > contained in line Z(n+1). This is correct for every finite n by > construction of the matrix. > This fact does never change. Therefore it is true for the whole > diagonal D, > Wrong. Therefore it is true for any D_n. D itself is no such D_n. Of course it is. You must not intermingle my n with a fixed value. It stands for any conceivable natural number. Otherwise you could not construct IN by means of the Peano axioms. These axioms are the basis of the complete infinite set of all natural numbers. Should their application in complete induction be restricted to a smaller set? > unless it contains elements with non-natural numbers as indices. > No. The diagonal _entries_ contains only natural numbers as indices, > numbers that also form the indices of the elements of all the D_n. So > if you prove some property holding for all elements of every D_n, then > this property will also hold for all elements of D. That is my opinion too, and nothing more is required. But that does not > make D itself inherit any properties from the D_n. In particular, D > is not the image of any Z(n). Wrong (see above). Again: n is not a fixed value. n represents any natural number, and there is no natural number which is not represented by n. You can see this fact best, if you try to find a natural number which is in IN but is not covered by my proof. === Subject: Re: CantorÕs diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> The diagonal squence D_n constructed by aÕ_nn = 1 + a_nn is always >> contained in line Z(n+1). This is correct for every finite n by >> construction of the matrix. >> This fact does never change. Therefore it is true for the whole >> diagonal D, >> Wrong. Therefore it is true for any D_n. D itself is no such D_n. > Of course it is. So which n are you talking about in particular? > You must not intermingle my n with a fixed value. In D_n it stands for that. > It stands for any conceivable natural number. Quite so. And for any conceivable natural number n, D is different from D_n. So D itself is no such D_n. > Otherwise you could not construct IN by means of the Peano > axioms. You are babbling. One has absolutely nothing to with the other. >> No. The diagonal _entries_ contains only natural numbers as indices, >> numbers that also form the indices of the elements of all the D_n. So >> if you prove some property holding for all elements of every D_n, then >> this property will also hold for all elements of D. > That is my opinion too, and nothing more is required. Wrong. For your proof, more is required. >> But that does not make D itself inherit any properties from the >> D_n. In particular, D is not the image of any Z(n). > Wrong (see above). Above was nothing but nonsense and babbling. > Again: n is not a fixed value. n represents any natural number, You know the difference between any and all, right? > and there is no natural number which is not represented by n. Rather which could not be assumed by n. That is right. And for no natural number n we have D_n = D. > You can see this fact best, if you try to find a natural number > which is in IN but is not covered by my proof. Your proof gives us a sequence D_n = (1-10^{-n})/9, and a value D=1/9. For any given value of n, we have D-D_n = 10^{-n}/9, so for any given value of n, D is different from D_n. The value of the difference, of course, depends on n, and there is no _fixed_ positive value epsilon such that _all_ D_n differ by more than that _fixed_ epsilon from D. But that is not required, unless you are suffering from operator dyslexia. It _does_ lead to the observation that D is the _limit_ of D_n. But it is never assumed by any of the D_n, nevertheless. It is my guess that you are also incapable of grasping the difference between continuous and uniformly continuous functions, right? Similar problem. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? Sorry, but you are mistaken! Cantor does not (only) intend to show > that the exchanged diagonal differs from the indicated limit, i. e. > the former value of the diagonal, but he has to show that the > exchanged diagonal is in fact different from any real number given in > his original Cantor-list. And that proof is valid only, if the > exchanged diagonal differs from any real number of the original > Cantor-list by at least one digit. Any consideration of an indicated > limit is completely irrelevant in this respect. And it does because by definition the exchanged diagonal differs > from *every* real in the list by at least one position. It differs from real[k] in position k, for every k. because it is > constructed like this. And as there are no reals in the list but those > who have a number k, it differs from *every* real in the list. > Your arguing is true only for finite sequences D_k of the diagonal. I stated for every k it is true that the element #k of the sequence D is different from the element #k of the sequence a_k for every k it is true that D is different from a_k because element #k of D differs from element #k of a_k induces D differs from k because two sequences differ whenever they differ by at least one element. There is no k such that a_k = D Please point out _exactly_ which one of my two implications is logically wrong in your system and why ? Can two sequences be identical and still differ by at least one element? -- Horst === Subject: Re: CantorÕs diagonal proof wrong? Your arguing is true only for finite sequences D_k of the diagonal. > I stated > for every k it is true that > the element #k of the sequence D is different from > the element #k of the sequence a_k > - for every k it is true that > D is different from a_k > because element #k of D differs from element #k of a_k induces D > differs from k because two sequences differ whenever they differ by > at least one element. > - There is no k such that a_k = D > Please point out _exactly_ which one of my two implications is > logically wrong in your system and why ? Can two sequences be > identical and still differ by at least one element? You seem to assume that k is a fixed number. It isnÕt. It can take any value of a natural number. There is no natural number which it could not represent. This meaning becomes clear if you consider the Peano axioms. In particulare the induction axiom (2) states that with n also its successor n+1 is a natural number. This axiom is the basis of the infinite set of all natural numbers. Should its application in complete induction be restricted to a smaller set? I proved that for n of IN there are elements of {2, 4, 6, ..., 2n} surpassing its cardinal number. The true inversion of this statement is: If there are no elements of {2, 4, 6, ..., 2n} surpassing its cardinal number, then n is not a natural number. But, of course, we are talking about natural numbers only. It is completely meaningless, whether there is an infinite set or no last element. The only assumption required for my proof is that IN consists of natural, hence finite, numbers. === Subject: Re: CantorÕs diagonal proof wrong? Your arguing is true only for finite sequences D_k of the diagonal. I stated for every k it is true that the element #k of the sequence D is different from > the element #k of the sequence a_k -> for every k it is true that D is different from a_k > because element #k of D differs from element #k of a_k induces D > differs from k because two sequences differ whenever they differ by > at least one element. > -> There is no k such that a_k = D > Please point out _exactly_ which one of my two implications is > logically wrong in your system and why ? Can two sequences be > identical and still differ by at least one element? > You seem to assume that k is a fixed number. It isnÕt. You are pulling my leg, arenÕt you ? When I say for every k then I assume that k is a fixed number ? > It can take any > value of a natural number. There is no natural number which it could > not represent. ThatÕs what my sentence for every k it is true that the element #k of the sequence D is different from the element #k of the sequence a_k said. You donÕt have to explain it to me. Which part of this sentence donÕt you understand? -- Horst === Subject: Re: CantorÕs diagonal proof wrong? > Sorry, but you are mistaken! Cantor does not (only) intend to show > that the exchanged diagonal differs from the indicated limit, i. e. > the former value of the diagonal, but he has to show that the > exchanged diagonal is in fact different from any real number given in > his original Cantor-list. And that proof is valid only, if the > exchanged diagonal differs from any real number of the original > Cantor-list by at least one digit. Any consideration of an indicated > limit is completely irrelevant in this respect. And it does because by definition the exchanged diagonal differs from *every* real in the list by at least one position. It differs from real[k] in position k, for every k. because it is constructed like this. Therefore it differs from real[k] for every k, because differing in one position is enough to be different. And as there are no reals in the list but those who have a number k, it differs from *every* real in the list. Where exactly is your problem? [To avoid possible duplication problems we assume a *decimal* representation and we assume that every real in the list is represented by a decimal expansion which ends by 9999.... and we assume that the exchanged diagonal is selected in a way such that it, too, does not end by 9999...., - which is always possible] -- Horst === Subject: Re: CantorÕs diagonal proof wrong? > Curt, This should be an easy one for you, then. I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. What is the first digit? > Another example, there are algorithms for the construction of Pi = > 3.14159... ad infinitum, non repeating, truly irrational, but cannot > be as easily proven as the number stated above. > Excuse me, could you tell me whether the digit number 10^10^100 of pi > (or of your number given above) is an even one or an odd one? Do you > believe such a number to exist, although no one will ever be able to > answer this question? With 0.123456789101112131415... I may possibly be able to tell you if 10^Googool is even or odd - after 30 minutes I will either know, or be confident that after another 30 minutes I will, or give up, but with Pi, no, cannot tell you. However, I understand modern numerical methods compute the digits of Pi by doubling the number of digits every pass, so itÕs not a matter of getting the job done one by one, for which there is not enough time, and I therefore believe there may be hope for answering your question. At any rate, I certainly believe that the digits exist and in both cases are one of the set {0,1,2,3,4,5,6,7,8,9}, hence are either odd or even with an even chance to be either. All of the above, except that it can ever be computed, I believe of the digits number Googool ^ Googool. Sorry, what was your question? RN > At one > stage I grasped Ōinfinityas a serious of steps, which I could choose > to let never end (as a mind game), and I have since been perfectly at > ease with PeanoÕs axioms and inductive reasoning, > That is the so-called potential infinity, which marks a direction, but > never causes us to leave the finite domain. Agreed RN > Lets create a table of integers like this: ...000000 > ...000001 > ...000002 ...000010 ...000123 ItÕs just a normal list of integers, but instead of following the normal > convention of leaving off the leading zeros (which we all know are implied > even if we donÕt write them) I include them in that table. > Your ideas are perfectly reasonable, if you add the remark that an > infinite sequence of digits (which would not form a natural number) > could only emerge from a list with at least one line enumerated by > omega or infinity. All other numbers are natural numbers and, hence, > are finite. But in a line enumerated by omega, also Cantor could not > find and exchange the diagonal element. Th.92s very idea of the diagonal > of the naturals can be found in my preprint > arxiv.org/abs/math.GM/0305310 Why can I not have an infinite sequence of digits in second place, rather than in the place omega? Besides, I can understand omega (by which I presume you mean the ordinal equivalent of the cardinal aleph, the size of the set of naturals ?), but I cannot imagine it as describing the position of the last item in an infinite list. RN So lets use CantorÕs logic on this table and see if we can construct a > number which is not in the table. We take the numbers from the diagonal, > and construct the number ...111111 just like we did above. Since we construct this number by changing a digit from every row, we know, > by CantorÕs logic, that the resulting number can not be in the table. > Therefore, with the wisdom of Cantor, IÕve proved that the number of > integers is greater than the number of integers. There are some integers > which are simply not in the list of all integers. Ok, so if Cantor was wrong, why was he wrong? The answer is one already well known to mathematicians. They just never > realized how it applied here. You canÕt use infinity as if it existed. It > doesnÕt exist. infinity is only a name for something which can not > exist. > In particular actual infinity, the same as finished infinity, is > obviously a contradiction. > Has any one else put forth this same argument (or others) that CantorÕs > proof is invalid? > Here are some other preprints: arxiv.org/abs/math.GM/0408089 > arxiv.org/abs/math.GM/0403238 === Subject: Re: CantorÕs diagonal proof wrong? Curt, This should be an easy one for you, then. I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. What is the first digit? > Another example, there are algorithms for the construction of Pi = > 3.14159... ad infinitum, non repeating, truly irrational, but cannot > be as easily proven as the number stated above. Excuse me, could you tell me whether the digit number 10^10^100 of pi > (or of your number given above) is an even one or an odd one? Do you > believe such a number to exist, although no one will ever be able to > answer this question? > With 0.123456789101112131415... I may possibly be able to tell you if > 10^Googool is even or odd - after 30 minutes I will either know, or be > confident that after another 30 minutes I will, or give up, but with > Pi, no, cannot tell you. However, I understand modern numerical > methods compute the digits of Pi by doubling the number of digits > every pass, so itÕs not a matter of getting the job done one by one, > for which there is not enough time, and I therefore believe there may > be hope for answering your question. At any rate, I certainly believe > that the digits exist and in both cases are one of the set > {0,1,2,3,4,5,6,7,8,9}, hence are either odd or even with an even > chance to be either. All of the above, except that it can ever be > computed, I believe of the digits number Googool ^ Googool. Sorry, > what was your question? > RN It is not a matter of time. You could employ your children and grandchildren. It is a matter of hardware. The whole universe (accessible to us) has less than 10^80 protons and surely less than WM > At one > stage I grasped Ōinfinityas a serious of steps, which I could choose > to let never end (as a mind game), and I have since been perfectly at > ease with PeanoÕs axioms and inductive reasoning, That is the so-called potential infinity, which marks a direction, but > never causes us to leave the finite domain. > Agreed > RN > Lets create a table of integers like this: ...000000 > ...000001 > ...000002 ...000010 ...000123 ItÕs just a normal list of integers, but instead of following the normal > convention of leaving off the leading zeros (which we all know are implied > even if we donÕt write them) I include them in that table. Your ideas are perfectly reasonable, if you add the remark that an > infinite sequence of digits (which would not form a natural number) > could only emerge from a list with at least one line enumerated by > omega or infinity. All other numbers are natural numbers and, hence, > are finite. But in a line enumerated by omega, also Cantor could not > find and exchange the diagonal element. Th.92s very idea of the diagonal > of the naturals can be found in my preprint > arxiv.org/abs/math.GM/0305310 > Why can I not have an infinite sequence of digits in second place, > rather than in the place omega? Besides, I can understand omega (by > which I presume you mean the ordinal equivalent of the cardinal aleph, > the size of the set of naturals ?), but I cannot imagine it as > describing the position of the last item in an infinite list. > RN Any line enumerated by a natural number has a finite number of lines as predecessors. Hence, the diagonal of the line numbers 111...111 does exist as a finite, hence natural number. But here is another puzzle: As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains larger numbers than its cardinal number is. If {2, 4, 6, ..., 2n} does not contain larger numbers than its cardinal number is, then n cannot be finite. > So lets use CantorÕs logic on this table and see if we can construct a > number which is not in the table. We take the numbers from the diagonal, > and construct the number ...111111 just like we did above. Since we construct this number by changing a digit from every row, we know, > by CantorÕs logic, that the resulting number can not be in the table. > Therefore, with the wisdom of Cantor, IÕve proved that the number of > integers is greater than the number of integers. There are some integers > which are simply not in the list of all integers. Ok, so if Cantor was wrong, why was he wrong? The answer is one already well known to mathematicians. They just never > realized how it applied here. You canÕt use infinity as if it existed. It > doesnÕt exist. infinity is only a name for something which can not > exist. In particular actual infinity, the same as finished infinity, is > obviously a contradiction. Has any one else put forth this same argument (or others) that CantorÕs > proof is invalid? Here are some other preprints: arxiv.org/abs/math.GM/0408089 > arxiv.org/abs/math.GM/0403238 > === Subject: Re: CantorÕs diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Any line enumerated by a natural number has a finite number of lines > as predecessors. Hence, the diagonal of the line numbers 111...111 > does exist as a finite, hence natural number. Nonsense. The word Hence is used completely groundless. > But here is another puzzle: > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. Where is the puzzle? Except in incredibly shoddy wording? First you make a claim about finite n, then you make some hypothetical claim without even specifying what values n is allowed to assume, then you wildly assume the existence of some set {2..2n} _still_ without specifying your base set of n, make a statement that refers to the previous set and find it contradictory given the original base set. This merely proves that you did something invalid. It does neither prove nor disprove the existence of some set {2..2n} with the desired property, in particular since you donÕt specify what base set n is allowed to be in in the first place! -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? > Any line enumerated by a natural number has a finite number of lines > as predecessors. Hence, the diagonal of the line numbers 111...111 > does exist as a finite, hence natural number. > Nonsense. The word Hence is used completely groundless. Do you object to the wording only or to the meaning too? > But here is another puzzle: > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. > Where is the puzzle? Except in incredibly shoddy wording? First you > make a claim about finite n, then you make some hypothetical claim > without even specifying what values n is allowed to assume, My claim is proven by complete induction, hence valid for all finite n. I dont know of other possibilities. But if my claim should be wrong, then, obviously, some other numbers must get involved. then you > wildly assume the existence of some set {2..2n} _still_ without > specifying your base set of n, make a statement that refers to the > previous set and find it contradictory given the original base set. > This merely proves that you did something invalid. It does neither > prove nor disprove the existence of some set {2..2n} with the desired > property, in particular since you donÕt specify what base set n is > allowed to be in in the first place! n is of IN, i.e. n is a natural number. Contradictory is only the assumption that the cardinal number of all positive even numbers is larger than any even number. Contradictory would be the reversed assumption. Hence, we must conclude that the set of all positive even numbers does not have any cardinal number (because it does not actually exist). === Subject: Re: CantorÕs diagonal proof wrong? !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Any line enumerated by a natural number has a finite number of >> lines as predecessors. Hence, the diagonal of the line numbers >> 111...111 does exist as a finite, hence natural number. >> Nonsense. The word Hence is used completely groundless. > Do you object to the wording only or to the meaning too? The meaning of hence is to indicate an implication. There is no implication: the right has nothing to do with the left. >> But here is another puzzle: >> As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains >> larger numbers than its cardinal number is. >> >> If {2, 4, 6, ..., 2n} does not contain larger numbers than its >> cardinal number is, then n cannot be finite. >> Where is the puzzle? Except in incredibly shoddy wording? First >> you make a claim about finite n, then you make some hypothetical >> claim without even specifying what values n is allowed to assume, > My claim is proven by complete induction, hence valid for all finite > n. You are again wafßing around. For somebody with your background, this is quite unnecessary. So since you refuse to cast your shoddy stuff into clear words, I will do this for you: For every set of the form S_n = {2k | k in N, k<=n} (with n in N) there exists a number x(n) in S_n with x(n)>card{S_n}. This can be proven by induction over n, by trivially choosing x(n)=2n. So far, so good. Now you think this implies a statement about the set E = {2k | k in N} (which happens to be the smallest superset for the set of all S_n with n in N, although not of the form S_n itself), namely that there exists a number x in E with x>card{E}. But this does not follow at all, certainly not by induction. And that is because E is not of the form S_n, and so induction on S_n tells you exactly nothing about E. That E is the smallest superset of all S_n (without being an S_n itself) will help you derive properties about its _elements_ (all of which are also elements of some S_n). But it does not help with properties of the set itself. And it does not generalize in the way you want. And if we get to the bottom of this, we find that your confusion is based on the following: For all x in E there exists an n depending on x such that x in S_n. This is true, and this makes it possible to deduce things about E. But what is not true is that: There exists an n such that for all x in E we also have x in S_n. And you are confusing the two. Permanently, repeatedly, and even if people rub your nose in it for hours. In some sort of quantifier dyslexia. And if you are unable to overcome this dyslexia, or control it, or at least be aware of it, you should not be teaching math. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: CantorÕs diagonal proof wrong? Curt, This should be an easy one for you, then. I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. What is the first digit? > Another example, there are algorithms for the construction of Pi = > 3.14159... ad infinitum, non repeating, truly irrational, but cannot > be as easily proven as the number stated above. Excuse me, could you tell me whether the digit number 10^10^100 of pi > (or of your number given above) is an even one or an odd one? Do you > believe such a number to exist, although no one will ever be able to > answer this question? With 0.123456789101112131415... I may possibly be able to tell you if > 10^Googool is even or odd - after 30 minutes I will either know, or be > confident that after another 30 minutes I will, or give up, but with > Pi, no, cannot tell you. However, I understand modern numerical > methods compute the digits of Pi by doubling the number of digits > every pass, so itÕs not a matter of getting the job done one by one, > for which there is not enough time, and I therefore believe there may > be hope for answering your question. At any rate, I certainly believe > that the digits exist and in both cases are one of the set > {0,1,2,3,4,5,6,7,8,9}, hence are either odd or even with an even > chance to be either. All of the above, except that it can ever be > computed, I believe of the digits number Googool ^ Googool. Sorry, > what was your question? > RN It is not a matter of time. You could employ your children and > grandchildren. It is a matter of hardware. The whole universe > (accessible to us) has less than 10^80 protons and surely less than > WM > At one > stage I grasped Ōinfinityas a serious of steps, which I could choose > to let never end (as a mind game), and I have since been perfectly at > ease with PeanoÕs axioms and inductive reasoning, That is the so-called potential infinity, which marks a direction, but > never causes us to leave the finite domain. Agreed > RN Lets create a table of integers like this: ...000000 > ...000001 > ...000002 ...000010 ...000123 ItÕs just a normal list of integers, but instead of following the normal > convention of leaving off the leading zeros (which we all know are implied > even if we donÕt write them) I include them in that table. Your ideas are perfectly reasonable, if you add the remark that an > infinite sequence of digits (which would not form a natural number) > could only emerge from a list with at least one line enumerated by > omega or infinity. All other numbers are natural numbers and, hence, > are finite. But in a line enumerated by omega, also Cantor could not > find and exchange the diagonal element. Th.92s very idea of the diagonal > of the naturals can be found in my preprint > arxiv.org/abs/math.GM/0305310 Why can I not have an infinite sequence of digits in second place, > rather than in the place omega? Besides, I can understand omega (by > which I presume you mean the ordinal equivalent of the cardinal aleph, > the size of the set of naturals ?), but I cannot imagine it as > describing the position of the last item in an infinite list. > RN > Any line enumerated by a natural number has a finite number of lines > as predecessors. Hence, the diagonal of the line numbers 111...111 > does exist as a finite, hence natural number. > But here is another puzzle: > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. The second sentence is meaningless as long as you no not specify what the text the sequence {2, 4, 6, ..., 2n} means for some n which is not finite, i.e. the condition {2, 4, 6, ..., 2n} does not contain larger numbers than its cardinal number cannot be assigned any of the truth values true or false for some 2n which is not finite because you didnÕt explain what the text {2,4,6,...2n} means for infinite n. If you use a logical statement using a variable n you have to specify *before* which universe this variable comes from, and you have to define the meaning fo any formula like {2,4,6,...,2n} for *every* element of this universe. A correct statement would be: Every *finite* subset of the set of even natural numbers >=2 contains an element which is larger than the cardinal number of this set. If a subset of the set of even natural numbers >=2 does not contain an element which is larger than the cardinal number of this subset then this subset is not finite. Note that in this version the meaning of the term subset *is* defined for *finite* and *not finite* because the assumed universe of subset is the collection of all subsets of {2,4,6,.,,,,} while you didnÕt even explain what 2n means for n not finite -- Horst === Subject: Re: CantorÕs diagonal proof wrong? > As long as n is finite, the sequence {2, 4, 6, ..., 2n} contains > larger numbers than its cardinal number is. > If {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number is, then n cannot be finite. > The second sentence is meaningless as long as you no not specify what > the text > the sequence {2, 4, 6, ..., 2n} > means for some n which is not finite, i.e. the condition > {2, 4, 6, ..., 2n} does not contain larger numbers than its > cardinal number > cannot be assigned any of the truth values true or false for some > 2n which is not finite because you didnÕt explain what the text > {2,4,6,...2n} means for infinite n. I do not pretend to know anything about the properties of non-natural numbers. Contrary, I am convinced, that they are phantoms, at best. Bust my statement remains true. In case that there are no non-natural numbers, it simply means, that it is true in any case. IsnÕt ist obvious, to conclude from the statement If n is in IN, then n is positive that any n was positive, in case there were no other numbers than n? > If you use a logical statement using a variable n you have to > specify *before* which universe this variable comes from, and you > have to define the meaning fo any formula like {2,4,6,...,2n} for > *every* element of this universe. In our universe of natural numbers, my statement is correct. If one wants to circumvent it, he should define an extended universe (and prove the consistency of his definition). > A correct statement would be: > Every *finite* subset of the set of even natural numbers >=2 contains > an element which is larger than the cardinal number of this set. > If a subset of the set of even natural numbers >=2 does not contain an > element which is larger than the cardinal number of this subset then > this subset is not finite. Of course we could do that, unless finity and infinity had not been intermingles by statements about the actually infinite set of finite numbers. In order to circumvent wrong wording, I stated: In every senquence of even numbers 2, 4, 6, ..., 2n, where n is any natural number ... Here we see clearer than in your version, that there are no actually infinite sets. But I do not start with that obvious assumption, because I have set out to prove it. > Note that in this version the meaning of the term subset *is* > defined for *finite* and *not finite* because the assumed universe of > subset is the collection of all subsets of {2,4,6,.,,,,} > while you didnÕt even explain what 2n means for n not finite I repeat: I consider exclusively finite n. For those my proof is correct. It may become incorrect for other numbers, but I am not interested in those numbers. === Subject: Re: CantorÕs diagonal proof wrong? Hi Bob, if that is your real name, ItÕs an infinite set, thereÕs always one more. Consider the natural/unit equivalency function. I make expansive statements. One representation of the binary expansions to represent a number is by the integer modulus, thatÕs the normal way, there are other ways to encode integers in a binary string, for example Gray coding, a non-weighted reßected binary code. In Gray coding, an increment changes one bit of the string. On many digital computers, the same register is a given power of two number of bits is used to represent the signed or unsigned machine integer quantity. Is not it neat to think that half of the integers are non-negative and the other half negative? Not including zero, if I tell you IÕve selected an integer and deemed it positive or negative by ßipping a coin, is not that fair? Half of all the infinitely many integers are even. About which antidiagonals you might create, in binary there is only one, and for EF it is dually represented or not on the range, and its addition to the list doesnÕt change the list, because of dual representation, because the pseudo-expansions of those n.s. real numbers would have an antidiagonal with infinitely many repeating terminating zeroes. For the decimal case or b>3, there is the inductive impasse, for any finite base there is a smaller interval thus represented, and then thus through a harsh undecomposable composition there is thus a trivial mapping to the rest of the real numbers. Reconsider the immediate and deferred. This is again with these nonstandard representations of all the real numbers. Can you not think abstractly? Again, as the integer part of a real number is a finite natural number, a single leading zero can be added thus that the antidiagonal is never on the range, inductively. ItÕs nice to know that there exist real numbers larger than zero. Consider the expansion as representation in a finite base, or radix or integral modulus, in an infinite base the expansion for the integer part is always one element. For your inductive antidiagonal algorithm to generate a list of real numbers, your list would have to satisfy the extensions of the nested intervals style proofs. EF does, with its nonstandard definition of the real number, and the antidiagonal algorithm of your choice generates a value that is never within a finite interval. Draw the line segment with the pencil, that involves continuous application of the pencil to the paper with the infinitely precise and hard marking pencil point and infinitely smooth markable paper. You can photocopy your line with magnification, but if you magnify a list of stippled points each applied individually, there would always be gaps, the reals are not complete without the line, taking into account the nature of the end- and non-endpoints on the line. There are only points on the line. Inductively, infinite sets are equivalent. That demands not denial nor acceptance but revision and extension of these naively true statements that contradict with other naively true statements. Infinite sets are equivalent. Skolemize: your model is countable. Hausdorff said: uncountable is countable union of countable. The set of all sets, a set in some set theories, would be its own powerset. In some theories, itÕs not its own powerset twice. Are there any applications solely attributable to the transfinite cardinals besides themselves? In some theories, the powerset is order type is successor. The line is a set of points. At the edge of theory all is nothing and nothing is all, and itÕs the beginning of the theory. Stare into the abyss. ItÕs a mirror. I remember one time on PBS there was a mystery with a lady, she was a member of the cult that stared at the Sun and went blind, she walked into the open elevator shaft. She stopped at the bottom. Concreteness is a goal. Borel and combinatorics can agree, they might need some mediation, and clarification of definitions. Ross === Subject: Re: CantorÕs diagonal proof wrong? > Hi Bob, if that is your real name, ItÕs not. Bob is a palindrome of his real name. > ItÕs an infinite set, thereÕs always one more. > Consider the natural/unit equivalency function. A function youÕve been talking about for a couple of years, without proving that it exists. And since your EF is a bijection from the natural numbers to the reals, thereÕs an easy proof that it doesnÕt exist. === Subject: Re: CantorÕs diagonal proof wrong? > Hi Bob, if that is your real name, > ItÕs not. Bob is a palindrome of his real name. Objection. Palindrome is a unary property: X is or is not a palindrome - it canÕt be a palindrome of something. I think you meant to say anagram. Brian Chandler === Subject: Re: CantorÕs diagonal proof wrong? >> Hi Bob, if that is your real name, >> ItÕs not. Bob is a palindrome of his real name. > Objection. Palindrome is a unary property: X is or is not a > palindrome - it canÕt be a palindrome of something. I think you > meant to say anagram. Unless you are a fan of Monty PythonÕs Flying Circus, in which case you are allowed to say things like, ŌBoltonis a palindrome of ŌNotlobÕ. -- Dave Seaman Judge YohnÕs mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: CantorÕs diagonal proof wrong? %IW48mQf3K=Ci&gZ7]]aazx@]Y-nq!r5{yH/#,?@lDdUDvOfByB2hVW0.@OM% {l/{cTÕ{w > So you just reverse the digits in the integer to create the real. I > claim this mapping is one to one and covers all the reals in that > range. For any real you give me, I can easily give you the matching > integer. > ************************************************************* ************ > ***** > ************************************************************* ************ > ***** Curt, This should be an easy one for you, then. You are about 300 messages too late. :) > I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. I just write it backwards following your same algorithm. Why do you think itÕs important to right left to right? :) > What is the first digit? Well, thatÕs what was made obvious to me. Integers are finite, and reals arenÕt. So even though I can write the natrual number backwards just as easy and just as fast as you can write the real, what I end up trying to write is not a natural number. And thatÕs where my argument falls ßat. But even beyond that, I later tried another variation of the argument to remove the definition of natrual numbers and reals from the picture. But that in the end doesnÕt work either. The reason my argument doesnÕt work is becuse I was using properties of the physical world to debate the validity of the proof. But I later came to see that mathematics is a study of the properties of language, and not a study of the properties of the physical world. And, within the domain of language, CantorÕs proof is consistent with all the other words it is based on. And consistency of the language is all that is required to make it a valid argument in mathematics. So even though it seems to take us to a place that is inconsistent with what we know about the universe, thatÕs just not important. To prove Cantor wrong in the scope of mathematics, we must find an inconsistency in the words. And that is not so easy to do. -- Curt Welch http://CurtWelch.Com/ curt@kcwc.com http://NewsReader.Com/ === Subject: Re: CantorÕs diagonal proof wrong? > So you just reverse the digits in the integer to create the real. I > claim this mapping is one to one and covers all the reals in that > range. For any real you give me, I can easily give you the matching > integer. > ************************************************************* ************ > ***** > ************************************************************* ************ > ***** Curt, This should be an easy one for you, then. > You are about 300 messages too late. :) > I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. > I just write it backwards following your same algorithm. Why do you think > itÕs important to right left to right? :) > What is the first digit? > Well, thatÕs what was made obvious to me. Integers are finite, and reals > arenÕt. Are you aware that the set of integers is a subset of the reals? 1, 2, 3, etc are each real numbers. What you mean is that integers have nothing but 0Õs to the right of the decimal point in their base-10 expansion. Also, you use the words finite and infinite in a nonstandard way. The number pi, which is irrational, is finite. === Subject: Re: CantorÕs diagonal proof wrong? > So you just reverse the digits in the integer to create the real. I > claim this mapping is one to one and covers all the reals in that > range. For any real you give me, I can easily give you the matching > integer. ************************************************************* ************ > ***** > ************************************************************* ************ > ***** Curt, This should be an easy one for you, then. You are about 300 messages too late. :) I give you the > (irrational) real 0.1 2 3 4 5 6 7 8 9 10 11 12 13 14... which is > constructed by concatanating all the integers after the decimal point, > ad infinitem. I firstly appeal to you to recognise that the length of > that (irrational) real is the kind of infinity you said you are > familiar with and which you accept. Secondly I appeal to you to > recognise that this is indeed an irrational, the recipe (algorithm) of > its construction ensures there is no repetition anywhere after the > decimal point (this can be proven rigorously but I donÕt think we need > to bother). Being an irrational I would now kindly ask you to write > this real down in your reverse fashion. I just write it backwards following your same algorithm. Why do you think > itÕs important to right left to right? :) What is the first digit? Well, thatÕs what was made obvious to me. Integers are finite, and reals > arenÕt. > Are you aware that the set of integers is a subset of the reals? 1, 2, > 3, etc are each real numbers. What you mean is that integers have > nothing but 0Õs to the right of the decimal point in their base-10 > expansion. > Also, you use the words finite and infinite in a nonstandard way. > The number pi, which is irrational, is finite. Agreed: (1) naturals a subset of the integers a subset of the rationals a subset of the reals, all of which may be, but need not be defined as infinite expansions using a base 2 or larger, for example 10. (2) I used infinite in the sense of Ōinfinite decimal expansionÕ === Subject: Re: do I understand co-variant vs. contra-variant? <4190db6a$12$fuzhry+tra$mr2ice@news.patriot.net> <41a55b60$17$fuzhry+tra$mr2ice@news.patriot.net> >I had in mind that for each ket in quantum mechanics, |X>, there >corresponds a bra, trivial that we barely notice it: itÕs essentially the same object >wearing two hats. Actually not, because from Dirac on physicists have been using labels for things that do not exist in the original space; that, in fact, is where the Dirac Delta Function comes in. The transformation exists, is natural and is trivial only when we stick to the original Hilbert space and its dual. >And unlike GR, where we can also interconvert between vector and >covector representations, we donÕt consider some objects as >natural vectors or covectors ThatÕs because we are dealing with a single vector space instead of a vector bundle over space-time. