mm-413 === Subject: Re: A (Non)Prime Conjecture I have reworked my original paper ( http://www.tln.net/~reriker/prime.h ) based on the feedback I have received. I am not sure that I have demonstrated it in a rigorous enough manner, but I have attempted to demonstrate that it is possible to generate all of the nonprime numbers. Once you have generated these, any gaps left are the prime numbers. Perhaps this is too unwieldy to be of any general use. However, perhaps it is of some interest (assuming, of course, that I have not made any blatant errors ; of course, if I have, I am sure that someone will point them out . > One way to approach the prime numbers is to look at a number and see > if it has any factors other than itself. Another way to approach the > problem is to look at all of the numbers that are NOT prime and > conclude that anything left over IS prime. This latter approach is the > basis of my conjecture. > See http://tln.net/~reriker/prime.h for more details. === Subject: Re: a/b+b/c+c/a=n > Except for n=142 and n=177, we have explicit solutions. (which is correct) and > except for the possibility that the values n=147 and n=177 must be added. (which contains a typo). The curve in question is the one for n=142. The minimal model has coefficients [0,1,1,-8471199,9484296625] (i.e it's y^2+y = x^3+x^2-8471199x+9484296625), and along the way to proving the rank is one, MWRANK identified just five (equivalent) homogeneous spaces which carry the generator: (5,4,6784,477604,6105436) (23,6,-9608,7586,1139277) (79,138,-7136,22622,385085) (1997,-2532,-8696,6260,11828) (2957,-3130,-8696,5064,7988) That is, it's sufficient to be able to solve at least one equation like 5 u^4 + 4 u^3v + 6784 u^2v^2 + 477604 uv^3 + 6105436 v^4 = z^2 None of the five has a solution with |u|,|v| < exp(13); for each of the five I currently have a machine looking for solutions with |u|,|v| < exp(14); no luck after about a week's CPU-power. If you'd like to check through height 15 or more, let me know. I have more homogenous spaces for 177, also being searched as we speak. (Its minimal form is [1,-1,0,-20448843,35591471693] in standard notation.) These are rank-1 curves and so they're amenable to Heegner-point computations, but I don't know how to do those. === Subject: Re: A problem from fundamental counting principle Spare me your suspicion, mister! They are NOT homework problem. Even if they are, if you CAN answer by giving real solution, then you are helping. What's wrong with giving the homework problem in this forum? If one presenting his homework problem, it means he has interest. Giving out the solution will stimulate his thinking. Anyway, this is my case. So, if anyone can give me the solution for the problem, fine. Actually I have already the solution, but I need some confirmation on several === Subject: Re: Axiomization of Number Theory |>> The (first-order) Peano axioms do not do the |>> job. |> |>Then take second-order :-) | |Right, why didn't I think of that. | |I've never been able to decide whether that should count. |The thing is, I'm not sure whether I'm badly confused about |second-order logic here - maybe someone who knows can |say: So you suffer from second-order confusion! I hope I can complement Oliver's explanation. |On the one hand it seems clear that the second-order PA |axioms characterize N. On the other hand it seems clear |that we can do second-order arithmetic in first-order |set theory... (my guess is that the point is that we're talking |about two different notions of second-order arithmetic |here, but when I assume that then I get all confused |over exactly what notion the first sentence refers to.) Second-order logic defines a satisfaction relationship between sentences in it and models. It involves quantifying over all subsets of the domain of the model, so for the relationship to make any sense, one has to be in a context where talking about all subsets makes sense. If you're going to be uneasy about the meaning of second-order axioms, this is the point at which to be uneasy. Then we have some associated concepts such as validity. A sentence in second order logic is valid if it holds in all models having the constants and relations mentioned in it. A sentence is categorical for a model if that model is the only model satisfying the sentence. The one thing we do NOT have is a formal system. The logical deduction relation isn't defined in terms of a concept of proof, let alone of mechanical verification of proofs. So second order logic isn't doing the same kind of job as formal systems are doing. In first-order logic, we have a similar set-up, but with the fortuitous advantage of having a complete axiomatization, in the sense that every valid statement can be proven just by following a certain set of mechanical rules (in principle). When we do second order logic in first-order set theory, we translate statements like the validity of a second-order deduction into first order language, without necessarily changing the intended meaning. We also pick out some facts in that intended meaning (of set theory) to use as axioms. But now we've produced a formal system, which generates *certain* of the facts about second order logic, but not all of them. This is not a new difficulty, since there wasn't a complete formal system for second order logic to begin with. As Oliver explained, this incompleteness (in spite of the completeness of first-order logic) corresponds to the existence of models of the first- order axioms for set theory which are incorrect. The first order axioms may have an intended meaning, but that doesn't mean that their models reflect the intended meaning. I may be thinking of epsilon as representing set membership, but the epsilon of a model of an axiom system for set theory might be some other relationship that just looks somewhat like it. Especially, by saying that there exists an x we might have in mind that there exists a set x, but the model doesn't contain all sets. I suspect I've already said as much as I had to say that was worth saying. Keith Ramsay === Subject: Re: Axiomization of Number Theory ># 1. Could you provide formal axiomatic proofs, rather than English >descriptions? For three of your first four, here's where you can find formal proofs. > 1. If A is a factor of B and B is a factor of A, then A=B. > Prop 23 in www.andrewboucher.com/papers/foea/VI1.pdf > 2. If A is a factor of B and B is a factor of C, then A is a factor of C. > Prop 24 in same paper > 4. There are an infinite number of prime numbers. > Prop 21 in www.andrewboucher.com/papers/foea/VII4.pdf I see a whole lot of undefined symbols and no explanation of the axioms and rules of inference in the above references. Perhaps you can simply give the axioms, rules, and proof here? However, 1 and 2 are so simple, only 4 is of interest at this point. You know, this idea of people saying See reference xyz and when you look you don't really find what you were talking about and then when you say Can you just give your result here? they call you lazy or something, isn't a new one. So, you're not going to try that old trick and call me lazy and refuse to actually give the axioms, rules and proof here, are you? Because, honestly, I don't see that you axiomized it in your papers, though, as I said, we are only talking about the first, simplest Theorems that I figured weren't that hard (about the most Peano Arithmetic can do) anyway. :) > (and incidentally > the last proposition, after yikes! 1800 pages of proof). Bet I can reduce it by a factor of 1800 (with the right axiomization). > At some point I may continue further. Hard cash would help. ;-) Darn, just when you got to the hard ones. Name your price. How about $1,000.00 if you prove the other theorems in Peano Arithmetic? ># 2. I listed the theorems in order of complexity, and you stopped >after the first few (simplest) ones. This is as I described: Peano >Arithmetic allows for the derivation of only the simplest of theorems. > Could you address the more complex theorems given later in the list? How would you even represent the statement N is a perfect number. in >Peano Arithmetic? > n is a perfect number can be translated as Sum{x : x | n} = n where Sum(P) refers to the x in (hope I got this right): [m][R][S] ( Seq(R,m,P) & Seq0(S,m) & S'm = x & S'0 = 0 & (i)(i < m =S'(i+1) = S'i + R'(i+1)) ). > Here: > [..] means there exists .. > Seq(R,m,P) means R is a sequence from {x : 1 <= x <= m} onto P > Seq0(S,m) means S is a sequence from {x : 0 <= x <= m} > S'z refers to the y for which Sz,y > The last two are exercises to define in second-order arithmetic. Sum(x:x|n)? Is that in Peano Arithmetic? Are you confusing the Predicate Calculus with some Programming Language? And do your rules apply to Sum(x:x|n)? I mean, if you're going to throw in arbitrary syntax and semantics, your rules will have to likewise be modified for this non-Predicate Calculus expression. What happened to the original Peano Arithmetic can do it? I said What is the formal wff? next just in case someone decided to drift away from Predicate Calculus and throw in some programming subroutine. Do you really maintain that all of the above is a Peano Arithmetic wff? Could we just stick to ^ v => ~ exists forall and relations, you know, Peano Arithmetic wffs? (I failed your two exercises, so I guess you'll just have to give a complete answer.) > So it's not difficult at all in PA2 (or a near-enough version of it). > Maybe you were thinking of PA1. I'll take either, as long as it doesn't make up new stuff along the way. >What is the formal wff? More fundamental than the >question of undecidable wffs being few and far between and the system >being sound (suggesting that true wffs are usually provable), is the >question of whether Peano Arithmetic can even represent the theorems >to be proved in the first place. In my eyes anyway, second-order PA does a good job of representing the > theorems. I'll believe it when I see it. >The Theory of Computation is axiomized with 3 axioms: TRUE(x) [the >universal set is recursively enumerable], YIT(I,J,K) [the >proof/halting YES predicate is recursive] and a single Incompleteness >Axiom -~YES(x,x) [The set of programs that do not halt YES on >themselves is not recursively enumerable.], which I actually prove by >going slightly outside of the system. > Cheater ! I'm supposed to prove the axioms within the system?? I only mentioned that one of the 3 axioms is almost a theorem as an aside. It's an axiom, ok? > Seriously, I don't think any of these three axioms are low enough > I certainly don't understand why I should accept them as true. Since when do we prove axioms? But if you insist, then tell me again - you doubt that: 1. The set of natural numbers is recursively enumerable? 2. The proof/halting YES predicate (Program A halts YES on input B at iteration C.) is recursive? 3. The set of programs that do not halt YES on themselves is not recursively enumerable? Honestly (no offense), but you seem to be grabbing at certain phrases slightly outside of the system was in the context of mentioning that one of the axioms is almost provable within the system (just an aside). The axioms are of course well-known facts. In each case, if you look at what I actually said, there's certainly no problem with these assertions. Can we just agree to: 1. No references without giving the result here as well (just in case one doesn't really find the result in the cited paper.) 2. No exercises. 3. No wffs that contain English. 4. No wffs that go outside of Peano Arithmetic (Predicate Calculus) syntax (unless you want to give a completely new system, which is ok if that is your intent, but then you need to say that and define the whole thing.) 5. No demanding money to give the answers. In other words, can we just stick to formal axioms, rules and proofs? BTW: I've been a reader of your internet publications for years. I may have even written you a fan mail sometime in the past! >Charlie Volkstorf >Cambridge, MA >axiomize at aol dot com === Subject: Re: Axiomization of Number Theory Hello all, >I am interested in axiomizing Number Theory. >Didn't Godel prove that that isn't possible? And din't Peano actually do it? >>[...] >>The (first-order) Peano axioms do not do the >>job. >Then take second-order :-) > Right, why didn't I think of that. > I've never been able to decide whether that should count. > The thing is, I'm not sure whether I'm badly confused about > second-order logic here - maybe someone who knows can > say: > On the one hand it seems clear that the second-order PA > axioms characterize N. On the other hand it seems clear > that we can do second-order arithmetic in first-order > set theory... (my guess is that the point is that we're talking > about two different notions of second-order arithmetic > here, but when I assume that then I get all confused > over exactly what notion the first sentence refers to.) You don't even need full-blown set theory. You can just turn the second order system into a multi-sorted first order one. Or if you want your ordinary single-sorted first order case, do the following Assume that L is a second order language and T is a theory in it. Assume you have some system of deduction for second order logic. Obviously, this system isn't complete, but assume it's sound with respect to the standard semantics. Assume also that it's effective. Define first a first order language L' from L by replacing the second order constants with first order constants and by adding an unary predicate Object and for every n in N add a predicate Predicate-of-arity-n and a predicate Applies-n. second order quantification by quantification restricted to the new predicate Predicate-of-arity-n. Replace all occurances of P(t 1,...,t n) (where P is either a predicate variable or constant and t i are terms) with Applies-n(P,t 1,...,t n). In addition, add to T' all the rules of inference pertaining predicate existence and application you have in the deductive system you're using for second order logic. These most likely include comprehension: for all formulae phi(x 1,...,x n), EP( Predicate(P) & Ax 1,...,x n(Applies-n(P,x 1,...,x n) <--> phi(x 1,...,x n)), choice and what not. Also, you'll have something like Applies-n(P,x 1,...,x n) --> Object(x 1) & Object(x 2) ... & Predicate-n(P). It should be obvious that whatever you can deduce with the second order deduction system can be done in this new system. Semantically, the situation is much worse. It's obvious that there are models M of T' for which the structure <{x | M |= Predicate-1(x) },Applies-1> is not isomorphic , as it should, if we want the standard second order semantics (reinterpreted on the first order level)! So while this sort of translation preserves deduction, it does not preserve semantics (in the sense that all models of T would be - modulo trivial details - isomorphic to models of T'). The fact that second order PA is categorical is due to semantics of second order logic, not its deductive structure (for which there is no canonical choice). Thus, in the above delineated sense whatever one can do in second order arithmetic can be done in a suitable extended first order arithmetic. This only holds if by doing we only mean deriving theorems from the axioms. Often we are not as intersted in this sort of properties but wish to concentrate on what can be correctly *expressed* in a language or logic. Which is exactly what e.g. the theory of model theoretic logics studies. As a sidenote, G.9adel first carried out the incompleteness results for omega-order PA, i.e. PA in the unramified type theory of PM. He was able to do this because he - in effect - considered omega-order PA as a first order system of the type described above. (You need to use a predicate Predicate-of-arity-n-of-order-k for all n,k and have axioms guaranteeing the type restrictions). === Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Axiomization of Number Theory > Since what is true but unprovable depends on the coding scheme (in > Godel's proof), it might be a question of choosing a system where we > avoid including our intended theorems (Number Theory) among the > undecidable sentences. But the coding scheme is quite arbitrary (with > simple requirements), pretty independent of the axiomatic system > itself. It doesn't depend on a single coding scheme, but on *all* coding schemes (for which a G.9adel type proof can be carried out). === Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Axiomization of Number Theory > When http:// is prefixed to address news browser recognizes it has web > link, highlights it and will open URL if clicked upon. > I'm not sure what you mean, but anyone who is interested is invited to > click on www.arxiv.org/h/cs.lo/0003071 Does you browser recognize that as a web address? > Mine and others don't as you didn't include http:// > nothing appears to click on and it sits there as an inert > ascii text going nowhere, doing nothing, meriting dismissal. Oooooooooooooooooooooooohhhhhhhhhhhhhhhhhhhhh === Subject: Re: Axiomization of Number Theory >Since what is true but unprovable depends on the coding scheme (in >Godel's proof), it might be a question of choosing a system where we >avoid including our intended theorems (Number Theory) among the >undecidable sentences. But the coding scheme is quite arbitrary (with >simple requirements), pretty independent of the axiomatic system >itself. > This is not right. . . . the set of true statements does not depend on > the coding (only on the intended model), Of course. Did I say otherwise? (The coding scheme is independent of the system, as I said.) > and the set of provable statements only > depends on the formal language and the axioms (and the underlying logic). Also of course. I guess I should have said The undecidable wff THAT WE BECOME AWARE OF due to Godel's proof depends on the coding scheme. > Michel. Charlie Volkstorf Cambridge, MA === Subject: Re: Axiomization of Number Theory > Since what is true but unprovable depends on the coding scheme (in > Godel's proof), it might be a question of choosing a system where we > avoid including our intended theorems (Number Theory) among the > undecidable sentences. But the coding scheme is quite arbitrary (with > simple requirements), pretty independent of the axiomatic system > itself. > It doesn't depend on a single coding scheme, but on *all* coding schemes > (for which a G del type proof can be carried out). Well, having a one-track mind, I only know how to consider one coding scheme at a time, myself. Charlie Volkstorf Cambridge, MA === Subject: Re: Axiomization of Number Theory > I've never been able to decide whether [saying arithmetic > is axiomatized by second-order PA] should count. > The thing is, I'm not sure whether I'm badly confused about > second-order logic here - maybe someone who knows can > say: On the one hand it seems clear that the second-order PA > axioms characterize N. On the other hand it seems clear > that we can do second-order arithmetic in first-order > set theory... (my guess is that the point is that we're talking > about two different notions of second-order arithmetic > here, but when I assume that then I get all confused > over exactly what notion the first sentence refers to.) >I don't see that you've asked any explicit question here, so I >have to kind of guess, but I suppose what you're asking is >something like ... so how come ZFC doesn't decide the >truth value of all sentences of arithmetic?. Yes, more or less. >One way of answering that is to note that not all first-order >models of ZFC correctly interpret the second-order semantics >of the natural numbers, because certain subsets of the natural >numbers of the model may not be elements of the model. (Here >perhaps elements of should be in quotes -- what I really mean is: >Suppose N^M is the set of all things that M believes to be natural numbers, >and suppose S is a subset of N^M, then there may be no object S^M >in M such that, for each element e of S, M thinks e is in S^M). >In fact, if N^M is not isomorphic to the real N, then N^M must >be illfounded, and its wellfounded part is a subset of N^M >that is not in M. Do you agree that nonetheless second-order PA _does_ decide the truth value of all sentences of arithmetic? And, sort of getting to my real question: given that models of set theory can be confused about what subsets of N exist (in a sense described more precisely above) do you see anything fuzzy or confusing about exactly what second-order PA actually means? === Subject: Re: Axiomization of Number Theory >|>> The (first-order) Peano axioms do not do the >|>> job. >||>Then take second-order :-) >| >|Right, why didn't I think of that. >| >|I've never been able to decide whether that should count. >|The thing is, I'm not sure whether I'm badly confused about >|second-order logic here - maybe someone who knows can >|say: >So you suffer from second-order confusion! I hope I can complement >Oliver's explanation. >|On the one hand it seems clear that the second-order PA >|axioms characterize N. On the other hand it seems clear >|that we can do second-order arithmetic in first-order >|set theory... (my guess is that the point is that we're talking >|about two different notions of second-order arithmetic >|here, but when I assume that then I get all confused >|over exactly what notion the first sentence refers to.) >Second-order logic defines a satisfaction relationship between sentences >in it and models. It involves quantifying over all subsets of the domain >of the model, so for the relationship to make any sense, one has to be in >a context where talking about all subsets makes sense. If you're going to >be uneasy about the meaning of second-order axioms, this is the point at >which to be uneasy. >Then we have some associated concepts such as validity. A sentence in >second order logic is valid if it holds in all models having the constants >and relations mentioned in it. A sentence is categorical for a model if >that model is the only model satisfying the sentence. >The one thing we do NOT have is a formal system. I really had no idea whether there existed formal systems characterizing validity in second-order logic the way standard systems do for FOL. If not then this gives the answer to Chapman's question So why not use second-order PA?. Answer: because that doesn't give us a formal proof system, which is exactly what we're looking for when we're trying to axiomatize arithmetic. >The logical deduction >relation isn't defined in terms of a concept of proof, let alone of >mechanical verification of proofs. So second order logic isn't doing the >same kind of job as formal systems are doing. In first-order logic, we >have a similar set-up, but with the fortuitous advantage of having a >complete axiomatization, in the sense that every valid statement can be >proven just by following a certain set of mechanical rules (in principle). >When we do second order logic in first-order set theory, we translate >statements like the validity of a second-order deduction into first order >language, without necessarily changing the intended meaning. We also pick >out some facts in that intended meaning (of set theory) to use as axioms. >But now we've produced a formal system, which generates *certain* of the >facts about second order logic, but not all of them. This is not a new >difficulty, since there wasn't a complete formal system for second order >logic to begin with. >As Oliver explained, this incompleteness (in spite of the completeness >of first-order logic) corresponds to the existence of models of the first- >order axioms for set theory which are incorrect. The first order axioms may >have an intended meaning, but that doesn't mean that their models reflect >the intended meaning. I may be thinking of epsilon as representing set >membership, but the epsilon of a model of an axiom system for set theory >might be some other relationship that just looks somewhat like it. >Especially, by saying that there exists an x we might have in mind that >there exists a set x, but the model doesn't contain all sets. >I suspect I've already said as much as I had to say that was worth saying. Was definitely worth saying, though. Oliver points out that he wasn't quite sure exactly what my question was - the reason I didn't state an explicit question is that I really knew nothing here, so a list of the explicit questions I had would not fit in the margin. Your comment about formal systems answers one of my questions, and a very basic important one. >Keith Ramsay === Subject: Re: Axiomization of Number Theory >> Hello all, > I am interested in axiomizing Number Theory. Didn't Godel prove that that isn't possible? > >>And din't Peano actually do it? > >[...] The (first-order) Peano axioms do not do the > job. >Then take second-order :-) I expressed confusion yesterday about the right answer to this comment. In case you missed it in the various informative replies to my post, note Keith Ramsay's comment: If we're doing real second-order logic then >The one thing we do NOT have is a formal system. Saying that second-order PA characterizes arithmetic does not give an axiomatization of the sort we're looking for, because there is no associated formal proof system. === Subject: Re: Axiomization of Number Theory > Saying that second-order PA characterizes arithmetic > does not give an axiomatization of the sort we're looking > for, because there is no associated formal proof system. Who are we? :-) === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Axiomization of Number Theory I've got a little question. Is the word axiomize? Because of some reason I thought it was axiomatize. === Subject: Re: Axiomization of Number Theory > > >># 1. Could you provide formal axiomatic proofs, rather than English >>descriptions? >> >For three of your first four, here's where you can find formal proofs. >1. If A is a factor of B and B is a factor of A, then A=B. >Prop 23 in www.andrewboucher.com/papers/foea/VI1.pdf > 2. If A is a factor of B and B is a factor of C, then A is a factor of C. >Prop 24 in same paper > 4. There are an infinite number of prime numbers. >Prop 21 in www.andrewboucher.com/papers/foea/VII4.pdf > >I see a whole lot of undefined symbols and no explanation of the >axioms and rules of inference in the above references. Perhaps you >can simply give the axioms, rules, and proof here? However, 1 and 2 >are so simple, only 4 is of interest at this point. Yikes! I did say that 4 was at the end of 1800 pages. So, I'm not going to give you the proof, other than in the paper as referenced. An introduction to the system can be found at: http://www.andrewboucher.com/papers/foea/I1.pdf Basically, if you read that, and you understand the meanings of the special symbols (which can usually be infered from the surroundings), you should be able to follow the proofs. The logical system is essentially natural-deduction. I wanted to make the language dynamic (allow for new symbols), and that complicated the definition of the language. In any case the axioms and definition of the language can be found in: http://www.andrewboucher.com/papers/foea/I11.pdf Specifically, the mathematical axioms are on pages 19-21. All symbols used are summarized on pages 30-32. A summary - all definitions and all propositions proved - can be found in: http://www.andrewboucher.com/papers/foea/FOEASummary.pdf >You know, this idea of people saying See reference xyz and when you >look you don't really find what you were talking about and then when >you say Can you just give your result here? they call you lazy or >something, isn't a new one. So, you're not going to try that old >trick and call me lazy and refuse to actually give the axioms, rules >and proof here, are you? Because, honestly, I don't see that you >axiomized it in your papers, though, as I said, we are only talking >about the first, simplest Theorems that I figured weren't that hard >(about the most Peano Arithmetic can do) anyway. :) > >(and incidentally >the last proposition, after yikes! 