> |> but it > |>takes a bit of work to prove the equivalence (the first proof is > |>apparently due to Gauss, around 1800). > | > |That sounds like the proof of the Fundamental Theorem of Arithmetic, > |rather than an explicit proof that all irreducibles have the prime > |divisor property (the prime divisor property, by the way, is that if p > |divides a*b then it divides a or it divides b, and that goes back to > |Euclid!). > Section II. 14. If neither a nor b can be divided by a prime number p, > the product ab cannot be divided by p. I don't see a definition of prime number given, but apparently he's > using the definition as irreducible. He claims Euclid had already proved this theorem in his Elements. > The citation is to Book VII, No. 32, but that doesn't seem right. That > proposition reads Any number either is prime or is measured by a > prime number. I assume the citation has a typo, but I don't see > where the actual proposition is. [ ... ] KRamsay is right, it is a typo (in the English translation, anyway). It should be Euclid VII.30: http://aleph0.clarku.edu/~djoyce/java/elements/bookVII/propVII30.html And Euclid's proof is actually simpler than the one in Gauss. So Arturo Magidin is right that the proof goes back to Euclid. ==== > >> Section II. 14. If neither a nor b can be divided by a prime number p, >> the product ab cannot be divided by p. >> >> I don't see a definition of prime number given, but apparently he's >> using the definition as irreducible. >> >> He claims Euclid had already proved this theorem in his Elements. >> The citation is to Book VII, No. 32, but that doesn't seem right. That >> proposition reads Any number either is prime or is measured by a >> prime number. I assume the citation has a typo, but I don't see >> where the actual proposition is. >> >> [ ... ] KRamsay is right, it is a typo (in the English translation, anyway). It >should be Euclid VII.30: >http://aleph0.clarku.edu/~djoyce/java/elements/bookVII/propVII30.html And Euclid's proof is actually simpler than the one in Gauss. So Arturo Magidin is right that the proof goes back to Euclid. The proof that irreducibles in the integers are primes, yes. Euclid also implicitly uses Unique Factorization (the Fundamental Theorem of Arithmetic) when he discusses Pythagorean Triples in Book X. An easier way to spot his use of Unique Factorization is to look at the explanation in http://math.nmsu.edu/~history/book/euclidpt.pdf since at one point he argues that if a product of two relatively prime integers is a square, then each must be a square. This requires unique factorization. The first explicit proof of Unique Factorization is due to Gauss, however, and that requires the use of the prime divisor property. It is Art. 16 of the _Disquisitiones Arithmeticae_, and in fact Gauss starts by saying that it is clear from elementary considerations that any composite number can be [factored] as a product of primes, and then uses the prime divisor property in the usual way to show that any two factorizations must in fact be equal. The elementary considerations are that you cannot have an infinite descending sequence of positive integers, and that if a positive number is composite, you can factor it into two strictly smaller positive integers. But this argument works for composite numbers, so long as you can prove that there is no infinite descending sequence of irreducibles. It is the prime divisor property that is a (possible) problem. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >Well, in a 3-connected planar graph you can define the faces uniquely in >graph-theoretic terms as non-separating cycles, so it would be almost >reasonable (but still not right) to use the word face to refer to >non-separating cycles in arbitrary graphs. Er... what is a non-separating cycle exactly? There's a little freedom to choose definitions that work here. One is a simple cycle such that any two edges outside the cycle can be connected by a path that has no internal vertices from the cycle. Or a chordless simple cycle such that the vertices outside the cycle induce a connected subgraph... -- David Eppstein http://www.ics.uci.edu/~eppstein/ Univ. of California, Irvine, School of Information & Computer Science ==== As often happens, I have managed to confused myself. Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers and assume S is countable. Since S is countable there is a function, f(), that maps S to the natural numbers. Using f() to order S, define a diagonal number, d. d is a computable number not in S. Therefore, S can't be the set of computable numbers. Cantor's powerset argument makes things even weirder. Let S be the set of all computable numbers (countable or uncountable). Let P(S) be the powerset of S. Cantor proves |S| < |P(S)|. This means that P(S) contains members that are not in S. These members must be noncomputable numbers. In fact, nearly all members of P(S) are noncomputable. So, the set of all computable numbers can't exist if it is countable and if it is uncountable, most of its powerset is noncomputable. What am I missing? Russell - the universe is one dimensional ==== [Strangely, only the r.o.mensa crosspost was added by Herc. Why is this thread crossposted to all these groups?] This is my point, We label rationals as countable, and irrationals as not countable! We label *the set of* rational numbers countable, and we label *the set of* irrational numbers uncountable. Correct. It's also true that the set of irrational numbers is uncountable *whether we label it as such or not.* This is an important point which we should not lose sight of. :) > Rational is just counting functions over_the_division_operation! [I'm assuming the _s indicate emphasis. I'd prefer to see _this_ style of emphasis, with the bars on the ends of the emphasized section, too -- but that's purely for my own convenience.] True, as long as you define what you mean by over the division operation. It sounds like you can have a pretty good definition worked out with a bit of thought, though: the set of rationals is the set you get by attempting to close the set of nonzero integers over the division operation, if you see what I mean. > SQRT(2) and pi are equally countable as rationals, in the same list > as rationals. Yes, the set {rationals, and SQRT(2), and pi} is also countable. > A lot of irrationals are countable! A *whole* lot of irrationals are countable -- in fact, if you give me a list of irrational numbers, no matter how long it is, even an infinitely long list, I *guarantee* you the number of elements in that list will be countable! :) However, it's impossible to count *all* the irrational numbers. You can't map R onto N. The *set of* irrational numbers is uncountable, unlike the *set of* rational numbers. > All the computable ones. > Why give division special precedence when we are concerned with countability > and forget the majority of useful numbers. These two phrases don't make sense on their own. > begin 666 countables.gif Rude. I've taken the time to extract and post your image at http://www.contrib.andrew.cmu.edu/~ajo/herc_is_rude.gif Note that your diagram is perfectly accurate: the set of rationals is countable; the set of irrationals is not; and the set of computable reals is. I don't understand what you mean by the notation NC numbers in the bottom left corner, nor what you mean by labeling the two halves of the chart Todays Understanding and Tomorrows Understanding -- are you hoping the OP will understand tomorrow? :) -Arthur ==== : Let S be the set of all computable numbers : and assume S is countable. What kinds of numbers are we talking about here? Rational? Real? Complex? Supernatural? Infinitesimal? -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America ==== > : Let S be the set of all computable numbers > : and assume S is countable. What kinds of numbers are we talking about here? > Rational? Real? Complex? Supernatural? Infinitesimal? > I have made the proof a little more rigorous. First, I define how to compute a computable number. I do this with a very non-standard type of Turing machine. Let's call these Russell machines (RM). A number is computable if it can be represented by the infinite binary string produced by a RM. A RM has a single, infinitely long tape. Every position of the tape is initially set to 0. A RM starts at the leftmost end of the tape. RMs can perform two operations, labeled 0 and 1. 0 = move write head one position to the right 1 = write a 1 to current location and move one position right RMs never read from the tape because they would always read 0. RMs never halt because they have no halt states. The state of an n-state RM can be specified with (1 + Ceil( log2( n ) ) ) bits where the first bit represents the operation to perform and the next Ceil(log(2(n)) bits represent which state to switch to. An n-state RM can be fully specified by listing all states which requires n * (1 + Ceil(log2(n))) bits. Some examples of RMs and their output: States RM Output 1 0 (0)... 1 1 (1)... 2 01 11 01(1)... 2 11 00 (10).... 3 001 010 100 (001)... The output of a RM has the following form: There is a finite, possibly empty initial string that is written once, and a finite, never empty repeating string that is written over and over. A computable number can be written as the initial string followed by the repeating string in parenthesis. Any number having this form can be produced by an RM. Some examples of computable numbers: (0) (1) 11(0) 01(110) The length of the initial string plus the length of the repeating string must be less than or equal to the number of states of the RM that produced the number. Now, prove that no set can contain every RM computable number. The standard diagonal argument has a problem - it is usually impossible to determine if the diagonal number produced is computable. So, I will define another method for finding a computable number that is not in a particular set of numbers. Let S be a set and every member of S be a RM computable number. Let s be a member of S. Let x be a RM computable number not in S. Let x initally be (0). If s has the form 111...111(0) and the length of the initial segment of s is longer or equal to the length of the initial segment of x then take the initial segment of s, append a 1, and make this new string the initial segment of x. Examples: s x (0) 1(0) 1(0) 11(0) 11(0) 111(0) Prove that x differs from every member of S. x differs from every s that does not end with (0). x differs from every s with an inital segment not of the form 111.111. x differs from every s with form 111...111(0). Therefore, x differs from every member of S. Prove that x is a RM computable number. Every s is a RM computable number. The initial segment of s has finite legth. Adding one character to a finite string results in a finite string. Since x has a finite initial string and a finite repeating string x must be RM computable. I have now shown that no set can contain every RM computable number. Russell - 2 many 2 count ==== Russell Easterly says... >The output of a RM has the following form: >There is a finite, possibly empty initial string that is written once, >and a finite, never empty repeating string that is written over and over. >A computable number can be written as the initial string followed by >the repeating string in parenthesis. >Let x initally be (0). >If s has the form 111...111(0) and the length of the initial segment of s >is longer or equal to the length of the initial segment of x then take the >initial segment of s, append a 1, and make this new string the initial >segment of x. Examples: s x >(0) 1(0) >1(0) 11(0) >11(0) 111(0) Prove that x differs from every member of S. You haven't given the definition of x. You've listed three different values for x: 1(0), 11(0), 111(0). Which one is x? Perhaps you are saying that x is some kind of limit of the sequence 1(0) 11(0) 111(0) But how do you prove that the limit exists? How do you prove that the limit is one of your RM computable numbers? This is in contrast to the usual diagonalization procedure, which produces a convergent sequence of reals. -- Daryl McCullough Ithaca, NY ==== > Let x initally be (0). > If s has the form 111...111(0) and the length of the initial segment > of s is longer or equal to the length of the initial segment of x then > take the initial segment of s, append a 1, and make this new string > the initial segment of x. Examples: s x > (0) 1(0) > 1(0) 11(0) > 11(0) 111(0) Prove that x differs from every member of S. x differs from every s that does not end with (0). > x differs from every s with an inital segment not of the form 111.111. > x differs from every s with form 111...111(0). > Therefore, x differs from every member of S. > Well ... let S be the set of al RM computable numbers of the form 1*(0) (any number of sequential ones followed by a repeating of zeros). In this way, you will not be able to compute a x which differs from every possible member of S (at least not using your technique). This is the kind of problems you run into when you're using infinite lists. -- Pento De wereld was soep, en het denken meestal een vork, tot smakelijk eten leidde dat zelden. - H. Mulisch ==== Let x initally be (0). > If s has the form 111...111(0) and the length of the initial segment > of s is longer or equal to the length of the initial segment of x then > take the initial segment of s, append a 1, and make this new string > the initial segment of x. Examples: s x > (0) 1(0) > 1(0) 11(0) > 11(0) 111(0) Prove that x differs from every member of S. x differs from every s that does not end with (0). > x differs from every s with an inital segment not of the form 111.111. > x differs from every s with form 111...111(0). > Therefore, x differs from every member of S. > Well ... let S be the set of al RM computable numbers of the form 1*(0) > (any number of sequential ones followed by a repeating of zeros). In this > way, you will not be able to compute a x which differs from every possible > member of S (at least not using your technique). > Why not? x will differ from every member of S and x will be a RM computable number. Russell - 2 many 2 count ==== >> Let x initally be (0). >> If s has the form 111...111(0) and the length of the initial >> segment of s is longer or equal to the length of the initial >> segment of x then take the initial segment of s, append a 1, and >> make this new string the initial segment of x. >> Examples: >> s x >> (0) 1(0) >> 1(0) 11(0) >> 11(0) 111(0) >> Prove that x differs from every member of S. >> x differs from every s that does not end with (0). >> x differs from every s with an inital segment not of the form >> 111.111. x differs from every s with form 111...111(0). >> Therefore, x differs from every member of S. >> Well ... let S be the set of al RM computable numbers of the form >> 1*(0) (any number of sequential ones followed by a repeating of >> zeros). In this way, you will not be able to compute a x which >> differs from every possible member of S (at least not using your >> technique). >> > Why not? > x will differ from every member of S and x will be a RM computable > number. Another way to look at his : if all the numbers in a set S or of the form 111...11(0), you can define a set S' (containing natural numbers) such that an element of S which has a sequence of m 1's corresponds to an element m of S'. Your x' would then be m+1. Now if you consider the set S I used in my previous post, it would correspond tot the entire set of natural numbers. As for each m, contained in this set, m+1 is also contained, there cannot exist an x' which does not belong to this set. -- Pento De wereld was soep, en het denken meestal een vork, tot smakelijk eten leidde dat zelden. - H. Mulisch ==== > Why not? > x will differ from every member of S and x will be a RM computable > number. And what number will x be ? Any number constructed of a sequence of 1's, followed by an infinite number of 0's will be in S, so x can't be of that form. -- Pento De wereld was soep, en het denken meestal een vork, tot smakelijk eten leidde dat zelden. - H. Mulisch ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Let x initally be (0). > If s has the form 111...111(0) and the length of the initial segment of s > is longer or equal to the length of the initial segment of x then take the > initial segment of s, append a 1, and make this new string the initial > segment of x. Examples: s x > (0) 1(0)* > 1(0)* 11(0) > 11(0) 111(0) Prove that x differs from every member of S. Isn't x just equal to a different s, note *. A new number has to be a function of the list, not a single member. Herc ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Any number having this form can be produced by an RM. > Some examples of computable numbers: (0) > (1) > 11(0) > 01(110) The length of the initial string plus the length of the repeating string > must be less than or equal to the number of states of the RM that > produced the number. > I don't see this. Here is a simplified conceptual diagram of a Russell Machine. s1 s2 s6 0 s5 --------------------- 1 s3 s4 s7 s8 There is one piece of information missing, which state s8 goes to. Add to this s8 -> s7 This RM will output 0 0 1 1 0 0 1 1 1 1 1 1 1 1 notation is 0 0 1 1 0 0 (1 1) If it can't halt then the states get reused. I would suggest : length(initial segment) + length(repeating segment) < 2 * number of states I think a RM is a generator of rational numbers, it can't do computation, but might outline a format for Cantor applied to computer generated lists. What's striking is that *any* list at any time is finite, it can grow but how do you assume you are manipulating an infinite list, absurd. Herc ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Any number having this form can be produced by an RM. > Some examples of computable numbers: (0) > (1) > 11(0) > 01(110) The length of the initial string plus the length of the repeating string > must be less than or equal to the number of states of the RM that > produced the number. > I don't see this. Here is a simplified conceptual diagram of a Russell Machine. s1 s2 s6 > 0 s5 > --------------------- > 1 s3 > s4 s7 > s8 There is one piece of information missing, which state s8 goes to. > Add to this s8 -> s7 This RM will output 0 0 1 1 0 0 1 1 1 1 1 1 1 1 > notation is 0 0 1 1 0 0 (1 1) If it can't halt then the states get reused. I assume that s1->s2->s3->s4->s5->s6->s7->s8->s8. The ouput is 001100(1). The initial seqment is length 6 and the repeating segment is length 1. 