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: group action and the fundamental group at 07:07 PM, anonymous@mathforum.org (anony) said: >Suppose S^1 acts on a space X, What kind of space? Your hypotheses are not sufficient to derive your conclusion. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: New countable infiniity logic <2tvccpF24gjr4U1@uni-berlin.de> <41a27d67$33$fuzhry+tra$mr2ice@news.patriot.net> at 12:32 PM, rem642b@Yahoo.Com (tinyurl.com/uh3t) said: >So letÕs name my property splittable, i.e. the reals under the >usual total ordering are splittable by the rationals. You seem to be >claiming that the concept of the whole set being splittable by some >particular subset, and the concept of that particular subset being >dense in the whole set, are the same concept. No, IÕm claiming that the concept of the reals being splittable by some particular subset, and the concept of that particular subset being dense in the reals, are the same concept. I made no claim about ordered sets other than the reals. In particular, I made no claim about subsets of the reals. >Note the above definition presumes weÕre dealing with a topology >which is simultaneously a totally ordered set such that the topology >and the total ordering are consistent (every open interval per the >ordering is an open set per the topology, and every open set per the >topology is a union of open intervals per the ordering; equivalently >the topology is generated by the ordering in the sense that the >topology is isomorphic to the set of all unions of open intervals >per the ordering). ThatÕs the topology one normally assumes for an ordered set. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: Torkel at Philosophy School (Was Re: Ozkural) <30lna1F30mb3bU1@uni-berlin.de> >What makes it any better than some other definition like my friendÕs >alternative definition? What makes Turkish grammar better than Chinese grammar? Nothing, but if I want people in Turkey to understand what I am saying, I need to share a common language with them. You are free to invent your own definition of G Major, but if you use that private definition in a news group devoted to Music, the other readers will, quite properly, regard you as a kook. In any discipline, communication requires a common vocabulary. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not === Subject: Re: TOE stop making up words. I alone have that authority. remember. >> I > Superman suffered kyrptonite. Civilization suffers wogite. > -- > Uncle Al > http://www.mazepath.com/uncleal/ > (Toxic URL! Unsafe for children and most mammals) > http://www.mazepath.com/uncleal/qz.pdf === Subject: Re: TOE >> I >Superman suffered kyrptonite. Civilization suffers wogite. Wogite? === Subject: Re: TOE In sci.math, abuse@mistral.net <1101674178.zqJlTNh4Kz1P+4h8dst7Rg@teranews>: I >>Superman suffered kyrptonite. Civilization suffers wogite. > Wogite? Trust me...you donÕt want to know. :-) -- #191, ewill3@earthlink.net ItÕs still legal to go .sigless. === Subject: TOE i === Subject: super difficult question!! IÕm looking for a continuous function f:R->R with discontinuity on irrational domain and continuous on Q. if u know the answer PLEASE send me.... === Subject: Re: super difficult question!! > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > if u know the answer > PLEASE send me.... If itÕs a continuous function it has no discontinuities! Perhaps what youÕre looking for is a function f:R -> R which is continuous at every rational and discontinuous at every irrational. Such a thing doesnÕt exist, because of a theorem: the set of points where a function is continuous forms a G_delta. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: super difficult question!! > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > if u know the answer > PLEASE send me.... > If itÕs a continuous function it has no discontinuities! > Perhaps what youÕre looking for is a function f:R -> R which is continuous > at every rational and discontinuous at every irrational. Such a thing > doesnÕt exist, because of a theorem: the set of points where a function is > continuous forms a G_delta. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada thank you all for responding my proffesor told the classroom that anyone who can find it will get automtic A. i guess he had a good reason.. === Subject: Re: super difficult question!! > IÕm looking for a continuous function f:R->R with discontinuity on > irrational domain and continuous on Q. > if u know the answer > PLEASE send me.... There is no such function. The set of points at which f is discontinuous must be an F_sigma set. The irrationals are not an F_sigma set. === Subject: Re: super difficult question!! R with discontinuity on > irrational domain and continuous on Q. > There is no such function. The set of points at which f is discontinuous > must be an F_sigma set. The irrationals are not an F_sigma set. I suggest then instead Maverick look at f(x) = 0 when x irrational = 1/n when rational x = m/n with coprime n,m in N Not super difficult... impossible. However, the reverse is possible; that is, you can find a function that is continuous on the irrationals and discontinuous on Q. === Subject: Re: super difficult question!! posting-account=xslVFw0AAACcw30rwkkdKYhs8AhLZVeJ Not super difficult... impossible. However, the reverse is possible; that is, you can find a function that is continuous on the irrationals and discontinuous on Q. === Subject: Re: measure on formal power series >Looking for a nontrivial measure on the set of all formal power series >R[[x]] where R is some finite set (actually, in my special case itÕs a >power set of some finite set). >Does anyone have an idea (except counting measure and zero-measure)? >>With respect to the natural topology (basis: cosets of x^m R[[x]]) >>this is a compact abelian group (if you put an abelian group structure >>on R). It has a Haar measure: in this case mu(a + x^m R[[x]]) = |R|^{-m}. > Even without group structure, you could call it product measure... > Since R is finite, use counting measure on R; then postulate > that the coefficients of the formal power series are independent. What do you mean by independent and how can it help? === Subject: Re: measure on formal power series >>Looking for a nontrivial measure on the set of all formal power series >>R[[x]] where R is some finite set (actually, in my special case itÕs a >>power set of some finite set). >>Does anyone have an idea (except counting measure and zero-measure)? > With respect to the natural topology (basis: cosets of x^m R[[x]]) > this is a compact abelian group (if you put an abelian group structure > on R). It has a Haar measure: in this case mu(a + x^m R[[x]]) = |R|^{-m}. Is a in R[[x]] or in R? In the first case, the set ((a+x^m R[[x]]) union (b+x^n R[[x]])) is not of the form (c+x^k R[[x]]), and in the second case, you get very few measurable sets... === Subject: Re: Category Theory books by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATGeCH20016; >>Hi all, >>IÕm intrested in buying a good book about category theory. I would like to >>buy (only)one of the following: >>1)Categories for the Working Mathematician >> Saunders Mac Lane; >>2)Conceptual Mathematics: A First Introduction to Categories >> F. William Lawvere, Stephen H. Schanuel . >>I need a good piece of advice... I would really appreciate it. >>TIA. >I donÕt know about the second, but the first is the standard classic >on the subject written by one of the founders, and it has had a recent >second edition. Certainly the Mac Lane book is very good, but itÕs a Graduate Text in Mathematics, which presupposes a certain amount of the reader, and category theory being what it is, it draws from a broad spectrum of mathematics for examples. The Lawvere-Schanuel book is also very good but rather different in orientation: it is meant to be readable at the undergraduate level (indeed, itÕs based on lectures to freshmen(!) at SUNY- Buffalo), and aims to impart an intuitive feeling for categories and what they might be good for. ItÕs not a systematic treatise on the subject, though. To the OP: if you must read only one book, IÕd go with the Mac Lane. The question might be easier to answer if you mentioned your interests, background, level of education, etc. But I have to wonder: *why on earth* are you limiting yourself to just one book?? Is that your MO for reading other mathematical subjects? Todd Trimble === Subject: AC question from e-mail Someone in this group (to whom I have bccÕd this post) sent me this in e-mail in response to a recent post in the Alef 0 thread. I think someone else here might better be able to post an answer. >Here is another question regarding AC. >IÕve been told of the following exaple where AC must >be called either way: >Partition the Reals by the following equivalence >relation: a~b <==> (a-b) is rational. >(Any pair of real numbers is quivalent iff their >differnce is rational.) >How do I prove that assuming impossibility to choose a >member of each equivalence class, is consistent with >the rest of the set theory axioms? -- Stephen J. Herschkorn sjherschko@netscape.net === Subject: Re: AC question from e-mail > Someone in this group (to whom I have bccÕd this post) sent me this in > e-mail in response to a recent post in the Alef 0 thread. I think > someone else here might better be able to post an answer. >> Here is another question regarding AC. >> IÕve been told of the following exaple where AC must >> be called either way: >> Partition the Reals by the following equivalence >> relation: a~b <==> (a-b) is rational. >> (Any pair of real numbers is quivalent iff their >> differnce is rational.) >> How do I prove that assuming impossibility to choose a >> member of each equivalence class, is consistent with >> the rest of the set theory axioms? A set consisting of one member from each equivalence class would be a nonmeasurable set, so a model of ZF in which all subsets of the reals are measurable would do it. I think thereÕs a Solovay forcing model in which ZF+AD holds, and all sets of reals would be measurable in that model. === Subject: Does it have a name? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATGpcZ21285; Does there exist a mathematical name for a mathematical Ōcreaturehaving all the properties of a (Lie) group, except the property of associativeness? An example of such a creature is exp(OÕ), where Ois the set of pure octonions, i.e., octonions having no real part. John Fredsted === Subject: Re: What do these numbers have in common? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATGplF21314; >At the weekend I was trying to solve the Times newspaper Listener >crossword. >I need to understand what the following numbers have in common, or >represent: >288, 728, 8280, 15875 and 86435 17^2 - 1, 27^2 - 1, 91^2 - 1, 126^2 - 1, 294^2 - 1. >The answer may or may not be mathematical. >Any help much appraciated. >Dr Matt Murphy >Lecturer >Dept of Materials Science >University of Liverpool === Subject: Re: What do these numbers have in common? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iATH9lJ23081; ... >numbers 288,728,8280,15875 and 86435 have in common ..the number 8 :) === Subject: What guarantees this quadratic form vanish? Let v(n) be an nx1 vector. Consider the nxn matrix Z(n) = I_n + v(n) v(n)Õ where I_n is the nxn identity matrix and Ō means transpose. Assume Z(n) is invertible and consider the quadratic form q = v(n)Zinv(n) v(n) where Zinv(n) is the matrix inverse of Z(n). What are necessary and sufficient conditions on v(n) such that q = 0 in the n -> infinity limit? Rodman === Subject: Re: What guarantees this quadratic form vanish? >Let v(n) be an nx1 vector. >Consider the nxn matrix > Z(n) = I_n + v(n) v(n)Õ >where > I_n is the nxn identity matrix and > Ō means transpose. >Assume Z(n) is invertible and consider the quadratic form > q = v(n)Zinv(n) v(n) >where Zinv(n) is the matrix inverse of Z(n). Assuming v(n) is a vector of real numbers, Zinv = I_n - v(n)v(n)Õ/(1+v(n)Õv(n)) and q = (v(n)Õv(n))/(1+v(n)Õv(n)) >What are necessary and sufficient conditions on v(n) >such that > q = 0 in the n -> infinity limit? Apply the above formula for q. -- 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: What guarantees this quadratic form vanish? >Let v(n) be an nx1 vector. >Consider the nxn matrix > Z(n) = I_n + v(n) v(n)Õ >where > I_n is the nxn identity matrix and > Ō means transpose. >Assume Z(n) is invertible and consider the quadratic form > q = v(n)Zinv(n) v(n) >where Zinv(n) is the matrix inverse of Z(n). >What are necessary and sufficient conditions on v(n) >such that > q = 0 in the n -> infinity limit? Start with evaluating the form of Zinv. Pretty straightforward. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: Re: What guarantees this quadratic form vanish? > Let v(n) be an nx1 vector. > Consider the nxn matrix > Z(n) = I_n + v(n) v(n)Õ > where > I_n is the nxn identity matrix and > Ō means transpose. > Assume Z(n) is invertible and consider the quadratic form > q = v(n)Zinv(n) v(n) > where Zinv(n) is the matrix inverse of Z(n). > What are necessary and sufficient conditions on v(n) > such that > q = 0 in the n -> infinity limit? > Rodman First of all, Z(n) is always invertible. Its inverse is |v> You should work out the steps to get this result. It should then be easy for you to find the condition on |v> that makes q = = 0 for any n. -- Julian V. Noble Professor Emeritus of Physics ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ For there was never yet philosopher that could endure the toothache patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1. === Subject: Re: What guarantees this quadratic form vanish? Let v(n) be an nx1 vector. > Consider the nxn matrix > Z(n) = I_n + v(n) v(n)Õ > where > I_n is the nxn identity matrix and > Ō means transpose. > Assume Z(n) is invertible and consider the quadratic form > q = v(n)Zinv(n) v(n) > where Zinv(n) is the matrix inverse of Z(n). What are necessary and sufficient conditions on v(n) > such that > q = 0 in the n -> infinity limit? Rodman > First of all, Z(n) is always invertible. Its inverse is > |v> Z^{-1} = I - ----------- > 1 + It should then be easy for you to find the condition on |v that makes q = = 0 for any n. === Subject: On the [Possible] Undecidability of GoldbachÕs Conjecture [in PA]. Introduction In the mathematics of SR (Special Relativity), there are 2 classes of 1st order expressions: a) Those that are true in all inertial frames of reference: the speed of light; the ratio Lenght(AB)/lenght(BC) = 1, where B is the midpoint of any a AC, are some examples. b) Those that are true in one inertial frame, but are wrong in another: m1 < m2, where m1, m2 are of the masses of, say, two protons in relative motion; the simultaneity of 2 events, are some example. One could easily suspect that inertial frame of reference is just a code name for model, for the formal system underlying SR. However, what could also seem to easily escape our attention is the fact relativity means that itÕs utterly meaningless to say m1 < m2 (or the other way around), or to say simultaneity(event1, event2) = true (or false). Relativity only means that: (1.a) If an expression of type b) is true in one inertial frame of reference, there will always be another inertial frame of reference where the same expression is false. (1.b) Therefore we donÕt need to know which frame the expression is true, and which one is false. In short, being relative of truth and falsehood, of expressions of type b), is what SR is really about. *** So what does all of this have to do with GC (GoldbachÕs Conjecture)? The answer is that, if what is going to be presented here is true, GC is in a special class of undecidable formulae of L(PA) that are relative, in the sense that each formula f in the class would have the following properties, in the meta level: (2.a) If f is true in one model, then there must exist another model in which f is false. (2.b) Likewise, if f is false in one model, then there must exist another model in which f is true. WeÕll show that GC would not be a theorem, because of (2.a). Similarly, ~GC would not be a theorem on the account of (2.b). Hence, GC would be an undecidable formula. And, as in SR, we need not [and perhaps can not] know exactly which model GC would be true, and which false. PA and the framework of FOL would _always_ guarantee the existences of 2 relative models, each of which would have the opposite truth-value for GC. The Proof In effect we must prove (2.a) and (2.b). And toward that end, weÕre going to construct what we could call as relative models of PA. Then, the strategy would be to prove that, in the meta level, GC would conform to (2.a) and (2.b), which would be re-written as: (2.aÕ) (GC is true in m1) => (GC is false in m2) (2.bÕ) (~GC is true in m1) => (~GC is false in m2) where m1, m2 are 2 relative models [to be defined later]. The *hypothetical* form of (2.a)Õs and (2.b)Õs and the relative generality of m1, m2 [to be explained later] is of significance: it renders as immaterial whether or not human reasoning knowledge could actually fully identify any model in which GC is true, or false. *Only the existence* of such 2 relative models is required to indict the undecidability on GC. WeÕll now define a sub class of PA models named as relative models. Let D be the familiar Von NeumanÕs (ZFC) set. D would serve as the domain from which there exist (uncountably) infinite numbers of successor functions. Each of these successor functions is naturally a _model_ of PA. In short, D is: D = {e0, e1, e2, .....}, where e0 = {}, e1 = {e0}, e2 = {e0,e1}, .... In D, considering the following models [i.e. successor functions]: f1: (e0,e1), (e1,e2), (e2,e3), (e3,e4), (e4,e5), (e5e6), (e6,e7), (e7,e8), (e8,e9),..... f2: (e0,e1), (e1,e2), (e2,e3), (e3,e4), (e4,e7), (e7e6), (e6,e5), (e5,e8), (e8,e9),..... f3: (e0,e1), (e1,e2), (e2,e3), (e3,e4), (e4,e5), (e5e8), (e8,e7), (e7,e6), (e6,e9),..... f4: (e0,e1), (e1,e2), (e2,e3), (e3,e4), (e4,e7), (e7e6), (e6,e5), (e5,e10), (e10,e9),(e9,e8), (e8,e11), (e11,e12),... ... These models have the following properties: (3.a) All even numbers in one model are also even numbers in the others (3.b) All prime numbers in one model are also prime numbers in the others. (3.c) There exists an non-zero even number [element] e_n such that the interval {e0, e1, ..., e_n} is interpreted as the arithmetic interval {0, 1, ..., n} respectively, for *all* of these models. This interval is named here as GI (Goldbach Interval). The only restriction on e_n is that it must be interpreted as greater than 4. [Of course, we could define an e_n whose interpreted value is so big that we could only know its existence, and which could not practically compute.] Using AC, letÕs choose 2 distinct models fm, fn, out of the collection of the models satisfying (3.a) - (3.c) above. The 2 models fm, fn then are named here as 2 relative models. Their being relative is also general: weÕve not yet imposed any specific restriction how relative they must be. For example, we could specify all the even numbers [elements] must be interpreted to be the same arithmetic even numbers, as exemplified by f1 and f2. [WeÕd name this as the same-even-value restriction and will user as part of the proof later]. Important: (3.b) requires that a prime number [element] e_p in any of these model must also be a prime number in any other model. But the interpreted arithmetic value/number of e_p could vary from one model to another. This observation, together with the design of (3.a) - (3.c) would serve one crucial purpose: so long as weÕre in the business making interpretation of GC over any 2 relative models - chosen by AC - itÕs impossible for us to know in the meta level which of these 2 relative models weÕre doing the interpretation of GC. And worse, as weÕll demonstrate later, if we suppose GC is true one of these model, there will be another [relative] model that GC would be false. And finally, we would be able to demonstrate similarly for ~GC. LetÕs now consider 2 closely related - syntactically and semantically - formulae in L(PA): (4.a) AnEp1,p2 [(n is even / n >=4) => ((p1, p2 are prime) / (p1 + p2 <= n)) ] (4.b) AnEp1,p2 [(n is even / n >=4) => ((p1, p2 are prime) / (p1 + p2 = n)) ] (4.a) is a piece-of-cake trivial theorem, while (4.b) is our GC. Naturally. The point of dragging (4.a) into discussion is: If a formula f is a theorem in PA, f must be independent of the differences between any 2 relative models: in the sense that f must be true in all models of PA. Therefore (4.a) is true in any 2 relative models from D. As weÕd soon show, the same could not be said of GC [i.e. (4.b)]: its truth value is relative - if itÕs true in one, itÕs going to be false in another relative model. And similarly for ~GC. And now weÕre ready to prove the undecidability of GC, by proving (2.aÕ) and (2.bÕ) in the meta level. Proof of (2.aÕ): Suppose GC [(4.b)] is a theorem of PA, then it must be true in all models of PA. But let m1, m2 be m2 relative models with the same-even-value restriction (as described above), so GC is true in m1 . So in making interpretation of GC in between the 2 models, n and p1, p2Õs being prime are semantically identical - in fact n is interpreted as the same arithmetic value in both m1 and m2. But with a suitable choosing of m2, we can have p1 interpreted as the same prime value in both model, while p2 would be interpreted as 2 different arithmetic prime values in m1 and m2. But since n is of the same arithmetical value in both model, and weÕve assumed (p1 + p2 = n) is true in m1, (p1 + p2 = n) must be false in m2 (since p2 is interpreted differently in m2). Therefore GC is false in m2, a contradiction to the assumption that GC is a theorem above. QED. Proof of (2.bÕ): Suppose ~GC is a theorem of PA, then ~GC must be true in all models of PA. But let m1, m2 be 2 relative models _without_ the same-even-value restriction. So ~GC must be true in m1. In other word, there exists an even number [element] n>=4 such that: (6) ~(p1 + p2 = n), for all prime numbers [elements] p1, p2. But since there is no same-even-value restriction between m1 and m2, and since GI (Goldbach Interval) only requires the same elements to be interpreted as 0, 2, 4 in both m1, m2; weÕd appropriately choose an m2 so that n in (6) would be interpreted the arithmetic number, say, 6. But since 6 = 3 + 3, so w.r.t. m2, there exist p1, p2 interpreted as 3 so that: (p1 + p2) = n So (6) is false in m2. Hence ~GC is also false in m2. QED. In brief, weÕve shown that GC can not be a theorem of PA, on the account of proof of (2.aÕ); and similarly, ~GC is not a theorem on the account of proof of (2.bÕ). Hence GC is undecidable. Conclusion Fundamentally, PA exhibits one peculiar feature not seen in some other 1st order systems [such as ZF]: from the primary n-ary relation (i.e. the successor function) mentioned in the axioms, we could come up with a different successor functions that satisfy the same set of axioms. But the multiplicity of successor functions seems to create a problem, in the meta level: where the functions overlap, we could find theorems that express whatÕs common amongst these functions. Where these functions differ (i.e. donÕt overlap) expressions that mean different arithmetic meanings, to different successor functions, are generally prone to undecidability. And in the case of GC, we seem to have shown that it is in deed undecidable. [Note that the 1st part of the conclusion of GC, i.e., (4.b) is: (p1, p2 are prime) but the *same* numbers/elements p1, p2 could mean different arithmetic primes depending on what successor function that is being chosen in the meta level.] Of course, the presentation above could be incorrect in some ways. In which case IÕd to thank in advance for any corrections and comments anyone might have. ---Nam Nguyen === Subject: Re: sin(z)=z, finally I found a nice and compact proof Or this: The function g(z) = sin(sqrt(z))/sqrt(z) - 1 (initially defined for real z>0, then extended to an entire function by its power series) is of order 1/2, which is not an integer, so it has to have infinitely many roots. (The order refers to the asymptotic behavior abs(g(z)) = O(exp(abs(z)^(1/2+eps)) as abs(z) goes to infinity true for all eps>0, and false for all eps<0.) Moreover, the exponent of convergence of the root sequence {z(n)} (respecting repetitions) is 1/2, so that (dropping a possible zero root) sum(1/abs(z(n))^(1/2+eps)) converges for all eps>0 and diverges for all eps<0. The non-zero roots of sin(z)-z are square roots of the non-zero roots of g above, with asymptotic behavior easy to adjust. (I looked it up in a chapter about the growth of entire functions in Saks&Zygmund, Analytic Functions.) A simple symmetry consideration shows that the non-zero roots of sin(z)-z come in blocks of four with the same absolute values. (And itÕs the first time that I considered a negative epsilon! :-)= > Instead of all of this, consider this elegant proof. > Let f(z) = sinz - z. Then f is entire and so, by PicardÕs Theorem, f takes every value in C, with a possible exception of one value, infinitely many times. Thus, either f(z) = 0 infinitely many times or f(z) = 2pi infinitely many times. > In the first case, we are done. > In the second case, note that sin(z+2pi)=sin(z) and so, > sin(z+2pi)-z=2pi infinitely many times. > Equivalently, sin(z)=z infinitely many times. > Remark: Note that this applies to periodic functions like e^z, cosz, etc. >
 How many solutions has the equation sin(z)=z (in
complex
>>numbers)?
>>Ok. Here is a nice proof of this result. Consider the
function:
>> g(z)= z/sinz
>>This function g is analytic in C(pi)Z and never vanishes.
You can
check
>>that the singularity at 0 is removeble since the limit as z
approaches
0
>>is 1. Thus, this function is meromorphic and it has a
sequence of poles
>>that go to infinity. Therefore, you can also check that when
this
>>happens, it follows that the function has a non-isolated
essential
>>singularity at infinity.
>Non-isolated indeed! But an essential singularity must be
isolated, so
>this is not an essential singularity. And therefore Picard
doesnÕt
apply.
>Consider what the function f(z) = sin(z) - z does on the
rectangular
>contour with corners 2n pi, (2n+1) pi, (2n+1) pi + m i, 2n
pi + m i
>for any positive integer n, where m is large...
>Robert Israel israel@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~
israelVancouver, BC, Canada V6T 1Z2
===
Subject: inverse fibonacci
hello:
does the <>, defined for this 
purpose as:
1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
converge and, if yes, to what value?
george szpiro
george@netvision.net.il
===
Subject: Re: inverse fibonacci
> hello:
> does the <>, defined for this 
purpose as:
> 1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
> converge and, if yes, to what value?
> george szpiro
> george@netvision.net.il
Yes it converges. The Fibonacci numbers grow approximately as
the powers of the Golden Section (1 + sqrt(5)0 / 2 which is
larger
than 1, so the reciprocals will converge - this is very easy
to make
of the Fibonacci Quarterly that is either from the late
sixties or the
early seventies or something like that which links this
particular sum
to values of elliptical functions. At the time I heard about
this
truth is truth no matter how ugly. Now I expect I would
probably find
it cool. At one point I came up with a method that might have
duplicated the result, or I might have been mistaken, but I
have
forgotten what I did.
You are probably aware that there is a closed form solution to
the sum
1/1 + 1/1 + 1/3 + 1/13 + 1/377 + .., the sum of the
reciprocals of the
2**nÕth Fibonacci numbers, and you can find it 
by searching
through
past postings to sci.math.
Achava
===
Subject: Re: inverse fibonacci
>does the <>, defined for this 
purpose as:
>1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
>converge and, if yes, to what value?
Yes, it converges. The ratio of consecutive terms approaches
1/phi, where
phi is the golden ratio, so if you want to approximate it,
calculate
several
terms and then use the formula for the sum of a geometric
series to
estimate
the rest.
1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21+1/34*1/(1-1/phi)
1/(1-1/phi) equal to 1+phi.
This approximation oscillates around the actual value as it
approaches, so
you can see how accurate your result is by comparing one
value to the next.
1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21+1/34*(1+phi) -
(1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21*(1+phi))
=1/34*(1+phi) - 1/21*phi
=-4.82378904e-05
So the calculated value of 3.35987646 is accurate to 4
decimal digits.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: inverse fibonacci
>does the <>, defined for this 
purpose as:
>1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
>converge and, if yes, to what value?
> Yes, it converges. The ratio of consecutive terms
approaches 1/phi,
where
> phi is the golden ratio, so if you want to approximate it,
calculate
several
> terms and then use the formula for the sum of a geometric
series to
estimate
> the rest.
> 1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21+1/34*1/(1-1/phi)
> 1/(1-1/phi) equal to 1+phi.
> This approximation oscillates around the actual value as it
approaches,
so
> you can see how accurate your result is by comparing one
value to the
next.
> 1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21+1/34*(1+phi) -
(1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21*(1+phi))
> =1/34*(1+phi) - 1/21*phi
> =-4.82378904e-05
> So the calculated value of 3.35987646 is accurate to 4
decimal digits.
> --Keith Lewis klewis {at} mitre.org
> The above may not (yet) represent the opinions of my
employer.
Another acceleration: for n = 2, 3, ...,
add 1/F(n-1) to the partial sum
1/F(1) + 1/F(2) + ... + 1/F(n)
and these approximations also oscillate around the limiting
value. Moreover, they remain rational.
How fast do they converge?
With some perseverance, you can show that the modified
sequence has the
same terms as the alternating (except for the offset 3) series
3 + 1/(F(1)*F(2)*F(3)) - 1/(F(2)*F(3)*F(4)) +
1/(F(3)*F(4)*F(5)) - ...
How did I discover this trick? I subjected the partial sums
to AitkenÕs
delta-square acceleration technique (found in older numerical
methods
books).
[Skip the rest if the previous trick is already good enough.]
Yet another kind of manipulation: throwing in a parameter.
Consider the function
f(x) = sum[n=1 to infinity] x^n / F(n)
(and you showed interest in f(1)).
The series converges for abs(x) < phi by Ratio Test (as in
the previous
reply), and the convergence can be accelerated by subtracting
the
singularity: the result is a functional equation
(*) f(x) = x*sqrt(5) / (psi - x) + f(-x/psi^2)
The series on the right side converges faster than the
original one
(and provides analytic continuation).
Exercise: Try f(8/5) from the definition (very slow) and from
(*)
(reasonably fast).
You can repeat the procedure many times; two steps look like
f(x) = x*sqrt(5)/(psi-x) - x*sqrt(5)/(psi^3+x) + f(x/psi^4),
still faster.
If we continue infinitely many times, we end up with a
partial fraction decomposition (a la Mittag-Lefßer)
f(x) =
x*sqrt(5) * sum[k=0 to infinity] (-1)^k / (phi^(2*k+1) -
(-1)^k * x)
(I hope I captured the pattern correctly).
Question: Has a functional equation of the type (*) been
studied?
(Type: linear non-homogeneous dilation equation of first
order.)
===
Subject: Re: inverse fibonacci
> hello:
> does the <>, defined for this 
purpose as:
> 1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
> converge and, if yes, to what value?
See

David
===
Subject: Re: inverse fibonacci
> does the <>, defined for this 
purpose as:
> 1/1 + 1/1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 ...
> converge and, if yes, to what value?
> See
>

BTW, I just noticed that the MathWorld link for Prevost
Constant, given
in the above, no longer exists. Apparently it is should now be
.
David Cantrell
===
Subject: Re: Personal conjectures
> Hello sci.math readers,
> This post is not about those, though. ItÕs about _your_
personal
> conjectures. IÕm sure that every mathematician has one or
more
> personal conjectures that he/she would give his/her soul to
prove.
> IÕd like to see the conjectures that you have thought up,
and that
> you think about whenever you have a spare moment.
> Asger.
I love Conjectures about prime numbers. This one I framed
many years
ago:
For each integer N >= 1 there is an X>=0 that produces a prime
number with the formula:
6(N - X^2) + 1
Also for each integer N>=1 there is an Y>=0 that produces a
prime with
the formula:
6(N - Y^2) - 1
That is, before X or Y attain sqr(N) it will result a prime
number .
Ludovicus
===
Subject: Re: Personal conjectures
> Hello sci.math readers,
> . IÕm sure that every mathematician has one or more
> personal conjectures that he/she would give his/her soul to
prove.
> IÕd like to see the conjectures that you have thought up,
and that
> you think about whenever you have a spare moment.
> Asger.
> I love Conjectures about prime numbers. This one I framed
many years
> ago:
> For each integer N >= 1 there is an X>=0 that produces a
prime
> number with the formula:
> 6(N - X^2) + 1
> Also for each integer N>=1 there is an Y>=0 that produces a
prime with
> the formula:
> 6(N - Y^2) - 1
> That is, before X or Y attain sqr(N) it will result a prime
number .
> Ludovicus
Naturally the value of X augments with N, but what about the
relation :
r = X / [log(N)]^2 for the first prime found? In this case my
record
in the formula 6(N - X^2) + 1 is 0.585 for the values N =
71160 , X
= 73.
And r = 0.528 in the formula 6(N - Y^2) - 1 N = 16892, Y = 50.
Ludovicus
===
Subject: Re: Personal conjectures
> ....
> This post is .... about _your_ personal
> conjectures. IÕm sure that every mathematician has one or
more
> personal conjectures that he/she would give his/her soul to
prove.
> IÕd like to see the conjectures that you have thought up,
and that
> you think about whenever you have a spare moment....
> IÕm not at all passionate about it in the way you 
describe,
but
> FWIW hereÕs a conjecture I made many years ago. It links
(perhaps) two
> completely different areas of mathematics.
Well, I guess that my belief that all mathematicians have my
kind of
personal conjectures where false...
> So my conjecture is the obvious generalization: the graph
defined
> as above for the set {1,2,3,...,n} can be represented as
the edge graph
> of a polytope which tessellates (n-1)-dimensional space.
This is exactly the kind of conjectures that I like; itÕs
easy to state and
it feels right :-)
Asger.
===
Subject: Re: group action and the fundamental group
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iATJ2KL02508;
> at 07:07 PM, anonymous@mathforum.org (anony) said:
>>Suppose S^1 acts on a space X,
>What kind of space? Your hypotheses are not sufficient to
derive your
>conclusion.
Why not? The only thing required is that the fundamental group
functor pi_1: Top_* --> Grp [from the category of pointed
topological spaces to the category of groups] preserves
products. See my
earlier post.
Todd Trimble
===
Subject: Re: group action and the fundamental group
at 07:02 PM, trimble1@optonline.net (Todd Trimble) said:
>Why not?
I overlooked the fact that he fixed a base point, which
eliminates the
objection I had in mind for his question. Sorry.
His second statement, however, is still not generally true.
IÕm not
sure whether he needs to assume connected for that or whether
a weaker
hypothesis is sufficient.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: Re Re Publication Scandal
> Now, let us put where our money where our mouth is and make
>this forum interesting. I make this challenge: DEFINE A REAL
NUMBER
>The definition must be original and different from mine. I
you post a
>correct definition, no ßaw, etc. IÕll send 
you a check for
$1000,
What the heck is a correct definition? For example, why
couldnÕt
we define real number to mean the big key on the keyboard that
moves the cursor all the way to the left?
Please use your real number more often.
You can donate my $1000 to charity.
dave
===
Subject: Re: Re Re Publication Scandal
>> Hi Folks,
>> Five years ago I was active in this forum. I have raised
some issues
>> that could not be settled here but they are now resolved
in the new
>> forum MathForge.net. Among them are the following: (1) the
>> Ullrich-Israel equation or dilemma 1 = 0.99; this turns
out to be
>> ill-defined in the real number system and, therefore, it is
nonsense.
>> Therefore, I call it the Ullrich-Israel nonsense (the
resolution is
>> now in the archives of MathForge.net although I have
resolved them
>> earlier in several of my papers). (2) Two of the axioms of
the real
>> number system, the completeness and dichotomy axioms are
false.
>> Counterexamples to them were constructed by Banach-Tarski
and
>> Brouwer, respectively. Therefore, the real number system
as presently
>> developed in, say, Roydenss Real Analysis, p. 31, is
nonsense. (3)
>> Mathematical precision I require that every mathematical
space be
>> well-defined by a set of axioms, that is, (a) the existence
and
>> properties of as well as relations among the concepts must
! be
>> specified by the axioms; the same holds for the structure
and other
>> properties of the mathematical space itself; (b) the rules
of
>> inference must be specific to the given mathematical space;
>> therefore, universal rules of inference such as formal
logic must be
>> rejected
> The nonsense of which you speak has been instrumental in
creating the
> science which the created the engineering which created the
computer
> systems into which you enter your nonsense.
> Any system of logic that does not lead to a contradiction
and produces
> the desired results is a good one.
Are you suggesting that his nonsense is a desired result, or
that the
system that produced his computer is a bad one?
===
Subject: Re: Coprime Grid Puzzle
> Start with an n-by-n grid.
> Place n^2 positive integers into the grid such that:
> Each integer is greater than the integer placed previously.
> Each integer is left of, right of, below, or above the
integer
> placed immediately previously into the grid.
> Every integer is coprime to every other integer in its
column
> and row, and the integers in both main diagonals are all
coprime with
> each other.
Clarification of the integers in both main diagonals are all
coprime
> with each other:
> I mean the integers in any main diagonal are all coprime
with the
> other integers in *that particular* diagonal.
> Anyone have any luck finding a 23 case?
> Leroy Quet
> Your goal is to have the final integer be as low as 
possible.
> For example,for n = 4, we can have:
1 2 3 29 17 19 22 23
> 7 5 4 27 13 2 3 25
> 8 13 17 25 and 11 1 5 26
> 9 11 19 23 10 9 7 29
> But can we do better than 29 for n=4?
Yes. I found by-hand a solution for n=4 that ends in a 23.
> (I leave finding a last-integer-is-23 solution for
rec.puzzles and
> sci.math readers as a puzzle.)
Is it possible to end on even a lower integer with n=4?
> (We can always have a solution for any n, just fill the grid
with
> increasing primes. Finding the best solution, however, is
trickier.)
Leroy Quet
I have been unable by-hand to find a solution to the 4-by-4
grid which
ends in an integer < 29 but > 23.
(And except for rotations/reßections of my 23-solution, I
cannot find
another 23-solution either.)
How would the sequence start where the n_th term is the lowest
possible high-integer for the n-by-n grid?
(Maybe someone could submit this sequence to the On-line
Encyclopedia
of Integer Sequences, if it is not already there.)
Leroy Quet
===
Subject: Re: Coprime Grid Puzzle
Leroy Quet  escribi.97:
> I have been unable by-hand to find a solution to the 4-by-4
grid which
> ends in an integer < 29 but > 23.
> (And except for rotations/reßections of my 23-solution, I
cannot find
> another 23-solution either.)
Found by Ruben in the mailing list Snark
7 - 5 - 3 - 2
9 - 22 - 23 - 1
10 - 21 - 19 - 17
11 - 13 - 14 - 15
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Coprime Grid Puzzle
> Solution:
> 3 1 23 22
> 4 11 13 21
> 5 9 14 19
> 7 8 15 17
> Are there any other 23-solutions (except for rotations and
reßections)?
> Yes there is:
> Exchanging both of 1) top and bottom rows, and 2) leftmost
and
> rightmost columns, will preserve all rows columns and
diagonals
> without it being a reßection or a rotation.
> - Risto -
If I understand you right, then exchanging the top and bottom
rows
and/or exchanging the left and right column does *not* get a
solution,
because the integers in ascending order would not form a
continuous
path.
(The 1 and 3 would not be next to each other, for example.)
Leroy Quet
===
Subject: Sum Involving Stirling Numbers & Harmonic Numbers
Let s(k,n) be an unsigned Stirling number of the first kind.
(s(1,1)=1;s(1,n)=0 if n not=1; s(k+1,n) = k*s(k,n) +s(k,n-1))
Let H(k) = sum{j=1 to k} 1/j.
Let zeta(k) = sum{j=1 to oo} 1/j^k.
Then, for n = a positive integer:
oo
---
s(k,n) H(k)
/ ------------- =
--- (k+1)!
k=1
n
---
/ zeta(k+1)
---
k=1
In linear mode:
sum{k=1 to oo} s(k,n) H(k)/(k+1)!