1800 pages of proof). > >Bet I can reduce it by a factor of 1800 (with the right axiomization). I bet you can. > >At some point I may continue further. Hard cash would help. ;-) > >Darn, just when you got to the hard ones. Name your price. How about >$1,000.00 if you prove the other theorems in Peano Arithmetic? You do realize that I was joking. > >># 2. I listed the theorems in order of complexity, and you stopped >>after the first few (simplest) ones. This is as I described: Peano >>Arithmetic allows for the derivation of only the simplest of theorems. >>Could you address the more complex theorems given later in the list? >>How would you even represent the statement N is a perfect number. in >>Peano Arithmetic? >> >n is a perfect number can be translated as Sum{x : x | n} = n where >Sum(P) refers to the x in (hope I got this right): >[m][R][S] ( Seq(R,m,P) & Seq0(S,m) & S'm = x & S'0 = 0 & (i)(i < m => S'(i+1) = S'i + R'(i+1)) ). >Here: >[..] means there exists .. >Seq(R,m,P) means R is a sequence from {x : 1 <= x <= m} onto P >Seq0(S,m) means S is a sequence from {x : 0 <= x <= m} >S'z refers to the y for which Sz,y >The last two are exercises to define in second-order arithmetic. > >Sum(x:x|n)? Is that in Peano Arithmetic? Are you confusing the >Predicate Calculus with some Programming Language? And do your rules >apply to Sum(x:x|n)? I mean, if you're going to throw in arbitrary >syntax and semantics, your rules will have to likewise be modified for >this non-Predicate Calculus expression. What happened to the original >Peano Arithmetic can do it? First, second-order Peano Arithmetic is not in the Predicate Calculus, which is first-order logic. Secondly, you have to accept abbreviations, or be willing to unbundle the abbreviating terms. So e.g. if you don't want to accept Sum{x : x | n} = n then I would write [m][R][S] ( Seq(R,m,P) & Seq0(S,m) & S'm = n & S'0 = 0 & (i)(i < m => S'(i+1) = S'i + R'(i+1)) ) for n is a perfect number, where someone (you can't and I won't) has to unbundle the Seq wffs. If < is not part of the Peano Arithmetic under consideration, it has to be rewritten as well (say in terms of addition). >I said What is the formal wff? next just in case someone decided to >drift away from Predicate Calculus and throw in some programming >subroutine. Do you really maintain that all of the above is a Peano >Arithmetic wff? Yes. >Could we just stick to ^ v => ~ exists forall and relations, you know, >Peano Arithmetic wffs? (I failed your two exercises, so I guess >you'll just have to give a complete answer.) I thought I stuck pretty well. > >So it's not difficult at all in PA2 (or a near-enough version of it). > Maybe you were thinking of PA1. > >I'll take either, as long as it doesn't make up new stuff along the >way. > >>What is the formal wff? More fundamental than the >>question of undecidable wffs being few and far between and the system >>being sound (suggesting that true wffs are usually provable), is the >>question of whether Peano Arithmetic can even represent the theorems >>to be proved in the first place. >> >In my eyes anyway, second-order PA does a good job of representing the >theorems. > >I'll believe it when I see it. > >>The Theory of Computation is axiomized with 3 axioms: TRUE(x) [the >>universal set is recursively enumerable], YIT(I,J,K) [the >>proof/halting YES predicate is recursive] and a single Incompleteness >>Axiom -~YES(x,x) [The set of programs that do not halt YES on >>themselves is not recursively enumerable.], which I actually prove by >>going slightly outside of the system. >> > >Cheater ! > >I'm supposed to prove the axioms within the system?? I only mentioned >that one of the 3 axioms is almost a theorem as an aside. It's an >axiom, ok? Ok ok. I'll take cheater back. > >Seriously, I don't think any of these three axioms are low enough > You can just assume the proposition itself, and that would do the trick as well. What you need is to have axioms that have some other justification that just that they prove what you're trying to prove. > >I certainly don't understand why I should accept them as true. > >Since when do we prove axioms? I didn't ask you to prove them. I wanted to know why I should accept them as true. You could, of course, prove your theorem from a contradiction, but that, I think you agree, would be boring. > But if you insist, then tell me again >- you doubt that: >1. The set of natural numbers is recursively enumerable? >2. The proof/halting YES predicate (Program A halts YES on input B at >iteration C.) is recursive? >3. The set of programs that do not halt YES on themselves is not >recursively enumerable? I wouldn't want to assume 2 or 3, certainly. But that could just be me. Maybe there are reasons for assuming 2 and 3, which I haven't thought about, and/or you are able to do something with these axioms other than just prove your theorem. >Honestly (no offense), but you seem to be grabbing at certain phrases >slightly outside of the system was in the context of mentioning that >one of the axioms is almost provable within the system (just an >aside). The axioms are of course well-known facts. In each case, if >you look at what I actually said, there's certainly no problem with >these assertions. >Can we just agree to: >1. No references without giving the result here as well (just in case >one doesn't really find the result in the cited paper.) No >2. No exercises. No >3. No wffs that contain English. Ok >4. No wffs that go outside of Peano Arithmetic (Predicate Calculus) >syntax (unless you want to give a completely new system, which is ok >if that is your intent, but then you need to say that and define the >whole thing.) Not sure. >5. No demanding money to give the answers. Are you kidding me ??!! >In other words, can we just stick to formal axioms, rules and proofs? >BTW: I've been a reader of your internet publications for years. I >may have even written you a fan mail sometime in the past! Glad to hear it! I need all the fans I can get ! > >>Charlie Volkstorf >>Cambridge, MA >>axiomize at aol dot com >> === Subject: Re: Axiomization of Number Theory > The logical system is essentially natural-deduction. I wanted to make > the language dynamic (allow for new symbols), and that complicated the > definition of the language. In any case the axioms and definition of > the language can be found in: > http://www.andrewboucher.com/papers/foea/I11.pdf >> Um sorry, that should be: http://www.andrewboucher.com/papers/foea/I2.pdf === Subject: Re: Axiom of choice: is it wrong? > The existence of undefinable reals is pretty weird, as was discussed > by Taylor and I a while ago. On the one hand, you can argue that > such things have to exist, by cardinality considerations. But you > can't begin to describe one. > Well, what about undefinable ordinals? It's easy to describe one, > e.g., the smallest one. Does that make them less weird, or weirder? You mean the smallest one above zero, or the smallest one completely? (BTW, Ordinals are things like 1, 2, 3 or a, b, c right?) I don't think it's weird at all. === Subject: Re: Axiom of choice: is it wrong? > The axiom of choice, that says it's always possible to choose one > element of each member of a family of sets, sounds intuitively right. The axiom of choice doesn't really say that. It says that whenever A is a family of non-empty sets, then there is a choice set C, s.t. for all a in A a / C = {z} (where / tries to denote intersection). It's a set existence axiom, and it's not about our ability to actually choose anything. For example, in case of a family of alpeh_omega disjoint sets, what could possible amount to our ability to choose a single member from each of these sets? Our ability to write down a 1st order formula to pick one from each? > Especially in the trivial case: surely it must be always possible to > choose one element out of one set. But I've just thought something > that seems to contradict this. > Let's consider the real numbers. We know it isn't a countable set. > Now, let's consider the subset of all the definable real numbers. > The subset of definable numbers > is countable. > Therefore, there are undefinable numbers, and what's more, practically > all real numbers are undefinable. > Now, let's consider the subset of all the undefinable real numbers. > There is no way of referring specifically to any one of them. How > could one possibly say that you have chosen one of them? You don't. That's part of the reason we have an *axiom* of choice. But here's a standard argument for the axiom of choice. Consider the iterative hierarchy V_0 = {} V_alpha+1 = Power-set(V_alpha) V_lambda = U_{alpha < lambda} V_alpha V = U_{alpha in On) V_alpha The family of sets was formed at some stage V_alpha, so all of its members were formed at some stage V_beta, beta < alpha (beta possible different for all of the members). But then at stage V_alpha+1 we should have also the choice set, since otherwise the power-set operation has missed out some subset V_alpha (as all the members of the members of the family of sets belong to V_alpha). === Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Axiom of choice: is it wrong? Fred says... > The existence of undefinable reals is pretty weird, as was discussed > by Taylor and I a while ago. On the one hand, you can argue that > such things have to exist, by cardinality considerations. But you > can't begin to describe one. >Well, what about undefinable ordinals? It's easy to describe one, >e.g., the smallest one. Does that make them less weird, or weirder? The problem is that the smallest undefinable ordinal *Is* a definition. Sort of. -- === Subject: Re: Axiom of choice: is it wrong? > Seaman says... > Now, let's consider the subset of all the undefinable real numbers. >> There is no way of referring specifically to any one of them. How about h = sum_{0^oo} h_k * 2^(-k) where h_k = 1 if the k-th TM halts with input k, > h_k = 0 otherwise. Then h is a specific real number. We know it's not computable (by the >unsolvability of the halting problem), but nevertheless, it exists. > Doly said undefinable rather than uncomputable. h is certainly > definable. > The existence of undefinable reals is pretty weird, as was discussed > by Taylor and I a while ago. On the one hand, you can argue that > such things have to exist, by cardinality considerations. But you > can't begin to describe one. Weird indeed! I'm fuzzy on the undefinable real thing. I can't find a proper definition for this phrase. Also, if a real number is thought to be an equivalence class of Cauchy sequences* of rational numbers then I would think if x is a real number there must be at least one Cauchy sequence of rationals that converges to it... If not, is it reasonable to conclude that x isn't a real number? * Or Dedekind cuts which I don't really understand yet... > -- > Daryl McCullough > Ithaca, NY === Subject: Re: Axiom of choice: is it wrong? > The existence of undefinable reals is pretty weird, as was discussed > by Taylor and I a while ago. On the one hand, you can argue that > such things have to exist, by cardinality considerations. But you > can't begin to describe one. > Weird indeed! > I'm fuzzy on the undefinable real thing. I can't find a proper > definition for this phrase. Also, if a real number is thought to be an > equivalence class of Cauchy sequences* of rational numbers then I > would think if x is a real number there must be at least one Cauchy > sequence of rationals that converges to it... If not, is it reasonable > to conclude that x isn't a real number? There is indeed a Cauchy sequence representing any given real number x. A whole lot of them, in fact. However, if x is undefinable, then any Cauchy sequence representing x is likewise undefinable. > * Or Dedekind cuts which I don't really understand yet... It doesn't particularly matter, because the same argument applies. Basically, it means there are more real numbers than there are definitions of real numbers. === Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Axiom of choice: is it wrong? fpluser says... > The existence of undefinable reals is pretty weird, as was discussed > by Taylor and I a while ago. On the one hand, you can argue that > such things have to exist, by cardinality considerations. But you > can't begin to describe one. >Weird indeed! >I'm fuzzy on the undefinable real thing. I can't find a proper >definition for this phrase. Well, definable isn't actually definable, formally. Informally, a real number is definable if there is some description of that number such that there is exactly one number meeting that description. For example: That number whose square is equal to 2 uniquely describes the number square-root(2). Since there are only countably many descriptions, there can be at most countably many definable real numbers. All the rest are undefinable. An undefinable real number would be one such that, no matter what true thing you can say about it, there will be another real (infinitely many, as a matter of fact) to which that description applies equally well. So there is absolutely nothing special about an undefinable real. (Being undefinable isn't special, since *most* reals are undefinable). -- === Subject: Re: Axiom of choice: is it wrong? > There is a big difference between set A can be well-ordered and formula phi(x, y) well-orders A (phi being a formula in ZFC, or > whatever your favourite set theory is). My point is: what if it is impossible to define any such formula? What is the meaning then of set A can be well-ordered? === Subject: Re: Axiom of choice: is it wrong? > Choosing one element from a single set has nothing to do with AC. It's > just the principle that if a set is nonempty, then it has a member. I don't deny that a nonempty set has at least one member. But the axiom of choice says that we can define a function. What if the function can't be defined? > How about h = sum_{0^oo} h_k * 2^(-k) > where h_k = 1tif the k-th TM halts with input k, > h_k = 0totherwise. > Then h is a specific real number. We know it's not computable (by the > unsolvability of the halting problem), but nevertheless, it exists. Undefinable as I was using the term is more restrictive than non computable. Some non computable numbers can be defined (you just put an example). === Subject: Re: Axiom of choice: is it wrong? > I'm fuzzy on the undefinable real thing. I can't find a proper > definition for this phrase. A real number that can't be defined with any finite mathematical expression. > Also, if a real number is thought to be an > equivalence class of Cauchy sequences* of rational numbers then I > would think if x is a real number there must be at least one Cauchy > sequence of rationals that converges to it... If not, is it reasonable > to conclude that x isn't a real number? Yes. But the sequence of rationals itself is infinite, and could be impossible to define in any finite way. === Subject: Re: Axiom of choice: is it wrong? > I'm fuzzy on the undefinable real thing. I can't find a proper > definition for this phrase. A real number that can't be defined with any finite mathematical expression. > Also, if a real number is thought to be an > equivalence class of Cauchy sequences* of rational numbers then I > would think if x is a real number there must be at least one Cauchy > sequence of rationals that converges to it... If not, is it reasonable > to conclude that x isn't a real number? Yes. But the sequence of rationals itself is infinite, and could be impossible to define in any finite way. === Subject: Re: Axiom of choice: is it wrong? > Choosing one element from a single set has nothing to do with AC. It's > just the principle that if a set is nonempty, then it has a member. > I don't deny that a nonempty set has at least one member. But the > axiom of choice says that we can define a function. What if the > function can't be defined? The axiom of choice doesn't apply to the situation you described. That was my point. The set of undefinable numbers is nonempty, and therefore we can choose a member (without invoking AC). > How about h = sum_{0^oo} h_k * 2^(-k) > where h_k = 1 if the k-th TM halts with input k, > h_k = 0 otherwise. > Then h is a specific real number. We know it's not computable (by the > unsolvability of the halting problem), but nevertheless, it exists. Undefinable as I was using the term is more restrictive than non > computable. Some non computable numbers can be defined (you just put > an example). Point taken, but I also added that even if we didn't have a definition or a description of any member of the set, it's still enough just to know that the set is nonempty. AC is not needed. === Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Axiom of choice: is it wrong? > The axiom of choice, that says it's always possible to choose one > element of each member of a family of sets, sounds intuitively right. > The axiom of choice doesn't really say that. It says that whenever A is > a family of non-empty sets, then there is a choice set C, s.t. for all a > in A a / C = {z} (where / tries to denote intersection). It's a set > existence axiom, and it's not about our ability to actually choose > anything. For example, in case of a family of alpeh_omega disjoint sets, > what could possible amount to our ability to choose a single member from > each of these sets? Our ability to write down a 1st order formula to > pick one from each? If a set can't be possibly defined in any way, what is the sense of saying that it exists? === Subject: Re: Axiom of choice: is it wrong? > There is a big difference between set A can be well-ordered and > formula phi(x, y) well-orders A (phi being a formula in ZFC, or > whatever your favourite set theory is). > My point is: what if it is impossible to define any such formula? What > is the meaning then of set A can be well-ordered? It is certainly possible for a set to be well-orderable without there being a formula which defines the well-ordering. For an example or proof you'll need one of the experts. GC === Subject: Re: Circles in complex space NNTP-Posting_Host: student-concordia-dorm.student.concordia.net NNTP-P0STING-HOST: 210.117.60.5 === >Subject: Circles in complex space. >In modern physics it's been explained that an object can rotate in >complex space and take 720 degrees to return back to its orginal >position. What sort of object?... If you're talking about how angular momentum generates rotations and how, for example, electrons have half integer spin-angular momentum so you end up with stuff that looks like: Exp[ i (1/2) x] then, it is true that when x goes from zero to four-Pi (two full rotations) you get back to the same thing... but nothing's really rotating around in physical space. >How is this possible, exactly? Obviously it's not the same >as taking the argument of a circle on a z plane for each point and >then figuring the area geometrically as if this were real space. figuring what area? Maybe you should think about Riemann (sp?) sheets. >What exactly is happening? Stuff... For a more detailed explaination, a good book is: Modern Quantum Mechanics by Sakurai >Also, does this mean you can take a measure through complex space and >get a negative or imaginary distance? Distance is a magnitude, right? Magnitude of the displacement vector is distance... so distance is always non-negative. In complex space magnitudes ( e.g. Sqrt[(z) (z*)] ) are still non-negative and real. adam === Subject: combinatorial tiling problem Problem: : Number of ways to tile a 3 X n region with 1 X 1, 2 X 2 and 3 X 3 tiles. Website: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi Formula: a(n)=a(n-1)+2a(n-2)+a(n-3) Question: 2 people at WSU are working on this problem and the recurrence. I easily solved it with mathematica using the traditional method of solving a cubic... They claim that sum k=0 to n binomial(2n-2k,k) is also a solution and it looks very believable. Can anyone prove this? === Subject: Re: combinatorial tiling problem > Problem: : Number of ways to tile a 3 X n region with 1 X 1, 2 X 2 and 3 > X 3 tiles. > Website: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi > Formula: a(n)=a(n-1)+2a(n-2)+a(n-3) > Question: 2 people at WSU are working on this problem and the recurrence. > I easily solved it with mathematica using the traditional method of > solving a cubic... They claim that sum k=0 to n binomial(2n-2k,k) is also > a solution and it looks very believable. Can anyone prove this? Use Wilf's Snake-Oil method. Let a_n = sum_k (2n-2k choose k) and let A(t) = sum_n a_n t^{2n}. (I take t^2n rather than t^n to avoid hassles later). Then A(t) = sum_k t^{2k} sum_n (2n-2k choose k) t^{2n-2k} = sum_k t^{2k} sum_m (2m choose k) t^{2m}. To do this inner sum, note that it's the even part of sum_r (r choose k) t^r = t^k/(1-t)^{k+1}. Hence 2 sum_m (2m choose k) t^{2m} = t^k/(1-t)^{k+1} + (-t)^k/(1+t)^{k+1}. We get 2 A(t) = sum_k [t^{3k}/(1-t)^{k+1} + (-t)^{3k}/(1+t)^{k+1} = 1/(1 - t - t^3) + 1/(1 + t + t^3) = 2/(1 - t^2 - 2t^4 - t^6) from which it is apparent that a_n = a_{n-1} + 2a_{n-2} + a_{n-3}. === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Complex Numbers |The fact that Riemann surfaces are orientable simply seemed too |irresistibly nice to not share. I hope my post is not too confusing. It's also true that complex manifolds of higher dimensions are orientable. Being able to multiply a tangent vector by i is enough to give us an orientation. The tangent space at a point has a basis of the form v_1, i*v_1, v_2, i*v_2, ..., v_n, i*v_n, and any two bases of the this form have the same orientation, so they provide a consistent orientation. Why do these bases have the same orientation always? In other words, why does a nonsingular linear transformation of C^n (linear over C) always preserve orientation? Well, I don't remember the most elegant proof, but one way to show it is to show first that each of them can be written as a composition of elementary linear transformations with complex coefficients, and then show that each kind preserves orientation. For instance, multiplying the first coordinate by 2+3i has the matrix (2 -3 0 0 ...) (3 2 0 0 ...) (0 0 1 0 ...) (0 0 0 1 ...) (... ) if we consider it as a transformation of the basis of C^n over the real numbers. (It has dimension 2n over the real numbers.) But that has a determinant of 2^2 + 3^3 > 0. Likewise, adding 2+3i times the first complex coordinate to the second coordinate has the matrix (1 0 0 0 ...) (0 1 0 0 ...) (2 -3 1 0 ...) (3 2 0 1 ...) (... ) whose determinant is 1. I guess in general if the determinant over the complex numbers is det_C(T), then the determinant over the reals is det_R(T) = |det_C(T)|^2. Yeah, I'm forgetting the elegant proof here. Anyhow, for a nonsingular complex-linear transformation the determinant as a real-linear transformation is always positive, which means that it preserves the orientation. Determining what conditions a (real) manifold has to satisfy in order for it to be possible to turn it into a complex manifold is one of the classic problems in the field. I forget what the general story is, but certainly orientability is one of the first criteria named. Keith Ramsay === Subject: Re: Complex Numbers Is there a special name for functions which are like analytic functions > except that they reverse the orientation (such as complex conjugation)? > I might call them conjugate-holomorphic. > Such > functions and analytic functions seem to be about equally fundamental. You think the function z (the identity function) and zbar (the > conjugation operation) are equally fundamental? Hard to say, but I think the group consisting of rotations and reflections of a one-dimensional real vector space is about as fundamental as the group consisting of rotations only. I think a mapping f: V->V from a complex vector space to itself which is such that f(v1+v2)=f(v1)+f(v2) and f(z.v)=(z^bar)f(v) (that is, an anti-linear transformation) is about as fundamental as a transformation g:V->V such that g(v1+v2)=g(v1)+g(v2) and g(z.v)=z.g(v) (that is, a linear transformation). I think a mapping from an analytic manifold to itself which is locally an anti-linear transformation is about as fundamental as a mapping which is locally a linear tranformation. By the way, if a mapping h from a complex vector space V onto another complex vector space W merely satisfies h(v1+v2)=h(v1)+h(v2) and some extremely natural assumptions, then either h(z.v)=(z^bar)h(v), h(z.v)=z.h(v). This means that linear and anti-linear transformations from a complex vector space V to a complex vector space W are essentially the group homomorfisms from V to W, a vector space being thought of as a group under addition. If z is a an integer the theorem is obvious, so let us consider the case where z=1/n for some natural number n. (n.(1/n))h(z)=h((n.(1/n))z)=h(((1/n)+(1/n)+...+(1/n))z)=h((1/n)z+(1/n)z+...+ (1/n)z)=h((1/n)z)+h((1/n)z)+...