6+1 <= 8. I should point out that the notation I use is not unique for a given string. For example, 0(10) = {01). The minimal representation is the one with the shortest initial segment. > I would suggest : > length(initial segment) + length(repeating segment) < 2 * number of states I think a RM is a generator of rational numbers, it can't do computation, > but might outline a format for Cantor applied to computer generated lists. I wanted a simple system that could generate unambiguous binary strings. I also didn't want to deal with whether the machine halted or not. I never expected RMs could do mathematics. > What's striking is that *any* list at any time is finite, it can grow but how > do you assume you are manipulating an infinite list, absurd. I wanted a system that created infinitely long strings. Time is not a factor. Russell - 2 many 2 count ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > : Let S be the set of all computable numbers > : and assume S is countable. What kinds of numbers are we talking about here? > Rational? Real? Complex? Supernatural? Infinitesimal? > Computables are a subset of real, he means computable real numbers. Herc ==== >Cantor's powerset argument makes things even weirder. Is it the power set argument or you? >Let S be the set of all computable numbers (countable or uncountable). >Let P(S) be the powerset of S. Cantor proves |S| < |P(S)|. Do you know what the power set *is*? These members must be noncomputable numbers. Set B must be some set of Chinese people. By your logic, one or more Presidents of the United States were in fact Chinese. >In fact, nearly all members of P(S) are noncomputable. In fact, nearly all Presidents of the United States have been Chinese. >What am I missing? The point? ==== >Cantor's powerset argument makes things even weirder. Is it the power set argument or you? > [...] >What am I missing? The point? Some of the comments I snipped indicate that you are in fact, honestly trying to help the OP, but the above quoted remarks are harsh. Is there any need to get your 'digs' in? And no, I'm not trying to be the etiquette police for sci.math; I know that's hopeless -- who polices the police anyway? ;-) ==== > As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. > Therefore, S can't be the set of computable numbers. > The problem is that there is no function f, as you describe, that is computable. Why? Well, you just provided the proof. In fact, you have provided a rather roundabout way of proving Turing's halting theorem (because there is an 'obvious' way to try to make a computable f - e.g. interprete the number as a computer program in ASCII code, and run that program). ==== > As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. > Therefore, S can't be the set of computable numbers. Well, perhaps you have confused scoping issues. In classical logic, identity is closely bound with universal quantification. This is particularly evident in the forcing methods of Cohen. The truth-functionality of ultrafilters is bound with Stone-Cech compactification. For a start, consider the problem of a least upper bound, E. Specker constructed an example of an increasing bounded algorithmic sequence of rational numbers which does not have a constructive limit; no constructive real number is the least upper bound of this sequence. This result is strengthened by I. D. Zaslavskir, who constructed an algorithmic sequence of rational numbers for which not only does there not exist a constructive least upper bound but for which there is also an algorithm which, for every constructive real number that is an upper bound of this sequence, permits us to find a smaller number which is also an upper bound of this sequence. --G. S. Ceitin The abstract for Surygin's Constructive sets with equality and their mappings reads: A constructive set of constructive objects, considered together with a constructively characterized relation of A classification of e-sets by well-defined criteria is introduced and its properties are studied. Some essential differences are discussed between the theory of e-sets and the theory of factor sets in classical mathematics. Constructive mappings from certain e-sets into others are studied and a number of phenomena are established which distinguish the theory of such mappings in an essentail way from the theory of mappings for factor sets in classical mathematics. I submit that Surygin's research on e-sets might have some answers for you. :-) mitch ==== I submit that Surygin's research on e-sets might have some answers for > you. :-) mitch > Now, I'm really scared. :-) Russell - Zeno was right. Motion is impossible. ==== I submit that Surygin's research on e-sets might have some answers for > you. :-) mitch > Now, I'm really scared. > :-) > There are some good reasons for restricting yourself to constructive mathematics. I have spent the last year being ridiculed on sci.logic for things I should not have been (hence the change in group lists on this thread). The theory I offered over there captured an intuitionist concept referred to as apartness. However, since I was unaware of that connection at the time,... well there is no need to repeat myself. Looking at Blackburn's Guide to Philosophical Logic I have discovered that this notion of apartness is fundamental to the super-semantics (sheaf semantics) of intuitionist logic. In particular, it seems to have some relation to measure theory in that apartness is needed to distinguish elements from zero for the intuitionistic theory of reals. There is work on constructive measures on constructive reals and work on solvable Boolean algebras. Moreover, when I turn to Blackburn's chapter on negation, it seems that the strongest definite concept of negation is associated with constructive concepts. So, when the discussion of sheaf semantics ends with an observation that most of the semantics is adequately captured by Hyland's description of an effective topos, I get the impression that there has been a great deal of good mathematics being ignored by curriculum committees. But, of course, we have to work with that of which we have knowledge. I wish I could direct you to specific papers, but as I hope you have discerned, I am only beginning to realize how scared I am myself. :-) I see you are getting many suggestions from others that should be helpful for avoiding confusion with your specific question. That is a good thing. :-) mitch ==== > As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. > > I have read that the set of computable numbers is countable. > > Let S be the set of all computable numbers > and assume S is countable. > > Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. You are using that 'd' is computable, but for 'd' to be computable, 'f' should be computable. But 'f' is not computable, so 'd' is not computable. Losely said, a number is computable if there is a finite algorithm that delivers that number. The number of finite algorithms is countable, so the number of computable numbers is countable. But you can not say for all finite algorithms in finite time whether they deliver a number or do not stop. So there always remain algorithms for which you do not know whether they must be in the list or not. The conclusion is that you can not compute the list. So, using your 'f' (which is not computable) you get a 'd' (which is also not computable). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >As often happens, I have managed to confused myself. >Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers >and assume S is countable. Since S is countable there is a function, f(), >that maps S to the natural numbers. >Using f() to order S, define a diagonal number, d. I don't really see how to do this. Do you know that every computable number has an infinite number of digits? If not, how do you know that f^{-1}(j) (i.e., the j-th computable number) has at least j digits, so that you can define a new number whose j-th digit is different from the j-th digit of f^{-1}(j)? >d is a computable number not in S. >Therefore, S can't be the set of computable numbers. Cantor's powerset argument makes things even weirder. Let S be the set of all computable numbers (countable or uncountable). >Let P(S) be the powerset of S. Cantor proves |S| < |P(S)|. >This means that P(S) contains members that are not in S. >These members must be noncomputable numbers. Actually, ALL members of P(S) are not computable numbers, since the elements of P(S) are not numbers at all: they are sets of numbers. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== : I don't really see how to do this. Do you know that every computable : number has an infinite number of digits? If not, how do you know that : f^{-1}(j) (i.e., the j-th computable number) has at least j digits, so : that you can define a new number whose j-th digit is different from : the j-th digit of f^{-1}(j)? It's a lot worse than that; he didn't originally know what he meant by number; by itself, number is simply too vague. If he meant real number then he really was obligated to SAY that. ==== >Let S be the set of all computable numbers >and assume S is countable. Since S is countable there is a function, f(), >that maps S to the natural numbers. >Using f() to order S, define a diagonal number, d. I don't really see how to do this. Do you know that every computable > number has an infinite number of digits? If not, how do you know that > f^{-1}(j) (i.e., the j-th computable number) has at least j digits, so > that you can define a new number whose j-th digit is different from > the j-th digit of f^{-1}(j)? This is really not a problem. Any integer (representation in any base) can be zero-padded from left to ensure it really has at least j digits. Of course, the mere existance of d does not make it computable, still. Ahto ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- >As often happens, I have managed to confused myself. >Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers >and assume S is countable. Since S is countable there is a function, f(), >that maps S to the natural numbers. >Using f() to order S, define a diagonal number, d. I don't really see how to do this. Do you know that every computable > number has an infinite number of digits? If not, how do you know that > f^{-1}(j) (i.e., the j-th computable number) has at least j digits, so > that you can define a new number whose j-th digit is different from > the j-th digit of f^{-1}(j)? d is a computable number not in S. >Therefore, S can't be the set of computable numbers. Cantor's powerset argument makes things even weirder. Let S be the set of all computable numbers (countable or uncountable). >Let P(S) be the powerset of S. Cantor proves |S| < |P(S)|. >This means that P(S) contains members that are not in S. >These members must be noncomputable numbers. Actually, ALL members of P(S) are not computable numbers, since the > elements of P(S) are not numbers at all: they are sets of numbers. > How about a simple practical halt test on each number to a finite number of digits, such that each has enough digits to make the diagonal. Here is the raw computable list after x computations of a parallel UTM. 1 1 1 2 2 2 2 2 3 3 4 5 5 5 5 5 5 5 5 The 'partly computable' list is constructed incrementally from that : 1 1 1 2 2 2 2 2 3 5 5 5 5 5 5 5 This guarantees a list of effectively computable numbers. Admitedly a large portion of processor time will be shared on non halting functions but the well behaved functions are still guaranteed to terminate with suitable processor scheduling. Herc ==== > As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. 47 12 487384 4738 74858587 343434 387498394838493 ... This is a countable list containing all computable numbers. Exactly what do you mean by the diagonal number? Do you mean a natural number whose n-th digit is different from the n-th digit of n-th member of the list? It seems that the diagonal has a digit in EVERY n-th place, and is therefore not even a natural number. Can you clarify your argument to explain what you mean by the diagonal number defined by the order f? Even if you are talking about computable reals, say between 0 and 1, you can then define a diagonal (an infinite decimal) but how do you know that diagonal number is computable? What rule generates it? You don't know, because you haven't made f explicit. You can't, because there are in fact an uncountable number of possible f's. That's because there are uncountably many bijections from N to N. ==== > As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. Why is that? Can you exhibit a Turing machine to compute d? In other words, you need a finite rule to compute this d. I don't think you're going to find one. > Therefore, S can't be the set of computable numbers. > [...] ==== As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. Why is that? Can you exhibit a Turing machine to compute d? Good question. I need to define what computable means. > In other words, you need a finite rule to compute this d. I don't > think you're going to find one. Maybe not. Let a computable number be defined as the unbounded, but finite output tape of a TM. Assume that each TM can be represented by a unique natural number. Since the number of TMs is countable, there must be a countable number of comptable numbers. We need to define the output tapes more precisely. Each output tape is unbounded, but finite and every position not written to by the TM is initially set to zero. Define the ordered set S as the set of numbers defined by S(i) = output of TM(i). Let d be the diagonal number. To compute d(i) take the ith position of S(i): If it is equal to 0 then d(i) = 1 else d(i) = 0. Can d can be written to an unbounded, but finite output tape? Probably not. The length of d is the largest i. Is the set of natural numbers a set of computable numbers? It seems computability implies countability, but countability does not imply computability. Russell - Merry Xmas ==== : Let a computable number be defined as the unbounded, but finite output tape : of a TM. That's possible but it's almost inverting the paradigm. The number is computable iff the TM says it is. Traditionally, in order for the TM to say something about the number, the number has to be the INPUT, NOT the output, of the TM. TMs are normally thought of as running on their input in order to answer a question about it. : Assume that each TM can be represented by a unique natural number. That does not need to be assumed; that is inherent in the definition of TM. : Since the number of TMs is countable, there must be a countable number of : comptable numbers. Right. Why isn't that just The Answer to the original question? Why didn't knowing this in advance prevent you, in advance, from asking the original question? : We need to define the output tapes more precisely. Far more importantly, you need to define the INPUT tapes precisely. The output tapes are potentially irrelevant. The *important* output of a TM is whether it halts or loops. : Each output tape is unbounded, but finite and every position : not written to by the TM is initially set to zero. The input and output tapes of a TM (viewed as variables, not constants; viewed as holes, not pegs) are, under the traditional definition, THE SAME (the same holes; but they can start with some pegs and finish with others). Since 0 is going to be inescapably needed as a true data-value in almost anybody's representation of numbers, You need to pick some OTHER symbol as your default. Most TM alphabets come with a built-in designated null symbol. When I was in college we spelled that as an upside-down delta. : Define the ordered set S as the set of numbers defined by : : S(i) = output of TM(i). TM(i) DOES NOT have a unique well-defined output! What TM(i)'s output is will VARY depending on what TM(i)'s INPUT was! Of course, you may, if you insist, make the input every bit as irrelevant as the output by assuming that it is always null, but you will wind up with a much bigger uglier collection of TM's to solve whatever problem you were originally planning to solve. : Let d be the diagonal number. : To compute d(i) take the ith position of S(i): : If it is equal to 0 then d(i) = 1 else d(i) = 0. It is going to dawn on you in a minute that since i can take every natural number as a value, and there are infinitely many of these, you have just defined a d with infinitely many digits. : Can d can be written to an unbounded, but finite output tape? : Probably not. The length of d is the largest i. But there IS NO largest i, OBVIOUSLY. : Is the set of natural numbers a set of computable numbers? Of course. Every natural number is finite and every finite number has the property that a TM can write it, once you decide on a digital representation scheme. -- --- It's difficult ... you need to be united to have any strength, but internal issues have to be addressed. --- E. Ray Lewis, on liberalism in America ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- : We need to define the output tapes more precisely. Far more importantly, you need to define the INPUT tapes > precisely. The output tapes are potentially irrelevant. > The *important* output of a TM is whether it halts > or loops. : Each output tape is unbounded, but finite and every position > : not written to by the TM is initially set to zero. The input and output tapes of a TM (viewed as variables, not constants; viewed > as holes, not pegs) are, under the traditional definition, THE SAME (the > same holes; but they can start with some pegs and finish with others). > Since 0 is going to be inescapably needed as a true > data-value in almost anybody's representation of numbers, You need to pick > some OTHER symbol as your default. Most TM alphabets come with a built-in > designated null symbol. When I was in college we spelled that as an upside-down > delta. : Define the ordered set S as the set of numbers defined by > : > : S(i) = output of TM(i). TM(i) DOES NOT have a unique well-defined output! > What TM(i)'s output is will VARY depending on what TM(i)'s INPUT was! > Of course, you may, if you insist, make the input every bit as irrelevant > as the output by assuming that it is always null, but you will wind up > with a much bigger uglier collection of TM's to solve whatever problem > you were originally planning to solve. > Hardly, just as ugly. TM's are not Finite State Automations that register the input systematically. The TM will in most cases have to produce the number itself to compare it to the input. The computable number at some stage of its defintion has to be output by a computer. Design a TM to parse pi! A TM to parse a number represented in binary will just output yes. Herc ==== As often happens, I have managed to confused myself. > Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers > and assume S is countable. Since S is countable there is a function, f(), > that maps S to the natural numbers. > Using f() to order S, define a diagonal number, d. > d is a computable number not in S. Why is that? Can you exhibit a Turing machine to compute d? Good question. I need to define what computable means. > In other words, you need a finite rule to compute this d. I don't > think you're going to find one. Maybe not. > Let a computable number be defined as the unbounded, but finite output tape > of a TM. Assume that each TM can be represented by a unique natural number. Since the number of TMs is countable, there must be a countable number of > comptable numbers. > Your conclusion is right, but your definition of computable needs work. See below. > We need to define the output tapes more precisely. > Each output tape is unbounded, but finite and every position > not written to by the TM is initially set to zero. > Here's the way Turing did it: every square on the tape is initially blank. The TM outputs 0s and 1s (and possibly other symbols, which we regard as 'garbage'). Any number between zero and one is called computable if there's a TM that starts with a blank tape and produces the binary expansion of the number (disregarding any garbage produced). This binary expansion is culled from the subsequence of the sequence of moves that produces 0s and 1s, so the 0s and 1s don't need to be printed out in the right order on the tape and actually the TM may overwrite some of the previously produced 0s and 1s. In general a computable number is one that differs by an integer from such a computable number between one and zero. Important point: a number with terminating binary expansion may be computed by a TM that *does NOT halt*. > Define the ordered set S as the set of numbers defined by S(i) = output of TM(i). Let d be the diagonal number. > To compute d(i) take the ith position of S(i): > If it is equal to 0 then d(i) = 1 else d(i) = 0. Can d can be written to an unbounded, but finite output tape? > Probably not. The length of d is the largest i. > Hmm...I'm not getting what you're saying. 