=
sum{k=1 to n} zeta(k+1).
I have posted many results like this years ago, and I would be
surprised if I had not already posted this or a
generalization.
Here are some similar results of mine:
Squares of Harmonic #s, Stirling #s, and Zeta Function:
Generalized harmonic numbers and Stirling number infinite:
Sum Involving Stirling Numbers, Zeta Functions, Harmonic
Numbers:
Leroy Quet
===
Subject: Re: Random numbers for C: End, at last?
> (as per RandyÕs suggestion, IÕm adding
sci.crypt.random-numbers
>> IÕve been using George MarsagliaÕs RNGs 
until now when IÕm
trying to
>> compile on 64 bit Linux. Here they are, with the tests -
>> http://tinyurl.com/4etu9 - that now fail. Some appropriate
places need
>> to be AND-ed?
> and Randy Howard replied:
> I suspect sci.crypt.random-numbers would be a better
choice, given all
> the broad cross-posting IÕm surprised it 
wasnÕt included.
> to which DZ replied:
>> I followed up to George MarsagliaÕs post made in 1999 to
these groups
> Looking at the code in MarsagliaÕs post at
http://tinyurl.com/4etu9, I
can
> see a few places where a 32-bit word was implicitly
required: SHR3, CONG,
> FIB, LFIB4, and SWB. So yes, if you want those to behave as
if they were
on
> a 32-bit machine, youÕll need to insert appropriate
AND-masking to make
it
> so.
IÕm going with an easy but non-portable solution that works
with GNU
g++/gcc. Is explicit AND-masking more efficient? Replace
typedef unsigned long UL;
in http://tinyurl.com/4etu9 with
#include 
typedef uint32_t UL;
DZ
===
Subject: Re: Random numbers for C: End, at last?
>> Looking at the code in MarsagliaÕs post at
http://tinyurl.com/4etu9, I
can
>> see a few places where a 32-bit word was implicitly
required: SHR3,
CONG,
>> FIB, LFIB4, and SWB. So yes, if you want those to behave
as if they were
on
>> a 32-bit machine, youÕll need to insert appropriate
AND-masking to make
it
>> so.
and DZ replied:
> IÕm going with an easy but non-portable solution that 
works
with GNU
> g++/gcc. Is explicit AND-masking more efficient? Replace
> typedef unsigned long UL;
> in http://tinyurl.com/4etu9 with
> #include  I basically looked over the paper Advanced Polynomial
Factorization,
> considered all the arguments and various comments from
journals,
and
> renamed it.
The result went to the Annals, which accepted it for review.
So, no, IÕm not concerned about sci.math posters coming 
later
and
> trashing the journal and yes IÕm assuming few problems.
After
all,
> IÕve gone over every step in the argument repeatedly to
quite a
few
> people, from Barry Mazur, to Ralph McKenzie at my alma mater
> in-person, to, yes, Usenet posters.
There is no error.
> Hmm. See below ...
> And also, IÕm not terribly concerned about how it will be
handled
by
> reviewers because I have so much information from what
happened
with
> the older paper.
Essentially, itÕs over, but IÕm sure many of 
you will cherish
your
> ability to continue to badmouth my work and act as if itÕs
nothing
for
> as long as you can because thatÕs really all you can do at
this
point.
I have included non-polynomial factorization, stepped out each
step,
> and noted the conßict with what follows from the ring of
algebraic
> integers, and noted it as an error within the discipline.
The paper is quite solid.
> It is?
I just wonder if you have revised it before it was submitted
> to the Annals. I mean, in the version which you posted
> very recently with A. Beckwith as a co-author, toward the
end
> you say
Now letting m = 1, f = sqrt(5), where I can let u = 1 as
> its value doesnÕt change the aÕs, I have
(m^3 f^6 - 3 m^2 f^4 + 3m) x^3 - 3(-1 + m f^2) x u^2 + u^3
= 65 x^3 - 12 x + 1.
But if you actually substitute m = 1, f = sqrt(5), u = 1
> into the left side, you get
53 x^3 - 12 x + 1,
so that cannot be right.
So you must have changed that part, huh?
Of course, if you go back farther in the paper, you see that
> you define
P(x) = f^2 [(m^3 f^4 - 3 m^2 f^2 + 3m) x^3
- 3 (-1 + mf^2) x u^2 + u^3 f],
and if you substitute m = 1, f = sqrt(5), u = 1 into THIS,
> you get
P(x) = 65 x^3 - 60 x + 5 sqrt(5) = 5(13 x^3 - 12 x + sqrt(5)),
so that doesnÕt work either. Hmm. I wonder what you and
> Beckwith actually meant to say here!?
Maybe the Annals can figure it out for you.
> Andrzej.
> And I thought that Nora Baron might be David Ullrich.
What ??? I thought he was Wilma Scranton!
> How many of you knew? Better yet, how many of you guys
thought the
> poster was actually a woman, eh?
You really donÕt know women if you did, now do you?
You have it wrong. ŌAndrzejis 
fictional.
> Really? So what was your purpose in ending with that name?
> Possibly you didnÕt do it deliberately but screwed up, as
itÕs easy to
> type oneÕs own name, now isnÕt it?
> And now wish to try and recover?
> How stupid do you think the undergrad readers are?
> (Sci.math readers are very gullible as long as posters
disagree with
> me, they seem willing to believe just about anything!)
> ŌHismath, however, is not. 
ItÕs clear you read to the
> end of Ōhismessage. I note no substantive 
response. My
> guess is, the version of ŌAPFyou have 
submitted to the
> Annals still contains the dimwit error to which 
ŌAndrzejÕ
refers.
> Further, I would guess you knew it even before it was
> submitted, and in any case you are not withdrawing it. Why
> otherwise would you continue to post the erroneous version
> by yourself and A. Beckwith? How stupid do you think Annals
> editors are?
> Hos stupid do you think readers are?
> You lie just as I said you lie, and now people can see it
now with
> this question about whether or not youÕre a guy or a girl.
> WhatÕs clear is that you are a liar.
Aah but you havenÕt considered that the 
fictional ŌAndrzejis
really Ann
Vandrzej (remember her?), AKA Nora Baron.
And she has been phenomenal in repeatedly providing clear
mathmatical
refutations of errors in your work.
KeithK
> As for my paper, I think the editors of the Annals can do
their jobs,
> so why should you worry about it?
> Are you scared Nora Baron/Andrzej?
> I mean, after all, the paper is at a journal. Why worry
about it now?
> Thus your paper with Beckwith is a lie, compounded by your
refusal
> to even try to fix it.
> The error that ŌAndrzejpoints out is not the 
main one.
> ItÕs likely too you have read to the end of *my* message
> in this same thread. Several people now, very likely
including
> yourself, understand what your central error is in 
ŌAPFÕ:
> misuse of your own definition of constant term. Yet you
> continue to maintain the lie that the paper has no error.
> The fact is, if that erroneous part is removed, there is
nothing left:
> no lemma, no partial result, no mathematics worth
mentioning.
> You need to read my post JSH: Understanding constant terms
and, in
> fact, if you have the *balls* to do it you can try to point
out an
> error in it!!!
> I mean, come on guy, youÕre a complete joke now!
> Think about all those posters who would come back to defend
you when
> IÕd tear into you, when they thought you were a girl, when
you are a
> guy, playing a girl!
> Now maybe some of them would like to tear into you, eh?
> You either know this already, or you are lying to yourself.
> I would guess the latter.
> Nora B.
> Ah, so this time you remembered to put Nora B., and have
decided to
> let Andrzej be the fictional character?
> I can guess why, as IÕm sure Andrzej will decide to retire
from
> posting, for some odd reason, eh?
> Some of you readers out there are really dumb clods to be
taken in by
> this fellow posting as two different people--one male, one
female.
> But, like IÕve said, youÕre ruled by your 
emotions.
> You need to learn to be ruled by logic, and hey, just
because you
> think youÕre part of some big gang here on sci.math where
youÕre
> backing each other up against the interloper--me--you 
donÕt
know these
> people.
> They are not really your friends.
> James Harris
===
Subject: Re: JSH: At the Annals
 me, they seem willing to believe just about anything!)
Hmm, fascinating. How do you know this? Since your supposed
audience
of sci.math readers never says anything, who are you to say
who or
what they believe?
Brian Chandler
http://imaginatorium.org
===
Subject: Re: JSH: At the Annals
you have to say it with conviction ... It!
> me, they seem willing to believe just about anything!)
> Hmm, fascinating. How do you know this? Since your supposed
audience
--Advice 0.05; free, if wrong, again!
http://tarpley.net/bush6.htm
===
Subject: Re: JSH: At the Annals
>  me, they seem willing to believe just about anything!)
> Hmm, fascinating. How do you know this? Since your supposed
audience
> of sci.math readers never says anything, who are you to say
who or
> what they believe?
> Brian Chandler
DonÕt say anything? They do, at times, speak up, which is 
how
I know.
And at times IÕve done surveys, and you can Google to see
those.
I know of what I speak, sci.math readers are remarkably
gullible, as
long as someone is disagreeing with me.
Besides the usual posters, at times, others pipe up, and are
quite
consistent in that way.
Like consider the situation here, where you managed to delete
out all
the mathematics (IÕm not surprised) and where Nora Baron has
been
revealed to be a guy playing a girl. In a little bit some
sci.mathÕer
will probably come to the defense of Nora Baron as if none of
that
happened, and claim that I donÕt reply to counter examples 
or
that I
donÕt answer objections because the sci.math readership is
rather
uniformly, stupid.
James Harris
===
Subject: Re: JSH: At the Annals
>>  (Sci.math readers are very gullible as long as
posters disagree with
>> me, they seem willing to believe just about anything!)
>> Hmm, fascinating. How do you know this? Since your
supposed audience
>> of sci.math readers never says anything, who are you to
say who or
>> what they believe?
>> Brian Chandler
>DonÕt say anything? They do, at times, speak up, which is
how I know.
>And at times IÕve done surveys, and you can Google to see
those.
Indeed. Every once in a while youÕll post a survey regarding
whether youÕre right about something. ItÕs 
always almost
unanimous,
usually literally unanimous, that youÕre wrong. Hard to see
why
youÕd mention this.
>I know of what I speak, sci.math readers are remarkably
gullible, as
>long as someone is disagreeing with me.
>Besides the usual posters, at times, others pipe up, and are
quite
>consistent in that way.
>Like consider the situation here, where you managed to
delete out all
>the mathematics (IÕm not surprised) and where Nora Baron 
has
been
>revealed to be a guy playing a girl. In a little bit some
sci.mathÕer
>will probably come to the defense of Nora Baron as if none
of that
>happened, and claim that I donÕt reply to counter examples
or that I
>donÕt answer objections because the sci.math readership is
rather
>uniformly, stupid.
>James Harris
************************
David C. Ullrich
===
Subject: Re: JSH: At the Annals
> Like consider the situation here, where you managed to
delete out all
> the mathematics (IÕm not surprised) and where Nora Baron
has been
> revealed to be a guy playing a girl.
Your obsession with the gender of Nora Baron has led me to
wonder of you are
facing some kind of gender crisis yourself.
Sure enough, although I have posted anagrams from the letters
of your name
before, I have always overlooked the importance
of your middle initial. If I include that in the anagram, we
have your
suspected true name: Ms. Harie Ass, Jr. revealing
that you are, in fact, a female.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: At the Annals
> The poster is not a she, itÕs a he. DonÕt 
you get it, even
now?
You moron, donÕt *you* get it, even now?
NO ONE BUT YOU *CARES* who Nora really is.
Are you so desperate to be right about *something* that
youÕll grasp at
any straw that ßoats by, even about something as 
insignificant
as this?
> Andrzej is just another of the usual suspects who tend to
argue with
> me, who apparently thought it a good idea to play at being
a female
> posting.
> For some reason he wanted people to know it I guess, as why
else
> finally end a Nora Baron post with his own name?
The name doesnÕt matter. The only thing that matters is that
Nora,
or Andrzej, or whomever is *RIGHT* and you are *WRONG*.
Period.
End of story.
Idiot.
--
Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you
improvise
fwbrown@bellsouth.net | if youÕre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: JSH: At the Annals
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ÕELIi
$t^
VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>>I basically looked over the paper Advanced Polynomial
Factorization,
>>considered all the arguments and various comments from
journals,
>>renamed it.
> Why would you rewrite it? The original version was correct,
right?
> People who said otherwise were lying - very hard to see why
a
> rewrite would be needed.
track record in that respect was pretty much abysmal.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: JSH: At the Annals
>> Andrzej.
> And I thought that Nora Baron might be David Ullrich.
> How many of you knew? Better yet, how many of you
> guys thought the poster was actually a woman, eh?
> You really donÕt know women if you did, now do you?
Okay, Sherlock, you may have uncovered something.
> Yet, in the same post and in her previous post, Nora Baron
> points out what she claims are errors in the steps of your
> argument, which you have not addressed. Your denials
> seem to be the only thing that remains constant.
> The poster is not a she, itÕs a he. DonÕt 
you get it, even
now?
> Andrzej is just another of the usual suspects who tend to
argue with
> me, who apparently thought it a good idea to play at being
a female
> posting.
> For some reason he wanted people to know it I guess, as why
else
> finally end a Nora Baron post with his own name?
> Now given a poster lying about his own sex, why are you
defending him
> with lies about errors?
> In the last couple of days IÕve stepped through the
argument in
> different ways to give people different looks and showed
how it boils
> down to accepting that constant terms are in fact, constant.
> ItÕs remarkable to me how bizarrely some of you here on
sci.math are,
> as here you are making a claim blown out of the water about
someone
> you call she who has just given up that they are a he.
> The story would be funny if it werenÕt so important in
addressing the
> issue of irrational behavior in people. Like with global
warming, how
> can anyone ever hope that a best solution can be found that
will be
> implemented, when in an absolute area like mathematics, an
important
> research result that depends on CONSTANTS being constant
not only
> faces strong opposition, but bizarre opposition as well,
like this he,
> she thing?
> James Harris
ThatÕs just another dodge. You still havenÕt 
responded to
NoraÕs
criticisms, or to N. SilverÕs salient point.
Ōcid Ōooh
===
Subject: Re: JSH: At the Annals
posting-account=JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ
Well are you ever going to respond to the MATHEMATICAL
criticisms of
your work? Ranting about whether one of your critics is male
or female
is just a dodge to avoid facing up to reality. Or maybe you
yourself
arenÕt real? The more I read your postings the more I think
thatÕs
true. If you are a real student of mathematics answer the
mathematical
criticisms.
===
Subject: Re: JSH: At the Annals
> Well are you ever going to respond to the MATHEMATICAL
criticisms of
> your work? Ranting about whether one of your critics is
male or female
> is just a dodge to avoid facing up to reality. Or maybe you
yourself
> arenÕt real? The more I read your postings the more I 
think
thatÕs
> true. If you are a real student of mathematics answer the
mathematical
> criticisms.
What mathematical criticisms? The poster that calls himself or
herself Nora Baron or Andrzej simply rehashes things IÕve 
shot
down long ago.
ItÕs an annoying habit. But itÕs also a way to 
see posters
like
yourself who are just instigators who donÕt give a damn 
whatÕs
actually going on.
Now then, if Nora Baron or Andrzej has a *mathematical*
objection
then they can reply in a pertinent thread like in JSH:
Understanding
contant terms or JSH: Understanding polynomials where I
explain IN
DETAIL.
Posting instead in a thread where I simply inform that a
paper has
gone to a journal is the dodge.
Hey, I LIKE talking about the mathematical argument, and this
weird
she-he thing is just silly anyway, so why not move on over to
the
threads I mentioned and try again?
Now for those of you who are still confused Nora Baron is a
palindrome, and the poster using that palindrome has ended a
Nora
Baron post with the name Andrzej, which is the kind of
accident
that can happen when youÕre using a pseudonym.
Challenged on it, the poster now claims that Andrzej is a
fictional
character, apparently trying to convince that the poster is a
she, and
not a he.
Either way, you now know that you were being lied to and if
youÕre
some guy who still wants to believe that Nora Baron is a
girl, then
maybe youÕre just really lonely. But then, why 
donÕt you go
to those
dating webpages?
And leave the mathematics to the real mathematicians?
James Harris
===
Subject: Re: JSH: At the Annals
> What mathematical criticisms? The poster that calls himself
or
> herself Nora Baron or Andrzej simply rehashes things IÕve
shot
> down long ago.
You could at least cite the posts containing your refutations.
> Now for those of you who are still confused Nora Baron is a
> palindrome, and the poster using that palindrome has ended
a Nora
> Baron post with the name Andrzej, which is the kind of
accident
> that can happen when youÕre using a pseudonym.
For those of you who are *still* confused James Harris is an
anagram for
M. Harie Ass Jr., which is the kind of accident
that can happen when youÕre an arrogant blowhard.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: At the Annals
Discussion, linux)
> Either way, you now know that you were being lied to and if
youÕre
> some guy who still wants to believe that Nora Baron is a
girl, then
> maybe youÕre just really lonely. But then, why 
donÕt you go
to those
> dating webpages?
> And leave the mathematics to the real mathematicians?
Duh. You explained it yourself.
Now Ōpure mathmakes sense as well as clearly 
itÕs a peacock
game,
where some of you see it as a way to show you as being highly
intelligent and thus more desirable to women. -- James S.
Harris
--
Jesse F. Hughes
And a journal can beg me for the right to publish it [...]
because
IÕd rather see it in People magazine [...]
--James Harris on his simple proof of FermatÕs last theorem
===
Subject: Re: JSH: At the Annals
posting-account=JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ
IÕve read what Nora Baron and the others say about your work
and they
are right. You havenÕt shot anything down and claiming you
have,
without addressing the real mathematical issues, doesnÕt
further your
case in the least. This, by the way, will be my last posting
on this
matter. I have no interest in arguing with someone who is
irrational.
===
Subject: Re: JSH: At the Annals
> Well are you ever going to respond to the MATHEMATICAL
criticisms of
> your work? Ranting about whether one of your critics is
male or female
> is just a dodge to avoid facing up to reality. Or maybe you
yourself
> arenÕt real? The more I read your postings the more I 
think
thatÕs
> true. If you are a real student of mathematics answer the
mathematical
> criticisms.
> What mathematical criticisms? The poster that calls himself
or
> herself Nora Baron or Andrzej simply rehashes things IÕve
shot
> down long ago.
Errr... Looks like you missed.
> ItÕs an annoying habit. But itÕs also a way 
to see posters
like
> yourself who are just instigators who donÕt give a damn
whatÕs
> actually going on.
> Now then, if Nora Baron or Andrzej has a *mathematical*
objection
> then they can reply in a pertinent thread like in JSH:
Understanding
> contant terms or JSH: Understanding polynomials where I
explain IN
> DETAIL.
> Posting instead in a thread where I simply inform that a
paper has
> gone to a journal is the dodge.
> Hey, I LIKE talking about the mathematical argument, and
this weird
> she-he thing is just silly anyway, so why not move on over
to the
> threads I mentioned and try again?
I agree. It doesnÕt matter whether Nora Baron is really 
Nora,
Andrezj,
Richard or David. You should ignore the topic, since it does
nothing to
advance your mathematical stature, and respond to the
mathematical points
brought up in threads like Undersanding constant terms.
So when are you going to respond to the mathematical
objections of William
Hughes (What you are saying here is that if 7 divides g(0)
then 7 must
divide g(x) for all x. This is simply not true.), Rupert
(Just because
the factors are divisible by 7 when x=0, it doesnÕt follow
that the factors
are divisible by 7 for all values of x), I.M.Davidson (If you
mean that
the product of the gÕs gives you P(x) and that the root or
roots of g_r(x)
=0 are algebraic integers then the gÕs must be monic and so
must P(x)),
Arturo Magidin (If and only if the values g_i(0) are algebraic
integers),
or even Nora Baron her(him)self (you defined 
Ōconstant termÕ
in a
perfectly reasonable way,and you should have stuck with your
definition.
Instead you got confused and assumed that whatever is in the
POSITION of
the
constant term *is* the constant term)?
===
Subject: Re: JSH: At the Annals
days. My association with the Department is that of an
alumnus.
[.snip.]
>So when are you going to respond to the mathematical
objections of William
>Hughes (What you are saying here is that if 7 divides g(0)
then 7 must
>divide g(x) for all x. This is simply not true.), Rupert
(Just
because
>the factors are divisible by 7 when x=0, it doesnÕt follow
that the
factors
>are divisible by 7 for all values of x), I.M.Davidson (If
you mean
that
>the product of the gÕs gives you P(x) and that the root or
roots of
g_r(x)
>=0 are algebraic integers then the gÕs must be monic and so
must P(x)),
>Arturo Magidin (If and only if the values g_i(0) are
algebraic
>integers),
My comment was directed at I.M. Davidson, not at the original
poster.
See
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: exposing elite ideas to students
> My focus on this issue is mainly math and science. Students
studying math
> and science are exposed to what the teacher presents.
Students studying
> math and science can only study it during the school day
within those 40
> or 50 minute class periods, and they donÕt really study 
it.
They are
> presented it mainly, especially in high school.
> When a student goes to college, they are exposed to elite
ideas and
> studies in math and science, that simply was not
communicated to the
> student in the lower grades. The library at the local
school probably
> wasnÕt that great, and students could not choose to study
in the library
> as opposed to being lectured at.
> How does one communicate elite ideas to the student? We
know that the
> teacher follows a curriculum, and then they embellish it a
bit, and if
> they themselves are good at the concepts, then they can
provide some
> additional direction to students, but the students still
have to attend
> the lectures, no matter their ability, and they all take
the same tests,
> and they all graduate from that class at the same time.
There is not much
> expected of students by the teacher, because they every
day, they present
> the same material to every student. They do not encourage
them to learn
> on their own in the classroom, simply because the students
have to attend
> the lecture, what the teacher presents. More insidiously,
students find
> that they have to be able to perform on the tests, as
grades are
> ultimately the only thing that matters in a school.
Students who have not
> learned the subject, or who were not presented the material
well, will
> simply regurgitate what was presented. Those that go beyond
the teachings
> of the lectures still have to regurgitate what the teacher
presented, and
> they find themselves having to be able to perform on the
test,
> regardless of the learning pattern the student finds is best
for himself.
> To get elite ideas to the student, the most practical
solution is to get
> the elite teachers to present the material in a fashion on
video. To
> communicate the thought process that the teacher/researcher
is thinking,
> he gives some impression to the student over the video. He
then
> supplements the video with good books, and allows the
students to engage
> in group discussion over the internet. The student does not
have to
> follow at the rate that everybody else is going. In fact,
there may be
> nobody but that one student studying the subject that the
researcher
> presented. The student gets to learn at his own pace, and
he is exposed
> to the elite ideas. The good thing about this is that the
teacher who
> presents the video only has to teach the class once, and
there doesnÕt
> need to be that many alternative/redundant teachers. That
means that the
> role of the teacher is to abandon the majority of his time
spent teaching,
> and spend it on research and real world work. Just like the
horse buggy
> makers, redundant teachers in every classroom is not
necessary. What is
> necessary is getting students to be able to think
independently. Sure
> they need to ask questions, and there can be online help
forums that
> answer the questions. But as of yet, students are not being
exposed to
> elite ideas in the lower grades, and this is holding them
back from later
> success. You find most students fearful of math, and
therefore, they
> donÕt want to go into research or work on projects that
require math and
> scientific investigation.
I think that you have something here. I got into a program to
teach math,
but I abandoned it when I saw that they were teaching the
same way that I
was taught, by show of force. I think that the authoritarian
method that
schools use are not so much to get students to learn, but to
preserve their
own employment and to reduce problems. I understand that one
person
cannot possibly manage a group of students who are all at
differeng
abilities and interests and knowledge at the moment. It is
all too much to
handle, and so, a strict set curriculum is presented, and the
students
either pass, or itÕs bad news buddy. I think that relieving
teachers of
their jobs for the most part will be a breath of fresh air.
The teachers no
longer will have ulcers. They could continue their learning
and perhaps
research through this very network of elite teachers and
researchers that
you speak of. It practically means that no matter what job
you have, as
long as the job does not place too much burden on you, you
could pull up a
research class perhaps sometime during the day or at night,
and once you
reach a proficiency level, you can work on a project. 
IÕm very
excited by
this, and I hope something like this is made available.
===
Subject: Re: exposing elite ideas to students
> To get elite ideas to the student, the most practical
solution is to
get
> the elite teachers to present the material in a fashion on
video. To
To add to this, I would make it even easier on the elite
teachres. Get
maybe 5 or 10 elite teachers (probably professors or
researchers) to share
the workload of producing the class, and then have some
support people as
well to make sure that everything runs smoothly. This means
that the video
of the class is preserved in time, and you practically donÕt
ever have to
teach that class again, anywhere in the nation. Imagine the
savings.
===
Subject: probability without replacement
posting-account=4k0nMAwAAABxwiOkL8XXi5jhypNRgivn
There are two boxes, each containing three balls numbered
from 1 to 3.
If i were to pick ball 1 from both the boxes (without
replacing it
back), followed by ball 3 from both the boxes (without
replacing it
back). How would probability of ball 1 and ball 3 be
calculated. I need
to write a software for this.
All your help is appreciated.
undbund
===
Subject: Re: probability without replacement
>There are two boxes, each containing three balls numbered
from 1 to 3.
>If i were to pick ball 1 from both the boxes (without
replacing it
>back), followed by ball 3 from both the boxes (without
replacing it
>back). How would probability of ball 1 and ball 3 be
calculated. I need
>to write a software for this.
Sorry -- I hit Ōsendprematurely.
The response I gave was for one box. Assuming that you pick
from the
two boxes independently, the probability of any given scenario
happening on both boxes is the square of the probability that
it
happens on one box.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: probability without replacement
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
> There are two boxes, each containing three balls numbered
from 1 to
3.
> If i were to pick ball 1 from both the boxes (without
replacing it
> back), followed by ball 3 from both the boxes (without
replacing it
> back). How would probability of ball 1 and ball 3 be
calculated.
I donÕt understand your question.
You state it as a given that you have picked ball 1 from
both boxes. Since all possible outcomes start with the
fact that you have picked ball 1 (twice), what do you
mean by the probability of ball 1?
If you ask, what is the probability of picking ball 1
from both boxes on the first draw, the answer is
(1/3)*(1/3) = 1/9.
For the second pick, there are only two possibilities
in each box. The probability of picking 3 from each
box is (1/2)*(1/2) = 1/4.
The probability of this sequence of events is
(1/9)*(1/4) = 1/36.
- Randy
===
Subject: Re: expectation value for ODE?
Hi Thomas
> What do you mean by value? Suppose you have a dynamical
system in
> IR^n given by x= f(x,m), where m is an n-dimensional
parameter. You
> may consider the initial value problem xÕ=f(x,m0), x(0)=x0
and compute
> x(T). Here x0,m0,T are fixed.
IÕm trying to figure out what is meant by the 
below.
The expected value of f(X) , where X is a random variable with
probability density function p(x) and support [a,b], where
-Infinity
<= a < b <= +Infinity , is
/b
E(f) = | f((x) p(x) dx
/a
which is supposed to be approximated by the sample mean
_ n
M= 1/n f(xi)
/
--i=1
> OK. LetÕs assume you want to find the 
distribution of x(T)
given a
> distribution of (x0,m0,T) in IR^(4n+1). If f is a nice,
sufficiently
> smooth function of x and m, then the standard theorems of
ODE theory
> guarantee smooth dependendence of x(T) depending on T, x0
and m0.
> So the distribution of x(T) is just the image of the
distribution of
> (x0,m0,T) under the ßow of your equation.
Where would I find a formal definition of uch 
theorems? ( I
canÕt go
over the entired literature on ODEs. is there a good textbook
on
this?, namely, theorems guaranteeing the smooth dependence of
x(t) on
T, x0, m0.
>is that same as the deterministic solution given an m?
>so...
>y(t) = e^(mt)* C
> In this case you have three random variables m, C and t
(but better
> think of t as fixed, randomly distributed value T) and you
have to
> define what the right distribution of them are.
> HTH, Thomas
===
Subject: Is a circle just a 2-dimensional sphere?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iATLOIR15119;
Is a circle just a 2-dimensional sphere?
===
Subject: Re: Is a circle just a 2-dimensional sphere?
> Is a circle just a 2-dimensional sphere?
No, circles and spheres are different things.
If you insist on the answer yes then proceed as follows.
Define an n-dimensional (for positive integer n) sphere to be
the set of
points
(x_1, x_2, ..., x_n)
such that
x_1^2 + x_2^2 + ... + x_n^2 = r^2
for some real r >= 0.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
>>Is a circle just a 2-dimensional sphere?
> No, circles and spheres are different things.
> If you insist on the answer yes then proceed as follows.
> Define an n-dimensional (for positive integer n) sphere to
be the set of
> points
> (x_1, x_2, ..., x_n)
> such that
> x_1^2 + x_2^2 + ... + x_n^2 = r^2
> for some real r >= 0.
This is traditionally known as the n-1 dimensional sphere of
radius r.
So the circle (presuming we mean just the circle, not the
disk) is the
1-dimensional sphere, the surface {x,y,z|x^2+y^2+z^2=r^2} is
the
2-dimensional sphere, etc. The reason for the number being
n-1 is the
dimension of the space described. The circle is a 1-manifold,
the
2-sphere is a 2-dimensional manifold, etc.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
>
>>Is a circle just a 2-dimensional sphere?
> No, circles and spheres are different things.
If you insist on the answer yes then proceed as follows.
Define an n-dimensional (for positive integer n) sphere to be
the set
of
> points
(x_1, x_2, ..., x_n)
such that
x_1^2 + x_2^2 + ... + x_n^2 = r^2
for some real r >= 0.
> This is traditionally known as the n-1 dimensional sphere
of radius r.
> So the circle (presuming we mean just the circle, not the
disk) is the
> 1-dimensional sphere, the surface {x,y,z|x^2+y^2+z^2=r^2}
is the
> 2-dimensional sphere, etc. The reason for the number being
n-1 is the
> dimension of the space described. The circle is a
1-manifold, the
> 2-sphere is a 2-dimensional manifold, etc.
And, just for the record, a 0-sphere consists of 2 points.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`ßY:3QYT$>dUwN^sm;MBV:
F7aL9x*q!`
ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/Õ>}Pc?@rl8cz]
d9RXOt Define an n-dimensional (for positive integer n) sphere to
be the set
of
> points
(x_1, x_2, ..., x_n)
such that
x_1^2 + x_2^2 + ... + x_n^2 = r^2
for some real r >= 0.
> This is traditionally known as the n-1 dimensional sphere
of radius r.
> So the circle (presuming we mean just the circle, not the
disk) is the
> 1-dimensional sphere, the surface {x,y,z|x^2+y^2+z^2=r^2}
is the
> 2-dimensional sphere, etc. The reason for the number being
n-1 is the
> dimension of the space described. The circle is a
1-manifold, the
> 2-sphere is a 2-dimensional manifold, etc.
> And, just for the record, a 0-sphere consists of 2 points.
To make things even clearer, let me point out that the
-1-sphere is the
empty set.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
>Is a circle just a 2-dimensional sphere?
No, a circle is just a 1-dimensional sphere.
Lee Rudolph
===
Subject: Re: Is a circle just a 2-dimensional sphere?
lr>
lr>>Is a circle just a 2-dimensional sphere?
lr>
lr>No, a circle is just a 1-dimensional sphere.
lr>
lr>Lee Rudolph
A circle includes its interior and is a 2-dimensional sphere;
a line segment is a 1-dimensional sphere,
a dot is a 0-dimensional sphere.
An n-dimensional sphere is the set of points with their n-sum
of
squared Cartesian coordinates (coordinate differences to the
center)
less or equal the squared radius.
Sphere surfaces and circle rims are another theme.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
ETAsAhQednPc1mwXks/KZ8X2AHnQLh+
NUgIUe2X02DgoB4t9MkQsvzlUkHTFIWI=
I think of a circle as the circumference of the area I call a
disk.
Likewise a sphere is the surface of a ball.
ThatÕs the way I see it, FWIW.
--OL
===
Subject: Re: Is a circle just a 2-dimensional sphere?


at 09:56 PM, mathar@amer.strw.leidenuniv.nl (Richard Mathar)
said:
>A circle includes its interior
No. A circle plus its interior is a disk. You are similarly
wrong on
the others.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: Is a circle just a 2-dimensional sphere?
>a line segment is a 1-dimensional sphere,
Is that right?
Sphere (3-D): x^2 + y^2 + z+2 = r^2
Circle (2-D): x^2 + y^2 = r^2
?????? (1-D): x^2 = r^2
To me that looks like two dots at equal and opposite
distances from
the origin, not a line segment. A line segment would be a one-
dimensional _ball_ to use David UllrichÕs term.
>An n-dimensional sphere is the set of points with their
n-sum of
>squared Cartesian coordinates (coordinate differences to the
center)
>less or equal the squared radius.
Maybe we went to different schools, but teh definitions I
learned all
had = and not <=.
--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
A: Maybe because some people are too annoyed by top-posting.
Q: Why do I not get an answer to my question(s)?
A: Because it messes up the order in which people normally
read text.
Q: Why is top-posting such a bad thing?
===
Subject: Re: Is a circle just a 2-dimensional sphere?
^OX9W/.#XpUmm`>TD2zNE-t}emfPkFR.Z5`ßY:3QYT$>dUwN^sm;MBV:
F7aL9x*q!`
ln!l}>Y6_45$%R|P7DSrBkEph@1-;P*s~F_28vO@e4p/Õ>}Pc?@rl8cz]
d9RXOt Sphere (3-D): x^2 + y^2 + z+2 = r^2
> Circle (2-D): x^2 + y^2 = r^2
> ?????? (1-D): x^2 = r^2
> To me that looks like two dots at equal and opposite
distances from
> the origin[...]
The zero dimensional sphere. Useful for topologists. S^0 is
two
points, S^1 is a circle, S^2 is a sphere, etc. The exponent
indicates
the topological dimension.
Oh, and S^{-1} is the empty set :-) Depending on how you set
up your
definition, this is the result for the round sphere in R^0.
Some might
consider it strange, but it makes some formulae come out
correctly in
situations where you might not care whether one set is empty.
For
example, the join formula for two spheres, S^{n+m+1} = S^n *
S^m holds.
===
Subject: Re: Is a circle just a 2-dimensional sphere?
>lrlr>>Is a circle just a 2-dimensional sphere?
>lrlr>No, a circle is just a 1-dimensional sphere.
>lrlr>Lee Rudolph
>A circle includes its interior
No, a circle plus its interior is a _disk_
>and is a 2-dimensional sphere;
which is a 2-dimensional _ball_.
>a line segment is a 1-dimensional sphere,
>a dot is a 0-dimensional sphere.
>An n-dimensional sphere is the set of points with their
n-sum of
>squared Cartesian coordinates (coordinate differences to the
center)
>less or equal the squared radius.
No, thatÕs an n-dimensional _ball_. The boundary of a ball 
is
a
sphere.
>Sphere surfaces and circle rims are another theme.
************************
David C. Ullrich
===
Subject: Re: JSH: Operator ambiguity, Escultura
> You see, the ambiguity in the square root operator still
remains,
> despite the convention.