+h((1/n)z)=n.h((1/n)z). Dividing both sides by n we obtain (1/n).h(z)=h((1/n)z), which is what we were looking for. The extension to rational numbers (p/q) is obvious. To get the result for real numbers r we need to make an extra assumption: h preserves topology. What makes this assumption so natural is that if you do not have it there is no reason for studying real and complex vector spaces in the first place: you could stick to rational and algebraic vector spaces. Next let k be a complex number with modulus 1 and an argument which is pi/2 in size, and let v be an arbitrary vector. Suppose we know what h(v) and h(k.v) are. We can then determine h(z.v) for an arbitrary complex number z. How do we detemine h(k.v)? Before we do, let us /drop/ some structure and consider W as a real vector space, and let h turn W into a complex vector space. The complex vector space so constructed will be such that h(k.v)=k'.h(v), where k' is a complex number with modulus 1 and an argument of size pi/2 (for some purposes, as we will see below, we can put k'=k). If W was not a real vector space but a complex vector space before we defined h, then W must either be such that h could have been used to define complex multiplication, or else we can say that h has not been righly defined. So how do we define complex multiplication in W? Let w be a vector in W which we wish to multiply by complex numbers. We then consider vectors in V which get mapped to w. Multiply each of them by k and apply h. We must now add as a requirement on h that each of them get mapped to the same element in W. That element will by definition be the result of multiplying w by k'. I now that this breakes all formal rules, but we could say that h(k)=k'. If V and W are totally unrelated it is convenient to put k=k', but if they are related, for example if they coinside, we could have k=-k'. Mattias === Subject: Re: Complex Numbers >>Is there a special name for functions which are like analytic functions >>except that they reverse the orientation (such as complex conjugation)? >I might call them conjugate-holomorphic. >>Such >>functions and analytic functions seem to be about equally fundamental. >You think the function z (the identity function) and zbar (the >conjugation operation) are equally fundamental? > Hard to say, but I think the group consisting of rotations and reflections > of a one-dimensional real vector space is about as fundamental as the group [*] > consisting of rotations only. I think a mapping f: V->V from a complex > vector space to itself which is such that f(v1+v2)=f(v1)+f(v2) and > f(z.v)=(z^bar)f(v) (that is, an anti-linear transformation) is about as > fundamental as a transformation g:V->V such that g(v1+v2)=g(v1)+g(v2) and > g(z.v)=z.g(v) (that is, a linear transformation). I think a mapping from an > analytic manifold to itself which is locally an anti-linear transformation > is about as fundamental as a mapping which is locally a linear > tranformation. [...] Mattias, Perhaps you intended two-dimensional real vector space in line [*]? I went looking for a difference between the group of (R-linear) rotations and reflections that act on the euclidean plane R^2, on the one part, and the group consisting solely of (R-linear) rotations acting on R^2. What I found is this: the rotations only form a commutative group, whereas the rotations and reflections combined form a group which is not commutative. For instance, consider the reflection R_x about the x-axis. Then, R_x(x,y) = (x, -y). and then another reflection, this time about the line x=y, which I denote R_xy: R_xy(x, y) = (y,x). An easy case to check is using the vector (1,1): R_x(R_xy((1,1))) = (1, -1) and on the other hand R_xy(R_x((1,1))) = (-1, 1) . David P.S. z |-> z^{bar}, is not an analytic map; z |-> z is. If there is a basic difference between +i and -i in complex analysis, I'd be interested to know... === Subject: Re: Complex Numbers >[...] Comments would be welcome as I am not an expert on Riemann surfaces. > Neither am I, but it sounds right to me. What I'm not sure about is > whether you're suggesting that this helps us decide which of the > two square roots of -1 is i and which is -i. >David C. Harris > >No, I am not suggesting that this tells us which square root >of -1 is i and which is -i. Orientable is not the same as oriented! >C is orientable but not oriented unless we make a choice of i or -i, >and either one will do as well as the other to give us an orientation! >The fact that Riemann surfaces are orientable simply seemed too >irresistibly nice to not share. I hope my post is not too confusing. Okie dokie. (See Ramsay's comments on the fact that the same thing works in higher dimensions...) >I agree with your post that says there is no way to tell i from -i. >It is true that one can add structure to C to make i and -i distinguishable, >but then C embeds into this new structure in two ways. >(At least that is all that I can see.) >David C. Harris === Subject: Re: Complex Numbers >... (i) For every z in C just one of the following holds: z > 0, z = 0, z < 0; You are, indeed, not paying attention. I have already said that: > z > 0 v z < 0 v z = 0, is not valid within complex numbers. In that case > is not an order on C, it's an order on a proper subset of it. GC === Subject: Re: Dense in the unit circle ? > Let U={z in C, |z|=1} be the unit circle in C. A classical result says that {exp(in), n in Z} is dense in U, and it's > not difficult to show that {exp(i sqrt(n)), n in Z} and {exp(i ln(n)), > n in Z} are dense in U too. But, can someone help me to show that {exp(i nî), n in Z} is dense in > U ? > The methods that prove the previous densities fail. > Have you had a look at the book by Kuipers and Niederreiter? No, I don't know this book. And, unfortunately, I'm far away from any library. Does noone have any idea ? === Subject: Re: Dense in the unit circle ? > No, I don't know this book. And, unfortunately, I'm far away from any > library. Does noone have any idea ? So it won't help if we give you the page number of the proof? Instead you want someone to type it in here? Why? Hint of idea: The difference (n+1)^2 - n^2 is FIRST DEGREE polynomial 2n+1, and we know the result for this. === Subject: Re: Dense in the unit circle ? > No, I don't know this book. And, unfortunately, I'm far away from any > library. Does noone have any idea ? > So it won't help if we give you the page number of the proof? Instead > you want someone to type it in here? Why? > Hint of idea: The difference (n+1)^2 - n^2 is FIRST DEGREE polynomial > 2n+1, and we know the result for this. And what about {n*sin(n), n integer}; is it dense in R ? (I found a paper where it is proved that this problem is equivalent to show that a_n -> +oo where Pi = [3;a_0;a_1;...a_n;...] (continued fraction for Pi). -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France === Subject: Differential Equations Hi guys I am looking for a differential equation software , for windows that can add exponential terms . Please help. Malcolm === Subject: Re: English language translated into math equations??? > Has anyone translated or attempted to translate English sentences into math > equations? For example, sky is blue translated into sky = blue. Yes, I > know that equation isn't accurate. It's just to give an idea of what I > mean. If someone has done so, I'd appreciate knowing the URLs on this topic > Scott Jensen In Linguistics, that would be called 'Formal Semantics', ie. the mathematical analysis and representation of the meaning of natural language expressions (in English, Polish, Japanese, etc.): A standard book on this issue is: An excelent manuscipt on the computational perspective is: http://www.comsem.org/ Cheers, D === Subject: Re: Factorial/Exponential Identity, Infinity > Assume the identity is correct. What good is it? I derived it from > assuming that half of the infinite binary strings have equal numbers > of zeros and ones. This assumption, depending on how you define 'half' in this context, is incorrect. > Consider the rational binary numbers of these forms: > .0101010101... > .1010101010... > I think those are 1/3 and 2/3 but I don't know. Let's see: Yes. > 1/4 + 1/16 + 1/64 + 1/256 + ... > 1/2 + 1/8 + 1/32+ 1/128 + ... > How do I evaluate those binary rational numbers? I want to know > reduced fractional forms for those and all other rationals between > zero and one with equal numbers of zeros and ones in their binary > expansion. Suppose you have .a1a2a3a4...ana1a2..., a binary number which repeats after n bits. Then it equals (.a1a2a3...an)(1 + 1/2^n + 1/2^2n + ...) The first bracket is a finite binary string and you can evaluate it. Call the second a(n). Then 2^n . a(n) = 2^n + a(n), so a(n) = 2^n / (2^n - 1). Yours, H... > Hi. > Ross === Subject: Re: Factorial/Exponential Identity, Infinity === Subject: Re: Factorial/Exponential Identity, Infinity > lim n->oo n! / (2 (n/2)! 2^n) = 1 >>By Sterling's approximation, n! has some resemblance to n^n in the >>large. Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo >Um, of course the formula he gives is wrong, but you haven't shown >that. (_If_ I have Stirling's formula right then > sqrt(n) (2n)!/(4^n (n!)^2) >has a finite limit, although your reckoning would show it also >tends to infinity.) Sterling's approximation: n! = (n/e)^n sqr(2n.pi) 2n! 2^2n = 2(n/e)^n 2^2n sqr(2n.pi) (2n)! = (2n/e)^2n sqr(4n.pi) = (n/e)^2n 2^2n sqr(4n.pi) (2n)! / (2n! 2^2n) = (sqr 1/2)(n/e)^n I reckon so. ---- === Subject: Re: Finite subgroups of GL(n,Z) > Is the same assertion true for the groupp GL(n,K) where K is a > number field ? I seriously doubt it. After all, GL(1,Q)=Q^* has free abelian groups of an arbitrary high rank as subgroups (freely generated by a chosen set of prime numbers) If you replace K with its ring of integers O, then the question again becomes interesting. I do think that we can reduce this to the case handled by professor Chapman. Let m=[K:Q]. We can represent multiplication in O (using a chosen integral basis) with integer m*m matrices. Thus GL(n,O) is isomorphic to a certain subgroup of GL(nm,Z) and we are done. Cheers, Jyrki === Subject: Re: Finite subgroups of GL(n,Z) > Is the same assertion true for the groupp GL(n,K) where K is a > number field ? Yes. We can easily reduce to the case K = Q. Each finite subgroup of GL(n,Q) fixes a lattice in Q^n, and that reduces it to GL(n,Z). === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Finite subgroups of GL(n,Z) > Is the same assertion true for the groupp GL(n,K) where K is a > number field ? Yes. We can easily reduce to the case K = Q. Each finite subgroup of GL(n,Q) fixes a lattice in Q^n, and that reduces it to GL(n,Z). === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Finite subgroups of GL(n,Z) > Is the same assertion true for the groupp GL(n,K) where K is a > number field ? Yes. We can easily reduce to the case K = Q. Each finite subgroup of GL(n,Q) fixes a lattice in Q^n, and that reduces it to GL(n,Z). === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Finite subgroups of GL(n,Z) > Is the same assertion true for the groupp GL(n,K) where K is a > number field ? I seriously doubt it. After all, GL(1,Q)=Q^* has free abelian groups > of an arbitrary high rank as subgroups (freely generated by a chosen > set of prime numbers) But these are infinite, and the OP asked about finite subgroups. > handled by professor Chapman. Let m=[K:Q]. We can represent multiplication I'm not, and I never have been, a professor. === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Finite subgroups of GL(n,Z) > But these are infinite, and the OP asked about finite subgroups. Quite. I attempted to cancel my post, when I realized my trivial mistake. I thought that the extra elbow room provided by the field coefficients would change the answer and was also confident in that the case of algebraic integer coefficients could be reduced to the case you handled. But as you have explained, in these cases it is irrelevant, whether we allow denominators in the matrix entries. > handled by professor Chapman. Let m=[K:Q]. We can represent multiplication > I'm not, and I never have been, a professor. I apologize to using a wrong form of address. Best , Jyrki === Subject: Re: Finite subgroups of GL(n,Z) > Is it true that for every n>=1 the group GL(n,Z) as only a finite number > of non-isomorphic finite subgroups ? > >Yes . >See : >J. Kuzmanovich and A. Pavlichenkov, Finite groups of matrices whose >entries are integers, Amer. Math. Monthly, 109 (2002), 173-186. A stronger result, due to Jordan, is that GL(n,Z) has finitely many conjugacy classes of finite subgroups - sorry, don't have a reference to hand! Derek Holt. === Subject: Gleichungen Hi! Kleines, einfaches Mathe-Problem. Hoffentlich kann mir jemand weiterhelfen. 3x + 2 (x - 4) = 8 (2x - 7) - 7 3x + 2x - 8 = 16x - 56 - 7 5x - 8 = 16x - 63 | - 16 x - 11x - 8 = - 63 | + 8 - 11x = - 55 | : 11 x = - 5 L = (- 5) gleiche Aufgabe anderer L.9asungsweg: 3x + 2 (x - 4) = 8 (2x - 7) - 7 3x + 2x - 8 = 16x - 56 - 7 5x - 8 = 16x - 63 | + 63 5x + 55 = 16 x | - 5 x 55 = 11 x | : 11 5 = x L = (+ 5) wie im L.9asungsheft angegeben. WARUM??? Die Defenitionsmenge wurde nicht angegeben, so das ich nicht wei¤ ob nun die negative L.9asungsmenge oder die positive die richtige ist, bevor ich im L.9asungsheft nachschlage... Mir ist klar, das ich beim ersten Versuch die -5 mit -1 multiplizieren kann... Vielen Dank Sabine === Subject: Re: Gleichungen > Kleines, einfaches Mathe-Problem. Hoffentlich kann mir jemand > weiterhelfen. > 3x + 2 (x - 4) = 8 (2x - 7) - 7 > 3x + 2x - 8 = 16x - 56 - 7 > 5x - 8 = 16x - 63 | - 16 x > - 11x - 8 = - 63 | + 8 > - 11x = - 55 | : 11 ^^ nicht ist 11, ist -11 > x = - 5 ^^^ so x = 5 > L = (- 5) > gleiche Aufgabe anderer L.9asungsweg: > 3x + 2 (x - 4) = 8 (2x - 7) - 7 > 3x + 2x - 8 = 16x - 56 - 7 > 5x - 8 = 16x - 63 | + 63 > 5x + 55 = 16 x | - 5 x > 55 = 11 x | : 11 > 5 = x > L = (+ 5) wie im L.9asungsheft angegeben. WARUM??? > Die Defenitionsmenge wurde nicht angegeben, so das ich nicht wei¤ ob > nun die negative L.9asungsmenge oder die positive die richtige ist, > bevor ich im L.9asungsheft nachschlage... Mir ist klar, das ich beim > ersten Versuch die -5 mit -1 multiplizieren kann... > Vielen Dank > Sabine === === Subject: Re: help needed with infinite series Did you test for convergence and plotted the function for n-> infinity ? Perhaps not every convergent series has a closed form, i.e., expressible in terms of transcendental/special functions, but sure can be always named. If the the series sum establishes some mathematical relationships, identities or links to significant scientific application, it can be named, gradually gains currency and then later it becomes closed form! === Subject: Re: help needed with infinite series >Hi , >I want to sum following infinite series to obtain a closed form expression. > sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0thanks in advance , >Sanjeev >sanjeev_bgp@yahoo.com === Subject: Re: help needed with infinite series > Hi , > I want to sum following infinite series to obtain a closed form expression. > sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0 Sanjeev > sanjeev_bgp@yahoo.com ******** If f(x) is the value of this series at x, then f satisfies the functional equation f(x^4) = f(x)/2. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: help needed with infinite series Hi , > I want to sum following infinite series to obtain a closed form expression. sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0 in advance , Sanjeev > sanjeev_bgp@yahoo.com > ******** > If f(x) is the value of this series at x, then f satisfies > the functional equation f(x^4) = f(x)/2. The function g(x) = k/sqrt(|ln(x)|) satisfies this functional equation, but that is not sufficient to say that f = g for some k. The best I can say is that there are values for k for which the function f-g satisfies this functional equation and has an infinite number of zeros, the set of which has 0 and 1 as limit points. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 === Subject: Re: help needed with infinite series > I want to sum following infinite series to obtain a closed form expression. > sum { ( x ^ ( 2 ^ 2n ) ) * 2^n } , n=-infinity to infinity , 0 ******** > If f(x) is the value of this series at x, then f satisfies > the functional equation f(x^4) = f(x)/2. The function g(x) = k/sqrt(|ln(x)|) satisfies this >functional equation, but that is not sufficient to say that >f = g for some k. The best I can say is that there are >values for k for which the function f-g satisfies this >functional equation and has an infinite number of zeros, the >set of which has 0 and 1 as limit points. Numerically it appears that f(x) sqrt(|ln(x)|) does oscillate, in a range between about 1.27563 and 1.28148. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Homo High thr91 scribbled the following: > http://www.msnbc.com/news/945134.asp ...State Conservative Party Chairman Long criticized the creation of > the school. > Is there a different way to teach homosexuals? Is there gay math? > This is wrong, Long said. There's no reason these children should be > treated separately. > Yes, I can think of plenty of gay math problems. (snip) Do the problems we teach heterosexual children consist solely of sex between men and women? No? Then why do you propose to teach homosexual children problems about sex between men and men, or between women and women? A homosexual person is not only his/her sexuality. There's a whole world outside what gender you happen to prefer. === /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ My absolute aspect is probably... - Mato Valtonen === Subject: Re: Homo High > Do the problems we teach heterosexual children consist solely of sex > between men and women? No? Then why do you propose to teach homosexual > children problems about sex between men and men, or between women and > women? 1. These are high school students, many of them sexually active, not children. 2. Homosexual high school students need the extra encouragement to develop their intellect in topics that often don't interest homosexuals, such as math. 3. Math is a political instrument, despite its pretenses to an objectivity devoid of context. === Subject: Re: Homo High thr91 scribbled the following: > Do the problems we teach heterosexual children consist solely of sex > between men and women? No? Then why do you propose to teach homosexual > children problems about sex between men and men, or between women and > women? > 1. These are high school students, many of them sexually active, not children. OK, if you want to nit-pick, change children to high-school students in my original reply and it still applies. If they're so sexually active, why aren't they being taught in a sexually active way? > 2. Homosexual high school students need the extra encouragement to develop > their intellect in topics that often don't interest homosexuals, such as > math. And what about the topics tht often don't interest heterosexuals? There's bound to be some. Would you recommend sexual provocation to make them more interesting to heterosexual students? > 3. Math is a political instrument, despite its pretenses to an objectivity > devoid of context. I disagree completely. Math is the most objective science there is. Its applications might be political instruments but math itself is another thing. You seem to still think that all there is to homosexuals is the fact that they're homosexual. Have you ever met any? If you have, have they ever done anything except have sex? === /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ The truth is out there, man! Way out there! - Professor Ashfield === Subject: Re: Homo High > Do the problems we teach heterosexual children consist solely of sex > between men and women? No? Then why do you propose to teach homosexual > children problems about sex between men and men, or between women and > women? > 1. These are high school students, many of them sexually active, not children. > 2. Homosexual high school students need the extra encouragement to develop > their intellect in topics that often don't interest homosexuals, such as > math. > 3. Math is a political instrument, despite its pretenses to an objectivity > devoid of context. My goodness, some people don't recognise humour when they're presented with it. Let's not forget heteroscedascity (I'm not sure about that spelling), homomorphisms and that gay favourite, the hairy ball theorem :) === Subject: Re: Homo High > 2. Homosexual high school students need the extra encouragement to develop > their intellect in topics that often don't interest homosexuals, such as > math. Yeah, and Turing never existed. *_PLONK_* Phil === Subject: Re: How to prove a true, but meaningless, theorem? > In a different thread of mine discussing operations upon elements of > the empty set mapping to other elements of the empty set, William Hale > asserted that the statement if a and b are prime numbers divisible by 4, then a+b is also a prime > number divisible by 4 > was true. > While I do not claim that it is false, I am interested in how one > would go about proving it true. > Actually there would be 2 different kinds of proofs I am interested > in: the first where we already know what we know about primes, and > the 2nd in some strange alternate universe where we have just now > defined primes and haven't yet even established that no primes > divisible by 4 exist. (In otherwords, a proof which does not assume > that no such primes exist) > Looking forward to your very wise insights in this matter, and I > assure you it is not homework. ***muah*** If a and b are prime numbers divisible by 4, then they are both 4. Then a + b is 8. However, if it wraps around at 7 and after that goes to -4, and you consider negatives of primes to also be primes, then that statement could be considered true. But this is a very trivial case. (...Starblade Riven Darksquall...) === Subject: Re: How to prove a true, but meaningless, theorem? > I claim that if a and b are prime numbers divisible by 4 then a+b >is also a > prime divisible by 4. > prime numbers, a and b, that are divisible by 4 BUT a + b is not >a prime > number divisible by 4! >[cut] >>No, it might be undecidable. >> >> No, you're confused about what undecidable means. I thought Charlie-Boo was correct. I'd typed a few paragraphs of definitions before I went back and > realized I was puzzled about what you meant. Since we're talking > about subtle things, it would probably be more efficient if you > clarified a few things before I tried to reply, cuz if I really don't > get what you're saying then I can't really answer it: > >If I have an undecibable statement S and you claim S is true, >then I can't convince you that you are wrong by showing >a particular instance where the statement is false. This sentence has me very puzzled. I mean I don't get it at > _all_ - of _course_ you could convince me that way? >I don't understand what you are questioning. >Steven says S is true. He gives his reason that is true >because I can't show him an instance where S is false. That explains the confusion. Nothing visible above says that he's giving that as his reason for saying that S is true. Possibly if I'd read earlier posts in the thread this would be clear. Possibly not. One might _infer_ from what's written above that that's his reason, but it's not explicitly stated... Anyway, of course I agree that that reason is insufficient. >I agree that I can't show him an instance where S is false. >Steven wants then to say that therefore S is true. >Charlie-Boo and I are saying that this does not prove that S is true. I didn't say that it did. > This > seems so clear that I can't help but wondering whether there > was a word missing or something? >I don't think I am missing any words. Remember that I am not >discussing whether I can determine S is decidable or not. >Charlie-Boo and I are discussing whether failure to be able >to exhibit the falsity of a statement is proof that the >statement is true. >But, this >does not make the statement S therefore true. My first impulse is to say I didn't say it did make S therefore > true. >But, we are talking about Steven's principle stated in his post. > But I'm going to wait on that, since the this > evidently refers to the previous sentence and I'm assuming > that I don't know what you meant to write there... >The statement under contention is >[i] THE ONLY WAY THAT YOU CAN CONVINCE ME >where S is false. and your paraphrase of the statement is >[ii] If I can't show an instance where S is false, >then S has to be true (and thus proved). Again: it doesn't look to me like [i] and [ii] are saying the same thing. Possibly [ii] is what was meant by [i], but [ii] is not what [i] says. The difference is the reason for my confusion about what you were saying - if [rightly or wrongly, justified by the rest of the thread or not] you're reading [i] as saying [ii] then I don't find your comments so mysterious. >Since S is undecidable, I can't show an instance where S is false. >But, I can't show an instance where not S is false either. >That is, S is neiter true or false: S is undecidable. Thus, >Steven's principle of proof does not always work. >P.S. Of course, Steven may not have meant this. I think the many >replies in this thread itself give the impression that there is >a principle that p implies q is true whenever p is false. >Of course that is true, but I don't like to think of it that way. >I gave a direct proof of the original poster's request. But, >maybe I show have given the proof by contradiction. I think >most (all?) are taking it is obvious that p is false. I think >that should be proved and it comes out automatically when >you work with the original statement that was to be proved. >If you throw away all the red herrings, you come to the >following statement to be proved: if 4 divides a then a is not prime. >-- Hale === Subject: How to show irreducibility of a polynomial Hi all, how can I show that the polynomial T^8 - 40 T^6 + 352 T^4 - 960 T^2 + 576 in Q[T] is irreducible over Q ist? (According to MuPAD (an algebra system) that is the case, but I need a calculation (that is a proof.)). PS: I got the polynomial by the following MuPAD-calculation (In case it helps...): fac := [ T+a*sqrt(2)+b*sqrt(3)+c*sqrt(5) $ a in [-1,1] $ b in [-1,1] $ c in [-1,1] ]; p := _mult ( i $ i in fac ); f:= expand (p); g := poly (f, [T], Dom::Rational); === Subject: Re: How to show irreducibility of a polynomial > Hi all, > how can I show that the polynomial > T^8 - 40 T^6 + 352 T^4 - 960 T^2 + 576 in Q[T] > is irreducible over Q ist? > PS: I got the polynomial by the following MuPAD-calculation > (In case it helps...): > fac := [ T+a*sqrt(2)+b*sqrt(3)+c*sqrt(5) > $ a in [-1,1] $ b in [-1,1] $ c in [-1,1] ]; > p := _mult ( i $ i in fac ); > f:= expand (p); > g := poly (f, [T], Dom::Rational); So its zeroes are +- sqrt(2) +- sqrt(3) +- sqrt(5). By Kummer theory, Q(sqrt(2), sqrt(3), sqrt(5)) has degree 8 over Q. It is a Galois extension, with Galois group of order 8 with elements g(u,v,w) where g(u,v,w): sqrt(2) -> u sqrt(2), sqrt(3) -> v sqrt(3), sqrt(5) -> u sqrt(5) where u, v and w run through {1, -1}. The roots of your polynomial are therefore the Galois conjugates of sqrt(2) + sqrt(3) + sqrt(5) over Q: the polynomial is its minimal polynomial, so irreducible. === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: How to show irreducibility of a polynomial >Hi all, >how can I show that the polynomial >T^8 - 40 T^6 + 352 T^4 - 960 T^2 + 576 in Q[T] >is irreducible over Q ist? >PS: I got the polynomial by the following MuPAD-calculation >(In case it helps...): >fac := [ T+a*sqrt(2)+b*sqrt(3)+c*sqrt(5) > $ a in [-1,1] $ b in [-1,1] $ c in [-1,1] ]; >p := _mult ( i $ i in fac ); >f:= expand (p); >g := poly (f, [T], Dom::Rational); > So its zeroes are +- sqrt(2) +- sqrt(3) +- sqrt(5). I have constructed the ponynomial to have that zeros. > By Kummer theory, Q(sqrt(2), sqrt(3), sqrt(5)) > has degree 8 over Q. It is a Galois extension, with Galois > group of order 8 with elements > g(u,v,w) where > g(u,v,w): sqrt(2) -> u sqrt(2), sqrt(3) -> v sqrt(3), sqrt(5) -> u sqrt(5) > where u, v and w run through {1, -1}. The roots of your polynomial > are therefore the Galois conjugates of sqrt(2) + sqrt(3) + sqrt(5) > over Q: the polynomial is its minimal polynomial, so irreducible. Nice argument - and indeed very short. I know you could not have known, but I wanted to show the irreducibility in order to show that L := Q(sqrt(2), sqrt(3), sqrt(5)) is a galois extension (we did not learn Kummer theory (yet)). My argumentation was planed to be as follows: Certainly [L : Q] <= 2 * 2 * 2 = 8. If we can prove that there are 8 pairwuise different Q-Autopmorphism, then we are ready. By some lemma of our tachings, it is sufficient to show that if alpha is a primitive element ( sqrt(2)+sqrt(3)+sqrt(5) turns out to be one ) then by sigma: alpha |-> any null of minpo_Q(alpha) there are all Q-Automorphisms given. So guess 8 pairwise different Q-Automorphisms of L, that map sqrt(2),sqrt(3) and sqrt(5) to the following values: sigma_1 : sqrt(2) |-> sqrt(2), sqrt(3) |-> sqrt(3), sqrt(6) |-> sqrt(6) sigma_2 : sqrt(2) |-> sqrt(2), sqrt(3) |-> sqrt(3), sqrt(6) |-> -sqrt(6) sigma_3 : sqrt(2) |-> sqrt(2), sqrt(3) |-> -sqrt(3), sqrt(6) |-> sqrt(6) sigma_4 : sqrt(2) |-> sqrt(2), sqrt(3) |-> -sqrt(3), sqrt(6) |-> -sqrt(6) sigma_5 : sqrt(2) |-> -sqrt(2), sqrt(3) |-> sqrt(3), sqrt(6) |-> sqrt(6) sigma_6 : sqrt(2) |-> -sqrt(2), sqrt(3) |-> sqrt(3), sqrt(6) |-> -sqrt(6) sigma_7 : sqrt(2) |-> -sqrt(2), sqrt(3) |-> -sqrt(3), sqrt(6) |-> sqrt(6) sigma_8 : sqrt(2) |-> -sqrt(2), sqrt(3) |-> -sqrt(3), sqrt(6) |-> -sqrt(6) Construct from them the minimal polynomial of sqrt(2)+sqrt(3)+sqrt(5) and forget the gueesed Q-Automorphisms -- they were just needed to get a candidate for the minimal polynomial quite easy. If T^8 - 40 T^6 + 352 T^4 - 960 T^2 + 576 in Q[T] would be irreducible, it certainly would be the minimal polynomial of sqrt(2)+sqrt(3)+sqrt(5) -- and by degree argument it is clear that sqrt(2)+sqrt(3)+sqrt(5) is a primitive element of the field extension. The reason for such unusual argumentation is that it is unneccessary to show that (1) sqrt(3) not in Q(sqrt(2)) (2) sqrt(5) not in Q(sqrt(3), sqrt(5)) (3) {...