'i' seems to be the way you are indexing the TMs. So what do you mean by largest 'i'? Of course, d is likely to be some irrational number and have infinite length, but that's not why it's not computable. The point is that a TM that computes d, according to your scheme, must first output S(i) to enough digits -- the i'th digit to be precise, in order to figure out the i'th digit of d-- or *halt* after a certain point before reaching the i'th digit so that you know the rest of the digits are zero. So picture this: your TM starts computing S(i), but maybe it takes a long time to get to the i'th digit; in any case you get it eventually. But the TM may chug on, i.e. it is a non-halting TM! Now of course, since this TM is really like a sub-TM of the one that computes d, you may say you'll just arrange it so that as part of the computing of d, you figure out if TM(i) halts or not. Well, now you (hopefully) see the difficulty involved. Actually, I dug out Turing's paper on computable numbers since my curiosity was piqued, and I'm struck by how readable it still seems. In fact, your very question is addressed! Since it seems you are not operating with a text, I suggest getting a copy of the paper; it's not only clear but has some of the most fundamental concepts of TMs in a fairly short form. It should be available at your university library. A.M. Turing, On Computable Numbers, with an Application to the Entscheidungsproblem., Proceedings of the London Mathematical Society, vol 42 (1936) p. 230-265. > Is the set of natural numbers a set of computable numbers? > Yes. I can compute them pretty easily ;-) Just remember that Turing's definition of computable number is a fairly natural one. It includes basically everything you would think of as 'computable' and even probably some more. > It seems computability implies countability, > but countability does not imply computability. I'm not sure what you're saying here. Of course we can always take a countable set of uncomputable numbers. > Russell > - Merry Xmas Happy Holidays. ==== >As often happens, I have managed to confused myself. >Any enlightenment will be appreciated. I have read that the set of computable numbers is countable. Let S be the set of all computable numbers >and assume S is countable. Since S is countable there is a function, f(), >that maps S to the natural numbers. >Using f() to order S, define a diagonal number, d. >d is a computable number not in S. To correctly conclude that d is a computable number not in S by Using f(), you would have to know that f is a computable function (in, essentially, whatever sense you are using computable in the phrase computable numbers). >Therefore, either >S can't be the set of computable numbers or f isn't computable. Correct conclusion: no such f is computable. Lee Rudolph ==== ... >I have read that the set of computable numbers is countable. Let S be the set of all computable numbers >and assume S is countable. Since S is countable there is a function, f(), >that maps S to the natural numbers. >Using f() to order S, define a diagonal number, d. >d is a computable number not in S. To correctly conclude that d is a computable number not in S > by Using f(), you would have to know that f is a computable > function (in, essentially, whatever sense you are using > computable in the phrase computable numbers). >Therefore, > either >S can't be the set of computable numbers > or f isn't computable. Correct conclusion: no such f is > computable. I don't agree that that is the problem per se. What is wrong is that d is not well-defined, ie, not shown to exist. It is not good enough to merely say define a diagonal number, d. If d is based on, for example, digit j of f(j), the process will fail by f(j) having fewer than j digits, which is the case for most computable numbers, expressed in any base. -jiw ==== > OK, what happens if we make ball n weigh 2^-n? > The number of balls in the bucket always increasing proponent would almost > certainly also accept a mass of balls in the bucket always decreasing > argument. Their therefore the count tends to infinity becomes therefore > the mass decreases to zero. And finally the therefore there are an infinite > number of balls at the end becomes therefore there are no balls at the end That's certainly a non sequitur. The premise of the calculation is > that the number of balls increases without limit, and the result > of the calculation is that their total mass diminishes with limit 0 . To conclude from this that the premise is thereby negated would > be to explode all logic, based merely on an appeal to physical > intuition. But the badthink is also physical intuition. I was trying to drop a block of physical intuation-based concrete into the flow of the badthink. Fight fire with fire. Phil -- Unpatched IE vulnerability: protocol control chars Description: Circumventing content filters Reference: http://badwebmasters.net/advisory/012/ Exploit: http://badwebmasters.net/advisory/012/test2.asp ==== > That's certainly a non sequitur. The premise of the calculation is > that the number of balls increases without limit, and the result > of the calculation is that their total mass diminishes with limit 0 . To conclude from this that the premise is thereby negated would > be to explode all logic, based merely on an appeal to physical > intuition. But the badthink is also physical intuition. I was trying to drop a > block of physical intuation-based concrete into the flow of the badthink. > Fight fire with fire. I knew you'd say that. The original objection, or rather demurral, is based on the nonconvergence of the sequence, N_n, the number of balls remaining after n steps. If one considers the infinite sequence of sets of remaining balls, { x_i_n } = { x_n+1, ... , x_10*n } , then it is uncontroversial to observe that for every m, however large, x_m is_not_an element_of { x_i_n } for all n >= m . The controversy is over the idea that this sequence can somehow be completed , resulting in the null set, because this idea flies in the face of the well-taught, and purely mathematical, notion of limits. Lew Mammel, Jr. ==== >I think that it is easy to prove that 0.9999999...= 1 or in other way that the limit of 1-10^(-n) when n-> infinity is equal to 1 . >But let thin about the floor of 0.9999999... >Is this equal to 1 or 0 ? Since 0.999... is equal to one, the floor of 0.999... is equal to the floor of one, or one. Doug ==== >I think that it is easy to prove that 0.9999999...= 1 or in other way that >the limit of 1-10^(-n) when n-> infinity is equal to 1 . >But let thin about the floor of 0.9999999... >Is this equal to 1 or 0 ? Since 0.999... is equal to one, the floor of 0.999... is equal to the floor > of > one, or one. > This example shows that the floor function does not commute with limits. In other words if you have a sequence a_n, lim floor(a_n) is not necessarily the same as floor lim(a_n). ==== >> >>I think that it is easy to prove that 0.9999999...= 1 or in other way that >>the limit of 1-10^(-n) when n-> infinity is equal to 1 . >>But let thin about the floor of 0.9999999... >>Is this equal to 1 or 0 ? >> >> Since 0.999... is equal to one, the floor of 0.999... is equal to the floor >> of >> one, or one. >This example shows that the floor function does not commute with limits. >In other words if you have a sequence a_n, lim floor(a_n) is not >necessarily the same as floor lim(a_n). Note that I never said that floor commutes with limits. Note *additionally* that your point has nothing to do with the argument at hand. Once you see why, get back to me. Doug ==== >I think that it is easy to prove that 0.9999999...= 1 or in other way that >>the limit of 1-10^(-n) when n-> infinity is equal to 1 . >>But let thin about the floor of 0.9999999... >>Is this equal to 1 or 0 ? > Since 0.999... is equal to one, the floor of 0.999... is equal to the floor >> of >> one, or one. > >This example shows that the floor function does not commute with limits. >In other words if you have a sequence a_n, lim floor(a_n) is not >necessarily the same as floor lim(a_n). Note that I never said that floor commutes with limits. Note *additionally* > that your point has nothing to do with the argument at hand. Once you see why, get back to me. Doug It certainly has nothing to do with the correct evaluation of [.99999...], but I think it goes to the core of what the poster is asking. Why does he ask about [.999999...] and not, say, [2]? Because one contains a limit (which he explicitly mentions) and the other one does not. If he's not asking about the difference between lim [] and [lim], what, notationally, is he wondering about, since he gives 0 and 1 as possible answers? He's already established that .99999... = 1, so unless he's suffered a sudden memory lapse, he can't be asking if [.9999...] = 0 because .99999... is in any way less than 1. Seems that commutation is the crux of the argument. Seems that fishfry was using your counterexample to make a more general statement for the o.p. to chew on, and wasn't really talking back at you specifically, right? ==== Communications in Contemporary Mathematics View table-of-contents and abstracts at http://www.worldscinet.com/ccm.html Refined Geometric Lp Hardy Inequalities G. Barbatis, S. Filippas and A. Tertikas Profile Of Solutions With Sharp Layers To Some Singularly Perturbed Quasilinear Dirichlet Problems Zongming Guo and J. R. L. Webb Global Deformations Of The Witt Algebra Of KricheveröNovikov Type Alice Fialowski And Martin Schlichenmaier The RogersöRamanujan Recursion And Intertwining Operators S. Capparelli, J. Lepowsky and A. Milas Stein Fillings Of Lens Spaces Wave Operators For The Nonlinear Schrodinger Equation With A Nonlinearity Of Low Degree In One Or Two Space Dimensions Kazunori Moriyama, Satoshi Tonegawa And Yoshio Tsutsumi For more information, go to http://www.worldscinet.com/ccm.html ==== Youre a nutcase living a fraudulent life. Rob > James Randi is interested enough to offer one million dollars to > anyone who can demonstrate any kind of paranormal ability under > mutually agreed controlled conditions. That there are no takers yet > speaks volumes. Then why do we have to put up with this? Help support the JREF through donations, > grants, gifts and memberships. Click here to learn more. If he can afford a million dollars to give away he can afford 1% of that > to do tests. I would place credentials on the offer if it wasn't such a > money making scheme. He made a few million doing magic shows did he? The only prizes on offer comes from a huge publicity money making venture and > a bunch of jerkoffs with their heads up their arses in Australia who wont even go > to Townsville in the north of the country to validate my **100,000** witnesses to > paranormal. The CEO said NO! That's it, no examination at all. > Here's the inevitable witnesses on the internet > You rule Truman. http://tinyurl.com/iky4 > Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! > >Is the truman living in Townsville? I've been hearing stuff, yeah. > http://groups.google.com/groups?selm=dksqpv41597ofbhr16mi7tdj2qfihvoldv%404a x.com > Hey Trueman... > what up, my uncle told me you post here... he's did some audio on your > secret show and gave me some tapes. love the show. http://groups.google.com/groups?selm=1epmdv8iqmnckeqhctnf17537rt8hevjeh%404a x.com > You rule Truman. http://www.google.com/groups?as_umsgid=9aqbbv85c53jobkenltkil2tbsdk7m2n3p@4a x.com > I was in Townsville over the weekend, and I heard him. > Very spooky! http://www.google.com/groups?as_umsgid=me7hbv4snqc6gnb1cktj3cvqtiibnpm6s9@4a x.com > I'm from Townsville and YOU ARE the Truman > I'm in Townsville. We're sick of you http://www.google.com/groups?selm=Pine.OSF.4.21.0309230859190.5384-100000%40 marlin.jcu.edu.au >Do you know if the truman is living in Townsville? > I've been hearing stuff, yeah. You can check these are legitimate, the last message id is from > marlin.jcu.edu.au James Cook University in Townsville. noone will consider it, hey maybe they're lying!! maybe they are????? lets dismiss it > in true skeptic style. get me a test, I've been harping for 2 years on it Herc > ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Youre a nutcase living a fraudulent life. Frauds don't have 100,000 witnesses you Ostrich. Frauds make offers and back down, like all skeptics. Sincere people are prepared to go ahead to work things out, frauds stick their heads in the sand. You Ostrich Herc > http://groups.google.com/groups?selm=dksqpv41597ofbhr16mi7tdj2qfihvoldv%404ax .com > Hey Trueman... > what up, my uncle told me you post here... he's did some audio on your > secret show and gave me some tapes. love the show. > http://groups.google.com/groups?selm=1epmdv8iqmnckeqhctnf17537rt8hevjeh%404ax .com > You rule Truman. > http://www.google.com/groups?as_umsgid=9aqbbv85c53jobkenltkil2tbsdk7m2n3p@4ax .com > I was in Townsville over the weekend, and I heard him. > Very spooky! > http://www.google.com/groups?as_umsgid=me7hbv4snqc6gnb1cktj3cvqtiibnpm6s9@4ax .com > I'm from Townsville and YOU ARE the Truman > I'm in Townsville. We're sick of you > http://www.google.com/groups?selm=Pine.OSF.4.21.0309230859190.5384-100000%40m arlin.jcu.edu.au >Do you know if the truman is living in Townsville? > I've been hearing stuff, yeah. You can check these are legitimate, the last message id is from > marlin.jcu.edu.au James Cook University in Townsville. ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- Youre a nutcase living a fraudulent life. Frauds don't have 100,000 witnesses you Ostrich. > Frauds make offers and back down, like all skeptics. Sincere people are prepared to go ahead to work things out, > frauds stick their heads in the sand. You Ostrich Herc Please find the appropriate help. Rob ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Youre a nutcase living a fraudulent life. Frauds don't have 100,000 witnesses you Ostrich. No, he's right; you're clearly mentally ill. Please seek competent help. --- ==== -- You rule Truman. http://tinyurl.com/iky4 Join www.theBanner.net Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. ----------------------------- <^> <(áÀá)> <^> ----------------------------- raven1 help. > TV and media has programmed you to act this way to maintain their control of you. Politics and media are one and the same, and they don't have room for Adam, real religion could deny a politic. That people think cameras follow them is an illness is a fallacy. There is NOONE on the internet except me claiming they are the Truman. There CAN be 1 Truman. You've fallen into a bunch of f*ckwit military controlled politicians bribing media to educate you into Satanism. You're a prize pupil too raven, keep watching your TV how it teaches you to recognise the bad politician speakers and silence them. Herc ==== I really hate to say it again, but you exhibit pretty much *every* symptom of schizophrenia. Please seek professional help. Your condition is treatable with proper medication. --- ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > I really hate to say it again, but you exhibit pretty much *every* > symptom of schizophrenia. Please seek professional help. Your > condition is treatable with proper medication. > why would hollywood promote skizophrennia by calling a show about a guy on camera who gets persecuted by the government TRUE you're telling me they're all signs of illness to believe in it, and they make a movie about it encouraging patients. why TRUEman raven.... think don't respond ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > I really hate to say it again, but you exhibit pretty much *every* >> symptom of schizophrenia. Please seek professional help. Your >> condition is treatable with proper medication. why would hollywood promote skizophrennia by calling a show >about a guy on camera who gets persecuted by the government TRUE you're telling me they're all signs of illness to believe in it, and they >make a movie about it encouraging patients. why TRUEman raven.... think don't respond Listen carefully and TRY to understand. Hollywood also makes motion pictures about Napoleon and promotes them as true. That does not prove YOU are Napoleon. The fact that there are, at any given time, probably several hundred psychiatric inpatients who believe that they ARE, in fact, Napoleon, should indicate something to you. The very notion that Hollywood would be promoting schizophenia by doing such a thing is clinically paranoid, based as it is on the notion that what motivates Hollywood is how thier product will effect a small group of mentally unstable people rather than how it will affect the millions of people who are (relatively) sane and pay to see it, thus recouping for them the investment of millions of dollars expended making the film. One more fact, easily checked by you and everyone else reading this, son... Neither Paramount Studios nor the film's screenwriter and author, Andrew Niccol EVER promoted The Truman Story as anything except comedic fiction and satire. The only person I have ever heard suggest that it is a true story is... YOU. ____________________________________________________________________________ ___ <><><><><><><> The Worlds Uncensored News Source <><><><><><><><> ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- ----------------------------- <^> <(áÀá)> <^> ----------------------------- > I really hate to say it again, but you exhibit pretty much *every* >> symptom of schizophrenia. Please seek professional help. Your >> condition is treatable with proper medication. why would hollywood promote skizophrennia by calling a show >about a guy on camera who gets persecuted by the government TRUE you're telling me they're all signs of illness to believe in it, and they >make a movie about it encouraging patients. why TRUEman raven.... think don't respond Listen carefully and TRY to understand. Hollywood also makes motion pictures about Napoleon and > promotes them as true. That does not prove YOU are Napoleon. The fact that there are, at any given time, probably several > hundred psychiatric inpatients who believe that they ARE, in fact, > Napoleon, should indicate something to you. The very notion that Hollywood would be promoting schizophenia > by doing such a thing is clinically paranoid, based as it is on the > notion that what motivates Hollywood is how thier product will effect > a small group of mentally unstable people rather than how it will > affect the millions of people who are (relatively) sane and pay to see > it, thus recouping for them the investment of millions of dollars > expended making the film. One more fact, easily checked by you and everyone else reading > this, son... Neither Paramount Studios nor the film's screenwriter and > author, Andrew Niccol EVER promoted The Truman Story as anything > except comedic fiction and satire. The only person I have ever heard > suggest that it is a true story is... YOU. > half a dozen people have posted on aus.tv they've heard of me, the truman. how many *prison* movies has *nic* *cage* done? hollywood twists names into the role all the time, there is no way they would call a show TRUE without *some* element to it. wake up! 100,000 witnesses and 100,000 usenetters, not 2 will talk. Jim Carrey costars Laurie Holden in Majestic in 2002. How did I predict that? I can show you the post from 2000. TOP 20 WEB RESULTS out of about 147,000 You could find 1000s of the people that call me Truman every day. Herc ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > I really hate to say it again, but you exhibit pretty much *every* >> symptom of schizophrenia. Please seek professional help. Your >> condition is treatable with proper medication. why would hollywood promote skizophrennia by calling a show >about a guy on camera who gets persecuted by the government TRUE you're telling me they're all signs of illness to believe in it, and they >make a movie about it encouraging patients. why TRUEman raven.... think don't respond It wouldn't be any use to explain that such bizarre, disordered thinking is characteristic of your affliction, so I'll just reiterate get help. --- ==== It wouldn't be any use to explain that such bizarre, disordered > thinking is characteristic of your affliction, so I'll just reiterate > get help. > and I will keep telling you to seek out the town of 100,000 witnesses to the truman show in townvsille australia. 