Your insistence that it is impossible to define a function for
sqrt(x),
i.e. to restrict the possible solutions to a single value,
invalidates
your own prime count function and disqualifies you from
further study of
algebra. Analogous situations arise in evaluating periodic
functions, in
finding Ōprincipal valuesÕ, in 
dealing with Ōbranch cutsand
elsewhere.
If you cannot understand how these issues are resolved, take
up a
different line of work. (Not sales, however. I doubt if you
could sell
manure to an organic farmer.)
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Operator ambiguity, Escultura
1 = 0 is a well known fact among insiders. There are
hundreds of proofs for this fact. The last one was
published by an Australian mathematician:
Jamie Simpson, A New Proof that 1=0,
JRM vol 32, Number 2, page 142
Richard Schorn
> So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
^^^^ ???
===
Subject: Re: JSH: Operator ambiguity, Escultura
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
I donÕt think this is the right definition of 
the imaginary
number i.
According to met i is defined as i^2 = -1. The function f(x)=
x^2 is
well defined because if x=y then x^2=y^2. However, the inverse
Ōfunctionis not really a function, because it 
can have more
than
one image f^-1(4) can be both 2 and -2.
So, by
convention
the inverse function of f is defined from the positive
orthonant of R
to the positive orthonant of R, so we have sqrt(4) = 2. and
sqrt(-1),
sadly enough is not defined. the
correspondence that you define, call this sqrt*) is not a
function
that has an element as an image, but a whole set as an image
(sqrt*(4)= {2,-2}).[/b]
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
> First some more preamble as *by convention* as has been
noted when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
Wait, youÕre not suggesting that sqrt(0) is positive are 
you?
Just
kidding.
Of course you arenÕt. As you well know, the convention that
sqrt(x) is
positive applies only to positive numbers x.
> So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
Hey! I remember thinking this was really cool. I did it with
5 = -5, but
it was the same idea. I showed it to my teacher (this was in
sixth grade),
and he said I couldnÕt do that, but he wasnÕt 
able to give me
a good reason
why not.
The reason it appealed to me so much was that once you have a
result like
that, you can prove anything. Wow! All you need to prove
anything is a
fuzzy definition like sqrt(n) is the thing that gives you n
when you
square
it. Provided youÕre willing to put your fingers 
in your ears
and say
la
la la la I am not lis en ing when mathematicians try to
explain why such
a
definition is invalid.
> Naively then, you may believe that you can just say, take
the positive
> of the square root but as Escultura showed, that doesnÕt
work:
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
Oh, this is the more sophisticated version. Usually itÕs
shown with a
few more steps:
i = sqrt(-1) = sqrt(1/-1) = sqrt(1)/sqrt(-1) = 1/i = -i.
^
ERROR!
> You see, the ambiguity in the square root operator still
remains,
> despite the convention.
Umm, no.
> It doesnÕt work to just try and always take the positive 
as
> EsculturaÕs example shows so clearly.
> Who has the resolution? IÕm curious as to whether or not
any of you
> think you can answer.
You ask this question like itÕs at the forefront of
mathematical knowledge
because (a) itÕs at the frontier of your mathematical
knowledge, and (b)
you believe that you are blazing new trails in mathematics.
Assumption (b) is whatÕs holding you back from learning 
math.
Before you
can learn, you have to realize that you donÕt know.
> James Harris
===
Subject: Re: JSH: Operator ambiguity, Escultura
hey, then 0 = 5 - 5 = -5 - 5 = -10;
you should turn that into a tutorial on Zero Divisors ... and
watch out for those chaotic rounding-off errors!
> Hey! I remember thinking this was really cool. I did it
with 5 = -5,
but
--Advice 0.05; Free, if wrong, again!
http://tarpley.net/bush6.htm
===
Subject: Re: JSH: Operator ambiguity, Escultura
> Assumption (b) is whatÕs holding you back from learning
math. Before you
> can learn, you have to realize that you donÕt know.
A Fundamental Truth.
===
Subject: Re: JSH: Operator ambiguity, Escultura
> Assumption (b) is whatÕs holding you back from learning
math. Before
you
> can learn, you have to realize that you donÕt know.
> A Fundamental Truth.
Oh please, the poster never answered the mathematical issue
raised but
instead turned to personal attacks. You deleted out all of the
context and made a reply that had nothing to do with the
issue.
For those who missed it, I used EsculturaÕs example of
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1, contradiction,
to show that itÕs naive to think that you can remove the
ambiguity
from the square root operator.
Rather than give a cogent answer the sci.math poster Jim
Ferry babbled
about his childhood, and simply *claimed* as if thatÕs all 
it
takes
that thereÕs no problem.
The regular sci.math poster Gib Bogle then came in to delete
out all
of the context and make his own reply.
ThatÕs how sci.mathÕers managed to paint me as 
a crank. They
do this
consistently, and sci.math readers cheer them on!
The sci.math readership is remarkably stupid and gullible.
James Harris
===
Subject: Re: JSH: Operator ambiguity, Escultura
Assumption (b) is whatÕs holding you back from learning 
math.
Before
you
> can learn, you have to realize that you donÕt know.
A Fundamental Truth.
> Oh please, the poster never answered the mathematical issue
raised but
> instead turned to personal attacks. You deleted out all of
the
> context and made a reply that had nothing to do with the
issue.
> For those who missed it, I used EsculturaÕs example of
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1,
contradiction,
> to show that itÕs naive to think that you can remove the
ambiguity
> from the square root operator.
> Rather than give a cogent answer the sci.math poster Jim
Ferry babbled
> about his childhood, and simply *claimed* as if thatÕs all
it takes
> that thereÕs no problem.
> The regular sci.math poster Gib Bogle then came in to
delete out all
> of the context and make his own reply.
> ThatÕs how sci.mathÕers managed to paint me 
as a crank.
They do this
> consistently, and sci.math readers cheer them on!
Can you give me an example showing sci.math readers cheering
them on?
> The sci.math readership is remarkably stupid and gullible.
Is it? How do you know?
Brian Chandler
http://imaginatorium.org
===
Subject: Re: JSH: Operator ambiguity, Escultura
...
> Oh please, the poster never answered the mathematical issue
raised but
> instead turned to personal attacks. You deleted out all of
the
> context and made a reply that had nothing to do with the
issue.
But some posters answered it.
> For those who missed it, I used EsculturaÕs example of
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1,
contradiction,
The contradiction exists because the assumption is made that
sqrt(a/b) = sqrt(a)/sqrt(b), which is false when in the
complex
numbers. Similarly log(a.b) = log(a) + log(b) is false when in
the complex numbers. Read a bit about Riemann surfaces and you
may understand.
--
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: JSH: Operator ambiguity, Escultura
> Read a bit about Riemann surfaces and you may understand.
He wonÕt.
--
--Tim Smith
===
Subject: Re: JSH: Operator ambiguity, Escultura
> ...
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1,
contradiction,
This is a reductio proof that sqrt(x/y) <> sqrt(x)/sqrt(y) .
This is a well-known and not very exciting fact.
===
Subject: Re: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
> First some more preamble as *by convention* as has been
noted when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
> So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
> Naively then, you may believe that you can just say, take
the positive
> of the square root but as Escultura showed, that doesnÕt
work:
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
> You see, the ambiguity in the square root operator still
remains,
> despite the convention.
> It doesnÕt work to just try and always take the positive 
as
> EsculturaÕs example shows so clearly.
> Who has the resolution? IÕm curious as to whether or not
any of you
> think you can answer.
Typical mistake of a beginner: In the complex numbers, sqrt
(a/b) = sqrt
(a) / sqrt (b) is not usually true. The convention decides
about the
sign of the result. This decision cannot possible made to be
consistent
in all cases for sqrt (a), sqrt (b) and sqrt (a/b).
Convert your complex numbers into polar coordinates, look at
what the
convention does to the phase angle, and it should be quite
obvious where
your mistake is.
===
Subject: Re: JSH: Operator ambiguity, Escultura
> It doesnÕt work to just try and always take the positive 
as
> EsculturaÕs example shows so clearly.
In general a complex number has two square roots, both
complex.
So it doesnÕt make sense to try to take the positive.
However, there are two cases
i) x^2 - a = 0 has two roots, one with positive real
part and one with negative real part.
ii) x^2 - a = 0 has two roots both with a real part of zero
So to get a unique square root.
i) take the root with positive real part
ii) take the root with positive imaginary part
Now x^2 +1 =0 has two roots, i and -i. Both have 0 real part.
So
by ii) we choose i and say the (principlal value of) the
square root
of -1 is i.
> Who has the resolution?
Me, No one else. Nobody ever though about this before me.
I deserve an Acadamy Award.
> IÕm curious as to whether or not any of you
> think you can answer.
> James Harris
Be curious no longer.
-William Hughes
===
Subject: Re: JSH: Operator ambiguity, Escultura
> ...
> of the square root but as Escultura showed, that doesnÕt
work:
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
You, or Escultura, whomever, thereby proves that
sqrt(x/y) != sqrt(x)/sqrt(y)
===
Subject: Re: JSH: Operator ambiguity, Escultura
>I would like to pull out and highlight something interesting
that E.
>E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
>talked about many times before, but I just noticed it and
think itÕs
>neat.
>First some more preamble as *by convention* as has been
noted when I
>brought up the subject of operator ambiguity before, sqrt(x)
is taken
>to be positive.
>So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
>if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
>which is not good.
>Naively then, you may believe that you can just say, take
the positive
>of the square root but as Escultura showed, that doesnÕt
work:
>i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
>You see, the ambiguity in the square root operator still
remains,
>despite the convention.
>It doesnÕt work to just try and always take the positive as
>EsculturaÕs example shows so clearly.
Of course you canÕt take the positive square root of a
negative
number - it doesnÕt _have_ a positive square root! You seem 
to
be under the impression that there is such a convention -
there
is not, the convention about the meaning of sqrt(x) is only
for x > 0.
You also seem to be under the impression that i is positive,
which is kind of funny - itÕs not. ItÕs only 
real numbers
that are positive or negative.
Finally, you seem to have missed the most important point
in the contradiction above, which is that itÕs impossible
to define sqrt(z) for complex z in such a way that
sqrt(zw) = sqrt(z) sqrt(w).
Except for totally missing the main points it was an
excellent post.
>Who has the resolution? IÕm curious as to whether or not 
any
of you
>think you can answer.
Guffaw. ThereÕs no contradiction to be resolved, because the
calculuation above is simply wrong, using properties of sqrt
that nobody ever claimed were valid.
>James Harris
************************
David C. Ullrich
===
Subject: Re: JSH: Operator ambiguity, Escultura
...
>Naively then, you may believe that you can just say, take
the positive
>of the square root but as Escultura showed, that doesnÕt
work:
>
>i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
>
>You see, the ambiguity in the square root operator still
remains,
>despite the convention.
...
> Finally, you seem to have missed the most important point
> in the contradiction above, which is that itÕs impossible
> to define sqrt(z) for complex z in such a way that
> sqrt(zw) = sqrt(z) sqrt(w).
He is missing one more point. That you can not algebraically
distinguish
i and -i is also a result from Galois theory (you can not
algebraically
distinguish the various roots of a primitive polynomial).
--
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: JSH: Operator ambiguity, Escultura
>...
>Naively then, you may believe that you can just say, take
the positive
>of the square root but as Escultura showed, that doesnÕt
work:
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
You see, the ambiguity in the square root operator still
remains,
>despite the convention.
>...
> Finally, you seem to have missed the most important point
> in the contradiction above, which is that itÕs impossible
> to define sqrt(z) for complex z in such a way that
> sqrt(zw) = sqrt(z) sqrt(w).
>He is missing one more point. That you can not algebraically
distinguish
>i and -i is also a result from Galois theory (you can not
algebraically
>distinguish the various roots of a primitive polynomial).
Well, yes, although thatÕs at least related to one I did
point out:
You also seem to be under the impression that i is positive.
************************
David C. Ullrich
===
Subject: Re: JSH: Operator ambiguity, Escultura
>...
>Naively then, you may believe that you can just say, take the
positive
>of the square root but as Escultura showed, that doesnÕt
work:
>
>i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
>
>You see, the ambiguity in the square root operator still
remains,
>despite the convention.
>...
> Finally, you seem to have missed the most important point
> in the contradiction above, which is that itÕs impossible
> to define sqrt(z) for complex z in such a way that
> sqrt(zw) = sqrt(z) sqrt(w).
>
>He is missing one more point. That you can not algebraically
distinguish
>i and -i is also a result from Galois theory (you can not
algebraically
>distinguish the various roots of a primitive polynomial).
> Well, yes, although thatÕs at least related to one I did
point out:
> You also seem to be under the impression that i is positive.
But being positive is not an algebraic property.
--
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: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
> First some more preamble as *by convention* as has been
noted when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
> So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
> Naively then, you may believe that you can just say, take
the positive
> of the square root but as Escultura showed, that doesnÕt
work:
> i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1.
Contradiction.
If we make the convention that sqrt(-1)=i, then itÕs not 
true
that
sqrt(1/-1)=sqrt(1)/sqrt(-1).
> You see, the ambiguity in the square root operator still
remains,
> despite the convention.
> It doesnÕt work to just try and always take the positive 
as
> EsculturaÕs example shows so clearly.
> Who has the resolution? IÕm curious as to whether or not
any of you
> think you can answer.
> James Harris
===
Subject: Re: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
First some more preamble as *by convention* as has been noted
when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
Naively then, you may believe that you can just say, take the
positive
> of the square root but as Escultura showed, that doesnÕt
work:
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
If we make the convention that sqrt(-1)=i, then itÕs not 
true
that
> sqrt(1/-1)=sqrt(1)/sqrt(-1).
So your assertion is that the substitution 1/-1 = -1 cannot
be made?
The resolution is that the sqrt() operator gives TWO answers.
It
always does, and convention canÕt force it to give only one
answer.
So i = +/-sqrt(-1), where the +/- in front is a nod to the
reality
that the result of using the square root operator is two
solutions.
ThatÕs the operator ambiguity.
Naively you may believe that if you simply say, take only ONE
answer
from the square root, and try to figure out some way to do it
that you
can succeed, but you will always have a contradiction lurking.
The ambiguity extends to other operators like the cuberoot
operator or
an infinity of other operators where you get more than one
answer.
James Harris
===
Subject: Re: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
First some more preamble as *by convention* as has been noted
when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4,
so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
Naively then, you may believe that you can just say, take the
positive
> of the square root but as Escultura showed, that doesnÕt
work:
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
If we make the convention that sqrt(-1)=i, then itÕs not 
true
that
> sqrt(1/-1)=sqrt(1)/sqrt(-1).
> So your assertion is that the substitution 1/-1 = -1 cannot
be made?
> The resolution is that the sqrt() operator gives TWO
answers. It
> always does, and convention canÕt force it to give only 
one
answer.
No, youÕre confusing the square root of a number (which has
both a positive
answer and itÕs negation) with the sqrt() operator, which is
_defined_ to
yield the positive root.
This definition prohibits the implied step above: i =
sqrt(1)/sqrt(-1),
because this equation is false since 1/sqrt(-1) = 1/i = -i.
> So i = +/-sqrt(-1), where the +/- in front is a nod to the
reality
> that the result of using the square root operator is two
solutions.
> ThatÕs the operator ambiguity.
No, i does not = +/-sqrt(-1). YouÕre tilting at 
definitions
again. The
_definition_ of Ōiis i = 
sqrt(-1).
KeithK
> Naively you may believe that if you simply say, take only
ONE answer
> from the square root, and try to figure out some way to do
it that you
> can succeed, but you will always have a contradiction
lurking.
> The ambiguity extends to other operators like the cuberoot
operator or
> an infinity of other operators where you get more than one
answer.
> James Harris
===
Subject: Re: JSH: Operator ambiguity, Escultura
> I would like to pull out and highlight something
interesting that E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think itÕs
> neat.
First some more preamble as *by convention* as has been noted
when I
> brought up the subject of operator ambiguity before,
sqrt(x) is taken
> to be positive.
So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) = 4,
so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4 = 0,
> which is not good.
Naively then, you may believe that you can just say, take the
positive
> of the square root but as Escultura showed, that doesnÕt
work:
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
> If we make the convention that sqrt(-1)=i, then itÕs not
true that
> sqrt(1/-1)=sqrt(1)/sqrt(-1).
> So your assertion is that the substitution 1/-1 = -1 cannot
be made?
No.
> The resolution is that the sqrt() operator gives TWO
answers. It
> always does, and convention canÕt force it to give only 
one
answer.
You can make a convention that it only gives one answer, but
you may
have to sacrifice identities like sqrt(a/b)=sqrt(a)/sqrt(b).

===
Subject: Re: JSH: Operator ambiguity, Escultura
> The resolution is that the sqrt() operator gives TWO
answers. It
> always does, and convention canÕt force it to give only 
one
answer.
Sure it can. Consider a square surface with an area of 4
square units.
Answer the question: What is the length
of one side? Are you saying that -2 units is a correct answer?
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Operator ambiguity, Escultura
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
> I would like to pull out and highlight something
interesting that
E.
> E. Escultura posted a few days ago, which IÕd guess 
heÕs
probably
> talked about many times before, but I just noticed it and
think
itÕs
> neat.
First some more preamble as *by convention* as has been noted
when I
> brought up the subject of operator ambiguity before,
sqrt(x) is
taken
> to be positive.
So, by the convention, sqrt(4) = 2, and thatÕs good as,
-2(-2) =
4, so
> if you say that sqrt(4) = 2 and sqrt(4) = -2, then 2 = -2,
and 4
= 0,
> which is not good.
Naively then, you may believe that you can just say, take the
positive
> of the square root but as Escultura showed, that doesnÕt
work:
i = sqrt(-1) = sqrt(1/-1) = 1/i, giving -1 = 1. Contradiction.
If we make the convention that sqrt(-1)=i, then itÕs not 
true
that
> sqrt(1/-1)=sqrt(1)/sqrt(-1).
> So your assertion is that the substitution 1/-1 = -1 cannot
be made?
sqrt(a/b) = sqrt(a)/sqrt(b) is not true for every real a, b:
a <> b
> The ambiguity extends to other operators like the cuberoot
operator
or
> an infinity of other operators where you get more than one
answer.
Can you show few examples with the other operators?
===
Subject: Re: JSH: Operator ambiguity, Escultura
Discussion, linux)
> The ambiguity extends to other operators like the cuberoot
operator
> or an infinity of other operators where you get more than 
one
> answer.
How does one get more than one answer for the cuberoot
operator (the
one that takes reals to reals, mind you)?
--
ItÕs an exercise in game theory[...] IÕve been 
brutally
logical in
my analysis on this point.[...] My analysis indicates that the
optimal strategy for mathematicians is to acknowledge the
result
today. --JSH gives practical reasons to accept his FLT proof
===
Subject: about complex analysis...
IÕve seen many times the proof of the fundamental theorem of
algebra using
LiouvilleÕs theorem. Is it possible to prove it using the
maximum principle?
Would it be similar to the one using LiouvilleÕs theorem?
===
Subject: Re: about complex analysis...
>IÕve seen many times the proof of the fundamental theorem 
of
algebra using
>LiouvilleÕs theorem. Is it possible to prove it using the
maximum
principle?
Yes.
>Would it be similar to the one using LiouvilleÕs theorem?
Kind of. Say P is a polynomial with no zero (and deg(P) > 0).
Then 1/P is bounded entire function, so Liouville shows itÕs
constant, contradiction. But if you think of it for a second,
the proof that 1/P is bounded actually shows that 1/P has
the stronger property _________, and so the maximum principle
shows that 1/P ________, a contradiction.
(This is homework, right?)
Never noticed this - in fact FTA is a simple consequence
of the maximum principle, so one wonders why they do it by
Liouville in all those books. Must be theyÕre just looking
for applications of Liouville.
************************
David C. Ullrich
===
Subject: FREE Nintendo DS--THIS REALLY WORKS!
posting-account=KapUSA0AAAD3Ymzw9IhehT-VYFoHZhR1
For a free Nintendo ds go to
http://www.ds4free.com/default.aspx?r=61306 ! This only
requires 4
referrals and is with a trusted company that has been around
for a long
time (offercentric)! Get in quick!
===
Subject: ignore my earlier post -- IÕve solved it!!!! BUT 
now
IÕm now stuck on a very important modification 
of the problem.
Let T(n) be some nxn matrix.
Let Z(n) = I(n) + T(n) T(n).
Let v(n) be the nx1 vector with all matrix elements equal to
1/n.
Define the quadratic form
q = v(n)T(n)Z(n)^(-1) T(n) v(n).
What are necessary and sufficient conditions on the matrix 
T(n)
such that
q -> 0 as n -> infinity.
(Hint: I believe it is true for ALL T(n), but I have trouble
proving it.)
Roddy
===
Subject: Re: ignore my earlier post -- IÕve solved it!!!! 
BUT
now
>IÕm now stuck on a very important modification 
of the problem.
>Let T(n) be some nxn matrix.
Real, I assume.
>Let Z(n) = I(n) + T(n) T(n).
>Let v(n) be the nx1 vector with all matrix elements equal to
1/n.
>Define the quadratic form
>q = v(n)T(n)Z(n)^(-1) T(n) v(n).
>What are necessary and sufficient conditions on the matrix
T(n)
>such that
>q -> 0 as n -> infinity.
>(Hint: I believe it is true for ALL T(n), but I have trouble
proving it.)
Yes, itÕs true.
For convenience, IÕll leave out all the 
(n)Õs.
Write T = B R where B is the positive semidefinite
square root of T Tand R is orthogonal. Then
TZ^(-1) T = RB (I+B^2)^(-1) B R so
||TZ^(-1) T|| <= 1
And then |q| <= ||v||^2 = 1/n
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: ignore my earlier post -- IÕve solved it!!!! 
BUT
now
>Let T(n) be some nxn matrix.
> Real, I assume.
>Let Z(n) = I(n) + T(n) T(n).
>Let v(n) be the nx1 vector with all matrix elements equal to
1/n.
>Define the quadratic form
>q = v(n)T(n)Z(n)^(-1) T(n) v(n).
>What are necessary and sufficient conditions on the matrix
T(n)
>such that
>q -> 0 as n -> infinity.
>(Hint: I believe it is true for ALL T(n), but I have trouble
proving
it.)
> Yes, itÕs true.
> For convenience, IÕll leave out all the 
(n)Õs.
> Write T = B R where B is the positive semidefinite
> square root of T Tand R is orthogonal. Then
> TZ^(-1) T = RB (I+B^2)^(-1) B R so
> ||TZ^(-1) T|| <= 1
> And then |q| <= ||v||^2 = 1/n
Rod
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Invariant Galilean Transformations (FAQ) On All
Laws
>> Androcles is a well-known fool, and his connection with
reality has
>> been served no useful purpose when you treat him as if he
was more
>> intelligent than he actually is, as you have done above.
It is interesting that she did not see fit to write explaining
what
specific objections to his points.
Androcles
===
Subject: Re: Invariant Galilean Transformations (FAQ) On All
Laws
Your sleazy-minded innuendoes are not appreciated. I do not
discuss
things with smutty-minded perverts.
Does your wife even know that she is married to a disgusting
pervert.
Maybe somebody should tell her what an absolute jerk you are.
*plonk*
To Androcles
> McAnally, shadows with the pseudonym Speicher,
> enjoy your tour into the psychotic world of
> ultra kook dom, theÕre trolls, by any other name.
> Ken
I have not posted to this newsgroup for some time, but your
comments
> make me want to puke.
> Yes, I read your post Judith, and I think I
> could arrange for a bit of morning sickness
> you want. When?
> Your disgusting and derisive comments were also made
without making
> any genuine contribution to the thread. Your empty-headed
tripe makes
> you a troll, because you waste everybodyÕs time with your
comment.
You also made your response knowing full-well that neither
Speicher
> nor McAnally would ever get to see it and that they would
not be able
> to defend themselves against your unwarranted abuse.
Speicher will
> not be able to read your post because he no longer reads
the group,
> and McAnally will not be able to read your post because he
has you
> kill filed (both of these are public knowledge). Attacking
people who
> have no recourse to respond is the action of a
yellow-bellied coward.
> So you interested in the color of my belly are you,
> IÕm beginning to like you.
> It is interesting that you did not see fit to write
explaining what
> objections to his points. Presumably you do not understand
what the
> discussion was about, and, therefore, were not able to any
worthwhile
> comments or contributions. It is interesting that you used
ŌtrollsÕ
> to rudely attack two people who have no recourse to answer
you, when
> you, yourself, are a troll of the worst kind. It is evident
that you
> are actually describing yourself and your own nature, when
you use the
> word Ōtroll(after all, you were the person 
who introduced
this
> epithet).
> Yes, I was terribly mean, IÕll make it up to you.
> Androcles is a well-known fool, and his connection with
reality has
> been served no useful purpose when you treat him as if he
was more
> intelligent than he actually is, as you have done above.
> Oh, thatÕs a sure sign of affection, you and I
> could gossip about the neighbours after were done
> getting you started on morning sickness.
> Is there any truth to the rumour that you changed one
letter in your
> name from an ŌFto a 
ŌTÕ?
> a woman e-travels 12,000 miles to ßirt with me.
> IÕd would have stuck with F but I got to many
> messages from all them beautiful starlets, but
> since the most beautiful women come from Australia
> I thought IÕd use Tucker, meaning I nourish the
> babes.
> When?
> Ken F Tucker
===
Subject: Re: Invariant Galilean Transformations (FAQ) On All
Laws
> Your sleazy-minded innuendoes are not appreciated. I do not
discuss
> things with smutty-minded perverts.
> Does your wife even know that she is married to a
disgusting pervert.
> Maybe somebody should tell her what an absolute jerk you
are.
Keep talkin dirty, youÕre turning me on.
Well Miss (goody two shoes) Hardy you started the
subject with this question, I quote you,
> Is there any truth to the rumour that you changed one
letter in your
> name from an ŌFto a 
ŌTÕ?
Which coming from a purported woman you sounded
like a horny slut, and thatÕs a life-style choice,
so I answered your question by letting you know
my name rhymes with the reputation, to satisfy
your evident curiousity about my sex life.
PS: Last time I heard that joke was in Public
School (I guess youÕd spell that Pubic School).
When?
Ken
===
Subject: Re: Mathworld errors
> I just got an email
> inviting me to send him a list of errors on mathworld.
> I havenÕt saved such a list, thought IÕd 
mention this
> in case anyone else has.
>Recently, on sci.math, you mentioned:
><to believe, you might at least include a question mark or
>something...
>(Today must be your first visit to mathworld - 
itÕs really
full of
>errors. I admit this is a good one.)>If you would be so kind
as to send us a list, Eric and I will be
>glad to fix them.
I have kept lists of errors, and used to submit them to Eric
every once in
a while. But it eventually became apparent to me that I could
spend a
large part of my life doing nothing but correcting MathWorld
errors...
As it happens, the good one mentioned above was on my list of
errors
yet
to be submitted, so thatÕs one I can remove from the list.
Just for the
heck of it, I decided to take care of the last four errors on
the list
now. Well, the last one, which concerned an equation in the
entry
Triangle
Wave, seems to have been taken care of already, so here are
the remaining
three:
1) In ,
the reference
Whittaker, J. M. On the Cardinal Function of Interpolation
Theory.
Proc.
Edinburgh Math. Soc. 2, 41-46, 1927.
is slightly incorrect. (The mistake seems to have been copied
from the
reference in the paper by McNamee et al.) Instead of volume
2, it should
be series 2, volume 1.
2) In 
absolute values are needed in the denominators.
3) In 
the correct spelling is casus irreducibilis, rather than
...lus.
[A copy of this post has been sent to Ed Pegg.]
David W. Cantrell
===
Subject: Re: Mathworld errors
> I just got an email
....
> 2) In  2) In  or in the numerators for radius of curvature.. right?
No. I meant what I said: In the entry for radius of
curvature, absolute
values are needed in the denominators. (Unless thereÕs a
usage which is
different from the one IÕm familiar with, we want the radius
of curvature
to be nonnegative always.)
I havenÕt looked at the entry for curvature itself.
David
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
thank you foryour answer.
I am interested by both questions:
Let T be an operator between two Banach spaces (or Hilbert
spaces),
what are the conditions on T and on these spaces such that T
admits a
Taylor series expansion, and how to perform this expansion.
I would appreciate any references about this.
>Hi all,
>is there an equivalent of Taylor series for operators
between Banach
>Spaces ?- if yes, what is the name of this equivalent, and
what are
>the conditions on this operators ?
> I can think of at least two things the question might mean
> (does an operator have a Taylor series/can we apply a Taylor
> series to an operator). Try to be a little more specific
about
> the question...
> ************************
> David C. Ullrich
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
>Let T be an operator between two Banach spaces (or Hilbert
spaces),
>what are the conditions on T and on these spaces such that T
admits a
>Taylor series expansion, and how to perform this expansion.
TaylorÕs theorem for functions from a Banach space to a
Banach space
is Theorem 6 in Nelson, Topics in Dynamics I: Flows,
Princeton U. Press
1969.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
>>Let T be an operator between two Banach spaces (or Hilbert
spaces),
>>what are the conditions on T and on these spaces such that
T admits a
>>Taylor series expansion, and how to perform this expansion.
>TaylorÕs theorem for functions from a Banach space to a
Banach space
>is Theorem 6 in Nelson, Topics in Dynamics I: Flows,
Princeton U. Press
>1969.
Huh. Can you give us a hint what sort of thing these are? I
can
only imagine two possibilities: Taylor series which really
amount to series for the function of one variable
f(t) = T(x + ty), or something like Taylor series in 
infinitely
many variables, by analogy with Taylor series in R^n.
>Robert Israel israel@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
************************
David C. Ullrich
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
>>TaylorÕs theorem for functions from a Banach space to a
Banach space
>>is Theorem 6 in Nelson, Topics in Dynamics I: Flows,
Princeton U.
Press
>>1969.
>Huh. Can you give us a hint what sort of thing these are? I
can
>only imagine two possibilities: Taylor series which really
>amount to series for the function of one variable
>f(t) = T(x + ty), or something like Taylor series in
infinitely
>many variables, by analogy with Taylor series in R^n.
For a function f from an open subset U of Banach space E to
Banach space
F, the Frechet derivative (if it exists) is a function Df
from U to
L(E,F) (the Banach space of bounded linear operators E -> F)
such that
f(x+y) = f(x) + Df(x) y + o(||y||). The Frechet derivative of
Df, if it
exists, is then a function D^2 f from U to L(E,L(E,F)), which
can be
identified with the space L(E x E, F) of bounded bilinear
forms from
E x E to F. We write D^2 f(x) y^2 for the value of the form
(D^2 f)(x)
at (y,y). Similarly for any positive integer k, D^k f, if it
exists, is a
function from U to the bounded k-linear forms from E^k to F,
and we write
D^k f(x) y^k for (D^k f)(x)(y,...,y) with k yÕs. Then this
version of TaylorÕs theorem says that if f is C^k on U
f(x+y) = f(x) + sum_{j=1}^k D^j f(x) y^j/j! + o(||y||^k)
Of course if E is R^n and F is R, this can be thought of as a
coordinate-free way of writing the multivariate Taylor series
of f to kÕth order.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
>TaylorÕs theorem for functions from a Banach space to a
Banach space
>is Theorem 6 in Nelson, Topics in Dynamics I: Flows,
Princeton U.
Press
>1969.
>>Huh. Can you give us a hint what sort of thing these are? I
can
>>only imagine two possibilities: Taylor series which really
>>amount to series for the function of one variable
>>f(t) = T(x + ty), or something like Taylor series in
infinitely
>>many variables, by analogy with Taylor series in R^n.
>For a function f from an open subset U of Banach space E to
Banach space
>F, the Frechet derivative (if it exists) is a function Df
from U to
>L(E,F) (the Banach space of bounded linear operators E -> F)
such that
>f(x+y) = f(x) + Df(x) y + o(||y||). The Frechet derivative
of Df, if it
>exists, is then a function D^2 f from U to L(E,L(E,F)),
Realized a minute too late that this was probably what was
>which can be
>identified with the space L(E x E, F) of bounded bilinear
forms from
>E x E to F. We write D^2 f(x) y^2 for the value of the form
(D^2 f)(x)
>at (y,y). Similarly for any positive integer k, D^k f, if it
exists, is a
>function from U to the bounded k-linear forms from E^k to F,
and we write
>D^k f(x) y^k for (D^k f)(x)(y,...,y) with k yÕs. Then this
>version of TaylorÕs theorem says that if f is C^k on U
>f(x+y) = f(x) + sum_{j=1}^k D^j f(x) y^j/j! + o(||y||^k)
>Of course if E is R^n and F is R, this can be thought of as a
>coordinate-free way of writing the multivariate Taylor series
>of f to kÕth order.
>Robert Israel israel@math.ubc.ca
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
************************
David C. Ullrich
===
Subject: Re: Functional Analysis: Equivalent of Taylor series
for operators
>thank you foryour answer.
>I am interested by both questions:
>Let T be an operator between two Banach spaces (or Hilbert
spaces),
>what are the conditions on T and on these spaces such that T
admits a
>Taylor series expansion, and how to perform this expansion.
The _other_ question is the topic of something called the
functional calculus, which you can read about in many
books on functional analysis.
This question makes less sense to me offhand. A Taylor
expansion for T would look something like
T(x) = sum c_n x^n,
but if x is in a Banach space thereÕs no such thing as
x^n.
>I would appreciate any references about this.
>>Hi all,
>>is there an equivalent of Taylor series for operators
between Banach
>>Spaces ?- if yes, what is the name of this equivalent, and
what are
>>the conditions on this operators ?
>> I can think of at least two things the question might mean
>> (does an operator have a Taylor series/can we apply a
Taylor
>> series to an operator). Try to be a little more specific
about
>> the question...
>> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: CantorÕs Theory: Mathematical creationism
> Given a closed set A, he [Cantor] considered the process of
> removing the isolated points to get a new closed set AÕ.
> If so, one can consider the intersection of all of them,
> A^(omega), which is of course also a closed set. Then
A^(omega)Õ
> =A^(omega+1) may still be different from A^(omega).
> ItÕs natural, and not the result of theological training,
to want to say
> keep going to this process.
YouÕre talking about a process which you repeat 
infinitely many
times, and then you keep going from there. That is certainly a
modified notion of process, which answers the following
question
you ask:
> |Both CantorÕs theory and Creationism theory are
pseudoscience. Both the
> |Creationists and the Cantorians impose upon their
disciples a world view
> |in which people must modify their thinking to incorporate
certain axioms
> |handed down from higher authority,
> Modify from what?
The intuitions we gain from living and observing the world we
live in.
===
Subject: Re: CantorÕs Theory: Mathematical creationism
<91g6q0dcfhd4frggimcoaj8m8a3glqkg0i@4ax.com>
<30htk9F31dsjcU1@uni-berlin.de>

at 12:28 PM, david_lawrence_petry@yahoo.com (David Petry)
said:
>As IÕve explained before, all of the math courses I took
before grad
>school were consistent with the idea that mathematics has
observable
>implications
Then they, and your Science courses, must have been extremely
shallow.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: CantorÕs Theory: Mathematical creationism
>> As an undergraduate, I took mathematics courses which I
thought
>> were relevant to my double major in physics and computer
science.
>> Nothing in those mathematics courses had anything to do
with
>> uncountability proofs.
> You never did advanced calculus?
