mm-3059 === Subject: Re: Old question: looking for prime number list > 1,000,000 On 02 Jan 2006 23:00:59 +0200, Phil Carmody >> ... >> What counts as a known prime number? Is 156593416582327342765171 >> known? I guess it is now, but it wasn't 5 minutes ago. >> The primest number I know is >> 19112684214957755703306290219340140859813072336321619 >> a number I have proven prime already at least a hundred times, so >> it must be prime. (I use it as input for my various implementations >> of ACL.) >That's nothing! >If you're looking for Cunningham Chains, and forget to double >the number (+/-1) that you're testing after finding a prime >starting point, then you may end up proving it several thousand >times before you realise that pressing ^C is probably a sensible >thing to do... We've all done that with respect to some algorithm or other. Those poor people without ^C -- they never break out. === Subject: Re: 2 GRE practise questions the mean of the resulting population, then set x to that mean, > which would minimize the resulting standard deviation (since (x-mean)^2 > in that case would be zero). > The mean in this case would of course still be 20. A pity I didn't > think of this first! :-) > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. Oh, there was a similar question to this stadrad deviation probelm in the exam. === Subject: Re: 2 GRE practise questions I look at this thread before heading for the exam this morning. === Subject: Re: 2 GRE practise questions <43b8c133$0$93567$892e7fe2@authen.yellow.readfreenews.net Q1) If n is an odd integer, the next larger odd integer is > - n^2 + 1 > No > - n^2 + 4 > yes > - n^2 + 2n + 1 > no > - n^2 + 4n + 4 > yes > - n^2 + n + 1 > yes > Which is lowest? 4, 4n+4 or n+1 if n=1 lowest odd? SO it is the last > answer. > Odd times odd is ODD; Odd times even number is EVEN, Odd + Even is Odd > To determing the *next* larger one, I would add 2 to n but tht is not > a choice. So does this mean the question has a problem? > Q 2) 10, 15, 25, 30, x > If d is the standard deviation of the numbers in the list above, for > which of the following of x, would the value of d be the least? > 0 , 5, 20, 30, 60 > # clostest to the average, which is about 20, so choose 20 > I only know using the steps to calculate the stdarad deviation. That > approach would be detrimenatlly time-coumsing. So can any one help me > with how to think a faster way? > Sure see above. > Anything else, anything harder? No. Took GRE this morning. Did very well. === Subject: Re: 2 GRE practise questions >> Q1) If n is an odd integer, the next larger odd integer is >> - n^2 + 1 >> No >> - n^2 + 4 >> yes >> - n^2 + 2n + 1 >> no >> - n^2 + 4n + 4 >> yes >> - n^2 + n + 1 >> yes >> Which is lowest? 4, 4n+4 or n+1 if n=1 lowest odd? SO it is the last >> answer. >> Odd times odd is ODD; Odd times even number is EVEN, Odd + Even is Odd >> To determing the *next* larger one, I would add 2 to n but tht is not >> a choice. So does this mean the question has a problem? >> Q 2) 10, 15, 25, 30, x >> If d is the standard deviation of the numbers in the list above, for >> which of the following of x, would the value of d be the least? >> 0 , 5, 20, 30, 60 >> # clostest to the average, which is about 20, so choose 20 >> I only know using the steps to calculate the stdarad deviation. That >> approach would be detrimenatlly time-coumsing. So can any one help me >> with how to think a faster way? >> Sure see above. >> Anything else, anything harder? > No. Took GRE this morning. Did very well. Congratulations! It is a tough test, I ace'd all the math, chem, physics, not quite as good on the English. You will love Grad school. Students there only want to learn as much as possible, all the slackers are gone. === Subject: Re: Ricker model >Hi all, >This is the first time for me to write about a math question, and I am >really not good at math. So please bear with me. >I ran into a passage that i don't understand. >it says >X (t+1) = b*X(t)*exp(-c*X(t)) >X(t) represents the density of individuals per unit habitat sie at time >t, b > 0 is the per capita birth rate, and exp(-c*X(t)) is the fraction >of offspring expected to survive one unit of time at population density >X(t). The exponential term that occurs here can be interpreted as >arising from an assumption of random contacts among individuals. >And the explanation about the exponential term is what I don't >don't feel comfortable about. >Could anyone explain about why the exponential term is in the >equation?? Consider an individual offspring. As it wanders around, the number of adults it encounters should be approximately proportional to X(t). In each such encounter it has a certain probability p of dying (this might be appropriate for certain fish species, where the adults like to eat the young of their own species). So the probability of surviving n encounters is (1-p)^n = exp(n ln(1-p)). With n proportional to X(t), you get exp(-c X(t)) for some positive constant c. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Ricker model === Subject: Re: ? crim and math > ... I'll just comment on the part below ... >I do find it alarming, simply because of the size of the population. I am >not even suggesting that you feel sorry for them, but simply that given the >size of this population that it's definately a serious problem. 2.5 million >people living in prison costs alot of money, and drains the GNP, there's an >economic impact, etc. >Just imagine a population the size of Chicago, all of them in prison, none >of them have jobs or do much work, and they have to be fed, clothed, housed, >etc etc. It gets expensive. The fact is, most prisoner's do work, and for less than minimum wage. The prisons are evolving into slave labor camps. Since the prisoners work for lower wages than workers on the outside, the effect is to take jobs away from the working class. The key to fighting crime is to build a large, secure middle and lower middle class, especially in the inner cities. This is what LBJ successfully began in the 1960s. By the late 1960s, with major amounts of federal money focused on rebuilding the infrastructure and providing key services in poorer neighborhoods, many such areas improved dramatically. A prime example was Harlem, NYC which was becoming quite respectable. All that quickly came to an end with advent of Reagan, who cut programs and services for the poor, shifted more than a trillion dollars in wealth to the ulta-wealthy in the form of tax cuts, all the while promising that the wealth thus created would trickle down to the poor. But the money never trickled down. Harlem regressed, and the South Bronx got so bad, it came to be regarded as a war zone. A vicious cycle ensues -- lack of jobs fuels a rise in crime, followed by an increase in the prison population, which then takes away still more jobs. Privately, the power elite doesn't mind high crime. The widespread fear of crime gives them a mandate to take greater control. Moreover, it makes the middle class fear the poor, and even the poor fear the poor, thus dividing the working class politically, and pulling the middle class to the right. At the same time, a large prison population working for next to nothing keeps wages low on the outside. Businesses like that. quasi === Subject: Re: ? crim and math You're exactly right. Having said that, we still have the situation that many prisoners belong where they are. Some really are too dangerous to be released. And when someone goes to prison who is actually innocent, chances are good that he might come out as a criminal - as in the case of a Wisconsin man who was exhonerated on a rape charge on DNA after doing almost 20 years - he gets out and murders someone. The situation needs to looked at very carefully, if not to fix the unfixable, certainly to avoid falling into it. === Subject: An amusing (amusing?) fact about Maplesoft I keep seeing during several hours http://maple.bug-list.org/maplesoft_com_jan_2_2006.jpg This is from http://www.maplesoft.com/ Compare with http://maple.bug-list.org/wolfram_com_jan_2_2006.jpg This is from http://www.wolfram.com/ http://maple.bug-list.org/derive_com_jan_2_2006.jpg This is from http://www.derive-europe.com/downloads.asp http://maple.bug-list.org/mupad_com_jan_2_2006.jpg This is from http://www.mupad.com/ Any comments? === Subject: Re: An amusing (amusing?) fact about Maplesoft Their site is still up, (see http://www.maplesoft.com/applications/ for example), so this looks like a configuration problem. === Subject: Re: An amusing (amusing?) fact about Maplesoft RP> this looks like a configuration problem. Actually, during at least 12 hours, *dozens* links at the Maplesoft's site does not work... including the main menu ones (!) Industry http://www.maplesoft.com/industry/index.aspx Academic http://www.maplesoft.com/academic/index.aspx Products http://www.maplesoft.com/products/index.aspx Maple http://www.maplesoft.com/products/maple/index.aspx MapleNet http://www.maplesoft.com/products/maplenet/index.aspx Maple T.A. http://www.maplesoft.com/products/mapleta/index.aspx Support http://www.maplesoft.com/support/ FAQ http://www.maplesoft.com/support/faqs/index.aspx Search http://www.maplesoft.com/search/index.aspx Contact http://www.maplesoft.com/contact/index.aspx Industry Home http://www.maplesoft.com/industry/index.aspx Application Center http://www.maplesoft.com/applications/app_center_mapleta.aspx Submit Your Application http://www.maplesoft.com/applications/submit_application_submit_form.aspx Tips/Techniques http://www.maplesoft.com/applications/app_center_techniques.aspx Application Center - Quick Search http://www.maplesoft.com/applications/app_center_quick_search.aspx?T=D Application Center - Advanced Search http://www.maplesoft.com/applications/app_center_advanced_search.aspx Community http://www.maplesoft.com/community/index.aspx Developers http://www.maplesoft.com/developers/index.aspx Resources http://www.maplesoft.com/resources/insideview/index.aspx Advanced Search http://www.maplesoft.com/applications/app_center_advanced_search.aspx Top Rated Applications http://www.maplesoft.com/applications/app_center_quick_search.aspx?T=T New Applications http://www.maplesoft.com/applications/app_center_quick_search.aspx?T=N Most Downloaded http://www.maplesoft.com/Applications/app_center_quick_search.aspx?T=D and lots more... All this is NOT a surprise for me as it fits perfectly the pattern I have mentioned here VB> Actually, Mr DeMarco is just a particular case. A part VB> of an amazing pattern to be demonstrated in all its VB> richness. VB> I would rather say, a part of Maple mythology. An integral part of Maplesoft is the Maple mythology. Best New Year wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: Re: An amusing (amusing?) fact about Maplesoft > RP> this looks like a configuration problem. > Actually, during at least 12 hours, *dozens* links at the > Maplesoft's site does not work... including the main menu > ones (!) I think that /application part is on a different computer - still it is not clear what happened. Alec === Subject: Re: infinity sci.math_20050407: I had already scanned over your M280, I do again. You seem to want to deny the existence of the continuum as an infinite set of points, individua. sci.math_20050409: Consider a well-ordering of the reals. Even in infinite induction it must meet the requirements of nested intervals (Cantor/Megill). Consider other characteristics that a well-ordering of the reals must have. sci.math_20050409: That's not to say that every single point I put forward about the infinite, or mathematical logic and theory, is absolutely correct. On the contrary, I stumble upon my own errors sometimes, in that to some extent researching my own posts gives me the most insight. On the other hand, when I was very much naive of set theory, although firmly grounded in mathematics, when asked for a bijection between the naturals and reals, I still use it. === Subject: Re: Dynamic Infinite sets > I will right natual number set N=0,1,2,3,4,...... at speed of one > unite( one natural number per unit time). I will call this set No . > That makes no sense. No matter how small your unit of time is, you can > never finish your task. I'm sure dozens of people have already tried to > explain this to you. > What I said is fairly logical, If you have to say this is logical, chances are it is not. Just as when a film advertisement tells me this film is hilarious I know it won't be funny at all. > two sets of natural numbers writtin at > different speeds > will not be the same set even if they run to infinity! simply because > the comparison > still holds even at infinite level. The comparison still holds at infinite level has no meaning. There is only one test for set equality: is every element of set A in set B, and is every element of set B in set A? If the answer to both of those sets is yes, they are the same set. - Randy === Subject: is a 1 manifold connected? How can I prove that a 1-manifold is connected if it is connected? === Subject: Re: is a 1 manifold connected? > How can I prove that a 1-manifold is connected if it is connected? Connected implies connected seems fairly immediate to me... === Subject: Re: is a 1 manifold connected? >How can I prove that a 1-manifold is connected if it is connected? If it's connected then it's connected, QED. === Subject: Re: is a 1 manifold connected? >>How can I prove that a 1-manifold is connected if it is connected? > If it's connected then it's connected, QED. what I meant to say is to prove that a 1-manifold is connected. I am just not sure though if it is actually true. I am reading a proof and I need this information for it to give meaning. === Subject: Re: is a 1 manifold connected? <43b9be1c$0$176$edfadb0f@dread11.news.tele.dk>How can I prove that a 1-manifold is connected if it is connected? > If it's connected then it's connected, QED. > what I meant to say is to prove that a 1-manifold is connected. I am just > not sure though if it is actually true. I am reading a proof and I need > this information for it to give meaning. Well, there are 1-manifolds that are _not_ connected, so you'll need to satisfy some further conditions to get what you seem to need. For example, what in your particular situation would rule out { (x,y) | x^2 + y^2 = 1 or 2 } , say ???? === Subject: Re: is a 1 manifold connected? <43b9be1c$0$176$edfadb0f@dread11.news.tele.dk> Or (0, 1) U (2, 3), for that matter. === Subject: Re: is a 1 manifold connected? >How can I prove that a 1-manifold is connected if it is connected? >> If it's connected then it's connected, QED. >what I meant to say is to prove that a 1-manifold is connected. I am just >not sure though if it is actually true. I am reading a proof and I need >this information for it to give meaning. OK, I think I what you were trying to ask is this:.. Given a 1-manifold, specified to you in some way. The problem is to prove that the manifold is connected. How should one go about proving this? Assuming the above question is what you intended, here's one basic approach ... The specification for the manifold will typically represented as the image of one or more piecewise functions, where for each piece, the associated function is defined and continuous on an interval. Since an interval is connected, and since the continuous image of a connected set is connected, it follows that each of the pieces is connected. Now it can be easily proved that the union of 2 connected sets which share a common point is connected. Hence, if 2 pieces have a point in common, then the union of those 2 pieces is connected. Thus, you can join up the pieces to form larger connected sets. If there are finitely many pieces, join them end to end, 2 at a time, each join creating a larger connected set until you have the whole manifold. If there are infinitely many pieces, do the same thing, but it may take a little more care to define what joins with what and to show that all the pieces join up into one happy component. quasi === Subject: Re: Hausdorff, not-first countable space where sequential closure = closure? operation is just the operation of adjoining to a set its > sequential limit points? Any sequential space satisfying > this condition would be Frechet-Urysohn. Were you aware that Franklin's two papers on sequential spaces are available on the internet? Stanley Phillip Franklin, Spaces in which sequences suffice, Fundamenta Mathematicae 57 (1965), 107-115. [MR 31 #5184; Zbl 132.17802] http://www.emis.de/cgi-bin/Zarchive?an=0132.17802 http://matwbn.icm.edu.pl/tresc.php?wyd=1&tom=57 Stanley Phillip Franklin, Spaces in which sequences suffice II, Fundamenta Mathematicae 61 (1967), 51-56. [MR 36 #5882; Zbl 168.43502] http://www.emis.de/cgi-bin/Zarchive?an=0168.43502 http://matwbn.icm.edu.pl/tresc.php?wyd=1&tom=61 If you want to look into the literature, some names to look up are: Josef Novak, Thomas W. Rishel, V. P. Zolotarev, R. M. Dudley, and Charles E. Aull. For a nice historical survey, see R. Fric, History of sequential convergence spaces, pp. 343-355 in C. E. Aull and R. Lowen (editors), HANDBOOK OF THE HISTORY OF GENERAL TOPOLOGY, Kluwer Academic Publishers, Volume 1, 1997. bibliography are available. http://tinyurl.com/dtwqv [p. 345] http://tinyurl.com/bodor [p. 348] http://tinyurl.com/bdfm6 [p. 350] http://tinyurl.com/94n9m [p. 351] http://tinyurl.com/8jfg9 [p. 352] http://tinyurl.com/7zo5d [p. 353] http://tinyurl.com/77q2p [p. 354] http://tinyurl.com/dskmg [p. 355] Dave L. Renfro === Subject: Absence of integer solutions Assertion-1 refers to the following four(4) equations under the conditions: odd k > 3 and all the variables in each equation are distinct and each > 1. a^k + b^k = c^k (1): d^k + e^k = 2f^k (2) g^2k + h^2k = m^2k (3): n^2k + p^2k = 2q^2k (4) Given situation: absence of integer solutions of (1) implies the absence of integer solutions of (2). Assertion-1: Absence of integer solutions of (3) also implies the absence of integer solutions of (4). Any comment upon the correctness of Assertion-1 will be highly appreciated. Any recent reference on this subject will be welcome. === Subject: Re: Can you embed the Beltrami plane in 3-space? > I am reading reading a book that asserts (with no proof) that it is > impossible. What I mean by an embedding is one for which the Beltrami > metric is just the ordinary arc length computed along the surface. It > seems to me that by just making a surface that crinkles more and more > as you go out from the center, you could embed. I know that someone > has figured out how to knit a portion of the plane in that way. You're probably referring to Daina Taiminia, a professor at Cornell, who has found a way to crochet a hyperbolic plane. I've tried this several times; it's illuminating. I teach a geometry course every winter, and offer my students the opportunity to crochet one as an alternative to one of their homework assignments. Once they have done it, they have a durable model that they then use to study additional properties of the hyperbolic plane. I believe that the result of Hilbert's is that there is no smooth embedding of the hyperbolic plane into R3. There's a later result about the existence of C1-embeddings. I don't know if it's been settled yet as to whether the crocheted surfaces are approximations of a C1 embedding or not. Here's the link to the Taiminia-Henderson paper about it: http://www.math.cornell.edu/~dwh/papers/crochet/crochet.html I also have pictures of some of my attempts at: http://chazjac.home.sg23.com/crochet.htm --charlie j === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth ... > Dik, what Moise and the wikipedia demonstrate is that you can have two > different approaches to the Axiomatics of Euclid or of Peano, but both > have arithmetic already embedded into the body of the axioms > themselves. They show nothigmng about Peano as they do not mention Peano. Of course you can do Peano using arithmetic but that is putting the cart before the horse. In that case you have to define arithmetic first. But you can use the Peano axioms without arithmetic defined, and define arithmetic in terms of the successor function from Peano. And that means that you have a sound basis for arithmetic. Indeed, you can do Eclid in two ways, analytical and synthetic. But when you do it in a synthetic way you are actually *not* doing arihtmetic. > themselves. You cannot have the Euclid Plane geometry without that > second group of five axioms as listed by Wikipedia above which is > arithmetic. It is those two groups of axioms the first five and second > five for ten in all that delivers Euclid geometry and it has arithmetic > as a majority of the axioms. Please read them again: > Euclid also had five common notions which can also be taken to be > axioms, though he later used other properties of magnitudes. > 1. Things which equal the same thing also equal one another. > 2. If equals are added to equals, then the wholes are equal. > 3. If equals are subtracted from equals, then the remainders are equal. > 4. Things which coincide with one another equal one another. > 5. The whole is greater than the part. For instance (4) is not sufficient, it should state that Things that can be made to coincide by rotation, mirroring and moving are equal, but that would be in current terminology. In that case (1) and (4) together define an equivalence relation. (5) defines a partial ordering (which can, with the use of (1) and (4) be made to a full ordering). So (1), (4) and (5) do *not* use arithmetic. I would not like to call (2) and (3) arithmetic either, they merely extend the equivalence relation to a new equivalence relation. Consider Thabit ibn Kurrah's proof of Pythagoras' theorem, it is purely geometric, no arithmetic involved. But it uses (4) (as modified by me and in this way it implies the parallel axiom) and (2). But it does *not* use (2) in any arithmetic sense. (And is, in my opinion, the most elegant geometric proof that exists, and it is already over 1000 years old.) And indeed the original formulation of the theorem was not arithmetical but geometric: The sum of the squares on the base sides of a triangle with right angle is equal to the square on the hypothenusa of that triangle. (And you may note that Pythagoras does not hold in non-Euclidean geometry.) > Now Moise shows the two approaches of a metric and a synthetic and I > suspect that Dik Winter is believing that the Peano axioms are all > synthetic approach where arithmetic is defined outside the Peano > axioms, but looking at (7) of Moise we see that arithmetic is embedded > in both approaches no matter if you go the synthetic or the metric. You are wrong in quite a number of ways. With the Peano axioms, arithmetic is *not* defined outside the Peano axioms. It is defined *using* the Peano axioms. And indeed they can be used with sets for which no arithmetic is defined. There is quite some difference. Moise only talks about Eclid's geometry, he says that addition is performed using congruence classes (I would call them equivalence classes, but that is just terminology). Nothing like that in the arithmetic based on the Peano axioms. But also in the use of congruence classes there is *no* use of arithmetic. > Thus, one can put Peano axioms into a same Table and call the Successor > postulate as a synthetic approach where it never mentions the distance > but the metric entry for Successor would state the Endless adding of > 1 And in line (7) Addition for the Peano axioms the synthetic approach > would say performed on Successors whereas the metric approach would > say performed on numbers In that case you have *first* to define what endless adding of 1 means, but not only that, but also what addition is. Without such definitions it makes no sense. So give me a clear definition of addition and endless adding of 1, without that we do not have a notion about what your metric approach to the Peano axioms mean. Try to define arithmetic without the Peano axioms as basis, please. Once you have done that you will find that implicitly you implied also the Peano axioms. > So Dik is > wrong when he says that Axiomatics of Euclid or Peano are void of > arithmetic. Both Euclid and Peano have arithmetic inside their > axiomatics and depending on what approach you take whether you want to > discuss Successors or want to discuss Endless adding of 1 to 0 and 1, > is up to you. I may also note that Euclid's arithmetic does not conform to the Peano axioms because there is no successor function. And as I stated before, and state again, the successor function is *not* necessarily the addition of 1. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth Dik, what Moise and the wikipedia demonstrate is that you can have two > different approaches to the Axiomatics of Euclid or of Peano, but both > have arithmetic already embedded into the body of the axioms > themselves. > They show nothigmng about Peano as they do not mention Peano. Of course > you can do Peano using arithmetic but that is putting the cart before the > horse. In that case you have to define arithmetic first. But you can > use the Peano axioms without arithmetic defined, and define arithmetic > in terms of the successor function from Peano. And that means that you > have a sound basis for arithmetic. > Indeed, you can do Eclid in two ways, analytical and synthetic. But > when you do it in a synthetic way you are actually *not* doing arihtmetic. > themselves. You cannot have the Euclid Plane geometry without that > second group of five axioms as listed by Wikipedia above which is > arithmetic. It is those two groups of axioms the first five and second > five for ten in all that delivers Euclid geometry and it has arithmetic > as a majority of the axioms. > Please read them again: > Euclid also had five common notions which can also be taken to be > axioms, though he later used other properties of magnitudes. > 1. Things which equal the same thing also equal one another. > 2. If equals are added to equals, then the wholes are equal. > 3. If equals are subtracted from equals, then the remainders are equal. > 4. Things which coincide with one another equal one another. > 5. The whole is greater than the part. > For instance (4) is not sufficient, it should state that Things that > can be made to coincide by rotation, mirroring and moving are equal, > but that would be in current terminology. In that case (1) and (4) > together define an equivalence relation. (5) defines a partial > ordering (which can, with the use of (1) and (4) be made to a full > ordering). So (1), (4) and (5) do *not* use arithmetic. I would > not like to call (2) and (3) arithmetic either, they merely extend the > equivalence relation to a new equivalence relation. > Consider Thabit ibn Kurrah's proof of Pythagoras' theorem, it is > purely geometric, no arithmetic involved. But it uses (4) (as > modified by me and in this way it implies the parallel axiom) and > (2). But it does *not* use (2) in any arithmetic sense. (And is, > in my opinion, the most elegant geometric proof that exists, and it > is already over 1000 years old.) And indeed the original formulation > of the theorem was not arithmetical but geometric: The sum of the > squares on the base sides of a triangle with right angle is equal to > the square on the hypothenusa of that triangle. (And you may note > that Pythagoras does not hold in non-Euclidean geometry.) > Now Moise shows the two approaches of a metric and a synthetic and I > suspect that Dik Winter is believing that the Peano axioms are all > synthetic approach where arithmetic is defined outside the Peano > axioms, but looking at (7) of Moise we see that arithmetic is embedded > in both approaches no matter if you go the synthetic or the metric. > You are wrong in quite a number of ways. With the Peano axioms, > arithmetic is *not* defined outside the Peano axioms. It is defined > *using* the Peano axioms. And indeed they can be used with sets for > which no arithmetic is defined. There is quite some difference. Moise > only talks about Eclid's geometry, he says that addition is performed > using congruence classes (I would call them equivalence classes, but > that is just terminology). Nothing like that in the arithmetic based > on the Peano axioms. But also in the use of congruence classes there > is *no* use of arithmetic. > Thus, one can put Peano axioms into a same Table and call the Successor > postulate as a synthetic approach where it never mentions the distance > but the metric entry for Successor would state the Endless adding of > 1 And in line (7) Addition for the Peano axioms the synthetic approach > would say performed on Successors whereas the metric approach would > say performed on numbers > In that case you have *first* to define what endless adding of 1 means, > but not only that, but also what addition is. Without such definitions > it makes no sense. So give me a clear definition of addition and > endless adding of 1, without that we do not have a notion about what > your metric approach to the Peano axioms mean. Try to define arithmetic > without the Peano axioms as basis, please. Once you have done that you > will find that implicitly you implied also the Peano axioms. > So Dik is > wrong when he says that Axiomatics of Euclid or Peano are void of > arithmetic. Both Euclid and Peano have arithmetic inside their > axiomatics and depending on what approach you take whether you want to > discuss Successors or want to discuss Endless adding of 1 to 0 and 1, > is up to you. > I may also note that Euclid's arithmetic does not conform to the > Peano axioms because there is no successor function. And as I stated > before, and state again, the successor function is *not* necessarily > the addition of 1. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Hi Dik. I was wondering if you could list some other Axiomatic systems. I happen to know of only Peano and Euclid. I am looking for an axiomatic system that is void of arithmetic altogether and only purely geometric. I was wondering if there is an axiomatic where it is void of geometry and purely only arithmetic. I can only relate to physics where you have Matrice operations, but I do not recall if Matrix theory was axiomatized. As for geometry, I do not recall if Topology has to be axiomatized or ever was, or whether topology comes from geometry with a few definitions. So I am looking to see if there exists only these two axiomatic systems of Euclid and Peano and all the rest of mathematics is appended definitions upon just those two axiomatic systems. Can you have matrix theory without any geometry? Can you have topology without any arithmetic? I am inclined to think that all of mathematics is based on just two Axiomatic Systems of Euclid and Peano and all the rest is add on definitions that extend just those two systems. Come to think of it, is Probability theory a piece of mathematics that has no geometry? And is it an axiomatic system? Come to think of it, Probability theory is the best candidate for a third axiomatic system. Any thoughts? Opinions, Dik? P.S. another thought comes to mind is Set theory, but that is an example of where it is an axiomatic system that has both geometry and arithmetic intertwined and embedded simultaneously. So Set theory would be a theory in support of my claims that you cannot have an Axiomatic system which is void or either arithmetic or geometry. Containment or membership is really a geometrical concept, is it not? And union and intersection are arithmetic concepts. So I think that where this is all going to end up is that every Axiomatics in mathematics has both arithmetic and geometry embedded into the axioms and that this notion that you can have a axiomatic system that is void of either geometry or arithmetic is fantasy. I do not know if there is a geometry embodied in Matrix theory, but then again I suppose matrix theory was never Axiomatized or Axiomatizable. I suppose Matrix theory is just a definition. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth Hi Dik. I was wondering if you could list some other Axiomatic systems. [...] There are tons; there are field axioms (which you should know), group axioms, equivalence axioms, Eilenberg-Steenrod axioms, Hausdorff axioms, Probability axioms, Zermelo-Fraenkel axioms (a version of set theory), etc. > So I am looking to see if there exists only these two axiomatic systems > of Euclid and Peano and all the rest of mathematics is appended > definitions upon just those two axiomatic systems. Nope. > Can you have matrix theory without any geometry? Sure. > Can you have topology without any arithmetic? Sure. --- Christopher Heckman === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth And it comes down to a question of can you have a Peano Axiom System void of arithmetic if it is the Counting Numbers that create Arithmetic? Similarly, and ridiculous would be to ask, can you have a Euclid Axiomatic System void of geometry and define geometry after you laid down the postulates. Sounds utterly ridiculous but then again to say that Peano axioms are void of arithmetic is kin to saying that Euclid axioms are void of geometry until you define them separately and later. And history should weigh in on this argument between Dik and I, where I take the stance that all axiomatic systems of mathematics have both geometry and arithmetic embedded simultaneously in their postulates, history shows us that Counting Numbers are older than man yet it took, what, until around 1890 before Counting Numbers were axiomatized. Can you have arithmetic and not have numbers? Can you have numbers yet no arithmetic. The answer is no. One creates the other simultaneously and you cannot have one without the other. So to think that you can have a Euclid axiomatics that is void of arithmetic or to have a Peano axiomatics that is void of arithmetic, is a false belief or understanding. Dik keeps telling me that no matter what sequence of Successors I cough up such as 0,5,15,20,.... or the Fibonacci sequence of 0,1,1,2,3,5,.... that once I define an arithmetic that all of those symbols become the sequence 0,1,2,3,4,..... But as a retort to that, I would say to Dik, that even before he defines a arithmetic, all he had to do was make his Successor postulate more clear and not obfuscation because when you take the Successor for what it truly is: the endless adding of 1, you bypass the unnecessary step of defining an entire Arithmetic just to turn the sequences of 0,5,15,20,.... and 0,1,1,2,3,5,.... into the Counting Numbers. So Dik would be happy by claiming there is no arithmetic in the Peano Axioms but once arithmetic is defined then you can separate out what are the Counting Numbers from what is not the Counting Numbers. But my retort is why wait to define Arithmetic to clear things up when the obfuscation in the first place was this term Successor when it really was endless adding of 1. And that Arithmetic was already embodied in the Axiomatics of the Natural Numbers. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth void of arithmetic if it is the Counting Numbers that create > Arithmetic? The counting numbers do not create arithmetic. The counting numbers and the successor function do. More dramatically: IN THE BEGINNING God created the natural numbers, a set of objects, and he saw they were good but unordered. And God created the Successor Function, to bring order to them, but he saw that they were not orderly enough. And God said: [the Peano Axioms], and he saw the Axioms and the Natural Numbers were good. And God created Peano, who brought forth addition, and the properties of addition, and multiplication, and the properties of multiplication, And the negative integers, and the rational numbers, and the real numbers, and the complex numbers, And Peano defined addition and multiplication for all of these numbers, and proved that they possessed useful properties, and God saw the numbers and operations were good. > Similarly, and ridiculous would be to ask, can you have a Euclid > Axiomatic System void of geometry and define geometry after you laid > down the postulates. It depends on whether you allow Projective Geometry ( http://mathworld.wolfram.com/ProjectiveGeometry.html ), another set of axioms I forgot to mention. > Sounds utterly ridiculous but then again to say > that Peano axioms are void of arithmetic is kin to saying that Euclid > axioms are void of geometry until you define them separately and later. I would say that the Euclidean axioms are certainly void of geometry of plane shapes and solids. > [...] The > history shows us that Counting Numbers are older than man yet it took, > what, until around 1890 before Counting Numbers were axiomatized. Can > you have arithmetic and not have numbers? Can you have numbers yet no > arithmetic. The answer is no. One creates the other simultaneously and > you cannot have one without the other. Which shows that you don't know anything about establishing foundations in mathematics. One question that led to the Peano axioms was: What is the minimum amount of structure needed on the set of natural numbers to guarantee that we end up with addition and multiplication that have the properties that they have? And: Is the commutative law of addition something which is God-given, or is there a good reason why it's commutative; i.e., can the commutative law be derived from something simpler? It turns out that with the Peano axioms, there is a good reason why x + y = y + x. It's not something that's arbitrary. > Dik keeps telling me that no matter what sequence of Successors I cough > up such as 0,5,15,20,.... or the Fibonacci sequence of 0,1,1,2,3,5,.... > that once I define an arithmetic that all of those symbols become the > sequence 0,1,2,3,4,..... Yes, because you're constructing a list of things you call natural numbers, where the position of the objects in the list is the most important part. The equation 1 + 2 = 3 boils down to: The 1st object in your list plus the 2nd object in your list equals the 3rd object in your list. (The first list item is called the 0th here.) It makes no difference whether your list is (0, 1, 2, 3, 4, ...) or (A, B, C, D, E, ...) or (zero, one, two, three, four, ...) or (null, eins, zwei, drei, vier, ...) or anything else. This is the sort of understanding you get by actually working with the Peano axioms. Once you actually do this, you realize that the names don't matter. AP is too lazy to take this road, however. > But as a retort to that, I would say to Dik, that even before he > defines a arithmetic, all he had to do was make his Successor postulate > more clear and not obfuscation because when you take the Successor for > what it truly is: the endless adding of 1, you bypass the unnecessary > step of defining an entire Arithmetic just to turn the sequences of > 0,5,15,20,.... and 0,1,1,2,3,5,.... into the Counting Numbers. BUT you also need to state that adding has certain properties, in particular, that it's commutative and associative, and that you never repeat the natural numbers. Otherwise, you may run into a situation where your addition has the property that 1 + 1 + 1 + 1 + 1 = 1, and then you only have four natural numbers. Now, once these properties are thrown in with the rest of AP's axioms, you do get an axiom system which is equivalent to Peano's; that is, if a set of numbers and its successor function satisfy Peano's axioms, then you can construct an addition which satisfies all of AP's axioms, and vice versa. > So Dik > would be happy by claiming there is no arithmetic in the Peano Axioms > but once arithmetic is defined then you can separate out what are the > Counting Numbers from what is not the Counting Numbers. But my retort > is why wait to define Arithmetic to clear things up when the > obfuscation in the first place was this term Successor when it really > was endless adding of 1. And that Arithmetic was already embodied in > the Axiomatics of the Natural Numbers. Arithmetic is embodied in the Peano Axioms, in the sense that AP was embodied in his body when he was one day old. It will eventually come out, but you can't just sit and wait for it. --- Christopher Heckman === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth Now I wonder if the concept of multiplication is embodied and embedded in the Peano Axioms. I wonder if Peano really wanted his Math Induction to be a postulate that is of multiplication. Can someone give a formulation of Math Induction that is of multiplication, rather than that of addition of suppose N and show for N+1. Now I know there are very many equivalent statements to Math Induction. But I wonder if there is one for which multiplication is clearly on center stage. So the real importance of Math Induction postulate is not this minimal set reason, but the most important feature should be that the Natural Numbers are intertwined with the arithmetic concept of multiplication. Now I know algebras can define multiplication from addition, but it is better to have both operations defined in the axiomatics. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth in the Peano Axioms. [...] Now you're beginning to think like a mathematician! The answer is yes, it is, but you need the right definition to make things work out in the end. The idea is to use the fact that multiplication is repeated addition. Here it is, once you've defined addition: If x and y are natural numbers, then x * 0 is defined to be 0 x * S(y) is defined to be (x * y) + x (Notice the parentheses around x * y, to avoid ambiguity.) So, for instance, if you wanted to calculate 1 * 2, it would look like: S(0) * S(S(0)) = [S(0) * S(0)] + S(0) = [(S(0) * 0) + S(0)] + S(0) = [0 + S(0)] + S(0) = S(0) + S(0) = S(S(0) + 0) = S(S(0)), which is 2 in normal English. This is the derivation of 1 * 2 = 2. Now it's not clear from the definition that x * y = y * x, so the normal course of events would be to prove that this is the case. Then the associative law (x * y) * z = x * (y * z) comes out, then the distributive law (x + y) * z = (x * z) + (y * z). This is all standard procedure, by the way. And it's a lot more accessible than p-adics. --- Christopher Heckman === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth Dik, what Moise and the wikipedia demonstrate is that you can have two > different approaches to the Axiomatics of Euclid or of Peano, but both > have arithmetic already embedded into the body of the axioms > themselves. > They show nothigmng about Peano as they do not mention Peano. Of course > you can do Peano using arithmetic but that is putting the cart before the > horse. In that case you have to define arithmetic first. But you can > use the Peano axioms without arithmetic defined, and define arithmetic > in terms of the successor function from Peano. And that means that you > have a sound basis for arithmetic. > Indeed, you can do Eclid in two ways, analytical and synthetic. But > when you do it in a synthetic way you are actually *not* doing arihtmetic. > themselves. You cannot have the Euclid Plane geometry without that > second group of five axioms as listed by Wikipedia above which is > arithmetic. It is those two groups of axioms the first five and second > five for ten in all that delivers Euclid geometry and it has arithmetic > as a majority of the axioms. > Please read them again: > Euclid also had five common notions which can also be taken to be > axioms, though he later used other properties of magnitudes. > 1. Things which equal the same thing also equal one another. > 2. If equals are added to equals, then the wholes are equal. And there it is, #2, the requirement for addition and thus arithmetic. I really doubt that Euclid only uses the fact that A = B and C = D implies A + C = B + D; i.e., well-definedness of addition. I think that commutatitivity of addition is required, for instance, at some point. > 3. If equals are subtracted from equals, then the remainders are equal. > 4. Things which coincide with one another equal one another. > 5. The whole is greater than the part. > For instance (4) is not sufficient, it should state that Things that > can be made to coincide by rotation, mirroring and moving are equal, > but that would be in current terminology. In that case (1) and (4) > together define an equivalence relation. (5) defines a partial > ordering (which can, with the use of (1) and (4) be made to a full > ordering). So (1), (4) and (5) do *not* use arithmetic. I would > not like to call (2) and (3) arithmetic either, they merely extend the > equivalence relation to a new equivalence relation. > Consider Thabit ibn Kurrah's proof of Pythagoras' theorem, it is > purely geometric, no arithmetic involved. But it uses (4) (as > modified by me and in this way it implies the parallel axiom) and > (2). But it does *not* use (2) in any arithmetic sense. (And is, > in my opinion, the most elegant geometric proof that exists, and it > is already over 1000 years old.) And indeed the original formulation > of the theorem was not arithmetical but geometric: The sum of the > squares on the base sides of a triangle with right angle is equal to > the square on the hypothenusa of that triangle. This actually is an arithmetic formulation of the Pythagorean Theorem. A purely geometric formulation would have to look like: Let ABC be a right triangle, with a right angle at C. Then construct squares whose sides have lengths AC, BC, and AB. Then the squares with sides of AC and BC can be cut into a finite number of pieces, translated, and rotated, as to entirely cover, without overlap or overhanging, the square whose side is AB. There's no reference to sum, equals, or squaring a length here. > Thus, one can put Peano axioms into a same Table and call the Successor > postulate as a synthetic approach where it never mentions the distance > but the metric entry for Successor would state the Endless adding of > 1 And in line (7) Addition for the Peano axioms the synthetic approach > would say performed on Successors whereas the metric approach would > say performed on numbers > In that case you have *first* to define what endless adding of 1 means, > but not only that, but also what addition is. Addition has to satisfy certain properties as well. For instance, if 1 + 1 + 1 = 1, then you don't end up with the natural numbers. > Without such definitions > it makes no sense. So give me a clear definition of addition and > endless adding of 1, without that we do not have a notion about what > your metric approach to the Peano axioms mean. Try to define arithmetic > without the Peano axioms as basis, please. Once you have done that you > will find that implicitly you implied also the Peano axioms. After I added the associative and commutative properties of addition, I commented that AP's axioms are in fact equivalent to Peano's Axioms. Furthermore, the Infinite Integers do not satisfy either set of axioms. --- Christopher Heckman > So Dik is > wrong when he says that Axiomatics of Euclid or Peano are void of > arithmetic. Both Euclid and Peano have arithmetic inside their > axiomatics and depending on what approach you take whether you want to > discuss Successors or want to discuss Endless adding of 1 to 0 and 1, > is up to you. > I may also note that Euclid's arithmetic does not conform to the > Peano axioms because there is no successor function. And as I stated > before, and state again, the successor function is *not* necessarily > the addition of 1. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth >This actually is an arithmetic formulation of the Pythagorean Theorem. >A purely geometric formulation would have to look like: >Let ABC be a right triangle, with a right angle at C. Then construct >squares whose sides have lengths AC, BC, and AB. Then the squares with >sides of AC and BC can be cut into a finite number of pieces, >translated, and rotated, as to entirely cover, without overlap or >overhanging, the square whose side is AB. The fact that the union of the two small squares has the same area as the large square does not a priori require the existence of a finite decomposition as you describe. Indeed, the equivalence of these notions was one of Hilbert's problems, and was proved in the negative. (IIRC it was the first of Hilbert's questions to be resolved.) dave === Subject: Re: axiom systems in math is like Linnaeus classification in biology and never gets to the center of truth >>This actually is an arithmetic formulation of the Pythagorean Theorem. >>A purely geometric formulation would have to look like: >>Let ABC be a right triangle, with a right angle at C. Then construct >>squares whose sides have lengths AC, BC, and AB. Then the squares with >>sides of AC and BC can be cut into a finite number of pieces, >>translated, and rotated, as to entirely cover, without overlap or >>overhanging, the square whose side is AB. >The fact that the union of the two small squares has the same area as >the large square does not a priori require the existence of a finite >decomposition as you describe. Indeed, the equivalence of these notions >was one of Hilbert's problems, and was proved in the negative. (IIRC it >was the first of Hilbert's questions to be resolved.) No indeed: the equivalence of those notions (even without rotation), *for polygonal regions in the Euclidian plane*, is true, and was presumably one of Hilbert's motivations to ask his problem, which was about *polyhedral regions in Euclidian 3-space* (that problem having, as you note, a negative answer). Lee Rudolph === Subject: Trick question Grapsh and more Imagine that you are the CEO of a company and you have caught you spouse cheating on you with one of your employees v0. You are determined to fire v0, but you are consered with the influence of the move on other people. In other words, every pair of employees have a connection valuef(vi, vj) which is lost if one of them is fired and every employee has a certain value g(vi) for the company. You want to fire a group of people S which contains v0, minimizing the total loss to the company. Any ideas for an algorithm? === Subject: Re: Trick question Grapsh and more > Imagine that you are the CEO of a company and you have caught > you spouse cheating on you with one of your employees v0. You are > determined to fire v0, but you are consered with the influence of the > move on other people. In other words, every pair of employees have a > connection valuef(vi, vj) which is lost if one of them is fired and > every > employee has a certain value g(vi) for the company. You want to fire a > group of people S which contains v0, minimizing the total loss to the > company. > Any ideas for an algorithm? Well, if valuef(vi, vj) >= 0 for all vi, vj, then you would only fire v0. --- Christopher Heckman === Subject: asians Some folks allege that asians are superior at math. I'm not sure. Anyone know why it is that no theorems are credited to asians? === Subject: Re: asians 01/02/2006 at 05:21 PM, bob@coolgroups.com said: >Some folks allege that asians are superior at math. I'm not sure. >Anyone know why it is that no theorems are credited to asians? Credited by whom? Theorems are certainly credited to Asians by other Mathematicians. There are also Mathematical constructs named after Asians, e.g., Chern classes. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: asians >Some folks allege that asians are superior at math. I'm not sure. >Anyone know why it is that no theorems are credited to asians? In probability, there is the Ito integral, and theorems credited to Ito. In Lie groups, there is the Iwasawa representation of semi-simple groups. If one counts India here, there are lots of them in probability and statistics. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: asians > Anyone know why it is that no theorems are credited to asians? Euopreans called something the Chinese Remainder Theorem because they could not remember the name of that Chinese mathematician they got it from ... :) Did you know you can list the mathematicians' biographies at http://www-groups.dcs.st-and.ac.uk/~history/index.html by place of birth? Try it! -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: asians >> Anyone know why it is that no theorems are credited to asians? >Euopreans called something the Chinese Remainder Theorem because they >could not remember the name of that Chinese mathematician they got it >from ... :) can be solved in this way, and the huge generalization of this is called the Chinese Remainder Theorem. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: asians > Anyone know why it is that no theorems are credited to asians? >>Euopreans called something the Chinese Remainder Theorem because they >>could not remember the name of that Chinese mathematician they got it >>from ... :) >can be solved in this way, and the huge generalization >of this is called the Chinese Remainder Theorem. I think you're talking about Sun Zi. He does treat one problem (where the moduli are 3, 5 and 7), but gives a method that works for arbitrary remainders, and the generalization to arbitrary pairwise coprime moduli is fairly obvious. On the other hand, 800 years later Qin Jiushao shows how to handle cases where the moduli are not pairwise coprime. See and I admit that I haven't seen the actual texts, but my impression is that while they may have been written in a style of stating and solving particular problems rather than stating and proving general theorems, Qin Jiushao at least did know the general result for integers. Of course they didn't think of generalizing to other rings. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: asians <030120060619529338%edgar@math.ohio-state.edu.invalid> Well, the OP is just wrong. Lots of major theorems have been proved by Asians. I'm sure S. S. Chern proved a few. As did Ramanujan. In global terms, mathematics is generally limited to the very wealthy. If you have to struggle to ward off malnutrition, you're not going to be thinking about deep maths problems. I don't know how many Asians have this type of luxurious middle-class lifestyle. (When I say I don't know, I mean that in the most literal sense of not having that knowledge.) Paul Epstein === Subject: Re: asians <030120060619529338%edgar@math.ohio-state.edu.invalid> Ramanujan was not rich, and he did some starving in his life. Many prominent mathematicians, from Riemann to Einstein, struggled with money and were not rich. === Subject: Re: asians > Some folks allege that asians are superior at math. I'm not sure. > Anyone know why it is that no theorems are credited to asians? They are good at multiplying. Tam === Subject: Re: asians Mathematics does not differentiate based on race. They are superior in high school, and maybe college. But once you get to more advanced courses, where just studying like a maniac - or getting tutored all day which is what they do in high school - does not work, it has very little to do with race. Not to say that there are no asians good at math. Tanyama - Shimura conjecture is perhaps the most famous theorem at the time and conjectured by two asians. The validity of TS implies that of FLT. Keep this in mind. When you choose to become a mathematician it is not something you do for a living, or a job. It is a life you choose, and the better you stick to that lifestyle the better your chances to become a good mathematician. On a separate note you should read the biography of Tanyama and Shimura. One of the commited suicide. Gives you a glimpse into the life of a mathematician. === Subject: Re: PhD in Math via distance learning? >Is it possible to obtain a (legitimate) doctorate in *mathematics* by >distance learning. I have been asked this question by someone who has an >MS and is active in mathematical research but for personal reasons >cannot travel. Any information on this subject would be appreciated. >Doing an Internet search on: distance learning PhD mathematics gets a >number of hits, but I couldn't see anything of interest. Perhaps someone >here has some useful information on this subject. > It is possible but difficult. There is a knowledge > requirement, which at most universities does not have to > be met before something used as a thesis or part of it is > published (but at some there is this stupid restriction), > and there is a requirement of a thesis. > The knowledge requirement can be a written or an oral exam, > or parts of this can be waived if there is enough evidence > from other grounds. Courses are rarely considered adequate > evidence, and from personal experience as a faculty member, > I can tell you that this is highly justified. > The defense of thesis can be in many ways, some of which > require the student to defend it in person, and others which > have internal and possibly external readers. > Any of the requirements of presence can be handled by > conference calls or video conferences. So I see no reason > why this should not be possible. Most jobs for Ph.D. mathematicians involve teaching. And most in-persion Ph.D. programs in mathematics therefore provide teaching experience. Maybe in the future some positions will involve distance-teaching only, but probably there are few (in any) such positions at the present time. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: PhD in Math via distance learning? > Is it possible to obtain a (legitimate) doctorate in *mathematics* by > distance learning. I have been asked this question by someone who has an > MS and is active in mathematical research but for personal reasons > cannot travel. Any information on this subject would be appreciated. > Doing an Internet search on: distance learning PhD mathematics gets a > number of hits, but I couldn't see anything of interest. Perhaps someone > here has some useful information on this subject. The Open University in the UK specialises in distance learning. A quick look on the OU site soon bought up some semi-promising links. Pure maths PhDs. http://puremaths.open.ac.uk/pmd_research.html In the UK at least, OU degrees are highly regarded - more so than that of the more typical polytechnics that have turned into universities. I doubt it is totally impossible to avoid travel, and in any case I personaly think presenting work at international conferences is all part of the PhD process. But with the OU the travel would be quite limited, although it might well have to be international travel if your friend is outside the UK. -- Dave K http://www.southminster-branch-line.org.uk/ Please note my email address changes periodically to avoid spam. It is always of the form: month-year@domain. Hitting reply will work for a couple of months only. Later set it manually. The month is always written in 3 letters (e.g. Jan, not January etc) === Subject: Re: PhD in Math via distance learning? <43b9d0ce@212.67.96.135 Is it possible to obtain a (legitimate) doctorate in *mathematics* by > distance learning. I have been asked this question by someone who has an > MS and is active in mathematical research but for personal reasons > cannot travel. Any information on this subject would be appreciated. > Doing an Internet search on: distance learning PhD mathematics gets a > number of hits, but I couldn't see anything of interest. Perhaps someone > here has some useful information on this subject. > The Open University in the UK specialises in distance learning. A quick > look on the OU site soon bought up some semi-promising links. > Pure maths PhDs. > http://puremaths.open.ac.uk/pmd_research.html > In the UK at least, OU degrees are highly regarded - more so than that > of the more typical polytechnics that have turned into universities. > I doubt it is totally impossible to avoid travel, and in any case I > personaly think presenting work at international conferences is all part > of the PhD process. But with the OU the travel would be quite limited, > although it might well have to be international travel if your friend is > outside the UK. > -- > Dave K It would be pertinent to see if Open University enrollment is permitted from the querent's country. David Ames === Subject: Re: PhD in Math via distance learning? >>The Open University in the UK specialises in distance learning. A quick >>look on the OU site soon bought up some semi-promising links. >>Pure maths PhDs. >>http://puremaths.open.ac.uk/pmd_research.html >>In the UK at least, OU degrees are highly regarded - more so than that >>of the more typical polytechnics that have turned into universities. >>I doubt it is totally impossible to avoid travel, and in any case I >>personaly think presenting work at international conferences is all part >>of the PhD process. But with the OU the travel would be quite limited, >>although it might well have to be international travel if your friend is >>outside the UK. >>-- >>Dave K > It would be pertinent to see if Open University enrollment is permitted > from the querent's country. > David Ames Very true. Generally (and I have no experience of the OU), UK universites are very keen to take foreign students, as they charge them a lot more than those from Europe. They are taken in preference to UK students in my experience, although nobody will admit that. As you rightly say, a student would need to contact the OU to discuss the special circumstances. The OU is the only university I know of (apart from the bogus ones that offer PhDs for $100), where the style of teaching is virtually all distance learning, so my guess is it would be a good place to investigate. -- Dave K http://www.southminster-branch-line.org.uk/ Please note my email address changes periodically to avoid spam. It is always of the form: month-year@domain. Hitting reply will work for a couple of months only. Later set it manually. The month is always written in 3 letters (e.g. Jan, not January etc) === Subject: Re: Thumber Theory Question ... >> Statement: >> Absence of non-zero integer solutions of (1) implies the absence of >> non-zero integer solutions of (2). >> >> x^(2k) + y^(2k) = z^(2k) (1) >> >> u^(2k) + v^(2k) = 2w^(2k) (2) > Here are some additional clarifications: > Odd k > 3 and x,y,z,u,v,w are all different from each other I have no idea how you can get from (1) to (2). Consider the following: p^2 + q^2 = 2.r^2. How many ways are there to write 2.r^2 as the sum of two squares? Let us first consider how many ways there are to write an arbitrary number n as the sum of two squares. Write n as n = 2^a_0 * p_1^(2.a_1) *...* p_m^(2.a_m) * q_1^b_1 *...* q_l^b_l where the p_i's are primes of the form 4t+3 and the q_i's are primes of the form 4t+1. Now define: B = (b_1 + 1)(b_2 + 1)...(b_l + 1) the number of ways is: 0 if any a_i is half an integer B/2 if all a_i are integer and B is even (B-(-1)^a_0))/2 if all a_i are integer and B is odd. The case u^(2k) + v^(2k) = 2w^(2k) is equivalent to (u^k)^2 + (v^k)^2 = 2.(w^k)^2, and so in n, a_0 is odd, all a_i are integer and all b_i are even, and so B is odd. This means we are in the third case with a_0 odd, so the number of ways is: (B + 1)/2. We need B > 1 (otherwise we have only the trivial solution). I have to think about it from here. But I still do not see a connection with (1). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Igor Severyanin You folks should be posting your poetry on http://www.poets.com/mystuff/portfolio.asp?S={59A83E7A-F152-447A-A457-352798 E4AAF7} That is where I put mine. You will get critiques for it from other amateur and some published poets on the site. Nice way to get feedback. ;) >Translations from Igor Severyanin > Envy not your friend if he's more handsome, > More intelligent or wealthier than you. > Let his merits and let his successes > Not tear up the laces on your shoe. > Move along your way without a care, > Smile still broader out of his success! > Maybe he'll face darkness and despair > And your porch will be adorned with bliss! > Laugh with him, and cry with his distresses: > Feel him with your heart, and for all time! > Do not block your friend from his successes: > Its a sin to do so! Truly, its a crime! > It took place at the sea, by the foam of the ocean, > Where the carriage of city rarely arrives. > In the tower of a palace the queen was playing Chopin, > And to sound of Chopin the page fell in love. > It was all very simple, it was all very dear: > The queen asked him to cut pomegranate in half, > And she gave him a half, and the page she did tire, > And to sound of sonatas the queen fell in love. > And she later submitted, submitted with thunder, > Like a slavegirl the queen slept all night till the day. > It took place at the sea, where the turquoise waves wander, > Where the page's sonatas and azure foam play. > This day of spring is hot and golden - > The city's blinded by the sun! > Im me again! I am emboldened! > Im in love, happy and Im young! > The soul sings and bursts for the fields and > I come to strangers and say hey. > What spaciousness I feel! What freedom! > What songs and flowers in my way! > Soon - vanish into the young meadows! > Soon - into snowdunes, full of bliss! > To look in pink faces of women, > Like friend, an enemy to kiss! > Make noise, the springtime forests mighty! > Bloom, lilac bushes! Grow tall, grass! > No sinners: Everyone is righteous > On a day so divinely blessed! > Champagne in a lily! Champagne in a lily! > With health and with wisdom it sparkles and shines! > A shot of Mignon with one of Escamillio > Champagne in a lily - a sacred wine. > Champagne in a lily bursting and sparkling - > The wine contained in a flowers cup. > I glory in rapture the Christ and the Antichrist > With soul deified in delight of a gulp! > A hawk and a mourning dove! Reichstag and Bastille - > The sleep and the wakefulness! Demon and Lord! > Lily in champagne and champagne in a lily - > The lighthouse of oneness in sea of discord! > You are ready from gloom to suicide, > Hang yourself, or shoot in the mouth. > Wait a while - and the spring will come to your side > After just three more snowy months. > Nightingales of the cherry will whistle, > Full of nightingales cherry will stand. > May go past you the shot from the pistol > And the rope fall apart in your hands. > With the fishing rods made of redwood > People will catch the fish on the hook, > And the swan with white breast and white feathers > Will swim lightly upon the lake. > Mounds will breathe with dampness and drown, > Will send redolence and be green, > And your neck, as it gives a way down, > Will become pouring with rain. > And the bushes under flooding river > Into lilac and cherry will bloom. > Noisy, singing, the spring will deliver > All your girlfriends and also - you. > And will love, and will bloom, and will spring again > All that dimmed in the winter from gloom. > All the dry will be cut by axe-wielding hand > And the juicy will bravely bloom. > Do not kill yourself, do hang your head, > Rather let your fantasy play. > We will live through these months however we can, > And soon afterwards - there is May! >Translated by Ilya Shambat. . Blessings be, Synn . By amending our mistakes, we get wisdom. By defending our faults, we betray an unsound mind. -The Sutra of Hui Neng === Subject: Re: correcting the Wikipedia page on Euclid's Infinitude of Primes proof <6r8e83-s9.ln1@sirius.tg00suus7038.net> <85u0coqj7w.fsf@lola.goethe.zz > The 5th axiom, the induction axiom, provides for infinity: > Wrong. > again via Mathworld: > if a set of numbers contains 0 and every successor of every number > in S, then every number is in S. > Are you saying the 5th axiom was 'injected' by Peano to ensure it > delivers an infinite set? > Axioms 1-4 guarantee infiniteness already. The fifth _limits_ the > scope of S by _excluding_ every number that is not reachable via > succession from 0. > And also, in that case, why doesn't the induction axiom > provide infinity? > The other axioms _already_ establish that 0 is in N, that the > successors of any elements of N are in N, that successors are unique > and 0 is not the successor of any number. This already establishes > infiniteness in the chain of succession starting at 0. > Yes, good correction. I noticed axiom 2 briefly, then said, 'Nah', > and plowed ahead. > When I first mentioned axiom #5 in this thread (or wherever), I said > that it was equivalent to the successor function being almost onto > (every number y except 0 has an element x such that > S(x) = y). After I posted, I realized that it said more than that. > I guess we're all learning about these axioms ... More than AP, in > fact! > The fifth only rules out _additional_ elements. > Interesting NEW point. > Actually it isn't new; someone talked about threads which are > possible if axiom #5 is removed. You could have a situation like: > N = {X, Y, Z, 0, 1, 2, 3, 4, ...} with > S(X) = Y, S(Y) = Z, S(Z) = X, > S(0) = 1, S(1) = 2, S(2) = 3, ... > which obeys axioms #1 - #4. The set of nonnegative real numbers is > another example. > --- Christopher Heckman I could have sworn I posted a reply to this in which I admitted forgetting about your prior, and quite detailed, if I remember, discussion of this very property of induction. If not, then this is it! My apologies for the belated post. To you Chris and/or Dik - I think he was also in on that. And somebody else? My memory is apparently pretty faulty. Ken === Subject: Re: Do these distributions have names? > That same old urn contains w white balls and b black balls. > Sequentially draw balls at random with replacement. Let X_r be the > number of black balls drawn before the rth white ball. Then X1 has > geometric distribution with parameter w/(w+b) and X_r has negative > binomial distribution with parameters r and w/(w+b). (Some authors > say 1+ X1 and r + X_r have respective geometric and negative > binomial distribution.) > Now draw balls *without* replacement. Let Y_r be the number of black > balls drawn before the rth white ball. Do the distributions of Y1 > and Y_r have established names? Hypergeometric? Try Googling on hypergeometric balls urns. Most of the 1st 10 hits look pretty standard. -- DW === Subject: Re: Do these distributions have names? Sequentially draw balls at random with replacement. Let X_r be the > number of black balls drawn before the rth white ball. Then X1 has > geometric distribution with parameter w/(w+b) and X_r has negative > binomial distribution with parameters r and w/(w+b). (Some authors > say 1+ X1 and r + X_r have respective geometric and negative > binomial distribution.) > Now draw balls *without* replacement. Let Y_r be the number of black > balls drawn before the rth white ball. Do the distributions of Y1 > and Y_r have established names? > Hypergeometric? > Try Googling on hypergeometric balls urns. > Most of the 1st 10 hits look pretty standard. No, the hypergeometric distribution applies to the number of white balls in the first n balls drawn without replacement. (Thus, the hypergoemetric corresponds to the binomial in the case of replacement.) I had looked around hypergeometric, which is clearly related. -SJH === Subject: polygon scaling problem Hi All, I'm working on a rescaling algorithm to make a shape auto-fit to the approximate screen size. I am applying a rescaling factor to all of my coordinates. This works great, and the shape resizes as expected. The problem is that the lengths of the sides of my polygon correspond to real-world dimensions (1 pixel = 1 inch) - when I rescale the polygon to 150% need the dimensions to still reflect the original length (not increased by 150%) I thought I could just divide the length by the scaling factor to get the original length. This works once, but then since the coordinates are modified on the first rescale, the subsequent attempts to get at the original length give me the length from the last rescale, not the original. I tried storing the value of the last x1 prior to the rescale so that I could then compare the post rescale x1 and ultimately keep track of my original points throughout the rescale operations, but this doesn't seem to work. This seems like a relatively trivial problem but I can't quite get it right. If anyone has any tips they would be greatly appreciated. Graham === Subject: Re: polygon scaling problem On 2 Jan 2006 19:41:57 -0800, Graham Street >Hi All, >I'm working on a rescaling algorithm to make a shape auto-fit to the >approximate screen size. I am applying a rescaling factor to all of >my coordinates. This works great, and the shape resizes as expected. >The problem is that the lengths of the sides of my polygon correspond >to real-world dimensions (1 pixel = 1 inch) - when I rescale the >polygon to 150% need the dimensions to still reflect the original >length (not increased by 150%) >I thought I could just divide the length by the scaling factor to get >the original length. This works once, but then since the coordinates >are modified on the first rescale, the subsequent attempts to get at >the original length give me the length from the last rescale, not the >original. >I tried storing the value of the last x1 prior to the rescale so that I >could then compare the post rescale x1 and ultimately keep track of my >original points throughout the rescale operations, but this doesn't >seem to work. >This seems like a relatively trivial problem but I can't quite get it >right. If anyone has any tips they would be greatly appreciated. Why not just store the original lengths somewhere? === Subject: Re: polygon scaling problem the problem there is that the shape can be resized by the user by dragging the edges.. so in that case the lengths have to adjust to be a new dimension (dist from point a to b / scaling factor) === Subject: Puthoff's theory compared with mine Searching for the Universal Matrix in Metaphysics Research News and Opportunities in Science and Theology 2, No. 8, p. 22 Templeton Foundation Press, April 2002 H. E. Puthoff, Ph.D.Institute for Advanced Studies at Austin Austin, Texas 78759 Throughout mankind's cultural history there has existed the metaphysical concept that man and cosmos are interconnected by a ubiquitous, all-pervasive sea of energy that undergirds, and is manifest in, all phenomena. This pre-scientific concept of a cosmic energy goes by many names in many traditions, such as ch'i, ki or qi (Taoism), prana (yoga), mana (Kahuna), barakah (Sufi), .8elan vital (Bergsonian metaphysics), and so forth. Sure why not? No objection. Complementary to the above metaphysical concept, contemporary physics similarly posits an all-pervasive energetic field called quantum vacuum energy, or zero-point energy, a random, ambient fluctuating energy that exists even in so-called empty space. Yes, but Hal & Co do not understand that the several kinds of Higgs fields that burst into being and becoming at decreasing energy scales as the universe expands is the cohering of this random energy. Indeed, the curvature field comes not from the random noise but from the phase of the coherent Higgs signal from the first Planck time ~ 5.4 10^-44 sec. The curvature tetrad field as an invariant Cartan 1-form is B(curved)= (Planck Length)[(dVacuum Phase1)(Vacuum Phase2) - (Vacuum Phase1)(dVacuum Phase2)] Einstein's basic local frame invariant formula is ds^2(curved) = (1 + B)(flat)(1 + B) = 1(flat)1 + B(flat)1 + 1(flat)B + B(flat)B When B = 0 we have special relativity without any real gravity and also no masses! B = 0 is the pre-inflation false vacuum that bursts into Level 2 bubbles of the Megaverse (Multiverse). Each bubble has an infinity of Level 1 universes like the one we happen to be stuck in like Flatlanders. This is consistent with Max Tegmark's theory for example as well as Lenny Susskind's Cosmic Landscape. The area flux density is the 2-form dA = 2(Planck Area)(dVacuum Phase1)/(dVacuum Phase2) The integral of dA over closed surfaces surrounding point defects in the Higgs field is quantized as integers x (Planck Area). The integers are called wrapping numbers, i.e. 2D winding numbers. This is essentially the Bekenstein black hole thermodynamics with quantization of areas derived from the second homotopy group being the integers. Note that d^2A = 0 but we define a NONLOCAL Bohm-Aharonov flux without flux as A = 3D volume integral of d^2A = closed surface integral of dA = IntegerxPlanck Area Provided there is at least one point singularity inside so that the closed surface is not a boundary but is a star gate portal to a parallel universe of the multiverse. This explains the NONLOCALITY OF GRAVITY ENERGY and why the Yilmaz theory is wrong. One can similarly have a local zero gravity energy density and still get a non-zero Poynting flux of gravity waves coming out of the closed non-bounding surface. That is, in the presence of interior Goldstone phase singularities of the Higgs coherent vacuum field, the integral of a local zero density integrand need not vanish! (The adjective zero-point means that such energy or activity exists even at a temperature of absolute zero where no thermal agitation effects remain.) Thus, even in the absence of matter, in the modern view annihilation processes, as well as so-called zero-point fluctuations of such fields as the electromagnetic field. Originally thought to be of significance only for such esoteric concerns as small corrections in atomic emission processes (e.g., the Lamb shift), it is now understood that vacuum fluctuation effects play a central role in large-scale phenomena of interest to technologists as well, such as the enhancement or inhibition of the spontaneous emission of light in atomic processes, the generation of short-range attractive forces between closely-spaced materials, and the possibility of extracting useful energy from vacuum fluctuations, the Holy Grail of energy research. Should we further consider the possibility that such random vacuum energy might be subject to influence by consciousness or intention, then, given that it is well understood by physicists that a restructuring or cohering of vacuum energy would have physical consequences for matter, animate or inanimate, such could provide a rational basis for healing or other processes that are part and parcel of the pre-scientific view. Fine, but where is Hal's math for this? He does not have any. There is nothing in his PV theory that has this effect. There is nothing in Haisch's theory either. But there is in mine. Very simple it is the phase-lock to the vacuum phases that generate curved space-time! B' ~ d(Vacuum Phase - Control Phase) Control Phase may be from the mental field of a brain. For example Geller effect. Hal has nothing like that in his formalism. I agree with Hal's intuition. I have been saying the same thing for a long time. Hal did not realize that the Higgs field is the clue. Nowhere does he mention the Higgs field in the context of cohering the random vacuum fluctuations into a smooth coherent field that gives gravity first, and later, another coherent field splits off to give inertia. In such fashion the similarities, differences and possible synthesis of the pre-scientific and modern concepts of an all-pervasive energy field can be considered. As a physicist specializing in fundamental quantum physics and yet interested in these issues, I have an abiding interest in pushing the envelope with regard to the present scientific paradigm. This includes the issue as to whether what we know of the life process itself can find rapprochement with modern quantum physics, or whether and how it needs to be extended. Given my own earlier decade-plus background as Director of the Cognitive Sciences program at SRI Intern'l in the '70's and early '80's, investigating remote viewing and other so-called paranormal phenomena, the life-science data I have to integrate all by themselves push the envelope (Proc. IEEE 64, 329 (1976); Jour. Sci. Exploration 10, 63 (1996). OK Unfortunately, as it now stands, mainstream physics reductionism is leading to an evermore complex picture of nature involving a forces, the implications of incorporating additional dimensions as in superstring theory, and so forth. Thus, in spite of efforts to develop a grand unified theory to simplify our picture of nature, the actual day-to-day work on this effort is complexifying faster than the hoped-for simplification. Therefore, not only are we missing holism on the grand scale, but a gratifying holism just for the physical sciences alone appears to be a rapidly accelerating goal post. Contemplation of such provocative issues in both the physical and life sciences led me into investigating an area of physics concerned with what is known as quantum vacuum fluctuations or zero-point energy, a universal background energy pervading all of space and associated with fluctuations of underlying space itself. Specifically, I began to consider the underlying quantum fluctuations as a fundamental stuff out of which a greater synthesis could be built. I hasten to add that I do not mean for such an approach to be simply reductionism on a grander scale, with no room for nonphysical factors to play a role. Rather, to the degree that energy is involved not only in physical but in nominally non- or para-physical phenomena (including, perhaps, such mundane phenomena as thought, charisma, etc., let alone psychokinesis), then such energy patterns might in principle emerge as a result of cohering or patterning the otherwise random, ambient zero-point energy. For me this hypothesis emerged when I considered how uneconomical Nature would have to be to posit, on the one hand, an all-pervading energetic field of ki or chi, as in the metaphysics of the martial arts and acupuncture, and, on the other hand, also posit an all-pervasive energetic field of quantum zero-point energy. It appeared to me to be more likely that we were dealing with a single underlying substructure which goes by various names in various cosmologies, depending on whether it is in its pre-manifest random form, or patterned at various hierarchical levels, including the purely material. In my professional area, I began with the pure physics side. In my first published study on the significance of zero-point energy for broad issues I showed that the basic stable states of matter are not merely inert, static structures, but rather depend on the presence of the underlying, sustaining zero-point energy which is continually being absorbed and re-emitted on a dynamic-balance basis. Pull the plug on the zero-point energy and all atomic structure would collapse (Physical Review D 35, p. 3266, 1987). In my second study I developed the idea, originally put forward as a hypothesis by the famous Russian physicist and human-rights advocate, Sakharov, that the gravity of Einstein's equations is not a separate, fundamental force, but rather is a pressure force derived from partial shielding of the ambient zero-point energy (Phys. Rev. A 39, p. 2333, 1989; 47, 3454, 1993). (A corollary to this view is that if one could cohere the otherwise random fluctuations of space, a host of interesting phenomena would follow.) What's wrong with that is that it is only dimensional analysis in the end with a lot of handwaving. It's simply imposing a cut off Lp^2 = hG/c^3 dressed up in fancy language of renormalization and regularization. My above formula is much better than what Sakharov suggested. He did not formulate metric elasticity precisely enough. In my third study (Phys. Rev. A 40, p. 4857, 1989; 44, pp. 3382 & 3385, 1991) I showed that on the cosmological scale a grand hand-in-glove dynamic equilibrium exists between the ever-agitated motion of matter on the quantum level and the surrounding zero-point energy field. One consequence of this is that we are literally, physically, in touch with the rest of the cosmos as we share with remote parts of the universe fluctuating zero-point-energy fields of even cosmological dimensions. Who is to say whether, for example, modulation of such fields might not carry meaningful information as in the popular concept of the Force? In a fourth study with colleagues from Lockheed, CIPA and Cal. State at Long Beach (Phys. Rev. A 49, 678,1994; The Sciences 34, 26, Nov/Dec 1994; Science 263, 612, 1994; Spec. in Sci. and Tech. 20, 99, 1997), we have shown that the simple property of inertia possessed by all bodies is simply resistance to being accelerated through the zero-point fluctuations, an extremely fundamental result in physics that provides an underpinning for Newton's Law of inertia. This last paragraph is wrong for reasons I gave earlier. In a fifth study (Spec. in Sci. and Tech. 13, 247, 1990; Phys. Rev. E 48, 1562, 1993) I examined the evidence that not only is zero-point energy at the base of a number of fundamental physical phenomena, but that in principle non-polluting energy can be extracted from the fluctuations so as to constitute a new energy source; a concept for which my research group has attracted seed funding, obtained encouraging laboratory evidence, applied for and obtained patents worldwide, and which is the focus of a present in-house program. Finally, in a sixth study (Phys. Essays 9, 156, 1996; Ad Astra 9, 42, 1997; Jour. Sci. Exploration 12, 295, 1998) I indicate how manipulation of the underlying zero-point-energy spacetime metric opens up the possibility for efficient interstellar propulsion, a concept well-received both in popular writings (e.g., Arthur C. Clarke) and by the mainstream (Air Force, NASA laboratories). Show us how? Where is the math? Where is a simple model? All of this characterizes the underlying, ambient, random quantum zero-point-energy sea as a blank matrix upon which coherent patterns can be written, such information constituting at the bottom end of the scale chauvinist like myself, an ascending ladder of possible other information structures, whether it be coherent electromagnetic field structures around living organisms, possibly non-biochemical components of memory, or other more esoteric aspects of Nature. If my goal for this research comes to full fruition, what would emerge would be an increased understanding that all of us are immersed, both as living and physical beings, in an overall interpenetrating and interdependent field in ecological balance with the cosmos as a whole, and that even the boundary lines between the physical and metaphysical would dissolve into a unitary viewpoint of the universe as a fluid, changing, energetic/information cosmological unity. Fine, but where is the math. Give one toy example of a coherent pattern. My model of that is B(curved)= (Planck Length)[(dVacuum Phase1)(Vacuum Phase2) - (Vacuum Phase1)(dVacuum Phase2)] Einstein's basic local frame invariant formula is ds^2(curved) = (1 + B)(flat)(1 + B) = 1(flat)1 + B(flat)1 + 1(flat)B + B(flat)B When B = 0 we have special relativity without any real gravity and also no masses! Where /zpf = (Effective Control Area)^-1cos(Vacuum Phase - Control Phase) Guv + /zpfguv = 0 is the basic metric engineering field equation. === Subject: Re: Puthoff's theory compared with mine Where's your website? PS: Tossy Google won't let me post to these groups- sci.space, sci.military === Subject: Re: Why are some constants not recognized where others of less importance are? > BTW, I don't think that the nth decimal place of X should be a > sequence in the OEIS, one reason being that the sequence is > base-dependent. > OEIS allows base-dependent sequences. I know. I consider that cheating, though. --- Christopher Heckman === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony Calculus to its rock bottom basic and raw fundamental is a interplay > between two numbers of 0 and infinity. Infinity is a process but zero > is certainly a number. They are inverses of one another [...] > They are NOT inverses of each other. If you find the formula 1/infinity > = 0 in your textbook (or even worse, 1/0 = infinity), you should burn > it immediately. > For instance, there is not one but two infinities associated with the > real numbers in Calculus: -infinity and +infinity (which is my > symmetric notation for infinity). > You only get a formula LIKE THIS when the idea of limits is introduced. > The limit of 1/x as x approaches +infinity is 0. > The limit of 1/x as x approaches -infinity is 0. > The limit of 1/x as x approaches 0 WHILE REMAINING POSITIVE is +infinity. > The limit of 1/x as x approaches 0 WHILE REMAINING NEGATIVE is -infinity. > The limit of 1/x as x approaches 0 DOES NOT EXIST. [...] Even worse is the fact that, if 1/infinity = 0, that means 0*infinity = 1. This is a problem because something which is approaching 0, multiplied by something getting big, MIGHT approach 1, but it might not. For instance, If f(x) = 1/x and g(x) = x, then as x approaches infinity, f(x) approaches 0 and g(x) approaches infinity, and f(x)*g(x) approaches 1. But if f(x) = 1/x^2 and g(x) = x, then f(x) still approaches 0 and g(x) still approaches infinity, but f(x)*g(x) approaches 0. This would imply that 0*infinity = 0. If f(x) = 2/x and g(x) = x, then f(x)*g(x) approaches 2, which means 0*infinity = 2. If f(x) = 1/x and g(x) = x^2, then f(x)*g(x) approaches infinity, which means 0*infinity is infinity. So 0*infinity does not have one specific value; it's what's called an indeterminate form. (This information can be found in the section on L'Hospital's Rule.) Another argument against 1/infinity being 0. --- Christopher Heckman === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony The reason calculus works is because 0 and infinity are inverses of one > another that as you shrink the tangent line to 0 > Shrinking is a process, like limits are a process. > In what sense limits are a process? > If they are computational process what is the algorithm? > ... the limit of 1/x, as x approaches infinity, is 0. > What means a real number to approach +infinity? A number is close to infinity if it is big in absolute value and positive. A variable approaches +infinity if it grows without bound. The best way to make sense of close to infinity is to draw the interval [-1,1], and to label the point 0 with the number 0, label the point 1/2 with the number 1, label the point 3/4 with the number 2, label 7/8 with 3, etc. (Label 1 - 1/2^n with the number n.) Now label 1 with +infinity (most authors actually just use infinity). Negative numbers in the interval [-1,1] get labeled with -1, -2, -3, ..., and -infinity, symmetric from the positive ones. Now two extended real numbers are close if they are close on this diagram. > To become closer and closer to +infinity? > If so, how close is 5 to +infinity? Literally, they are infinitely far away from each other. That's why in my classes, I talk about variables approaching infinity (i.e., getting bigger) instead of being close to infinity. --- Christopher Heckman === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony ... the limit of 1/x, as x approaches infinity, is 0. > What means a real number to approach +infinity? > A number is close to infinity if it is big in absolute value and > positive. A variable approaches +infinity if it grows without bound. > The best way to make sense of close to infinity is to draw the > interval [-1,1], and to label the point 0 with the number 0, label the > point 1/2 with the number 1, label the point 3/4 with the number 2, > label 7/8 with 3, etc. (Label 1 - 1/2^n with the number n.) Now label 1 > with +infinity (most authors actually just use infinity). Negative > numbers in the interval [-1,1] get labeled with -1, -2, -3, ..., and > -infinity, symmetric from the positive ones. Labels -2, -3, ... are outside the interval [-1;1], because 1 - 1/2^(-1) = -1. What is the label of point 2? > Now two extended real numbers are close if they are close on this > diagram. > To become closer and closer to +infinity? > If so, how close is 5 to +infinity? > Literally, they are infinitely far away from each other. That doesn't make sense in the real number system, because +/-infinity are not members of R. Jeffrey === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony > In sci.logic, ken.quirici@excite.com >> on 1 Jan 2006 11:22:53 -0800 >> [...] >> I too was mystified - but Wikipedia mentions precisely these >> expressions as adics - I think 10-adics. They even show how >> ...99999 = -1 because ....99999 + 1 = 0 (by standard >> addition algorithm, proceding place-by-place leftward >> carrying 1 infinitely many places left). >> http://en.wikipedia.org/wiki/Adic > No; that's something different. AP shuns p-adics entirely now. > I once asked him what the difference between an Infinite Integer and a > 10-adic was, and I never got an answer. >> I'll admit I'm not all that familiar with Hensel. >> Still, >> http://mathworld.wolfram.com/p-adicNumber.html >> does hint at a 1-1 correspondence between the p-adics >> and the form >> .....54321.0 >> or, for that matter, the form >> 0.12345... . >> So color me slightly puzzled. If there is indeed a 1-1 correspondence >> between the p-adics and the reals, then the p-adics truly are >> uncountable (as are the reals). However, I'm not so sure of >> the conclusion that >> ...999. = -1 >> in the 10-adic realm; [...] > Think of addition of 10-adics like the addition of base-10 integers, > with carries, and sums. The only difference is that the 10-adics have > an (countably) infinite number of places. > ...99999 > +1 > becomes > ...1111 > ...99999 > +1 > ...00000 > (view with a fixed-width font). > I'm not sure why anyone would want such beasties, in that case, > unless one wants to set up the ordering > -...123 < -123 < 0 < +123 < +...123 > for some reason. > What, precisely, are the rules for arithmetic for this notation? > For that matter, what is a formal definition of > this notation? An integer, for instance, can be > defined using Peano's, or can simply be defined as the > set of all nonempty finite strings from the alphabet > {0,1,2,3,4,5,6,7,8,9}, with the requirement that the first > digit must not be a zero unless that's the only digit. > (The alphabet can be modified if one wants to use different > bases; for example, hexadecimal tacks on 6 more letters > as digits.) The n-adics can be thought of in two ways, but they boil down to being built on countably infinite sequences of elements of {0, 1, ..., n-1}, which are really functions from the natural numbers to {0, 1, ..., n-1}. So an n-adic A can be thought of as consisting of the digits a(0), a(1), a(2), ..., which are usually arranged as ...a(3)a(2)a(1)a(0). Informally, this is a shortcut for a(0) + n a(1) + n^2 a(2) + n^3 a(3) + ... , whether the series actually converges or not. Formally, the n-adic A is a number that has the properties A == a(0) (mod n), A == a(0) + n a(1) (mod n^2), A == a(0) + n a(1) + n^2 a(2) (mod n^3), and so on. In either case, the n-adic ...000123 ends up being the same as the integer 123 (written in base n). Addition and multiplication of n-adics are done the same way as addition and multiplication of base n numerals, except you may have an infinite number of places to compute. The distance between two n-adics is defined in a strange way; basically A and B are close if a(i) = b(i) for all i from 1 up to some big number N, and the larger N is, the closer A and B are. --- Christopher Heckman === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony unless one wants to set up the ordering > -...123 < -123 < 0 < +123 < +...123 > for some reason. I dislike that notation. Here is a brief primer on the p-adics: http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant material begins on page 15. Basically, the idea is to consider a rational number x (in lowest terms) and figure out how many times a prime p divides the numerator. Call that number the order(p,x). Define a norm by |x|_p = p ^ (-order(p,x)) if x =/= 0, oo otherwise. The related metric is non-Archimedean metric on Q, and you can find a completion for Q under this metric. The complete space is called Q_p. Indeed, many results from classical real analysis carry over (as long as they don't require the the metric to be Archimedean) and by considering properties of the metric vis-a-vis the objects they metrize, you can analyze algebraic properties. > What, precisely, are the rules for arithmetic for this notation? I wouldn't want to figure it out. Using the traditional notation for dealing with rationals is straightforward. 