1 0 0 0 0 0 w i t n e s s e s this isn't getting through is it?? Herc please read! Hey Trueman... what up, my uncle told me you post here... he's did some audio on your secrete show and gave me some tapes. love the show. I'm from Townsville and YOU ARE the Truman I was in Townsville over the weekend, and I heard him. Very spooky! I'm in Townsville. We're sick of you You rule Truman. >Do you know if the truman is living in Townsville? I've been hearing stuff, yeah. http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 http://tinyurl.com/rv5f http://tinyurl.com/v1yf ==== > It wouldn't be any use to explain that such bizarre, disordered > thinking is characteristic of your affliction, so I'll just reiterate > get help. > and I will keep telling you to seek out the town of 100,000 witnesses > to the truman show in townvsille australia. 1 0 0 0 0 0 w i t n e s s e s this isn't getting through is it?? Herc please read! > Hey Trueman... > what up, my uncle told me you post here... he's did some audio on your > secrete show and gave me some tapes. love the show. I'm from Townsville and YOU ARE the Truman I was in Townsville over the weekend, and I heard him. > Very spooky! I'm in Townsville. We're sick of you You rule Truman. Do you know if the truman is living in Townsville? > I've been hearing stuff, yeah. > http://tinyurl.com/iky5 > http://tinyurl.com/iky8 > http://tinyurl.com/iky9 > http://tinyurl.com/iky4 > http://tinyurl.com/rv5f > http://tinyurl.com/v1yf Youre delusional. Please seek help. Rob ==== > Skeptic organisations are not interested in the scientific investigation of > the paranormal. > Skeptic organisations are interested in shutting down businesses that make > paranormal claims. > It doesn't matter what evidence or results the business is getting, simply having > a paranormal claim is enough for the 1000s of skeptic organisations to take > action to close them down. If everything that does not work were to be shut down the face of the world would be changed. But as only a very few do not have an interest in at least one thing that does not work, it is unclear who would be in charge of shutting. -- What do you get when you cross a gas chamber with a bottle of Manischewitz? The most expensive whine in the world. -- The Iron Webmaster, 2966 ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Fair enough, JREF is 1000 times ahead of Ozskeptics who don't have > a test to their name this century I don't think. The major current affairs program did a story on them as backing down > and denying applicants and degrading them. They wouldn't even agree > to examine the results of a test that a University was proposing to do for > the applicant. Check the coverage ACA did on Australia Skeptics : > http://www.skeptics.com.au/features/press/aca.htm > All the JREF applicants are funding their own transport to the tests, with > the phony Ozskeptics in my neighbourhood I'm left in the dark. Be good if Universities did preliminaries, like ghostbusters, part of a practical > for statistical analysis... There -is- a test for one of hercs claims. > The webpage is here http://www.twofromoz.freeserve.co.uk/sceptic/herctest/ > all they want now (evidently) is for herc to put up... a few weeks ago with only 10 members but I never heard back. Ideally an active forum to start with, with 100s of posts a day not contrived would definately work. If about 10 people wanted to promote my forum www.yeoldecoffeeshoppe.com we could be running in a couple months. Gets great CTR (click through rate), probably 50 people will visit the site just from this post, but with 5,000 visits it wasn't enough to take off. Once started you get about 1 post per 10 visits. Image just a dozen people with join YeOldeCoffeeShoppe in their sigs. THEN once we have a controlled forum at our disposal I can hand over the website to some commitee from sci.skeptic and they modify the forum so EVERYONE has to guess the author of 1st new replies to them, only a few hours coding to modify a forum script. Then its an even competition, we all play GuessTheNewbie for a couple months and you score a fraction over chance because of writing style or other clues, and I score heads and shoulders above everyone else BECAUSE PEOPLES REPLIES TO ME MATCH THEIR ALIAS MR BLACK, even though its an even competition for everyone. Herc ever tried getting ONE person to cooperate with a new idea on usenet? ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Skeptic organisations are not interested in the scientific investigation > of > the paranormal. Skeptic organisations are interested in shutting down businesses that make > paranormal claims. It doesn't matter what evidence or results the business is getting, simply > having > a paranormal claim is enough for the 1000s of skeptic organisations to > take > action to close them down. Thousands? Name 60 please. Or did you make that up? > Atleast one in every state of Australia, pop < 20 million http://www.skeptics.com.au/about/contact.htm State & Regional Branches New South Wales : Victoria : South Australia : ACT (Australian Capital Territory): Western Australia : Northern Territory : Queensland : Gold Coast (Queensland) Gold Fields (Ballarat, Victoria) : Borderline (Mitta Mitta, Albury, Wodonga): Hunter Valley (NSW) : Tasmania ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Skeptic organisations are not interested in the scientific investigation >> of >> the paranormal. >> Skeptic organisations are interested in shutting down businesses that make >> paranormal claims. >> It doesn't matter what evidence or results the business is getting, simply >> having >> a paranormal claim is enough for the 1000s of skeptic organisations to >> take >> action to close them down. >> Thousands? Name 60 please. Or did you make that up? Atleast one in every state of Australia, pop < 20 million http://www.skeptics.com.au/about/contact.htm >State & Regional Branches Herc's a nutjob. He's probably upset with the government because his dole payment was either late or up for review. New South Wales : >Victoria : >South Australia : >ACT (Australian Capital Territory): >Western Australia : >Northern Territory : >Queensland : >Gold Coast (Queensland) >Gold Fields (Ballarat, Victoria) : >Borderline (Mitta Mitta, Albury, Wodonga): >Hunter Valley (NSW) : >Tasmania -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. ==== -- You rule Truman. http://tinyurl.com/iky4 Join www.theBanner.net Hey Trueman...love the show. YOU ARE the Truman I heard him. Very spooky! >Is the truman living in Townsville? I've been hearing stuff, yeah. ----------------------------- <^> <(áÀá)> <^> ----------------------------- Herc's a nutjob. He's probably upset with the government because his > dole payment was either late or up for review. > What's your 'kook' count up to today Wally want me to check? 1580 wasn't it, 1460 in Novemeber.. Herc ==== People are gullible. Youre the worst type of fool. One thats convinced by his own fraud. Rob ----------------------------- <^> <(áÀá)> <^> ----------------------------- Skeptic organisations are not interested in the scientific investigation > of > the paranormal. Skeptic organisations are interested in shutting down businesses that make > paranormal claims. It doesn't matter what evidence or results the business is getting, simply > having > a paranormal claim is enough for the 1000s of skeptic organisations to > take > action to close them down. Thousands? Name 60 please. Or did you make that up? > Atleast one in every state of Australia, pop < 20 million http://www.skeptics.com.au/about/contact.htm > State & Regional Branches New South Wales : > Victoria : > South Australia : > ACT (Australian Capital Territory): > Western Australia : > Northern Territory : > Queensland : > Gold Coast (Queensland) > Gold Fields (Ballarat, Victoria) : > Borderline (Mitta Mitta, Albury, Wodonga): > Hunter Valley (NSW) : > Tasmania > ==== >Data encryption 360 degrees rotation document 90 degrees and >encryption >on every angel then change it two binary code and fold it over like a >piece of paper then having the one's and zero cancel each other out. if you written a very long letter and then change it two binary code >it would look like this 01010101010101010101010 >10010101010101010101010 >01010101001010101010010 >00010101000101010101010 >10010101010100101010101 would equal = 01 >01010101010100001100101 >01001010101010101010111 >11110111001101010101010 >01010101010101010101010 >10101010101010101010101 if you took the piece of paper and folded it and folded it and folded >it the 0 and 1 would cancel each out and if you keep folding the piece >of paper too the smallest you would have 4 numbers left if 1+1 = >nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 > 01 now if the key new the folding times you could send 2 bytes over >the internet and unzip a >100 zetabyte program you computer could store all the programs ever >written but just need the key to unzip then you could us this for SETI >for signals or can you imagine a computer processor that would be 1.8 >Hz but run like 100 million zeta hz you could use the new 64 bit >process second side to unzip while the front side processes. or use >this for the matrix or quantum computing or supercomputer. 64 bit. 1+1 >= nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 dont forrget to use < >and > signes 0>1+1<0 > > 00. (^ this is my protracted reply. I encoded it using your algorithm; please reverse it to get the full text back - it contains important details about the publication of your ground breaking work. Ground breaking in stupidity, that is.) ==== http://www.skeptics.com.au/features/challenge.htm Briefly, this is what you have to do. You first apply in writing, clearly stating what you are claiming to be able to do. ------------------------------------------ Dear Australian Skeptics, My name is xxx xxx, I would like to demonstrate a paranormal ability being able to guess peoples names from what they say, write or do. Contact: [private] xxx xxx ---------------------------------------------- >You have made no testable claim to Australian Skeptics and there will be no >further communication from us. The writer wanted to compete for the Australian Skeptics $100,000 Challenge, but was unable to actually say what it was that he could do > http://www.skeptics.com.au/features/press/aca.htm > Any charge of our backing down is blatantly untrue and neither Mr Puffet nor A Current Affair has any reason to suppose otherwise. ==== calculus(integration, differentiation) is contained in the sat exam range?? right?? sat exam seems to think real life-math problem. ==== Umm no. Unless you mean the Mathematics subject exam. If not , then perhaps a perusal of the verbal portion of the SAT general exam is in order? > calculus(integration, differentiation) is contained in the sat exam range?? right?? sat exam seems to think real life-math problem. ==== i think........ um.....USA high school math curriculum does not contain intergration, differentiation?? when students learn integration,differentiation?? ==== In sci.math, hot-girl : > i think........ um.....USA high school math curriculum does not contain intergration, > differentiation?? when students learn integration,differentiation?? > Ah, youth; back in the late 70's the high school math curriculum (at least, in my area of the US) did not require calculus. I got lucky, though, and was able to take it in my senior year using resources from a local college. The book was a fairly fat red one with a drawing of a lamp -- and of course now I can't find it. I don't remember when I first learned about actual integrals, though, although one can treat problems such as integral(1 to 5) (x^3 / 11) dx as simple formal exercises in algebraic manipulation (the result is (5^4 - 1) / 44). It's not clear to me whether that route is better than laying the groundwork as to what precisely a Riemannian integral is, and I've confused my personal history further by taking a course on Lebesgue integration, which uses a completely different theory to compute the above answer -- a theory which I've now forgotten most of the particulars of, although I still have to book (Rudin's _Real and Complex Analysis_, second edition). I am not hopeful that the schools have gotten much better in the meantime, and further complications now include dangerous weapons such as 9mm semiautomatics being smuggled in to attempt to kill a teacher one doesn't like. Having to worry about such issues detracts from actual learning, unfortunately; one can't learn calculus while cowering under a desk. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== learning, unfortunately; one can't learn calculus while >cowering under a desk. :-) Actually, that's exactly what I would do. 1. Takes my mind off stuff I have no control over and 2. What else are you going to do--count your navel? I always open a math book on days when the world is against me (remember those days when everything goes wrong?). Solving a problem gives me a small victory which didn't take much time and restored my self-confidence. Doing math is a series of small successes used to build a large house of knowledge. I never understood how people could ever dislike it. /BAH ==== > i think........ um.....USA high school math curriculum does not contain intergration, > differentiation?? In US secondary schools, a course in the calculus is frequently available, rarely required. > when students learn integration,differentiation?? Some in secondary school, some during their university years, many (most?) not at all. It is entirely possible to get an AB in non-technical / non-quantitative fields while knowing no mathematics beyond, roughly, elementary plane geometry, and algebra to the point of solving simultaneous linear equations and quadratic equations. ==== ... > years of posting here, incredible as that seems. Furthermore, believe > it or not, but it is not out of the bounds of reality for me to start > posting from a Yahoo address, now that I am hosting my Amateur > Mathematics web content on Yahoo. But please, feel free to believe whatever you want. The truth doesn't > matter to you, does it? acknowledging that this thread comes from JSH. <31db93f.0312250011.16178abc@posting.google.com> ==== > Implementing the MSN passport, if I can believe the protests of the > people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. Some of us don't care to sign up for anything Microsoft offers, given their less than stellar security records and interesting approaches to privacy concerns. Yes, I suppose one could give false information to sign up for an MS passport, but that seems a bit beside the point. Passport is intended as a one-password-for-many-sites system, and some of us just have enough excitement on our hands without trusting Microsoft with such power. There is a difference between technical difficulty and a refusal to get a passport for other reasons. But why should you understand such subtle hairsplitting? -- Even if [...] a communistic regime should come [to China], the old tradition [...] will break Communism and change it beyond recognition, rather than Communism [...] break the old tradition. It must be so. -- Lin Yutang on Socialism with Chinese characteristics in 1935 ==== > Implementing the MSN passport, if I can believe the protests of the >> people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. Despite his protestations to the contrary, it semms unlikely this is before for obvious forgeries, there was a post from Dec. 23 from the usual msn address, posted through google: http://groups.google.com/groups?selm=3c65f87.0312231952.542d6cb%40posting.go ogle.com and the description of the yahoo group in question reads: If you love to go on usernet and stir up controversy and generally cause trouble in other words you are a troll this group is for you. Also if you are a fan of trolling you are welcome. Share your favorite trolls, tips and the most gullible user groups. Finally trolls have a place to get together and share the tricks of the trade. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > Implementing the MSN passport, if I can believe the protests of the >> people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. Despite his protestations to the contrary, it semms unlikely this is Note that Arturo Magidin is an incompetent liar and a chronic newsgroup stalker. Not too long ago I made a post stepping through my proof of Fermat's Last Theorem for p=3 that's not that much different from my Part 2. I felt that it was clearly interesting enough mathematics that a competent mathematician or an honest one would have to admit the truth, so that's what I said. I ended that post by saying that only an incompetent mathematician or a liar could post against it. I said that because the argument is so simple, direct, and especially because it has to do with Fermat's Last Theorem. Yet Magidin posted against it. He insinuated that the argument was false, and continued that position through repeated posts, only to finally claim that he would not discuss things further as he was killfiling me. Some of you may think I call Magidin a liar because of the question of whether or not rational numbers find their way into the ring I use in my proof. I'm not such an amateur or so stupid that I will be pushed off of a correct proof by a meaningless objection. I maintain that Magidin is indeed either incompetent as a mathematician, or more likely a liar, or both. James Harris ==== > Implementing the MSN passport, if I can believe the protests of the >> people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. > Despite his protestations to the contrary, it semms unlikely this is Note that Arturo Magidin is an incompetent liar and a chronic > newsgroup stalker. > I am inclined to agree with Arturo that this is a troll imitating James Harris, perhaps making creative use of troll methodology. The wording in the sentence above sounds slightly off. Harris's latest shtick is to call his adversaries cranks rather than stalkers. Even if he has a yahoo website, why would he not be using his usual site, trollsareus is less than convincing. Plus as Arturo notes, the intro paragraph on the site indicates it is intended for nothing but trolling techniques. There is no hint of the usual Harrisian pseudo-math. More explanations below - > Not too long ago I made a post stepping through my proof of Fermat's > Last Theorem for p=3 that's not that much different from my Part 2. I have seen no such post in the recent past. And what is this reference to 'my Part 2' ? > I felt that it was clearly interesting enough mathematics that a > competent mathematician or an honest one would have to admit the > truth, so that's what I said. > Where? > I ended that post What post? > by saying that only an incompetent mathematician or > a liar could post against it. I said that because the argument is so > simple, direct, and especially because it has to do with Fermat's Last > Theorem. Yet Magidin posted against it. He insinuated that the argument was > false, and continued that position through repeated posts, only to > finally claim that he would not discuss things further as he was > killfiling me. > This looks like a snippet from something in the distant past. In the most recent interchange between Harris and Magidin, the latter offered to quit responding to the former's posts forever if the former would just say the word. Harris with some profanity accepted the offer. Magidin did not mention killfiling anyone, and did not promise to quit posting in Harris threads though not directly in response to Harris. And in fact that is what has happened. > Some of you may think I call Magidin a liar because of the question of > whether or not rational numbers find their way into the ring I use in > my proof. > This is not an issue in the most recent arguments. It was previously. Again, this looks like a snippet from a long-past Harris post. > I'm not such an amateur or so stupid that I will be pushed off of a > correct proof by a meaningless objection. > Harris does make inordinate use of the word pushed, as in pushed out of the ring of algebraic integers or pushed into a field, etc. This is characteristic of Harris's impoverished vocabulary. It is similar to his overuse of the words break and force: short, primitive physical imagery belying his wish to push mathematicians off their pedestal, to force them to acknowledge his genius, to break their iron grip which prevents his acceptance and validation. This is also likely a direct quote, but again not from a recent post. > I maintain that Magidin is indeed either incompetent as a > mathematician, or more likely a liar, or both. > This too sounds very Harris-like from some time in the past. My guess is, this is a somewhat half-hearted and amateurish troll-job. To be really successful at it, to be a master troll as the real jstevh undeniably is, you have to delude yourself into a belief in what you're saying. Of course this time the troll worked in the sense that several people, including Arturo M. and Nora B., responded. But I think on the whole it is a flop. ===================================================================== I did a little searching after writing the above. Here is an excerpt from a Harris post of September 29, 2001 (Why I call Magidin an incompetent liar): > I was reading through replies to my posts on the newsgroup, when I > noticed a reply from Arturo Magidin. He started the post by > explaining how he was replying since he'd claimed to have killfiled > me, so he wouldn't see my posts. Some have noticed I've been particularly hard on this person. I > continually maintain that he's either incompetent as a mathematician, > a liar or both. James Harris The same post contains the following: > But I'm not such an amateur or so stupid that I will be pushed > off of a correct proof by a meaningless objection. ========================================================================= This seems to settle the question pretty thoroughly, wouldn't say, troll-boy ? Nora B. ==== > Implementing the MSN passport, if I can believe the protests of the >> people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. Despite his protestations to the contrary, it semms unlikely this is Note that Arturo Magidin is an incompetent liar and a chronic > newsgroup stalker. Not too long ago I made a post stepping through my proof of Fermat's > Last Theorem for p=3 that's not that much different from my Part 2. I > felt that it was clearly interesting enough mathematics that a > competent mathematician or an honest one would have to admit the > truth, so that's what I said. I ended that post by saying that only an incompetent mathematician or > a liar could post against it. I said that because the argument is so > simple, direct, and especially because it has to do with Fermat's Last > Theorem. Yet Magidin posted against it. He insinuated that the argument was > false, and continued that position through repeated posts, only to > finally claim that he would not discuss things further as he was > killfiling me. Some of you may think I call Magidin a liar because of the question of > whether or not rational numbers find their way into the ring I use in > my proof. I'm not such an amateur or so stupid that I will be pushed off of a > correct proof by a meaningless objection. I maintain that Magidin is indeed either incompetent as a > mathematician, or more likely a liar, or both. > James Harris No, I think you are incompetent because you do not answer all objections and when you can't refute an objection, you result to insults. Anyone with some sense would refute it POLITELY. -- David Moran Chief Meteorologist Oklahoma Storm Team ==== > Implementing the MSN passport, if I can believe the protests of the > people replying to me here, is incredible difficult. >>As usual, you do not understand the protests of others. >> Despite his protestations to the contrary, it semms unlikely this is >> Note that Arturo Magidin is an incompetent liar and a chronic >> newsgroup stalker. This paragraph seems to be original. >> Not too long ago I made a post stepping through my proof of Fermat's >> Last Theorem for p=3 that's not that much different from my Part 2. I >> felt that it was clearly interesting enough mathematics that a >> competent mathematician or an honest one would have to admit the >> truth, so that's what I said. >> I ended that post by saying that only an incompetent mathematician or >> a liar could post against it. I said that because the argument is so >> simple, direct, and especially because it has to do with Fermat's Last >> Theorem. >> Yet Magidin posted against it. He insinuated that the argument was >> false, and continued that position through repeated posts, only to >> finally claim that he would not discuss things further as he was >> killfiling me. >> Some of you may think I call Magidin a liar because of the question of >> whether or not rational numbers find their way into the ring I use in >> my proof. >> I'm not such an amateur or so stupid that I will be pushed off of a >> correct proof by a meaningless objection. >> I maintain that Magidin is indeed either incompetent as a >> mathematician, or more likely a liar, or both. But the paragraphs above were originally posted in http://groups.google.com/groups?selm=3c65f87.0109291024.133073d3%40posting.g oogle.com on a thread called Why I call Magidin an incompetent liar, on Sept. 29 2001. The first four paragraphs are identical, down to the line breaks, to the ones in that post. The next paragraph, the one that starts I'm not such an amateur... is edited. The original started I'm an amateur. I can make small mistakes. I don't mind it when honest, competent people point that out to me. >No, I think you are incompetent because you do not answer all objections and >when you can't refute an objection, you result to insults. Anyone with some >sense would refute it POLITELY. This is not the real James Harris. -- ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu <31db93f.0312250011.16178abc@posting.google.com> <874qvotj7s.fsf@phiwumbda.org> ==== > Implementing the MSN passport, if I can believe the protests of the >> people replying to me here, is incredible difficult. As usual, you do not understand the protests of others. Some of us > don't care to sign up for anything Microsoft offers, given their less > than stellar security records and interesting approaches to privacy > concerns. Yes, I suppose one could give false information to sign up for an MS > passport, but that seems a bit beside the point. Passport is intended > as a one-password-for-many-sites system, and some of us just have > enough excitement on our hands without trusting Microsoft with such > power. There is a difference between technical difficulty and a refusal to > get a passport for other reasons. But why should you understand such > subtle hairsplitting? Damn. I hadn't looked closely at the cited material or the posting address. I reckon IHBT. Silly me. -- Mathematicians are rather important in the infrastructures of many organizations that protect civilization. I've determined that they are a consistent security risk, and seem to have other agendas, other loyalties beyond loyalty to their respective nations. -- James Harris ==== int_0^1 dx/(x^x) = 1 + 1/2^2 + 1/3^3 + 1/4^4 + ... This is Exercise 62 in the chapter about the gamma and beta functions in Spiegel's _Advanced Calculus_ from 1963. But I get the feeling that there is some other demonstration which is obvious _after_ we see it. Anybody know of any? Any remarks about a relationship with the integral of exp(-x^2), if there is one, would also be appreciated. TIA, Larry ==== >int_0^1 dx/(x^x) = 1 + 1/2^2 + 1/3^3 + 1/4^4 + ... >This is Exercise 62 in the chapter about the gamma and beta functions in >Spiegel's _Advanced Calculus_ from 1963. But I get the feeling that there is >some other demonstration which is obvious _after_ we see it. Anybody know >of any? writing 1/x^x = exp(-x ln(x)) and using the Taylor series for exp. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Oh well. James started one of his latest post with a polynomial with factorisation (citations are not exact): > P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) = > = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) with > b_3(x) = a_3(x) - 3 using this substitution we get the factorisation: > = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) where he says that the a's are roots of: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) so far so good. A lot of conclusions are drawn from this factorisation, it is however the consequence of another factorisation. Set y = 49x, we get: P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 = = (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7) where the a's are roots of: a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y) So although the constant term of two of the factors are divisible by 7, and the third is co-prime to it (it is 22), in general *none* of the factors are divisible by 7, as P(y) is only divisible by 7 in a limited number of cases. So I am still wondering what JSH is trying to show with his constant terms. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== In sci.math, Dik T. Winter : > Oh well. James started one of his latest post with a polynomial with > factorisation (citations are not exact): > P(x) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) = > = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) > with > b_3(x) = a_3(x) - 3 > using this substitution we get the factorisation: > = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7) > where he says that the a's are roots of: > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) > so far so good. A lot of conclusions are drawn from this factorisation, it is however > the consequence of another factorisation. Set y = 49x, we get: > P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 = > = (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7) > where the a's are roots of: > a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y) So although the constant term of two of the factors are divisible by > 7, and the third is co-prime to it (it is 22), in general *none* of the > factors are divisible by 7, as P(y) is only divisible by 7 in a limited > number of cases. So I am still wondering what JSH is trying to show > with his constant terms. So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7 are algebraic integers, but admittedly it's not a priority. It's clear that generally speaking, they are not, although a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. Your manipulation is interesting and simplifies the problem considerably. :-) However, now one has to deal with y being a multiple of 49, if x was originally an algebraic integer. (I'm also curious as to the rest of his proof; this is only a small snippet thereof. It's a bit like examining a small area (sans the actual hole) of a flat tire to try to figure out why the car won't move.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > In sci.math, Dik T. Winter > > : ... > A lot of conclusions are drawn from this factorisation, it is however > the consequence of another factorisation. Set y = 49x, we get: > P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 = > = (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7) > where the a's are roots of: > a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y) > > So although the constant term of two of the factors are divisible by > 7, and the third is co-prime to it (it is 22), in general *none* of the > factors are divisible by 7, as P(y) is only divisible by 7 in a limited > number of cases. So I am still wondering what JSH is trying to show > with his constant terms. > > So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7 > are algebraic integers, but admittedly it's not a priority. You first are required to show what a_1, a_2 and a_3 are. Luckily that is possible... (*) > It's > clear that generally speaking, they are not, although > a_1(0) = 0, a_2(0) = 0, a_3(0) = 3. > > Your manipulation is interesting and simplifies the problem > considerably. :-) However, now one has to deal with y being > a multiple of 49, if x was originally an algebraic integer. But it also shows that the factors are not divisible by 7 for *all* y, so it shows that divisibility properties are erratic. > (I'm also curious as to the rest of his proof; this is > only a small snippet thereof. It's a bit like examining > a small area (sans the actual hole) of a flat tire to try > to figure out why the car won't move.) The rest of his proof hinges on the fact that exactly two factors are (FLT proof) or should be (definition error) divisible by 7. ---- (*) A very nice showing of that I found while looking around at the solutions of cubic equations. All expositions I have seen miss a very basic fact (I think), but the exposition at mathworld is closest (this Let's calculate the roots of z^3 + a.z^2 + b.z + c = 0. Let us have the following definitions: Q = (12.b - 4.a^2) R = 36.a.b - 108.c - 8.a^3 K1 = cbrt(R + sqrt(Q^3 + R^2))/2 K2 = cbrt(R - sqrt(Q^3 + R^2))/2 W = (-1 + i.sqrt(3))/2, W^2 = (-1 -i.sqrt(3))/2, W^3 = 1 then we have the following roots: z_1 = (-a + W .K1 + W^2.K2)/3 z_2 = (-a + W^2.K1 + W .K2)/3 z_3 = (-a + K1 + K2)/3 filling in all stuff (and hoping I did not make a basic arithmetic mistace), we find that the z's correspondend to James' a's in order. The beauty of this presentation is (I think) how three cubic roots of two different values are used. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: george_cox@btinternet.com X-Punge: Micro$oft ==== >elegance of hexadecimal How is hexadecimal any more elegant than any other base? >Unless you're examining bowling scores, use hexadecimal. Real programmers use octal. All of our units of measurement are scaled in decimal. Headecimal is only reasonable for equipment that deals with numbers in bytes that are multiples of 4 bits. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Real programmers use octal. All of our units of measurement are scaled > in decimal. Headecimal is only reasonable for equipment that deals > with numbers in bytes that are multiples of 4 bits. Isn't that all equipment? -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== >Message-id: <3febbc35$8$fuzhry+tra$mr2ice@news.patriot.net >elegance of hexadecimal How is hexadecimal any more elegant than any other base? >Unless you're examining bowling scores, use hexadecimal. Real programmers use octal. All of our units of measurement are scaled >in decimal. Headecimal is only reasonable for equipment that deals >with numbers in bytes that are multiples of 4 bits. Which explains why DEC used octal for 16-bit registers and 20-bit addressing? -- > Shmuel (Seymour J.) Metz, SysProg and JOAT >not reply to spamtrap@library.lspace.org -- Mensanator Ace of Clubs ==== To paraphrase Berzeluis. To elaborate, I often found it confusing to associate particular properties, algebraic objects and other such things with the name they are given. Surely it only adds a level of complexity to a field of study which does not need one! What are your opinions on having things called Schurian, Jacobian, or Gaussian? Do you feel that names of things in mathematics should greater reflect information about the said things? Do you feel it retards mathematical progress (on a personal, or collective level)? What is the worst named object you can think of? I hope the quote alludes to the fact that naming systems aren't only a problem in mathematics, but perhaps it would be productive to restrict the discussion to math despite this obvious fact. ==== maneesh > To elaborate, I often found it confusing to associate particular > properties, algebraic objects and other such things with the name they > are given. Surely it only adds a level of complexity to a field of > study which does not need one! > What are your opinions on having things called Schurian, Jacobian, or > Gaussian? The medical boys have long had a practice of avoiding all eponyms, in favour of words which give some clue to the meaning. E.g. the eustachian tube is now called the salpinx (trumpet). I agree with you and with them, but it might be tough to think up replacements for Vandermonde determinant and so many others. LH ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > To elaborate, I often found it confusing to associate particular > properties, algebraic objects and other such things with the name they > are given. Surely it only adds a level of complexity to a field of > study which does not need one! > What are your opinions on having things called Schurian, Jacobian, or > Gaussian? > The medical boys have long had a practice of avoiding all eponyms, in favour > of words which give some clue to the meaning. E.g. the eustachian tube is > now called the salpinx (trumpet). I agree with you and with them, but it > might be tough to think up replacements for Vandermonde determinant and so > many others. > LH > Wait for formalsim. Godel made a self evident truth and everyone assumed noone could formalise mathematics. We could rewrite maths terms now but what names are the *most* suitable? Say a heirarchy of functions, does the term refer to its place among the list, its function among its peers, an analogy to a sister domain? Is integration a function like addition? a construct like parenthesis? a fundamental operation? a relation? an inverse of differentiation? a type of function that loses information? a topogical construct? a statistical method? ..... which name which name Herc ==== What do you expect your public opinion poll, answered by any clown who wants to respond, to mean? And so what if they all agree with you? Then what happens? >To paraphrase Berzeluis. To elaborate, I often found it confusing to associate particular >properties, algebraic objects and other such things with the name they >are given. Surely it only adds a level of complexity to a field of >study which does not need one! >What are your opinions on having things called Schurian, Jacobian, or >Gaussian? >Do you feel that names of things in mathematics should greater reflect >information about the said things? Do you feel it retards >mathematical progress (on a personal, or collective level)? What is >the worst named object you can think of? I hope the quote alludes to the fact that naming systems aren't only a >problem in mathematics, but perhaps it would be productive to restrict >the discussion to math despite this obvious fact. ==== What do you expect your public opinion poll, answered by any clown > who wants to respond, to mean? And so what if they all agree with you? > Then what happens? > This clown just thought it was an interesting question, and that he might find some meaning in another clown's answers. Why not just let people answer and see if anything interesting comes up, that's what a discussion group is for isn't it? ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > What do you expect your public opinion poll, answered by any clown > This clown just thought it was an interesting question, and that he > might find some meaning in another clown's answers. Why not just let > people answer and see if anything interesting comes up, that's what a > discussion group is for isn't it? He was talking about me, apparently proof of god is off topic in mathematics. We did maths logic semesters, computer science logic semesters, applied maths logic semesters, and theoretical computing logic semesters. Each used a different syntax for A -> B. Two exams had the same question and in computers the answer wanted Naturals to include 0, so reflexive was the answer, in maths the Naturals start at 1 so irreflexive was the answer. The sciences go through periods of breadth of terminolgy then unify to have greater depth. Herc ==== > What is the worst named object you can think of? > The one's where different writers have the same name for different things and different names for the same things. Especially the ones that are just the opposite of the others. For example. A space is T4 when disjoint closed set separable by open disjoint sets. A space is normal when it's T2 and T4 while other authors say normal is the above T4 and T4 is the above normal. ==== > To paraphrase Berzeluis. To elaborate, I often found it confusing to associate particular > properties, algebraic objects and other such things with the name they > are given. Surely it only adds a level of complexity to a field of > study which does not need one! > What are your opinions on having things called Schurian, Jacobian, or > Gaussian? > Do you feel that names of things in mathematics should greater reflect > information about the said things? Do you feel it retards > mathematical progress (on a personal, or collective level)? What is > the worst named object you can think of? I hope the quote alludes to the fact that naming systems aren't only a > problem in mathematics, but perhaps it would be productive to restrict > the discussion to math despite this obvious fact. Naming is a matter of historical accident, historical inaccuracy, and human whim. Most theorems (can I say most? at least some.) were first proved by someone other than their associated name. Fermat's Last Theorem being the classic case. Then of course why is it Zorn's Lemma, the Axiom of Choice, and the Well-Ordering Principle? What is the difference among lemma, axiom, and principle, if in this case they are used to label logical propositions known to be equivalent to one another? You might as well campaign for geographical names to make logical sense. ==== > To paraphrase Berzeluis. To elaborate, I often found it confusing to associate particular > properties, algebraic objects and other such things with the name they > are given. Surely it only adds a level of complexity to a field of > study which does not need one! > What are your opinions on having things called Schurian, Jacobian, or > Gaussian? > Do you feel that names of things in mathematics should greater reflect > information about the said things? Do you feel it retards > mathematical progress (on a personal, or collective level)? What is > the worst named object you can think of? I hope the quote alludes to the fact that naming systems aren't only a > problem in mathematics, but perhaps it would be productive to restrict > the discussion to math despite this obvious fact. Why are you named Maneesh? Shouldn't your name reflect what you do? Gib ==== Why are you named Maneesh? Shouldn't your name reflect what you do? Gib > sheesh they're all named after men! Herc Gib's 1st reply to God (me) that in theory should give away his name : http://tinyurl.com/3e942 Where you are concerned this seems to be true. I think I could tell one of your posts without your name attached. Gib ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- > To paraphrase Berzeluis. To elaborate, I often found it confusing to associate particular > properties, algebraic objects and other such things with the name they > are given. Surely it only adds a level of complexity to a field of > study which does not need one! > What are your opinions on having things called Schurian, Jacobian, or > Gaussian? > Do you feel that names of things in mathematics should greater reflect > information about the said things? Do you feel it retards > mathematical progress (on a personal, or collective level)? What is > the worst named object you can think of? I hope the quote alludes to the fact that naming systems aren't only a > problem in mathematics, but perhaps it would be productive to restrict > the discussion to math despite this obvious fact. what better name for proving uncountability than CantOr who could devise a unit number other than our phallic historic figure Peno who could formulate the unprovable other than GODel. what distribution allows us to filter the population results down to sample? Gaussian Mathematicians aren't just part of our history, they form a giant platonic jigsaw puzzle. some others Bill Gates ~ Billionaire Ronald Ray Gun ~ Star Wars program Lady Di ~ dies Tiger Woods ~ golf wood Nic Cage ~ prison movies Herc ==== > Uncle Al's Linda caught a piece and spent a week near death. Uncle Al > caught it and had a few days of runny nose - Pauling's Vitamin C > titration works. A bottle of 100 500 mg Vitamin C tablets and a roll > of paper towels is not so bad. The Mexican Americans (i.e. just about everybody in your neighborhood, Al) use a bottle of vitamin E and garlic pills (or garlic cloves swallowed whole) at the first sign of catching flu, and they swear it works everytime. I don't know about the garlic, but is vitamin E as good as C? -- sartarAshi ro az sare kachale mA mikhAd yAd begireh. ==== > Uncle Al's Linda caught a piece and spent a week near death. Uncle Al >> caught it and had a few days of runny nose - Pauling's Vitamin C >> titration works. A bottle of 100 500 mg Vitamin C tablets and a roll >> of paper towels is not so bad. The Mexican Americans (i.e. just about everybody in your >neighborhood, Al) use a bottle of vitamin E and garlic pills >(or garlic cloves swallowed whole) at the first sign of >catching flu, and they swear it works everytime. I don't >know about the garlic, but is vitamin E as good as C? Exuding garlic odors would keep other people away. Thus, the bug stops here. One of most appalling things is nowhere do news reports about the flu say, Stay home if you're sick. If your kids are sick, don't take them to the grocery store. In addition, they're spraying the damned flu mist in the grocery stores (one-stop shopping). Doesn't anybody think twice about aerating flu bugs over other people's food? /BAH ==== > Uncle Al's Linda caught a piece and spent a week near death. Uncle Al > caught it and had a few days of runny nose - Pauling's Vitamin C > titration works. A bottle of 100 500 mg Vitamin C tablets and a roll > of paper towels is not so bad. >>The Mexican Americans (i.e. just about everybody in your >>neighborhood, Al) use a bottle of vitamin E and garlic pills >>(or garlic cloves swallowed whole) at the first sign of >>catching flu, and they swear it works everytime. I don't >>know about the garlic, but is vitamin E as good as C? Exuding garlic odors would keep other people away. Thus, > the bug stops here. Meanwhile, those other Mexican Americans who think like you, they place and keep the garlic cloves inside their navel cavities :-) Hehe :) This is true! I know it must be the vitamin E doing the trick, as the garlic, when consumed, at most would slow down the movement of the intestines a bit. I don't think this would help one much with flu (it helps with diarrhea). But why is that they go for vitamin E and not C which is cheaper and perhaps more effective. Is it the amount they have to take? E is stored in the body (I think) and C is not. Would this mean a lower amount of E would work like a large amount of C? -- sharmemAn bAd ze pashmineye Aludeye khish gar bedin fazlo honar nAme karAmAt barim fetne mibArad azin saghfe mogharnas barkhiz tA be meykhAne panAh az hame AfAt barim ghadre vaght ar nashenAsad delo kAri nakonad bass khejAlat ke azin hAsele owghAt barim dar biyAbAne fanA gomshodan Akhar tA key? rah beporsim magar pey be mohemmAt barim Hafez ==== ----------------------------- <^> <(áÀá)> <^> ----------------------------- Uncle Al's Linda caught a piece and spent a week near death. Uncle Al > caught it and had a few days of runny nose - Pauling's Vitamin C > titration works. A bottle of 100 500 mg Vitamin C tablets and a roll > of paper towels is not so bad. >>The Mexican Americans (i.e. just about everybody in your >>neighborhood, Al) use a bottle of vitamin E and garlic pills >>(or garlic cloves swallowed whole) at the first sign of >>catching flu, and they swear it works everytime. I don't >>know about the garlic, but is vitamin E as good as C? Exuding garlic odors would keep other people away. Thus, > the bug stops here. Meanwhile, those other Mexican Americans who think like you, > they place and keep the garlic cloves inside their navel > cavities :-) Hehe :) This is true! I know it must be the vitamin E doing the trick, as the > garlic, when consumed, at most would slow down the movement > of the intestines a bit. I don't think this would help one > much with flu (it helps with diarrhea). But why is that they > go for vitamin E and not C which is cheaper and perhaps more > effective. Is it the amount they have to take? E is stored > in the body (I think) and C is not. Would this mean a lower > amount of E would work like a large amount of C? > Its what works for you. This would be an impossible answer to find amongst the billion dollars of alternative medicine information. Vitamin E is an antioxidant?, (so is C I think) like a chemical anti virus. Herc ==== > ----------------------------- <^> <(áÀá)> <^> ----------------------------- > Uncle Al's Linda caught a piece and spent a week near death. Uncle Al > caught it and had a few days of runny nose - Pauling's Vitamin C > titration works. A bottle of 100 500 mg Vitamin C tablets and a roll > of paper towels is not so bad. >>The Mexican Americans (i.e. just about everybody in your >>neighborhood, Al) use a bottle of vitamin E and garlic pills >>(or garlic cloves swallowed whole) at the first sign of >>catching flu, and they swear it works everytime. I don't >>know about the garlic, but is vitamin E as good as C? Exuding garlic odors would keep other people away. Thus, > the bug stops here. Meanwhile, those other Mexican Americans who think like you, > they place and keep the garlic cloves inside their navel > cavities :-) Hehe :) This is true! I know it must be the vitamin E doing the trick, as the > garlic, when consumed, at most would slow down the movement > of the intestines a bit. I don't think this would help one > much with flu (it helps with diarrhea). But why is that they > go for vitamin E and not C which is cheaper and perhaps more > effective. Is it the amount they have to take? E is stored > in the body (I think) and C is not. Would this mean a lower > amount of E would work like a large amount of C? Its what works for you. This would be an impossible answer to find amongst > the billion dollars of alternative medicine information. Vitamin E is an > antioxidant?, (so is C I think) like a chemical anti virus. I have found several Chinese medicines that work much better than the Western medicines for certain things. When I young, and got colds, I took those multi-purpose cold remedies, that combined decongestants, antihistamines, antitussives, expectorants, pain relievers, etc. in one pill. As I wised up, I just took the specific ingredient that I needed. I also tried many grams of vitamin C with no results, although I did find that zinc tablets sometimes helped. I found the Chinese cold medicines worked better than Western medicine medicines, and that ginger was a common ingredient, so I bought some ginger, and when I feel a cough or the sniffles coming on, I snack on the ginger, and it seems to work very well. I have also noticed that it relieves the coughing caused by smog and bad air, so it might be good for asthma. Try it. It might help you. -- Tom Potter http://tompotter.us ==== > ----------------------------- <^> <(áÀá) <^> ----------------------------- > I have found several Chinese medicines that work much better > than the Western medicines for certain things. When I young, and got colds, > I took those multi-purpose cold remedies, > that combined decongestants, antihistamines, antitussives, > expectorants, pain relievers, etc. in one pill. > As I wised up, I just took the specific ingredient > that I needed. I also tried many grams of vitamin C with no results, > although I did find that zinc tablets sometimes helped. I found the Chinese cold medicines worked better > than Western medicine medicines, and that ginger > was a common ingredient, so I bought some ginger, > and when I feel a cough or the sniffles coming on, > I snack on the ginger, and it seems to work very well. You snack on RAW ginger? Brave, you are! I have heard that people boil it, like a concentrated broth and drink it (back home); I have never done that. One other way *might* be put a BUNCH of ginger (overwhelmingly) in boiling meat (chicken, beef) in making soup. That's food my brother makes; I don't know whether it would work as a cold remedy. I have also noticed that it relieves the coughing > caused by smog and bad air, so it might be good > for asthma. Try it. > It might help you. ==== > Uncle Al's Linda caught a piece and spent a week near death. Uncle Al > caught it and had a few days of runny nose - Pauling's Vitamin C > titration works. A bottle of 100 500 mg Vitamin C tablets and a roll > of paper towels is not so bad. >>The Mexican Americans (i.e. just about everybody in your >>neighborhood, Al) use a bottle of vitamin E and garlic pills >>(or garlic cloves swallowed whole) at the first sign of >>catching flu, and they swear it works everytime. I don't >>know about the garlic, but is vitamin E as good as C? >> Exuding garlic odors would keep other people away. Thus, >> the bug stops here. >> One of most appalling things is nowhere do news reports about >> the flu say, Stay home if you're sick. If your kids are sick, >> don't take them to the grocery store. >> In addition, they're spraying the damned flu mist in the grocery >> stores (one-stop shopping). Doesn't anybody think twice about >> aerating flu bugs over other people's food? >This work ethic throughout having a cold is >idiotic. I got a terrible cold every >year from my teens up to 26. You can spot >the people you're stuck in the same room >with spreading it around, you know you'll get >it then. If it's a virus, keep wetting your nose. That's how I managed to avoid catching one whenever we went on an airplane. > ..Every year 1 to 2 months of pain. In my case, I now catch it at least 8 times. So, for a disease whose cycle is 3-4 weeks, I have to plan on being sick for half a year if I survive the first round. >I've been unemployed 6 years and not 1 cold. In Japan they wear a paper mask if they have to go out with the flu. When JMF was getting chemo, the nurses and receptionists would brag about how they came in even though they were sick. Patient advocates would come in and sit in the chemo production line room sneezing and coughing. These people would be given a mask to wear, which they would remove when they had to sneeze and/or cough. Somewhere within the last two or three decades basic hygiene has become politically incorrect. /BAH ==== In sci.physics, jmfbahciv@aol.com <3fec496e$0$4745$61fed72c@news.rcn.com>: > > Uncle Al's Linda caught a piece and spent a week near death. Uncle > Al >> caught it and had a few days of runny nose - Pauling's Vitamin C >> titration works. A bottle of 100 500 mg Vitamin C tablets and a roll >> of paper towels is not so bad. The Mexican Americans (i.e. just about everybody in your >neighborhood, Al) use a bottle of vitamin E and garlic pills >(or garlic cloves swallowed whole) at the first sign of >catching flu, and they swear it works everytime. I don't >know about the garlic, but is vitamin E as good as C? Exuding garlic odors would keep other people away. Thus, > the bug stops here. One of most appalling things is nowhere do news reports about > the flu say, Stay home if you're sick. If your kids are sick, > don't take them to the grocery store. In addition, they're spraying the damned flu mist in the grocery > stores (one-stop shopping). Doesn't anybody think twice about > aerating flu bugs over other people's food? > >>This work ethic throughout having a cold is >>idiotic. I got a terrible cold every >>year from my teens up to 26. You can spot >>the people you're stuck in the same room >>with spreading it around, you know you'll get >>it then. If it's a virus, keep wetting your nose. That's how I managed > to avoid catching one whenever we went on an airplane. OK, I gotta ask: is this something that's done using a small atomizer (e.g., a squirt bottle) undetectably, or is this something that one can visibly see dripping and people think you've got the plague and avoid you