Nor studied measure theory, Lebesgue integration or general
topology I presume. And I thought dumbing-down in US education
was a recent phenomenon :-(
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: CantorÕs Theory: Mathematical creationism
> Nor studied measure theory, Lebesgue integration or general
> topology I presume. And I thought dumbing-down in US
education
> was a recent phenomenon :-(
Can somebody enumerate all different integration tecniques?
IÕve heard now about Lebesgue integration, Stieltjes 
integral
..
As a physicist, I always calculated integrals the common way,
by employing the main theorem of calculus. Numerically
sometimes.
WhatÕs the beef with these definitions? And how 
many are there?
Han de Bruijn
===
Subject: Re: CantorÕs Theory: Mathematical creationism
>>
>> CantorÕs theory is purely graduate level mathematics.
>> Maybe it was to you. I first met uncountability proofs in
formal
>> education in the first week of my BA course.
> I first encountered the basic uncountability proof in junior
high
> school. I saw it as something of a game which had no
connection
> to any of the other math I was learning, and I never
encountered
> it again until graduate school.
ItÕs a shame that none of the math courses you studied
had any relevance to mathematics :-(
> I was shocked when I learned that graduate level mathematics
> incorporates uncountability in such a fundamental way.
Shocked! Poor you :-(
IÕm shocked that you didnÕt actually study 
mathematics
until graduate school. IÕm sorry it was too much for you
and your preconceived prejudices.
> It is,
> after all, just a fantasy world,
No.
> which someday will be expunged
> from mathematics.
No.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: CantorÕs Theory: Mathematical creationism
> So, whatÕs happened to Cantor is that, according to
contemporary
> Platonists, that he *discovered* a beautiful theory,
something
> foundational, but got beaten for that, and even got sick.
ThatÕs why
> heÕs a hero, a martyr, a prophet.
> Find me a mathematician that treats Cantor as a hero, a
martyr, a
> prophet. I donÕt know of any.
I may have exaggerated it, but I remind HilbertÕs wonderful
quote
about Cantor:
No one shall expel us from this paradise Cantor created for us
Why this affection for something that is not even necessary
for doing
analysis or any concrete mathematics (as it turned out
later?) ?
(Speaking of course about the *whole* theory) I imagine part
of it may
be because Cantor did suffer for having vigorously defended
his
theory. As a true mathematician! I do think the debate had a
personal
ßavor, and in the end Cantor won.
Maybe Hilbert was concerned with not all of CantorÕs theory
(about
those things that could not even be named!), but just the
part that
looked properly axiomatic for him, that it gave the hope that
things
could be formal enough. Just an idea of mine. ThatÕs another
guess
into the past, from what I know about HilbertÕs program.
I wouldnÕt want to make up history, IÕm no 
revisionist. That
is all I
can tell from what I read of his life. If there are people
who know
better, please step up and correct my interpretation.
[What I think I do know is, the correlation between his
mental illness
and his theory does not show a causation. ItÕs too weak. I 
am
certain
the poster who made the connection was inventing things. 
ItÕs
a rather
indirect relation. YouÕd grow paranoid, if people tried to
prevent you
from publishing your work or getting a job, etc. I donÕt 
know
what
precisely happened to Cantor, but the way itÕs written it
doesnÕt
sound nice. Imagine for a while that you really had a
foundational
idea, and that your critics suppressed you for eternity. If
you had
the tendency, you could go crazy under such pressure.]
I think, if Hilbert saw a modern programming language, he may
have
thought differently about CantorÕs theory. Because 
todayÕs
programming is just like the formalization he dreamt of. We
should
really talk about what programs we can write, rather than
immaterial
things.
--
Eray Ozkural
===
Subject: Re: CantorÕs Theory: Mathematical creationism
<30htk9F31dsjcU1@uni-berlin.de>
<87u0rbp8d9.fsf@phiwumbda.org>
Discussion, linux)
>> Find me a mathematician that treats Cantor as a hero, a
martyr, a
>> prophet. I donÕt know of any.
> I may have exaggerated it, but I remind HilbertÕs 
wonderful
quote
> about Cantor:
> No one shall expel us from this paradise Cantor created for
us
> Why this affection for something that is not even necessary
for doing
> analysis or any concrete mathematics (as it turned out
later?) ?
Mathematics with infinity is considerably easier than
mathematics
without infinity, no? I know that the program of reverse
mathematics
shows that standard mathematics can get by with rather less
than ZFC,
but
(1) IÕm not sure which bits they drop. If powerset and the
axiom of
infinity is common to many of the theories studied by reverse
mathematics folk, then Cantor is still with us.
(2) As I understand it, one *can* get by with less, but not
for free.
Use a smaller theory than ZFC and the proofs become rather
more
difficult, no?
How do you justify the claim that the powerset and axiom of
infinity
are not even necessary for doing analysis or any concrete
mathematics?
> (Speaking of course about the *whole* theory) I imagine
part of it may
> be because Cantor did suffer for having vigorously defended
his
> theory. As a true mathematician! I do think the debate had
a personal
> ßavor, and in the end Cantor won.
Yeah? I imagine it was because of the pro-Jewish conspiracy
and a
little confusion over CantorÕs name. My theory is about as
plausible
as yours.
> I wouldnÕt want to make up history, IÕm no 
revisionist.
That is all I
> can tell from what I read of his life. If there are people
who know
> better, please step up and correct my interpretation.
Well then, why not stop *guessing* what happened and look it
up
instead? IÕm not a historian of mathematics either. So know
what I
do to try to avoid error? I try not to expound on the history
of
mathematics---at least not without references. I find it
preferable
to making things up and waiting for corrections.
But to each his own.
> I think, if Hilbert saw a modern programming language, he
may have
> thought differently about CantorÕs theory. Because 
todayÕs
> programming is just like the formalization he dreamt of. We
should
> really talk about what programs we can write, rather than
immaterial
> things.
Please.
--
This page contains information of a type (text/html) that can
only be
viewed with the appropriate Plug-in. Click OK to download
Plugin.
--- Netscape 4.7 error message
===
Subject: Re: CantorÕs Theory: Mathematical creationism
> We should
> really talk about what programs we can write, rather than
immaterial
> things.
So progams are material things? And so are integers with an
infinite
number of digits?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: CantorÕs Theory: Mathematical creationism
<41a556ab$15$fuzhry+tra$mr2ice@news.patriot.net>
at 10:17 PM, examachine@gmail.com (Eray Ozkural exa) said:
>I see that you are a priest.
ROTF,LMAO! You seem to have a rather active imagination, and
are
seeing things that donÕt exist.
>What the hell is Qaballah, then?
ItÕs a type of mysticism that has nothing to do with 
cardinal
numbers
or set theory.
>What reason is there to take such silly mythology seriously?
What reason is there to beat your wife? You donÕt have to 
take
something seriously in order to distinguish it from something
else.
Neither accepting Qaballah not rejecting Qaballah has any
relevance to
the theory of transfinite cardinals.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: CantorÕs Theory: Mathematical creationism
> at 10:17 PM, examachine@gmail.com (Eray Ozkural exa) said:
>I see that you are a priest.
> ROTF,LMAO! You seem to have a rather active imagination,
and are
> seeing things that donÕt exist.
You are right. Had you been that informed, you would know
better about
Kabbalah.
>What the hell is Qaballah, then?
> ItÕs a type of mysticism that has nothing to do with
cardinal numbers
> or set theory.
The description of Kabbalah you opposed to defined clearly the
reason
how Kabbalah was related to CantorÕs theory. Exactly in 
which
way do
you object to that? Are you claiming that Kabbalah did not
assert the
existence of a higher order of infinity, or other nonsense
like that?
You may want to read the below excerpt from the wikipedia
entry. The
poster expressed well, and said nothing wrong about Kabbalah!
And it
is certain that the below features about the notion of God in
Kabbalah
are directly relevant to the transfinite cardinals in 
CantorÕs
set
theory, which is not science.
http://en.wikipedia.org/wiki/Kabbalah
Kabbalists speak of the first aspect of God as Ein Sof; this 
is
translated as the infinite, or that which has no limits. In
this
view, nothing can be said about this aspect of God. This
aspect of God
is impersonal. Kabbalists speak of the second aspect of God
as being
seen by the universe as ten emanations from God; these
emanations are
called sefirot.
The sefirot mediate the interaction of the ultimate unknowable
God
with the physical and spiritual world. Some explain the
sefirot as
stages of the creative process whereby God, from His own
infinite
being, created the progression of realms which culminated in
our
finite and physical universe. Others suggest that the 
sefirot
may be
thought of as analogous to the fundamental laws of physics.
Just as
gravity, electro-magnetism, the strong nuclear force, and the
weak
nuclear force allow for interactions between matter and
energy, the
ten sefirot allow for interaction between God and the 
Universe.
http://www.jewfaq.org/kabbalah.htm
According to Kabbalah, the true essence of G-d is so
transcendent that
it cannot be described, except with reference to what it is
not. This
true essence of G-d is known as Ein Sof, which literally means
without end, which encompasses the idea of His lack of
boundaries in
both time and space. In this truest form, the Ein Sof is so
transcendent that It cannot have any direct interaction with
the
universe. The Ein Sof interacts with the universe through ten
emanations from this essence, known as the Ten Sefirot.
http://www.aish.com/spirituality/kabbala101/Kabbala_2_-_
Perceiving_the_Infin
ite.asp
>What reason is there to take such silly mythology seriously?
> What reason is there to beat your wife? You donÕt have to
take
> something seriously in order to distinguish it from
something else.
> Neither accepting Qaballah not rejecting Qaballah has any
relevance to
> the theory of transfinite cardinals.
Well, it may actually have something to do with their
*reality*, not
that you can conceive them. Surely, you can think of fiction.
I can conceive of a world, which has no causation, but that
is not our
world. If I formalized my notion of such a world, would I
have made my
fantasy real?
--
Eray Ozkural
===
Subject: Re: CantorÕs Theory: Mathematical creationism
<41a556ab$15$fuzhry+tra$mr2ice@news.patriot.net>
<41a96cd2$7$fuzhry+tra$mr2ice@news.patriot.net>
at 06:03 PM, examachine@gmail.com (Eray Ozkural exa) said:
>The description of Kabbalah you opposed to defined clearly
the reason
>how Kabbalah was related to CantorÕs theory.
No.
>Exactly in which way do you object to that?
In that it is false.
>Are you claiming that Kabbalah did not assert the existence
of a
>higher order of infinity, or other nonsense like that?
The issue is not a higher order of infinity; the issue is
cardinal
numbers, defined in terms of 1-1 correspondences. That has
nothing to
do with Qaballah, or with the way that infinity[1] is used in
Qaballah.
But I am claiming that Qaballah is not concerned about
infinity, but
rather with concepts that do not translate well into English.
>You may want to read the below excerpt from the wikipedia
entry.
Hardly a reliable source. Please quote something from
Qaballah that
refers to cardinal numbers.
>Kabbalists speak of the first aspect of God as Ein Sof; this
is
>translated as the infinite, or that which has no limits.
As I said, it doesnÕt translate well. And it 
isnÕt a cardinal
number.
>The sefirot mediate the interaction of the ultimate
unknowable God
>with the physical and spiritual world.
K3wl. What does that have to do with cardinal numbers?
>According to Kabbalah, the true essence of G-d is so
transcendent
>that it cannot be described, except with reference to what
it is
>not. This true essence of G-d is known as Ein Sof, which
literally
>means without end,
Closer, but still not accurate.
>Well, it may actually have something to do with their
*reality*, not
>that you can conceive them. Surely, you can think of fiction.
Indeed, but I would never claim that Gandalf was a character
in a Tale
of Two Cities, nor that Sidney Carton was one of the nazgul.
>I can conceive of a world, which has no causation, but that
is not
>our world. If I formalized my notion of such a world, would
I have
>made my fantasy real?
YouÕre ducking the question; the specific issue 
is whether
CantorÕs
concept of cardinal numbers came from Qaballah, not whether
either is
really. If you formalize your fantasy and Joe down the block
formalizes his, that does not mean that your fantasy derives
from
JoeÕs.
[1] It isnÕt, but I can see how someone might incorrectly
translate
various distinct phrases into the single English word 
infinity.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
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===
Subject: number of serieses
Hi!
I would like to know the number of serieses consisting of n
non-negative
integers where the sum of a series is below a predefined
positive integer,
K.
Is it possible to find it in a closed formula?
===
Subject: Re: number of serieses
>I would like to know the number of serieses consisting of n
non-negative
>integers where the sum of a series is below a predefined
positive integer,
I hope you mean positive integers, because if you include 0
there are an
infinite number of series.
Look up Stirling numbers of the second kind. Is that what you
want?
http://mathworld.wolfram.com/
StirlingNumberoftheSecondKind.html
>Is it possible to find it in a closed formula?
Yes.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: number of serieses
>>I would like to know the number of serieses consisting of n
non-negative
>>integers where the sum of a series is below a predefined
positive
integer,
>>K.
> I hope you mean positive integers, because if you include 0
there are an
> infinite number of series.
> Look up Stirling numbers of the second kind. Is that what
you want?
he is counting (0,0), (0,1), (0,2), (0,3), (1,0), (1,1),
(1,2), (2, 0)
(2,1) and (3,0). The answer in general is the binomial
coefficient (n+K choose n).
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Cantor reloaded
We consider a matrix in the form:
m_p1 m_p2 m_p3 m_p4 ... m_pi ...
: : : : :
m_1q: m_11 m_12 m_13 m_14 ... m_1i ...
m_2q: m_21 m_22 m_23 m_24 ... m_2i ...
m_3q: m_31 m_32 m_33 m_34 ... m_3i ...
m_4q: m_41 ...
.
.
.
m_jq: m_j1 ... m_ji ...
.
.
.
The m_ji are numerals (for example 0 ... 9 in the decimal
system).
Every m_pi and m_jq should be associated to a real number
betweeen 0
and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should be
infinite in both directions (horizontal and vertical): while
the
irrational numbers are infinite in decimal places, the columns
and the
rows must be infinite.
The matrix containes all real numbers.
So this matrix is a equivalent to a ŌlistÕ, in 
which the real
numbers
are numbered with real numbers.
Like a list of natural numbers which ist numbered with natural
numbers, this list of real numbers must be quadratic.
This matrix (or list on other considerations) canÕt contain
itÕs
anti-diagonal.
Albrecht S. Storz
===
Subject: Re: Cantor reloaded
> Do you think whether it is possible to put infinite things 
or
objects,
> especially mathematical objects, in a row? If the objects
are
infinite
> like real numbers, is it also possible? And how do these
things
behave
> in the infinite at the point at which the natural numbers
end?
These
> questions are very silly, but they can lead to another
understanding
> of the things.
> Why do you think, a matrix, filled with numerals of real
numbers
does
> not exist?
Of course you can make a list of all real numbers, by
application of
the well-ordering principle.
However, the numbers of rows in your list must be equal to the
cardinality of the real number system. So assume that you
have such a
list.
> This matrix (or list on other considerations) canÕt 
contain
itÕs
> anti-diagonal.
This is where the problem lies. How can you be certain that
such an
antidiagonal exist.
The construction should take the form of an argument by
transfinite
inducition. Correspond with each entry in a row an ordinal
number.
For finite ordinals one can take the procedure of Cantor,
however if
you have limit ordinals, things get complicated. Take a limit
ordinal alpha. Assume that there exist a real number (probably
more), different from every entry in a row gamma less then
alpha.
How can you be sure that there is a real number different
from every
ordinal number less or equal to alpha.
If you can solve out above problem, your assertion is valid.
But I think your problem can easily be falsified. Assume that
the list
of real numbers has an antidiagonal. This antidiagonal is the
limit of
a rational number. Just look at the construction of the
antidiagonal.
This limit must be a real number. (by definition, a rational
number
is the limit of a sequence of real numbers). Therefore the
antidiagonal is a real number. By your assertion the
antidiagonal is
not in the list of real numbers, hence it cannot be a real
number.
Therefore, we have a contradiction. Either, the antidiagonal
is in
the list of real numbers or the list of real numbers does not
have an
antidiagonal.
So if your assertion is true: the list of real numbers does
not
contain its anti-diagonal then the result follows: the
antidiagonal
of the list of real numbers does not exist.
thomas
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: Cantor reloaded
> But I think your problem can easily be falsified. Assume
that the list
> of real numbers has an antidiagonal. This antidiagonal is
the limit of
> a rational number. Just look at the construction of the
antidiagonal.
> This limit must be a real number. (by definition, a rational
number
> is the limit of a sequence of real numbers).
You donÕt mean this in the other way: a real number is the
limit of a
sequence of rational numbers?
> Therefore the
> antidiagonal is a real number. By your assertion the
antidiagonal is
> not in the list of real numbers, hence it cannot be a real
number.
> Therefore, we have a contradiction. Either, the
antidiagonal is in
> the list of real numbers or the list of real numbers does
not have an
> antidiagonal.
> So if your assertion is true: the list of real numbers does
not
> contain its anti-diagonal then the result follows: the
antidiagonal
> of the list of real numbers does not exist.
So you would argue in the same way: The conclusion of 
CantorÕs
diagonal proof (on reals) is either: no (complete) list
exists, or:
the antidiagonal does not exists. ?
Albrecht
> thomas
> *-----------------------*
> www.GroupSrv.com
> *-----------------------*
===
Subject: Re: Cantor reloaded
>> Do you think whether it is possible to put infinite things
or
> objects,
>> especially mathematical objects, in a row? If the objects
are
> infinite
>> like real numbers, is it also possible? And how do these
things
> behave
>> in the infinite at the point at which the natural numbers
end?
> These
>> questions are very silly, but they can lead to another
> understanding
>> of the things.
>> Why do you think, a matrix, filled with numerals of real
numbers
> does
>> not exist?
Of course you can make a list of all real numbers, by
application of
> the well-ordering principle.
Why not just observe that the identity map from R to R is a
list in the
same sense? In the context of CantorÕs proof, a list is
generally
taken to be a map defined on the naturals, but if you are
going to allow
lists with uncountable domains, you may as well just choose R.
> However, the numbers of rows in your list must be equal to
the
> cardinality of the real number system. So assume that you
have such a
> list.
>> This matrix (or list on other considerations) canÕt
contain itÕs
>> anti-diagonal.
There is no such thing. The anti-diagonal is a function
a: R^N -> R (*)
such that, for each f in R^N (meaning f: N -> R is a list),
there is a
number x = a(f) in R such that x not in ran(f).
If your list is the identity map on R, then it is a member of
R^R, not
R^N, and is not in the domain of a, as indicated by (*).
Hence, the
anti-diagonal of the identity on R (or of any mapping whose
domain is
not N) has no meaning.
--
Dave Seaman
Judge YohnÕs mistakes revealed in Mumia Abu-Jamal ruling.

===
Subject: Re: Cantor reloaded
at 03:08 PM, albstorz@gmx.de (albrecht) said:
>The matrix containes all real numbers.
No.
>this list of real numbers must be quadratic.
What do you mean by quadratic?
>This matrix (or list on other considerations) canÕt contain
itÕs
>anti-diagonal.
Which tells you that it doesnÕt contain all real numbers.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: Cantor reloaded
The matrix containes all real numbers.
> So this matrix is a equivalent to a ŌlistÕ, in 
which the
real
numbers
> are numbered with real numbers.
Using CantorÕs diagonalization proof one can show that there
must be a
real number that is not in your list. Hence a matrix of all
real
numbers does not exist.
[/quote]Like a list of natural numbers which ist numbered with
natural
numbers, this list of real numbers must be quadratic.
[/quote]
What do you mean with quadratic. So: The set consisting of
the numbers
{1,2,3,4} can be written as.
1 2
3 4
Is this set quadratic???? (whatever it may mean)
thomas
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: Cantor reloaded
> The matrix containes all real numbers.
> So this matrix is a equivalent to a ŌlistÕ, in 
which the
real
> numbers
> are numbered with real numbers.
> Using CantorÕs diagonalization proof one can show that
there must be a
> real number that is not in your list. Hence a matrix of all
real
> numbers does not exist.
Do you think whether it is possible to put infinite things or
objects,
especially mathematical objects, in a row? If the objects are
infinite
like real numbers, is it also possible? And how do these
things behave
in the infinite at the point at which the natural numbers end?
These
questions are very silly, but they can lead to another
understanding
of the things.
Why do you think, a matrix, filled with numerals of real
numbers does
not exist?
> [/quote]Like a list of natural numbers which ist numbered
with
> natural
> numbers, this list of real numbers must be quadratic.
> [/quote]
> What do you mean with quadratic. So: The set consisting of
the numbers
> {1,2,3,4} can be written as.
> 1 2
> 3 4
> Is this set quadratic???? (whatever it may mean)
We consider a list or a matrix. If the list or the matrix
isnÕt
quadratic then the question whether an anti-diagonal number is
contained or not is not so easy to answer. If the list or the
matrix
is quadratic, the diagonal and the anti-diagonal number is of
the same
kind as the numbers in rows and columns. Your matrix
contained the
diagonal number 1 4 and an anti-diagonal, for example 8 6.
Both
contained two digits such as your rows and columns.
A. S. Storz
> thomas
> *-----------------------*
> www.GroupSrv.com
> *-----------------------*
===
Subject: Re: Cantor reloaded
> We consider a matrix in the form:
> m_p1 m_p2 m_p3 m_p4 ... m_pi ...
> : : : : :
> m_1q: m_11 m_12 m_13 m_14 ... m_1i ...
> m_2q: m_21 m_22 m_23 m_24 ... m_2i ...
> m_3q: m_31 m_32 m_33 m_34 ... m_3i ...
> m_4q: m_41 ...
> m_jq: m_j1 ... m_ji ...
> The m_ji are numerals (for example 0 ... 9 in the decimal
system).
> Every m_pi and m_jq should be associated to a real number
betweeen 0
> and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should
be
> infinite in both directions (horizontal and vertical): while
the
> irrational numbers are infinite in decimal places, the
columns and the
> rows must be infinite.
> The matrix containes all real numbers.
If the rows and columns of this matrix are indexed by the
integers,
then the claim that it can contain all real numbers requires
proof.
> So this matrix is a equivalent to a ŌlistÕ, in 
which the
real numbers
> are numbered with real numbers.
> Like a list of natural numbers which ist numbered with
natural
> numbers, this list of real numbers must be quadratic.
> This matrix (or list on other considerations) canÕt 
contain
itÕs
> anti-diagonal.
Such a matrix can easily be reordered into the more standard
form of
list, from which as many anti-diagonals as one wishes.
> Albrecht S. Storz
===
Subject: Re: Cantor reloaded
We consider a matrix in the form:
m_p1 m_p2 m_p3 m_p4 ... m_pi ...
> : : : : :
> m_1q: m_11 m_12 m_13 m_14 ... m_1i ...
> m_2q: m_21 m_22 m_23 m_24 ... m_2i ...
> m_3q: m_31 m_32 m_33 m_34 ... m_3i ...
> m_4q: m_41 ...
> .
> .
> .
> m_jq: m_j1 ... m_ji ...
> .
> .
> .
The m_ji are numerals (for example 0 ... 9 in the decimal
system).
> Every m_pi and m_jq should be associated to a real number
betweeen 0
> and 1 (as 0 . m_11 m_12 m_13 m_14 ...) . The matrix should
be
> infinite in both directions (horizontal and vertical): while
the
> irrational numbers are infinite in decimal places, the
columns and the
> rows must be infinite.
> The matrix containes all real numbers.
> If the rows and columns of this matrix are indexed by the
integers,
> then the claim that it can contain all real numbers
requires proof.
You are right. I should not use natural numbers as indices.
But their
function is only, to have labels to talk about. They are not
necessary
for the construction. I should label it in a better way with
alpha,
beta and so on. If we want indexing, can we use the Powerset
of the
natural numbers?
> So this matrix is a equivalent to a ŌlistÕ, in 
which the
real numbers
> are numbered with real numbers.
> Like a list of natural numbers which ist numbered with
natural
> numbers, this list of real numbers must be quadratic.
This matrix (or list on other considerations) canÕt contain
itÕs
> anti-diagonal.
> Such a matrix can easily be reordered into the more
standard form of
> list, from which as many anti-diagonals as one wishes.
>
I think this matrix could not be converted to a usual list. A
list
represents a bijection between a set X and the set of the
natural
numbers. I try to have a bijection between Y and R.
Albrecht S. Storz
===
Subject: Re: Cantor reloaded
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
> The matrix containes all real numbers.
What makes you think so?
- Randy
===
Subject: Re: Cantor reloaded
> The matrix containes all real numbers.
> What makes you think so?
> - Randy
If the rows and the columns are filled with (different)
irrational
numbers without end, the matrix should represent all real
numbers.
Do you know the diagonal proof of G. Cantor? His proof
consider every
list of rational numbers.
So I do for every matrix with real numbers. Also for a
completely
matrix with all reals. I donÕt need to have this matrix. But
if this
matrix exists it canÕt contain the anti-diagonal.
Albrecht
===
Subject: Re: Cantor reloaded
Correction:
We consider a matrix in the form:
m_palpha m_pbeta m_pgamma ....m_pi
: : : : :
m_alphaq: m_alphaalpha m_alphabeta m_alphagamma ... m_alphai
...
m_betaq: m_betaalpha m_betabeta m_betagamm ... m_betai ...
m_gammaq: m_gammaalpha m_gammabeta m_gammagamma ... m_gammai
...
m_deltaq: m_deltaalpha ...
.
.
.
m_jq: m_jalpha ................... m_ji ...
.
.
.
The m_ji are numerals (for example 0 ... 9 in the decimal
system).
Every m_pi and m_jq should be associated to a real number
betweeen 0
and 1 (as 0 . m_alphaalpha m_alphabeta m_alphagamma
m_alphadelta
...).
The indices must be taken out of a appropriate set (perhaps
Powerset
of |N?).
But the indices are not necessary for the consideration.
Essentially
is only the matrix itself.
The matrix should be infinite in both directions (horizontal
and
vertical): since the irrational numbers are infinite in 
decimal
places, the columns and the
rows must be infinite.
The matrix is able to contain all real numbers.
So this matrix is a equivalent to a ŌlistÕ, in 
which the real
numbers
are numbered with real numbers (or with something with the
same
cardinality).
This matrix (or list on other considerations) canÕt contain
itÕs
anti-diagonal.
Albrecht S. Storz
===
Subject: Re: Cantor reloaded
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
> The matrix containes all real numbers.
> What makes you think so?
> - Randy
> If the rows and the columns are filled with (different)
irrational
> numbers without end, the matrix should represent all real
numbers.
Only if the irrational numbers are countable.
> Do you know the diagonal proof of G. Cantor? His proof
consider every
> list of rational numbers.
His proof says, suppose I have a list of real numbers
and goes on to show that there exists a number not on
the list.
> So I do for every matrix with real numbers.
Have you seen the proof of the countability of the
rationals? It lays the rational numbers out in a matrix
very much like yours, and shows that you can trace a
path through that matrix that associates a natural
number with every position in the matrix. That is, that
the number of elements in the matrix is countable.
You canÕt start by assuming the irrationals are countable.
By CantorÕs method, as soon as you tell you 
youÕve got
a matrix of irrationals, IÕll find you one not 
on the
list. You have infinitely many reals, but you still
donÕt have ALL the reals.
- Randy
===
Subject: Re: Cantor reloaded
> The matrix containes all real numbers.
> What makes you think so?
- Randy
> If the rows and the columns are filled with (different)
irrational
> numbers without end, the matrix should represent all real
numbers.
> Only if the irrational numbers are countable.
Why? The matrix consist of the numerals of real numbers in
rows. It
has so many rows as there exists real numbers. So they are not
enumerable. Why canÕt this matrix exist?
If this matrix is thinkable it leads to the same conclusion
like
CantorÕs proof but for the impossibility of the bijection R
--> R.
> Do you know the diagonal proof of G. Cantor? His proof
consider every
> list of rational numbers.
> His proof says, suppose I have a list of real numbers
> and goes on to show that there exists a number not on
> the list.
> So I do for every matrix with real numbers.
> Have you seen the proof of the countability of the
> rationals? It lays the rational numbers out in a matrix
> very much like yours, and shows that you can trace a
> path through that matrix that associates a natural
> number with every position in the matrix. That is, that
> the number of elements in the matrix is countable.
> You canÕt start by assuming the irrationals are countable.
I donÕt assume like this.
Albrecht
> By CantorÕs method, as soon as you tell you 
youÕve got
> a matrix of irrationals, IÕll find you one not 
on the
> list. You have infinitely many reals, but you still
> donÕt have ALL the reals.
> - Randy
===
Subject: topology.....---
hello........doctor~
if X is path-connected, then X is locally path-connected.
--------------------------------------------
i think.....
counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....}
this is path-connected, but this is not locally
path-connected.
um....is this right counter-example ??
===
Subject: Re: topology.....---
> hello........doctor~
> if X is path-connected, then X is locally path-connected.
> --------------------------------------------
> i think.....
> counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....}
> this is path-connected, but this is not locally
path-connected.
> um....is this right counter-example ??
I donÕt believe so. what point doesnÕt have a 
path connected
neighborhood? Given a point (x,x/n) one can certainly find 
some
neighborhood that doesnÕt intersect the line y=x/(n-1) or
y=x/(n+1).
Perhaps you want to expand this set somewhat to make it
non-locally
path-connected. (Hint: what is the closure of this set?)
===
Subject: Re: topology.....---
> hello........doctor~
> if X is path-connected, then X is locally path-connected.
> --------------------------------------------
> i think.....
> counter-example is {(x,y) | 0<= x <=1, y = x/n, n=1,2,....}
> this is path-connected, but this is not locally
path-connected.
> um....is this right counter-example ??
> I donÕt believe so. what point doesnÕt have 
a path connected
> neighborhood? Given a point (x,x/n) one can certainly find
some
> neighborhood that doesnÕt intersect the line y=x/(n-1) or
y=x/(n+1).
> Perhaps you want to expand this set somewhat to make it
non-locally
> path-connected. (Hint: what is the closure of this set?)
-------------------------------------------------------------
-----------
um........
HereÕs a compact Hausdorff (and hereditarily normal, but not
metrizable)
counterexample. Define the long circle using the following
steps.
Step 1: Let X be the set of all ordinals that are less than
or equal to
the first uncountable ordinal *:=omega_1 = aleph_1. Given the
order
topology, X is a non-metrizable compact Hausdorff space. Step
2: Form
the quotient space Y := (X x [0,1])/~, where the points (a,1)
are
identified with (a+1,0), for all countable ordinals a. Y is a
long
interval, a connected, locally connected compact ordered
space. ItÕs
almost what you want, but it fails to be path connected. We
remedy
that by identifying the first and last points. This space is
now
path connected and locally connected, but not locally path
connected
at the bad point with no countable neighborhood base.
-------------------------------------------------------------
--------
-------------------------------------------------------------
--------
> I am searching for a subset of the Euclidean plane that is
path
> connected but not locally connected in *any* point.
Consider {(1+ r e^{it})^2: 0 <= r < infinity, 0 < t < Pi, t
rational}
in the complex plane.
-------------------------------------------------------------
--------
um......by Robert B. Israel.
more complex........oh....my god.
===
Subject: Test statistic for a coin toss
I want to form a statistical hypothesis built around the
likelihood that a
tossed coin will land on heads approximately 50% of the time.
Could anyone
help me get started? I mainly need help deciding which test
statistic to use
(Z, T, F, or chi squared).
Any help would be great.
===
Subject: Re: Test statistic for a coin toss
> I want to form a statistical hypothesis built around the
likelihood that
> a
> tossed coin will land on heads approximately 50% of the
time. Could
> anyone
> help me get started? I mainly need help deciding which test
statistic to
> use
> (Z, T, F, or chi squared).
You are interested in the binomial distribution, which can be
approximated by a Gaussian distribution. Do you know of
Bernoulli trials?
--
Mostly economics:

r c
v s a Whether strength of body or of mind, or wisdom, or
i m p virtue, are found in proportion to the power or
wealth
e a e of a man is a question fit perhaps to be discussed
by
n e . slaves in the hearing of their masters, but highly
@ r c m unbecoming to reasonable and free men in search of
d o the truth. -- Rousseau
===
Subject: A079586
njas:
The first %H was moved.
%I A079586
%S A079586
3,3,5,9,8,8,5,6,6,6,2,4,3,1,7,7,5,5,3,1,7,2,0,1,1,3,0,2,9,1,8,
9,2,7,1,
%T A079586
7,9,6,8,8,9,0,5,1,3,3,7,3,1,9,6,8,4,8,6,4,9,5,5,5,3,8,1,5,3,2,
5,1,3,0,
%U A079586
3,1,8,9,9,6,6,8,3,3,8,3,6,1,5,4,1,6,2,1,6,4,5,6,7,9,0,0,8,7,2,
9,7,0,4
%N A079586 Decimal expansion of sum(k>=1,1/F(k)) where F(k)
is the k-th
Fibonacci number A000045(k).
%H A079586 Eric WeissteinÕs World of Mathematics, Recipr
ocal Fibonacci Constant
%H A079586 Eric WeissteinÕs World of Mathematics, 
Fibonacci
Number
%F A079586 3.35988566624....
%Y A079586 Cf. A084119.
%K A079586 cons,nonn
%O A079586 1,1
--
Don Reble djr@nk.ca
===
Subject: re:Determining number of roots of a polynomial
I canÕt remember the details, but I remember having to use 
an
algorithm involving Sturm sequences. The algorithm gave the
number
of real roots (which may include multiplicities) greater than
some
given number. Thus if you knew a lower bound for the roots,
you
could use it to get all real roots.
Added note: Look up Sturm sequence using Google - it will get
you
the necessary information.
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: Uncle assAl: (SR) Lorentz tÕ, x
= Intervals
I donÕt currently have time to discuss this right now, as
much as I
would like to, but I have a few question for you.
1) I read your clock analysis, do you think the HK experiment
is
possible. You said the counter is perfect in a clock
(usually) but the
oscillator is not, so is HK possible to perform and get
accurate results?
2) I still have some difficult with your light example,
because a
component of the lightÕs velocity is in the direction of
ship, it seems
like that should matter. In the other version (GarderÕs) the
light moves
perpendicular to ship. Seems like this should matter, I 
canÕt
quite put
my finger on it though. Do you see what I am saying? Is there
any
validity to it?
Peter
===
Subject: Re: Uncle assAl: (SR) Lorentz tÕ, x
= Intervals
>I donÕt currently have time to discuss this right now, as
much as I
>would like to, but I have a few question for you.
> 1) I read your clock analysis, do you think the HK
experiment is
> possible. You said the counter is perfect in a clock
(usually) but the
> oscillator is not, so is HK possible to perform and get
accurate
> results?
No.
> 2) I still have some difficult with your light example,
because a
> component of the lightÕs velocity is in the direction of
ship, it
> seems like that should matter. In the other version
(GarderÕs) the
> light moves perpendicular to ship. Seems like this should
matter, I
> canÕt quite put my finger on it though. Do you 
see what I am
saying?
No.
Is there any
> validity to it?
No.
> Peter
===
Subject: Re: Uncle assAl: (SR) Lorentz tÕ, x
= Intervals
>> HeÕs an example that IÕll describe with 
Newtonian physics
OK?
> Assume that Stella runs away from Terrance in a straight
line at 4/5
> the speed of X m/s, then turns around real fast and runs
back at the
> same speed 5 seconds. SheÕs gone for 10 seconds, right?
> It computes but isnÕt ßat said. IFF X m/s is wrt 
Terrance.