} is a Q-Basis of L (1)+(2) is normally needed in order to show that Q(sqrt(2), sqrt(3), sqrt(5)) has degree 8 over Q. In our lecture, it turned out to be wuite hard to proof (2), so I looked for an alternative way of prooving it -- and therefore point (2) was ommitted. === Subject: Re: How to show irreducibility of a polynomial > The reason for such unusual argumentation is that it is unneccessary to > show that > (2) sqrt(5) not in Q(sqrt(3), sqrt(5)) > In our lecture, it turned out to be wuite hard to proof (2), so I looked > for an alternative way of prooving it -- and therefore point (2) was > ommitted. Kummer theory is the machine that reduces these to trivialities. But if you want a naive argument try this. Lemma If L = K(sqrt(r)) where K is a field of characteristic not 2, and s in K is a square in L then either s is a square in K or rs is a square in K. Proof: s = (a + b sqrt(r))^2 = a^2 + b^2 r + 2ab sqrt(r). There's nothing to prove if sqrt(r) is in K so assume not. Then 2ab = 0, so b = 0 or a = 0 and s = a^2 or s = b^2 r. QED If 5 is a square in Q(sqrt(2), sqrt(3)) then 5 or 15 is a square in Q(sqrt(2)) do 5, 15, 10 or 30 is a square in Q. Not!! === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: How to start with this problem? Let A,B and C be sets such that AUB=AUC and A intersection B = A intersection C, show that B=C. Well, it seems obious to me that B=C, but mathematically, where do I start from? Sure its a homework question, but can any one just please give me the first one or two steps to begin? msk === Subject: Re: How to start with this problem? 996109a6.0307290411.1b3bea0b@posting.google.com... > Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. For such problems, don't think too much, use the indicator function instead :) First condition gives: (1 - I_A)*(I_B - I_C) = 0 Second condition gives: I_A*(I_B - I_C) = 0 Adding those two equalities gives you I_B = I_C, hence B = C. -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France === Subject: Re: How to start with this problem? > Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. > Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? The usual way is to show that the left-hand side is a subset of the right-hand side, and vice versa. To show that B is a subset of C, you can go back to the definitions of unions. Take an element of B. What do you know about it? Can you prove from the givens above that it must lie in C? === Subject: Re: How to start with this problem? > Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. > Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? > msk If you want to prove that two sets are equal, you can usually do it by proving that they are subsets of each other. If you want to prove that a set is a subset of another set, you can usually do it by taking an arbitrary element from the first set and prove that it is an element of the second set. Try this for a while, and then look at the spoiler below... spoiler coming... spoiler coming... spoiler coming... spoiler coming... spoiler arrived! For all x: x in B ==> x in AuB because B is a subset of AuB ==> x in AuC because AuB = AuC Now we have 2 possibilities: 1) x not in A ==> x in C because x in AuC but not in A 2) x in A ==> x in AnB because we started with x in B ==> x in AnC because AnB = AnC ==> x in C because AnC is a subset of C so in both cases we have x in B ==> x in C so B is a subset of C. You should now be able to complete the second part. Dirk Vdm === Subject: Re: How to start with this problem? > Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? > msk > If you want to prove that two sets are equal, you can > usually do it by proving that they are subsets of each other. > If you want to prove that a set is a subset of another > set, you can usually do it by taking an arbitrary element > from the first set and prove that it is an element of > the second set. > Try this for a while, and then look at the spoiler below... > spoiler coming... > spoiler coming... > spoiler coming... > spoiler coming... > spoiler arrived! > For all x: > x in B > ==> x in AuB because B is a subset of AuB > ==> x in AuC because AuB = AuC > Now we have 2 possibilities: > 1) x not in A > ==> x in C because x in AuC but not in A > 2) x in A > ==> x in AnB because we started with x in B > ==> x in AnC because AnB = AnC > ==> x in C because AnC is a subset of C > so in both cases we have > x in B ==> x in C > so > B is a subset of C. > You should now be able to complete the second part. Actually, due to symmetry, you are done. However it would be advisable for the OP to work it out. Dirk Vdm === Subject: Re: How to start with this problem? Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? > msk If you want to prove that two sets are equal, you can > usually do it by proving that they are subsets of each other. > If you want to prove that a set is a subset of another > set, you can usually do it by taking an arbitrary element > from the first set and prove that it is an element of > the second set. Try this for a while, and then look at the spoiler below... spoiler coming... spoiler coming... spoiler coming... spoiler coming... spoiler arrived! For all x: > x in B > ==> x in AuB because B is a subset of AuB > ==> x in AuC because AuB = AuC > Now we have 2 possibilities: > 1) x not in A > ==> x in C because x in AuC but not in A > 2) x in A > ==> x in AnB because we started with x in B > ==> x in AnC because AnB = AnC > ==> x in C because AnC is a subset of C so in both cases we have > x in B ==> x in C > so > B is a subset of C. You should now be able to complete the second part. > Actually, due to symmetry, you are done. However it would be advisable for > the OP to work it out. Hush!.... That was part of the lesson - he should have come up with that himself ;-) Hopefully not too late at this stage... Dirk Vdm === Subject: Re: How to start with this problem? Dirk Van de moortel Let A,B and C be sets such that AUB=AUC and > A intersection B = A intersection C, show that > B=C. Well, it seems obious to me that B=C, but mathematically, > where do I start from? Sure its a homework question, > but can any one just please give me the first one or > two steps to begin? > msk If you want to prove that two sets are equal, you can > usually do it by proving that they are subsets of each other. > If you want to prove that a set is a subset of another > set, you can usually do it by taking an arbitrary element > from the first set and prove that it is an element of > the second set. Try this for a while, and then look at the spoiler below... spoiler coming... spoiler coming... spoiler coming... spoiler coming... spoiler arrived! For all x: > x in B > ==> x in AuB because B is a subset of AuB > ==> x in AuC because AuB = AuC > Now we have 2 possibilities: > 1) x not in A > ==> x in C because x in AuC but not in A > 2) x in A > ==> x in AnB because we started with x in B > ==> x in AnC because AnB = AnC > ==> x in C because AnC is a subset of C so in both cases we have > x in B ==> x in C > so > B is a subset of C. You should now be able to complete the second part. Actually, due to symmetry, you are done. However it would be advisable for > the OP to work it out. > Hush!.... > That was part of the lesson - he should have come up > with that himself ;-) > Hopefully not too late at this stage... Sorry. I just slapped my hand. I will be quite, I promise (at least until the next one comes along). > Dirk Vdm === Subject: Re: How to start with this problem? Hush!.... > That was part of the lesson - he should have come up > with that himself ;-) > Hopefully not too late at this stage... > Sorry. I just slapped my hand. I will be quite, I promise (at least until > the next one comes along). Okay, one for you then :-) Dirk Vdm === Subject: Interesting question. Is it conceivable in any mathematical system that the smallest ordinal - assuming that the ordinals begin after the additive identity - is NOT the same as the multiplicative identity? This would be equivilant of asking if 1 times a number would not be equal to that number. But let's not confuse ourselves by using the number '1' here. Let's just think in terms of the abstract. Also, can the multiplicative identity change depending on its orientation in n space in any conceivable manifold? That is, multiplying 1 * 2 = 2 would not be the same as 1 * 1 * 2, which might be, oh, say, 4. But again, let's not confuse ourselves with numbers like 1, 2, and 4. (...Starblade Riven Darksquall...) === Subject: Re: Interesting question. > Is it conceivable in any mathematical system that the smallest ordinal > - assuming that the ordinals begin after the additive identity - is > NOT the same as the multiplicative identity? I can't even begin to understand what you're asking, but maybe this will interest you: In the ring 2Z / 18Z, whose elements can be given as {0, 2, 4, 6, 8, 10, 12, 14, 16}, the additive identity is 0 and the multiplicative identity is 10. === Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Interesting question. charset=Windows-1252 > In the ring 2Z / 18Z, whose elements can be given as > {0, 2, 4, 6, 8, 10, 12, 14, 16}, > the additive identity is 0 and the multiplicative identity > is 10. What are the rules for a ring (versus a field)? I noticed that 6x6 = 6x12 = 12x12 = 0, so quotients are not unique within this ring. === Subject: Re: inverse of a matrix sum? Content-transfer-encoding: 8bit > Given A and B, each an n x n matrix with non-zero determinant, and > which I have already calculated the inverses A^-1 and B^-1, > is there a simple formula for the inverse of (A+B)? Is there > a computational method of calculating (A+B)^-1 that is less > computational effort because A^-1 and B^-1 are known? [...] I've been waiting for someone to point out that A + B may not be invertible. I guess it's up to me. A + B may not be invertible. === === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can Americans have had it too easy for too long. Like domesticated animals who are provided everything, they loose their survival skills, their willingness to work for their needs. Have hard work to do, like roofing? Look at the crews. How many are Mexican? Have hard work to do, like engineering? Where have American kids gone? To business college. Without competitors, an organism gets flabby and lazy. We're out of shape and competitors have arrived... === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can > ny Can't Add There's a lot you can't do, ny boya. Most of all, you can't be a decent human being. Got mangoes, ny boya? === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can charset=Windows-1252 Regarding Asians, a lot of first and second generation Asians don't speak US English that well, so they have a tendency to focus on the sciences. Also, a high percentage of the first generate Asian immigrants were well educated, upper class citizens before they came here to the US. This created a foundation and culture for future Asian immigrants. The same is true of many immigrants from India, we're picking the cream of the crop from the applicants to naturalize. I'm not sure what the percentage representation there is in science and industry for caucasion USA citizens, but my guess is that in sales, non-technical management, accounting, and other similar jobs, that the percentages of immigrants will be less. I'm also curious as to the distribution in the fields of architecture, civil engineering, chemical engineering, mechanical engineering are, as these are the older forms of engineering. === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can >ny Can't Add >But Suresh Venktasubramanian Can >By Fred Reed >None of this is original with me. In 1999, the National >Academy of Sciences released a study noting that over >half of U.S. engineering doctorates are awarded to >foreign students. Where are Smith and Jones? At the bar having a cool one. >Why are members of these very small groups doing so much >of the important research for the United States? That's >easy. They're smart, they go into the sciences, and they >work hard. Potatoes are more mysterious. It's not >affirmative action. They produce. The qualifications of >these students can easily be checked. They have them. The >question is not whether these groups perform, or why, but >why the rest of us no longer do. What has happened? Simple - there are no rewards for that kind of performance. We reward hard work by ... more hard work. Successful scientists and engineers are blessed with 60 hour weeks and no life. >It is not an easy question, Of course it is. >but a lot of it, I think, is >the deliberate enstupidation of American education. But it can't be, for reasons that will become evident. >Again, the idea is not original with me. Said the >American Educational Research Association of the NAS >report, Serious deficiencies in American pre-college >education, No. >along with wavering support for basic research, Yes. >were cited by the panel as major contributors to this problem. >Consider mathematics. In the mid-Sixties I took freshman >chemistry at Hampden-Sydney College, a solid school in >Virginia but not nearly MIT. It was assumed-assumed >without thought-that students knew algebra cold. They had >to. You can't do heavy loads of highly mathematical >homework, or wrestle with ideas like integrating >probability densities over three-space, or do endless >gas-law and reaction-rate calculations, if you aren't >sure how exponents work. >Remedial mathematics at the college level was unheard of. You just weren't aware of it. There was remedial college math in the 60s. Indeed, there were college students who had never taken Algebra, though they were unlikely to be in a solid school. Remember that even if Algebra is required for high school graduation, and for admission to college X, that requirement can be fulfilled by having taken the class as a freshman and getting a D-, then never using any algebra again for the rest of high school. What are the odds that such a kid 4 or more years later will know algebra? Of course such kids wouldn't have any interest in integrating probability densities over three-space, or do endless gas-law and reaction-rate calculations, and in fact they would not end up in the classes that the author was taking. >The assumption was that people who weren't ready for >college work should be somewhere else. Not necessarily. They could be at a lesser college, or they could be in a non-technical major. Why does an English Lit major need to know algebra? >No one thought >about it. Today, remedial classes in both reading and >math are common at universities. We seem to be dumbing >ourselves to death. Missing from this is that many of the people taking these remedial classes are people returning to school after years in the workforce. >I recently had children go through the high schools of >Arlington, Va., a suburb of Washington. I watched them >come home with badly misspelled chemistry handouts from >half-educated teachers, watched them do stupid, make-work >science projects that taught them nothing about the >sciences but used lots of pretty paper. And yet Arlington high schools produce some of the top students in the country. >The extent of scholastic decline is sometimes >astonishing. So help me, I once saw, in a middle school >in Arlington, a student's project on a bulletin board >celebrating Enrico Fermi's contributions to Nucler >Physicts (Scripps-Howard National Spelling Bee >2001, Sean Conley; 2000, George Thampy; 1999, Nupur >Lala). So some Americans can't spell, or rather, don't care whether they can spell? A different and somewhat unrelated problem. I am sure that some other student spelled the word correctly, but the teachers don't put only the best students' work on the wall these days. >It appears that a few groups are keeping their standards >up and the rest of us are drowning our children in self- >indulgent social engineering, political correctness, and >feel-good substitutes for learning. But if it were the education system, then the groups could not keep their standards up. Those spelling bee champions, despite their foreign sounding last names, were probably all American kids, born in this country and most of them attending public schools. It is not that the other students aren't being taught, but rather that they are the horses led to water who cannot be made to drink. They'd rather revel in the hormones of teenhood, rebelling against parents and authority and chillin' with their homies when they aren't tying to get inside the pants of the opposite gender. And for the most part, American parents indulge their kids by allowing this. But some groups' parents are less likely to, so their kids excel in high school and lead the way in college. So the problem is American values, not the American education system. >Some of our growing dependency is hidden. We do not >merely rely on small industrious groups in America and on >foreigners working here. Increasingly the United States >contracts out its technical thinking to Asia. But it isn't dependency, it is economics. Asian labor is cheaper. US corporations can hire several Indian or Thai workers in those countries for the price of one worker in this country, and they don't have to worry about American labor laws. It isn't necessarily that the workers are better - they might be but they don't have to be - but that usually two workers can do more work than one, no matter how good the one worker is. >If you read technically aware publications like Wired >magazine (and how many people do?) Not many. You can't make money or get laid by reading, and those are the two primary American pastimes. >you find that major >American corporations have more and more of their >computer programming done by people in, for example, >India. In cities like Bombay, large colonies of Indians >work for U.S. companies by Internet. This again means >that counting names at American institutions >underestimates the growth of intellectual dependence. That presumes that computer programming is a mark of intellectual superiority. >This too we would be wise to ponder. Americans often >think of India chiefly as a land of ghastly poverty. >Well, yes. It is also a country with about three times >our population and a lot of very bright people who want >to get ahead. They're professionally hungry. We no longer >are. That's a good summary. >People speak of globalization. This is it, and it's just >beginning. Where will it take us? How long can we >maintain a technologically dominant economy if we are, as >a country, no longer willing to do our own thinking? If >we rely heavily on less than 10 percent of our own >population while employing more and more foreigners >abroad? The answer is simple: when competition starts to hurt American workers, the American workers will act smarter. They don't need to yet. >It's not them. It's us. I've heard the phrase, the Asian >challenge to the West. I don't think so. When Sally Chen >gets a doctorate in biochemistry, she's not challenging >America. She's getting a doctorate in biochemistry. Those >who study have no reason to apologize to those who don't. Why would anyone think that getting a doctorate is a reason to apologize. Of course, if Sally Chen was born in this country, she is an American and not an Asian - a problem that too many analysts who engage in name-counting forget. >The Mathematical Association of America runs a contest >for the extremely bright and prepared among high-school >students. It is called the United States of America >Mathematics Olympiad, and it provides a means of >identifying and encouraging the most creative secondary >mathematics students in the country. >An unedited section of a list of those recently chosen: >Sharat Bhat, Tongke Xue, Matthew Peairs, Wen Li, Jongmin >Baek, Aaron Kleinman, David Stolp, Andrew Schwartz, Rishi >Gupta, Jennifer Laaser, Inna Zakharevich, Neil Chua, >Jonathan Lowd, Rubinsteinsalze, Joshua Batson, >Jimmy Jia, Jichao Qian, Dmitry Taubinsky, David Kaplan, >Erica Wilson, Kai Dai, Julian Kolev, Jonathan Xiong, >Stephen Guo. >Q.E.D. Since his premise is not clearly stated, it isn't clear what he thinks he's proven. There's every reason to believe that all those kids are American kids, even though their names aren't Smith and Jones. And this is the central fact that the author is forgetting: Smith and Jones are no longer a dominant majority in this country, so of course other names will appear. And perhaps those families, because they are more recent immigrants, have more ambitious standards - that is a mark of the families, and not the schools. But it is not a sign of weakness in this country - America has always prospered by its immigrants. It's just that the blue-bloods like to pretend that they were the ones who made this country succeed because they were here first (which is also wrong, since the Amerinds beat them here). lojbab === lojbab lojbab@lojban.org LeChevalier, Founder, The Logical Language Group (Opinions are my own; I do not speak for the organization.) Artificial language Loglan/Lojban: http://www.lojban.org === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can Consider mathematics. In the mid-Sixties I took freshman >chemistry at Hampden-Sydney College, a solid school in >Virginia but not nearly MIT. It was assumed-assumed >without thought-that students knew algebra cold. They had >to. You can't do heavy loads of highly mathematical >homework, or wrestle with ideas like integrating >probability densities over three-space, or do endless >gas-law and reaction-rate calculations, if you aren't >sure how exponents work. Remedial mathematics at the college level was unheard of. > You just weren't aware of it. There was remedial college math in the > 60s. Indeed, there were college students who had never taken Algebra, > though they were unlikely to be in a solid school. By 1889, 80% of our colleges and universities had well established preparatory/remedial programs in place. Today that number is 81%. The % was similar in the 60s. > Not necessarily. They could be at a lesser college, or they could be > in a non-technical major. Why does an English Lit major need to know > algebra? Yes, and, well, sometimes people change majors. >No one thought >about it. Today, remedial classes in both reading and >math are common at universities. We seem to be dumbing >ourselves to death. > Missing from this is that many of the people taking these remedial > classes are people returning to school after years in the workforce. Someone elsewhere shared numbers on this recently. I may have saved that post. I'll have to go look. Meanwhile, many are older, returning students. Some are also immigrants. I had a student a couple years ago who moved in during her senior year. She handled math quite well but hadn't had time to get much of a handle on her English. The University wisely accepted this very bright and determined girl - but on the condition she enroll in remedial reading/writing. === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can 60s. Indeed, there were college students who had never taken Algebra, > though they were unlikely to be in a solid school. This was posted by Sam Lubell a few days ago: Yes, but nearly half (46%) of freshmen in remedial courses were over 22 (meaning they most likely finished high school three years before enrolling) and 27% were over 30. I would add that for the 46%, not only would it mean it was at least 3 years since they finished h.s., it could have 6 or 7 or more since they took algebra which, at least in my experience, is usually a h.s. freshman course. (http://www.ihep.com/Pubs/PDF/Remediation.pdf) === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can >The Mathematical Association of America runs a contest for the extremely >bright and prepared among high-school students. It is called the United >States of America Mathematics Olympiad, and it provides a means of >identifying and encouraging the most creative secondary mathematics >students in the country. >An unedited section of a list of those recently chosen: Sharat Bhat, >Tongke Xue, Matthew Peairs, Wen Li, Jongmin Baek, Aaron Kleinman, David >Stolp, Andrew Schwartz, Rishi Gupta, Jennifer Laaser, Inna Zakharevich, >Neil Chua, Jonathan Lowd, Rubinsteinsalze, Joshua Batson, Jimmy >Jia, Jichao Qian, Dmitry Taubinsky, David Kaplan, Erica Wilson, Kai Dai, >Julian Kolev, Jonathan Xiong, Stephen Guo. >Q.E.D. > Since his premise is not clearly stated, it isn't clear what he thinks > he's proven. There's every reason to believe that all those kids are > American kids, even though their names aren't Smith and Jones. And this > is the central fact that the author is forgetting: Smith and Jones are no > longer a dominant majority in this country, so of course other names will > appear. And perhaps those families, because they are more recent > immigrants, have more ambitious standards - that is a mark of the > families, and not the schools. [...] Yes, I was puzzled by this also. I've known many with names that would sound very similar to those above, and many of them went through the same ol' crappy educational system as every other American. The assumption above is that Stephen Guo, for example, is either a recent immigrant or something of that sort, when for all we know, Stephen Guo's great grandfather worked on the railroads in California. This kind of assumption ends up meaning little but revealing some of the biases that people who write this kind of material. Perhaps the author was unfamiliar with ny Venktasubramanian. === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can -----BEGIN PGP SIGNED MESSAGE----- >ny Can't Add >But Suresh Venktasubramanian Can >By Fred Reed >Why are members of these very small groups doing so much >of the important research for the United States? That's >easy. They're smart, they go into the sciences, and they >work hard. Potatoes are more mysterious. It's not >affirmative action. They produce. The qualifications of >these students can easily be checked. They have them. The >question is not whether these groups perform, or why, but >why the rest of us no longer do. What has happened? > Simple - there are no rewards for that kind of performance. > We reward hard work by ... more hard work. Successful > scientists and engineers are blessed with 60 hour weeks and no life. I was talking to my brother-in-law about this social phenomena. Young people today get conflicting messages: 21st-century capitalism broadcasts what I call leisure/consumption values at them 24/7 as part of their ad campaigns. And yet the underpinnings of said capitalism rest on progress in the sciences and mechanical arts, and mastering those fields requires behaviors nearly completely antithetical to that of the consumer/leisure message. It's little wonder then that kids from families the least indoctrinated in the consumer/leisure values are overrepresented. What can be done? A la Marvin Harris, the same thing that was done when the previous incarnation of when this happened: make conspicious consumption bad form. And the best way to do this is obliquely, to have a progressive income tax system with high rates on upper income earners. As Harris noted, once the display of wealth got you identified as someone who *should* be paying high taxes, the kids of the rich too started wearing tattered blue jeans, too. Equalitarian societies suppress displays of individual power and wealth, while hierarchical societies revel in them. Given the economic demographics since the 1980s, it's little wonder that as our society becomes more economically unequal, gaining social prestige by overt consumerism and displays of wealth have gotten more and more widespread. And the long-term consequences and effects aren't good. Stewart -----BEGIN PGP SIGNATURE----- Version: 6.5.8ckt http://www.ipgpp.com/ iQCVAwUBPyayYzSyh6XxoDgRAQF4gQQApYSIMKOmSJ59rJpT8WV55h6MEzraYPR/ ZUjw9xXazLQLwnhjlA9i3k6w+laTuNxQNVPw5qfgFCmWAh2Nv6fCBgV4BtMCBVD1 oqvNpncTLBJ7SrPcBzww/n/agvaNRFkY3wuuRNeDpLXXhW+X64BY+5HousEVTYPN nj+kvZFVU+k= =QCwJ -----END PGP SIGNATURE----- === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can >Regarding Asians, a lot of first and second generation Asians >don't speak US English that well, so they have a tendency >to focus on the sciences. Also, a high percentage of the >first generate Asian immigrants were well educated, upper >class citizens before they came here to the US. This created >a foundation and culture for future Asian immigrants. >The same is true of many immigrants from India, we're picking >the cream of the crop from the applicants to naturalize. >I'm not sure what the percentage representation there is in >science and industry for caucasion USA citizens, but my guess >is that in sales, non-technical management, accounting, and >other similar jobs, that the percentages of immigrants will >be less. >I'm also curious as to the distribution in the fields of >architecture, civil engineering, chemical engineering, >mechanical engineering are, as these are the older forms >of engineering. It is quite possible that ny can add, but it is very unlikely that ny has any idea that addition is more than the mechanics of base 10 plug-and-chug. The same holds for anything else in which more than memorization and routine are involved; the schools make it very difficult, if not in many cases impossible, for it to be otherwise. The teachers are in this category. Memorization and routine does not make it any easier to understand anything whatever. === This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Deptartment of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can > ny Can't Add > But Suresh Venktasubramanian Can In my country too there seems to be a rising antiintellectualism. Certainly in the schools it is well known that for boys it is uncool to work hard. Also here, asian families respect hard work, intellectual and other. > .... Where are Smith and Jones? Hiding out from the bounty hunters? GC === Subject: Re: lrange/string range Originator: claird@lairds.com (Cameron Laird) ># . ># . ># . ># >Why is it O! = 1 instead O! = 0? ># # >Some decisions are arbitrary; the person setting the precedent makes the ># . ># . ># . ># Why is ># 0! = 1 ># ? To agree with the gamma # mathworld.wolfram.com/GammaFunction.h >. >Actually Gamma was derived to match factorial, which had already decided 0! = 1. I was poking fun, of course. Chronologically, yes, certainly, it (along with everything else), and, by 1816, Arbogast had established our n! notation, although the old |_n notation continued in use well into the twentieth century. Euler only introduced the gamma in 1729, and it was Gauss at the beginning of the next century who established that name for it. HOWEVER, even if someone early on had thought 0! := 0 a superior convention, it wouldn't have survived long. Taylor's formula works better with 0! := 1, and Stirling in 1730 is already hinting at the gamma with his analytic generalizations. If you think of the factorial purely in terms of combinatorics, well then, yes, an evaluation at zero is a matter of conven- tion. Historically, though, probabilistic calculations were *not* what brought attention to the factorial; it was the analytic work of people like Taylor and Euler. In that sense, factorial *had* to agree with the gamma. === Cameron Laird Business: http://www.Phaseit.net Personal: http://phaseit.net/claird/home.h === Subject: Re: lrange/string range > . > . > . >Why is it O! = 1 instead O! = 0? Because n! is the number of permutations of n objects, and there is exactly one way to order the empty set. === Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: lrange/string range >HOWEVER, even if someone early on had thought 0! := 0 a >superior convention, it wouldn't have survived long. Taylor's >formula works better with 0! := 1, and Stirling in 1730 is >already hinting at the gamma with his analytic generalizations. >If you think of the factorial purely in terms of combinatorics, >well then, yes, an evaluation at zero is a matter of conven- >tion. On the contrary: if you think of the factorial purely in terms of combinatorics, and think of combinatorics as the study of the category of finite sets, then 0! = 1 is no convention at all, it's just a case of the general definition. Given a (finite) set X (which might be empty), let X! denote the set of bijections from X to X (an eminently combinatorial notion), then prove (utterly trivially) that the cardinality of X! depends only on the cardinality of X, and so justify defining n! in the usual way for n a (finite) definition 0! := n with n not equal to 1 would not only be a convention, it would be a really, really stupid convention, in the face of the theorem 0! = 1 that I just proved. Lee Rudolph === Subject: Re: lrange/string range >HOWEVER, even if someone early on had thought 0! := 0 a >superior convention, it wouldn't have survived long. Taylor's >formula works better with 0! := 1, and Stirling in 1730 is >already hinting at the gamma with his analytic generalizations. >If you think of the factorial purely in terms of combinatorics, >well then, yes, an evaluation at zero is a matter of conven- >tion. > On the contrary: if you think of the factorial purely in terms > of combinatorics, and think of combinatorics as the study of > the category of finite sets, then 0! = 1 is no convention at all, > it's just a case of the general definition. Given a (finite) set > X (which might be empty), let X! denote the set of bijections from > X to X (an eminently combinatorial notion), then prove (utterly > trivially) that the cardinality of X! depends only on the cardinality > of X, and so justify defining n! in the usual way for n a (finite) > definition 0! := n with n not equal to 1 would not only be a > convention, it would be a really, really stupid convention, in > the face of the theorem 0! = 1 that I just proved. And similar considerations give 0^0 = 1. === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Mapping of integers to reals ~ Cantor's disproof [...] > What about mapping every irrational onto a rational? Can that be done? No. Irrationals have cardinality greater than aleph-null since they're not countable. > Or better yet: Mapping every transcendental number onto a, uh, what is > the nickname or nontranscendental? algebraic? Gemoetric? Polynomialic? > Whatever. > (What is the nickname for nontranscendental anyways?) Algebraic I think. > Well can you map every transcendental to a... uh... that? I'm not sure. You may be able to prove equivalency of the two cardinalities, but I don't think you can decide which would be larger, since that would imply a cardinality lying between aleph-null and c which has already been shown to be not proveable. === New definition of irony: 'Today's liberal Democrats are like the supporters of the Third Reich of the '30's and '40's - they absolutely trusted the government to make things right. ' -Comment made on the internet by an ardent GW Bush supporter. === Subject: Re: Mapping of integers to reals ~ Cantor's disproof |-|erc says... >I'm using the fact that H(n) is noncomputable to prove that H(555) cannot >be equal to 1 *and* that H(555) cannot be equal to 0. > Well, that doesn't make any sense. The fact that H is a noncomputable > function doesn't imply that H(555) is noncomputable. >But H(555) can be noncomputable, No, it can't. H(555) is some natural number, and every natural number is computable. >the analysis is perfectly valid using that assumption, and there >certainly exist n where H(n) is non computable. No, there doesn't. When we say that H (as a function) is noncomputable, we mean that there is no *single* function that can compute every value of H. There is a function that can compute H(0), and a *different* function that can compute H(1), and yet another function that can compute H(2), etc. Some computable function computes H(555), but it can't be the *same* function that computes H(0), H(1), H(2), etc. >Which makes pi/4 a contender for r since you used the same proof >technique. Except that your proof relies on H(555) being noncomputable, which is provably false, while my proof relies on H (the function) being noncomputable, which is provably true. > The same proof goes through for every computable real. > No computable real can be equal to r. >Nice work, but I suggest its a convention that well defined programs >terminate, Well, obviously if there is no program that can compute r, then no well-defined program can compute r, either. -- === Subject: Mathematicians, helping each other? Some of you don't realize that having rather basic algebra in my work means that I now know what mathematicians are capable of, when it comes to lying about even basic mathematics. So I can look over the entire field recognizing just what is possible from mathematicians themselves, with information from my own experiences to see exactly how it can be done. Basically in mathematics the community aspect allows mathematicians to help each other. You don't necessarily need a correct paper or a proper argument. You just need your group. === Subject: Re: Mathematicians, helping each other? > Some of you don't realize that having rather basic algebra in my work > means that I now know what mathematicians are capable of, when it > comes to lying about even basic mathematics. > So I can look over the entire field recognizing just what is possible > from mathematicians themselves, with information from my own > experiences to see exactly how it can be done. > Basically in mathematics the community aspect allows mathematicians to > help each other. > You don't necessarily need a correct paper or a proper argument. > You just need your group. > You still haven't answered the objections raised. I have made several posts with errors. They were corrected and I either explicitly or tacitly accepted the correction or explained why I disagreed. Eventually, a consensus was reached. Others here have done the same. You have not. It isn't about mathematicians having a group hug. It's about everyone using the same set of rules for how to define what is being discussed, move from one premise to the next, and avoid errors by drawing conclusions without clear support. As long as you do not follow those rules, you will not be doing mathematics above the level of simple algebra. I posted my objections to your paper. Please either address them or retract your paper. Others have done the same. Please address their objections as well. Changing the subject does not count. Hoping we'll forget does not count. === === Subject: Re: Mathematicians, helping each other? > ... > Basically in mathematics the community aspect allows mathematicians to > help each other. > ... Of course mathematicians help each other. The mathematicians in sci.math would help you if only you'd let them. I recall David Libert going to an enormous amount of trouble to help you in one of your earlier attempts. Where you grateful and willing to learn? No. Where you rude and obnoxious? Yes GC === Subject: Re: Math in my favor, short FLT proof found > What is amazing is how many of you seem to be taken in by bad > arguments for bogus mathematics as I've given Why do you lead by insulting your audience? Only a troll would do that. === Subject: Re: Math in my favor, short FLT proof found > What is amazing is how many of you seem to be taken in by bad > arguments for bogus mathematics as I've given Only one trolling ineducable boring idiot is in the basket, Harris - you. http://w0rli.home.att.net/youare.swf http://b5.sdvc.uwyo.edu/bab5/snds/argcstpd.wav NOBODY has ever agreed with you. > So here you have my claim that I've found a *short* proof of Fermat's > Last Theorem, where the methods used involve factoring polynomials > into non-polynomial factors, which you can't find in all of > mathematics outside of my work, No doubt. > meanwhile I face a lot of hostility > over my work, from people who can't show an error within the work > itself. NOBODY has ever failed to show the invalidty of your spews. -- Uncle Al http://www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: Math in my favor, short FLT proof found > and see an example where my methods work. Those methods are > extraordinarily basic, but they are radical as I factor polynomials > into non-polynomial factors. What's radical about that? Let me do the same thing for you: x^3 - 1 = (x^(3/2) - 1)(x^(3/2) + 1) Sorry to break the news to you, but there's nothing extraordinary about it. What IS extraordinary is that you can't recognize simple errors in your own work. Let me show you something just as extroadinary by factoring the integer 10 into non-integer factors. 10 = (Pi)*(Sqrt[2] / Pi)*(5Sqrt[2]) Wow, some of the factors are even transcendental. If you want I can even throw in some POLYNOMIAL factors. Amazing! In fact, I can include other things in there too if you like as long as multiplication by real numbers is defined on them. > That's it. Think carefully, what I do that's so extraordinary is > factor polynomials into non-polynomial factors. Why haven't > mathematicians done that already? Why would they fight correct > mathematics which uses such a fascinating approach? > The paper in near current form went by the New York Journal of > Mathematics without claim of error, as the chief editor said it didn't > fit their format. It's currently at another journal and I'm waiting > to hear more from them. What that means is we didn't waste time finding your errors because it doesn't even fit our format. > So here you have my claim that I've found a *short* proof of Fermat's > Last Theorem, Isn't it like 10 pages or so now? Probably more. You call that short? Show me any textbook with a theorem which takes 10+ pages to prove. where the methods used involve factoring polynomials > into non-polynomial factors, which you can't find in all of > mathematics outside of my work, meanwhile I face a lot of hostility > over my work, from people who can't show an error within the work > itself. People show errors in your work every day. However, as has been the case for the past 7 (going on 8) years, you don't read them. === Subject: Re: Minimal conditions for linearity over the reals Bil and Larry: There is probably more than one definition in use for a vector subspace. I'll explain this below. However, this does not help me to understand which set Robert was referring to in his post. Please, Robert, could you provide an explicit (preferably set-theoretic) definition of the sets you're referring to and the operations that make them vector spaces? I apologize in advance for being a little thick. and would very much appreciate it if you indulge me. A disparity in definitions of vector subspace may revolve around the question of whether or not the field of scalar multiplication is permitted to vary between the superspace and the subspace. I only have a few references here, but the ones I've looked at agree with my recallection: that a vector subspace must be a vector space over the same field as the superspace. I am happy to concede that another, more general, (and useful) definition might allow a subspace to be a vector space over some subfield of the superspace's scalar field. In the case of the reals and rationals, the rationals aren't even closed under multiplication by the scalars of the reals, and so under the more restrictive definition, can not qualify as a vector subspace. So that must not be what Robert means. But even if I accept the more general definition, and suppose he means by reals over the rationals an algebra I hadn't even considered before: the ring of real addition with scalar multiplication by rationals, I have another problem: Since this thing doesn't qualify as a subspace according to my functional analysis book, I'm unsure if I can apply the Axiom of Choice to conclude that it has a Hamel basis. I would guess that it's still a theorem, but it will take work... My references are: A Survey of Modern Algebra, 4th ed, G. Birkhoff, S MacLane, 1977 Introductory Functional Analysis with Applications, Erwin Kreyszig, 1978 >Robert, >I've been puzzling over what you mean by the reals over the rationals. >I guess you mean to somehow decompose the reals into algebraically closed >subsets. Do you mean some kind of quotient space? I know what is meant >by the quotient of a vector space with a vector subspace, but the >rationals aren't a vector subspace of the reals (in any sense I know), >Can you expand on this please? >> >> For example, let B be a Hamel basis of the reals over the rationals, >> take some b_1 in B and define f(sum_{b in B} r_b b) = r_{b_1}. >> > I think you have missed something essential. The reals are a vector > space over the rationals. > Check the definition of a vectors space over a field. > Larry > (this space unintentially left blank ..... > make obvious deletion for email === To reply by e-mail, remove 'nospam' from the e-mail address. === Subject: Re: Minimal conditions for linearity over the reals > In the case of the reals and rationals, the rationals aren't even closed > under multiplication by the scalars of the reals, and so under the more > restrictive definition, can not qualify as a vector subspace. The rationals don't need to be closed under multiplication by reals. The claim is that: (1) R is a vector space over Q i.e., the vectors are R, the scalars are Q (2) Q is a subspace of R i.e., the vectors are Q, the scalars are Q In case (2), Q is closed under (vector) addition, and it's closed under (scalar) multiplication by elements of Q. (That's all that is required: closure under multiplication by elements of the scalar field, which is Q, not R.) So Q is a subspace of R. I'm not sure if you're confused here by the fact that the vectors in this space are also fields. What is a vector space other than an abelian group (of vectors, under addition) and a field of scalars obeying the distributive laws with respect to the vectors? Well, both R and Q are fields, so they're automatically abelian groups under addition, so they can certainly be used as vectors. The distributive laws are then satisfied because R and Q also happen to have multiplication defined, compatible with the scalar field Q. === Subject: Re: Minimal conditions for linearity over the reals >There is probably more than one definition in use for a vector subspace. No, just one. >However, this does not help me to understand which set Robert was >referring to in his post. >Please, Robert, could you provide an explicit (preferably set-theoretic) >definition of the sets you're referring to and the operations that make >them vector spaces? I apologize in advance for being a little thick. >and would very much appreciate it if you indulge me. The real numbers R are a vector space over the field of rationals Q. Addition is as usual, scalar multiplication is the ordinary multiplication (by members of Q). >A disparity in definitions of vector subspace may revolve around the >question of whether or not the field of scalar multiplication is permitted >to vary between the superspace and the subspace. For the purpose of this discussion, the scalars are the rationals. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: nasty complicated equations prominently featuring the number 13 > Subject says it all, I need a very unlucky-looking > equation to torture a friend. Maybe something from > Ramanujan's number-theory work? I don't have access > to a library at the moment, and have done a lot of web > searching with no luck. Can anyone help me out here? How about the Riemann-Weil explicit formula? http://www.maths.ex.ac.uk/~mwatkins/zeta/weilexplicitformula.htm -mb === Subject: Re: Need a REALLY simple example of mathematicial induction >>Can someone please provide an example of a VERY simple theorem -- something >>trivial NOT involving sums, divisibility, modular arithmetic, etc. -- that > THEOREM: All even numbers are divisible by two. > PROOF: Consider the first even number, 2. 2 is clearly divisible by 2, >There is no first even number because if n is even so is n - 2 . > Depends on your definition of even number, now doesn't it? > If I might, you're being a bit pedantic, since the OP was only asking for I am unapologetic. I can't say for sure but my guess is that since the op asked for a very simple theorem he may not be a mathematics expert and if so he will be in danger of being confused. > a very simple induction proof, which I provided. If you're going to be > an asshole about it, In any case there's no need for vulgarity. Ladies may be present. GC > then fine: all POSITIVE even numbers are divisible > by two. Happy? > Doug === Subject: Nonprimes & Primes (All of Them) I have reworked my original paper ( http://www.tln.net/~reriker/prime.h ) based on feedback from other sci.math posters. I am not sure that I have demonstrated it in a rigorous enough manner, but I have attempted to demonstrate that it is possible to generate all of the nonprime numbers. Once you have generated these, any gaps left are the prime numbers. Perhaps this is too unwieldy to be of any general use. However, perhaps it is of some interest (assuming, of course, that I have not made any blatant errors ; of course, if I have, I am sure that someone will point them out . === Subject: Re: Nonprimes & Primes (All of Them) > I have reworked my original paper ( > http://www.tln.net/~reriker/prime.h ) based on feedback from other > sci.math posters. > I am not sure that I have demonstrated it in a rigorous enough manner, > but I have attempted to demonstrate that it is possible to generate > all of the nonprime numbers. Once you have generated these, any gaps > left are the prime numbers. An alternative way of generating the non-primes is to take products ab where a and b are at least two. Once you have generated these, any gaps left are the prime numbers. === Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.h His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Nonprimes & Primes (All of Them) That was poorly worded on my part. I guess that is what I get for writing something when I am tired. I hope that this is not as poorly written (although there is no guarantee, considering that I am still tired). The goal had been to cut down on the amount of calculation necessary to generate the nonprimes. Obviously, this is relatively easy for the even components of a and b that are at least two. I had hoped that what I was working on would be some improvement for the odd ones. Unfortunately, it was only a very marginal, if somewhat personally interesting, improvement. If nothing else, I learned a bit from the process as well as from the feedback. > An alternative way of generating the non-primes is to take products > ab where a and b are at least two. Once you have generated these, any gaps > left are the prime numbers. === Subject: Re: n-prime sets >I believe five 1's and nineteen 2's will work for n=24. Upon further inspection, it looks like every n > 105 can be done using just 1's and 2's. There are just 16 values of n up to 10000 that are infeasible: 6, 7, 10, 16 through 23, 38, and 102 through 105. Some notables: n=9736 can be done using as few as 43 1's. n=9405 can be done using only 34 2's. Empirically it looks as though these minimum numbers grow with n (for what I'm sure are good reasons), albeit very slowly. Of course even if it's true it's hopelessly unprovable, since it would imply something analogous to Bertrand's postulate, but for prime k-tuples. -- Erick === Subject: Re: n-prime sets charset=iso-8859-1 > I believe five 1's and nineteen 2's will work for n=24. I missed my train for work, so I have got some spare time:-) 1+2=3, 1+2+2=5, 1+2+2+2=7, 1+1+1+2+2=7, 1+2+2+2+2+2=11, ... Better denote that as 1/1, 1/2, 1/3, 3/2, 1/5 with the understanding a/b is short for a 1's and b 2's such that a+2b is prime. The sum a+b shall match any numbers from 2 to 24. Aha, to get from a/b to the next constellation a'/b' will need an operation [x,y], which increases a by x and b by y [x,y](a/b) := (a+x)/(b+y) It is easy to get along by [0,1], as long as a+2b=p and p+2 is prime. But circumstances will make it necessary to use other operations such as [2,-1] or [-2,3]. Nice. And thanks for the examples given so far. David? Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Please recommend some entertainment math books > Abbot EA. Flatland. (Stewart also has an annotated version, which I > have not yet read.) No, Stewart's book is a sequel, not an annotation. It's called Flatterland. You needn't have read Flatland first. It belongs on the list. === Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Please recommend some entertainment math books > >> Abbot EA. Flatland. (Stewart also has an annotated version, which I >> have not yet read.) > >No, Stewart's book is a sequel, not an annotation. It's called >Flatterland. You needn't have read Flatland first. It belongs >on the list. Incorrect, Sir. In addition to Flatterland, there is another book entitled The Annotated Flatland. Perseus Publishing: Cambridge, MA, 2002. It is sitting right here on my shelf. === Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: probably a stupid question I am currently trying to familiarize myself with tensor calculus. However I have come across a question to which I cannot find an answer. Namely, in the book I am using (D.Kay Tensor Calculus p.37) it is stated that ( _ is subscript ): If T_i are the components of covariant vector T, then S_ij = T_i T_ j - T_ j T_i are the components of a skew symmetric covariant tensor S. As far as I can understand each S_i_ j should equal 0 ( T_i T_ j = T_ j T_i). But how can a tensor, having all of it's components equal to 0, be skew symmetric? Or is it a tensor at all? But it is said to be a tensor. So should I deduce that multiplication of tensor components is not commutative? === Subject: Re: probably a stupid question > I am currently trying to familiarize myself with tensor calculus. However I > have come across a question to which I cannot find an answer. Namely, in the > book I am using (D.Kay Tensor Calculus p.37) it is stated that ( _ is > subscript ): > If T_i are the components of covariant vector T, then S_ij = T_i T_ j - T_ j > T_i are the components of a skew symmetric covariant tensor S. > As far as I can understand each S_i_ j should equal 0 ( T_i T_ j = T_ j > T_i). But how can a tensor, having all of it's components equal to 0, be > skew symmetric? Or is it a tensor at all? But it is said to be a tensor. So > should I deduce that multiplication of tensor components is not commutative? Yes, I think you are right. S_{i,j} will be the zero tensor. Of course, that does make it a skew symmetric tensor. === Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Re: reference for liouville's boundedness theorem sought === Subject: Re: reference for liouville's boundedness theorem sought >Now if we were lucky enough to have p(z) = 1 + az^m, with no >higher order terms, then it's easy to see we have a contradiction. >Because for any r > 0, the map 1 + az^m takes D(0,r) onto >D(1,|a|r^m), and D(1,r^m) contains points of absolute value < 1. >Unfortunately life is not that simple, and we have to deal with >q(z). But note the following: If |z| is small, then |q(z)| will be >much smaller than |az^m|. So the strategy is this: Choose t such >that exp(imt) = -|a|/a. Then you'll see that for small r > 0, >|p(rexp(it))| < 1 - (|a|r^m)/2, and that contradiction proves FTA. >> Let k > 0 be the smallest with ak /= 0, >> c = max{ 0, |aj| : k < j <= n } >> for |z| < 1 >> |p(z)| <= |a0 + ak z^k| + sum(j=k+1..n) |aj| |z|^j >> < |a0 + ak z^k| + c(n-k)|z| >That last line of your looks questionable. Indeed, I gave too much slack. Yes, a0 = 1. |p(z)| < |a0 + ak z^k| + c(n-k)|z|^(k+1) Let z^k = -|ak|r/ak and 0 < r < 1 |p(z)| < | a0 - |ak|r | + c(n-k)r|z| = a0 - |ak|r + c(n-k)rr^(1/k) Choose r so that |ak|r < a0 we also need c(n-k)r^(1/k) < |ak|, or when k < n r < |ak|^k / (c(n-k))^k >You have |p(z)| <= |1 + az^m| + |q(z)|, and |q(z)|/|z|^m -> 0 as z -> >0. Therefore |q(z)| <= (|a||z|^m)/2 for small z. So |p(rexp(it))| <= >1 - |a|r^m + (|a|r^m)/2 < 1 for small r > 0. Hurray thanks. Time to proclaim 'QED'? >What you want to remember though is the basic intuition: Nonconstant >polynomials take open sets to open sets. We don't need to prove all >of that, but that's the basic idea. If you know that, then FTA is >clear. And this is obvious for 1 + az^m; you're going from discs to >discs. Now you add on q(z), which is small compared to z^m. How could >that ruin things? It can't. That's unique to the complex plane. Things would go wrong were |z| not a midget. >PS: When you respond without skipping a line, like below, it's hard >to read. Ok, I'll try to remember. ---- === Subject: Re: reference for liouville's boundedness theorem sought >> Mean value theorem quickly proves odd degree p in R[x] has real >> root. What algebraist can prove that with even twice or thrice >> the labors? >The algebraist's proof uses >every odd degree real polynomial has a root and >every complex number has a squareroot. >Then, with no more analysis, proceeds to prove >every nonconstant complex polynomial has a root. Oh? I've heard the proof is much longer than an analysis proof, such as the one presented by World Wide Wade, and involves Galois theory. ---- === Subject: Re: reference for liouville's boundedness theorem sought > Oh? I've heard the proof is much longer than an analysis proof, such > as the one presented by World Wide Wade, and involves Galois theory. I like the analysts' proof, too. But whether one proof is shorter depends on what you already know. The analysts' proof will be pretty long if in has to include a proof of the Bolzano-Weierstrass theorem (in order to show that a continuous function on a closed bounded set achieves its minimum)... === G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re How to start with this problem? I was waiting for my post to appear at google groups so that i can post the second part worked out, as a reply, but it took so long, so I am posting it here.Second Part: For all x: x in C: ==> x in AuC because C is subset of AuC ==> x in AuB because AuB=AuC Now we have two possibilites 1) x not in A ==> x in B because x in AuB but not in A 2) x in A ==> x in AnC because we already started with x in C ==> x in AnB because AnC=AnB ==> x in B because AnB is subset of B Thus in both cases we conclude that x in C ==> x in B Thus C is subset of B B=C. === Subject: Re: Re How to start with this problem? > I was waiting for my post to appear at google > groups so that i can post the second part worked out, > as a reply, but it took so long, so I am posting it here.Second Part: > For all x: > x in C: > ==> x in AuC because C is subset of AuC > ==> x in AuB because AuB=AuC > Now we have two possibilites > 1) x not in A > ==> x in B because x in AuB but not in A > 2) x in A > ==> x in AnC because we already started with x in C > ==> x in AnB because AnC=AnB > ==> x in B because AnB is subset of B > Thus in both cases we conclude that > x in C ==> x in B > Thus C is subset of B > B=C. Have you really worked on the second part, or did you merely blindly swap the letters B and C of the first part? You could have done that without any problem, you know... It would have been perfectly valid to make a remark about the symmetry of B and C, to immediately write down that C is a subset B. Be careful, they don't always come with this nice symmetry property. In fact, they usually don't ;-) Cheers, Dirk Vdm === Subject: Re how to start with this question. ----- Original Message ----- === Subject: Re: Re How to start with this problem? > Have you really worked on the second part, or > did you merely blindly swap the letters B and C > of the first part? > You could have done that without any problem, > you know... It would have been perfectly valid > to make a remark about the symmetry of B and > C, to immediately write down that C is a subset > B. > Be careful, they don't always come with this nice > symmetry property. In fact, they usually don't ;-) > Cheers, > Dirk Vdm Respectable Dirk V dm! I was expecting these remarks, but be asure that I have fully it down exactly in your style was just a copy of style. The solution was easy enough for me to grab. My text book that has this question gives a hint regarding this question, but I hate such hints, so I wanted to find out some other ideas. And be asure that I have worked it out and now know how to atleast start up with such questions. :) === Subject: Reimanns Hypothesis has been proved Kaida Shi: A Geometric Proof of Riemann Hypothesis: http://front.math.ucdavis.edu/math.GM/0307136 === Subject: REQ: nasty complicated equations prominently featuring the number 13 Subject says it all, I need a very unlucky-looking equation to torture a friend. Maybe something from Ramanujan's number-theory work? I don't have access to a library at the moment, and have done a lot of web searching with no luck. Can anyone help me out here? Fred === Subject: Roots and Vectors This post offers no solution, but skirts around some possibilities. The idea is to use the dot product form of a polynomial to extract something towards a solution with vector analysis. n a[0]+SUM a[p]t^p =0 nth degree polynomial p=1 if N=(a[1],a[2],...,a[n] and P[t]=(t,t^2,t^3,...,t^n) then a[0] + N*P[t] = 0 in general n a[0] + SUM a[p]x[p] = 0 p=1 is the equation of the plane with normal N Let Q[2] be any point that satisfies the equation of the plane. Then the minimum distance from the plane to the origin becomes, N*Q[2] -------- |N| But N*Q[2] = -a[0] and the position vector Q[1] of the point on the plane of minimal magnitude (perpendicular to the plane) from the origin becomes, -a[0] Q[1] = ------- N |N|^2 If the plane is translated to intersect the origin while maintaining the same normal N, then the equation of the new plane becomes, n SUM a[p]x[p]=0 p=1 The orthogonal projection of the space curve P[t] onto this plane becomes the curve S[t] in the plane, P[t]*N N S[t] = P[t] - -------- ----- |N| |N| described as the position vector of the curve subtracted from its projection on the normal. Some points on P[t] may not be in the plane, but all points of S[t] are in the plane. The plane is restored to its original place where the plane curve S[t] becomes the plane curve R[t], P[t]*N N R[t] = Q[1] + P[t] - -------- ----- |N| |N| Note that when R[t]=P[t], P[t]*N N ------- ----- = Q[1] and the identity holds |N| |N| The plane containing S[t] is rotated to superimpose with the plane formed by the x[1],x[2] axes having unit normals n[3],n[4],n[5],...,n[n]. The planes intersect along the line joining the two points (a[1],0,0,..,0) and (0,a[2],0,0,..,0). This line forms the axis of rotation by the angle N*(0,0,1,0,..,0) N*(0,0,0,1,..,0) A = arccos ---------------- , ----------------- , ..... |N| |N| rotating in the plane parallel to the plane of (z[1],z[2],..,z[n])*(-a[1],a[2],0,0,..)=0 Once the two planes are superimposed, the same operations are performed on the curve P[t], for which all but two components will be zero and the solution more readily obtained. === Subject: Re: Search algorithm needed for specially structured data > I have a table containing numbers: > a1,a2,a3,a4,a5,a6,a7,... > b1,b2,b3,b4,b5,b6,b7,... > c1,c2,c3,c4,c5,c6,c6,... ... > Typically, these tables have 1000 columns and 1500 rows. ... > a1 < a2 < a3 < a4 ... < an ... that is, row cells are in increasing order. > Also, the column cells are in increasing order, so that > a1 < b1 < c1 ... ... > How can I efficiently search for a given number? You haven't provided sufficient information for conclusive answers. How many tables? How many searches per table? Distinct integer numbers in limited range? Is answer just Present/Not Present, or nearest number, or coordinates? Source of data? If the range is quite limited and Present/Not Present is good enough, make a bit map. For k numbers, setup will take O(k) time and each search will take O(1) time. If you have one table that you are going to search over and over, you could put all the numbers in an array, sort into order, and use binary search. For k numbers, setup takes O(k log k) time and each search takes O(lg k) time. If you have many tables and won't do many searches per table: (a) Of course you could binary search columns until you get a match or exhaust the columns, using O(m(lg n)) time for n rows and m columns. Or (b) you can do a binary binary search since if element i,j is less than your target it rules out all elements p,q with p<=i and q<=j, and if greater, all elements p,q with p>=i and q>=j; then divide and recurse. This takes O(m(lg m) + n(lg n)) time, obviously worse. Or (c) you could follow an isoline, using O(m+n+(lg n)) time, obviously far better than (a) or (b). -jiw === Subject: Re: Sets producing primes. |Hardy and Littlewood tell me that I can go on for ever, and I trust them on |that. Is it a theorem of theirs? How does it go? Keith Ramsay === Subject: Re: Sets producing primes. > This seems like a fairly standard-looking question, though I don't recall > seeing it before. > How big can be a set of natural numbers, so that one plus the product of > ANY TWO of them is a prime? A very crude estimation of densities suggests that the cardinalities of such > sets must be bounded. > Hmmm, really? > Here is such a set of cardinality 5:- {1, 2, 6, 18, 30} (The pair-products are 2 6 18 30 12 36 60 108 180 540; each + 1 is a prime) > I don't see why there should not exist some n such that n+1, 2n+1, 6n+1, > 18n+1 and 30n+1 are all prime, although such an n may be hard to find. > Heuristically, the density should be something like > k / (ln x)^5 > and the integral converges. ^^^^^^^^^ diverges > How much bigger can they be? Theory or data both welcome. I see that I am too late to be the first to improve on cardinality 5, but perhaps it is still interesting to compare the smallest examples (largest element minimized, then second largest etc.) for different cardinalities: 1 1 2 1 2 6 2 5 8 14 4 10 24 25 28 2 8 9 14 30 44 2 8 9 14 30 44 54 8 14 30 44 54 119 152 170 6 21 26 110 146 170 182 356 390 I did not find any examples with cardinality 10 and each element <= 1000. --- J K Haugland http://www.neutreeko.com === Subject: Re: Sets producing primes. > |Hardy and Littlewood tell me that I can go on for ever, and I trust them on > |that. > Is it a theorem of theirs? How does it go? Nope, a conjecture. A generalisation of the twin prime conjecture. They came up with two famous conjectures, both of which sounded on teh surface to be believable: a) No set of primes is denser than the set {2..n} b) Any admissible tuplet of primes will occur infinitely often However, they're contradictory statements. It's possible to construct an admissible tuplet (with over a thousand terms, IIRC) which has a density greater than that of the primes in {2..n}. If such a tuplet of primes exists, then it contradicts the density conjecture. (a)'s suspected of being false for this reason, but the proof may not be found in my lifetime. There's more on the Hardy & Littlewood conjectures in the glossary at Professor Caldwell's Prime Pages. http://primepages.org/ Phil === Subject: Solving equation Hi all, I've been pondering about the following equation, but cannot solve it in general: N = (A + 4*x) + (B + 4*y) where N is known and A and B are known constants and the unknowns x and y have known upper bounds. Everything is positive integers. How do I find x and y in general? For instance, A=34 and B=66 and N=1215, upper bounds are 8 for x and 16 for y. Solution is x=3 and y=3. Can I solve this efficiently in general? How would one implement a solver in a computer and what would the running time be? What if x and y are huge numbers, as in several hundred digits? I'm happy if I can solve it for small numbers, but just curios... Any help is greatly appreciated! === Subject: Re: Solving equation > I've been pondering about the following equation, but cannot solve it > in general: > N = (A + 4*x) + (B + 4*y) > where N is known and A and B are known constants and the unknowns x > and y have known upper bounds. Everything is positive integers. How do > I find x and y in general? Gee, why not Simplify? 4x + 4y = N-A-B, thus 4 divides N-A-B and N-A-B > 0 c = (N-A-B)/4, is a positive integer. So you want to find positive integer solutions to x + y = c for some positive integer c. All the solutions x,y are 1,c-1; 2,c-2;... c-2,2; c-1,1 and the bounds for x,y are lower 1 and upper c-1 === Subject: Re: Solving equation > Hi all, > I've been pondering about the following equation, but cannot solve it > in general: > N = (A + 4*x) + (B + 4*y) > where N is known and A and B are known constants and the unknowns x > and y have known upper bounds. Everything is positive integers. How do > I find x and y in general? > For instance, A=34 and B=66 and N=1215, upper bounds are 8 for x and > 16 for y. > Solution is x=3 and y=3. (34 + 4*3) + (66 + 4*3) = (34+12)+(66+12) = 46+78 = 124, not 1215 As the previous poster stated, for the equation given, one could easily simplify to x + y = (N-A-B)/4. Presumably, however, you have mis-written the equation. --------------------------------------------------------------------- | Good and evil both increase at compound Ben Hocking, Grad Student | interest. That is why the little hocking@cs.virginia.edu | decisions you and I make every day are of | such infinite importance. - C. S. Lewis --------------------------------------------------------------------- === Subject: Solving equation (corrected?) Dear sci.math'ers. I posted the attached message earlier today, but as Ashlie pointed out, the equation was wrong. Here is the corrected version: N = (A * (x-1)) + ((B + (4*(x-1)) * y-1) As before everything is integers and upper bounds for x is 8 and 16 for y. Lower bound for both x and y are 1. Let N=552, A=66 and B=34. The solution in this case is unique: x=3 and y=11 How do I solve this in general? Do I need to minimize using linear programming? I only need one solution if that make things easier. Can I solve this in a computer program efficiently, even if A,B and N are huge (hundreds of digits) and x and y are less than, say, 5000? (I know C and Java, not Mathematica, etc) -han > Hi all, > I've been pondering about the following equation, but cannot solve it > in general: > N = (A + 4*x) + (B + 4*y) > where N is known and A and B are known constants and the unknowns x > and y have known upper bounds. Everything is positive integers. How do > I find x and y in general? > For instance, A=34 and B=66 and N=1215, upper bounds are 8 for x and > 16 for y. > Solution is x=3 and y=3. === Subject: sqrt(x) not unique!? (was: Re: Tell me your paradoxes!) >-1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... The last step, where you take the sqr(1) is false. It is = +/- 1. This > isn't really a paradox at all... > He's right, you're wrong. The square root of one is 1. Just 1. No, actually sqrt(1) is defined to be any number which when multiplied > by itself equals 1. 1 has two square roots, 1 and -1. In the above paradox, > 1^(1/2)=-1. Are you sure of this!? I have been lead to believe that the square root is only one (otherwise it wouldn't be a function). The square root of x is the positive number that multiplied by itself yields x. Supposedly this has made people refrain from defining a square root for arbitrary complex numbers. I believe, though, that on occasion a complex square root has been defined (where the argument determines which of the two candidates is the square root). The fact that neither candidate is usually more important than the other for imaginary numbers may also be a reason why people choose not to define a complex square root. === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) >Huh? That's not a definition No it isn't, because T1 occurs on both sides. >and it's confusing. Do you mean > I / G / (G/T1) ? But it's not confusing for this reason - the fact that A or B and C means A or (B and C) is absolutely standard; no more confusing than the fact that 1 + 2*3 means 1 + (2*3). >I presume / or, / and. > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. >Ditto. === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. >Huh? I opened this message expecting a sentence in FOL. But T1 and >T2 are predicates, and you want to show that they're the same... But >the FOL doesn't have rules of inference for doing this. Huh??? One can certainly use rules of inference in FOL to show that two predicates are equivalent. >I suspect >you're using a second-order logic. And I suspect that you must >disambiguate your sentences, as wedges and vees don't mix well without >parentheses. And I suspect that what you're trying to prove isn't >called a tautology, since it doesn't seem to follow from the meanings >of the truth functional operators in your ambiguous wffs. >Alex Solla >Junior >Reed College === Subject: Re: stuck proving simple(?) tautology >Given a binary relation G(x,y) and T1 defined as: >T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) This cannot be the _definition_ of T1, because T1 appears on both sides. Did you mean to say just that T1 satisfies this identity? >and T2 defined as: >T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) >where I is identity relation, prove that T1 = T2. === Subject: Re: stuck proving simple(?) tautology >Given a binary relation G(x,y) and T1 defined as: >T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) > This cannot be the _definition_ of T1, because T1 > appears on both sides. Did you mean to say just > that T1 satisfies this identity? Yes, T1 that satisfies the equation. The theory says that this constraint alone may be not enough to define T1 unambiguously, so that they add some minimal condition. I'm not sure I understand this because we can express T1 as a series T1(x,y) = I / G(x,y) / G(x,y) & G(x,y) / G(x,y) & G(x,y) & G(x,y) / ... which looks unambiguous to me. (BTW, what symbol is better for graphic challenged NG: / or &?) In the other message you mentioned that you can infer the equality by pure logic means (which implies that you don't need this minimal condition:-). === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. > Huh? I opened this message expecting a sentence in FOL. But T1 and > T2 are predicates, and you want to show that they're the same... But > the FOL doesn't have rules of inference for doing this. I suspect > you're using a second-order logic. And I suspect that you must > disambiguate your sentences, as wedges and vees don't mix well without > parentheses. And I suspect that what you're trying to prove isn't > called a tautology, since it doesn't seem to follow from the meanings > of the truth functional operators in your ambiguous wffs. OK, let's make it second-order sentence then (for any T1(x,y)) (for any T2(x,y)) ( T1 = I / G / G / T1 & & T2 = I / G / T2 / T2 => T1 = T2 ) Could it be proved/ refuted? === Subject: Re: stuck proving simple(?) tautology > OK, let's make it second-order sentence then > (for any T1(x,y)) (for any T2(x,y)) ( T1 = I / G / G / T1 & > & T2 = I / G / T2 / T2 => T1 = T2 ) > Could it be proved/ refuted? Clearly, this was incomplete sentence, as I have to add definition of I too: (for any T1(x,y)) (for any T2(x,y)) (for any G(x,y)) (for any I(x,y)) ((I(x,y)<=>x = y) & T1 = I / G / G / T1 & T2 = I / G / T2 / T2 => T1 = T2 ) === Subject: Re: Sum over all unordered pairs === Subject: Re: Sum over all unordered pairs >> { {x,y} | x,y in O; x /= y } >But how to write a sum over such a collection? >I've seen the sum over all unordered pairs of numbers being written as: > sum f(i, j) > iwhere > f(i, j) = f(j, i) Doing a range of 0,1,2 sum = f(0,1) + f(0,2) + f(1,2) > The formula has i and to prevent pairs of a number with itself > But what if instead of numbers, i and j are objects with no > [natural] order? In that case, I cannot write i < j. But perhaps I > could write something like: > sum f(i, j) > {i, j} in {{x, y} | x, y in O; x /= y} = sum { f(i, j) | i,j in O, i /= j } = f(0,1) + f(0,2) + f(1,2) + f(1,0) + f(2,0) + f(2,1) Includes both f(i,j) and f(j,i) as f is function over ordered pairs. Now when f(i,j) = f(j,i), you may like (1/2) sum { f(i, j) | i,j in O, i /= j } > where O is the set of all objects. With index variables would be the standard way of writing sums and > products, but introducing dummy ordering of the objects seems a bit > inelegant to me. When f isn't symmetrical, how you divide OxO into two sets, one summed the other excluded, will cause difference in sum; but your notion wants for some sort of division of aye's and nay's. So in general, your notion isn't as simple as you wish. ---- === Subject: Re: Taking Calculus without taking precalcu charset=Windows-1252 I would recommend skipping the pre-calculus class. Pre-calculus didn't even exist when I was in high school, but then again, in 1969, our high school had just added Calculus to their curriculum. There were only 11 students in that first class (out of 900 in 12th grade), including one advanced 11th grader (I was in 12th grade during this class). The book we used was Crowell and Slesnik Calculus and Analytic Geometry, (only one version plublished, 1968), but then my 3rd semester of Calculus (in college) used the popular book by the same name, which has had numerous versions published (mine was the 3rd version). Junior colleges, then high schools later added pre-calculus classes, mostly for the non-math oriented majors. In your case, with an A in trig, I wouldn't bother taking the pre-calculus class. I'm not sure why a school would make you take a pre-requisite course or challenge to take Calculus if you've completed algebra and trig. Geometry is nice, but it's more of logic class (at least what I took) than the analytic geometry that is part of some Calculus classes. ( publishes two versions of books, with and without Analytic Geometry). === Subject: Re: Taking Calculus without taking precalcu In my opinion, it all depends on the value of that 'A' in trigonometry. I have had many students with such grades, who also took Calculus in High School, who test into remedial algebra. For our sequence, the real knowledge of algebra and trig is in Calculus II (series and methods of integration). --irascible since 1957 === Subject: Re: Taking Calculus without taking precalcu >My friends that have taken precalculus have said they pretty much knew >everything and that it was a waste and that they had wished they had >gone straight to calculus. I've been thinking about doing the same, I >have a firm grounding in trigonometry got a A in it. Would you >recommend against me just outright skipping precalculus? The only >catch is I would have to write a challenge letter requesting I skip >the prerequisite of precalculus. Any suggestions on what to say in >there a math professor would be looking at the letter You should have strong algebra skills -- can easily manipulate exponent and radicals, for instance, and can reliably factor complicated expressions, combine rational expressions, and so forth -- and you should have strong skills in analytic trigonometry, especially proving trig identities by devising useful substitutions. If that is true then I see no reason why you need precalculus. Heck, if even _half_ of that is true you'll be better off than most of my calculus students. I thought a giveaway problem on yesterday's quiz was to integrate sin^3(x) / [ 1 - cos^2(x) ] dx All you have to do is rewrite 1-cos^2(x) as sin^2(x), but a third of the class couldn't do it. Many of them still think sqrt(x^2-25) = x-5. === Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. -- Barry === Subject: Re: Taking Calculus without taking precalcu >But anyway, here is a simple test. If you can sit down right now, and >without books or notes, and taking all the time you want, derive a formula >for tan(a+b+c) in terms of tan(a), tan(b), and tan(c), then you don't need >precalculus. Otherwise, I would say you do. I'm curious: What fraction of students who finished precalculus within the past two months, with a C or better, could do that? Maybe my perspective is jaundiced by the students I see, but my guess is under 1%. === Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. -- Barry === Subject: Re: Taking Calculus without taking precalcu charset=Windows-1252 > In my opinion, it all depends on the value of that 'A' in trigonometry. I have > had many students with such grades, who also took Calculus in High School, > who test into remedial algebra. I could claim that most students who take calculus without taking pre-calculus probably do better than those that do take pre-caculus. This ignores the reason why the students who took pre-calculus choose to do so. Then again, I remember a math major who fogot how to derive the quadratic equation using the completing the square method. === Subject: Re: Tell me your paradoxes! A and B play a game called the metagame. The rules of the metagame are that A plays first by picking a two player game that is guaranteed to end in finitely many moves, and that they play the game named by A, with B playing first. The metagame seems guaranteed to end in finitely many moves, because whatever game A chooses will end in finitely many moves. So A starts by picking a game: Let's play the metagame. Since this means that B has to make the first move in the metagame, B has to pick a game. B says Let's play the metagame. They repeat this. Is the metagame guaranteed to end? If it is, then it seems they're allowed to choose the metagame as the first move in the metagame. On the other hand, if they're allowed to pick the metagame, the above line of play is possible, which means the game isn't guaranteed to end. But how can't it be guaranteed to end, given the reasoning described above. This isn't my original idea of course, although I've forgotten whose idea it was originally. Keith Ramsay === Subject: Re: Tell me your paradoxes! charset=iso-8859-1 > (2) In the town of Placerville (CA) the barber shaves everyone who doesn't > shave himself. Who shaves the barber? > She shaves herself, obviously. Statement (2) is not exclusive, it doesn't limit the barber from shaving only those who don't shave themselves (2) is true even if the barber shaves everyone in town, even those who do shave themselves. However, if it were worded so it was exclusive (only those perons who don't shave himself), then she shaves herself is a good answer. === Subject: Re: Tell me your paradoxes! charset=Windows-1252 > A and B play a game called the metagame. The rules > of the metagame are that A plays first by picking a two > player game that is guaranteed to end in finitely many > moves, and that they play the game named by A, with > B playing first. The metagame seems guaranteed to > end in finitely many moves, because whatever game A > chooses will end in finitely many moves. > So A starts by picking a game: Let's play the metagame. > Since this means that B has to make the first move in the > metagame, B has to pick a game. B says Let's play the > metagame. They repeat this. > Is the metagame guaranteed to end? If it is, then it seems > they're allowed to choose the metagame as the first move > in the metagame. On the other hand, if they're allowed to > pick the metagame, the above line of play is possible, which > means the game isn't guaranteed to end. But how can't it > be guaranteed to end, given the reasoning described above. A plays first by picking a two player game that is guaranteed to end in finitely many moves. Since A picks the metagame, there is either a limit to how many times the game picking sequence can be repeated, or A is not following the rules. However, assuming A and B have limited lifespans, the sequence will end once one of them dies, so A is not violating the rules in this case, the number of moves is finite. === Subject: Re: Tell me your paradoxes! > A and B play a game called the metagame. The rules > of the metagame are that A plays first by picking a two > player game that is guaranteed to end in finitely many > moves, and that they play the game named by A, with > B playing first. The metagame seems guaranteed to > end in finitely many moves, because whatever game A > chooses will end in finitely many moves. > So A starts by picking a game: Let's play the metagame. > Since this means that B has to make the first move in the > metagame, B has to pick a game. B says Let's play the > metagame. They repeat this. > Is the metagame guaranteed to end? If it is, then it seems > they're allowed to choose the metagame as the first move > in the metagame. On the other hand, if they're allowed to > pick the metagame, the above line of play is possible, which > means the game isn't guaranteed to end. But how can't it > be guaranteed to end, given the reasoning described above. > This isn't my original idea of course, although I've forgotten > whose idea it was originally. J H Conway, I believe. > Keith Ramsay Andrew Taylor === Subject: Re: Tell me your paradoxes! > -1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... (a^m)^n = a^(m*n) only holds when m and n are integers. === Subject: Re: Tell me your paradoxes! > -1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... (a^m)^n = a^(m*n) only holds if m and n are integers. === Subject: Re: Tell me your paradoxes! > -1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... >(a^m)^n = a^(m*n) only holds when m and n are integers. That's not true. For example (2^2)^(1/2) = 2^(2 * 1/2) = 2. Derek Holt. === Subject: Re: Tell me your paradoxes! >-1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... The last step, where you take the sqr(1) is false. It is = +/- 1. This > isn't really a paradox at all... > He's right, you're wrong. The square root of one is 1. Just 1. No, actually sqrt(1) is defined to be any number which when multiplied by itself equals 1. 