at all costs? :-) I'll admit I could see how moistening the mucous membrane might help in flushing the bug down into the gastrointestinal system (where, if one's lucky, it'll be digested), although I can also see some risk (how much, I've no idea) of it getting lodged in the lungs instead. Of course visibly dripping has its advantages, too, by keeping unwanted neighbors away... :-) It just sounds more uncomfortable... > >> ..Every year 1 to 2 months of pain. In my case, I now catch it at least 8 times. So, for a disease > whose cycle is 3-4 weeks, I have to plan on being sick for > half a year if I survive the first round. Ew. > >>I've been unemployed 6 years and not 1 cold. >>In Japan they wear a paper mask if they have to go out with the flu. When JMF was getting chemo, the nurses and receptionists would brag > about how they came in even though they were sick. Patient advocates > would come in and sit in the chemo production line room sneezing > and coughing. These people would be given a mask to wear, which > they would remove when they had to sneeze and/or cough. Somewhere within the last two or three decades basic hygiene has > become politically incorrect. Not to mention extremely confused. Ideally, the sick person would have a set of little baggies that they'd hand out to everyone; the baggie would contain the guaranteed sterile (unless opened) paper mask. Of course, that's probably a bit much to ask -- although the waiting room receiptionist or emergency room tech could stockpile them, perhaps. I'll admit colds for me are extremely rare, although I will also admit to washing my hands frequently after using the restroom facilities, and using my elbows, feet, or arms instead of my hands for opening the doors thereto and within. /BAH > -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >In sci.physics, jmfbahciv@aol.com >: If it's a virus, keep wetting your nose. That's how I managed >> to avoid catching one whenever we went on an airplane. OK, I gotta ask: is this something that's done using a small >atomizer (e.g., a squirt bottle) undetectably, Nah, I just asked for a glass of water, dipped my finger in and rubbed the outside of the nostrils. > . or is this >something that one can visibly see dripping and people think >you've got the plague and avoid you at all costs? :-) I'll admit I could see how moistening the mucous membrane might >help in flushing the bug down into the gastrointestinal system >(where, if one's lucky, it'll be digested), although I can also >see some risk (how much, I've no idea) of it getting lodged >in the lungs instead. The purpose is to counter the lack of humidity in the airplane air. Of course visibly dripping has its advantages, too, by keeping >unwanted neighbors away... :-) It just sounds more uncomfortable... Just damp will do :-). admit to washing my hands frequently after using the restroom >facilities, and using my elbows, feet, or arms instead of my hands >for opening the doors thereto and within. Exactly. One of my tricks is to wear long-sleeved sweatshirts and cover my hand before opening a door. While out, not touching eyes and nose (that's when they always decide to get itchy) also helps. Since women think it's cute to allow their sick kids to touch everything in the grocery store, I wash everything I can before I put them away. Always buy packaged produce. Make your own hambuger after washing the meat. /BAH ==== > I'm missing these conjectures, where subjective numbers are given > for the estimated probability that they are valid. I'm missing probability estimates that complicated mathematical proofs are > indeed correct. In math there is only : true or false > but often statements like : true with probability 90% are also useful, > but people rarely dare to give such estimates. That's because you don't know until the final step that a theorem is true (and that's assuming you've done everything correctly). Number theory has some hilarious examples where we were nearly certain (and wrong) that a theorem was true because we'd checked it to the largest integers our computers could store; the counterexamples were far beyond what we had ever considered checking. Let's say a proposed proof of a conjecture has 10,000 initial steps, invoking all sorts of well-established results, leading in the general direction of the desired result. All that's left, we think, is one more step, but, unfortunately, our conjecture turns out to be false. What good is an estimate in this case? Were I in that situation, the only plausible number I can come up with is that we were over 99% sure that this conjecture was correct. If you can give an algorithm which estimates the veracity of a statement better than a human can intuit, which is rarely a good way to judge complex theorems, you'll be making news. Besides, probability is usually a description of what will happen to frequencies when an experiment is run over and over again; I can't think of any interpretation of probability which does not involve repeated trials(*). How does this model apply to a theorem, which we know is either true or false (even if we know we can't prove it either way)? It's not as if it's true on Tuesday and false all other days (some obvious exceptions aside). You might be interested, though, in fuzzy logic and AI computing. There are systems which emulate human thinking in that they assign probability to truth statements and make decisions based on that. Justin Davis (*) Doesn't the probability distribution of actual events come from what the frequency distributions will look like after an infinite number of trials? ==== You might be interested, though, in fuzzy logic and AI computing. There are systems which emulate human thinking in that they assign probability to truth statements and make decisions based on that. Let's be clear that fuzzy set membership and fuzzy truth values are not probabilities. See the FAQ for comp.ai.fuzzy on Usenet. -Will Dwinnell http://will.dwinnell.com ==== >> I'm missing these conjectures, where subjective numbers are given >> for the estimated probability that they are valid. >> >> I'm missing probability estimates that complicated mathematical proofs are >> indeed correct. >> >> In math there is only : true or false >> but often statements like : true with probability 90% are also useful, >> but people rarely dare to give such estimates. > >That's because you don't know until the final step that a theorem is >true (and that's assuming you've done everything correctly). Number you have a better feeling about this than the reader can have. Why not transmit this extra information ? >theory has some hilarious examples where we were nearly certain (and >wrong) that a theorem was true because we'd checked it to the largest >integers our computers could store; the counterexamples were far >beyond what we had ever considered checking. sounds like one of these rare counterexamples. You may just remember them better than all the other boring normalities >Let's say a proposed proof of a conjecture has 10,000 initial steps, >invoking all sorts of well-established results, leading in the general >direction of the desired result. All that's left, we think, is one >more step, but, unfortunately, our conjecture turns out to be false. >What good is an estimate in this case? Were I in that situation, the >only plausible number I can come up with is that we were over 99% sure >that this conjecture was correct. If you can give an algorithm which that's OK. It's not so much dishonorable to be wrong with a probability estimate, provided you state it as this. >estimates the veracity of a statement better than a human can intuit, >which is rarely a good way to judge complex theorems, you'll be making >news. > >Besides, probability is usually a description of what will happen to >frequencies when an experiment is run over and over again; I ca >think of any interpretation of probability which does not involve >repeated trials(*). How does this model apply to a theorem, which we >know is either true or false (even if we know we can't prove it either >way)? It's not as if it's true on Tuesday and false all other days you may use your experience with other theorems >You might be interested, though, in fuzzy logic and AI computing. >There are systems which emulate human thinking in that they assign >probability to truth statements and make decisions based on that. > >Justin Davis > >(*) Doesn't the probability distribution of actual events come from >what the frequency distributions will look like after an infinite >number of trials? this is how we are trying to interpret probabilities. But we still can and do assign probabilities to events which will (or will not) occur only once. We compare with similar situations in the past or try to calculate it from deduced rules. ==== > For fun i just had a look and on a medium sized PC a > timing could be ~ 1 min for 10^6 digits and ~ 20 min > for 16*10^6 digits based on FFT and AGM (Ooura). So > its likely MMA uses lenghty precomputed values. While you are playing with that maybe you could help me with this which came up in a puzzle thread about pi on nz.general: >> Linkname: Currently publisized PI record (to be updated >> URL: http://www.super-computing.org/pi_current.html >>[...] >> (First digit '3' for pi or '0' for 1/pi is not included in the above count.) >>Frequency distribution for pi-3 up to 200,000,000,000 decimal places: >> '0' : 20000030841; '1' : 19999914711; '2' : 20000136978; '3' : 20000069393 >> '4' : 19999921691; '5' : 19999917053; '6' : 19999881515; '7' : 19999967594 >> '8' : 20000291044; '9' : 19999869180; Chi square = 8.09 >>Sorry, that was as far as the quote went, then it changes to me, for what >>it is worth: >>Order of frequencies: >>8,2,3,0,7,4,5,6,9 Sorry I missed the one. 8,2,3,0,7,4,5,1,6,9 Now the first odd number to occur, 3, is half the last even number, 6. Taken in pairs 8-2=6, 3-0=3, 7-4=3, 5-1=4, 6-9=-3. 4 and 6 are the strangers. (Though 6 being twice 3.) Then the next pairs 2-3=-1, 0-7=-7, 4-5=-1, 1-6=-5, 9-8=-1, with 7 & 5 strangers. So the strangers in those two procedures together are 4,5,6,7. 1 & 3 are the `commoners' in those procedures, leaving 0,2,8, which but for the interspersed 3 would also be the commonest digits in pi up to 200 billion places. >>with even numbers more represented, occurring 100000262069 times as >>opposed to odd numbers 99999737931 times >>0 to 4 occurring 100000073614 times in total >>5 to 9 occurring 99999926386 times in total >>And if my arithmetic is not too faulty the average of pi's digits >>(without the first digit (3)) so far >>comes to 4.499999410475 > Cool! And on that basis, I'd predict the expected value of pi's digits > in pairs to be 49.5. And by three digits to be 499.5. Or n digits the > expected value would be (10^n - 1)/2. Expressed as fractions of 10^n > that would be: > 1 digit -> 0.45 > 2 digits -> 0.495 > 3 digits -> 0.4995 > n digits -> 0.4999.....99995 (or near enough .5) > So the expected value of any infinitely long sub-string of pi (of > which there will be an infinite number of course) will be 0.5 > expressed as a fraction. >>But don't we have (10^200,000,000,000-1)/2, then, and only 4.4999995? > Yes, there is a flaw in the logic somewhere. But what? ==== When I using X=AB to solve the equation AX=B, where A and B are matrix, the computer generate the following mistake: ??? Error using ==> maple Error, (in expand/bigdiv/EuclideanNorm) integer too large in context Error in ==> C:MATLAB6p1toolboxsymbolic@symmldivide.m On line 25 ==> X = maple('linsolve',char(A),char(B),'''_rank'''); Is there anybody know what is the problem of this? How to solve this problem? Any except X=AB, X=inv(A)*B and X=linsolve(A,B), is there any methods to solve the symbolic linear equations? Many thanks!! ==== > Hallo, I wanted to know, whether my solution of the following task is right: > ### > task: > ------- > Examine for which t from |R the following system of equations has > solutions above |R and determine all solutions, if necessary. x_1 + 4x_2 - 2x_3 + 3x_4 = t > 3x_1 + 5x_2 + 2x_4 = 5 > 7x_2 - 6x_3 + 7x_4 = 13 my solution: > ------------------ > / 1 4 -2 | t-3s > | 3 5 0 | 5-2s | > 0 7 -6 | 13-7s / After several transformations we get: / 1 4 - 2 | t-3s > | 0 -7 -6 | 5+7s-3t | > 0 0 0 | 18-3t / => 18-3t = 0 => t = 6 / 1 4 -2 | 6-3s > 0 -7 -6 | 7s-13 / We choose x_3 as another free parameter k > (As we have done it with x_4 as s before.) > And get the following after some transformations: / 1 0 s + 38k/7 - 10/7 > 0 1 13/7 - s - 6k/7 / Therewith if t = 6 the set of solutions is: > |L = {(s + 38k/7 - 10/7; 13/7 - s - 6k/7)} > ### Is my approach right? Und are the results correct? > Karl. > For t = 6, only 2 of the equations are 'independent', and you get a two parameter set of solutions for the x_i, and the parameters can be any two of the x_i, but would customarily be x_3 and x_4. For t <> 6, the 3 equations are 'independent', and you get a one parameter set of solutions for the x_i, and the parameter can be any one of the x_i, but would customarily be x_4. ==== I have a computationally expensive fuction of three variables F(p1,p2,p3) which I need to minimize. This problem is solved with three-dimensional search algorithm in the cube with the parties p1 [min_p1.. max_p1]-p2 [min_p2.. max_p2]-p3 [min_p3.. max_p3]. A step of search on p1 - dp1, on p2 - dp2, p3 - dp3. How to compute in this case the confidence interval of the values received by search. If to take this interval as the half of step of digitization (for example), it will be constant value - independent of quantity of measurements. thanks in advice, Konst ==== >>Consider >> {(1+ r e^{it})^2: 0 <= r < infinity, 0 < t < Pi, t rational} >>in the complex plane. > Because any neighbourhood of 1 contains pieces of curves (for t near Pi) > in the lower half plane that only connect back to 1 by looping around 0. Oh, I didn't notice the parentheses. Ren.8e. -- Ren.8e Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China ==== >> 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. Isn't that subset locally connected in z=1? I mean all the rays with rational angle t have this point in common. Ren.8e. -- Ren.8e Meyer Student of Physics & Mathematics Zhejiang University, Hangzhou, China ==== Dear NG, Could someone here write the sketch of the proof of the (weaker) Bezout theorem, i.e. given two curves (say one being irreducible.. to make matters a bit easier) C1 and C2 defined by functions f1 and f2, the interesection between these two curves is less or equal deg(f1)*deg(f2). I have not so deep knowledge in algebraic geometry, my understanding is as far as the first few chapters of Shafarevich goes. If there is an easy proof of Bezout theorem I would be glad to see it (I have already seen the proof of the fact that C1 and C2 interesect finitely without using Bezout, so maybe that could be implemented). Sincerely, Jose Capco PS: I don't mind if the proof is on P^2 (ie projective plane of three coordinate, or extended projective plane of the affine plane of Bezout theorem on P^2, but unfortunately it made a lot of assumptions and referred a few books for the detailed proof (which I dont have). A sketch would suffice and I will work out the rest. ==== > Dear NG, Could someone here write the sketch of the proof of the (weaker) Bezout > theorem, i.e. given two curves (say one being > irreducible.. to make matters a bit easier) C1 and C2 defined by functions > f1 and f2, the interesection between these > two curves is less or equal deg(f1)*deg(f2). I'm not an algebraic geometer (far from it), but I would prove it by showing that the resolvant (or resultant) R(f1,f2) of f1(x,y),f2(x,y), regarded as polynomials in y over k[x], is a polynomial of degree <= mn in x. There is a minor issue of dealing with the case where 2 points of intersection have the same x-coordinate, but this is easily dealt with by applying an affine transformation. Similarly the projective case can be dealt with by applying a projective transformation so that the line at infinity does not contain any point of intersection. -- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 ==== says... >I'm not an algebraic geometer (far from it), >but I would prove it by showing that the resolvant (or resultant) R(f1,f2) >of f1(x,y),f2(x,y), regarded as polynomials in y over k[x], >is a polynomial of degree <= mn in x. > Ok, I am not well versed in the resolvent.. but let me verify if I understand this correctly. You are telling that if I work with f1 and f2 as functions of k(y)[x] (we had like k(y) to be a field, thus fraction field), and solve the resolvent we can arrive to the Bezout theorem. Am I correct that the resolvent determines the common roots of f1 and f2 (ie if the resolvent is zero at a point (x,y) then f1 and f2 are zero at those points). I just want to verify if Sincerely, Jose Capco ==== >For any non-zero cardinal A, is there a group of cardinality A? This cannot be proven in ZF, since there are models of ZF (without the Axiom of Choice) in which there exist cardinals A such that no group can have cardinality A (i.e. no set of cardinality A can be endowed with the structure of a group). David McAnally -------------- ==== > [EL] > Is that the entry of your CV? > Do not forget to include the rest of your fabulous qualities such as: > Rude. > Insolent. > Fraud. > Imaginative villain. And then you may add that you are ignorant and arrogant at the end of > the list of course. The big Surprise IS: > That you have been criticising a mainstream mathematical notation that > I COPIED from a college edition of Probability and Statistics by > Murray R. Spiegel who is Ph. D. Professor and Chairman of Mathematics > at Rensselaer Polytechnic Institute of Connecticut. > So call him, and take it out with him, mother-fucker. I am expecting you to disappear in shame if there was one drop of > blood remaining in your face. > You can pretend all you wish to be the all knowing and fool the fools > who subscribe to your fraudulence. > But this time you swallowed the bait and the line whole. Goodbye Jerk. > :-) > EL > Not a ghost of a chance of a red face, EL. > Dinky van der Mumble is the original Troll and useless idiot that preys on > unsuspecting intellectuals as they crudely try out their ideas on s.p.r, > prior to committing to a paper. After all, it is better to have someone else > check your real errors before you get egg on your face. > Dinky jeers and sneers at everything, publishing what he considers as > errors. I've even set him up with V = (c+v)/(1-cv/c^2) and he took the bait > hook line and sinker, highlighting the minus sign and publishing. This is > because I caught his goof in his own equation, which he shrugged off as a > typo. > Anyone else reading will know, if they are reasonable intelligent, so I get > to have my work published on Dinky's site so long as I include an obvious > error. > That's quite nice, really. The only problem is, nobody remotely intelligent > will bother reading Dinky's site. > Androcles [EL] > :) > If you see horror movies you can read Dirk's posts, they could be > both, horrifically comic, bloody and full of gore. > Now this is not a joke so read on. > A dirk is a dagger especially designed for backstabbing; seriously, > and if you do not believe me look up a dictionary. LOL! I'm English and live in Britain, EL. I need no