X m/s relative to Terrance, who is stationary OK?
> Well what if
> every 1/10 of a second Terrance sends an american dollar to
Stella at
> the speed of X m/s, then sheÕd get all 100 USD right? And
sheÕd get
> the first $10 before turning around and the 
final $90 on her
way back,
> right?
> Unless sent only at the end of each 1/10th second, in which
case still
close
> enuf. WeÕll give you the tie.
Usually I prefer to send one every millionth of a second so
that +/-
isnÕt seen as such a big deal.
> What if every 1/6 of a second she sent Terrance a canadian
> dollar at the speed of X m/s (so that means she doesnÕt
have to throw
> it as hard on her way back, but on her way out she has to
throw the
> canadian dollar harder than Terrance throws the american
dollar),
> X m/s wrt Terrance, ŌobviouslyÕ.
Yup.
> so
> Terrance gets all 60 CAD, but he gets the first 30 CAD
during the
> first 9 seconds and the final 30 during the last 
1 second.
> ok
Yeah.
> So if Terrance compares the RATE at which he THROWS dollars
versus
> RECIEVES dollars, then for 9 seconds he gets them at 1/3
the rate he
> sends them, and for 1 second he gets them at 3/1 the rate
he sends
> them. And for Stella, she notes that for 5 seconds she GETS
dollars
> at 1/3 the rate she sends them (oh no! they are BOTH
getting dollars
> at a rate less than they are sending, either Newtonian
physics must be
> wrong or our notion that a two people canÕt both be giving
money to
> each and both me losing money is wrong, oh no! See how
silly that
> is!) and for 5 seconds she GETS dollars them at 3/1 the
rate she sends
> them.
> rofßmfao!
> What a terribly unfair setup!
> For the first tenth of a second he sends dollars at an
infinitely greater
> rate than he gets them!
You can have both Terrance and Stella send money W times as
often, and
thereby reduce this initial period of time to be as small as
you like.
But if you consider the bank to be the medium, itÕs the
classical
situation where if you each send each other money, the bank
getÕs some
interest.
> So much for Newton and Einstein! (Apparently, according to
your ŌlogicÕ.)
> rofßmfao!
> How on earth could a non-psychotic imagine that
fiddling-with-receiving
> babble (however correct) has any contradictory point to
make about the
> sending/etc facts?
That was my opinion about Androcles, thatÕs why I 
constructed
this
example.
> Ok now letÕs draw a picture of the situation.
F
|
E
|
C D
| /
B /
|/
A
> Where A is the point when Stella is about to run away from
Terrance.
> Where B is the point when Terrance is sending Stella his
10th dollar.
> Where C is the point when Terrance is sending Stella his
50th dollar.
> Where D is the point when Stella is sending Terrance her
30th dollar.
> Where D is the point when Stella turns around.
> Where D is the point when Stella is recieving TerranceÕs
10th dollar.
> Where E is the point when Terrance is sending Stella his
90th dollar.
> Where E is the point when Terrance is recieving StellaÕs
30th dollar.
> Where F is when Stella returns.
> Where F is the point when Stella is recieving TerranceÕs
100th dollar.
> Where F is the point when Terrance is recieving StellaÕs
60th dollar.
> Not that at C and D Stella and Terrance have both RECIEVED
less
> dollars than they are sent, this is the fact analogous to
time
> dilation notice how it is STUPID to think is MEANS
anything, it just
> means that some dollars have been sent and not yet recieved.
> rofßmfao!
> The name of the stupidity is probably UIT - Unnecessarily
Ignorant
> Theorist.
Androcles at least canÕt get the math right for the SR twin
paradox
so I though showing the math for the financial twin paradox
should
clear things up.
> The only correct analogy at work here is that you are
asserting your
> Ōobservables onlythesis, treating as 
unobservables the
velocities and
> sending rates you specified. The 
receiving/Õobservingrates
have
absolutely
> no meaning in regard to the givens you stated although they
can be
> calculated therefrom.
No meaning? You COUNT. If between events A and B you sent 10
dollars
and only got 1/3 as many, thatÕs a relative amount. You
compare
dollars to dollars. IÕm just trying to see if you can get 
the
math
without twisting it around. Once you can demonstrate that you
can
handle the math we can start talking about special relativity.
> A theory/theorist that takes the relative primitives -
those numbers you
> specifed - and treats them as irrelevant is deranged.
They arenÕt irrelevant. They simply arenÕt 
observable to
Terrance and
Stella. They donÕt have clocks, only tickers. They 
donÕt have
meter
sticks, but they can look at their bank accounts.
> rofßmfao!
> Hey, UIT, just what is it in set theory that says one must
act in
unnecesary
> ignorance?
Set theory says what exists in the material model. It is up
to the
physical theory to say what corresponds to observables. But
observables are ALWAYS events, and it isnÕt set theory that
says that,
itÕs the fact that is all we can observe.
> Just what is it about set theory in your thinking that says
Einstein was
> wrong to posit clocks (and consequently observers) at every
point in
space?
He said to imagine clocks and observers. And itÕs only
pedagogically wrong if people misunderstand what he meant, as
I think
YOU do. The better method is to STICK to observables.
> Just what is it about set theory in your thinking that
demands ignorance?
Clarity. You must remain ignorant of anything you donÕt
understand so
that you can STICK to things you understand and build up the
things
you didnÕt understand FROM the things that you do.
> The idea that you must treat the only relevant knowables -
the givens in
> your gedanken experiment - as irrelevant is worse than
silly.
They arenÕt irelevant to US the modellers, but they are
irrelevant to
Stella and Terrance. Stella can just run again and again at
different
efforts until she gets TerranceÕs money coming at 1/3 the
rate she
sends it. She doesnÕt need anything else. Stella can run
until she
recieves 10 dollars, so she doesnÕt need a clock to tell her
when to
turn around. The point is that I want to show you that if two
people
have tickers and that one moves so that each recieves ticks
from the
other oneÕs ticker at 1/3 the rate of their ticker snds
ticks, and
that one turns around after getting 3K ticks for their own
ticker and
turns back in such a way as to get ticks from the other 
oneÕs
ticker
at 3/1 the rate of their own ticker sends ticks, then when
they get
back together, one ticker will have ticked 6K ticks, and the
other
ticker will have ticked 10K ticks. Can you handle that math.
If
doesnÕt depend on X, as long the ticker has X programmed in
it right
and that ticks moves from one ticker to the other person at a
speed of
X.
> You imagine lengths and time as unobservables but then
imagine you can
> caluclate observables on their basis.
Sure if you want to do that. IÕm fine with 
drawing lines in an
abstract Minkowski plane and counting line segments, but
since I donÕt
know if you know how to do that, IÕll base the lines on the
unmeasureable. The point is that once you are done, the WHOLE
experiment CAN be described in terms of ticks and relative
counts of
ticks. Here it goes.
Terrance and Stella start out together with identical
tickers. Stella
and Terrance separate such that they each recieve ticks from
the other
oneÕs ticker at a rate that is 1/3 the rate that they 
recieve
ticks
from their own ticker. After 9K ticks from his own ticker,
Terrance
starts getting ticks from StellaÕs ticker faster, at a rate
of 3/1 the
rate he recieves ticks from his own ticker. After K more
ticks of his
own ticker, Stella is rejoined with Terrance. Terrance
clearly got 6K
ticks from StellaÕs ticker. And Stella clearly got 10K ticks
from
TerranceÕs ticker. So the ONLY thing we need to compute is
HOW MANY
ticks Stella got from Terrance before the rate changed. If
the answer
is 9K, then we WOULD have a paradox, but because ticks take a
while to
get to someone, we can actually figure out that Stella got
only 1K
ticks at the 1/3 rate, and so she got 1K ticks from Terrance
at 1/3
the rate she sent the first 3K ticks to Terrance. And she got
9K
ticks from Terrance at 3/1 the rate she sent the second 3K
ticks to
Terrance. No paradox. AND no reference to meters, seconds,
durations, distances, anything unobervable, just ticks and
rates
actual sending and recieving. Do you get it? Are you ready
for SR
yet?
===
Subject: Re: Uncle assAl: (SR) Lorentz tÕ, x
= Intervals
> HeÕs an example that IÕll describe with 
Newtonian physics
OK?
>> Assume that Stella runs away from Terrance in a straight
line at 4/5
>> the speed of X m/s, then turns around real fast and runs
back at the
>> same speed 5 seconds. SheÕs gone for 10 seconds, right?
> It computes but isnÕt ßat said. IFF X m/s is wrt 
Terrance.
>> Well what if
>> every 1/10 of a second Terrance sends an american dollar
to Stella at
>> the speed of X m/s, then sheÕd get all 100 USD right? And
sheÕd get
>> the first $10 before turning around and the 
final $90 on her
way
>> back,
>> right?
> Unless sent only at the end of each 1/10th second, in which
case still
> close
> enuf. WeÕll give you the tie.
>> What if every 1/6 of a second she sent Terrance a canadian
>> dollar at the speed of X m/s (so that means she doesnÕt
have to throw
>> it as hard on her way back, but on her way out she has to
throw the
>> canadian dollar harder than Terrance throws the american
dollar),
> X m/s wrt Terrance, ŌobviouslyÕ.
>> so
>> Terrance gets all 60 CAD, but he gets the first 30 CAD
during the
>> first 9 seconds and the final 30 during the 
last 1 second.
> ok
>> So if Terrance compares the RATE at which he THROWS
dollars versus
>> RECIEVES dollars, then for 9 seconds he gets them at 1/3
the rate he
>> sends them, and for 1 second he gets them at 3/1 the rate
he sends
>> them. And for Stella, she notes that for 5 seconds she
GETS dollars
>> at 1/3 the rate she sends them (oh no! they are BOTH
getting dollars
>> at a rate less than they are sending, either Newtonian
physics must
>> be
>> wrong or our notion that a two people canÕt both be 
giving
money to
>> each and both me losing money is wrong, oh no! See how
silly that
>> is!) and for 5 seconds she GETS dollars them at 3/1 the
rate she
>> sends
>> them.
> rofßmfao!
> What a terribly unfair setup!
> For the first tenth of a second he sends dollars at an
infinitely
> greater
> rate than he gets them!
> So much for Newton and Einstein! (Apparently, according to
your
> ŌlogicÕ.)
> rofßmfao!
> How on earth could a non-psychotic imagine that
> fiddling-with-receiving
> babble (however correct) has any contradictory point to
make about the
> sending/etc facts?
>> Ok now letÕs draw a picture of the situation.
>> F
>> |
>> E
>> |
>> C D
>> | /
>> B /
>> |/
>> A
>> Where A is the point when Stella is about to run away from
Terrance.
>> Where B is the point when Terrance is sending Stella his
10th dollar.
>> Where C is the point when Terrance is sending Stella his
50th dollar.
>> Where D is the point when Stella is sending Terrance her
30th dollar.
>> Where D is the point when Stella turns around.
>> Where D is the point when Stella is recieving TerranceÕs
10th dollar.
>> Where E is the point when Terrance is sending Stella his
90th dollar.
>> Where E is the point when Terrance is recieving StellaÕs
30th dollar.
>> Where F is when Stella returns.
>> Where F is the point when Stella is recieving TerranceÕs
100th
>> dollar.
>> Where F is the point when Terrance is recieving StellaÕs
60th dollar.
>> Not that at C and D Stella and Terrance have both RECIEVED
less
>> dollars than they are sent, this is the fact analogous to
time
>> dilation notice how it is STUPID to think is MEANS
anything, it just
>> means that some dollars have been sent and not yet
recieved.
> rofßmfao!
> The name of the stupidity is probably UIT - Unnecessarily
Ignorant
> Theorist.
> The only correct analogy at work here is that you are
asserting your
> Ōobservables onlythesis, treating as 
unobservables the
velocities
> and
> sending rates you specified. The 
receiving/Õobservingrates
have
> absolutely
> no meaning in regard to the givens you stated although they
can be
> calculated therefrom.
> A theory/theorist that takes the relative primitives -
those numbers
> you
> specifed - and treats them as irrelevant is deranged.
> rofßmfao!
> Hey, UIT, just what is it in set theory that says one must
act in
> unnecesary
> ignorance?
> Just what is it about set theory in your thinking that says
Einstein
> was
> wrong to posit clocks (and consequently observers) at every
point in
> space?
> Just what is it about set theory in your thinking that
demands
> ignorance?
> The idea that you must treat the only relevant knowables -
the givens
> in
> your gedanken experiment - as irrelevant is worse than
silly.
> You imagine lengths and time as unobservables but then
imagine you can
> caluclate observables on their basis.
> eleaticus
I gave up with the idiot, he canÕt count to 10. Good luck
with him.
Androcles.
===
Subject: Re: Uncle assAl: (SR) Lorentz tÕ, x
= Intervals
> HeÕs an example that IÕll describe with 
Newtonian physics
OK?
>> Assume that Stella runs away from Terrance in a straight
line at 4/5
>> the speed of X m/s, then turns around real fast and runs
back at the
>> same speed 5 seconds. SheÕs gone for 10 seconds, right?
> It computes but isnÕt ßat said. IFF X m/s is wrt 
Terrance.
>> Well what if
>> every 1/10 of a second Terrance sends an american dollar
to Stella at
>> the speed of X m/s, then sheÕd get all 100 USD right? And
sheÕd get
>> the first $10 before turning around and the 
final $90 on her
way
>> back,
>> right?
> Unless sent only at the end of each 1/10th second, in which
case still
> close
> enuf. WeÕll give you the tie.
>> What if every 1/6 of a second she sent Terrance a canadian
>> dollar at the speed of X m/s (so that means she doesnÕt
have to throw
>> it as hard on her way back, but on her way out she has to
throw the
>> canadian dollar harder than Terrance throws the american
dollar),
> X m/s wrt Terrance, ŌobviouslyÕ.
>> so
>> Terrance gets all 60 CAD, but he gets the first 30 CAD
during the
>> first 9 seconds and the final 30 during the 
last 1 second.
> ok
>> So if Terrance compares the RATE at which he THROWS
dollars versus
>> RECIEVES dollars, then for 9 seconds he gets them at 1/3
the rate he
>> sends them, and for 1 second he gets them at 3/1 the rate
he sends
>> them. And for Stella, she notes that for 5 seconds she
GETS dollars
>> at 1/3 the rate she sends them (oh no! they are BOTH
getting dollars
>> at a rate less than they are sending, either Newtonian
physics must
>> be
>> wrong or our notion that a two people canÕt both be 
giving
money to
>> each and both me losing money is wrong, oh no! See how
silly that
>> is!) and for 5 seconds she GETS dollars them at 3/1 the
rate she
>> sends
>> them.
> rofßmfao!
> What a terribly unfair setup!
> For the first tenth of a second he sends dollars at an
infinitely
> greater
> rate than he gets them!
> So much for Newton and Einstein! (Apparently, according to
your
> ŌlogicÕ.)
> rofßmfao!
> How on earth could a non-psychotic imagine that
> fiddling-with-receiving
> babble (however correct) has any contradictory point to
make about the
> sending/etc facts?
>> Ok now letÕs draw a picture of the situation.
>> F
>> |
> E
>> |
> C D
>> | /
> B /
>> |/
>> A
>> Where A is the point when Stella is about to run away from
Terrance.
>> Where B is the point when Terrance is sending Stella his
10th dollar.
>> Where C is the point when Terrance is sending Stella his
50th dollar.
>> Where D is the point when Stella is sending Terrance her
30th dollar.
>> Where D is the point when Stella turns around.
>> Where D is the point when Stella is recieving TerranceÕs
10th dollar.
>> Where E is the point when Terrance is sending Stella his
90th dollar.
>> Where E is the point when Terrance is recieving StellaÕs
30th dollar.
>> Where F is when Stella returns.
>> Where F is the point when Stella is recieving TerranceÕs
100th
>> dollar.
>> Where F is the point when Terrance is recieving StellaÕs
60th dollar.
>> Not that at C and D Stella and Terrance have both RECIEVED
less
>> dollars than they are sent, this is the fact analogous to
time
>> dilation notice how it is STUPID to think is MEANS
anything, it just
>> means that some dollars have been sent and not yet
recieved.
> rofßmfao!
> The name of the stupidity is probably UIT - Unnecessarily
Ignorant
> Theorist.
> The only correct analogy at work here is that you are
asserting your
> Ōobservables onlythesis, treating as 
unobservables the
velocities
> and
> sending rates you specified. The 
receiving/Õobservingrates
have
> absolutely
> no meaning in regard to the givens you stated although they
can be
> calculated therefrom.
> A theory/theorist that takes the relative primitives -
those numbers
> you
> specifed - and treats them as irrelevant is deranged.
> rofßmfao!
> Hey, UIT, just what is it in set theory that says one must
act in
> unnecesary
> ignorance?
> Just what is it about set theory in your thinking that says
Einstein
> was
> wrong to posit clocks (and consequently observers) at every
point in
> space?
> Just what is it about set theory in your thinking that
demands
> ignorance?
> The idea that you must treat the only relevant knowables -
the givens
> in
> your gedanken experiment - as irrelevant is worse than
silly.
> You imagine lengths and time as unobservables but then
imagine you can
> caluclate observables on their basis.
> eleaticus
> I gave up with the idiot, he canÕt count to 10. Good luck
with him.
> Androcles.
You canÕt count 10 lines and 6 lines. You think if there 
must
be 10
lines going right, that there must always be 10 lines going
left. It
is your unwarranted conditions that create paradoxes, not
other
peopleÕs math.
===
Subject: Course****R/S System: Advanced Programming, San
Francisco***Seattle, December 20-21
R/S Advanced Programming course in Seattle on
December 20-21
Please check out the full description and Agenda of
the 2-day class on the website:
www.xlsolutions-corp.com/Radv.htm
And let us know if we should hold seats for you.
Ask for group discount!
HereÕs the outline:
Course outline:
- Overview of R/S fundamentals: Syntax and Semantics
- Class and Inheritance in R/S-Plus
- Concepts, Construction and good use of language objects
- Coercion and efficiency
- Object-oriented programming in R and S-Plus
- Advanced manipulation tools: Parse, Deparse, Substitute,
etc.
- How to fully take advantage of Vectorization
- Generic and Method Functions; S4 (S-Plus 6)
- Search path, databases and frames Visibility
- Working with large objects
- Handling Properly Recursion and iterative calculations
- Managing loops; For (S-Plus) and for() loops
- Consequences of Lazy Evaluation
- Efficient Code practices for large computations
- Memory management and Resource monitoring
- Writing R/S-Plus functions to call compiled code
- Writing and debugging compiled code for R/S-Plus system
- Connecting R/S-Plus to External Data Sources
- Understanding the structure of model fitting functions in
R/S-Plus
- Designing and Packaging efficiently
This course will also deal with lots of S-Plus efficiency
issues and
any special topics from participants is welcome.
Please let us know if you and your colleagues are interested
in this
class to take advantage of group discount. Register now to
secure your
seat
in this course!
Elvis Miller, PhD
Manager Training.
XLSolutions Corporation
206 686 1578
www.xlsolutions-corp.com
===
Subject: Why is Q not locally compact?
Every point in Q is compact, and every point is open as well
because of
discrete topology, so around every point x, there is a
compact set (in this
case it is x) which contains an open set (in this case x)
containing x.
WhatÕs wrong with this reasoning?
Isaac
===
Subject: Re: Why is Q not locally compact?
>Every point in Q is compact, and every point is open as well
because of
>discrete topology, so around every point x, there is a
compact set (in this
>case it is x) which contains an open set (in this case x)
containing x.
>WhatÕs wrong with this reasoning?
The topology of Q is not the discrete topology.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Why is Q not locally compact?
===
Subject: Course ****R/S System: Advanced Programming,
Seattle***December
20-21
R/S Advanced Programming course in Seattle on
December 20-21
Please check out the full description and Agenda of
the 2-day class on the website:
www.xlsolutions-corp.com/Radv.htm
And let us know if we should hold seats for you.
Ask for group discount!
HereÕs the outline:
Course outline:
- Overview of R/S fundamentals: Syntax and Semantics
- Class and Inheritance in R/S-Plus
- Concepts, Construction and good use of language objects
- Coercion and efficiency
- Object-oriented programming in R and S-Plus
- Advanced manipulation tools: Parse, Deparse, Substitute,
etc.
- How to fully take advantage of Vectorization
- Generic and Method Functions; S4 (S-Plus 6)
- Search path, databases and frames Visibility
- Working with large objects
- Handling Properly Recursion and iterative calculations
- Managing loops; For (S-Plus) and for() loops
- Consequences of Lazy Evaluation
- Efficient Code practices for large computations
- Memory management and Resource monitoring
- Writing R/S-Plus functions to call compiled code
- Writing and debugging compiled code for R/S-Plus system
- Connecting R/S-Plus to External Data Sources
- Understanding the structure of model fitting functions in
R/S-Plus
- Designing and Packaging efficiently
This course will also deal with lots of S-Plus efficiency
issues and
any special topics from participants is welcome.
Please let us know if you and your colleagues are interested
in this
class to take advantage of group discount. Register now to
secure your
seat
in this course!
Elvis Miller, PhD
Manager Training.
XLSolutions Corporation
206 686 1578
www.xlsolutions-corp.com
===
Subject: JSH: Final exam
In arguing with me posters continually push that x=0 is a
special
case.
Well, hereÕs yet another answer and I want you to pay 
careful
attention to what the sci.mathÕers do now.
Remember
49(300125x^3 - 18375 x^2 - 360 x + 22) =
(5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
where the aÕs are the roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and I find out what the constant terms are by using x=0, but
now IÕll
consider x = 7.
Then I have
a^3 + 3(342)a^2 - 49(816361) = 0
which is irreducible over Q, for the aÕs and
49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
where youÕll notice that 102040002 is coprime to 7.
So 7 divides TWO and only TWO of the factors. How do you know?
Because when x=0, two of the aÕs equal 0, 
thatÕs why.
x = 0, is equivalent to x = 0 mod 7.
Notice though that the cubic defining the aÕs is 
still
irreducible
over Q, and irreducibility has no impact.
However, in the ring of algebraic integers NONE of the aÕs
have 7 as a
factor, and each has a non-unit factor in common with 7.
You will get the same result with x = 7k, where k is an
integer,
including k=0, which is not a special case after all.
IÕm curious to see if any of the sci.mathÕers 
wish to argue
over this
point, as, of course, I can add more detail.
You see, the sci.math readership is stupid, and will believe
just
about anything if certain people say, especially if they
argue with
me. So those posters are used to saying stupid things which
are
mathematically incorrect. I find it interesting to see what
they will
try now.
They are stupid, after all. So they will try.
James Harris
===
Subject: Re: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
> Well, hereÕs yet another answer and I want you to pay
careful
> attention to what the sci.mathÕers do now.
> Remember
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and I find out what the constant terms are by using x=0, but
now IÕll
> consider x = 7.
> Then I have
> a^3 + 3(342)a^2 - 49(816361) = 0
> which is irreducible over Q, for the aÕs and
> 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> where youÕll notice that 102040002 is coprime to 7.
> So 7 divides TWO and only TWO of the factors. How do you
know?
You donÕt. Since the polynomial is irreducible, 7 divides
either NONE a_1(7), a_2(7), or a_3(7), or ALL of them.
Suppose A is a solution of
a^3 + 3(342)a^2 - 49(816361) = 0
and A is divisible by 7 in the algebraic integers. That
is, A = 7 * B, where B is an algebraic integer. Then B
satisfies the equation
7 B^3 + 3(342) B^2 - 816361 = 0,
which is irreducible and non-monic and primitive. None of the
roots of this latter equation are algebraic integers.
Therefore
none of a_1(7), a_2(7), or a_3(7) are divisible by 7.
However it is equally true that each of them share a nonunit
factor with 7 in the ring of algebraic integers.
> Because when x=0, two of the aÕs equal 0, 
thatÕs why.
> x = 0, is equivalent to x = 0 mod 7.
x = 0 is indeed a special case, because then the polynomial
that
the aÕs satisfy is reducible. That is not true when x = 7
as you note above.
> Notice though that the cubic defining the aÕs 
is still
irreducible
> over Q, and irreducibility has no impact.
Wrong. See above.
> However, in the ring of algebraic integers NONE of the aÕs
have 7 as a
> factor, and each has a non-unit factor in common with 7.
True, as I just proved. But if that is true, how can you
conclude
that exactly TWO of
(5 a_1(7) + 7), (5 a_2(7) + 7), and (5 a_3(7) + 7)
are divisible by 7 ?
[Of course you have slyly neglected to specify in what ring
your
original claims about divisibility were being made. The
natural
assumption is that it is the ring of algebraic integers.
However,
from what you say above, you evidently realize that this must
not
be true. This leads me to conclude that since you have not
specified
the ring, your statement is essentially vacuous.]
[Note too here that the statement you just made is tantamount
to
conceding what we have said from the beginning about the main
conclusion of ŌAdvanced Polynomial 
FactorizationÕ: i.e., the
main
conclusion is false. And it only took you a year and a half to
get it.]
> You will get the same result with x = 7k, where k is an
integer,
> including k=0, which is not a special case after all.
> IÕm curious to see if any of the 
sci.mathÕers wish to argue
over this
> point, as, of course, I can add more detail.
Go for it.
> You see, the sci.math readership is stupid, and will
believe just
> about anything if certain people say, especially if they
argue with
> me. So those posters are used to saying stupid things which
are
> mathematically incorrect. I find it interesting to see what
they will
> try now.
> They are stupid, after all. So they will try.
It may turn out that you are wrong, as it has very often in
the
past and even now, as you are finally admitting regarding the
conclusion of
ŌAPFÕ. In that case, who is stupid? Who passes 
the final exam ?
Nora B.
> James Harris
===
Subject: Re: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
> Well, hereÕs yet another answer and I want you to pay
careful
> attention to what the sci.mathÕers do now.
> Remember
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and I find out what the constant terms are by using x=0, but
now IÕll
> consider x = 7.
> Then I have
> a^3 + 3(342)a^2 - 49(816361) = 0
> which is irreducible over Q, for the aÕs and
> 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> where youÕll notice that 102040002 is coprime to 7.
> So 7 divides TWO and only TWO of the factors.

No. In this case, NONE of the factors are divisible by 7
in the algebraic integers.
If any of 5a_i + 7 were divisible by 7, then for that
factor, weÕd have to have a_i divisible by 7 so then
b_i = a_i/7 would have to be an algebraic integer.
Notice that since the aÕs satisfy
a^3 + 3*342*a^2 - 49*816361 = 0
then, rearranging and factoring 816361,
a^3 = -3*342*a^2 + 49*7*13*8971
and dividing through by 7^3 we have
(a^3)/(7^3) = -(3*342/7)((a^2)/(7^2)) + 13*8971
or, using b = a/7 we would have
b^3 + (3*342/7)b^2 - 13*8971 = 0
so the bÕs all satisfy
7b^3 + 3*342b^2 - 7*13*8971 = 0
The left side is a non-monic irreducible polynomial,
showing that b cannot be an algebraic integer, from
which we conclude that none of the aÕs are divisible
by 7.
Rick
===
Subject: Re: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
Yup, I will argue.
> Well, hereÕs yet another answer and I want you to pay
careful
> attention to what the sci.mathÕers do now.
> Remember
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and I find out what the constant terms are by using x=0, but
now IÕll
> consider x = 7.
> Then I have
> a^3 + 3(342)a^2 - 49(816361) = 0
> which is irreducible over Q, for the aÕs and
> 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> where youÕll notice that 102040002 is coprime to 7.
Yup.
> So 7 divides TWO and only TWO of the factors. How do you
know?
The question is *why*?
> Because when x=0, two of the aÕs equal 0, 
thatÕs why.
Oh, not very convincing. When x = 0, sqrt(x) = 0, so sqrt(x)
is divisible
by 7? It would be true if the aÕs were polynomials. They are
not.
> x = 0, is equivalent to x = 0 mod 7.
Using equivalence classes mod 7 is only useful when you are
doing
additions and multiplications. With other functions it just
does
not work. You are actually saying that sqrt(7) should be 0
mod 7.
Which means that sqrt(7) is divisible by 7, which means that
1/7
is an element of the ring you are in, and so 7 is a unit in
that ring.
> Notice though that the cubic defining the aÕs 
is still
irreducible
> over Q, and irreducibility has no impact.
Yup, and so the aÕs are not polynomials and that has a
terrific impact.
> However, in the ring of algebraic integers NONE of the aÕs
have 7 as a
> factor, and each has a non-unit factor in common with 7.
And the non-unit factors multiply together to give a multiple
of 49. So,
what is the problem?
--
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: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
> Yup, I will argue.
Here readers can get a good dose of pseudo-math from one the
most
obnoxious posters on sci.math as this guy not only lies about
basic
mathematics, as you can see in his post, but he copied from
my Usenet
posts without my permission on to his own webpage, where he
added in
negative commentary!
And refused to quit using my Usenet posts in violation of
both the
spirit and letter of international copyright law.
And he lies about math.
> Well, hereÕs yet another answer and I want you to pay
careful
> attention to what the sci.mathÕers do now.
> Remember
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and I find out what the constant terms are by using x=0,
but now IÕll
> consider x = 7.
> Then I have
> a^3 + 3(342)a^2 - 49(816361) = 0
> which is irreducible over Q, for the aÕs and
> 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> where youÕll notice that 102040002 is coprime to 7.
> Yup.
> So 7 divides TWO and only TWO of the factors. How do you
know?
> The question is *why*?
> Because when x=0, two of the aÕs equal 0, 
thatÕs why.
Notice that you have
49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
with 49 on the left side, and it has to divide through in
some way.
For two of the aÕs you know they result from functions that
equal 0,
when
x = 0 mod 7 as they equal 0, when x = 0
and you canÕt get any better than that!
The third equals 3 when x = 0, so it is blocked from havinng
factors
in common with 7, when x has 7 itself as a factor, like when
x = 7.
ItÕs basic. To believe otherwise you need to challenge
algebra.
> Oh, not very convincing. When x = 0, sqrt(x) = 0, so
sqrt(x) is
divisible
> by 7? It would be true if the aÕs were polynomials. They
are not.
Now notice, the poster canÕt deny the result when the 
aÕs are
polynomials as then you can physically SEE how it works, so
he handles
that right off the bat by asserting that whether or not they
are
polynomials matters.
Also he tossed in a weird assertion that doesnÕt follow as
thereÕs no
reason to suggest that sqrt(7) is divisible by 7. HeÕs just
trying to
throw up smoke as one other tactic is to confuse readers to
the point
that they just give up and trust that if someone is
disagreeing with
me, then I must be wrong.
> x = 0, is equivalent to x = 0 mod 7.
> Using equivalence classes mod 7 is only useful when you are
doing
> additions and multiplications. With other functions it just
does
So now congruences only matter with additions and
multiplications,
according to the sci.mathÕer, so isnÕt it 
interesting what
math this
person is teaching you?
But, I didnÕt necessarily phrase my own sentence well, as my
point is
that
x=0 is equivalent to x = 0 mod 7, with respect to having
factors of 7
so my using x = 0 mod 7, is a neat way to show that the
result at x=0
is NOT a special case at all, as the constant terms work as
IÕve
explained repeatedly, while the claims of people arguing with
me fall
ßat.
If this werenÕt such a HUGE issue with massive repercussions
for
mathematicians worldwide, then maybe these sci.math-ers might
finally
give up and admit the truth, but wait, IÕm giving them too
much
credit.
It could be a minor issue with few repercussions and these
people,
from what IÕve seen over the years would still lie to you.
Lying is in their nature.
> not work. You are actually saying that sqrt(7) should be 0
mod 7.
No IÕm not.
> Which means that sqrt(7) is divisible by 7, which means
that 1/7
> is an element of the ring you are in, and so 7 is a unit in
that ring.
Desperation.
> Notice though that the cubic defining the aÕs 
is still
irreducible
> over Q, and irreducibility has no impact.
> Yup, and so the aÕs are not polynomials and that has a
terrific impact.
This poster hardly even tries, but why should he?
These TACTICS WORK on sci.math, as IÕve noted for years now.
The sci.math readership is like some kind of religious cult,
willing
to believe just about anything as long as posters disagree
with me.
They are true believers on sci.math, and itÕs freaky.
> However, in the ring of algebraic integers NONE of the
aÕs have 7 as a
> factor, and each has a non-unit factor in common with 7.
> And the non-unit factors multiply together to give a
multiple of 49. So,
> what is the problem?
Oh, hey, itÕs all ok if it works out in the end, right?
If you want the full picture you need to read my original
post, and my
other reply to the other sci.math-er.
The story here is truly sad as the problem I found *was* just
a
historical one, where math professors today could simply
point out
that they followed what they were taught.
But now as the days turn into months and the months move
toward years,
thereÕs little reason to believe that such a huge result
hasnÕt
traveled through the math community, so what was a fascinating
historical error, is now a case of modern fraud.
Math professors who go out today to teach ßawed algebraic
number
theory to their students may be criminals and it may be
possible to
prosecute them under anti-fraud laws, but that sounds so
removed from
reality that they will probably continue without real fear.
But, for those of you listening to them, the impact is very
real.
You are being taught false information, and being forced to do
homework, and take tests to regurgitate that false
information and God
help you if you come back saying that itÕs false based on
what you
learned on newsgroups from a guy called James Harris.
I feel sorry for you, but if you look around you can see how
often
society does things that inside you feel are wrong, but
people just go
on about their business as thatÕs what theyÕve 
learned is
necessary.
And many of you, unfortunately, will be like them, today, and
the next
day.
So, in a great field, where truth is paramount, you will
knowingly
learn false things, and maybe even pride yourself on the
grades you
get from professors who are compromised.
But youÕre living in a mirage...
James Harris
===
Subject: Re: JSH: Final exam
Your cubic polynomial in Ōahas 
coefficients which are
functions of ŌxÕ:
Q(a,x) = a^3 + 3(-1 + 49x)a^2 - 49(2401x^3 - 147x^2 + 3x)
Therefore its roots are functions of ŌxÕ.
Your polynomial in Ōxis:
P(x) = 14706125x^2 - 900375x^2 - 17640x + 1078
which you have factored in terms of the roots of Q(a,x):
P(x) = (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
You claim that since the roots of Q(a,x) evaluated at x = 0,
are 0, 0, and 3
that
P(0) = (0 + 7)(0 + 7)(5*3 + 7) = (7)(7)(22) = 1078
from which you somehow conclude that two of the terms in
parentheses are
divisible by 7 for any value
of ŌxÕ. Or have I misunderstood?
If that is your claim, it is surely false. We have
Q(a,x) @ x=0 : a^3 - 3a^2
Q(a,x) @ x=1 : a^3 + 144a^2 - 110593
Q(a,x) @ x=2 : a^3 + 291a^2 - 912674
Q(a,x) @ x=3 : a^3 + 438a^2 - 3112137
The roots of Q(a,x) are a_1(x), a_2(x) and a_3(x), so
a_1(0) = 0
a_2(0) = 0
a_3(0) = 3
a_1(1) = -31.32993946314472877...
a_2(1) = -138.2104344584112440...
a_3(1) = 25.54037392155592731...
a_1(2) = -63.31229381879031244...
a_2(2) = -279.3003544412207553...
a_3(2) = 51.61264826001106776...
a_1(3) = -31.65614690939515622...
a_2(3) = -420.3902390936727750...
a_3(3) = 77.68498899747098938...