'cid 'ooh === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony In sci.logic, poopdeville@gmail.com on 2 Jan 2006 16:28:35 -0800 >> I'm not sure why anyone would want such beasties, in that case, >> unless one wants to set up the ordering >> -...123 < -123 < 0 < +123 < +...123 >> for some reason. > I dislike that notation. Here is a brief primer on the p-adics: > http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant > material begins on page 15. These aren't p-adics. (Unless 'p' stands for Plutonium. :-) ) I'm not sure what they are, really. Probably meaningless crap at this point. > Basically, the idea is to consider a rational number x (in lowest > terms) and figure out how many times a prime p divides the numerator. > Call that number the order(p,x). Define a norm by |x|_p = p ^ > (-order(p,x)) if x =/= 0, oo otherwise. The related metric is > non-Archimedean metric on Q, and you can find a completion for Q under > this metric. The complete space is called Q_p. Yes, I'm vaguely aware of the metric. I'll admit I don't see what it does apart from measuring elements of Q in a novel fashion. But then, I'm not familiar with all this. > Indeed, many results from classical real analysis carry over (as long > as they don't require the the metric to be Archimedean) and by > considering properties of the metric vis-a-vis the objects they > metrize, you can analyze algebraic properties. >> What, precisely, are the rules for arithmetic for this notation? > I wouldn't want to figure it out. Using the traditional notation for > dealing with rationals is straightforward. Aye. > 'cid 'ooh -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony unless one wants to set up the ordering > -...123 < -123 < 0 < +123 < +...123 > for some reason. > I dislike that notation. Here is a brief primer on the p-adics: > http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant > material begins on page 15. > Basically, the idea is to consider a rational number x (in lowest > terms) and figure out how many times a prime p divides the numerator. > Call that number the order(p,x). Define a norm by |x|_p = p ^ > (-order(p,x)) if x =/= 0, oo otherwise. The related metric is > non-Archimedean metric on Q, and you can find a completion for Q under > this metric. The complete space is called Q_p. It seems to be more than notation. Is there any way of representing ....9999999 as a product of primes*? Also, don't you allow denominator prime factors, whose metric would be > 1 (unlike numerator prime factors)? I vaguely remember Mathworld's p-adic treatment mentioning denominator factors as also 'amenable' to the metric you describe. Ken *actually now that I think of it these beasties are so weird there may very well be prime factors of 'representations' like....8973, under some definition of multiplication - altho Wikipedia I think applies standard long-multiplication to 'em as 'multiplication'. > Indeed, many results from classical real analysis carry over (as long > as they don't require the the metric to be Archimedean) and by > considering properties of the metric vis-a-vis the objects they > metrize, you can analyze algebraic properties. > What, precisely, are the rules for arithmetic for this notation? > I wouldn't want to figure it out. Using the traditional notation for > dealing with rationals is straightforward. > 'cid 'ooh === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony I'm not sure why anyone would want such beasties, in that case, > unless one wants to set up the ordering > -...123 < -123 < 0 < +123 < +...123 > for some reason. > I dislike that notation. Here is a brief primer on the p-adics: > http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant > material begins on page 15. > Basically, the idea is to consider a rational number x (in lowest > terms) and figure out how many times a prime p divides the numerator. > Call that number the order(p,x). Define a norm by |x|_p = p ^ > (-order(p,x)) if x =/= 0, oo otherwise. The related metric is > non-Archimedean metric on Q, and you can find a completion for Q under > this metric. The complete space is called Q_p. > It seems to be more than notation. Is there any way of representing > ....9999999 as a product of primes*? I was immediately turned off to that notation when I studied these things. Honestly, I've seen it before, and I know it's supposed to represent a p-adic, but I have no idea what the correspondence is anymore. Anybody know? > Also, don't you allow denominator > prime factors, whose metric would be > 1 (unlike numerator prime > factors)? I vaguely remember Mathworld's p-adic treatment mentioning > denominator factors as also 'amenable' to the metric you describe. You're right. First define the order of an integer: order(p, x) is the biggest power of p that divides x. The definition of the order of a rational m/n is then order(p, m) - order(p, n). The metric is defined as before. I mistakenly thought that the stipulation that the fraction was in lowest terms took care of this issue. 'cid 'ooh === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony I'm not sure why anyone would want such beasties, in that case, > unless one wants to set up the ordering > -...123 < -123 < 0 < +123 < +...123 > for some reason. > I dislike that notation. Here is a brief primer on the p-adics: > http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant > material begins on page 15. > Basically, the idea is to consider a rational number x (in lowest > terms) and figure out how many times a prime p divides the numerator. > Call that number the order(p,x). Define a norm by |x|_p = p ^ > (-order(p,x)) if x =/= 0, oo otherwise. The related metric is > non-Archimedean metric on Q, and you can find a completion for Q under > this metric. The complete space is called Q_p. > It seems to be more than notation. Is there any way of representing > ....9999999 as a product of primes*? > I was immediately turned off to that notation when I studied these > things. Honestly, I've seen it before, and I know it's supposed to > represent a p-adic, but I have no idea what the correspondence is > anymore. Anybody know? ...4321, as a 10-adic, is a formal object X with the following properties: X mod 10 = 1 X mod 100 = 21 X mod 1000 = 321 X mod 10000 = 4321, etc. If the digits in the p-adic are eventually 0, X can be represented by a natural number. Other strings of digits will sometimes yield rational numbers, square roots, etc., if they satisfy the appropriate properties. For instance, ...6667 (as a 10-adic) is equal to 1/3 since it satisfies the equation 3 * N = 1. But not all fractions are in the n-adics; for instance, there is no 10-adic which acts like 1/2. (Which isn't too difficult to prove, unless your name is Archimedes Plutonium.) One way to think of it is to think of writing it as a decimal, 0.1234..., except that we don't have equalities like 0.1999... = 0.2000... That also makes the p-adic metric easier to digest. --- Christopher Heckman > Also, don't you allow denominator > prime factors, whose metric would be > 1 (unlike numerator prime > factors)? I vaguely remember Mathworld's p-adic treatment mentioning > denominator factors as also 'amenable' to the metric you describe. > You're right. > First define the order of an integer: order(p, x) is the biggest power > of p that divides x. The definition of the order of a rational m/n is > then order(p, m) - order(p, n). The metric is defined as before. I > mistakenly thought that the stipulation that the fraction was in lowest > terms took care of this issue. > 'cid 'ooh === Subject: Re: new way of teaching the Limit in Calculus; infinity the inverse is 0; why Cantor transfinites are phony In sci.logic, poopdeville@gmail.com on 2 Jan 2006 19:49:24 -0800 >> I'm not sure why anyone would want such beasties, in that case, >> unless one wants to set up the ordering >> -...123 < -123 < 0 < +123 < +...123 >> for some reason. >> I dislike that notation. Here is a brief primer on the p-adics: >> http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf -- the relevant >> material begins on page 15. >> Basically, the idea is to consider a rational number x (in lowest >> terms) and figure out how many times a prime p divides the numerator. >> Call that number the order(p,x). Define a norm by |x|_p = p ^ >> (-order(p,x)) if x =/= 0, oo otherwise. The related metric is >> non-Archimedean metric on Q, and you can find a completion for Q under >> this metric. The complete space is called Q_p. >> It seems to be more than notation. Is there any way of representing >> ....9999999 as a product of primes*? > I was immediately turned off to that notation when I studied these > things. Honestly, I've seen it before, and I know it's supposed to > represent a p-adic, but I have no idea what the correspondence is > anymore. Anybody know? I'm not entirely certain myself, but I for one would think that ....654321 . would correlate to the real number in base p: 0.123456... since the p-adics are Q with an unusal metric. >> Also, don't you allow denominator >> prime factors, whose metric would be > 1 (unlike numerator prime >> factors)? I vaguely remember Mathworld's p-adic treatment mentioning >> denominator factors as also 'amenable' to the metric you describe. > You're right. > First define the order of an integer: order(p, x) is the biggest power > of p that divides x. The definition of the order of a rational m/n is > then order(p, m) - order(p, n). The metric is defined as before. I > mistakenly thought that the stipulation that the fraction was in lowest > terms took care of this issue. Are you sure you didn't mean the exponent of the biggest power of p that divides x? Or perhaps order(p,n) / order(p,m)? The metric for the rational number p^a * r / s, where a is a signed integer, p and r are relatively prime and p and s are relatively prime (presumably r and s are also relatively prime but it doesn't matter as much) is p^(-a). That much I know regarding p-adics. :-) > 'cid 'ooh -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: A Question on Graphs > Let T be a tree of order k+1. Given a graph G of order rk+1 s.t. any > r+1 vertices span at least one edge. Show that G contains a subgraph > isomorphic to T. > I have showed that every graph of minimum degree k contains a copy of > T, so I want to show that G has a subgraph with minimum degree >=k. But A graph, in which every subgraph has minimum degree less than k, is k-colorable. Your assumption on G imply that G is not k-colorable. === Subject: Re: A Question on Graphs === Subject: Re: A Question on Graphs > Let T be a tree of order k+1. Given a graph G of order rk+1 s.t. any > r+1 vertices span at least one edge. Show that G contains a subgraph > isomorphic to T. > I have showed that every graph of minimum degree k contains a copy of > T, so I want to show that G has a subgraph with minimum degree >=k. But Induction. For the general case, delete a leaf of the tree and apply induction. --- Christopher Heckman === Subject: Re: Graduate Level Math Books For Sale There's still a bunch of books available. Here's the updated link: http://www.craigslist.org/eby/bks/120730267.html Daniel Giaimo === Subject: Re: Transitive group action In fact your argument is what the Burnside lemma boils down to in this situation. What I was wondering is the following: * A group never is the union of two proper subgroups * A finite group which is not cyclic trivially is the union of a finite number of proper subgroups (take the cyclic groups sitting inside). But can you choose the subgroups to be conjugate? If this wasn't possible, this would answer the original question too. I read parts of Serre's book on trees where the HNN construction is used to produce these infinite groups which only have two conjugacy classes, but these groups appear to be quite huge indeed. Are there smaller constructions? >The result reamins true for any group G (not necessarily finite) acting >transitively on a finite set X, since you can reduce to the finite case >by working in the image of G in the symmetric group on X. The obvious generalizations are left to the reader :-) === Subject: Re: Transitive group action >In fact your argument is what the Burnside lemma boils down to in this >situation. What I was wondering is the following: >* A group never is the union of two proper subgroups >* A finite group which is not cyclic trivially is the union of a finite >number of proper subgroups (take the cyclic groups sitting inside). But >can you choose the subgroups to be conjugate? >If this wasn't possible, this would answer the original question too. It is equivalent to the original question, so the answer is no. >used to produce these infinite groups which only have two conjugacy >classes, but these groups appear to be quite huge indeed. >Are there smaller constructions? It depends what you mean by smaller. AFAIK, it is not known whether there exist finitely generated infinite groups with exactly two conjugacy classes, but there do exist finitely generated infinite groups with finitely many conjugacy classes. I have been told that there exist Tarski Monsters (i.e. infinite groups in which all nontrivial proper subgroups have prime order p, for some sufficiently large prime number p) in which all of the subgroups of order p are conjugate, and there are exactly p conjugacy classes of elements. They are small in sense of finitely generated, but monsters are not usually regarded as being small! Derek Holt. infinite groups which >>The result reamins true for any group G (not necessarily finite) acting >>transitively on a finite set X, since you can reduce to the finite case >>by working in the image of G in the symmetric group on X. >The obvious generalizations are left to the reader :-) === Subject: Re: Transitive group action OK, I got it now: The following two statements are equivalent: (A) If a group G transitively acts on a set X with |X|>1, then there is a g in G which acts without fixed point. (B) Let H be a proper subgroup of a group G. Then G is not a finite union of conjugates of H. A implies B: Assume otherwise. Then G acts transitively on the finite set of conjugates of H in G (which has cardinality > 1 since H is not normal). But every element has a fixed point (the conjugate of H in which it is contained), contrary to (A). B implies A: Since G acts transitively, point stabilizers are proper subgroups, which are all conjugate. Thx everyone for listening. === Subject: Re: Transitive group action >OK, I got it now: >The following two statements are equivalent: >(A) If a group G transitively acts on a set X with |X|>1, then there is >a g in G which acts without fixed point. >(B) Let H be a proper subgroup of a group G. Then G is not a finite >union of conjugates of H. >A implies B: Assume otherwise. Then G acts transitively on the finite >set of conjugates of H in G >(which has cardinality > 1 since H is not normal). But every element >has a fixed point (the conjugate of H in which it is contained), >contrary to (A). >B implies A: Since G acts transitively, point stabilizers are proper >subgroups, which are all conjugate. B implies A is only true if you assume X to be finite. Derek Holt. === Subject: laser detection system I'm making some program such as scanning program using laser, It will be used on the highway. The result is 2D. We have 2 lasers for measure how the car's shape is. one is projecting vertical direction(projecting ground) the other is not so vertical, It has some angle. So , we project this onto the car, After that, we get the result and draw the graph(using MS excel) However, The results(2 graphs) which are from 2 lasers are littlebit different because of the angle between 2 laser beems. In my opinion, because the angle is different, the shape is also will be different. I want to match the shape as possible. If you know, please reply me. thank you very much ..^^ === Subject: Re: laser detection system > I'm making some program such as scanning program using laser, > It will be used on the highway. > The result is 2D. > We have 2 lasers for measure how the car's shape is. > one is projecting vertical direction(projecting ground) the other is > not so vertical, It has some angle. > So , we project this onto the car, After that, we get the result and > draw the graph(using MS excel) > However, The results(2 graphs) which are from 2 lasers are littlebit > different because of the angle between 2 laser beems. > In my opinion, because the angle is different, the shape is also will > be different. > I want to match the shape as possible. > If you know, please reply me. > thank you very much ..^^ The error between the two is your correction factor, you can put it in a look up table if you need to. Or calibrate types of Cars. Laser must be eyesafe, can reflect off crome and hit drivers eye. === Subject: generalized gravity center Let A be a bounded subset of the plane and for any natural n let A_n denote the 1/n-neighbourhood of A. Let x_n be the gravity center of A_n and I(A) denote the accumulation set of the sequence x_n. Then I(A) may be viewed as a generalized gravity center of A. My question is: Is it possible that I(A)=A and A is not a one-point set? Another question: Is it possible that I(A) contains an open disk? vedran === Subject: simple question Why rational number is infinite repeating decimal but irrational number is infinite non repeating number? How to prove it? === Subject: Re: simple question >Why rational number is infinite repeating decimal but irrational number >is infinite non repeating number? How to prove it? In the long division to compute the decimal representation, how many remainders can you have? What happens when the same remainder appears twice? Suppose we have the repeating decimal .123123123123123... x = .123123123123123... 1000 x = 123.123123123123... 999 x = 123 So x = 123/999 = 41/333. Can you do this for any repeating decimal? Rob Johnson take out the trash before replying === Subject: Re: simple question <20060103.034701@whim.org> What you represented only shows that finite repeating decimals are rational numbers, how to show the inverse conclusion, namely, how to show that rational number p/q will result in finite remainders when long division. And then how to show the irrational number? === Subject: Re: simple question <20060103.034701@whim.org> sunnewton escreveu: > What you represented only shows that finite repeating decimals are > rational numbers, how to show the inverse conclusion, namely, how to > show that rational number p/q will result in finite remainders when > long division. > And then how to show the irrational number? Suppose you divide p by q , p and q positive integers. At each step of the long division, you get a remainder, which is a number in the finite set {0, 1.....q-1}. If at some step you get the remainder 0, then you're done and there's no repetition. But it may happen that you never get 0. Since there are only q-1 possibilities for the remainder, if the process proceeds indefinetely, then at some step you must get a same remainder you had got before (pigeon hole principle). And so, you have a same finite sequence of digits over, and over and over.....Here's the repetition. This may not be a rigorous proof, but that's how it works. So you see that every rational has either a finite decimal representation or is represented by an infinite sequence of finite blocks of repeating digits (something like 2.37429429429....). And the converse is also true. Using a contrapositive argument, you see every irrational has an infinite and non- repeating decimal representation, and the converse is also true. Amanda === Subject: Re: simple question Please quote that to which you are responding. Not everyone uses Google to read usenet news, and they will have no idea to what you are responding. >Why rational number is infinite repeating decimal but irrational number >is infinite non repeating number? How to prove it? >>In the long division to compute the decimal representation, how many >>remainders can you have? What happens when the same remainder appears >>twice? >>Suppose we have the repeating decimal .123123123123123... >> x = .123123123123123... >>1000 x = 123.123123123123... >> 999 x = 123 >>So x = 123/999 = 41/333. Can you do this for any repeating decimal? >What you represented only shows that finite repeating decimals are >rational numbers, how to show the inverse conclusion, namely, how to >show that rational number p/q will result in finite remainders when >long division. >And then how to show the irrational number? Read up on contrapositives. The answers to the questions I posed show that a rational number has a terminating or repeating decimal (using long division) and that a terminating or repeating decimal represents a rational number (by computing the rational number represented). This means that an irrational number has an infinite, non-repeating decimal (if it has a finite or repeating decimal, it is rational), and any number with an infinite, non-repeating decimal is irrational (since a rational has a finite or repeating decimal). Rob Johnson take out the trash before replying === Subject: abt diffeomorphism If a mapping is a diffeomorphism, [ y_1,y_2,...,y_n]^T=f([x_1,x_2,...x_n]^T): R^n->R^n, if y_i bounded, is it that x_i is also bounded? === Subject: Re: abt diffeomorphism On 3 Jan 2006 03:53:21 -0800, sunnewton y_1,y_2,...,y_n]^T=f([x_1,x_2,...x_n]^T): R^n->R^n, if y_i bounded, is >it that x_i is also bounded? Does a diffeomorphism have an inverse? If yes, is the inverse continuous? ************************ David C. Ullrich === Subject: Re: abt diffeomorphism According to the definition of diffeomorphism, it of course has inverse. and we can assume the mapping is sufficient smooth === Subject: Re: abt diffeomorphism Ok, I see, they should be bounded === Subject: Re: abt diffeomorphism (-pi/2,pi/2): x |--> arctan(x) ... === Subject: simplicial abelian groups; projective object of the form.. Happy new year to all of you. I wondered if someone could give me a hint about the following problem: Given the category of simplicial abelian groups (has enough projectives). To show that every projective is of the form (up to iso): direct sum_i of {free abelian groups on the standard simplex n_i}. === Subject: Re: Is the set N of natural numbers well defined? > For any set S, by definition S = {All y : y e S}, > including the empty > set, i.e., S = {}! > So, N ={all y} is > NON SEQUITUR, since the empty set satisfies the same > condition as N, > unless you are claiming that the empty set is > everything. You are right.I should have put the argument as follows: The set N is defined as, N ={0} U {All S(x): x e N}. But,All S(x):x e N <-->All y : y!=0 and y e N. So, N ={0} U {All y: y !=0 and y e N } But y e N is tautological. So,N = {0} U {All y: y !=0} or, N = {All y} Thus N is the set of all ordinals, which we know, does not satisfy the axiom of foundation. The key issue is whether the ZFaxioms,solely by themselves, allow us to infer the existence of more than one inductive set. I would think they do not. -Apoorv === Subject: Analytic function on B(0,2) Suppose f is analytic on the open disk of radius 2 centered at 0, B(0,2). Suppose f takes only real values on { z : | z | = 1 }. How can I show that f is constant on B(0,2)? James === Subject: Re: Analytic function on B(0,2) <29333536.1136295353849.JavaMail.jakarta@nitrogen.mathforum.org>, > Suppose f is analytic on the open disk of radius 2 centered at 0, B(0,2). > Suppose f takes only real values on { z : | z | = 1 }. How can I show that f > is constant on B(0,2)? On the unit circle, sum (n=0,oo) a_n*e^(int) = a_0 + sum (n=1,oo) (a_n/2)*e^(int) + (conj(a_n)/2)*e^(-int), where sum a_n*z^n is the Taylor series of f at 0. Equate coefficients. === Subject: Re: Analytic function on B(0,2) >Suppose f is analytic on the open disk of radius 2 centered at 0, B(0,2). Suppose f takes only real values on { z : | z | = 1 }. How can I show that f is constant on B(0,2)? For a start, note that the imaginary part of f is harmonic, and hence vanishes for |z| < 1. >James ************************ David C. Ullrich === Subject: Re: Analytic function on B(0,2) Ah. Then, Max Mod theorem for harmonic functions, then Open Mapping theorem, then Identity theorem. === Subject: Re: Analytic function on B(0,2) > Suppose f is analytic on the open disk of radius 2 centered at 0, B(0,2). Suppose f takes only real values on { z : | z | = 1 }. How can I show that f is constant on B(0,2)? > James Consider the function g(z)= f(z*)*, where '*' stands complex conjugation. It is easily seen that g(z) is also analytic in B(0,2). Now, if z is on the unit circle, then z*=1/z, so by your assumption g(z)=f(1/z) for all the z on the unit circle. This formula gives an analytic continuation of g(z) to the whole plane. Furthermore, as f(z) has a limit, when z->0, we easily see that g(z) is bounded in a neighborhood of infinity. A compactness argument then shows that g(z) is bounded everywhere, and Liouville's theorem does the rest. Jyrki Lahtonen, Turku, Finland === Subject: Differential operator introduces orthogonality In a recent homework, the first part of a question was A vector field B is everywhere orthogonal to a family of surfaces f(x) = constant. Show that B = p(x)*(Df) where p(x) is a scalar field (note x here denotes vector x, D denotes del or upside down capital delta symbol). I struggled with it for a while, then after the handin I looked at the solutions and it said The surfaces f = constant are everywhere orthogonal to (Df). So B must be everywhere parallel to (Df), or B = p(x)*(Df). I understand the second part of this, I just dont get why The surfaces f = constant are everywhere orthogonal to (Df). === Subject: Re: Differential operator introduces orthogonality > In a recent homework, the first part of a question was A vector field B is everywhere orthogonal to a family of surfaces f(x) = constant. Show that B = p(x)*(Df) where p(x) is a scalar field (note x here denotes vector x, D denotes del or upside down capital delta symbol). > I struggled with it for a while, then after the handin I looked at the solutions and it said The surfaces f = constant are everywhere orthogonal to (Df). So B must be everywhere parallel to (Df), or B = p(x)*(Df). > I understand the second part of this, I just dont get why The surfaces f = constant are everywhere orthogonal to (Df). Let x0 be an arbitrary point, so you want to show that (Df)(x0) is orthogonal to the surface S={x | f(x) = f(x0)}. This means that (Df)(x0) is perpendicular to any vector v that's tangent to the surface at x0. Let v be such a vector and let p: (-e,e) --> S be a smooth path with p(0) = x0, p'(0) = v (for some e >0 sufficiently small). Then <(Df)(x0),v> = (f o p)'(0) = 0 since (f o p) is a _constant_ function. That proves the claim ... Well, I'd put it just a tad more strongly -- f _defines_ the family of surfaces ... === Subject: Re: Differential operator introduces orthogonality > In a recent homework, the first part of a question > was A vector field B is everywhere orthogonal to a > family of surfaces f(x) = constant. Show that B = > p(x)*(Df) where p(x) is a scalar field (note x here > denotes vector x, D denotes del or upside down > capital delta symbol). > I struggled with it for a while, then after the > handin I looked at the solutions and it said The > surfaces f = constant are everywhere orthogonal to > (Df). So B must be everywhere parallel to (Df), or B > = p(x)*(Df). > I understand the second part of this, I just dont > get why The surfaces f = constant are everywhere > orthogonal to (Df). > Let x0 be an arbitrary point, so you want to show > how that (Df)(x0) is > orthogonal to the surface S={x | f(x) = f(x0)}. > )}. This means that > (Df)(x0) > is perpendicular to any vector v that's tangent to > to the surface at > x0. > Let v be such a vector and let p: (-e,e) --> S be > be a smooth path > with p(0) = x0, p'(0) = v (for some e >0 >0 sufficiently small). With you up to here... > Then > <(Df)(x0),v> = (f o p)'(0) Why? > = 0 since (f o p) is a > _constant_ > function. > That proves the claim ... > it is just a function which happens to have the > property that if it has the same value for two > different points then they lie on the same surface in > the family. > Well, I'd put it just a tad more strongly -- f > - f _defines_ the family > of surfaces ... === Subject: Re: Differential operator introduces orthogonality <24965512.1136309603060.JavaMail.jakarta@nitrogen.mathforum.org > In a recent homework, the first part of a question > was A vector field B is everywhere orthogonal to a > family of surfaces f(x) = constant. Show that B = > p(x)*(Df) where p(x) is a scalar field (note x here > denotes vector x, D denotes del or upside down > capital delta symbol). > I struggled with it for a while, then after the > handin I looked at the solutions and it said The > surfaces f = constant are everywhere orthogonal to > (Df). So B must be everywhere parallel to (Df), or B > = p(x)*(Df). > I understand the second part of this, I just dont > get why The surfaces f = constant are everywhere > orthogonal to (Df). > Let x0 be an arbitrary point, so you want to show > how that (Df)(x0) is > orthogonal to the surface S={x | f(x) = f(x0)}. > )}. This means that > (Df)(x0) > is perpendicular to any vector v that's tangent to > to the surface at > x0. > Let v be such a vector and let p: (-e,e) --> S be > be a smooth path > with p(0) = x0, p'(0) = v (for some e >0 >0 sufficiently small). > With you up to here... > Then > <(Df)(x0),v> = (f o p)'(0) > Why? Hmmm ... depends on the definitions you're using, I suppose. Another approach: (Df)(x0) is the linear map that's obtained in just the way that I indicated: if v is a tangent vector at x0 and you want to find the corresponding tangent vector at f(x0), you choose a path p that has v as its tangent vector at x0 and calculate the tangent vector to the curve (f o p) at f(x0). _That_'s (Df)(x0)(v) ... and, in the present case, it's always the zero vector if v is tangent to S. > = 0 since (f o p) is a > _constant_ > function. > That proves the claim ... > it is just a function which happens to have the > property that if it has the same value for two > different points then they lie on the same surface in > the family. > Well, I'd put it just a tad more strongly -- f > - f _defines_ the family > of surfaces ... > > === Subject: Maplesoft is committed to providing the highest level of support for the products it sells. ... when we, the Maplesoft's owners and top officers, are not so much cocktailed that even our main site http://www.maplesoft.com had been showing Server Error in '/' Application during almost 24 hours in line: http://maple.bug-list.org/maplesoft_com_jan_2_2006.jpg http://maple.bug-list.org/maplesoft_com_jan_3_2006.jpg ................................................................. Perfect. And absolutely not surprising for us. This is just yet another element of the Maplesoft/Maple crisis we have predicted with the help of our unique VM machine. http://maple.bug-list.org/maple-crisis.php Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ ................................................................. === Subject: Re: Maplesoft is committed to providing the highest level of support for the products it sells. >... when we, the Maplesoft's owners and top officers, are not >so much cocktailed that even our main site >http://www.maplesoft.com >had been showing Server Error in '/' Application during >almost 24 hours in line: >http://maple.bug-list.org/maplesoft_com_jan_2_2006.jpg >http://maple.bug-list.org/maplesoft_com_jan_3_2006.jpg Sheesh. Presumably they had a problem on one of their servers over the holiday. These things happen. Get over it. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Uniform Dirichlet distribution for 3 dimensions I'd looked at that already, actually, but overlooked the fact that the simplex is of length sqrt(2) and not 1. =) -Ken === on 22 Nov 2005 00:29:09 in a message titled Maplesoft bugs, Or, Trust our expertize, buy Mathematics and Mathematical Compuation for $125 only -------------^^^^^^^^^^--------------- VB> Hello Maple 10 potential and actual customers over the world, VB> There are dozen thousands bugs in Maple 10 itself. But with VB> Maplesoft'2005 one can bang against a fault at ANY point. http://webstore.maplesoft.com/Product.aspx we read A New Twist to Fourier Tranforms -----------------------^^^^^^^^^ Not enough? Then visit these pages: http://www.maplesoft.com/applications/app_center_view.aspx?AID=1540&CID=14&S CID=120 Lesson 32 -- Using Laplace Tranforms to Solve Initial Value Problems http://www.maplesoft.com/applications/app_center_browse.aspx?CID=14&SCID=120 Lesson 32 -- Using Laplace Tranforms to Solve Initial Value Problems http://www.maplesoft.com/applications/app_center_browse.aspx?SCID=120&CID=14 Lesson 32 -- Using Laplace Tranforms to Solve Initial Value Problems Boom-boom-boom! To hell all spell-checkers, why care! Isn't it sufficient to drink a liter of bourbon & exclaim loudly http://www.maplesoft.com/support/ Maplesoft is committed to providing the highest level of support for the products it sells. ? Of cause, on Nov 22, 2005 I saw BOTH typos. But, intentionally, reported publicly only ONE of these two. So, do you realize what's happened? This is a piece of a PATTERN: how Maplesoft fixes the bugs. ................................................................. However, to tell the truth, all these 'Compuation's & 'Tranform's are completely dwarfed by the unparalleled Maple bug stuff our VM machine discovered in Maple. Best New Year wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director Cyber Tester, LLC http://www.cybertester.com/ http://maple.bug-list.org/ http://www.CAS-testing.org/ === Subject: elliptic function How to write the following equation in the form of Jacobi elliptic functions: dphi/dxi = ( a(k) + b(k)q^2 + c(k)q^4 ) / ( (1 - q^2) (1 - k^2 q^2) ) where a(k), b(k), and c(k) are even polynomials with at most of order 4, and q is function of phi and phi is function of xi: q = q(phi(xi)). N. Karjanto === Subject: Re: Well Ordering the Reals Ross A. Finlayson said: >> If the set AB contains a degenerate interval g(o) = [a, a] for >> some ordinal o and the function g() well-ordering AB, then >> the domain of g has a maximal element. > construction we have discussed before, you have a sequence of > monotonically increasing a_n's and a sequence of monotonically > decreasing b_n's, where all the a_n's are less than all the b_n's. Are > you using a different construction than this? The degenerate interval > [a,a] has a in between the a_n's and b_n's. What do you mean by > maximal?? In the example I gave, the degenerate interval is [pi,pi] > at the omega-th ordinal. How is pi maximal? > None of the rest of your post makes any sense whatsoever to me, until > you explain what you mean by maximal. >> ...then AB is countable, and then only a countable >> sequence of nested intervals would have no element >> within each interval and endpoint of none, not leading >> to Cantor's conclusion. > None of this violates Cantor's conclusion. Cantor's conclusion is > based on the fact that there is a real that maps to no (finite) natural > number. Whether or not reals map to countable ordinals is undecidable, > since that is essentially the Continuum Hypothesis. Even if you assume > CH, then you can map the reals to the countable ordinals, which is > perfectly acceptable since there are uncountably many countable > ordinals. >> The reals are a complete ordered field, but the >> projectively extended reals don't have hyperreals, >> just a point at infinity. > What are projectively extended reals? Please use standard > terminology, as none of this amkes any sense when you use obscure or > artificial names for these constructs. >> Similarly the naturals may have a one-point >> compactification, but then you might consider they are >> already compact, and so the compactification is implicit. > What the hell is one-point compactification? Please see above note > about artificial names. >> There is no universe in ZF. > Please, for God's sakes, DEFINE what you mean by this! What is a > universe? >> There is no maximal element of the ordinals in ZF. > Here we have a reference to maximal element, and it seems to mean > something entirely different than suggested earlier by degenerate > intervals. If by this you mean there is no ordinal X greater than > every other ordinal, then yes, you are correct. >> If the universe exists, and the only objects in ZF are sets, >> and there is no universal set in ZF, where the universe >> does exist and doesn't exist, .... > If by universe you mean the class of all sets, then no it does not > live in ZF, it lives in NBG. Or it can live in any number of non-well > founded theories, each of which has ZF as a sub-theory. >> The universe is infinite. > Not necessarily. Consider a universe of sets with no axioms. Then > there are no sets, and the universe is empty. Consider an axiomatic > theory with only a single axiom: there exists the empty set. In this > universe, only one set exists (with no other axioms to generate other > sets). I think it depends on the theory. > Jonathan Hoyle > Eastman Kodak > Happy New Year. Happy New Year to you too, and to all!! > I guess first is to consider the maximality of dom(g) where g is a > function well-ordering the nested intervals of set AB, where the single > endpoint auxiliary pairs or reals and ordinals are in some superset, > and g is a bijection between the intervals of AB and an ordinal o+1. > The maximal element of the set o+1 is an ordinal o, or, say, m. So, m > is the maximal element of dom(g) if there exists a degenerate inteval > in AB. If there doesn't exist a degenerate interval in AB, then f is > not a well-ordering of the reals. So, if there does exist a degenerate > interval in AB then m is a maximal element of dom(g), and g(m) is a > degenerate interval. > If m is a countable ordinal c, then a countable sequence of nested > intervals is considered. An infinite sequence doesn't have a maximal > element. The well-ordering f constructing AB leads to a well-ordering > g of AB where if dom(g), the domain of the function g well-ordering the > nested intervals, has a maximal element m else there are elements in > the reals not in the range of f, the putative function well-ordering > the reals. So if m = c for some countable ordinal c, it's not apparent > how to interconvert the sequence and the well-ordered countable set > with minimal and maximal element, where it is infinite. Perhaps you > might consider that AB is finite. > If m is an uncountable ordinal o, including the consideration that it > is the least uncountable ordinal, then AB contains uncountably many > non-degenerate intervals, and as the reals are the complete ordered > field, each of those contains a rational number. Ahhhh...... Now I see where you are going. If there are an uncountably infinite set of nested intervals, and between any two reals is a rational, then there is a rational in each subinterval, and the rationals are uncountable. Very good. I am not sure I see any problem yet with this argument. Interesting point. Hmmm... > If you can consider that AB is finite, then there are adjacent points > in the reals, in their normal ordering. A finite number of reals in the interval? Er, uh..... is that what you mean? > Onwards to later points, projectively extended reals and one-point > compactification are each rather standardized terms, basically > snubbing Cauchy/Weierstrass and the blinders of limit evaluation, which > is very useful. > The universe is infinite. Infinite sets are equivalent. > There is only one null axiom theory. > So, well-order the reals. > Happy New Year! > Ross -- Smiles, Tony === Subject: Re: Well Ordering the Reals Jonathan Hoyle said: >> If the set AB contains a degenerate interval g(o) = [a, a] for >> some ordinal o and the function g() well-ordering AB, then >> the domain of g has a maximal element. > construction we have discussed before, you have a sequence of > monotonically increasing a_n's and a sequence of monotonically > decreasing b_n's, where all the a_n's are less than all the b_n's. Are > you using a different construction than this? The degenerate interval > [a,a] has a in between the a_n's and b_n's. What do you mean by > maximal?? In the example I gave, the degenerate interval is [pi,pi] > at the omega-th ordinal. How is pi maximal? > None of the rest of your post makes any sense whatsoever to me, until > you explain what you mean by maximal. I don't know if this has been answered to your satisfaction, but it was immediately obvious to me what Ross meant, I do believe (correct me if I'm wrong). He means that c, or pi, or whatever the value that the interval converges on in this way, has the highest index in the sequence of reals, for any value in that original interval. Where a_n=b_n, there is no other real in the interval with a higher index, or n. Now, this seemed correct to me for a minute or two, but is not. While c has a higher index than any of the a_n's or b_n's that marked the intermediate endpoints of the shrinking interval, there are values between successive a_n's and between successive b_n's which may very well have a higher index than the final point of convergence between a's and b's. We select subsequent values for a and b by finding the next real in the sequence which lies within the current interval. In so choosing subsequent a's and b's, we are skipping over elements with values between a_n and a_n+1 which have higher indexes than n+1. They may also have higher indexes than c. So, c is not necessarily the real number with the highest index between a_0 and b_0. >> ...then AB is countable, and then only a countable >> sequence of nested intervals would have no element >> within each interval and endpoint of none, not leading >> to Cantor's conclusion. > None of this violates Cantor's conclusion. Cantor's conclusion is > based on the fact that there is a real that maps to no (finite) natural > number. Whether or not reals map to countable ordinals is undecidable, > since that is essentially the Continuum Hypothesis. Even if you assume > CH, then you can map the reals to the countable ordinals, which is > perfectly acceptable since there are uncountably many countable > ordinals. (sigh) I'm sorry, but limit ordinals are a mistake. >> The reals are a complete ordered field, but the >> projectively extended reals don't have hyperreals, >> just a point at infinity. > What are projectively extended reals? Please use standard > terminology, as none of this amkes any sense when you use obscure or > artificial names for these constructs. Google it: http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html >> Similarly the naturals may have a one-point >> compactification, but then you might consider they are >> already compact, and so the compactification is implicit. > What the hell is one-point compactification? Please see above note > about artificial names. Try Google: http://mathworld.wolfram.com/One-PointCompactification.html Please try to remember that, just because you haven't heard of something, it doesn't mean it doesn't exist. >> There is no universe in ZF. > Please, for God's sakes, DEFINE what you mean by this! What is a > universe? An all-encompassing domain that includes everything. A grand structure. >> There is no maximal element of the ordinals in ZF. > Here we have a reference to maximal element, and it seems to mean > something entirely different than suggested earlier by degenerate > intervals. If by this you mean there is no ordinal X greater than > every other ordinal, then yes, you are correct. In this case, what you think and what he means are both true. He means maximal in the the ordering, index-wise. Ross, am I wrong? >> If the universe exists, and the only objects in ZF are sets, >> and there is no universal set in ZF, where the universe >> does exist and doesn't exist, .... > If by universe you mean the class of all sets, then no it does not > live in ZF, it lives in NBG. Or it can live in any number of non-well > founded theories, each of which has ZF as a sub-theory. >> The universe is infinite. > Not necessarily. Consider a universe of sets with no axioms. Then > there are no sets, and the universe is empty. Consider an axiomatic > theory with only a single axiom: there exists the empty set. In this > universe, only one set exists (with no other axioms to generate other > sets). I think it depends on the theory. What about the actual Universe as a whole? Finite or infinite? > Jonathan Hoyle > Eastman Kodak -- Smiles, Tony === Subject: Re: Well Ordering the Reals > I don't know if this has been answered to your satisfaction, but it was > immediately obvious to me what Ross meant The blind leading the blind! === Subject: Re: Well Ordering the Reals Jonathan Hoyle said: >> It's as before where I say: well-order the reals and somebody >> says, OK, here, f is a well-ordering of the reals. Then, as >> above as we discuss, that leads to a variation on Cantor's >> nested intervals where f is not a bijection to any ordinal O', >> where the existence of that function is the same thing as >> there being a well-ordering of the reals. > I think I see where your proof fails, and the best way to see that is > to compare and contrast it with the original Cantor proof. > Essentially, Cantor proved that for any open interval I = (a,b) in R > and any sequence of real numbers X = { x_n } for n in N, there are > elements of I that are not found in X. > Cantor's First Proof of Uncountability: > Suppose X = { x_n } is a sequence of all the real numbers for all i in > N, with each x_n unique. Let a_0 = x_i and b_0 = x_j where i is the > smallest integer such that x_i is in (a,b) and j is the smallest > integer such that x_j is in (a_0, b). Now for each n>=0 construct > (a_n+1,b_n+1) such that a_n+1 = x_i where i is the smallest integer > such that a_n such that a_n+1 So we have a_n's monotonically increasing and the b_n's monotonically > decreasing with all the a_n's less than all the b_n's. Since these are > open intervals, this process continues infinitely. Now there are one > of two possibilities: Either lim |b_n - a_n| = 0 or lim |b_n - a_n| 0. If lim |b_n - a_n| > 0, then there are an infinite number of values > not in X and we are done. If lim |b_n - a_n| = 0, then essesntially > the a_n's and b_n's form a Dedekind cut for some real number c so that > lim a_n = lim b_n = c. Since each (a_n,b_n) is an open interval, we > know that c is not found in any previous a_n or b_n. QED. When you talk about the limit, you mean as n goes to oo. Now, what does it mean if lim(n->oo: a_n)=c=lim(n->oo: b_n)? It means it takes an infinite number of steps to converge to this intermediate value. So, indeed, the set of reals is uncountably infinite, since it requires an actually infinite number of iterations to get to some (actually the vast majority) of real numbers, an likewise an infinite number of bits to represent those values. It doesn't mean those values don't exist in the sequence. It means the sequence is infinite, which is what we should expect. > Okay, now Ross, you propose to replace X = { x_n } for n in N with any > arbitrary well ordering of R, call it M = { m_o } for o in O' where O' > is the set of all ordinals with cardinality less than |R|. Is this > correct? The problem you run into here is that your process is defined > only for successor ordinals, not for limit ordinals. For example, how > do you propose to define a_omega and b_omega? Presumably, you want to > choose an a_omega to be m_i where i is the least ordinal greater than > all the previous a_n's and b_omega to be m_j where j is least ordinal > less than all the previous b_n's, but you do not know that you can > choose such an a_omega and b_omega. The limit ordinals are a mistake and a kludge. The predecessor discontinuities, introduced by the declaration that the set of finites is countably infinite and the set of countably infinite ordinals is uncountably infinite, are the only thing that tricks you into believing you have a well ordering of an uncountably infinite set. I am sorry, but von Neumann ordinals, while a cute construction, are a misguided notion when it comes to dealing with oo. > For example, suppose by chance your a_n's looked like this: { 3, 3.1, > 3.14, 3.141, 3.1415, ... } and your b_n's looked like this: { 4, 3.2, > 3.15, 3.142, 3.1416, ... }. Then the only value you can choose for > both a_omega and b_omega is pi, yet you cannot have the open interval > (pi, pi). And it's not just omega you have to worry about, this is a > problem for *EACH* limit ordinal less than |R|, of which there are an > uncountable number. You have to choose only those well orderings in > which this does not happen, and to do that you must supply a well > ordering. This is incorrect. IF the nested interval does NOT converge on the intermediate value, THEN the ordering does not include all reals. If, after any infinite number of iterations, as n->oo, the two values a_n and b_n DO converge on a value, then indeed it is a type of Dedekind cut, and you DO have a sequential ordering of the reals, as demonstrated by the H-riffic numbers. In order for it to be a well-ordering, it must not have any infinite descending chains, but avoiding them is clearly impossible in an uncountably infinite set without introducing predecessor discontinuities as with the von Neumann ordinals. A sequential ordering of the real numbers has been offered, but of course, it is not a well ordering. It requires an infinite number of iterations to achieve c. > So, time to turn the tables on you. Go ahead Ross, well order the > reals. :-) All positive reals: 1: real(1) 2: real(x) -> (real(2^x) and real(2^-x)) > Jonathan Hoyle > Eastman Kodak -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Cantor's First Proof of Uncountability: > > Suppose X = { x_n } is a sequence of all the real numbers for all i in > N, with each x_n unique. Let a_0 = x_i and b_0 = x_j where i is the > smallest integer such that x_i is in (a,b) and j is the smallest > integer such that x_j is in (a_0, b). Now for each n>=0 construct > (a_n+1,b_n+1) such that a_n+1 = x_i where i is the smallest integer > such that a_n such that a_n+1 > So we have a_n's monotonically increasing and the b_n's monotonically > decreasing with all the a_n's less than all the b_n's. Since these are > open intervals, this process continues infinitely. Now there are one > of two possibilities: Either lim |b_n - a_n| = 0 or lim |b_n - a_n| 0. If lim |b_n - a_n| > 0, then there are an infinite number of values > not in X and we are done. If lim |b_n - a_n| = 0, then essesntially > the a_n's and b_n's form a Dedekind cut for some real number c so that > lim a_n = lim b_n = c. Since each (a_n,b_n) is an open interval, we > know that c is not found in any previous a_n or b_n. QED. > When you talk about the limit, you mean as n goes to oo. Now, what does it > mean > if lim(n->oo: a_n)=c=lim(n->oo: b_n)? lim(n->oo: a_n)=c means that for every positive real number epsilon, {n : |a_n - c| < epsilon} contains all but finitely many n. Similarly for lim(n->oo: b_n)=c. > It means it takes an infinite number of > steps to converge to this intermediate value. This is circular, defining /converging/ in terms of /converging/. > I am sorry, but von Neumann ordinals, while a cute construction, are > a misguided notion when it comes to dealing with oo. So somehow adding one element to an infinite set of elements destroyes everything? === Subject: Re: Well Ordering the Reals Virgil said: > Cantor's First Proof of Uncountability: > > Suppose X = { x_n } is a sequence of all the real numbers for all i in > N, with each x_n unique. Let a_0 = x_i and b_0 = x_j where i is the > smallest integer such that x_i is in (a,b) and j is the smallest > integer such that x_j is in (a_0, b). Now for each n>=0 construct > (a_n+1,b_n+1) such that a_n+1 = x_i where i is the smallest integer > such that a_n such that a_n+1 > So we have a_n's monotonically increasing and the b_n's monotonically > decreasing with all the a_n's less than all the b_n's. Since these are > open intervals, this process continues infinitely. Now there are one > of two possibilities: Either lim |b_n - a_n| = 0 or lim |b_n - a_n| > 0. If lim |b_n - a_n| > 0, then there are an infinite number of values > not in X and we are done. If lim |b_n - a_n| = 0, then essesntially > the a_n's and b_n's form a Dedekind cut for some real number c so that > lim a_n = lim b_n = c. Since each (a_n,b_n) is an open interval, we > know that c is not found in any previous a_n or b_n. QED. > > When you talk about the limit, you mean as n goes to oo. Now, what does it > mean > if lim(n->oo: a_n)=c=lim(n->oo: b_n)? > lim(n->oo: a_n)=c means that for every positive real number epsilon, > {n : |a_n - c| < epsilon} contains all but finitely many n. > Similarly for lim(n->oo: b_n)=c. Do you even know what the topic is? I didn't ask what a limit meant. > It means it takes an infinite number of > steps to converge to this intermediate value. > This is circular, defining /converging/ in terms of /converging/. I am not defining converging. I am talking about the uncountability of the reals ala Cantor's first proof thereof. > I am sorry, but von Neumann ordinals, while a cute construction, are > a misguided notion when it comes to dealing with oo. > So somehow adding one element to an infinite set of elements destroyes > everything? No, defining the set of finites as countably infinite, and the set of countable infinities as uncountably infinite, is misguided. You seem to recall that I said it's an error of 1. That's good. You're not entirely brain-dead. When you count from 1, you see easily that every natural is in the position denoted by itself, so the nth natural is n. This clearly shows that the set cannot be infinite, because n would then be an infinite position in the set, and also an infinite value in the set, which you do not allow. If all values are finite, then so are all positions, and so is the size of the set. Now, if you start at 0, then you say there are always more elements than the largest value in the set, since the nth element is n-1. This leads to the notion that the size of the set of finite naturals is larger than all finite naturals, and you say it's infinite. But, how much larger is it? It's always 1 larger. So, if you think that a value that is 1 larger than the largest value is infinite, when all values are finite, then you must think that adding 1 to a finite can yield an infinite. Clearly, the addition of one element does not turn a finite set into an infinite one, and the addition of a single element should not change your conclusion about the size of the overall set from finite to infinite. The logic behind the von Neumann ordinals is radically flawed, for this reason, as I have explained, and as you have continued to ignore, having nothing to say in response. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > > Cantor's First Proof of Uncountability: > > Suppose X = { x_n } is a sequence of all the real numbers for > all i in N, with each x_n unique. Let a_0 = x_i and b_0 = x_j > where i is the smallest integer such that x_i is in (a,b) and j > is the smallest integer such that x_j is in (a_0, b). Now for > each n>=0 construct (a_n+1,b_n+1) such that a_n+1 = x_i where i > is the smallest integer such that a_n where j is the smallest integer such that a_n+1 > So we have a_n's monotonically increasing and the b_n's > monotonically decreasing with all the a_n's less than all the > b_n's. Since these are open intervals, this process continues > infinitely. Now there are one of two possibilities: Either lim > |b_n - a_n| = 0 or lim |b_n - a_n| > 0. If lim |b_n - a_n| > > 0, then there are an infinite number of values not in X and we > are done. If lim |b_n - a_n| = 0, then essesntially the a_n's > and b_n's form a Dedekind cut for some real number c so that > lim a_n = lim b_n = c. Since each (a_n,b_n) is an open > interval, we know that c is not found in any previous a_n or > b_n. QED. > > When you talk about the limit, you mean as n goes to oo. Now, > what does it mean if lim(n->oo: a_n)=c=lim(n->oo: b_n)? > > lim(n->oo: a_n)=c means that for every positive real number > epsilon, {n : |a_n - c| < epsilon} contains all but finitely many > n. Similarly for lim(n->oo: b_n)=c. > Do you even know what the topic is? I didn't ask what a limit meant. TO asks Now, what does it mean if lim(n->oo: a_n)=c=lim(n->oo: b_n)? Now he says he didn't ask that! > > > It means it takes an infinite number of steps to converge to this > intermediate value. > > This is circular, defining /converging/ in terms of /converging/. > I am not defining converging. I am talking about the uncountability > of the reals ala Cantor's first proof thereof. When TO says that something means something, how is that not defining anything? > > > I am sorry, but von Neumann ordinals, while a cute construction, > are a misguided notion when it comes to dealing with oo. > > > So somehow adding one element to an infinite set of elements > destroys everything? > No, defining the set of finites as countably infinite, and the set of > countable infinities as uncountably infinite, is misguided. I would rather be guided by those who say so than misguided enough to swallow TO's nonsense. You seem > to recall that I said it's an error of 1. That's good. You're not > entirely brain-dead. When you count from 1, > The logic behind the von Neumann ordinals is > radically flawed, for this reason, as I have explained, and as you > have continued to ignore, having nothing to say in response. I have met von Neumann, and studied a number of his works, and find then uniformly of immensely better quality that anything that TO has produced here. If TO ever has a record of accomplishments anywhere near as impressive as von Neumann's, only then will his criticisms of von Neumann be anything but sour grapes. === Subject: Re: Well Ordering the Reals Cantor's First Proof of Uncountability: > Suppose X = { x_n } is a sequence of all the real numbers for all i in > N, with each x_n unique. Let a_0 = x_i and b_0 = x_j where i is the > smallest integer such that x_i is in (a,b) and j is the smallest > integer such that x_j is in (a_0, b). Now for each n>=0 construct > (a_n+1,b_n+1) such that a_n+1 = x_i where i is the smallest integer > such that a_n such that a_n+1 So we have a_n's monotonically increasing and the b_n's monotonically > decreasing with all the a_n's less than all the b_n's. Since these are > open intervals, this process continues infinitely. Now there are one > of two possibilities: Either lim |b_n - a_n| = 0 or lim |b_n - a_n| > 0. If lim |b_n - a_n| > 0, then there are an infinite number of values > not in X and we are done. If lim |b_n - a_n| = 0, then essesntially > the a_n's and b_n's form a Dedekind cut for some real number c so that > lim a_n = lim b_n = c. Since each (a_n,b_n) is an open interval, we > know that c is not found in any previous a_n or b_n. QED. > When you talk about the limit, you mean as n goes to oo. Now, what does it > mean > if lim(n->oo: a_n)=c=lim(n->oo: b_n)? > lim(n->oo: a_n)=c means that for every positive real number epsilon, > {n : |a_n - c| < epsilon} contains all but finitely many n. > Similarly for lim(n->oo: b_n)=c. > Do you even know what the topic is? I didn't ask what a limit meant. > It means it takes an infinite number of > steps to converge to this intermediate value. > This is circular, defining /converging/ in terms of /converging/. > I am not defining converging. I am talking about the uncountability of the > reals ala Cantor's first proof thereof. > I am sorry, but von Neumann ordinals, while a cute construction, are > a misguided notion when it comes to dealing with oo. > So somehow adding one element to an infinite set of elements destroyes > everything? > No, defining the set of finites as countably infinite, and the set of countable > infinities as uncountably infinite, is misguided. You seem to recall that I > said it's an error of 1. That's good. You're not entirely brain-dead. When you > count from 1, you see easily that every natural is in the position denoted by > itself, so the nth natural is n. This clearly shows that the set cannot be > infinite, because n would then be an infinite position in the set, and also an > infinite value in the set, which you do not allow. If all values are finite, > then so are all positions, and so is the size of the set. Now, if you start at > 0, then you say there are always more elements than the largest value in the > set, since the nth element is n-1. This leads to the notion that the size of > the set of finite naturals is larger than all finite naturals, and you say it's > infinite. But, how much larger is it? It's always 1 larger. So, if you think > that a value that is 1 larger than the largest value is infinite, when all > values are finite, then you must think that adding 1 to a finite can yield an > infinite. Clearly, the addition of one element does not turn a finite set into > an infinite one, and the addition of a single element should not change your > conclusion about the size of the overall set from finite to infinite. The logic > behind the von Neumann ordinals is radically flawed, for this reason, as I have > explained, and as you have continued to ignore, having nothing to say in > response. > -- > Smiles, > Tony Here's a shout out to Tim James, the Pyro Korean. Well, you can say an uncountable set is a countable union of countable sets. There's ample agreement in the literature with that to say what you want about it. Of course, knowing where that is actually involves learning those things. So, it is possible to appear uninformed arguing uninformedly either way. Ross === Subject: Re: Well Ordering the Reals Here's a shout out to Tim James, the Pyro Korean. >> Well, you can say an uncountable set is a countable union of >> countable sets. There's ample agreement in the literature with >> that to say what you want about it. I know I am not the person you are directing this to, but just a correction here: a countable union of countable sets is still countable. Instead of union, do you perhaps mean product? In that case, a finite product of countable sets is countable, but a denumerable product of countable sets can be uncountable. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals Here's a shout out to Tim James, the Pyro Korean. >> Well, you can say an uncountable set is a countable union of >> countable sets. There's ample agreement in the literature with >> that to say what you want about it. > I know I am not the person you are directing this to, but just a > correction here: a countable union of countable sets is still > countable. Instead of union, do you perhaps mean product? In that > case, a finite product of countable sets is countable, but a > denumerable product of countable sets can be uncountable. > Jonathan Hoyle > Eastman Kodak I have a science fiction idea. Instead of traveling light years to distant solar systems, teleport some around here. Say, half a light-year away. Happy new year. You know what, I've read more science fiction than, uh, anybody I know. Algebra can be kind of boring when you're reading a book about the entire ing cosmos. I thought that up from figuring out how to turn out the Sun like a light. I haven't figured it out, but I got to thinking about it. Heh heh: fast..., slow. Blinkin' Christmas lights. Luxeon, which is an Agilent/Phillips production, just dropped the price of power LEDs 75%. The LEDs, they are current devices. It's good to have current limiting power supplies on current devices, because they dissipate power, into say, light in the case of these Light Emitting Diodes. A laser runs through an optical crystal and the power supply pumps the laser diode, or other similar device. You can just use a resistor, but as the current actually wants to flow through the current device, the power is dissipated as heat in the resistor, as current is actually trying to run through the current device as current increases as its reciprocal the resistance in the current device, a device that's based on electrical current, has forward bias on the diode junction, and its capability to autodestruct given enough current requires not allowing the current in the diode exceeding specification. The LED is not the Laser Diode. It is similar. Overdrive an LED relatively gently and it'll briefly emit colours rather higher in the spectrum than the one it's meant to produce, and then die, with a very small sad sizzling noise. The thing you are left with after this may be referred to as a friode. -- http://www.dansdata.com/caselight.htm There are some wavelength spectrums of light that affect mitochondria which exist in lots of cells in animals and vegetables. Photobiology is studying how the light directly affects the physical process of the mitochondria and associated machinery, which is mind-boggling. So, to avoid mind-boggling current to the LED, the circuit design involves devices besides the LED that don't obey Ohm's law. Components like basically resistors obey Ohm's law. When you get into other devices then they actually mostly obey Ohm's law, current, voltage and resistance. They just tend to do so in ways that are known so their physical effects can be utilized for things. If the LED is cooled to absolute zero, you can run as much current through it as you want, and more. So maybe a standardized LED casing with a cooling jacket would be useful, but I guess that's what they call high power lasers. For example, if you know a lot of mathematics, you can design a computer circuit that uses electricity. Ross === Subject: Re: Well Ordering the Reals Here's a shout out to Tim James, the Pyro Korean. >> Well, you can say an uncountable set is a countable union of >> countable sets. There's ample agreement in the literature with >> that to say what you want about it. I know I am not the person you are directing this to, but just a correction here: a countable union of countable sets is still countable. Instead of union, do you perhaps mean product? In that case, a finite product of countable sets is countable, but a denumerable product of countable sets can be uncountable. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals David R Tribble said: > David R Tribble said: >> Your alleged mapping between !N and P(!N) uses all of the members >> of !N to map to the subsets of !N containing only finite naturals. >> Your mapping is incomplete, because there are no more members >> in !N left to map to the remaining unmapped members in P(!N). >> In that case, you will have used all of the members of !N (finite and >> infinite) to map to some of the subsets of !N, some containing finite >> naturals and some containing infinite naturals. >> But you're still going to have a boatload of subsets that don't get >> mapped by any naturals from !N. Specifically, the subset !N itself >> (our old favorite) will not get mapped because there is no natural >> left in !N that isn't already mapped to some other subset, and >> because there is no natural x in !N with enough bits (that's right, >> with not enough bits) to map to every natural in subset !N including >> x itself. > If .....11111, an uncountably infinite unending string of bits, both denotes > the entire set !N and its largest element, then how many bits does each take, > so that it breaks? > That's the problem. Using your mapping, there can be no element in !N > that simultaneously is the largest element x of !N and that maps to the > entire set !N. > For this to be true, x would have to have a length as long as its own > value. The value of x is mapped by the bit in x at position log(x), > but since x must be the largest element in !N in order to map to the > entire set !N, there cannot be a log(x)+1 bit that maps a number > larger than x, so log(x) must also be the largest number, so > log(x) = x. This is clearly impossible for any x > 1, no matter how > many bits wide x is. > I've given this proof several times, and you have never shown that > you comprehend it. I fully comprehend that point, and that is the point I am making through the description of this bijection between !N and P(!N). YOU are the one that fails to grasp the significance of what you yourself are saying. Yes, the number of bits matters, and this is not only true for a bijection with the power set, but for any bijection between sets, which your standard theory willfully ignores. When you map x to 2x and claim the evens are equinumerous with the naturals, does it not occur to you that every time you double a value the string requires one more bit? When you listed all the naturals, and they required ALL the finite bit positions, where does this extra bit position come from? Is there another finite bit position that you weren't using before, which suddenly becomes available? If you were originally using all finite bit positions to list all the finite naturals, then are you using some first infinite bit position? Were you NOT using all finite bit positions? What is your answer to this question? When you map x to x^2, and say the number of squares of naturals is equinumerous to the set of naturals, are you forgetting that squaring a number causes it to require twice as many bits? Were you using all finite bit positions to list all the naturals? If so, did the number of finite bit positions suddenly double? Where does your other aleph_0 bits come from? You pretend to have some number, aleph_0, which represents the number of finite naturals. Does this number grow or shrink by 1, or by an order of magnitude, at will or whim? You claim to understand the problem when it comes to the power set, even though this bijection has no identifiable point of failure. Well, your other bijections have no identifiable point of failure, and yet, they suffer from the same malady. When you start being as careful with your other power set of an uncountably infinite set, then you will start getting a handle on why the T-riffics and the declaration of N as a potentially infinite variable are required for the proper treatment of such infinite sets. Indeed, rather that being unable to grasp what you are saying, I am leading this part of the discussion, in order to nail down for you just how crucial what you say above really is, not just for the power set bijection, but for every bijection where your theory willfully ignores this very aspect of the mathematics. That is my point, as I have stated multiple times. You do, of course, recall my repeatedly saying that I AGREE that the power set is always larger than the root set. What you do NOT seem to recall is that my point is that the existence of a bijection alone is not sufficient to declare two infinite sets the same size. Got it? Finally? Doubt it. Count your bits. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > David R Tribble said: > David R Tribble said: >> Your alleged mapping between !N and P(!N) uses all of the members >> of !N to map to the subsets of !N containing only finite naturals. >> Your mapping is incomplete, because there are no more members >> in !N left to map to the remaining unmapped members in P(!N). >> In that case, you will have used all of the members of !N (finite and >> infinite) to map to some of the subsets of !N, some containing finite >> naturals and some containing infinite naturals. >> But you're still going to have a boatload of subsets that don't get >> mapped by any naturals from !N. Specifically, the subset !N itself >> (our old favorite) will not get mapped because there is no natural >> left in !N that isn't already mapped to some other subset, and >> because there is no natural x in !N with enough bits (that's right, >> with not enough bits) to map to every natural in subset !N including >> x itself. > > If .....11111, an uncountably infinite unending string of bits, both > denotes > the entire set !N and its largest element, then how many bits does each > take, > so that it breaks? > > That's the problem. Using your mapping, there can be no element in !N > that simultaneously is the largest element x of !N and that maps to the > entire set !N. > > For this to be true, x would have to have a length as long as its own > value. The value of x is mapped by the bit in x at position log(x), > but since x must be the largest element in !N in order to map to the > entire set !N, there cannot be a log(x)+1 bit that maps a number > larger than x, so log(x) must also be the largest number, so > log(x) = x. This is clearly impossible for any x > 1, no matter how > many bits wide x is. > > I've given this proof several times, and you have never shown that > you comprehend it. > > > I fully comprehend that point, But then > When you map x to 2x and claim the evens are equinumerous with the naturals, > does it not occur to you that every time you double a value the string > requires > one more bit? Since no 'strings' are required in construction of the naturals, the number of bits is irrelevant. But even in the case of binary representation, there is a bijection between the strings representing x's and those representing 2*x's, and that is sufficient to show that they are of equal cardinality. > When you listed all the naturals, and they required ALL the > finite bit positions, where does this extra bit position come from? TO is assuming, contrary to fact, that there must be some natural that uses all the bit positions in its representation. Since each natural uses only finitely many places, there is always another finite place available for each natural. > Is there > another finite bit position that you weren't using before, which suddenly > becomes available? Assumes conditions contrary to fact. > If you were originally using all finite bit positions to > list all the finite naturals, then are you using some first infinite bit > position? No! > Were you NOT using all finite bit positions? No single standard natural requires more than a finite number of bit positions for its binary representation, so for each natural, one can doulbe the value of its binary representation by merely appending 0 to the binary string representing it. > What is your answer to > this question? See above! Does TO claim that there is any finite natural whose binary representation is /not/ doubles by appending a 0 to it? > When you map x to x^2, and say the number of squares of naturals is > equinumerous to the set of naturals, are you forgetting that squaring a > number > causes it to require twice as many bits? 0^2 = 0 and 1^2 = 1, 10^2 = 100, etc., so we are 'forgetting' only what is not true. > Were you using all finite bit > positions to list all the naturals? A list of all finite naturals will use all bit positions, but no single finite natural needs more that a finite number of them. If we consider the set of all one ended binary bit-strings (with the end digit at the right end) then the set of finite integers are represented by the strings with only finitely many 1's, or, equivalently, with a left most 1 in them. > If so, did the number of finite bit > positions suddenly double? No! There is no finite natural whose square is not also a finite natural requiring only a finite number of bits. > Where does your other aleph_0 bits come from? Where does any such requirement spring from? It does not occur in standard arithmetic. > You pretend to have some number, aleph_0, which represents the number of > finite > naturals. If one is to allow the set of finite naturals to have a number, one must call it something. Would you prefer that we call it George? > Does this number grow or shrink by 1, or by an order of magnitude, > at > will or whim? Not in standard set theory. What idiocies go on in TO land is not our responsibility. > You claim to understand the problem when it comes to the power > set, even though this bijection has no identifiable point of failure. For any function, f, from any set S to its power set P(S), the set {x in S: x not in f(x)} in P(S) is not in the image of f, so that no such function can be a surjection, much less a bijection. So that when TO claims to have such a bijection, it must be that his S is a proper class and not a set at all. > Well, > your other bijections have no identifiable point of failure, and yet, they > suffer from the same malady. When you start being as careful with your other > power set of an uncountably infinite set, then you will start getting a > handle > on why the T-riffics and the declaration of N as a potentially infinite > variable are required for the proper treatment of such infinite sets. Such propriety lies only in the eye of the willfully blind. > Indeed, rather that being unable to grasp what you are saying, I am leading > this part of the discussion You may think you lead, but you have a dearth of followers. > , in order to nail down for you just how crucial > what you say above really is, not just for the power set bijection, but for > every bijection where your theory willfully ignores this very aspect of the > mathematics. That is my point, as I have stated multiple times. You do, of > course, recall my repeatedly saying that I AGREE that the power set is always > larger than the root set. What you do NOT seem to recall is that my point is > that the existence of a bijection alone is not sufficient to declare two > infinite sets the same size. Got it? Finally? Doubt it. Count your bits. Since the /size/ measurement by which P(S) is adjudged larger than S is only that of the Cantor cardinality, nothing else is relevant. === Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: > David R Tribble said: >> Your alleged mapping between !N and P(!N) uses all of the members >> of !N to map to the subsets of !N containing only finite naturals. >> Your mapping is incomplete, because there are no more members >> in !N left to map to the remaining unmapped members in P(!N). >> In that case, you will have used all of the members of !N (finite and >> infinite) to map to some of the subsets of !N, some containing finite >> naturals and some containing infinite naturals. >> But you're still going to have a boatload of subsets that don't get >> mapped by any naturals from !N. Specifically, the subset !N itself >> (our old favorite) will not get mapped because there is no natural >> left in !N that isn't already mapped to some other subset, and >> because there is no natural x in !N with enough bits (that's right, >> with not enough bits) to map to every natural in subset !N including >> x itself. > > If .....11111, an uncountably infinite unending string of bits, both > denotes > the entire set !N and its largest element, then how many bits does each > take, > so that it breaks? > > That's the problem. Using your mapping, there can be no element in !N > that simultaneously is the largest element x of !N and that maps to the > entire set !N. > > For this to be true, x would have to have a length as long as its own > value. The value of x is mapped by the bit in x at position log(x), > but since x must be the largest element in !N in order to map to the > entire set !N, there cannot be a log(x)+1 bit that maps a number > larger than x, so log(x) must also be the largest number, so > log(x) = x. This is clearly impossible for any x > 1, no matter how > many bits wide x is. > > I've given this proof several times, and you have never shown that > you comprehend it. > > > > I fully comprehend that point, > But then > > When you map x to 2x and claim the evens are equinumerous with the naturals, > does it not occur to you that every time you double a value the string > requires > one more bit? > Since no 'strings' are required in construction of the naturals, the > number of bits is irrelevant. But even in the case of binary > representation, there is a bijection between the strings representing > x's and those representing 2*x's, and that is sufficient to show that > they are of equal cardinality. There is likewise a bijection between the strings in !N and those in P(!N). If you refute this, do so with a specific example. > When you listed all the naturals, and they required ALL the > finite bit positions, where does this extra bit position come from? > TO is assuming, contrary to fact, that there must be some natural that > uses all the bit positions in its representation. Since each natural > uses only finitely many places, there is always another finite place > available for each natural. No, I assume that if you are listing all the naturals with bits in finite positions, then you have used all the finite positions, otherwise you have not listed all the finite naturals to begin with. Do you or do you not have a complete set using the complete set of bits? > Is there > another finite bit position that you weren't using before, which suddenly > becomes available? > Assumes conditions contrary to fact. Such as? The fact this follows from is the fact that all naturals using all finite bit positions is equal to the set of evens, which uses all finite bit positions, except the first. You seem to be of the opinion that the first bit is insignificant, but that's really an adjective pertaining to that opinion. > If you were originally using all finite bit positions to > list all the finite naturals, then are you using some first infinite bit > position? > No! No, of course not. You are only using finite bit positions, every single one ...... except the first. > Were you NOT using all finite bit positions? > No single standard natural requires more than a finite number of bit > positions for its binary representation, so for each natural, one can > doulbe the value of its binary representation by merely appending 0 to > the binary string representing it. So, indeed, you have added one bit, shifting all others left by one position. So, were there unused finite positions to the left of the string? Could you have had more to begin with, and if so, did you really have all the finite naturals to begin with? You must see there is a contradiction in here. Look again at David's objection to the bijection between !N and P(!N). Do you agree with his reasoning? > What is your answer to > this question? > See above! Does TO claim that there is any finite natural whose binary > representation is /not/ doubles by appending a 0 to it? Of course not, given any particular finite number. Shift all values to the left one bit, and the value doubles. But, that's not the question. Is there a fixed set of aleph_0 bits in the set of all naturals, or can we add and remove bits, while still calling it the complete set? > > When you map x to x^2, and say the number of squares of naturals is > equinumerous to the set of naturals, are you forgetting that squaring a > number > causes it to require twice as many bits? > 0^2 = 0 and 1^2 = 1, 10^2 = 100, etc., so we are 'forgetting' only what > is not true. 0 and 1 are multiplicative hole and identity. The product of two numbers in any base with x and y bits, respectively, has at least x+y-1 bits. Do you actually argue that fact? Does the potential removal of 1 bit change the nature of the argument? If you think so, then you must think this is the first infinite bit, and the preceding the last finite bit, but you don't think that, do you? > Were you using all finite bit > positions to list all the naturals? > A list of all finite naturals will use all bit positions, but no single > finite natural needs more that a finite number of them. If we consider > the set of all one ended binary bit-strings (with the end digit at the > right end) then the set of finite integers are represented by the > strings with only finitely many 1's, or, equivalently, with a left most > 1 in them. As long as no 1 is in an infinite bit position. Now, you have the set of allfinite naturals in binary form. Can you still add another bit? If so, didn't you just double the number of finite naturals? > If so, did the number of finite bit > positions suddenly double? > No! There is no finite natural whose square is not also a finite natural > requiring only a finite number of bits. No kidding, but squaring a number means it has twice as many bits, perhaps minus 1 bit. Were you using only half of the complete set of finite bit positions to begin with? > Where does your other aleph_0 bits come from? > Where does any such requirement spring from? It does not occur in > standard arithmetic. It most certainly does, in any base. A number of base b with n digits is at least equal to b^(n-1), so its square is at least equal to b^(2n-2), and therefore has at least 2n-1 digits. That's a fact of standard arithmetic. Do you disagree? > > You pretend to have some number, aleph_0, which represents the number of > finite > naturals. > If one is to allow the set of finite naturals to have a number, one must > call it something. Would you prefer that we call it George? I would prefer to acknowledge that it's not a fixed number that can be treated like a completed quantity. > Does this number grow or shrink by 1, or by an order of magnitude, > at > will or whim? > Not in standard set theory. What idiocies go on in TO land is not our > responsibility. Actually, this IS what standard transfinite set theory does, which I am trying to address, and which I avoid in my theory. > You claim to understand the problem when it comes to the power > set, even though this bijection has no identifiable point of failure. > For any function, f, from any set S to its power set P(S), the set > {x in S: x not in f(x)} in P(S) is not in the image of f, so that no > such function can be a surjection, much less a bijection. Yes, that is a last element argument, which boils down in the case of !N with whatever the last element is, which doesn't exist. In the case of the power set, such a construction can be derived. In other cases it is not possible to construct such a description for obvious reasons, and yet, the same reasoning put forth by David applies nonetheless. > So that when TO claims to have such a bijection, it must be that his S > is a proper class and not a set at all. Well, as far as I can tell, classes are an outgrowth of the inherent instability of transfinite set theory due to the introduction of the von Neumann ordinals and misuse of induction. > Well, > your other bijections have no identifiable point of failure, and yet, they > suffer from the same malady. When you start being as careful with your other > power set of an uncountably infinite set, then you will start getting a > handle > on why the T-riffics and the declaration of N as a potentially infinite > variable are required for the proper treatment of such infinite sets. > Such propriety lies only in the eye of the willfully blind. You should know. You should see what you have been stepping in. > > Indeed, rather that being unable to grasp what you are saying, I am leading > this part of the discussion > You may think you lead, but you have a dearth of followers. Well, you are a dearth, that's true, but you seem to follow me all the time. > , in order to nail down for you just how crucial > what you say above really is, not just for the power set bijection, but for > every bijection where your theory willfully ignores this very aspect of the > mathematics. That is my point, as I have stated multiple times. You do, of > course, recall my repeatedly saying that I AGREE that the power set is always > larger than the root set. What you do NOT seem to recall is that my point is > that the existence of a bijection alone is not sufficient to declare two > infinite sets the same size. Got it? Finally? Doubt it. Count your bits. > > Since the /size/ measurement by which P(S) is adjudged larger than S is > only that of the Cantor cardinality, nothing else is relevant. Buit, it is not. The relationship between any set and its power set is exactly described by 2^n due to combinatorics. When applying this to N, and saying the power set of the naturals has size 2^N, this is not figurative, but a precise and correct statement. Likewise, N^2 is also larger than N, as is 2N, and N+1. This is the approach I am advocating, and your refusal to even consider the idea is a real mistake, but not my loss. It's never too late, you know.... -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > When you map x to 2x and claim the evens are equinumerous with > the naturals, does it not occur to you that every time you double > a value the string requires one more bit? > > Since no 'strings' are required in construction of the naturals, > the number of bits is irrelevant. But even in the case of binary > representation, there is a bijection between the strings > representing x's and those representing 2*x's, and that is > sufficient to show that they are of equal cardinality. > There is likewise a bijection between the strings in !N and those in > P(!N). If you refute this, do so with a specific example. Assuming any f: !N -> P(!N), there is no x in !N for which f(x) = {y in !N: y not in f(y)}. Thus, for every f: !N -> P(!N), f is NOT even a surjection, and thus not a bijection either. > > When you listed all the naturals, and they required ALL the > finite bit positions, where does this extra bit position come > from? > > TO is assuming, contrary to fact, that there must be some natural > that uses all the bit positions in its representation. Since each > natural uses only finitely many places, there is always another > finite place available for each natural. > No, I assume that if you are listing all the naturals with bits in > finite positions, then you have used all the finite positions, They do not wear out. One can use them again for other naturals, and since no one natural uses them all, there is always another bit position to use for its successor or its double or whatever. > otherwise you have not listed all the finite naturals to begin with. To seems not to understand that digit positions do not get consumed by use. > Do you or do you not have a complete set using the complete set of > bits? We have a complete and infinite set of finite naturals in which each natural uses only finitely many of infinitely many available bits. What TO has is self delusion. > > Is there another finite bit position that you weren't using > before, which suddenly becomes available? > > Assumes conditions contrary to fact. > Such as? That any finite set of naturals uses up more that finitely many bits. The next natural can always be constructed still using only finitely many bits. > > > Were you NOT using all finite bit positions? > > No single standard natural requires more than a finite number of > bit positions for its binary representation, so for each natural, > one can double the value of its binary representation by merely > appending 0 to the binary string representing it. > So, indeed, you have added one bit, shifting all others left by one > position. So, were there unused finite positions to the left of the > string? Could you have had more to begin with, and if so, did you > really have all the finite naturals to begin with? You must see there > is a contradiction in here. Look again at David's objection to the > bijection between !N and P(!N). Do you agree with his reasoning? > > > What is your answer to this question? > > See above! Does TO claim that there is any finite natural whose > binary representation is /not/ doubled by appending a 0 to it? > Of course not, given any particular finite number. Shift all values > to the left one bit, and the value doubles. But, that's not the > question. Is there a fixed set of aleph_0 bits in the set of all > naturals, or can we add and remove bits, while still calling it the > complete set? The set of bits is in one to one correspondences with the set of naturals which is in one to one correspondence with the set of even naturals which is in one to one correspondence with a host of other countably infinite sets. All such sets are equally complete. > > When you map x to x^2, and say the number of squares of naturals > is equinumerous to the set of naturals, are you forgetting that > squaring a number causes it to require twice as many bits? > > 0^2 = 0 and 1^2 = 1, 10^2 = 100, etc., so we are 'forgetting' only > what is not true. > 0 and 1 are multiplicative hole and identity. The product of two > numbers in any base with x and y bits, respectively, has at least > x+y-1 bits. Do you actually argue that fact? TO's statement is that the square of a natural requires twice as many bits to express as the naturals itsself. I merely pointed out that TO was, as usual, wrong! > Does the potential > removal of 1 bit change the nature of the argument? It