dictionary, I've known what a dirk is since I learned to read. > France for the ancient emigrants that came to Europe through Morocco, > and they were mixed Arabs and Berber who invaded Spain in the 8th > century. > Those ancient invaders settled in Moor of Spain and those emigrants to > France were Moorish alchemists who gained some money and respect due > to the knowledge of alchemy. However, the Moortels are a special branch of the family that was > banished for black magic practices and for their bloodlust and evil > manners. Here we have a yet special offspring that was named as Dirk to > define his destiny to be a backstabbing Evil Moorish mother fucker. So no matter how much his black-magic knowledge is, he is still a > lowly creature and a parasite. He is a typical hopeless case that has no cure else than a heap of > straw and a high cross with angry men carrying flames. He knows it, > and he must be expecting to be flamed all the time. EL LOL, yep! Except I just cannot be bothered, it is easier to kill file-him and let him rant. He'll go on doing it anyway. What he really wants is attention, to be noticed and recognized as a follower of relativity, and attempts to display his loyalty in the hope of catching a bone of gratitude, thrown his way occasionally by some other devout follower of his faith. I rather think that ignoring him inflames him even more than insulting him. Androcles ==== > LOL, yep! Except I just cannot be bothered, it is easier to kill file-him > and let him rant. He'll go on doing it anyway. What he really wants is > attention, to be noticed and recognized as a follower of relativity, and > attempts to display his loyalty in the hope of catching a bone of gratitude, > thrown his way occasionally by some other devout follower of his faith. I > rather think that ignoring him inflames him even more than insulting him. > Androcles [EL] Consider it done. :) EL ==== [EL] > Is that the entry of your CV? > Do not forget to include the rest of your fabulous qualities such as: > Rude. > Insolent. > Fraud. > Imaginative villain. And then you may add that you are ignorant and arrogant at the end of > the list of course. The big Surprise IS: > That you have been criticising a mainstream mathematical notation that > I COPIED from a college edition of Probability and Statistics by > Murray R. Spiegel who is Ph. D. Professor and Chairman of Mathematics > at Rensselaer Polytechnic Institute of Connecticut. > So call him, and take it out with him, mother-fucker. I am expecting you to disappear in shame if there was one drop of > blood remaining in your face. > You can pretend all you wish to be the all knowing and fool the fools > who subscribe to your fraudulence. > But this time you swallowed the bait and the line whole. Goodbye Jerk. > :-) > EL > Not a ghost of a chance of a red face, EL. > Dinky van der Mumble is the original Troll and useless idiot that preys on > unsuspecting intellectuals as they crudely try out their ideas on s.p.r, > prior to committing to a paper. After all, it is better to have someone else > check your real errors before you get egg on your face. > Dinky jeers and sneers at everything, publishing what he considers as > errors. I've even set him up with V = (c+v)/(1-cv/c^2) and he took the bait > hook line and sinker, highlighting the minus sign and publishing. This is > because I caught his goof in his own equation, which he shrugged off as a > typo. > Anyone else reading will know, if they are reasonable intelligent, so I get > to have my work published on Dinky's site so long as I include an obvious > error. Let's hope they are *reasonably* intelligent indeed: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LoadCrap.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CrapHuh.html What a poor loser :-) Dirk Vdm ==== [snip] Reproduce the double maxima of the nova, or give a > simple explanation for it, using your wave model. My > explanation is a simple as it can be. > Employ your own fully fledged loyalty to wave theory, > and research it yourself. Nobody is stopping you. > There is no scientific terrorism. Just prejudice. We've > always had that. > Androcles [EL] > I visited your web-page and do not see any equations > or data tables. > Those curves mean nothing since you failed to include > any explanation. > Please help me to conclude that I did not waste my time. El, I think the time has come for you to use > your secret Lady Sanity weapon on Androcles. > I'm sure he will appreciate it. > Good luck ;-) Dirk Vdm [EL] > Dirk, I have no secret weapons, Indeed, it isn't much of a secret anymore. > Let's call it *silly* weapons, okay? but I am beginning to understand the > motivations behind your Fumble's collection. :) Bravo, you are a fast learner. > You have read the top two lines of the page, right? > Keywords: Ignorance and Arrogance. > [EL] > Is that the entry of your CV? > Do not forget to include the rest of your fabulous qualities such as: > Rude. > Insolent. > Fraud. > Imaginative villain. And then you may add that you are ignorant and arrogant at the end of > the list of course. The big Surprise IS: > That you have been criticising a mainstream mathematical notation that > I COPIED from a college edition of Probability and Statistics by > Murray R. Spiegel who is Ph. D. Professor and Chairman of Mathematics > at Rensselaer Polytechnic Institute of Connecticut. HAHAHAHAHAHAHAHAHAHAHA! That must be a real ignorant Ph. D. Professor and Chairman of Mathematics then. Probably not nearly as arrogant as you, so I won't hold anything against him :-) So, you are lying, calling authority, misquoting, and still having no idea what you are talking about. Dirk Vdm > So call him, and take it out with him, mother-fucker. I am expecting you to disappear in shame if there was one drop of > blood remaining in your face. > You can pretend all you wish to be the all knowing and fool the fools > who subscribe to your fraudulence. > But this time you swallowed the bait and the line whole. And you included sci.math. Brilliant :-) Here you go then, since you *really* insist: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/MolBioBra2.html > Goodbye Jerk. > :-) > EL Dirk Vdm ==== [snip] Reproduce the double maxima of the nova, or give a > simple explanation for it, using your wave model. My > explanation is a simple as it can be. > Employ your own fully fledged loyalty to wave theory, > and research it yourself. Nobody is stopping you. > There is no scientific terrorism. Just prejudice. We've > always had that. > Androcles [EL] > I visited your web-page and do not see any equations > or data tables. > Those curves mean nothing since you failed to include > any explanation. > Please help me to conclude that I did not waste my time. El, I think the time has come for you to use > your secret Lady Sanity weapon on Androcles. > I'm sure he will appreciate it. > Good luck ;-) Dirk Vdm [EL] > Dirk, I have no secret weapons, Indeed, it isn't much of a secret anymore. > Let's call it *silly* weapons, okay? but I am beginning to understand the > motivations behind your Fumble's collection. :) Bravo, you are a fast learner. > You have read the top two lines of the page, right? > Keywords: Ignorance and Arrogance. > [EL] > Is that the entry of your CV? > Do not forget to include the rest of your fabulous qualities such as: > Rude. > Insolent. > Fraud. > Imaginative villain. And then you may add that you are ignorant and arrogant at the end of > the list of course. The big Surprise IS: > That you have been criticising a mainstream mathematical notation that > I COPIED from a college edition of Probability and Statistics by > Murray R. Spiegel who is Ph. D. Professor and Chairman of Mathematics > at Rensselaer Polytechnic Institute of Connecticut. HAHAHAHAHAHAHAHAHAHAHA! > That must be a real ignorant Ph. D. Professor and Chairman > of Mathematics then. Probably not nearly as arrogant as you, > so I won't hold anything against him :-) > So, you are lying, calling authority, misquoting, and still having > no idea what you are talking about. Dirk Vdm [EL] I am only delighted to show *all the readers* your participation in a practical definition of ignorance and arrogance combined. You are an ignorant insect, who has absolutely no value, yet you accuse Professor Murray R. Spiegel whom I *copied* to the letter, his notation published in authentic publications for education, and you accuse him of being ignorant! How more arrogant can you be! You are the ultimate fumble of all fumbles and in your hole that you dug you stumble. I gave the readers an authentic name that they can verify any time and I gave the readers the position and rank and address of work of that authority of course. Yes I call authority because you are an arrogant ignorant arsehole. I did not lie at all and I did not misquote. So, my word is against yours, while I am an academic emeritus and you are nobody. > So call him, and take it out with him, mother-fucker. I am expecting you to disappear in shame if there was one drop of > blood remaining in your face. > You can pretend all you wish to be the all knowing and fool the fools > who subscribe to your fraudulence. > But this time you swallowed the bait and the line whole. And you included sci.math. Brilliant :-) [EL] Indeed I did, because I have nothing to hide. I wanted to show all the readers, *who* is the arrogant ignorant arsehole. Naturally, I do have grate confidence in a highly reputed colleague Professor such as Professor Murray R. Spiegel, while I have zero confidence in a pest like yourself. > Here you go then, since you *really* insist: > http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/MolBioBra2.html [EL] Oh my! If it only had any significance I might have celebrated, but unfortunately that URL points to an astrologist / numerologist who descends from taro card and tealeaves' readers' family {{{ {{{ *_* Sterling Van De Moortel Sterling coaches people to invent their lives through their language? motto is: See it; say it; let it be so! *_* Boris Van de Moortel http://ourworld.compuserve.com/homepages/koenvandemoortel/borisvdm.htm *_* Koen Van de Moortel & Dragana Van de Moortel http://ourworld.compuserve.com/homepages/koenvandemoortel/homepage.htm *_* Immigrants: Moortel http://www.geocities.com/Heartland/Flats/8762/page145.htm }}} Notice that many of the Moortels ended up in Amsterdam as astrologists and Taro Card Readers. Our Noble Dirk is using cyber-space to create an illusion of superiority to live within while he is quite aware of his origins and his chaotic fate of being nobody. }}} > Goodbye Jerk. > :-) > EL Dirk Vdm ==== [snip] > The big Surprise IS: > That you have been criticising a mainstream mathematical notation that > I COPIED from a college edition of Probability and Statistics by > Murray R. Spiegel who is Ph. D. Professor and Chairman of Mathematics > at Rensselaer Polytechnic Institute of Connecticut. HAHAHAHAHAHAHAHAHAHAHA! > That must be a real ignorant Ph. D. Professor and Chairman > of Mathematics then. Probably not nearly as arrogant as you, > so I won't hold anything against him :-) > So, you are lying, calling authority, misquoting, and still having > no idea what you are talking about. Dirk Vdm [EL] > I am only delighted to show *all the readers* your participation in a > practical definition of ignorance and arrogance combined. > You are an ignorant insect, who has absolutely no value, yet you > accuse Professor Murray R. Spiegel whom I *copied* to the letter, his > notation published in authentic publications for education, and you > accuse him of being ignorant! > How more arrogant can you be! > You are the ultimate fumble of all fumbles and in your hole that you > dug you stumble. I gave the readers an authentic name that they can verify any time and > I gave the readers the position and rank and address of work of that > authority of course. > Yes I call authority because you are an arrogant ignorant arsehole. > I did not lie at all and I did not misquote. HAHAHAHAHAHAHAHAHAHA! > So, my word is against yours, while I am an academic emeritus and you > are nobody. Hehe, another appeal to authority? Or did you mean inferiority? Besides, *your* word against mine? What do you think you can still buy with *your* word after this: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/LadySanity.html I don't need *my* words - these are *yours* ;-) I bet you were the poorest loser of the family when you were a kid. In order to feel good and win a game, you even cheated on your little brother, didn't you? Dirk Vdm ==== In a group G, we can define two series of characteristic subgroups. The upper central series: 1 < Z[1] < Z[2] .... where Z[i+1]/[i] is the center of G/Z[i] and the lower central series: G > L[1] > L[2] > ... where L[i+1] is the subgroup generated by commutators of the form (x,y) with x in G and y in L[i]. I believe (?) I've managed to prove that: If either series is finite, so is the other, and both series have the same length (this allows a nilpotent group to be defined using either series). Each group of the L series is contained in the corresponding group of the Z series (but not in the one immediately before it). My (obvious questions) are : * Is this correct ? * If both series are finite, are they identical ? ----------------------------------------------------------- Jacques Willekens ==== > In a group G, we can define two series of characteristic subgroups. The upper central series: 1 < Z[1] < Z[2] .... where Z[i+1]/[i] is the center of G/Z[i] You m ean Z[i+1]/Z[i]. > and the lower central series: G > L[1] > L[2] > ... where L[i+1] is the subgroup generated by commutators of the form (x,y) with > x in G and y in L[i]. And L[0]=G... You might want to be careful there... Some authors have L[1]=G. For example, I am used to a group being nilpotent of class k if and only if L[k+1] is trivial, (which means L[1]=G, L[2] is the commutator subgroup, etc). > I believe (?) I've managed to prove that: If either series is finite, so is the other, and both series have the same > length (this allows a nilpotent group to be defined using either series). Correct. > Each group of the L series is contained in the corresponding group of the Z > series (but not in the one immediately before it). Depends on what you mean by corresponding. For example, in a nilpotent group of class 3, the third term of the lower central series (the subgroup ((G,G),G)) is contained in the first center; the second term in the second center, and the first term in the third center. So, presumably you mean that if both series are of length n, then L[i] is contained in Z[n-i+1], or some similar thing... This follows from the general property that leads to naming the first series upper central series and the second lower central series. In general, the L[i] will be contained in the corresponding term of any central series, and the terms of any central series will be contained in the corresponding Z[i]. > My (obvious questions) are : * Is this correct ? Yes. Well, the result is. Don't know about your particular proof. (-: > * If both series are finite, are they identical ? Certainly not. Take any group G, nilpotent of class k>1, and take G+C, where C is cyclic and finite. Then the lower central series of G+C gives you the same subgroups as the lower central series of G, but the upper central series starts with a larger center, equal to Z(G)+C. So, even if they were identical for G, they would not be identical for G+C. If you want a different example, consider the upper triangular 3x3 matrices with 1's in the diagonal, whose (1,2) and (2,3) entries are integer modulo p^2, and whose (1,3) entry is an integer modulo p. The commutator subgroup is generated by all matrices with 0's in the (1,2) and (2,3) entries, but the center consists of all matrices whose (1,2) and (2,3) entries are 0 modulo p. Arturo Magidin, sans .sig ==== >In a group G, we can define two series of characteristic subgroups. The upper central series: 1 < Z[1] < Z[2] .... where Z[i+1]/[i] is the center of G/Z[i] and the lower central series: G > L[1] > L[2] > ... where L[i+1] is the subgroup generated by commutators of the form (x,y) with >x in G and y in L[i]. I believe (?) I've managed to prove that: If either series is finite, so is the other, and both series have the same >length (this allows a nilpotent group to be defined using either series). Each group of the L series is contained in the corresponding group of the Z >series (but not in the one immediately before it). My (obvious questions) are : * Is this correct ? Yes - you can find all of this in many books on Group Theory under the general theory of nilpotent groups. >* If both series are finite, are they identical ? Not necessarily. Try the direct product of the dihedral group of order 8 and a cyclic group of order 2. Derek Holt. ==== etc..... Its been a while since I laughed this hard. What?! Who gave you the authority to laugh? ==== crap] > And every word you utter simply continues to confirm what all the readers of > this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being domindated by ignorant irritating people. These people force decent people of the web due to their slimey nature. I.e. they flame them off the web. That means that jerks like you remain behind and fill it up. So I expect that all newsgroups are fill with jerks like you, varney, bilgem, speicher etc. And when I say jerk I'm being *very* gracious. ==== > crap] And every word you utter simply continues to confirm what all the readers of > this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being > domindated by ignorant irritating people. These people force decent > people of the web due to their slimey nature. I.e. they flame them off > the web. That means that jerks like you remain behind and fill it up. Sorta like saying only really stupid people post on the net, in a post on the net... So I expect that all newsgroups are fill with jerks like you, varney, > bilgem, speicher etc. And when I say jerk I'm being *very* gracious. ==== > crap] And every word you utter simply continues to confirm what all the readers of > this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being > domindated by ignorant irritating people. These people force decent > people of the web due to their slimey nature. I.e. they flame them off > the web. That means that jerks like you remain behind and fill it up. So I expect that all newsgroups are fill with jerks like you, varney, > bilgem, speicher etc. And when I say jerk I'm being *very* gracious. Useless troll. Waste of time. Plonk. ==== crap] And every word you utter simply continues to confirm what all the > readers of > this ng think of you. Do you really not realise that? Wow. What an ignont comment. The internet is well known for being > domindated by ignorant irritating people. These people force decent > people of the web due to their slimey nature. I.e. they flame them off > the web. That means that jerks like you remain behind and fill it up. So I expect that all newsgroups are fill with jerks like you, varney, > bilgem, speicher etc. And when I say jerk I'm being *very* gracious. > Useless troll. > Waste of time. Plonk. Any reason that you think you need to butt into a conversation between two people? And if plonk means that you've killfiled me - thank you. Otherwise please do so. Pmb ====