Hence, Ōx = 0is indeed a Ōspecial 
casefrom which you
argue to a false
generalization.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Final exam
...
> So 7 divides TWO and only TWO of the factors. How do you
know?
>
> The question is *why*?
>
> Because when x=0, two of the aÕs equal 0, 
thatÕs why.
>
> Notice that you have
> 49(102040002) = (5a_1(7) + 7)(5a_2(7) + 7)(5a_3(7) + 7)
> with 49 on the left side, and it has to divide through in
some way.
> For two of the aÕs you know they result from functions 
that
equal 0,
> when
> x = 0 mod 7 as they equal 0, when x = 0
This is wrong. They are 0 when x = 0, they are not 0 when x
!= 0,
whether x = 0 mod 7 or not. Note that the aÕs are the roots
of:
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
for any a to be 0, 2401 x^3 - 147 x^2 + 3x must be 0. This is
(with
integer x) only the case when x = 0. Talking about false
math...
> The third equals 3 when x = 0, so it is blocked from
havinng factors
> in common with 7, when x has 7 itself as a factor, like
when x = 7.
This is also wrong. Because if x has 7 as a factor a3(x) does
not
necessarily have 7 as a factor. You are at least consistent in
your errors. You assume that if
1. f(0) = 0
2. x = 0 mod 7
you can conclude that
f(x) = 0 mod 7.
That conclusion is false in general and holds only when f(x)
is a
polynomial in x.
> ItÕs basic. To believe otherwise you need to challenge
algebra.
Yup, it is basic, and to believe otherwise you need to
challenge algebra.
> Oh, not very convincing. When x = 0, sqrt(x) = 0, so
sqrt(x) is
divisible
> by 7? It would be true if the aÕs were polynomials. They
are not.
> Now notice, the poster canÕt deny the result when the 
aÕs
are
> polynomials as then you can physically SEE how it works, so
he handles
> that right off the bat by asserting that whether or not
they are
> polynomials matters.
> Also he tossed in a weird assertion that doesnÕt follow as
thereÕs no
> reason to suggest that sqrt(7) is divisible by 7.
Lets check your assertion above with sqrt(x) as function:
1. sqrt(0) = 0? yes.
2. 7 = 0 mod 7? yes.
sqrt(7) = 0 mod 7? no.
So your conclusion above is not a valid conclusion. It is your
logic that suggests that sqrt(7) is divisible by 7.
> x = 0, is equivalent to x = 0 mod 7.
>
> Using equivalence classes mod 7 is only useful when you are
doing
> additions and multiplications. With other functions it just
does
> So now congruences only matter with additions and
multiplications,
> according to the sci.mathÕer, so isnÕt it 
interesting what
math this
> person is teaching you?
What other operations do you think it works with? Not with
the sqrt
function. Also not with division.
> x=0 is equivalent to x = 0 mod 7, with respect to having
factors of 7
> so my using x = 0 mod 7, is a neat way to show that the
result at x=0
> is NOT a special case at all, as the constant terms work as
IÕve
> explained repeatedly, while the claims of people arguing
with me fall
> ßat.
Can you please explain why your inference work *must* work
for your
function, but does not work for sqrt?
--
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: JSH: Final exam
Originator: richard@cogsci.ed.ac.uk (Richard Tobin)
>You assume that if
> 1. f(0) = 0
> 2. x = 0 mod 7
>you can conclude that
> f(x) = 0 mod 7.
>That conclusion is false in general and holds only when f(x)
is a
>polynomial in x.
To be pedantic, there are other classes of function that it
holds for.
For example, it holds for all functions that are zero at all
the
integers.
-- Richard
===
Subject: Re: Final exam
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and I find out what the constant terms are by using x=0, but
now IÕll
> consider x = 7.
Where do you get this silly arithmetic? What are you choosing
this
particular equation as your starting point. Does it have any
significance?
Where do you get this silly arithmetic? What are you choosing
this
particular equation as your starting point. Does it have any
significance?
Did it come to you in a Kekule style benzene dream? Did god
give it to you?
Is it special, or is it random?
===
Subject: Re: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
> Well, hereÕs yet another answer and I want you to pay
careful
> attention to what the sci.mathÕers do now.
> Remember
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
When x = 0, the above is 1078, which is divisible by 11
When x = 1, the above is 13789188, which is not divisible by
11.
Certainly there are at least *some* properties where x=0
leads to
different results than x=1.
Of course, x=1 could be the special case, but either way, the
properties
are not the same.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: JSH: Final exam
> In arguing with me posters continually push that x=0 is a
special
> case.
Well, hereÕs yet another answer and I want you to pay 
careful
> attention to what the sci.mathÕers do now.
Remember
49(300125x^3 - 18375 x^2 - 360 x + 22) =
> When x = 0, the above is 1078, which is divisible by 11
> When x = 1, the above is 13789188, which is not divisible
by 11.
> Certainly there are at least *some* properties where x=0
leads to
> different results than x=1.
> Of course, x=1 could be the special case, but either way,
the properties
> are not the same.
Told you theyÕd fight.
Consider also that the poster deleted out the relevant
information, an
old sci.math tactic, and now consider the facts
49(300125x^3 - 18375 x^2 - 360 x + 22) =
(5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
where the aÕs are the roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
and setting x=0, letÕs you see that two of the 
aÕs go to 0,
at that
point as you have
49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) =
(5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7)
and
a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0)) = 0
so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the 
aÕs
equal 0,
while one equals, 3, at x=0.
Since I have the correct mathematics I can endlessly explain
it in
different variations which is fun for me, but a headache for
those who
lie about my work as notice that I simply went to using x = 0
mod 7,
and in my original post specifically used x = 7 to show that
you get
an irreducible over Q cubic for the aÕs, which shows that in
the ring
of algebraic integers NONE of the aÕs have 7 as a factor.
But, from the result at x=0, you know that one of the aÕs
cannot have
non-unit factors in common with 7 for x=0 mod 7, but they do
in the
ring of algebraic integers.
It turns out what are non-unit factors in the ring of
algebraic
integers are actually unit factors that are not units in the
ring of
algebraic integers because they are not roots of some monic
polynomial
with integer coefficients.
You see that definition focusing on roots of monic polynomials
with
integer coefficients gives a ring that has these quirks. You
can
figure out the quirks, if you trust algebra and use logic, but
if
youÕre a ßawed human being desperate to have an easy tool 
to
prove
things, then you will fight it.
Much of the supposedly great accomplishments in algebraic
number
theory over the last hundred years plus are simply vapor.
Now to believe the sci.mathÕers, you need to start 
fiddling
with your
understanding of congruence, and IÕm sure some of them will
come
forward to lie to you about it.
ThatÕs just the fun of the game I play on sci.math and this
newsgroup.
WhatÕs not a game is that some of you probably will go to
class today
and some math professor is going to confidently teach you
mathematical
information that is wrong.
When you go to class today, if your professor talks about
algebraic
number theory, or misuses Galois Theory (given what youÕve
learned
here) I want you to carefully notice how you feel.
Hold on to that feeling so that you never forget it.
James Harris
===
Subject: Re: JSH: Final exam
> When you go to class today, if your professor talks about
algebraic
> number theory, or misuses Galois Theory (given what youÕve
learned
> here) I want you to carefully notice how you feel.
If any students think about you in class at all (which I
doubt), IÕm
certain theyÕll be telling their professors something like,
Wait till
you hear what that idiot Harris said *this* time!
--
Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you
improvise
fwbrown@bellsouth.net | if youÕre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: JSH: Final exam
>In arguing with me posters continually push that x=0 is a
special
>case.
>Well, hereÕs yet another answer and I want you to pay 
careful
>attention to what the sci.mathÕers do now.
>Remember
>49(300125x^3 - 18375 x^2 - 360 x + 22) =
>>When x = 0, the above is 1078, which is divisible by 11
>>When x = 1, the above is 13789188, which is not divisible
by 11.
>>Certainly there are at least *some* properties where x=0
leads to
>>different results than x=1.
>>Of course, x=1 could be the special case, but either way,
the properties
>>are not the same.
> Told you theyÕd fight.
> Consider also that the poster deleted out the relevant
information, an
> old sci.math tactic, and now consider the facts
> 49(300125x^3 - 18375 x^2 - 360 x + 22) =
> (5a_1(x) + 7)(5a_2(x) + 7)(5a_3(x) + 7)
> where the aÕs are the roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> and setting x=0, letÕs you see that two of the 
aÕs go to 0,
at that
> point as you have
> 49(300125(0)^3 - 18375 (0)^2 - 360 (0) + 22) =
> (5a_1(0) + 7)(5a_2(0) + 7)(5a_3(0) + 7)
> and
> a^3 + 3(-1 + 49(0))a^2 - 49(2401 (0)^3 - 147 (0)^2 + 3(0))
= 0
> so a^3 - 3a^2 = 0, which is a^2(a - 3) = 0, so two of the
aÕs equal 0,
> while one equals, 3, at x=0.
> Since I have the correct mathematics I can endlessly
explain it in
> different variations which is fun for me, but a headache
for those who
> lie about my work as notice that I simply went to using x =
0 mod 7,
> and in my original post specifically used x = 7 to show that
you get
> an irreducible over Q cubic for the aÕs, which shows that
in the ring
> of algebraic integers NONE of the aÕs have 7 as a factor.
> But, from the result at x=0, you know that one of the aÕs
cannot have
> non-unit factors in common with 7 for x=0 mod 7, but they
do in the
> ring of algebraic integers.
And it doesnÕt occur to you that you are making an argument
that 7 is in
the same category of special cases as 0 for the purposes of
dividing by
7? Many people have shown that your arguments are simply
false when
x=1. They didnÕt do anything fancy, just cranked out the
numbers and
showed they donÕt have the properties you claim they do.
Did I change what was being talked about? Yes. Did you miss
the point?
Yes. The burden of proof is always on you to show that x=0 is
*not* a
special case. The best you have done is to argue that x=0 mod
7 has a
special property. That is *still* a special case.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: JSH: Just playing
You are stupid compared to me, but thatÕs not a surprise.
If you realized that early, would you have played with me?
Instead IÕve had years of fun, but now I need to move on to
other
games.
So I now tell you outright that you are stupid, and show it
to you
with more mathematics.
You will fight, and I will break you.
I could have before, but then you wouldnÕt have played with
me any
more.
James Harris
===
Subject: Re: JSH: Just playing
> You are stupid compared to me, but thatÕs not a surprise.
> If you realized that early, would you have played with me?
> Instead IÕve had years of fun, but now I need to move on 
to
other
> games.
> So I now tell you outright that you are stupid, and show it
to you
> with more mathematics.
> You will fight, and I will break you.
> I could have before, but then you wouldnÕt have played 
with
me any
> more.
And hereÕs the runny-nosed little gutter-snipe again, 
shaking
his
fist at the rest of the kids and telling them through his
tears that
he doesnÕt want to be invited to their stinky old clubhouse,
anyway.
Most people outgrow that sort of thing eventually...
--
Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you
improvise
fwbrown@bellsouth.net | if youÕre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: JSH: Just playing
> You are stupid compared to me, but thatÕs not a surprise.
> If you realized that early, would you have played with me?
> Instead IÕve had years of fun, but now I need to move on 
to
other
> games.
> So I now tell you outright that you are stupid, and show it
to you
> with more mathematics.
> You will fight, and I will break you.
> I could have before, but then you wouldnÕt have played 
with
me any
> more.
> James Harris
Stop......Hammer time.
===
Subject: Re: JSH: Just playing
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ÕELIi
$t^
VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>> You are stupid compared to me, but thatÕs not a surprise.
>> If you realized that early, would you have played with me?
>> Instead IÕve had years of fun, but now I need to move on
to other
>> games.
>> So I now tell you outright that you are stupid, and show
it to you
>> with more mathematics.
>> You will fight, and I will break you.
>> I could have before, but then you wouldnÕt have played
with me any
>> more.
> Stop......Hammer time.
I am wondering about the alcoholic content of this Hammer.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: JSH: Just playing
>You are stupid compared to me, but thatÕs not a surprise.
>If you realized that early, would you have played with me?
>Instead IÕve had years of fun, but now I need to move on to
other
>games.
Facinating, the way we continually swing back and forth
between
the current youÕre just ants for mighty me to toy with and
bitter complaints about how the awful treatment you get here
has ruined your life.
>So I now tell you outright that you are stupid, and show it
to you
>with more mathematics.
>You will fight, and I will break you.
>I could have before, but then you wouldnÕt have played with
me any
>more.
>James Harris
************************
David C. Ullrich
===
Subject: Re: JSH: Just playing
> You are stupid compared to me, but thatÕs not a surprise.
> If you realized that early, would you have played with me?
> Instead IÕve had years of fun, but now I need to move on 
to
other
> games.
> So I now tell you outright that you are stupid, and show it
to you
> with more mathematics.
> You will fight, and I will break you.
> I could have before, but then you wouldnÕt have played 
with
me any
> more.
Is this addressed to anyone in particular? Or everyone?
Figuring there
are tens or hundreds of thousand of sci.math readers, itÕs
statistically
unlikely that you are smarter than ALL of them. Feynman had a
remark
along these lines in one of his popular essays. He pointed
out that in a
room full of people, he couldnÕt be sure he was the smartest
person, but
he was reasonably certain that he was smarter than the
average of all
the people.
Personally, I donÕt consider myself smarter than you.
Frankly, not even
saner, just crazy in a different dimension. And not nearly as
entertaining. But still, you donÕt quite understand what
constitutes a
mathematical proof; and you refuse to take the time to study
enough
abstract algebra to communicate with anyone. YouÕve clearly
put a lot of
time and energy into your work, and youÕve brought awareness
of the
algebraic integers to many of us who never knew about them
before. You
bring a lot of entertainment to this newsgroup.
If youÕre about to take one of your periodic vacations from
posting, I
for one will miss you. See you next time around.
===
Subject: Re: JSH: Just playing
> You are stupid compared to me, but thatÕs not a surprise.
> If you realized that early, would you have played with me?
> Instead IÕve had years of fun, but now I need to move on 
to
other
> games.
YouÕve said this before. You came back.
> So I now tell you outright that you are stupid, and show it
to you
> with more mathematics.
> You will fight, and I will break you.
YouÕve said this before as well, it didnÕt 
happen.
> I could have before, but then you wouldnÕt have played 
with
me any
> more.
Perhaps if you learned to relax you wouldnÕt say things like
the above.
Why do you feel this is a competition?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Just playing
> You are stupid compared to me, but thatÕs not a surprise.
Narcissistic Personality Disorder
A pervasive pattern of grandiosity (in fantasy or behavior),
need for
admiration, and lack of empathy, beginning by early adulthood
and present
in
a variety of contexts, as indicated by five (or more) of the
following:
(1) has a grandiose sense of self-importance (e.g.,
exaggerates
achievements
and talents, expects to be recognized as superior without
commensurate
achievements)
(2) is preoccupied with fantasies of unlimited success,
power, brilliance,
beauty, or ideal love
(3) believes that he or she is special and unique and can
only be
understood by, or should associate with, other special or
high-status
people
(or institutions)
(4) requires excessive admiration
(5) has a sense of entitlement, i.e., unreasonable
expectations of
especially favorable treatment or automatic compliance with
his or her
expectations
(6) is interpersonally exploitative, i.e., takes advantage of
others to
achieve his or her own ends
(7) lacks empathy: is unwilling to recognize or identify with
the feelings
and needs of others
(8) is often envious of others or believes that others are
envious of him
or
her
(9) shows arrogant, haughty behaviors or attitudes
Reprinted with permission from the Diagnostic and Statistical
Manual of
Mental Disorders, fourth Edition. Copyright 1994 American
Psychiatric
Association
***
Perhaps instead of spewing your useless nonsense, you should
study this
disorder, go see a doctor and then drug and drink yourself to
you pass out.
Research into your delusional fits of grandeur would be a more
appropriate vocation for you!
Think about, eh!
===
Subject: Re: Just playing
ßip_alpha@safebunch.com says...
> You are stupid compared to me, but thatÕs not a surprise.
> Narcissistic Personality Disorder
> A pervasive pattern of grandiosity (in fantasy or
behavior), need for
> admiration, and lack of empathy, beginning by early
adulthood and present
in
> a variety of contexts, as indicated by five (or more) of the
following:
> (1) has a grandiose sense of self-importance (e.g.,
exaggerates
achievements
> and talents, expects to be recognized as superior without
commensurate
> achievements)
James Harris nPhd.
===
Subject: finite maze solving algorithm
I was wondering if anyone knows if all possible topologies of
finite
2d mazes can be solved by a finite algorithm. For example, we
know
that all fully connected mazes can be solved by picking a
wall and
exhaustively following it. Can a general solution work for
all mazes
including the ones that are piecewise disconnected? If this is
possible, is the general solution a solved problem?
===
Subject: Re: finite maze solving algorithm
> I was wondering if anyone knows if all possible topologies
of finite
> 2d mazes can be solved by a finite algorithm.
Yes, they can.
> For example, we know
> that all fully connected mazes can be solved by picking a
wall and
> exhaustively following it. Can a general solution work for
all mazes
> including the ones that are piecewise disconnected? If this
is
> possible, is the general solution a solved problem?
Yes it is. Look for the Pledge Algorithm. It is
well-described in
the book Turtle Geometry, including a proof for how it solves
every
finite maze.
The algorithm works by keeping track of the total amount of
turning
you do as you move throughout the maze. Using this purely
local
information (you donÕt need breadcrumbs, or to mark walls
etc.) it is
possible to escape.
It is rather simple:
1. Initialize a counter to zero, and define an arbitrary
direction to
be ŌnorthÕ.
2. Move straight north until an obstacle is met.
3. Turn left and follow the obstacle. Keep track of Ōtotal
turningÕ
and increment the counter by +1 for every full 360 degree turn
clockwise, and -1 for every full 360 turn anti-clockwise.
4. Leave the obstacle when it is possible to move straight
north and
the counter reads zero. Goto step 2.
Steven
===
Subject: Re: finite maze solving algorithm
> I was wondering if anyone knows if all possible topologies
of finite
> 2d mazes can be solved by a finite algorithm. For example,
we know
> that all fully connected mazes can be solved by picking a
wall and
> exhaustively following it. Can a general solution work for
all mazes
> including the ones that are piecewise disconnected? If this
is
> possible, is the general solution a solved problem?
Interesting. How are you defining a maze, here? Usually, I
think of a
maze as just a graph, which you can solve using something
simple like
depth-first search, regardless of whether it is planar. It is
probably
better to think in terms of connectedness of rooms than
connectedness of
walls.
(I hope I am not being ignorant here, but you seem to be using
topology and piecewise disconnected in strange ways.
Obviously if
the topological space is disconnected, there will be no path
possible
between points in different components - in the discrete
topology, if
youÕre not there, you canÕt get there.)
HTH
--
Kevin
===
Subject: Re: finite maze solving algorithm
> I was wondering if anyone knows if all possible topologies
of finite
> 2d mazes can be solved by a finite algorithm. For example,
we know
> that all fully connected mazes can be solved by picking a
wall and
> exhaustively following it. Can a general solution work for
all mazes
> including the ones that are piecewise disconnected? If this
is
> possible, is the general solution a solved problem?
> Interesting. How are you defining a maze, here? Usually, I
think of a
> maze as just a graph, which you can solve using something
simple like
> depth-first search, regardless of whether it is planar. It
is probably
> better to think in terms of connectedness of rooms than
connectedness of
> walls.
> (I hope I am not being ignorant here, but you seem to be
using
> topology and piecewise disconnected in strange ways.
Obviously if
> the topological space is disconnected, there will be no
path possible
> between points in different components - in the discrete
topology, if
> youÕre not there, you canÕt get there.)
> HTH
I am probably using those terms in an unusual way, being
entirely a
layman. When I refer to piecewise disconnnected I mean the
walls not
the paths. I understand that an unconnected path can not be
traversed.
To look at this problem interms of paths (which I guess is
the graph
theory approach you refer to) you would need to posit a set
of graphs
that included loops, dead ends, routes that deadend in loops,
and
various structures of loops nested and sharing common
branches. I wish
usenet was friendlier to hand drawn diagrams embedded in
text. That
would make explaining this a lot easier.
===
Subject: Re: finite maze solving algorithm
> I was wondering if anyone knows if all possible topologies
of finite
> 2d mazes can be solved by a finite algorithm. For example,
we know
> that all fully connected mazes can be solved by picking a
wall and
> exhaustively following it. Can a general solution work for
all mazes
> including the ones that are piecewise disconnected? If this
is
> possible, is the general solution a solved problem?
A maze with n walls is homeomorphic to the n-times punctured
plane.
So the method of cuts used by Cauchy to derive a
simply-connected
domain can be applied. The cuts in this instance become
barriers
joining the walls in some sequence, the last one getting a
final
barrier going off to infinity. Then any two points in the maze
can be
joined by a path which is homotopically unique.
to the same thing but with the barriers introduced on the run
rather
than in advance.
===
Subject: Re: finite maze solving algorithm
 I was wondering if anyone knows
if all possible topologies of finite
>> 2d mazes can be solved by a finite algorithm. For example,
we know
>> that all fully connected mazes can be solved by picking a
wall and
>> exhaustively following it. Can a general solution work for
all mazes
>> including the ones that are piecewise disconnected? If
this is
>> possible, is the general solution a solved problem?
>Interesting. How are you defining a maze, here? Usually, I
think of a
>maze as just a graph, which you can solve using something
simple like
>depth-first search, regardless of whether it is planar. It is
probably
>better to think in terms of connectedness of rooms than
connectedness of
>walls.
The basic problem with a depth-first search is the chance
youÕll get into a
loop going around and around a disconnected piece. The
problem can be
avoided by making a rule to never cross your own path.
Or is there something more that IÕm missing?
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: finite maze solving algorithm
> I was wondering if anyone knows if all possible topologies
of finite
> 2d mazes can be solved by a finite algorithm. For example,
we know
> that all fully connected mazes can be solved by picking a
wall and
> exhaustively following it. Can a general solution work for
all mazes
> including the ones that are piecewise disconnected? If this
is
> possible, is the general solution a solved problem?
>>Interesting. How are you defining a maze, here? Usually, I
think of a
>>maze as just a graph, which you can solve using something
simple like
>>depth-first search, regardless of whether it is planar. It
is probably
>>better to think in terms of connectedness of rooms than
connectedness of
>>walls.
> The basic problem with a depth-first search is the chance
youÕll get into
a
> loop going around and around a disconnected piece. The
problem can be
> avoided by making a rule to never cross your own path.
> Or is there something more that IÕm missing?
If you keep going in circles, then you are not doing a
depth-first
search. Depth-first search will systematically explore *any*
maze
in its entirety.
In depth-first-search, you keep traversing down previously
untraversed
paths until either you reach a place where all possible paths
have
already been traversed or you reach a dead end. In these
situations,
you backtrack along the path youÕve just traversed until
reach a
place where there is an unexplored corridor, and then you
head down that
corridor.
To find your way through an actual maze (i.e. not one written
on paper),
you need some way of marking passages. A 19Õth century
algorithm for
traversing a maze is as follows. Bring along a pile of
pennies to
mark passages as you enter and leave junctions. During the
algorithm,
passages without pennies havenÕt been traversed yet, those
with one
penny have been traversed exactly once, and those with two
pennies have
been traversed and then backtracked along in the opposite
direction.
Enter the maze. When you encounter a junction that you
havenÕt seen
before (no passageways with pennies), then choose a passage
at random
(remember to drop a penny in the passage you just left and in
the
passage you just entered). If you hit a dead end, turn around
and go
back. If you are walking down a passage for the first time and
you
encounter a junction youÕve seen before, then turn around 
and
head
back (remember to drop two pennies, one for entering the
junction
and one for leaving). If you are walking down a passage for
the
second time and encounter a junction, then take a new passage
if
there is one, otherwise take an old passage (marked with one
penny).
When you reach the solution, passages marked exactly once will
indicate a direct path back to the start. If there is no
solution,
then you will end up back at the start with all passages
marked twice.
===
Subject: Re: finite maze solving algorithm
>> I was wondering if anyone knows if all possible topologies
of finite
>> 2d mazes can be solved by a finite algorithm. For example,
we know
>> that all fully connected mazes can be solved by picking a
wall and
>> exhaustively following it. Can a general solution work for
all mazes
>> including the ones that are piecewise disconnected? If
this is
>> possible, is the general solution a solved problem?
>Interesting. How are you defining a maze, here? Usually, I
think of a
>maze as just a graph, which you can solve using something
simple like
>depth-first search, regardless of whether it is planar. It is
probably
>better to think in terms of connectedness of rooms than
connectedness of
>walls.
> The basic problem with a depth-first search is the chance
youÕll get into
a
> loop going around and around a disconnected piece. The
problem can be
> avoided by making a rule to never cross your own path.
> Or is there something more that IÕm missing?
> --Keith Lewis klewis {at} mitre.org
> The above may not (yet) represent the opinions of my
employer.
I donÕt think that rule would be sufficient, How 
would it
handle the
case where a path deadended in a lopp. If you could never
cross your
path again youÕd be stuck. Certainly there are lots of 
graphs
that can
not be traced if you donÕt allow, jumping, retracing or
crossing?
Maybe I am misunderstanding your suggestion.
===
Subject: Re: finite maze solving algorithm
>>I was wondering if anyone knows if all possible topologies
of finite
>>2d mazes can be solved by a finite algorithm. For example,
we know
>>that all fully connected mazes can be solved by picking a
wall and
>>exhaustively following it. Can a general solution work for
all mazes
>>including the ones that are piecewise disconnected? If this
is
>>possible, is the general solution a solved problem?
>Interesting. How are you defining a maze, here? Usually, I
think of a
>maze as just a graph, which you can solve using something
simple like
>depth-first search, regardless of whether it is planar. It is
probably
>better to think in terms of connectedness of rooms than
connectedness of
>walls.
>>The basic problem with a depth-first search is the chance
youÕll get into
a
>>loop going around and around a disconnected piece. The
problem can be
>>avoided by making a rule to never cross your own path.
>>Or is there something more that IÕm missing?
>>--Keith Lewis klewis {at} mitre.org
>>The above may not (yet) represent the opinions of my
employer.
> I donÕt think that rule would be sufficient, 
How would it
handle the
> case where a path deadended in a lopp. If you could never
cross your
> path again youÕd be stuck. Certainly there are lots of
graphs that can
> not be traced if you donÕt allow, jumping, retracing or
crossing?
> Maybe I am misunderstanding your suggestion.
OK, in laymanÕs terms, assume you have a long rope and a 
loaf
of bread.
Tie one end of the rope to a hitch outside, and keep it
reasonably
taut. Whenever you enter a new room, drop a bread crumb. Now,
determine which adjacent rooms have bread in them. If all
adjacent
rooms have bread, follow the rope back to the last room. If
any of them
have no bread, go in that room, drop a crumb, check the
adjacent rooms
for crumbs, and so on. You are then guaranteed to visit every
room in
the maze, assuming the rooms are all connected somehow.
HTH
--
Kevin
===
Subject: Re: finite maze solving algorithm
>I was wondering if anyone knows if all possible topologies
of finite
>2d mazes can be solved by a finite algorithm. For example, we
know
>that all fully connected mazes can be solved by picking a
wall and
>exhaustively following it. Can a general solution work for
all mazes
>including the ones that are piecewise disconnected? If this
is
>possible, is the general solution a solved problem?
>>Interesting. How are you defining a maze, here? Usually, I
think of a
>>maze as just a graph, which you can solve using something
simple like
>>depth-first search, regardless of whether it is planar. It
is probably
>>better to think in terms of connectedness of rooms than
connectedness of
>>walls.
> The basic problem with a depth-first search is the chance
youÕll get into
a
> loop going around and around a disconnected piece. The
problem can be
> avoided by making a rule to never cross your own path.
> Or is there something more that IÕm missing?
> --Keith Lewis klewis {at} mitre.org
> The above may not (yet) represent the opinions of my
employer.
I was under the impression that rule was part of DFS on
arbitrary
graphs, though it isnÕt needed for trees. Otherwise cycles
are a problem.
===
Subject: Re: topology.....+++
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iAU32mj11249;
>hello......doctor~
>let X be a topolotical space and A, B be subsets of X with B
in A.
>We give the subspace topology on A.
>(a) if B is dense in A and A is dense in X,
>then show that B is dense in X.
>(b) Given an example of a dense proper subset of Q, the set
of
>all rational number.
>------------------------------------------------------------
>i can do (a).
>but i canÕt find the example of (b).
>i only know that Q is dense subset R.
>so, i need your adivice.
>thank you very much for your advice.
Unless IÕm missing something, (b) seems trivial. Just leave
out one point
from Q. The rest is a dense proper subset of Q.
===
Subject: Unstoppable Force vs Immovable Object
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iAU32nX11264;
What happens when an unstoppable force collides with a
immovable object?
This question definitely lies within the realm of the
theoretical but there
is a definite solution that one may establish if such an event
could happen.
Just use basic properties of physics! The Normal Force of the
unstoppable
force is equal to infinity ( F=oo ). Then you take the Strong
Force
that makes any object solid- the reason you do not fall to
the center of the
earth- which is a set number and therefore less then infinity.
Now Normal
Force exceeds Strong Force and the unstoppable force simply
passes through
the unmovable object.
I may be in error but I have been thinking about this and
come to this
solution. If you have another expiation then write it, this
has been plaguing
my mind for awhile.
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an irresistable force collides with a
immovable object?
There is an inconceivable concussion!
.
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable
> object?
The same thing that happens when you go further North than
the North Pole.
This is a silly, purely verbal claptrap question.
Bob Kolker
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable
object? This question definitely lies within the realm of the
theoretical but there is a definite solution that one may
establish if
such an event could happen. Just use basic properties of
physics! The
Normal Force of the unstoppable force is equal to infinity (
F=oo ). Then you take the Strong Force that makes any object
solid-
the reason you do not fall to the center of the earth- which
is a set
number and therefore less then infinity. Now Normal Force
exceeds
Strong Force and the unstoppable force simply passes through
the
unmovable object.
> I may be in error but I have been thinking about this and
come to this
solution. If you have another expiation then write it, this
has been plaguing
my mind for awhile.
It is logically impossible for an unstoppable force to meet an
immovable object. When a force meets an object, of logical
necessity
either the object moves or fails to move. So, of logical
necessity,
either the object isnÕt immovable or the force 
isnÕt
unstoppable.
===
Subject: Re: Unstoppable Force vs Immovable Object
> It is logically impossible for an unstoppable force to meet
an
> immovable object. When a force meets an object, of logical
necessity
> either the object moves or fails to move. So, of logical
necessity,
> either the object isnÕt immovable or the force 
isnÕt
unstoppable.
Any finite mass on which a net force is exerted will
accelerate in the
direction of the net force. NewtonÕs Second Law.
Bob Kolker
===
Subject: Re: Unstoppable Force vs Immovable Object
It is logically impossible for an unstoppable force to meet an
> immovable object. When a force meets an object, of logical
necessity
> either the object moves or fails to move. So, of logical
necessity,
> either the object isnÕt immovable or the force 
isnÕt
unstoppable.
> Any finite mass on which a net force is exerted will
accelerate in the
> direction of the net force. NewtonÕs Second Law.
> Bob Kolker
Yes, I know. In that case, itÕs also physically impossible
that there
should be an immovable object. I was ignoring the laws of
physics and
going for logic alone.
===
Subject: Re: Unstoppable Force vs Immovable Object
> Yes, I know. In that case, itÕs also physically impossible
that there
> should be an immovable object. I was ignoring the laws of
physics and
> going for logic alone.
Logic, as such, has very little to say about physical forces.
As I said
you are just playing a word game.
Bob Kolker
===
Subject: Re: Unstoppable Force vs Immovable Object
> Yes, I know. In that case, itÕs also physically impossible
that there
> should be an immovable object. I was ignoring the laws of
physics and
> going for logic alone.
> Logic, as such, has very little to say about physical
forces. As I said
> you are just playing a word game.
> Bob Kolker
Why? WhatÕs wrong with the argument?
===
Subject: Re: Unstoppable Force vs Immovable Object
> Why? WhatÕs wrong with the argument?
Force has a physical meaning, not a logical meaning. A force
is some
kind of action that changes the momentum of a body.
Bob Kolker
===
Subject: Re: Unstoppable Force vs Immovable Object
Why? WhatÕs wrong with the argument?
> Force has a physical meaning, not a logical meaning. A
force is some
> kind of action that changes the momentum of a body.
> Bob Kolker
I donÕt think youÕve identified a 
problem with my reasoning.
===
Subject: Re: Unstoppable Force vs Immovable Object
> I donÕt think youÕve identified 
a problem with my reasoning.
Your reasoning addresses a physical concept in a non-physical
way. As
I said, you are playing word games.
Bob Kolker
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable
> object?
They married, gave birth to me and then divorced shortly
afterward having caused sufficient trouble.
Bart
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable object?
You get a tedious riddle.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable
> object? This question definitely lies within the realm of 
the
> theoretical but there is a definite solution that one may
establish
> if such an event could happen. Just use basic properties of
physics!
> The Normal Force of the unstoppable force is equal to
infinity
> ( F=oo ). Then you take the Strong Force that makes any
object solid-
> the reason you do not fall to the center of the earth-
which is a set
> number and therefore less then infinity. Now Normal Force
exceeds
> Strong Force and the unstoppable force simply passes
through the
> unmovable object.
Then itÕs not really unmovable, is it?
In the real world, there are no such things as unstoppable
forces or
immovable objects. So invoking physics to solve this problem
is
inappropriately mixing the hypothetical world of the puzzle
with the real
world. In the hypothetical world of the puzzle, if there is
such a thing
as
an unstoppable force, then there cannot exist such a thing as
an immovable
object, and vice versa. This puzzle is just as trivial as
saying Joe is
taller than Jim, and Jim is taller than Joe. Which one is
taller? The
given conditions are contradictory.
--Mark
===
Subject: Re: Unstoppable Force vs Immovable Object
> What happens when an unstoppable force collides with a
immovable
> object? This question definitely lies within the realm of 
the
> theoretical but there is a definite solution that one may
establish
> if such an event could happen. Just use basic properties of
physics!
> The Normal Force of the unstoppable force is equal to
infinity
> ( F=oo ). Then you take the Strong Force that makes any
object solid-
> the reason you do not fall to the center of the earth-
which is a set
> number and therefore less then infinity. Now Normal Force
exceeds
> Strong Force and the unstoppable force simply passes
through the
> unmovable object.
> I may be in error but I have been thinking about this and
come to
> this solution. If you have another expiation then write it,
this has
> been plaguing my mind for awhile.
I have two responses, both with the same impact.
IÕll preface it all by acknowledging that some of what 
IÕve
written may
seem a bit harsh, but I sincerely mean no offense. I just
think youÕre
being a bit self-indulgent, and (if this has *really* been
plaguing your
mind) perhaps you have too much time on your hands. Have you
considered
taking up masturbation?
First (quickÕnÕdirty): This is nearly as 
stupid as the
dungheap
that has been erected by JSH over the past 8 years. Get a
life.
Second (on the technical side):
The strong force is not the reason objects are solid. It is
the reason
atomic nuclei hold together, but at the distances atoms
relate to one
another, the strong interaction is ineffective. If it were
the strong
force at play, youÕd see a lot more nuclear fusion than is
common today.