1 has two square roots, 1 and -1. In the above paradox, 1^(1/2)=-1. === Subject: Re: Tell me your paradoxes! >-1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... The last step, where you take the sqr(1) is false. It is = +/- 1. This > isn't really a paradox at all... > He's right, you're wrong. The square root of one is 1. Just 1. No, actually sqrt(1) is defined to be any number which when multiplied > by itself equals 1. 1 has two square roots, 1 and -1. In the above paradox, > 1^(1/2)=-1. The two roots of x^2 - 1 = 0 are given by x = +/- sqrt(1). If sqrt actually had two values, then it would not be necessary to write +/- when writing sqrt, but in fact, I think you will find that it's a universal practice when both roots are intended. Notice that a root of x^2 + 1 = 0 is not the same thing as the square root of 1. === Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Tell me your paradoxes! > Have you got any paradoxes to share? Write them here! suppose you got two envelopes with some money in each of them. Lets assume that you are told that in one of them there is the double amount of money as in the other. Of course so far the situation is totally symmetric in both envelopes, so you cannot distinguish them in any way. Now you may choose one ot them and look into it. Say the amount of money therein is x. Now you are given the chance - either to take x - or to close the envelope again and take the money of the other In the first case, of course, what you get is x. In the second case you have a fifty-fifty chance for either x/2 or 2x, so your expectation value is 3x/2, which is more than in the first case. Obviously it is better to take the other envelope. So the situation is NOT symmetrical anymore although the only thing you did is to look into one envelope and see some arbitrary amount x in there. === Subject: Re: Tell me your paradoxes! > The two roots of x^2 - 1 = 0 are given by x = +/- sqrt(1). > If sqrt actually had two values, then it would not be necessary to write +/- when writing sqrt, but in fact, I think you will find that it's a > universal practice when both roots are intended. Notice that a root of > x^2 + 1 = 0 is not the same thing as the square root of 1. > -- > Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > === Subject: Re: Tell me your paradoxes! > The two roots of x^2 - 1 = 0 are given by x = +/- sqrt(1). > If sqrt actually had two values, then it would not be necessary to write +/- when writing sqrt, but in fact, I think you will find that it's a > universal practice when both roots are intended. Notice that a root of > x^2 + 1 = 0 is not the same thing as the square root of 1. > -- > Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > Sorry, I sent an empty reply. So I suppose I was right in my sqrt(x) not unique!? posting then. === Subject: Re: Tell me your paradoxes! >Have you got any paradoxes to share? Write them here! > suppose you got two envelopes with some money in each of them. Lets > assume that you are told that in one of them there is the double > amount of money as in the other. Of course so far the situation is > totally symmetric in both envelopes, so you cannot distinguish them in > any way. > Now you may choose one ot them and look into it. Say the amount of > money therein is x. Now you are given the chance > - either to take x > - or to close the envelope again and take the money of the other > In the first case, of course, what you get is x. In the second case > you have a fifty-fifty chance for either x/2 or 2x, so your > expectation value is 3x/2, which is more than in the first case. > Obviously it is better to take the other envelope. So the situation is > NOT symmetrical anymore although the only thing you did is to look > into one envelope and see some arbitrary amount x in there. Actually, if x=$2.33, then the other envelop must contain $4.66 (unless half pennies are allowed). Also note that you have changed the discussion from the probability of getting the larger value to whether you should change envelopes *based* on that probability. === === Subject: Re: Tell me your paradoxes! >> -1 = (-1)^3 = (-1)^(6/2) = ((-1)^6)^(1/2) = 1^(1/2) = 1 ... >(a^m)^n = a^(m*n) only holds when m and n are integers. >That's not true. For example (2^2)^(1/2) = 2^(2 * 1/2) = 2. If we define x^y as exp(y Log(x)) where Log is the principal branch of the natural logarithm, i.e. Im(Log(x)) in (-pi, pi], then a correct statement is (a^m)^n = exp(n Log(a^m)) = exp(n (m Log(a) + 2 pi i k)) = a^(mn) exp(2 pi i n k) where k is the integer such that -pi < Im(m Log(a)) + 2 pi k <= pi . In particular, (a^m)^n = a^(mn) is always true in the following cases: 1) n is an integer 2) m in (-1,1] 3) a > 0 4) -pi < Im(m Log(a)) <= pi Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Tell me your paradoxes! What is the formal definition of a paradox? === Subject: Re: Tell me your paradoxes! >Have you got any paradoxes to share? Write them here! I have put two sealed envelopes in a bowl. Each of them has money in it; one has exactly twice the amount as the other. You may choose one and keep the money. Your choice is not final until you open the envelope. You cannot tell how much money is in the envelope by feeling it. You choose. Then you realize that, if the envelope you have chosen has x dollars, the expected amount in the other envelope is 1/2 * x/2 + 1/2 * 2x = 5x/4 dollars. Therefore, you decide to switch. By this reasoning, you should never settle on any one envelope, but should continually switch to the the other. === Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Tell me your paradoxes! > The two roots of x^2 - 1 = 0 are given by x = +/- sqrt(1). > If sqrt actually had two values, then it would not be necessary to write > +/- when writing sqrt, but in fact, I think you will find that it's a > universal practice when both roots are intended. Notice that a root of > x^2 + 1 = 0 is not the same thing as the square root of 1. > Sorry, I sent an empty reply. So I suppose I was right in my sqrt(x) not > unique!? posting then. If you mean that sqrt(x) denotes the principal value of the square root, then yes, I agree. And by the way, I meant x^2 - 1 = 0 in that last sentence above. === Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Tell me your paradoxes! charset=iso-8859-1 > I have put two sealed envelopes in a bowl. Each of them has money in > it; one has exactly twice the amount as the other. You may choose one > and keep the money. Your choice is not final until you open the > envelope. You cannot tell how much money is in the envelope by feeling it. > You choose. Then you realize that, if the envelope you have chosen has > x dollars, the expected amount in the other envelope is 1/2 * x/2 + > 1/2 * 2x = 5x/4 dollars. Therefore, you decide to switch. However, when you choose the first envelope, you had the same odds. One of the envelopes has x dollars. No matter which one you choose, the probably outcomes are 1/2 * x/2 + 1/2 * 2x = 5x/4. So there's no point in switching, as the first envelope has the same expected return as the second. === Subject: Re: The basic idea behind my great forthcoming proof > Cantor introduced that idea that the sizes of > infinite sets can be compared. > He gave a definition of cardinality, one which apparently seemed > natural to Frege, Dedekind and other contemporaries. > Surely you're not suggesting that his *definition* is the source of > all problems? You could certainly say that making the definition magnified the problem. Before he made his definition, all uses of infinity in mathematics were consistent with the idea that infinity had a potential existence, rather than an actual existence. === Subject: Re: The basic idea behind my great forthcoming proof > As long as it is logically consistent, it is valid > mathematics. That is a very dangerous and very wrong belief. Logical consistency alone doesn't give us a way to distinguish reality from fantasy, or science from mysticism. === Subject: Re: what is the eigenvalue of a linear transform? > I have a question about linear transform. What is the definition of the > eigenvalue of linear transform? How to show the linear transformation T: > f(x) -> integrate[f(t), t from 0 to x] has no eigenvalues? the answer to your first question can be found in any source on linear algebra. Concerning your second question: Have a look at the polynomials x^k, k=0,... . What is the action of T on these functions? How would the infinite dimensional matrix corresponding to this transformation look like? Alois === Subject: Re: what is the eigenvalue of a linear transform? > I have a question about linear transform. What is the definition of the > eigenvalue of linear transform? How to show the linear transformation T: > f(x) -> integrate[f(t), t from 0 to x] has no eigenvalues? the answer to your first question can be found in any source on linear > algebra. > Concerning your second question: Have a look at the polynomials > x^k, k=0,... . > What is the action of T on these functions? How would the infinite > dimensional matrix corresponding to this transformation look like? An alternate way: convert the eigenvalue equation to a differential equation, solve it explicitly using well-known methods, deduce the only solution is f(x) = 0. === Subject: Re: Why are martingales called martingales? ... >> Jacqueline Pichoche's _Dictionnaire Etymologique du francais_ >> (Robert, 1979) clarifies this, and, I think, settles where >> the betting system got its name. My translation: [snip def supporting above] Maybe it settles it, maybe not. The previously-suggested >bridle idea is supported by Webster's 1913 Dictionary >(according to http://www.hyperdictionary.com anyhow): >3. (Gambling) The act of doubling, at each stake, that which has >been lost on the preceding stake; also, the sum so risked; -- >metaphorically derived from the bifurcation of the martingale >of a harness. [Cant] --Thackeray. That is, hyperdictionary's >entry makes no mention of this usage of martingale deriving from >Martigues, but instead says it derives from martingale as harness. My 1924 printing of the 1909 edition of the Merriam-Webster > New International Dictionary (commonly called Webster's First > International) doesn't have that citation from Thackeray. I > don't know which Thackeray it was, but assume it to be the > English novelist and literary man of that name. In that case > certainly (and in any case, maybe with less certainty) I would > greatly favor the scholarly opinion of Jacqueline Picoche > (a professional lexicographer writing in 1979) over the > (very likely less scholarly, and in any case many decades > earlier) opinion of Thackeray. The metaphorical derivation > given by Thackeray has all the earmarks of folk-etymology > (if not his own fertile imagination). Also, though Picoche > doesn't give a citation for jouga a la martegalo, she does > date it to the 18th century, whereas the Oxford English > Dictionary's earliest citation for martingale in the > gaming sense is from 1815 (ah, and followed by Thackeray-the- > novelist, in a novel, 1854) as a noun, and from 1823 as > a verb (whose etymology asks the reader to compare to French > martingaler without positively claiming that that's its source; > the OED etymology of the *noun* martingale is even less help). > And to think that for all these years I thought the game-theorist > David Gale had named it in honor of his brother, Martin.... Groan. This is the best explantion so far! GC 872 === Subject: Re: a/b+b/c+c/a=n The following is a complete list of the values of n under 200... (We can know that as soon as we have explicit solutions for these two n. Is it 142, or is it 147? === Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Axiomization of Number Theory <3df1e59f.0307272023.23967ab3@posting.google.comWrite http://www.arxiv.org/h/cs.lo/000307 so we can link to it but even > better would be to state them with abstract and publication date and > journal to interest us in reading your arxiv. When http:// is prefixed to address news browser recognizes it has web link, highlights it and will open URL if clicked upon. > I'm not sure what you mean, but anyone who is interested is invited to > click on www.arxiv.org/h/cs.lo/0003071 Does you browser recognize that as a web address? Mine and others don't as you didn't include http:// nothing appears to click on and it sits there as an inert ascii text going nowhere, doing nothing, meriting dismissal. === Subject: Re: Bush unveils new economic plans based on the Theorems of Cantor Miracle Bush Quadruples National Treasury Every Month. Amazing, how did he do it? He drove US into bankruptcy. But then the treasury would be broke. Exactly, four times nothing is nothing. > In a press conference today the honorable George W. Bush unveiled a > new economic plan which is based on the indisputable principals of > Cantor. In this new plan, said the president, 99% of the weight of taxes > will fall upon the 20 poorest percent of the population. But wait! > he said, silencing a cry of outrage which arose. According to the > principals of the transfinitude, in the long run this is a fair > program as it will tax all citizens equally. You see, the set of all > natural numbers has the same magnitude as the set of all natural > multiples of 100. This shows that in the long run, the rich will end > up paying just as much as the poor and all will be fair and just. > The president then went on to justify his planned $6ion tax relief > cut for the next fiscal year. You see, said G.W., Cantor has shown > us that the magnitude of an infinite set is not affected by the > addition or removal of a finite number of elements. Therefor, in the > long run, this $6B tax refund will have no adverse effects > whatsoever. > The president then changed the topic to that of the war in Iraq. To > save on spending, we will no longer issue our soldiers firearms or > helmets. While I understand that this will increase their casualty > rate tremendously, please bear in mind that, if we assume we will > forever be at war somewhere or other, then again the theorems of > Cantor assure us that this will have no impact whatsoever on the > magnitude of the set of our casualties. For, if this increases the > death rate by a factor of 500, Cantor assures us that the set of > naturals has the same magnitude as the set of natural multiples of > 500. === Subject: Circles in complex space. In modern physics it's been explained that an object can rotate in complex space and take 720 degrees to return back to its orginal position. How is this possible, exactly? Obviously it's not the same as taking the argument of a circle on a z plane for each point and then figuring the area geometrically as if this were real space. What exactly is happening? Also, does this mean you can take a measure through complex space and get a negative or imaginary distance? (...Starblade Riven Darksquall...) === Subject: Re: Complex Numbers >David, >After reading your posts it occurred to me that perhaps all Riemann >surfaces(= 1 dimensional >analytic manifolds) are orientable. I went to the library and found that >yes this is true. >Furthermore the proof is very simple: >Definition: A differentiable manifold is orientiable if there is a choice >of atlas such that for >every pair of coordinate charts (f,U) and (g,V) in the atlas, the Jacobian >of the map >fg^(-1) is positive. >Theorem. Every Riemann surface is orientable. >Proof: Let f and g be given as above with also fg^(-1) analytic per the >definition of Riemann surface. >Let fg^(-1) = u + iv. The Jacobian of fg^(-1) is det([u_x, u_y], [v_x, >v_y]). The Cauchy-Riemann equations >say that u_x = v_y and u_y = -v_x. Thus we have that the Jacobian is >(u_x)^2 +(u_y)^2 >= 0. We cannot have = 0 because h^(-1) exists. QED. >Thus not only is it the case that an appropriate atlas exists, but in fact >_every_ atlas >shows that a Riemann surface is orientable! >Consider the case of the mobius strip. Intuitively if we follow coordinate >charts around the strip, >the orientation is reversed and if one uses analytic functions then (after >appropriate identifications) >following the coordinate charts around the mobius strip changes the identity >function to >complex conjugation. Complex conjugation is NOT an analytic function as may >be easily >verified from the definition of derivative. >Comments would be welcome as I am not an expert on Riemann surfaces. > Neither am I, but it sounds right to me. What I'm not sure about is > whether you're suggesting that this helps us decide which of the > two square roots of -1 is i and which is -i. >David C. Harris > > No, I am not suggesting that this tells us which square root of -1 is i and which is -i. Orientable is not the same as oriented! C is orientable but not oriented unless we make a choice of i or -i, and either one will do as well as the other to give us an orientation! The fact that Riemann surfaces are orientable simply seemed too irresistibly nice to not share. I hope my post is not too confusing. I agree with your post that says there is no way to tell i from -i. It is true that one can add structure to C to make i and -i distinguishable, but then C embeds into this new structure in two ways. (At least that is all that I can see.) David C. Harris === Subject: Re: Factorial/Exponential Identity, Infinity > I like to think about infinity, and the mathematical work around > infinity, because I enjoy deriving novel results and especially > extending them to show them non-trivial. > I derived this some time ago: > lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n! > By Sterling's approximation, n! has some resemblance to n^n in the large. > Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) = oo > Also when n = 2 > 4 = (2 (n/2)! 2^n)/n! /= 1 > What it is there is an identity relating the factorial to the binary > exponent. What it says there is that > lim n-> oo n! / (2(n/2)!) = 2^n > Mathematical that makes no sense as > lim... is a constant and 2^n a variable. I think the identity is correct, actually. It's obvious that n! > 2^n in the large. That starts at n=4 where 24>16, for n=5 120>32, and for all following integers. What happens though is you divide n! by (n/2)! and 2. About the limit and having the variable on both sides of the equation, the idea there is that each side of that equality could be multiplied or divided by the corresponding sides of some other equality with n diverging. lim_n->oo n!/(2(n/2)!) = lim_n->oo 2^n There's only one variable, n. If it was x on the left and y on the right then it wouldn't be true uness x=y. The expresion n! has a value less than n^n for any integer n greater than one. I'm interested in this Stirling function(s) and approximation(s), I am not familiar with them and read about them. Assume the identity is correct. What good is it? I derived it from assuming that half of the infinite binary strings have equal numbers of zeros and ones. Consider the rational binary numbers of these forms: .0101010101... .1010101010... I think those are 1/3 and 2/3 but I don't know. Let's see: 1/4 + 1/16 + 1/64 + 1/256 + ... 1/2 + 1/8 + 1/32+ 1/128 + ... How do I evaluate those binary rational numbers? I want to know reduced fractional forms for those and all other rationals between zero and one with equal numbers of zeros and ones in their binary expansion. Hi. Ross === Subject: Re: Factorial/Exponential Identity, Infinity > lim_n->oo n!/(2(n/2)!) = lim_n->oo 2^n > There's only one variable, n. If it was x on the left and y on the > right then it wouldn't be true uness x=y. Wrong! The variable on each side is a bound (or dummy) variable which has n meaning outside of the limit statement, so changing the variable within either limit statement has no effect on the meaning of that limit statement. === Subject: Re: Finite subgroups of GL(n,Z) > Is it true that for every n>=1 the group GL(n,Z) as only a finite number > of non-isomorphic finite subgroups ? > If G is a subgroup of GL(n,Z) then G acts faithfully on Z^n > which is contained in R^n. There is a positive definite quadratic > form Q on R^n (an inner product) which is invariant under G > (take any positive definite form, and sum its images under G). > We can represent G as a subgroup of O(n), the orthogonal group > and then G fixes a lattice L (corresponding to Z^n). > Suppose first that G acts irredcibly on R^n. In that case let > v be a nonzero vector of shortest length in L. The images > of v under G must span R^n as a vector space, as otherwise > their R-span would be a G-invariant subspace of R^n (contrary > to irreducibility of the G-action). This means that R^n is spanned > by the set L_1 of the nonzero vectors of minimal length in L. > The group G permutes L_1 and only the identity acts trivially > on L_1 since L_1 spans R^n. Hence G is isomorphic to a subgroup > of Sym(L_1), i.e., |G| <= |L_1|! > I claim that the size of L_1 is bounded by a function of n. > This follows from the kissing number bounds in the theory of lattices. > If v and w lie in L_1 and |v-w| < |v| = |w| then as v-w in L this > contradicts v being a shortest nonzero vector in L. Hence the spherical caps on the sphere of radius |v| centred at the points > of L_1 and having radius pi/6 don't meet. These have positive area on the sphere and so there's an upper bound for |L_1| > (depending on n). We conclude that for each n there > is a bound on the order of finite G acting faithfully on GL(n,Z). > What about non-faithful G? If G acts nonfaithfully, there is > a subgroup A of Z^n that it preserves with A isomorphic to Z^m > 0 < m < n. Then also G preserves A' = QA intersect Z^n. Now > Z^n/A' is isomorphic to Z^{n-m}. G acts on A' (isomorphic to Z^m) > and Z^n/A' (isomorphic to Z^{n-m}). An element of G acting trivially > on both must be the identity. Inductively there is a bound C_m > on the size of the image of G in the action on A' and a bound C_{n-m} > on the size of the image of G in the action on Z^n/A'. Then > |G| <= C_m C_{n-m}. So there is a bound on |G| for all finite subgroups of > GL(n,Z). > The answer is yes. Is the same assertion true for the groupp GL(n,K) where K is a number field ? === Subject: Homo High http://www.msnbc.com/news/945134.asp ...State Conservative Party Chairman Long criticized the creation of the school. Is there a different way to teach homosexuals? Is there gay math? This is wrong, Long said. There's no reason these children should be treated separately. Yes, I can think of plenty of gay math problems. e.g. the hetero male spreads Human Papilloma Virus (Cervical Cancer) to 3 girls. The girls have an average of 2.75 partners. Cervical cancer has a 15% fatality rate. Determine how many straight girls will die of cervical cancer after 10 generations. Q1: In a nightclub full of 10 het men and 2 het women, determine how many pairings are possible. A1: 10 * 9 = 90. Q2: In a nightclub full of 10 gay men and 2 lesbian women, determine how many pairings are possible. A2: (9 * 7 * 5 * 3 * 1) + (1) = 945 + 1 = 946. Q3: In the het night club, a female has a 40% chance of being attracted to any of the poorly dressed het men. If an unpaired het male has a 2% chance of going on a shooting rampage, what is the likelihood of a shooting rampage occurring in the het club? A3: Case 1: Both women find an attractive het. The chances of this are 0.4 * 0.4 = 16%. Then the other 8 men * 0.02 = .16 Case 2: One of the women finds an attractive het. The chances of this are 2(0.4 * 0.6) = 48%. Then the other 9 men * 0.02 = .18 Case 3: Neither woman finds an attractive het. The chances of this are 0.6 * 0.6 = 36%. Then the 10 men * 0.02 = .2 Adding up, (.16)(.16) + (.48)(.18) + (.36)(.2) = .0256 + .0864 + .072 = 18.4% of a shooting rampage. Q4: A child of a straight couple has a 3% chance of having a major birth defect. A person with a major birth defect has a 15% chance of becoming a future violent criminal or rapist. A straight couple with a monster baby has a 17% chance of getting a divorce. A child of a single mother has a 47% chance of becoming a future violent criminal or rapist. Determine the overall odds of a monster baby born to a straight couple to develop into a future predator. A4: (.17)(.47) + (1 - .17)(.15) = .0799 + .1245 = about 20%. === Subject: Re: ny Can't Add But Suresh Venktasubramanian Can A lot of the problem with America's primary and secondary school is that parents refuse, in the main, to back up teachers. Parents do not want there children to get bad grades. So, teachers teach down to their students. A slippery slope! In addition to that, school psychologists are too concerned fact, American childern suffered from inflated self-esteem! Moreover, American society in general disdains intellectuals. It is cooler, if not outright funny, to be stupid. People coming from other parts of the world value education. They are used to competing with many more students/people than the average American is. To stand out in some of the Asian countries where there is such a ridiculously large over-population, takes a lot of work. Lurch > ny Can't Add > But Suresh Venktasubramanian Can > By Fred Reed > The other day I went to the Web site of Bell Labs, one of > the country's premier research outfits. I clicked at > random on a research project, Programmable Networks for > Tomorrow. The