Probably more than youÕd like to see, if you get my drift.
You should read about the fundamental forces.
Next, the irresistable force/immovable object conundrum is
nothing
more than a classical paradox, in which
mutually-contradictory entities
are proposed, after which one attempts to proclaim ones own
smarty-
pantsitude. ItÕs a game of infantile logic grappling with
semantics.
As they say: Philosophy is the wading pool of the mind
Is that what you mean to do with your life? Why not solve the
existence
of Evil in a world created by a Perfect God? What about the
nature of
human hope and the problem of personal mortality?
Well, maybe a problem that isnÕt just us gnashing our teeth:
how about
the problem of human suffering? Immense wealth and abysmal
poverty?
Getting JSH to pull his head out of his own ass?
The point is that if you are an intelligent person, you are
doing
yourself a monumental disservice by wasting your limited time
in
this lifecycle on semantic quibbling. If this problem is truly
plaguing your mind, then you need to get off it, and do
something
that matters.
If youÕre a dimbulb, then it doesnÕt matter 
what the hell you
do,
just as long as you stay out of the way of others.
Suit yourself, of course, but donÕt expect everyone to be
wowed
by your savoir-faire at the bar when youÕre explaining it
over a beer.
Dale.
===
Subject: Re: Unstoppable Force vs Immovable Object
>What happens when an unstoppable force collides with a
immovable object?
This
>question definitely lies within the realm of the theoretical
but there is
a
>definite solution that one may establish if such an event
could happen.
Just
>use basic properties of physics! The Normal Force of the
unstoppable force
is
>equal to infinity ( F=oo ). Then you take the Strong Force
that
makes
>any object solid- the reason you do not fall to the center
of the earth-
>which is a set number and therefore less then infinity. Now
Normal Force
>exceeds Strong Force and the unstoppable force simply passes
through the
>unmovable object.
>I may be in error but I have been thinking about this and
come to this
>solution. If you have another expiation then write it, this
has been
plaguing
>my mind for awhile.
f/a = pressure = p1 and p2
When considering this situation you have to consider the area
that the
force
is applied which is the pressure. ( And assume they both use
the same level
of
hardness and material, for the unstoppable force and
unmovable object)
unstoppable force
|
p1 v p2
f/a--->|O|<---f/a ( The object has to have
^ the balancing f/a)
|
Immovable
Object
If the area of the immovable object comes into contact with
the same area
of
the unstoppable force and they remain in balance, then the
result is just
zero
movement. p1=p2, p1 - p2 = 0, the result is like no force
applied, and they
just sit there.
If the unstoppable force has a smaller area than the
Immovable object,
then
even though the immovable object never moves, it could be
penetrated by the
unstoppable force. If the hardness level changes in regard to
the specific
material used, this could also change things. If the
unstoppable force is
made
of a brittle applied material as compared to the unmovable
object, the
unstoppable force would fall apart, and the unmovable object
just stays in
place unchanged.
This can also be applied to self-defense logic. Even though a
person may
be
unmovable, a knife could still penetrate that person, unless
that person
uses a
shield that can counter the applied knife, like a knife proof
vest to
prevent
penetration. Once the unstoppable force of the knife reaches
the
immovable-impenetrable object, they just sit there and
nothing happens.
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: Unstoppable Force vs Immovable Object
>>What happens when an unstoppable force collides with a
immovable object?
>This
>>question definitely lies within the realm of the theoretical
but there is
a
>>definite solution that one may establish if such an event
could happen.
Just
>>use basic properties of physics! The Normal Force of the
unstoppable
force
>>equal to infinity ( F=oo ). Then you take the Strong Force
that
makes
>>any object solid- the reason you do not fall to the center
of the earth-
>>which is a set number and therefore less then infinity. Now
Normal Force
>>exceeds Strong Force and the unstoppable force simply
passes through the
>>unmovable object.
>>I may be in error but I have been thinking about this and
come to this
>>solution. If you have another expiation then write it, this
has been
>plaguing
>>my mind for awhile.
>
> f/a = pressure = p1 and p2
> When considering this situation you have to consider the
area that the
>force
>is applied which is the pressure. ( And assume they both use
the same
level
>hardness and material, for the unstoppable force and
unmovable object)
>unstoppable force
> |
> p1 v p2
>f/a--->|O|<---f/a ( The object has to have
> ^ the balancing f/a)
> |
> Immovable
> Object
> If the area of the immovable object comes into contact with
the same
area
>the unstoppable force and they remain in balance, then the
result is just
>zero
>movement. p1=p2, p1 - p2 = 0, the result is like no force
applied, and
they
>just sit there.
> If the unstoppable force has a smaller area than the
Immovable object,
then
>even though the immovable object never moves, it could be
penetrated by
the
>unstoppable force. If the hardness level changes in regard
to the specific
>material used, this could also change things. If the
unstoppable force is
>made
>of a brittle applied material as compared to the unmovable
object, the
>unstoppable force would fall apart, and the unmovable object
just stays in
>place unchanged.
> This can also be applied to self-defense logic. Even though
a person
may
>unmovable, a knife could still penetrate that person, unless
that person
uses
>shield that can counter the applied knife, like a knife
proof vest to
prevent
>penetration. Once the unstoppable force of the knife reaches
the
>immovable-impenetrable object, they just sit there and
nothing happens.
Another interesting thought about this is, you have heard it
said,
those
that live by the sword will die by the sword, but those that
live by the
shield
will continue to live by the shield. A shield doesnÕt attack
anyone or
hurt
anyone. The best offense is a good defense. A person that did
happen to
attack
by a sword would get tired trying to hurt that person, while
the other
person
with the shield just sits there and takes it easy.
And another thought ....
The best way to win a nuclear war, is to have a good
defensive shield
and
have a good high energy EM pulse directed beam shield that
would shut down
the
electronics used in the nuclear weapon, before it even had a
chance to
explode.
I would say that these timers and switches used in nuclear
bombs are
electronics and IC circuits. If a strong enough EM pulse beam
is directed
to
the missle, it should shut down, even the guidence system
wouldnÕt work.
>SmartÕs Alt. Physics News Group
>http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
>S. Enterprize (Science Journal)
>http://smart1234.s-enterprize.com/
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: Unstoppable Force vs Immovable Object
 that live by the sword will die by the sword, but those
that live by the
shield
> will continue to live by the shield. A shield doesnÕt
attack anyone or
hurt
> anyone.
Hogwash. A shield can be a useful offensive weapon. Perhaps
you should
crosspost to one of the SCA groups where IÕm sure lots of
people would
be willing to point out just how wrong you are.
The best offense is a good defense. A person that did happen
to attack
> by a sword would get tired trying to hurt that person,
while the other
person
> with the shield just sits there and takes it easy.
Hogwash^2.

Rick
===
Subject: Definition of Z
> Some people want to ask: Why is the definition of Z 
fixed?
What makes
> it fixed? What is it fixed *relative* to? What 
does this
definition
> *necessarily* depend on?
> These are philosophical questions, something I believe you
are not
> good at.
> This is just silliness, not philosophical questions. Had you
> understood the usual elementary definition, you could have
explained
> to your misguided friend why the representation of the
integers in
> base n, for any n, is finite.
Fine. You think these are not philosophical questions, but
silliness.
ThatÕs okay. My friend was misguided, and I 
didnÕt learn
anything
about the elementary definitions in the secondary school about
sets
and integers, or later in my life. ThatÕs quite nice. I
wonder then,
perhaps it is not right to question these things? If
mathematics is so
well defined, and so beyond doubt, why is there a need for
philosophy
of mathematics at all? Apparently none. Not any more. We only
need
history of philosophy of mathematics, you know all the silly
questions
like RussellÕs paradox, until it was finally 
settled forever
by modern
set theory. ItÕs just silliness, all of it. Since 
foundations
of
mathematics is rock solid today, then I think all those
philosophers
and mathematicians who work on this silly stuff can pack up
and go on
a really long vacation.
--
Eray Ozkural
===
Subject: Re: Definition of Z
> why is there a need for philosophy
> of mathematics at all? Apparently none. Not any more.
Good, now that youÕve finally grasped that, 
perhaps you can
pack
your bags and go home.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Definition of Z
> If mathematics is so
> well defined, and so beyond doubt, why is there a need for
philosophy
> of mathematics at all? Apparently none.
Your comments are based on a misapprehension, not as to the
significance of the philosophy of mathematics, but regarding
the quality
of your own contributions.
===
Subject: Nonlocally compact groups, representations,
applications
What are some examples of nonlocally compact groups whose
representations
are important for either physics or other areas?
Still wondering about harmonic analysis on these guys. In
compact groups,
representations are very useful because of Peter-Weyl. But
could it be the
case that we donÕt need to think about representations of
non-compact groups
when we try to do harmonic analysis on non-compact groups?
Could it be the
case that something totally different than representations is
needed to
study this?
Isaac
===
Subject: Finding unique sums. Need Help
I was reading from the initial posting and was facing the same
problem
I am not a maths geeks but would love to have a solution from
the
Maths Gurus in this forum.
Its very convincing that 2^n series is the smallest series
when
dealing with non negative numbers...
I am repeating a previous question that was posted
For example, if the sum of random 10 elements of a
series is given,
is it possible to get the individual elements.
Example such a series of positive integers 2^n growing
exponentially.
(1+2+4+8+16+32+64+ 128+256+508....etc)
Here is the actual problem
I have customers punching in numbers corresponding to the
selection
that they make based on 2^n series..Below is the example
1-Beer
2-soda
4-pizza
8-coke
16-Mushroom(topping)
32-Pepers(topping)
64-Onions(topping)
128-chicken(topping)
When the customer orders for example Beer(1)
+Pizza(4)+Mushroom(16)
I sum the value which come to 21(1+4+16)
Now given the sum how would I be able to find the individual
elements
to fulfill the order....??
I hope I was able to clearly explain the scenario....
I would greatly apprecaite your contribution to this problem
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: Finding unique sums. Need Help
| jack asked:
| I was reading from the initial posting and was facing the
same
| problem
| I am not a maths geeks but would love to have a solution
from the
| Maths Gurus in this forum.
|
| Its very convincing that 2^n series is the smallest series
when
| dealing with non negative numbers...
| I am repeating a previous question that was posted
| For example, if the sum of random 10 elements of a
| series is given,
| is it possible to get the individual elements.
| Example such a series of positive integers 2^n growing
| exponentially.
| (1+2+4+8+16+32+64+ 128+256+508....etc)
|
| Here is the actual problem
| I have customers punching in numbers corresponding to the
selection
| that they make based on 2^n series..Below is the example
|
| 1-Beer
| 2-soda
| 4-pizza
| 8-coke
| 16-Mushroom(topping)
| 32-Pepers(topping)
| 64-Onions(topping)
| 128-chicken(topping)
|
| When the customer orders for example Beer(1)
+Pizza(4)+Mushroom(16)
|
| I sum the value which come to 21(1+4+16)
|
| Now given the sum how would I be able to find the individual
elements
| to fulfill the order....??
|
| I hope I was able to clearly explain the scenario....
|
| I would greatly apprecaite your contribution to this problem
Express the 21 in binary (base 2): 10101 ---- each 1 would
reßect what is ordered, with the right-most 1 being beer, the
next 1 digit would be soda (which wasnÕt ordered, so the 2nd
right-
most digit was a 0), etc...
____________________________Gerard S.
===
Subject: How To Win A First Strike Nuclear War
The best way to win a nuclear war, is to have a good
defensive shield
and
have a good high energy EM pulse directed beam shield, that
would shut down
the electronics used in the nuclear weapon circuitry, before
it even has a
chance to explode. I would say that these timers and switches
used in
nuclear
bombs ( or nuclear missles) are mostly electronics and IC
circuits. If a
strong
enough EM pulse beam is directed to the missle, it should
theoretically
shut
down, and even the guidence system wouldnÕt work.
So instead of matching nuclear bomb with nuclear bomb
deterrence strategy
as
a way to stop nuclear war, the various countries could use a
good defensive
shield strategy, which would guarantee a win and no nuclear
war conßict. I
think the Russians were working on a newer faster nuclear
missle system
that
wouldnÕt be able to be matched in regard to a deterrence
strategy, but it
still
takes more time for that multi-warhead missle to reach itÕs
target. The
Strong
EM pulse shield works just as soon as you press the button.
So a good
defense
is much better than a good offense.
Then you may say what about conventional warfare. Well the
same
strategy
could be used. Just have a good shield with high EM pulse
shield that would
knock out the electronics. But what about if they try to
knock the EM pulse
beam shield? Well just place it deep underground where it
could be EM
pulse
protected.
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: How To Win A First Strike Nuclear War
> The best way to win a nuclear war, is to have a good
defensive shield
and
> have a good high energy EM pulse directed beam shield, that
would shut
down
> the electronics used in the nuclear weapon circuitry,
before it even has
a
> chance to explode. I would say that these timers and
switches used in
nuclear
> bombs ( or nuclear missles) are mostly electronics and IC
circuits. If a
strong
> enough EM pulse beam is directed to the missle, it should
theoretically
shut
> down, and even the guidence system wouldnÕt work.
> So instead of matching nuclear bomb with nuclear bomb
deterrence
strategy as
> a way to stop nuclear war, the various countries could use
a good
defensive
> shield strategy, which would guarantee a win and no nuclear
war conßict.
I
> think the Russians were working on a newer faster nuclear
missle system
that
> wouldnÕt be able to be matched in regard to a deterrence
strategy, but it
still
> takes more time for that multi-warhead missle to reach 
itÕs
target. The
Strong
> EM pulse shield works just as soon as you press the button.
So a good
defense
> is much better than a good offense.
> Then you may say what about conventional warfare. Well the
same
strategy
> could be used. Just have a good shield with high EM pulse
shield that
would
> knock out the electronics. But what about if they try to
knock the EM
pulse
> beam shield? Well just place it deep underground where it
could be EM
pulse
> protected.
Sounds as a stupid idea to me. In order to destroy
*non-shielded*
electronics on a distance of over
100 miles, youÕll need an EMP in the order of many GigaWatt 
I
guess. Many
radarsystems already use
pulses in the order of several MegaWatt (momentary power
within the pulse),
and if you sit on a
distance of 100 meters with your 2 dollar FM-radio, you may
not even notice
that the radar is
functioning! So youÕll need a nuclear explosion to create
your EMP....
If the electronics is shielded adequatly, then it will never
work for
obvious reasons. If it does
work (just hypothetically), then electronics may be shut down
but mechanical
navigation systems can
hold the missile on track quit nicely and we only have time
left to think
why doesnÕt the missile
stop before we all return to ashes.
The best way to win is not to strike at all!
Jeroen
===
Subject: Re: How To Win A First Strike Nuclear War
>> The best way to win a nuclear war, is to have a good
defensive shield
>and
>> have a good high energy EM pulse directed beam shield,
that would shut
down
>> the electronics used in the nuclear weapon circuitry,
before it even has
a
>> chance to explode. I would say that these timers and
switches used in
>nuclear
>> bombs ( or nuclear missles) are mostly electronics and IC
circuits. If a
>strong
>> enough EM pulse beam is directed to the missle, it should
theoretically
>shut
>> down, and even the guidence system wouldnÕt work.
>> So instead of matching nuclear bomb with nuclear bomb
deterrence
strategy
>> a way to stop nuclear war, the various countries could use
a good
defensive
>> shield strategy, which would guarantee a win and no
nuclear war conßict.
I
>> think the Russians were working on a newer faster nuclear
missle system
>that
>> wouldnÕt be able to be matched in regard to a deterrence
strategy, but
it
>still
>> takes more time for that multi-warhead missle to reach
itÕs target. The
>Strong
>> EM pulse shield works just as soon as you press the
button. So a good
>defense
>> is much better than a good offense.
>> Then you may say what about conventional warfare. Well the
same
>strategy
>> could be used. Just have a good shield with high EM pulse
shield that
would
>> knock out the electronics. But what about if they try to
knock the EM
pulse
>> beam shield? Well just place it deep underground where it
could be EM
>pulse
>> protected.
>Sounds as a stupid idea to me. In order
ThatÕs because you donÕt think much.
You can make a 3D hologram laser web high energy EM field with
many high
energy lasers. The sky would look like laser light net.
Anything passing
through it would have itÕs electronic circuitry burned up.
It would be like placing a IC chip in the microwave oven for
about 2
seconds. The internal making of the chip is burned up,
without using that
much
energy to totally
destroy the IC chip. You use only the enegy needed to shut
down the missle
circuitry. You may say that the electronic shielding would
protect it, but
a
microwave oven heats things up on the inside too, not just
the outside.
Example of a High Energy Field EM Pulse Laser Web
This field could be spread out for hundreds of miles and be an
effective
shield. This field could also be made of gamma ray frequency
range and
gamma
OOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOO
OOOOOOOOOOOOOOOO
So as the missle passes through this, it is shut down before
reaching
itÕs
target. There isnÕt a need to aim at each particular 
missile,
thousands of
missiles could be shut down all at the same time as they
reach the shield.
to destroy *non-shielded* electronics
>on a distance of over
>100 miles, youÕll need an EMP in the order of many GigaWatt
I guess. Many
>radarsystems already use
>pulses in the order of several MegaWatt (momentary power
within the
pulse),
>and if you sit on a
>distance of 100 meters with your 2 dollar FM-radio, you may
not even
notice
>that the radar is
>functioning! So youÕll need a nuclear explosion to create
your EMP....
>If the electronics is shielded adequatly, then it will never
work for
obvious
>reasons. If it does
>work (just hypothetically), then electronics may be shut
down but
mechanical
>navigation systems can
>hold the missile on track quit nicely and we only have time
left to think
>why doesnÕt the missile
>stop before we all return to ashes.
>The best way to win is not to strike at all!
> Jeroen
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: How To Win A First Strike Nuclear War
>> The best way to win a nuclear war, is to have a good
defensive
shield
>and
>> have a good high energy EM pulse directed beam shield,
that would shut
down
>> the electronics used in the nuclear weapon circuitry,
before it even
has a
>> chance to explode. I would say that these timers and
switches used in
>nuclear
>> bombs ( or nuclear missles) are mostly electronics and IC
circuits. If
a
>strong
>> enough EM pulse beam is directed to the missle, it should
theoretically
>shut
>> down, and even the guidence system wouldnÕt work.
>> So instead of matching nuclear bomb with nuclear bomb
deterrence
strategy
>as
>> a way to stop nuclear war, the various countries could use
a good
defensive
>> shield strategy, which would guarantee a win and no
nuclear war
conßict. I
>> think the Russians were working on a newer faster nuclear
missle
system
>that
>> wouldnÕt be able to be matched in regard to a deterrence
strategy, but
it
>still
>> takes more time for that multi-warhead missle to reach
itÕs target.
The
>Strong
>> EM pulse shield works just as soon as you press the
button. So a good
>defense
>> is much better than a good offense.
>> Then you may say what about conventional warfare. Well the
same
>strategy
>> could be used. Just have a good shield with high EM pulse
shield that
would
>> knock out the electronics. But what about if they try to
knock the EM
pulse
>> beam shield? Well just place it deep underground where it
could be EM
>pulse
>> protected.
>Sounds as a stupid idea to me. In order
> ThatÕs because you donÕt think much.
> You can make a 3D hologram laser web high energy EM field
with many
high
> energy lasers. The sky would look like laser light net.
Anything passing
> through it would have itÕs electronic circuitry burned up.
Not easy, almost impossible when a lot of missiles attack at
the same time.
With lasers you look for
complete destruction of the missile (burning a hole in it),
not Ōfryingits
electronics. ThatÕs
entirely different. The US already does research on this, and
it was shown
that it is exceptionally
difficult. Even if the shield 
Ōknowsthe exact position of
the missile,
then it was difficult to
destroy the missile.
> It would be like placing a IC chip in the microwave oven
for about 2
> seconds. The internal making of the chip is burned up,
without using that
much
> energy to totally
> destroy the IC chip. You use only the enegy needed to shut
down the
missle
> circuitry. You may say that the electronic shielding would
protect it, but
a
> microwave oven heats things up on the inside too, not just
the outside.
Apparantly weÕre back to Ōnormal
EM-fields (optics/laser is
also EM). Ever
heard of the Faraday
cage or the (true!) assumption that youÕre safe within a car
when lightning
strikes? Shielding
against EM-fields is very simple, but you seem to completely
deny these
facts of physics. Your
microwave oven story is a laugh. Do you really want to
torture your
microwave in a home-test? Then
put a piece of cake in a metal box and put it in your
microwave. The metal
may get heated, but the
cake doesnÕt. Shielding!
> Example of a High Energy Field EM Pulse Laser Web
> This field could be spread out for hundreds of miles and be
an
effective
> shield. This field could also be made of gamma ray frequency
range and
gamma
> OOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOO
> OOOOOOOOOOOOOOOO
high energy on large distances? Energy enough to shut
electronics down?
> So as the missle passes through this, it is shut down
before reaching
itÕs
> target. There isnÕt a need to aim at each particular
missile, thousands
of
> missiles could be shut down all at the same time as they
reach the
shield.
And were do the missiles fall down? And yes, youÕll have to
aim for each
missile. If you want a grid
fine enough to catch evere missile, youÕll need 
a grid of
several inches.
Now take the total surface
area of the shield and calculate how many focussed beams
youÕll have to
make. Can you imagine that
mankind does not have the ability to generate the power
needed for such a
shield? YouÕll just have
to aim for each missile.
> to destroy *non-shielded* electronics
>on a distance of over
>100 miles, youÕll need an EMP in the order of many GigaWatt
I guess.
Many
>radarsystems already use
>pulses in the order of several MegaWatt (momentary power
within the
pulse),
>and if you sit on a
>distance of 100 meters with your 2 dollar FM-radio, you may
not even
notice
>that the radar is
>functioning! So youÕll need a nuclear explosion to create
your EMP....
>If the electronics is shielded adequatly, then it will never
work for
obvious
>reasons. If it does
>work (just hypothetically), then electronics may be shut
down but
mechanical
>navigation systems can
>hold the missile on track quit nicely and we only have time
left to
think
>why doesnÕt the missile
>stop before we all return to ashes.
>The best way to win is not to strike at all!
> Jeroen
> SmartÕs Alt. Physics News Group
>
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
> S. Enterprize (Science Journal)
> http://smart1234.s-enterprize.com/
===
Subject: Re: How To Win A First Strike Nuclear War
> So instead of matching nuclear bomb with nuclear bomb
deterrence
> strategy as
> a way to stop nuclear war, the various countries could use
a good
> defensive shield strategy, which would guarantee a win and
no nuclear
> war conßict. I think the Russians were working on a newer
faster
> nuclear missle system that wouldnÕt be able to be matched
in regard to
> a deterrence strategy, but it still takes more time for that
> multi-warhead missle to reach itÕs target. The Strong EM
pulse shield
> works just as soon as you press the button. So a good
defense is much
> better than a good offense.
Now just convince the rest of the world you are right (which
you have a
good track record of doing). However, I am willing to bet
that once the
first nuclear missle is launched, most people 
wonÕt be around
to see how
the war turns out.
> SmartÕs Alt. Physics News Group
>
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
> S. Enterprize (Science Journal)
> http://smart1234.s-enterprize.com/
- Tim
--
Timothy M. Brauch
NSF Fellow
Department of Mathematics
University of Louisville
email is:
news (dot) post (at) tbrauch (dot) com
===
Subject: Re: How To Win A First Strike Nuclear War
> So instead of matching nuclear bomb with nuclear bomb
deterrence
> strategy as
> a way to stop nuclear war, the various countries could use
a good
> defensive shield strategy, which would guarantee a win and
no nuclear
> war conßict. I think the Russians were working on a newer
faster
> nuclear missle system that wouldnÕt be able to be matched
in regard to
> a deterrence strategy, but it still takes more time for that
> multi-warhead missle to reach itÕs target. The Strong EM
pulse shield
> works just as soon as you press the button. So a good
defense is much
> better than a good offense.
Now just convince the rest of the world you are right (which
you have a
> good track record of doing). However, I am willing to bet
that once the
> first nuclear missle is launched, most people 
wonÕt be
around to see how
> the war turns out.
Only the inhuman that is impervious to radiation could win
the nuclear
war. There is no winning for us in nuclear war.
--
Eray Ozkural
===
Subject: Re: KnuthÕs division and multiplication algorithms
> I need KnuthÕs division algorithm 4.3.1 and multiplication
algorithm 4.3.3
from Volume 2.
> The problem is that I donÕt have that Volume.
> Are there any links with those algorithms?
Unknown. Do you have a local technical library?
--
Randy Howard (2reply remove FOOBAR)
For some reason most people seem to be born without the part
of the brain that understands pointers. -- Joel Spolsky
===
Subject: Re: KnuthÕs division and multiplication algorithms
> I need KnuthÕs division algorithm 4.3.1 and multiplication
algorithm
4.3.3
> from Volume 2.
> The problem is that I donÕt have that Volume.
Are there any links with those algorithms?
> Unknown. Do you have a local technical library?
I am sure www.amazon.com or www.amazon.co.uk will get you a
copy for a
very reasonable price.
===
Subject: Re: November 25 is Infinite Clause day!!
> Where is THE LIST?
> Just donÕt ask him to send it as an attachment to an
e-mail; it will
> take forever to open.
ItÕs worse than that! ItÕs infected with a 
virus that makes
you unable
to understand simple logic, such as proof by contradiction,
and that
virus infects you after only about twenty terms of THE LIST
have been
read into memory, so even if you abort as soon as you realize
itÕll
take forever to open, itÕs already too late.
(P.S. This is all a joke, except for that last Never...
line!!)
===
Subject: Re: November 25 is Infinite Clause day!!

<9kq772-t9i.ln1@sirius.athghost7038suus.net>

<96c872-j7j.ln1@sirius.athghost7038suus.net>
Discussion, linux)
> DonÕt tell Xanthian... getting a group consensus is
unmathematical, we
> have to duke it out!
> Barb Knox, George Green, Jesse James, John Savard, Dave B,
you,
> Ullrich, Will and dozens of others have all been vocally
opposed to me
> in their defence of Cantors diag proof. Now they are all
silent? is
> it apathy?
In my case, yes. I havenÕt read your posts in a few days. I
caught
GeorgeÕs followup today and hence this response.
What did you think? That your brilliant exposition was
obviously
correct and I was too embarrassed to admit it? YouÕre so 
cute.
> This is the bait :
>> An infinite number of people toss a coin 
infinite times
each. Can you
>> guarantee a new sequence of Heads and Tails?
> Herc : no, thatÕs silly run some simulations and
extrapolate to
> infinity.
> Ghost : take the diag as in CantorÕs uncountabble real
proof.
> Jesse : hope they all toss heads.
No. I said that there is nothing in the statement of your
problem
which is inconsistent with the outcome that they all toss
heads.
Unless, of course, we interpret the probabilities involved in
the
frequentist sense, in which case each of the individual
sequences
should satisfy that
lim_{n->oo} (Number of H in initial n tosses/n) = 1/2
In that case, *none* of them could toss all heads, and
thereÕs your
new sequence.
What the hell do you mean can we guarantee a new sequence
anyway?
(1) Is there a sequence of H and T not on the list? Yes,
obviously.
Use CantorÕs diagonal argument. None of these arguments 
based
on
probability are really needed, but your question is so
stupidly posed
that I chose to use them instead of the standard Cantor
argument.
(2) Is it necessary that, when we toss the coin, the result
will be a
sequence not on the list? Obviously not, but the probability
that the
result is not on the list appears to be one, near as I can
figger.
That the probability is one does not make the outcome
*necessary*
however.
Running simulations and extrapolating to infinity is a
remarkable
new method of proof. What a bold reinvention of mathematics
we have
here. I canÕt wait for the coming revolution.
In any case, by calling me Jesse James, you have shown a
subtle wit
of the sort I havenÕt seen since, oh, fourth grade or so.
--
What you want with a hen? What you want with a woman
WonÕt cackle when she lays when she wonÕt do 
nothinI say?
What you want with a hen? -- Charlie Patton,
WonÕt cackle when she lays Banty Rooster Blues
===
Subject: Re: November 25 is Infinite Clause day!!


<9kq772-t9i.ln1@sirius.athghost7038suus.net>

<96c872-j7j.ln1@sirius.athghost7038suus.net>
<87653og9oe.fsf@phiwumbda.org>
posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L
huh! Go figger!
Herc
===
Subject: Spline
I have a set of 3D control points and want
to generate a spline (cubic possibly) passing
_through_ them.
What makes it challenging is that I donÕt know
the control point positions (other than for the
first point) but instead their tangential angles
(v.s. z- and x-axis respectively) and each points
distance (along the curve) from the previous one.
Any clues or pointers?
===
Subject: Embedding theorem and uniqueness?
Hi-
According to the embedding theorem, any manifold can be
represented as a surface in a higher dimensional space (ie-
the
2-manifold with constant curvature a^2 can be represented as
the
sphere of radius a in three dimensional space). My question
is- is
the surface unique (up to trivial transformations such as
translation,
rotation, and curling up in a yet higher dimensional space
such as
when one rolls up a plane to a cylinder). For instance, is
there
another surface in three dimensional space that has constant
curvature
a^2? Are there other solutions which have constant curvature
but
eventually have a singularity at global locations?
-I
===
Subject: Re: Embedding theorem and uniqueness?
> According to the embedding theorem, any manifold can be
> represented as a surface in a higher dimensional space (ie-
the
> 2-manifold with constant curvature a^2 can be represented
as the
> sphere of radius a in three dimensional space).
Which embedding theorem? -- there are several. Note I am not
expert in
this area, but have some knowledge.
This also depends on what type of embedding do you mean:
topological or
isometric? A topological emebdding merely maps the topology
of the
original manifold into that of the larger manifold. An
isometric
embedding does that and also requires that the metric of the
original
manifold be the projection of the larger manifoldÕs metric
onto the
embedded image.
For any embedding to be possible, the original manifold must
have a
topology compatible with the topology of the region of the
larger
manifold occupied by the image. And the dimension of the
larger manifold
must be greater or equal to the dimension of the original
manifold.
For isometric embeddings the requirements are far more
strict....
For example, for an arbitrary 3+1-dimension Lorentzian
manifold to be isometrically embedded into an N+M-dimension
ßat manifold, the lowest known dimension is N=88,M=2 (!).
> My question is- is
> the surface unique (up to trivial transformations such as
translation,
> rotation, and curling up in a yet higher dimensional space
such as
> when one rolls up a plane to a cylinder).
[rolling up a plane into a cylinder is a major topological
change, and should not be included in your list -- that is
not trivial at all.]
This depends on the type of embedding (see above), on the
topologies of
the manifolds, and for isometric embeddings also on their
metrics.
For instance: consider embedding S^2 into E^3-(one point).
There are two
distinct ways to do it, one encloses the deleted point and
the other
does not. With suitable metrics either could be isometric.
Of course any two isometric embeddings will be the same in
the sense
of there being an isometry between them. And any two
topological
embeddings will be the same in the sense of there being a
topological
mapping between them. By them I mean a region of the larger
manifold
enclosing the embedded image. So your question about
uniqueness needs to
be more precise about what unique means -- specifically do you
really
mean unique up to isometry/topological-mapping?. I think that
is a
more precise statement of your up to trivial
transformations... (for
instance, in E^3 both rotations and translations are
isometries of the
space to itself). If so, then clearly any such embedding is
unique.
But note this also implies that embedding S^2 into E^3 and
into E^100
are the same....
But look back 2 paragraphs -- those two embeddings of S^2 are
quite
clearly different, and yet the previous paragraph showed the
embedding
is unique up to isometry/topological-mapping. So it should be
clear that
this entire discussion is not rigorous enough.... Here there
be dragons.
This is directly related to local vs global issues....
> For instance, is there
> another surface in three dimensional space that has
constant curvature
> a^2?
As you mention curvature, IÕll assume you mean an isometric
embedding.
If you mean embedding into ßat space E^3, then IÕm 
pretty
sure any pair
of such embeddings will be congruent, and it is therefore
unique in your
sense. Buf if, for instance, one embeds into E^3-(one point),
then there
are two different mappings; etc.
> Are there other solutions which have constant curvature but
> eventually have a singularity at global locations?
If you mean embedding into ßat space E^3, then I donÕt 
think
so. But I
am not knowledgeable enough to be sure.
Tom Roberts tjroberts@lucent.com
===
Subject: Re: Embedding theorem and uniqueness?
Hi-
questions in the response.
> Which embedding theorem? -- there are several. Note I am
not expert in
> this area, but have some knowledge.
> This also depends on what type of embedding do you mean:
topological or
> isometric?
Sorry. I definitely meant isometric.
> For isometric embeddings the requirements are far more
strict....
> For example, for an arbitrary 3+1-dimension Lorentzian
> manifold to be isometrically embedded into an N+M-dimension
> ßat manifold, the lowest known dimension is N=88,M=2 (!).
That is terrible. Is this lower bound considered
conservative, or are there particular examples that are known
to need
this?
> My question is- is
> the surface unique (up to trivial transformations such as
translation,
> rotation, and curling up in a yet higher dimensional space
such as
> when one rolls up a plane to a cylinder).
> [rolling up a plane into a cylinder is a major topological
> change, and should not be included in your list -- that is
> not trivial at all.]
Good point. I was just trying to point out that if you are
given a surface with a given curvature (like a plane with
curvature=0
everywhere), you can always increase the number of dimensions
and roll
up the manifold while presenving the curvature- a scroll
would have
been a better visual, or perhapse something simpler like a
speed bump
extending to infinity. Another example would be that of one
dimensional manifolds. They always have zero curvature
everywhere,
regardless of the shape of the curve. This is to say that I
know of
examples of different surfaces with the same curvature, but
suspect
that if you ignore these trivial examples I suspect the
result is
unique.
> This depends on the type of embedding (see above), on the
topologies of
> the manifolds, and for isometric embeddings also on their
metrics.
> For instance: consider embedding S^2 into E^3-(one point).
There are two
> distinct ways to do it, one encloses the deleted point and
the other
> does not. With suitable metrics either could be isometric.
> Of course any two isometric embeddings will be the same in
the sense
> of there being an isometry between them. And any two
topological
> embeddings will be the same in the sense of there being a
topological
> mapping between them. By them I mean a region of the larger
manifold
> enclosing the embedded image. So your question about
uniqueness needs to
> be more precise about what unique means -- specifically do
you really
> mean unique up to isometry/topological-mapping?. I think
that is a
> more precise statement of your up to trivial
transformations... (for
> instance, in E^3 both rotations and translations are
isometries of the
> space to itself). If so, then clearly any such embedding is
unique.
> But note this also implies that embedding S^2 into E^3 and
into E^100
> are the same....
Part of the problem is that I am still trying to determine the
more rigerous definition. Basically I am trying to build an
intuitive
view of GR, by picturing metrics as surfaces in higher
dimensional
space. For instance I know that the robertson walker metric
at a
particular time slice has constant curvature everywhere. I can
picture this as a sphere in a n+1 dimensional space. I like
this
better than the metric description because an n+1 dimensional
space
has many different metric descriptions which differ by choice
of
coordinates. I think I might be able to get around this ugly
guage
invariance by reformulating for my own purposes GR in the
higher
dimensional space, but if in fact the set of curves with
constant
curvature everywhere is not unique, then a new guage
invariance makes
this idea less attractive.
-I
===
Subject: Re: Embedding theorem and uniqueness?