scientists working on the project were > Gisli Hjalmstysson, Nikos Anerousis, Pawan Goyal, K. K. > Ramakrishnan, Jennifer Rexford, Kobus Van der Merwe, and > Sneha Kumar Kasera. > Clicking again at random, this time on the Information > Visualization Research Group, the research team turned > out to be Ellson, Emden Gansner, Mocenigo, > Stephen North, Jeffery Korn, Eleftherios Koutsofios, Bin > Wei, Shankar Krishnan, and Suresh Venktasubramanian. > Here is a pattern I've noticed in countless organizations > at the high end of the research spectrum. In the > personnel lists, certain groups are phenomenally over- > represented with respect to their appearance in the > general American population: Chinese, Koreans, Indians, > and, though it doesn't show in the above lists, Jews. > What the precise statistical breakdown across the world > of American research might be, I don't know. An awful lot > of personnel lists look like the foregoing. > Think about this: Asians make up a small percent of the > population, yet there are company directories in Silicon > Valley that read like a New Delhi phone book. Many of our > premier universities have become heavily Asian, with many > of these students going into the sciences. If Chinese > citizens and Americans of Chinese descent left tomorrow > for Beijing, American research, and graduate schools in > the sciences and engineering, would be crippled. > Jews are two or three percent of the population. On the > rough-cut assumption that Goldstein is probably Jewish, > and Ferguson probably isn't, it is evident that Jews are > doing lots more than their share of research-and, given > that people named Miller may well be Jewish, the name- > recognition approach probably produces a substantial > undercount. I asked a friend, researching a book on > Harvard, the percentage of Asian and Jewish students. > Answer: Asians close to 20%. Jews close to 25%- > unofficial, because you are allowed to list by gender, > ethnicity, geography, but not religion. Our last taboo. > None of this is original with me. In 1999, the National > Academy of Sciences released a study noting that over > half of U.S. engineering doctorates are awarded to > foreign students. Where are Smith and Jones? > Why are members of these very small groups doing so much > of the important research for the United States? That's > easy. They're smart, they go into the sciences, and they > work hard. Potatoes are more mysterious. It's not > affirmative action. They produce. The qualifications of > these students can easily be checked. They have them. The > question is not whether these groups perform, or why, but > why the rest of us no longer do. What has happened? > It is not an easy question, but a lot of it, I think, is > the deliberate enstupidation of American education. > Again, the idea is not original with me. Said the > American Educational Research Association of the NAS > report, Serious deficiencies in American pre-college > education, along with wavering support for basic > research, were cited by the panel as major contributors > to this problem. > Consider mathematics. In the mid-Sixties I took freshman > chemistry at Hampden-Sydney College, a solid school in > Virginia but not nearly MIT. It was assumed-assumed > without thought-that students knew algebra cold. They had > to. You can't do heavy loads of highly mathematical > homework, or wrestle with ideas like integrating > probability densities over three-space, or do endless > gas-law and reaction-rate calculations, if you aren't > sure how exponents work. > Remedial mathematics at the college level was unheard of. > The assumption was that people who weren't ready for > college work should be somewhere else. No one thought > about it. Today, remedial classes in both reading and > math are common at universities. We seem to be dumbing > ourselves to death. > I recently had children go through the high schools of > Arlington, Va., a suburb of Washington. I watched them > come home with badly misspelled chemistry handouts from > half-educated teachers, watched them do stupid, make-work > science projects that taught them nothing about the > sciences but used lots of pretty paper. > The extent of scholastic decline is sometimes > astonishing. So help me, I once saw, in a middle school > in Arlington, a student's project on a bulletin board > celebrating Enrico Fermi's contributions to Nucler > Physicts (Scripps-Howard National Spelling Bee > 2001, Sean Conley; 2000, George Thampy; 1999, Nupur > Lala). > It appears that a few groups are keeping their standards > up and the rest of us are drowning our children in self- > indulgent social engineering, political correctness, and > feel-good substitutes for learning. > Some of our growing dependency is hidden. We do not > merely rely on small industrious groups in America and on > foreigners working here. Increasingly the United States > contracts out its technical thinking to Asia. > If you read technically aware publications like Wired > magazine (and how many people do?), you find that major > American corporations have more and more of their > computer programming done by people in, for example, > India. In cities like Bombay, large colonies of Indians > work for U.S. companies by Internet. This again means > that counting names at American institutions > underestimates the growth of intellectual dependence. > The Indians, and others, have discovered the suddenly > important principle that intellectual capital is > separable from physical capital. To program for Boeing, > you don't have to be anywhere near Seattle. Nor do you > need an aircraft plant. All you need is a $700 computer, > a book called something like How to Program in C++, and a > fast Internet connection. Crucial work like circuit- > design can now be done abroad by bright people who don't > need chip factories. They need workstations, the > Internet, and engineering degrees. > This too we would be wise to ponder. Americans often > think of India chiefly as a land of ghastly poverty. > Well, yes. It is also a country with about three times > our population and a lot of very bright people who want > to get ahead. They're professionally hungry. We no longer > are. > People speak of globalization. This is it, and it's just > beginning. Where will it take us? How long can we > maintain a technologically dominant economy if we are, as > a country, no longer willing to do our own thinking? If > we rely heavily on less than 10 percent of our own > population while employing more and more foreigners > abroad? > It's not them. It's us. I've heard the phrase, the Asian > challenge to the West. I don't think so. When Sally Chen > gets a doctorate in biochemistry, she's not challenging > America. She's getting a doctorate in biochemistry. Those > who study have no reason to apologize to those who don't. > The Mathematical Association of America runs a contest > for the extremely bright and prepared among high-school > students. It is called the United States of America > Mathematics Olympiad, and it provides a means of > identifying and encouraging the most creative secondary > mathematics students in the country. > An unedited section of a list of those recently chosen: > Sharat Bhat, Tongke Xue, Matthew Peairs, Wen Li, Jongmin > Baek, Aaron Kleinman, David Stolp, Andrew Schwartz, Rishi > Gupta, Jennifer Laaser, Inna Zakharevich, Neil Chua, > Jonathan Lowd, Rubinsteinsalze, Joshua Batson, > Jimmy Jia, Jichao Qian, Dmitry Taubinsky, David Kaplan, > Erica Wilson, Kai Dai, Julian Kolev, Jonathan Xiong, > Stephen Guo. > Q.E.D. > Fred Reed > f.v.reed@att.net > First appeared in The American Conservative. > - - - - - - - > Fred Reed,columnist for The Washington Times, former > Marine, streety police writer, occasional terrified war > correspondent,and afficionado of raffish bars, offers > weekly his unique, often satirical and arguably > opinionated views on ...everything. You'll grind your > teeth. (He denies that he gets a kickback from the dental > lobby, though no one believes him.) But you'll think. I'm an equal-opportunity irritant, says Fred > democratically. Visit his website here: > http://www.fredoneverything.net/ > Source - http://mensnewsdaily.com/archive/r/reed/03/reed072703.htm > Jai Maharaj > http://www.mantra.com/jai > Om Shanti > Shubhanu Nama Samvatsare Dakshinaya Nartana Ritau > Kark Mase Krishna Pakshe Indu Vasara Yuktayam > Pushya Nakshatr Vajr-Siddhi Yog > Chatushpad-Naag Karan Amavasya Yam Tithau > Hindu Holocaust Museum > http://www.mantra.com/holocaust > Hindu life, principles, spirituality and philosophy > http://www.hindu.org > http://www.hindunet.org > The truth about Islam and Muslims > http://www.flex.com/~jai/satyamevajayate > o Not for commercial use. Solely to be fairly used for the > educational purposes of research and open discussion. The contents of > this post may not have been authored by, and do not necessarily represent > the opinion of the poster. The contents are protected by copyright law > and the exemption for fair use of copyrighted works. > o If you send private e-mail to me, it will likely not be read, > considered or answered if it does not contain your full legal name, > current e-mail and postal addresses, and live-voice telephone number. > o Posted for information and discussion. Views expressed by others > are not necessarily those of the poster. === Subject: Re: lambda calculus > It seems to me that the lambda calculus is the formulation that makes the > Church-Turing thesis most plausible. How come I've never seen a book that uses > it to prove the unsolveability of the halting problem? Does anybody know of a > source like that? How so? There's basically 3 models of computation shown to be equivalent, TMs, recursive functions, lambda abstractions (almost 2GL, 3GL, 4GL). But each methods' approach to the Halting problem only requires that functions are enumerated and compute a parameter, an actual prototype calculation of the method used is not required of the proof. Hence we write TM(n) = xyz just as easily as Lambda x. utm n. Would be a nice book if it exists, but most lamdba calculus work is on optimisation, so that one day we might use it! Herc === Subject: Re: Mapping of integers to reals ~ Cantor's disproof > > : : : : >> Cute TM, but I think you miss the point of the problem. You need to > : >> find a function (or TM) that can map every irrational to a unique > : >> integer. > : : >> That's easy: I can define a function that maps every irrational to the > : >> unique integer 17. > : : >> Wow, you did it... I'm in awe of your genius. > : : >Nothing to it. Mapping every irrational to a *unique* integer is > : >trivial: it's inherent in the idea of a function that every element of > : >the domain is mapped to a unique element of the range. Now, if you'd > : >wanted to map every irrational to a *different* integer, that would be > : >much harder. Impossible, in fact. > > : Huh? I just mapped every irrational to 18. > > : That *is* a different integer from 17, isn't it? > What would be impossible is to map every irrational to an integer such > that for any two unequal irrationals, the integers to which they are > mapped in that mapping are also unequal. > Of course, both of you realized that already... > hello , > This is part of the claim of the thread have you read the whole thread? > The integers they map to are the Turing Machine numbers that generate > them. Two different outputs from 2 machines implies the machines > are different, as TM's are deterministic. > Herc What about mapping every irrational onto a rational? Can that be done? Or better yet: Mapping every transcendental number onto a, uh, what is the nickname or nontranscendental? algebraic? Gemoetric? Polynomialic? Whatever. (What is the nickname for nontranscendental anyways?) Well can you map every transcendental to a... uh... that? (...Starblade Riven Darksquall...) === Subject: Re: Mapping of integers to reals ~ Cantor's disproof >I'm using the fact that H(n) is noncomputable ... > That *fact* is wrong. H(n) *is* computable (for every n). H(n) is simply > an integer; in fact, quite restricted: a member of {0, 1}. But H (the > function) is NOT computable. > Daryl said it well: you need a *different* computable function (a constant > function will do) for each n. (In fact, you only need two such computable > functions). Computability of a *function* (as opposed to a value in that > function's range) is a different matter. > All you're saying is that every real number is computable, because we have > ten constant functions into {0,1,2,3,4,5,6,7,8,9} that together permit any > digit of any real number to be computed -- and those ten functions are all > computable! Wowee. > Michel. Actually, by that logic, it can be done just for 0 and 1, since every real number can be expressed as an infinitely long set of sums of binary fractions. Rational numbers would be repeated, and irrational numbers must be as well, since each irrational number lies between two rational numbers. Take two rational numbers A and B, and one irrational number C, along with an integer n. If you take A Just don't ask me for a 24-prime set, I don't have one yet :). >Nor I, but I've found the following n-prime sets: >n = 17: eight 2's and nine 3's >n = 28: 15 2's and 13 1's I believe five 1's and nineteen 2's will work for n=24. -- Erick === Subject: Re: Please recommend some entertainment math books > Please recommend some entertainment math books. > My undergraduate major was math, but, no graduate training, I had > graduate training in electrical engineering. > I am looking for some entertainment math books for my background. The > books should cover some new and depth results. Most entertainment math > books are for general public, too simple and too shallow. I am looking > for some easy and fun to read books with depth, and suitable for math > undergraduate major. (Not trivial subject of chaos or fractal, or > things like those). > Ideally the book can cover some recent active research results, but, > in someway, people without graduate training can understand and enjoy. Conway, The Sensual Quadratic Form. Julian Havil, Gamma. Robin Wilson, Four Colors Suffice. Donald Saari, Chaotic Elections. Ian Stewart, Another Fine Math You've Got Me Into. David Gale, Tracking the Automatic Ant. Colin Adams, The Knot Book. Charles Livingston, Knot Theory. Alf van der Poorten, Notes on Fermat's Last Theorem. === Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Please recommend some entertainment math books >Please recommend some entertainment math books. >[...] >I am looking for some entertainment math books for my background. The >books should cover some new and depth results. Most entertainment math >books are for general public, too simple and too shallow. I am looking >for some easy and fun to read books with depth, and suitable for math >undergraduate major. How about Edwards HM. Riemann's Zeta Function. ? The following do not exactly fit the as far as the depth you request, but I think you will find them very enertaining nonetheless. I certainly did. Abbot EA. Flatland. (Stewart also has an annotated version, which I have not yet read.) Aczel AD. Fermat's Last Theorem Davis PJ, Hersch R. The Mathematical Experience. Really, absolutely delightful. Not mathematics, but with some mathematical content: Carroll L. The Annotated Alice. (Annotated by M. Gardner. Again, delightful.) If you want to teach yourself some basic mathematics with which you are not yet familiar, and which will allow yourself to understand more advanced mathematics, you might take a look at any of these short textbooks. As someone else has put it, these are not exactly bedtime reading. Bartle RG. The Elements of Integration. Halmos PR. Naive Set Theory. Lindley DV. Making Decisions. Moore TO. Elementary General Topology. (Actually, Munkres JR, Topology is a better, in-depth book, but it is not short.) Ross SM. Applied Probability Models with Optimization Applications. (Again, Ross's Intro. to Probability Models is a more fleshed-out text.) Ross SM. An Introduction to Mathematical Finance. Spivak M. Calculus on Manifolds. Perhaps you will enjoy Nasar S. A Beautiful Mind. Struik DJ. A Concise History of Mathematics. === Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Please recommend some entertainment math books > Please recommend some entertainment math books. > My undergraduate major was math, but, no graduate training, I had > graduate training in electrical engineering. > I am looking for some entertainment math books for my background. The > books should cover some new and depth results. Most entertainment math > books are for general public, too simple and too shallow. I am looking > for some easy and fun to read books with depth, and suitable for math > undergraduate major. (Not trivial subject of chaos or fractal, or > things like those). > Ideally the book can cover some recent active research results, but, > in someway, people without graduate training can understand and enjoy. David Wells' _Curious and Interesting Geometry_ is lots of fun. Read it at bedtime and you will be getting back up to find a pencil and paper. Eric Weisstein's MathWorld is available as a book, published by CRC press. I would urge you to buy it directly from Eric's site. === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. Huh? I opened this message expecting a sentence in FOL. But T1 and T2 are predicates, and you want to show that they're the same... But the FOL doesn't have rules of inference for doing this. I suspect you're using a second-order logic. And I suspect that you must disambiguate your sentences, as wedges and vees don't mix well without parentheses. And I suspect that what you're trying to prove isn't called a tautology, since it doesn't seem to follow from the meanings of the truth functional operators in your ambiguous wffs. Alex Solla Junior Reed College === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. A couple of questions: Then I represents x=y ? What's with the repeated (redundant) G(x,y) in the definition of T1 and T2(x,y) in the definition of T2? At this point, I am not so sure that T1 and T2 are uniquely defined (can tell better after the above questions are answered), though I guess T1 could = T2 if they are equally ambiguous! Charlie Volkstorf Cambridge, MA === Subject: Re: stuck proving simple(?) tautology > Given a binary relation G(x,y) and T1 defined as: > T1(x,y) = I / G(x,y) / G(x,y) / T1(x,y) Huh? That's not a definition and it's confusing. Do you mean I / G / (G/T1) ? I presume / or, / and. > and T2 defined as: > T2(x,y) = I / G(x,y) / T2(x,y) / T2(x,y) > where I is identity relation, prove that T1 = T2. Ditto. === Subject: Taking Calculus without taking precalcu My friends that have taken precalculus have said they pretty much knew everything and that it was a waste and that they had wished they had gone straight to calculus. I've been thinking about doing the same, I have a firm grounding in trigonometry got a A in it. Would you recommend against me just outright skipping precalculus? The only catch is I would have to write a challenge letter requesting I skip the prerequisite of precalculus. Any suggestions on what to say in there a math professor would be looking at the letter === Subject: Re: Taking Calculus without taking precalcu > My friends that have taken precalculus have said they pretty much knew > everything and that it was a waste and that they had wished they had > gone straight to calculus. I've been thinking about doing the same, I > have a firm grounding in trigonometry got a A in it. Would you > recommend against me just outright skipping precalculus? The only > catch is I would have to write a challenge letter requesting I skip > the prerequisite of precalculus. Any suggestions on what to say in > there a math professor would be looking at the letter I always thought precalc was for students who had algebra II (quadratic equations, graphing polynomials, etc.) but not trig. I seem to remember that precalc covers the trig functions, the exp and log function, rules for exponents, and the concept of inverse functions (as in exp and log). If you feel good about most of that, you should take calculus directly. === Subject: Re: Taking Calculus without taking precalcu > My friends that have taken precalculus have said they pretty much knew > everything and that it was a waste and that they had wished they had > gone straight to calculus. I've been thinking about doing the same, I > have a firm grounding in trigonometry got a A in it. Would you > recommend against me just outright skipping precalculus? The only > catch is I would have to write a challenge letter requesting I skip > the prerequisite of precalculus. Any suggestions on what to say in > there a math professor would be looking at the letter There is no subject of precalculus; it is something invented by bureaucrats and is to be avoided if at all possible. Your A in trig, unfortunately, is meaningless without further information. A's are handed out in high school these days to students who can pass the metal detector test. The most important thing is your actual understanding and facility with trig, algebra, and geometry. By all means write the challenge letter if you are sure of your abilities in these areas. === Subject: Re: Taking Calculus without taking precalcu > My friends that have taken precalculus have said they pretty much knew > everything and that it was a waste and that they had wished they had > gone straight to calculus. I've been thinking about doing the same, I > have a firm grounding in trigonometry got a A in it. Would you > recommend against me just outright skipping precalculus? The only > catch is I would have to write a challenge letter requesting I skip > the prerequisite of precalculus. Any suggestions on what to say in > there a math professor would be looking at the letter > There is no subject of precalculus; it is something invented by bureaucrats > and is to be avoided if at all possible. Your A in trig, unfortunately, is > meaningless without further information. A's are handed out in high school > these days to students who can pass the metal detector test. The most > important thing is your actual understanding and facility with trig, > algebra, and geometry. By all means write the challenge letter if you are > sure of your abilities in these areas. With great respect for Wade and his many excellent posts, I must disagree on this one. Precalculus is basically college algebra + trigonometry + sometimes an introduction to transcendental functions. Show me a student who called it a waste, and nine times out of ten, I'll show you a student who can't graph a logarithm function or even something like y = sin(x-pi). One of the best students I ever had took every course twice because of her feelings of insecurity. For whatever reason, it works great. But anyway, here is a simple test. If you can sit down right now, and without books or notes, and taking all the time you want, derive a formula for tan(a+b+c) in terms of tan(a), tan(b), and tan(c), then you don't need precalculus. Otherwise, I would say you do. === Subject: Re: Taking Calculus without taking precalcu > Would you >recommend against me just outright skipping precalculus? The only >catch is I would have to write a challenge letter requesting I skip >the prerequisite of precalculus. Any suggestions on what to say in >there a math professor would be looking at the letter How is your retention of Intermediate Algebra? You performed excellently in Trigonometry, but how is your retention of that main topic? If your school has a qualification examination for skipping PreCalculus, you could take advantage of it to see if you could be recommended to skip to Calculus. If your retention is very good of Intermed Algebra and the Trigonometry, then PreCalculus may very well be a waste of time; you could later review any subtopic you need from any old PreCalculus textbook. Calculus will mostly just require some basic trigonometry, and all of intermediate algebra. You may find that the study of Calculus itself will force you THEN to review what you need. Your friends generally have the right idea IF you retain your Intermediate Algebra and Trigonometry knowledge. G C === Subject: Re: Tell me your paradoxes! >(2) In the town of Placerville (CA) the barber shaves everyone who doesn't >shave himself. Who shaves the barber? Not a paradox. The barber can shave himself without violating anything you've said. Doug === Subject: Re: Tell me your paradoxes! >shave himself. Who shaves the barber? >Not a paradox. The barber can shave himself without violating anything >you've >said. Yes. I suppose the barber would have to only shave those who don't shave themselves. That nobody in Placerville shaves probably won't help me any. Rich === Subject: Re: Tell me your paradoxes! >Have you got any paradoxes to share? Write them here! > (1) How many kinds of infinity are there? An infinite number? If so, what > kind of infinity? Is this really a paradox, or just mindbooglingly confusing without actually being paradoxical? > (2) In the town of Placerville (CA) the barber shaves everyone who doesn't > shave himself. Who shaves the barber? She shaves herself, obviously. Yours, H... === Subject: what is the eigenvalue of a linear transform? , I have a question about linear transform. What is the definition of the eigenvalue of linear transform? How to show the linear transformation T: f(x) -> integrate[f(t), t from 0 to x] has no eigenvalues? -Walala === Subject: Re: Why are martingales called martingales? > ... > Jacqueline Pichoche's _Dictionnaire Etymologique du francais_ > (Robert, 1979) clarifies this, and, I think, settles where > the betting system got its name. My translation: > >[snip def supporting above] >Maybe it settles it, maybe not. The previously-suggested >bridle idea is supported by Webster's 1913 Dictionary >(according to http://www.hyperdictionary.com anyhow): 3. (Gambling) The act of doubling, at each stake, that which has >been lost on the preceding stake; also, the sum so risked; -- >metaphorically derived from the bifurcation of the martingale >of a harness. [Cant] --Thackeray. That is, hyperdictionary's >entry makes no mention of this usage of martingale deriving from >Martigues, but instead says it derives from martingale as harness. > My 1924 printing of the 1909 edition of the Merriam-Webster > New International Dictionary (commonly called Webster's First > International) doesn't have that citation from Thackeray. I > don't know which Thackeray it was, but assume it to be the > English novelist and literary man of that name. In that case > certainly (and in any case, maybe with less certainty) I would > greatly favor the scholarly opinion of Jacqueline Picoche > (a professional lexicographer writing in 1979) over the > (very likely less scholarly, and in any case many decades > earlier) opinion of Thackeray. The metaphorical derivation > given by Thackeray has all the earmarks of folk-etymology > (if not his own fertile imagination). Also, though Picoche > doesn't give a citation for jouga a la martegalo, she does > date it to the 18th century, whereas the Oxford English > Dictionary's earliest citation for martingale in the > gaming sense is from 1815 (ah, and followed by Thackeray-the- > novelist, in a novel, 1854) as a noun, and from 1823 as > a verb (whose etymology asks the reader to compare to French martingaler without positively claiming that that's its source; > the OED etymology of the *noun* martingale is even less help). And to think that for all these years I thought the game-theorist David Gale had named it in honor of his brother, Martin.... === Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)