means that TO's statement is false, rather than true, which in a /mathematical/ discussion is of some importance, how ever little importance TO may give it. > > > Were you using all finite bit positions to list all the naturals? > > A list of all finite naturals will use all bit positions, but no > single finite natural needs more that a finite number of them. If > we consider the set of all one ended binary bit-strings (with the > end digit at the right end) then the set of finite integers are > represented by the strings with only finitely many 1's, or, > equivalently, with a left most 1 in them. > As long as no 1 is in an infinite bit position. Now, you have the set > of allfinite naturals in binary form. Can you still add another bit? > If so, didn't you just double the number of finite naturals? But adding another bit, whether one appends a '1' or a '0' to all the strings one has, does not create any new strings that are not already present. > > > If so, did the number of finite bit positions suddenly double? > > No! There is no finite natural whose square is not also a finite > natural requiring only a finite number of bits. > No kidding, but squaring a number means it has twice as many bits, > perhaps minus 1 bit. Were you using only half of the complete set of > finite bit positions to begin with? To seems to think he has a point in all this garbage, but never makes it. > > Where does your other aleph_0 bits come from? > > Where does any such requirement spring from? It does not occur in > standard arithmetic. > It most certainly does, in any base. A number of base b with n digits > is at least equal to b^(n-1), so its square is at least equal to > b^(2n-2), and therefore has at least 2n-1 digits. That's a fact of > standard arithmetic. Do you disagree? Irrelevancies do not carry much weight. > > You pretend to have some number, aleph_0, which represents the > number of finite naturals. > > > If one is to allow the set of finite naturals to have a number, one > must call it something. Would you prefer that we call it George? > I would prefer to acknowledge that it's not a fixed number that can > be treated like a completed quantity. Depends on what one means by number. If one means cardinality, then the number of finite naturals is a fixed cardinality and is as completed, whatever than means, as any other cardinality. > > > > > Does this number grow or shrink by 1, or by an order of > magnitude, at will or whim? > > Not in standard set theory. What idiocies go on in TO land is not > our responsibility. > Actually, this IS what standard transfinite set theory does, which I > am trying to address, and which I avoid in my theory. Does TO insist that the set of naturals cannot have the same cardinality as some of its proper subsets, such as, for example, the set of even naturals? > > You claim to understand the problem when it comes to the power > set, even though this bijection has no identifiable point of > failure. > > For any function, f, from any set S to its power set P(S), the set > {x in S: x not in f(x)} in P(S) is not in the image of f, so that > no such function can be a surjection, much less a bijection. > Yes, that is a last element argument WRONG! Nowhere is there any allusion to any last element in this proof of TO's errors. > which boils down in the case of > !N with whatever the last element is, which doesn't exist. Where is there any reference, even indirectly, to any last element in the proof that f: s -> P(S) cannot be a surjection? > In the > case of the power set, such a construction can be derived. In other > cases it is not possible to construct such a description for obvious > reasons, and yet, the same reasoning put forth by David applies > nonetheless. > > So that when TO claims to have such a bijection, it must be that > his S is a proper class and not a set at all. > Well, as far as I can tell, classes are an outgrowth of the inherent > instability of transfinite set theory due to the introduction of the > von Neumann ordinals and misuse of induction. So that one sees that TO cannot tell much. Proper classes are one response to the problems posed by Russell's paradox. And TO is the only one here misusing anything. > > > Well, your other bijections have no identifiable point of > failure, and yet, they suffer from the same malady. When you > start being as careful with your other bijections, in the manner > uncountably infinite set, then you will start getting a handle on > why the T-riffics and the declaration of N as a potentially > infinite variable are required for the proper treatment of such > infinite sets. > > Such propriety lies only in the eye of the willfully blind. > You should know. You should see what you have been stepping in. It is TO that has been producing all of it. And I have good boots. > > Indeed, rather that being unable to grasp what you are saying, I > am leading this part of the discussion > > You may think you lead, but you have a dearth of followers. > Well, you are a dearth, that's true, but you seem to follow me all > the time. On the contrary, following TO is quite impossible for those who give anything more than lip service to logic > > > , in order to nail down for you just how crucial what you say > above really is, not just for the power set bijection, but for > every bijection where your theory willfully ignores this very > aspect of the mathematics. That is my point, as I have stated > multiple times. You do, of course, recall my repeatedly saying > that I AGREE that the power set is always larger than the root > set. What you do NOT seem to recall is that my point is that the > existence of a bijection alone is not sufficient to declare two > infinite sets the same size. Got it? Finally? Doubt it. Count > your bits. > > Since the /size/ measurement by which P(S) is adjudged larger than > S is only that of the Cantor cardinality, nothing else is relevant. > Buit, it is not. The relationship between any set and its power set > is exactly described by 2^n due to combinatorics. When applying this > to N, and saying the power set of the naturals has size 2^N, this is > not figurative, but a precise and correct statement. Likewise, N^2 is > also larger than N, as is 2N, and N+1. This is the approach I am > advocating, and your refusal to even consider the idea is a real > mistake, but not my loss. It's never too late, you know.... It seems to be for TO. TO seems by now to have become invincibly ignorant on all issues of logic and mathematics. === Subject: Re: Well Ordering the Reals David R Tribble said: > David R Tribble said: >> [...] An infinite sequence has no end, so at first >> glance it's hard to see what an infinite position might be. But I >> comprehend the infinite ordinals, so I can, in fact, visualize an >> infinite ordinal index followed by more infinite ordinal indices. >> A simple example is the set of naturals, reordered so that all the >> odds follow all of the evens: >> N2 = { 0, 2, 4, 6, ..., 1, 3, 5, 7, ...} > In your mind that's infinite, but it's still not, so you really hve no concept > of infinity as far as I can tell. > For the nth time, if it's not an infinite set/sequence, where does it > stop? For what finite value of L (for your N = S^L) does it end? > Do you have an answer to that or not? For the 2^nth time, I have already answered this. Of course there is no identifiable end to the finite naturals, but that doesn't mean there are an infinite NUMBER of them. Any infinite number of naturals will include infinite values, since any two naturals, a and b, with an infinite number of naturals between, them will differ by an infinite value, meaning that |a-b| is infinite, so that either a or b (or both) is infinite. If all values in the set are finite, then the difference between any pair of them is finite, and there are only an infinite number of intermediate values between any two. You have a finite set, there, pal. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > David R Tribble said: > David R Tribble said: >> [...] An infinite sequence has no end, so at first glance it's >> hard to see what an infinite position might be. But I >> comprehend the infinite ordinals, so I can, in fact, visualize >> an infinite ordinal index followed by more infinite ordinal >> indices. >> A simple example is the set of naturals, reordered so that all >> the odds follow all of the evens: >> N2 = { 0, 2, 4, 6, ..., 1, 3, 5, 7, ...} > > In your mind that's infinite, but it's still not, so you really > hve no concept of infinity as far as I can tell. > > For the nth time, if it's not an infinite set/sequence, where does > it stop? For what finite value of L (for your N = S^L) does it > end? Do you have an answer to that or not? > > > For the 2^nth time, I have already answered this. Of course there is > no identifiable end to the finite naturals, but that doesn't mean > there are an infinite NUMBER of them. If the set of finite naturals doesn't end, then it is Dedekind infinite, which is the only infinite that matters to anyone except TO. > Any infinite number of naturals > will include infinite values, Only in TOmatics. In the real world it doesn't happen that way. > since any two naturals, a and b, with > an infinite number of naturals between But TO can't identify any such pair of natural numbers, since there is never more than a finite difference between any two of them. > If all values in the set are finite, then the > difference between any pair of them is finite, and there are only an > infinite number of intermediate values between any two. Now that is stupid even for TO to say. Between any two finite naturals there are only /finitely/ many other naturals, even in TOmatics. === Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: > David R Tribble said: >> [...] An infinite sequence has no end, so at first glance it's >> hard to see what an infinite position might be. But I >> comprehend the infinite ordinals, so I can, in fact, visualize >> an infinite ordinal index followed by more infinite ordinal >> indices. >> A simple example is the set of naturals, reordered so that all >> the odds follow all of the evens: >> N2 = { 0, 2, 4, 6, ..., 1, 3, 5, 7, ...} > > In your mind that's infinite, but it's still not, so you really > hve no concept of infinity as far as I can tell. > > For the nth time, if it's not an infinite set/sequence, where does > it stop? For what finite value of L (for your N = S^L) does it > end? Do you have an answer to that or not? > > > For the 2^nth time, I have already answered this. Of course there is > no identifiable end to the finite naturals, but that doesn't mean > there are an infinite NUMBER of them. > If the set of finite naturals doesn't end, then it is Dedekind infinite, > which is the only infinite that matters to anyone except TO. > Any infinite number of naturals > will include infinite values, > Only in TOmatics. In the real world it doesn't happen that way. > since any two naturals, a and b, with > an infinite number of naturals between > But TO can't identify any such pair of natural numbers, since there is > never more than a finite difference between any two of them. > If all values in the set are finite, then the > difference between any pair of them is finite, and there are only an > infinite number of intermediate values between any two. > Now that is stupid even for TO to say. Between any two finite naturals > there are only /finitely/ many other naturals, even in TOmatics. That was a typo, but you seem to have gotten the point, only because I stated the opposite, Mary Mary Quite Contrary. So, if there are only finitely many naturals between any two naturals, how are there an overall infinite number. Are there naturals outside all other naturals, between which those other naturals do not reside? Virgil, you must admit, this at least sounds contradictory. Is there no point in trying to resolve this logically, even for the sake of argument? Or, is argument conducted for its own sake? Art for art's sake is dreck. Math for math's sake is no better. For god's sake, ask some new questions. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > > David R Tribble said: > David R Tribble said: >> [...] An infinite sequence has no end, so at first glance it's >> hard to see what an infinite position might be. But I >> comprehend the infinite ordinals, so I can, in fact, visualize >> an infinite ordinal index followed by more infinite ordinal >> indices. >> A simple example is the set of naturals, reordered so that all >> the odds follow all of the evens: >> N2 = { 0, 2, 4, 6, ..., 1, 3, 5, 7, ...} > > In your mind that's infinite, but it's still not, so you really > hve no concept of infinity as far as I can tell. > > For the nth time, if it's not an infinite set/sequence, where does > it stop? For what finite value of L (for your N = S^L) does it > end? Do you have an answer to that or not? > > > For the 2^nth time, I have already answered this. Of course there is > no identifiable end to the finite naturals, but that doesn't mean > there are an infinite NUMBER of them. > > If the set of finite naturals doesn't end, then it is Dedekind infinite, > which is the only infinite that matters to anyone except TO. > > > > Any infinite number of naturals > will include infinite values, > > Only in TOmatics. In the real world it doesn't happen that way. > > > since any two naturals, a and b, with > an infinite number of naturals between > > But TO can't identify any such pair of natural numbers, since there is > never more than a finite difference between any two of them. > > > If all values in the set are finite, then the > difference between any pair of them is finite, and there are only an > infinite number of intermediate values between any two. > > Now that is stupid even for TO to say. Between any two finite naturals > there are only /finitely/ many other naturals, even in TOmatics. > > That was a typo, but you seem to have gotten the point, only because I stated > the opposite, Mary Mary Quite Contrary. So, if there are only finitely many > naturals between any two naturals, how are there an overall infinite number. TO seems to be assuming that there is an interval of naturals that contains all naturals. That requires both a least and a largest. But if there is no largest, then there are always more beyond any so far listed, and that is where the always another keeps occuring. > Are there naturals outside all other naturals, between which those other > naturals do not reside? Is TO asserting that there is an interval of naturals in which all naturals must reside? That would require the existence of a largest one, as well as a smallest one, for which there could be no successor. But every natural must have a successor. > Virgil, you must admit, this at least sounds > contradictory. Your arguments, certainly do! > Is there no point in trying to resolve this logically, even > for > the sake of argument? Or, is argument conducted for its own sake? I merely state what one can derive from the axioms of one or another of the set theories in general use, and object to TO's requiring properties contradictory to those axioms to be dedcible from them. > Art for art's sake is dreck. Math for math's sake is no better. For > god's sake, ask some new questions. If TO cannot bear math for math's sake, he should stop posting his dreck in sci.math, and if he cannot bear logic for logic's sake, he should stop posting in sci.logic. And why should I bother with new questions when I have not received satisfactory answers to my old ones? === Subject: Re: Well Ordering the Reals David R Tribble said: >> and yet, you refuse to entertain >> infinite bit strings because you have no concept of an infinite position in a >> sequence. That's not my shortcoming, but yours. > David R Tribble said: >> Quite the contrary. An infinite sequence has no end, so at first >> glance it's hard to see what an infinite position might be. But I >> comprehend the infinite ordinals, so I can, in fact, visualize an >> infinite ordinal index followed by more infinite ordinal indices. > Well, goody! Then what is your problem with my uncountably infinite bit > strings, and why do you keep inisting that all bit positions are finite? > Okay then, you have infinite naturals composed of an uncountable > infinite number of bits. Now, was that so hard? > So now tell us, how do you add 1 to one of those infinite numbers? You find the rightmost 0 to the left of the digital point, and invert all bits from that bit to the 0 bit (1's place), just like in regular numbers. Duh. > How do you even write one of those numbers? After all, a notation > like ...111 does not suffice, because it implies a countable number > of digits. That implies an infinite number of bits, not necessarily countable or uncountable, and that is adic notation, not T-riffic. I can place the limit points of my T-riffics wherever I please, at aleph_0, aleph_1, c, 2^N, n^2, or even a million, and the T-riffics continue to function without flaw. Let's put our limit point at 2^N, generally agreed to be an uncountable number, and we get 1:000...000, or 2^(2^N). What's so hard about that? -- Smiles, Tony === Subject: Re: Well Ordering the Reals > David R Tribble said: >> and yet, you refuse to entertain infinite bit strings because >> you have no concept of an infinite position in a sequence. >> That's not my shortcoming, but yours. > > David R Tribble said: >> Quite the contrary. An infinite sequence has no end, so at >> first glance it's hard to see what an infinite position might >> be. But I comprehend the infinite ordinals, so I can, in fact, >> visualize an infinite ordinal index followed by more infinite >> ordinal indices. > > Well, goody! Then what is your problem with my uncountably > infinite bit strings, and why do you keep inisting that all bit > positions are finite? > > Okay then, you have infinite naturals composed of an uncountable > infinite number of bits. > > So now tell us, how do you add 1 to one of those infinite numbers? > You find the rightmost 0 to the left of the digital point, and invert > all bits from that bit to the 0 bit (1's place), just like in > regular numbers. Duh. That assumes that in this uncountable mess there is such a rightmost 0, which is not at all obvious. And it ignores, among many other problems, the problem of how to add one to a number which has no 0's in any position but has 1's in all possible positions. === Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: >> and yet, you refuse to entertain infinite bit strings because >> you have no concept of an infinite position in a sequence. >> That's not my shortcoming, but yours. > > David R Tribble said: >> Quite the contrary. An infinite sequence has no end, so at >> first glance it's hard to see what an infinite position might >> be. But I comprehend the infinite ordinals, so I can, in fact, >> visualize an infinite ordinal index followed by more infinite >> ordinal indices. > > Well, goody! Then what is your problem with my uncountably > infinite bit strings, and why do you keep inisting that all bit > positions are finite? > > Okay then, you have infinite naturals composed of an uncountable > infinite number of bits. > > So now tell us, how do you add 1 to one of those infinite numbers? > You find the rightmost 0 to the left of the digital point, and invert > all bits from that bit to the 0 bit (1's place), just like in > regular numbers. Duh. > That assumes that in this uncountable mess there is such a rightmost > 0, which is not at all obvious. And it ignores, among many other > problems, the problem of how to add one to a number which has no 0's in > any position but has 1's in all possible positions. Well, I did address this, although that is not a well-formed T-riffic as I first defined them. Again, treat them like 2's complement. You can either have an unending string of 0's to the left, representing a positive value, in which case there will be a rightmost one which is left of the digital point. The other possibility is that you have an unending string of 1's to the left, which is what you are talking about with your string of all 1's forever. In 2's complement, a 1 in the leftmost bit indicates a negative value. When we invert a digital number to get its negative, we simply invert all bits. In traditional 2's complement, we add a 1 after inverting the bits, because we are dealing with integers which have an unending string of 0's to the right of the digital point, which become all 1's and therefore equal to 0.11111......, or an additional 1. So, your unending string of all 1's (and zeroes to the right of the digital point), is equal to -1. You may confirm this by adding 1 to it, and seeing that you get an infinite string of all 0's. You may also note that by adding 1 in the form of 0.111....., you get ....1111.1111...., which is as equal to 0 as ....000.000..... Where we DO run into problems is with patterns that have both 0's and 1's and repeat forever, in which case we can't know whether they are positive or negative, can't always compare two values, and can't do normal arithmetic on them. These are the problematic p-adics, not the T-riffics. Perhaps you have another objection to the T-riffics? Maybe some thoughts on the H-riffic reals? Nice to see you're still here, old buddy. Hope you had a nice holiday. Mine sucked. ;) -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > > David R Tribble said: >> and yet, you refuse to entertain infinite bit strings >> because you have no concept of an infinite position in a >> sequence. That's not my shortcoming, but yours. > > David R Tribble said: >> Quite the contrary. An infinite sequence has no end, so at >> first glance it's hard to see what an infinite position >> might be. But I comprehend the infinite ordinals, so I can, >> in fact, visualize an infinite ordinal index followed by >> more infinite ordinal indices. > > Well, goody! Then what is your problem with my uncountably > infinite bit strings, and why do you keep inisting that all > bit positions are finite? > > Okay then, you have infinite naturals composed of an > uncountable infinite number of bits. > > So now tell us, how do you add 1 to one of those infinite > numbers? > You find the rightmost 0 to the left of the digital point, and > invert all bits from that bit to the 0 bit (1's place), just like > in regular numbers. Duh. > > That assumes that in this uncountable mess there is such a > rightmost 0, which is not at all obvious. And it ignores, among > many other problems, the problem of how to add one to a number > which has no 0's in any position but has 1's in all possible > positions. > Well, I did address this, although that is not a well-formed T-riffic > as I first defined them. TO's T-erribles are not in any sense /well formed/. Again, treat them like 2's complement. You > can either have an unending string of 0's to the left, representing a > positive value, in which case there will be a rightmost one which is > left of the digital point. This would require that the index set for the digits to the right of the digital point be well ordered but uncountable, thus it must contain something like limit ordinals. But adding one to something with a zero at any limit ordinal one will never generate a carry to that limit ordinal. > The other possibility is that you have an > unending string of 1's to the left, which is what you are talking > about with your string of all 1's forever. In 2's complement, a 1 in > the leftmost bit indicates a negative value. When we invert a digital > number to get its negative, we simply invert all bits. In traditional > 2's complement, we add a 1 after inverting the bits, because we are > dealing with integers which have an unending string of 0's to the > right of the digital point, which become all 1's and therefore equal > to 0.11111......, or an additional 1. So, your unending string of all > 1's (and zeroes to the right of the digital point), is equal to -1. > You may confirm this by adding 1 to it, and seeing that you get an > infinite string of all 0's. You may also note that by adding 1 in the > form of 0.111....., you get ....1111.1111...., which is as equal to 0 > as ....000.000..... Then one has a situation in which x + 1 < x, which is not allowed in naturals numbers. > Where we DO run into problems is with patterns that have both 0's and > 1's and repeat forever, in which case we can't know whether they are > positive or negative, can't always compare two values, and can't do > normal arithmetic on them. These are the problematic p-adics, not the > T-riffics. Does TO say that none of his T-erribles can have both infinitely many 0's and infinitely many 1's? If not how do they avoid the problems he himself points out? And if so, doesn't that make his number system incomplete, so that, for example, the additiona of two of his numbers is not always defineable. > Perhaps you have another objection to the T-riffics? That they require properties that sets cannot have in any standard set theory, such as allowing surjections from things TO claims are sets to what TO claims are the power sets of such sets. At best, TO's sets are proper classes, not sets at all. And, as such, do not have power sets. Maybe some > thoughts on the H-riffic reals? Nice to see you're still here, old > buddy. Hope you had a nice holiday. Mine sucked. ;) === Subject: Re: Well Ordering the Reals David R Tribble said: [...] > No finite step moves from the finite to the infinite, but infinite steps do. > If incrementing finite values eventually creates an infinite value, > then there must be an infinite number of finite values. Otherwise, > how would you get an infinite value? As soon as you increment an infinite number of times, according to infinite series, you have an infinite value. Any sum of a finite number of finite values is finite. Any sum of an infinite number of constant finite values is infinite. That set theory contradicts this obvious fact is at the root of its problems, the fruit being such nonsensical notions as Banach-Tarski, where cutting aball into five pieces causes it comehow to multiply its volume. > You've already halfway admitted it, when you stated that it takes an > infinite number of increments to get an infinite value. So by your > logic, there must be an infinite number of finite values being > incremented to get to that infinite value. Otherwise, how could > you apply an infinite number of increments if there are only a > finite number of finite values to be incremented? By YOUR logic, apparently, since an infinite number of finite increments produces an infinite value, you must have an infinite number of finite naturals, despite the fact that any infinite number of them would necessarily include infinite values, due to the infinite number of incrmeents involved. Since an infinite number of increments procudes an infinite value, a set with all finite values must include only a finite number of increments, otherwise it would include infinite values. Your failure to follow this very simple, non- convoluted logical statement is yours, not mine. > Those kind of > inductive arguments regarding finiteness of the naturals break down at n=oo, > And yet you are the one who keeps trying to treat infinite values as > though they are finite values, obeying all the same rules and acting > like them. They can be treated very similarly, as with the T-riffic numbers, with which no one has found any flaw, despite your repeated attempts to treat them like adic numbers and attribute the shortcomings of the adics to them. The T-riffic numbers do not make claims of finiteness due to inductive arguments that only hold for finite numbers. That's YOUR bag. I suggest you empty it before it backs up into your colon. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > David R Tribble said: No Largest Finite!!!! (GONG!!!) Huyah Huyah Ommmmm........ > (jingle jingle) > [...] No finite step moves from the finite to the infinite, but > infinite steps do. > > If incrementing finite values eventually creates an infinite value, > then there must be an infinite number of finite values. Otherwise, > how would you get an infinite value? > As soon as you increment an infinite number of times, according to > infinite series, you have an infinite value. Any sum of a finite > number of finite values is finite. Any sum of an infinite number of > constant finite values is infinite. Zero is a constant finite value! > That set theory contradicts this > obvious fact is at the root of its problems Set theory does not contradict any obvious facts, but TO often does. > > You've already halfway admitted it, when you stated that it takes > an infinite number of increments to get an infinite value. So by > your logic, there must be an infinite number of finite values being > incremented to get to that infinite value. Otherwise, how could > you apply an infinite number of increments if there are only a > finite number of finite values to be incremented? > By YOUR logic, apparently, since an infinite number of finite > increments produces an infinite value, you must have an infinite > number of finite naturals Quite so. > despite the fact that any infinite number > of them would necessarily include infinite values, If there were any such infinite values, there would have to be a first one, which means that its immediate predecessor would have to be finite. That is a far worse anomaly than anything produced by standard set theories. > Since an infinite number of > increments procudes an infinite value Or an infinite number of different finite values! > a set with all finite values > must include only a finite number of increments, otherwise it would > include infinite values. Your failure to follow this very simple, > non- convoluted logical statement is yours, not mine. But following TO's delusions would mean accepting that there must be a first (smallest) infinite, since every non-empty set of naturals has a first (smallest) member. And that leads to worse chaos than anything that standard set theories produce. > > > Those kind of inductive arguments regarding finiteness of the > naturals break down at n=oo, > > And yet you are the one who keeps trying to treat infinite values > as though they are finite values, obeying all the same rules and > acting like them. > They can be treated very similarly, as with the T-riffic numbers, > with which no one has found any flaw, That TO does not acknowledge any of the manifold flaws in his number systems that have been so clearly and repeatedly pointed out is only evidence of TO's willful blindness, not of any validity in TOmatics . === Subject: Re: Well Ordering the Reals Virgil said: > David R Tribble said: No Largest Finite!!!! (GONG!!!) Huyah Huyah Ommmmm........ > (jingle jingle) > [...] No finite step moves from the finite to the infinite, but > infinite steps do. > > If incrementing finite values eventually creates an infinite value, > then there must be an infinite number of finite values. Otherwise, > how would you get an infinite value? > > As soon as you increment an infinite number of times, according to > infinite series, you have an infinite value. Any sum of a finite > number of finite values is finite. Any sum of an infinite number of > constant finite values is infinite. > Zero is a constant finite value! Hi Virgil. Happy New year. Yes, in general, people consider 0 a finite value, though in my theory, it is not a finite value, but an absolute infinitesimal value. So, indeed, a sum of an infinite number of zeroes could be zero, or some finite value, or possibly infinite. I meant finite in the non-zero sense. Do you agree, now? > That set theory contradicts this > obvious fact is at the root of its problems > Set theory does not contradict any obvious facts, but TO often does. So, adding 1 an infinite number of times to 1 will result in a finite value? That's obviously qrong, making its negation obviously right, in contradiction with the notion that the set of finite naturals contains an infinite number of elements. > > > You've already halfway admitted it, when you stated that it takes > an infinite number of increments to get an infinite value. So by > your logic, there must be an infinite number of finite values being > incremented to get to that infinite value. Otherwise, how could > you apply an infinite number of increments if there are only a > finite number of finite values to be incremented? > > By YOUR logic, apparently, since an infinite number of finite > increments produces an infinite value, you must have an infinite > number of finite naturals > Quite so. Uh, yeah, right, except you claim to have no infinite values in your naturals, remember? > despite the fact that any infinite number > of them would necessarily include infinite values, > If there were any such infinite values, there would have to be a first > one, which means that its immediate predecessor would have to be finite. > That is a far worse anomaly than anything produced by standard set > theories. There is no reason to consider a first infinite number, omega notwithstanding, any more than there is reason to consider a last finite number. Everyone knows the chasm between the finite and infinite cannot be crossed in a single finite step. > Since an infinite number of > increments procudes an infinite value > Or an infinite number of different finite values! Every new natural is generated by incrementing the previous one, so the nth natural is the result of n increments. If n is infinite, so is the nth natural. > a set with all finite values > must include only a finite number of increments, otherwise it would > include infinite values. Your failure to follow this very simple, > non- convoluted logical statement is yours, not mine. > But following TO's delusions would mean accepting that there must be a > first (smallest) infinite, since every non-empty set of naturals has a > first (smallest) member. And that leads to worse chaos than anything > that standard set theories produce. No, the assumption that every non-empty set of naturals has a smallest natural apparently leads to that conclusion. That's why I don't assume that. When two ideas contradict each other it's worthwhile to consider them both, and exactly what their sticking point is. That assumption on your part follows directly from the assumption that all naturals are finite, so of course it contradicts ideas following from the idea of infinite naturals. If you shelve that assumption for a moment, that contradiction disappears. > > > > Those kind of inductive arguments regarding finiteness of the > naturals break down at n=oo, > > And yet you are the one who keeps trying to treat infinite values > as though they are finite values, obeying all the same rules and > acting like them. > > They can be treated very similarly, as with the T-riffic numbers, > with which no one has found any flaw, > That TO does not acknowledge any of the manifold flaws in his number > systems that have been so clearly and repeatedly pointed out is only > evidence of TO's willful blindness, not of any validity in TOmatics . The only flaws pointed out have either been the result of confusing the T- riffics with the adics, or have been answered directly with counterexamples and explanations as to how to perform the operations in question. If you think you have some objection which does not fall into either category, state it now, because you haven't yet. Have a nice day, V. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Virgil said: > > David R Tribble said: > No Largest Finite!!!! (GONG!!!) Huyah Huyah Ommmmm........ > (jingle jingle) > [...] No finite step moves from the finite to the infinite, but > infinite steps do. > > If incrementing finite values eventually creates an infinite value, > then there must be an infinite number of finite values. Otherwise, > how would you get an infinite value? > > As soon as you increment an infinite number of times, according to > infinite series, you have an infinite value. Any sum of a finite > number of finite values is finite. Any sum of an infinite number of > constant finite values is infinite. > > Zero is a constant finite value! > Hi Virgil. Happy New year. > Yes, in general, people consider 0 a finite value, though in my theory, it is > not a finite value, but an absolute infinitesimal value. So, indeed, a sum of > an infinite number of zeroes could be zero, or some finite value, or possibly > infinite. I meant finite in the non-zero sense. Do you agree, now? No! No number of zeros adds to anything but zero, at least in any standard arithmetic.. > > That set theory contradicts this > obvious fact is at the root of its problems > > Set theory does not contradict any obvious facts, but TO often does. > So, adding 1 an infinite number of times to 1 will result in a finite value? One can only add 1 to 1 once, after that one is adding 1 to something else. > That's obviously qrong, making its negation obviously right There is no necessity that adding infinitely many 1's together is even possible, much less meaningful. > in contradiction with the notion that the set of finite naturals > contains an infinite number of elements. The continuing delusion of TO's that the set of finite naturals can be finite without having a last or largest member is the root of all his errors. > > > > > > You've already halfway admitted it, when you stated that it takes > an infinite number of increments to get an infinite value. So by > your logic, there must be an infinite number of finite values being > incremented to get to that infinite value. Otherwise, how could > you apply an infinite number of increments if there are only a > finite number of finite values to be incremented? > > By YOUR logic, apparently, since an infinite number of finite > increments produces an infinite value, you must have an infinite > number of finite naturals > > Quite so. > Uh, yeah, right, except you claim to have no infinite values in your > naturals, > remember? It is only the conclusion that we have infintiely many finite naturals that I was agreeing with, not TO's delusion that there are any infinite finite naturals. > > > despite the fact that any infinite number > of them would necessarily include infinite values, > > If there were any such infinite values, there would have to be a first > one, which means that its immediate predecessor would have to be finite. > That is a far worse anomaly than anything produced by standard set > theories. > There is no reason to consider a first infinite number, omega > notwithstanding, > any more than there is reason to consider a last finite number. Everyone > knows > the chasm between the finite and infinite cannot be crossed in a single > finite > step. Then why does TO keep insisting that a process which involves only single finite steps crosses that chasm? > > Since an infinite number of > increments procudes an infinite value > > Or an infinite number of different finite values! > Every new natural is generated by incrementing the previous one, so the nth > natural is the result of n increments. If n is infinite, so is the nth > natural. But n is never infinite, if it is a natural number, since every natural is generated by a process that involves only a single finite stepat any stage. > > > a set with all finite values > must include only a finite number of increments, otherwise it would > include infinite values. Your failure to follow this very simple, > non- convoluted logical statement is yours, not mine. > > But following TO's delusions would mean accepting that there must be a > first (smallest) infinite, since every non-empty set of naturals has a > first (smallest) member. And that leads to worse chaos than anything > that standard set theories produce. > No, the assumption that every non-empty set of naturals has a smallest > natural > apparently leads to that conclusion. That's why I don't assume that. Then TO is denying that the naturals are well ordered? > When two ideas contradict each other it's worthwhile to consider them > both, and exactly what their sticking point is. WE have done so, and find that TO's ideas at fault. > That assumption on your part follows directly > from the assumption that all naturals are finite, so of course it contradicts > ideas following from the idea of infinite naturals. If you shelve that > assumption for a moment, that contradiction disappears. AS (1) No mathematician finds any contradiction in the infiniteness of the set of finite integers and (2) according to the definitions of mathematics, the set of finite naturals HAS to be infinite, TO is way off base still. > > > > Those kind of inductive arguments regarding finiteness of the > naturals break down at n=oo, > > And yet you are the one who keeps trying to treat infinite values > as though they are finite values, obeying all the same rules and > acting like them. > > They can be treated very similarly, as with the T-riffic numbers, > with which no one has found any flaw, > > That TO does not acknowledge any of the manifold flaws in his number > systems that have been so clearly and repeatedly pointed out is only > evidence of TO's willful blindness, not of any validity in TOmatics . > The only flaws pointed out have either been the result of confusing > the T- riffics with the adics, or have been answered directly with > counterexamples and explanations as to how to perform the operations > in question. If you think you have some objection which does not fall > into either category, state it now, because you haven't yet. There is that blindness again! One such flaw is TO's insistence on having a set which surjects to its power set, when it has been proved that no such sets can exist. What TO may have is a proper class which is not a set, in which case, there is no such thing as a power set for it. But my take is that TO has nothing as all except self delusion. === Subject: Re: Well Ordering the Reals > As soon as you increment an infinite number of times, according to infinite > series, you have an infinite value. Any sum of a finite number of finite values > is finite. Any sum of an infinite number of constant finite values is infinite. Dead wrong. Sum up all the negative powers of two and you get 1. Do you know what a convergent infinite series is? Bob Kolker === Subject: Re: Well Ordering the Reals Robert J. Kolker said: > > As soon as you increment an infinite number of times, according to infinite > series, you have an infinite value. Any sum of a finite number of finite values > is finite. Any sum of an infinite number of constant finite values is infinite. > Dead wrong. Sum up all the negative powers of two and you get 1. Do you > know what a convergent infinite series is? > Bob Kolker Hi Bob - Happy New year. Yes, of course, you are technically correct, when it comes to negative powers of 2, but I was referring to the set of natural numbers, as you well know. I SHOULD have said that any sum of an infinite number of finite values, which differ by a constant finite value, is infinite. That is to say that, if we start with a finite number and increment it by a constant finite value, a finite number of such increments will result in a finite value, while an infinite number of such increments yields an infinite value. Do you disagree with that statement? I hope not. -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Robert J. Kolker said: > > > As soon as you increment an infinite number of times, according to > infinite > series, you have an infinite value. Any sum of a finite number of finite > values > is finite. Any sum of an infinite number of constant finite values is > infinite. > > Dead wrong. Sum up all the negative powers of two and you get 1. Do you > know what a convergent infinite series is? > > Bob Kolker > > > Hi Bob - > Happy New year. > Yes, of course, you are technically correct, when it comes to negative powers > of 2, but I was referring to the set of natural numbers, as you well know. I > SHOULD have said that any sum of an infinite number of finite values, which > differ by a constant finite value, is infinite. That is to say that, if we > start with a finite number and increment it by a constant finite value, a > finite number of such increments will result in a finite value, while an > infinite number of such increments yields an infinite value. Do you disagree > with that statement? I hope not. An infinite number of increments if all of the same positive value, does not yield anything at all except divergence. === Subject: Re: Well Ordering the Reals C'mon, you professors. You won't get anything through to this crank. That is only one definition of infinite, for sets, which does not jibe with the notion of infinite quantity, as I've explained. When I say that set is unbounded, but finite, I am not failing to grasp anything. I am applying logic which is ignored generally, and pointing out an inconsistency between infinity as defined by sets in the standard manner, and quantitative infinity, which I see as more fundamental, but you fail to grasp this. OT, this thread dramatically illustrates Hammick's Law of Thread Length: The length of a thread is inversely proportional to its value. Next week's lesson: Hammick's Law of Thread Decay. === Subject: Re: Well Ordering the Reals Larry Hammick said: > C'mon, you professors. You won't get anything through to this crank. > That is only one definition of infinite, for sets, which does not jibe with the notion of infinite quantity, as I've explained. When I say that set is unbounded, but finite, I am not failing to grasp anything. I am applying logic which is ignored generally, and pointing out an inconsistency between infinity as defined by sets in the standard manner, and quantitative infinity, which I see as more fundamental, but you fail to grasp this. > OT, this thread dramatically illustrates Hammick's Law of Thread Length: The length of a thread is inversely proportional to its value. > Next week's lesson: Hammick's Law of Thread Decay. What an insightful comment. Until you have something interesting to say, why don't you just shut up and read? If you have an objection, state it, if you think you can do so in an intelligible manner, which I rather doubt. Apparently, you like to make up laws regarding nothing and name them after yourself. Talk about cranks....... -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Larry Hammick said: > C'mon, you professors. You won't get anything through to this crank. > That is only one definition of infinite, for sets, which does not jibe > with the notion of infinite quantity, as I've explained. When I say that > set is unbounded, but finite, I am not failing to grasp anything. I am > applying logic which is ignored generally, and pointing out an > inconsistency between infinity as defined by sets in the standard manner, > and quantitative infinity, which I see as more fundamental, but you fail to > grasp this. > > OT, this thread dramatically illustrates Hammick's Law of Thread Length: > The length of a thread is inversely proportional to its value. > > Next week's lesson: Hammick's Law of Thread Decay. > > What an insightful comment. Until you have something interesting to say, why > don't you just shut up and read? His comment is quite a propos. > If you have an objection, state it Why should Larry bother to object when he has oceans of evidence that no serious objection to TO's idiocies is taken seriously. , if you > think you can do so in an intelligible manner, which I rather doubt. If TO doubts it, it may be a sure thing. > Apparently, you like to make up laws regarding nothing and name them after > yourself. Talk about cranks....... Look who's talking!!! === Subject: Re: Well Ordering the Reals Happy New Year to you too, Ross. Hope 2006 finds you well. >> I guess first is to consider the maximality of dom(g) where g >> is a function well-ordering the nested intervals of set AB, >> where the single endpoint auxiliary pairs or reals and >> ordinals are in some superset, and g is a bijection between >> the intervals of AB and an ordinal o+1. The maximal element >> of the set o+1 is an ordinal o, or, say, m. So, m is the >> maximal element of dom(g) if there exists a degenerate >> inteval in AB. If there doesn't exist a degenerate interval in >> AB, then f is not a well-ordering of the reals. So, if there >> does exist a degenerate interval in AB then m is a maximal >> element of dom(g), and g(m) is a degenerate interval. >> If m is a countable ordinal c, then a countable sequence of >> nested intervals is considered. Okay so far. >> An infinite sequence doesn't have a maximal element. I have given you an example of an infinite sequence that does: [3,4], [3.1,3.2], [3.14,3.15], ... This sequence is infinite but degenerates at ordinal omega with degenerate interval [pi,pi]. >> The well-ordering f constructing AB leads to a well-ordering g >> of AB where if dom(g), the domain of the function g >> well-ordering the nested intervals, has a maximal element m >> else there are elements in the reals not in the range of f, the >> putative function well-ordering the reals. Okay again. >> So if m = c for some countable ordinal c, it's not apparent >> how to interconvert the sequence and the well-ordered >> countable set with minimal and maximal element, where it >> is infinite. Perhaps you might consider that AB is finite. I don't know what you mean by interconvert. Can you exaplain using my example of the sequence of nested intervals? >> If m is an uncountable ordinal o, including the consideration >> that it is the least uncountable ordinal, then AB contains >> uncountably many non-degenerate intervals, and as the reals >> are the complete ordered field, each of those contains a >> rational number. This is not a problem though as the intervals are nested, and thus overlapping. You can of course have uncountably many intervals reusing the same rationals. It is only when you have disjoint (non-degenerate) intervals that they must be no more than countably many. >> If you can consider that AB is finite, then there are adjacent >> points in the reals, in their normal ordering. In my example, AB cannot be finite. I think that makes that suggestion moot. Besides there is an abundance of proofs (unrelated to this) that show that the reals are not sequential in their normal ordering, and those proofs don't just go away because we want them to. >> Onwards to later points, projectively extended reals and >> one-point compactification are each rather standardized >> terms, basically snubbing Cauchy/Weierstrass and the >> blinders of limit evaluation, which is very useful. My apologies then if they are standardized terms, as I guess this shows my ignorance not running across them before. I did find a definition of one-point compactification. Could you please define projectively extended reals for me? Perhaps this is related to what I know as the extended real line? This is used in Real Analysis, which is defined as the real line plus the ideal point of infinity, denoted R with oo as a subscript, or simply written as R U { oo }. The extended real line is topologically equivalent to the circle, and some standard operation can be defined for it. A similar procedure is done in Complex Analysis to create the extended complex plane, which is topologically eqivalent to the sphere. In this latter case, lines in the complex plane can be thought of a circles containing the point oo. This is useful as certain analytic functions can be said to preserve circles. >> The universe is infinite. This depends of course on the definition of universe. >> Infinite sets are equivalent. Provably false. >> There is only one null axiom theory. Yes, it is the theory with no theorems. >> So, well-order the reals. Provably undecidable in an analytic way. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals Jonathan Hoyle said: > Happy New Year to you too, Ross. Hope 2006 finds you well. >> I guess first is to consider the maximality of dom(g) where g >> is a function well-ordering the nested intervals of set AB, >> where the single endpoint auxiliary pairs or reals and >> ordinals are in some superset, and g is a bijection between >> the intervals of AB and an ordinal o+1. The maximal element >> of the set o+1 is an ordinal o, or, say, m. So, m is the >> maximal element of dom(g) if there exists a degenerate >> inteval in AB. If there doesn't exist a degenerate interval in >> AB, then f is not a well-ordering of the reals. So, if there >> does exist a degenerate interval in AB then m is a maximal >> element of dom(g), and g(m) is a degenerate interval. >> If m is a countable ordinal c, then a countable sequence of >> nested intervals is considered. > Okay so far. >> An infinite sequence doesn't have a maximal element. > I have given you an example of an infinite sequence that does: [3,4], > [3.1,3.2], [3.14,3.15], ... This sequence is infinite but degenerates > at ordinal omega with degenerate interval [pi,pi]. >> The well-ordering f constructing AB leads to a well-ordering g >> of AB where if dom(g), the domain of the function g >> well-ordering the nested intervals, has a maximal element m >> else there are elements in the reals not in the range of f, the >> putative function well-ordering the reals. > Okay again. >> So if m = c for some countable ordinal c, it's not apparent >> how to interconvert the sequence and the well-ordered >> countable set with minimal and maximal element, where it >> is infinite. Perhaps you might consider that AB is finite. > I don't know what you mean by interconvert. Can you exaplain using > my example of the sequence of nested intervals? >> If m is an uncountable ordinal o, including the consideration >> that it is the least uncountable ordinal, then AB contains >> uncountably many non-degenerate intervals, and as the reals >> are the complete ordered field, each of those contains a >> rational number. > This is not a problem though as the intervals are nested, and thus > overlapping. You can of course have uncountably many intervals reusing > the same rationals. It is only when you have disjoint (non-degenerate) > intervals that they must be no more than countably many. Actually, you do indeed have an uncountably infinite set of disjoint nested intervals. If n is oo where a_n=b_n, then you have an uncountably infinite set of disjoint intervals of the form [a_n,a_n+1] and an uncountably infinite set of disjoint intervals of the form [b_n+1,b_n]. Within each of these disjoint intervals lies a rational number, does it not? Then, according to Ross's logic, there would have to be an uncountably infinite set of rationals between a_0 and b_0. The only way to avoid this, and this may be valid, is to say that there exists a rational between any two reals which are finitely different, meaning that, once the number of subintervals achieves actual infinity, rather than being a countably infinite set of finite values for n, then there is no guarantee that a rational exists in the interval. In fact, in that case, one can probably say it is impossible for a rational to exist in the infinitesimal interval between two irrational reals, though I am not sure how one would prove this. >> If you can consider that AB is finite, then there are adjacent >> points in the reals, in their normal ordering. > In my example, AB cannot be finite. I think that makes that suggestion > moot. Besides there is an abundance of proofs (unrelated to this) that > show that the reals are not sequential in their normal ordering, and > those proofs don't just go away because we want them to. Do those proofs address the concept of infinitesimals? >> Onwards to later points, projectively extended reals and >> one-point compactification are each rather standardized >> terms, basically snubbing Cauchy/Weierstrass and the >> blinders of limit evaluation, which is very useful. > My apologies then if they are standardized terms, as I guess this shows > my ignorance not running across them before. I did find a definition > of one-point compactification. Could you please define projectively > extended reals for me? I tossed you a link from Mathworld. > Perhaps this is related to what I know as the extended real line? > This is used in Real Analysis, which is defined as the real line plus > the ideal point of infinity, denoted R with oo as a subscript, or > simply written as R U { oo }. The extended real line is topologically > equivalent to the circle, and some standard operation can be defined > for it. A similar procedure is done in Complex Analysis to create the > extended complex plane, which is topologically eqivalent to the sphere. > In this latter case, lines in the complex plane can be thought of a > circles containing the point oo. This is useful as certain analytic > functions can be said to preserve circles. Yes, that is precisely what Ross means by the projectively extended reals, but I believe you are incorrect in saying the extended complex plane is topologically equivalent to the sphere, by which I assume you mean the surface of the sphere. I used to think that too, but we all make mistakes. The extended plane with circular axes, complex or real, does not make a sphere, sorry. The extended plane is topologically equivalent to the surface of the torus. If it were a sphere, then the point directly opposite the origin would be simultaneously 0 and oo in each dimension. >> The universe is infinite. > This depends of course on the definition of universe. >> Infinite sets are equivalent. > Provably false. >> There is only one null axiom theory. > Yes, it is the theory with no theorems. >> So, well-order the reals. > Provably undecidable in an analytic way. And thus a wild goose chase, since it is inherently dependent on predecessor discontinuities, being an uncountably infinite set. > Jonathan Hoyle > Eastman Kodak -- Smiles, Tony === Subject: Re: Well Ordering the Reals > Actually, you do indeed have an uncountably infinite set of disjoint nested > intervals. The maximal possible number of disjoint nested intervals is 1, unless one allows the empty set to be called an interval, in which case it is 2. === Subject: Re: Well Ordering the Reals > I guess first is to consider the maximality of dom(g) where g >> is a function well-ordering the nested intervals of set AB, >> where the single endpoint auxiliary pairs or reals and >> ordinals are in some superset, and g is a bijection between >> the intervals of AB and an ordinal o+1. The maximal element >> of the set o+1 is an ordinal o, or, say, m. So, m is the >> maximal element of dom(g) if there exists a degenerate >> inteval in AB. If there doesn't exist a degenerate interval in >> AB, then f is not a well-ordering of the reals. So, if there >> does exist a degenerate interval in AB then m is a maximal >> element of dom(g), and g(m) is a degenerate interval. >> If m is a countable ordinal c, then a countable sequence of >> nested intervals is considered. > Okay so far. >> An infinite sequence doesn't have a maximal element. > I have given you an example of an infinite sequence that does: [3,4], > [3.1,3.2], [3.14,3.15], ... This sequence is infinite but degenerates > at ordinal omega with degenerate interval [pi,pi]. >> The well-ordering f constructing AB leads to a well-ordering g >> of AB where if dom(g), the domain of the function g >> well-ordering the nested intervals, has a maximal element m >> else there are elements in the reals not in the range of f, the >> putative function well-ordering the reals. > Okay again. >> So if m = c for some countable ordinal c, it's not apparent >> how to interconvert the sequence and the well-ordered >> countable set with minimal and maximal element, where it >> is infinite. Perhaps you might consider that AB is finite. > I don't know what you mean by interconvert. Can you exaplain using > my example of the sequence of nested intervals? >> If m is an uncountable ordinal o, including the consideration >> that it is the least uncountable ordinal, then AB contains >> uncountably many non-degenerate intervals, and as the reals >> are the complete ordered field, each of those contains a >> rational number. > This is not a problem though as the intervals are nested, and thus > overlapping. You can of course have uncountably many intervals reusing > the same rationals. It is only when you have disjoint (non-degenerate) > intervals that they must be no more than countably many. >> If you can consider that AB is finite, then there are adjacent >> points in the reals, in their normal ordering. > In my example, AB cannot be finite. I think that makes that suggestion > moot. Besides there is an abundance of proofs (unrelated to this) that > show that the reals are not sequential in their normal ordering, and > those proofs don't just go away because we want them to. >> Onwards to later points, projectively extended reals and >> one-point compactification are each rather standardized >> terms, basically snubbing Cauchy/Weierstrass and the >> blinders of limit evaluation, which is very useful. > My apologies then if they are standardized terms, as I guess this shows > my ignorance not running across them before. I did find a definition > of one-point compactification. Could you please define projectively > extended reals for me? > Perhaps this is related to what I know as the extended real line? > This is used in Real Analysis, which is defined as the real line plus > the ideal point of infinity, denoted R with oo as a subscript, or > simply written as R U { oo }. The extended real line is topologically > equivalent to the circle, and some standard operation can be defined > for it. A similar procedure is done in Complex Analysis to create the > extended complex plane, which is topologically eqivalent to the sphere. > In this latter case, lines in the complex plane can be thought of a > circles containing the point oo. This is useful as certain analytic > functions can be said to preserve circles. >> The universe is infinite. > This depends of course on the definition of universe. >> Infinite sets are equivalent. > Provably false. >> There is only one null axiom theory. > Yes, it is the theory with no theorems. >> So, well-order the reals. > Provably undecidable in an analytic way. > Jonathan Hoyle > Eastman Kodak For a set of nested, non-degenerate intervals, none sharing both endpoints, they can be put in a 1-1 correspondence with disjoint, non-degenerate intervals. So, m, the maximal element of dom(g) the domain of a function bijecting an ordinal o+1 to the set AB of nested intervals with mobile endpoints, with each interval sharing no endpoints, generated by induction over a well-ordering f of the reals, f a function from an ordinal O' to R, m can not be an uncountable ordinal, else there exists a distinct rational in each of uncountably many disjoint intervals. The maximal element m of dom(g) exists, else there are no degenerate intervals in AB, and thus there exists an element of R not in the range of f. Is that agreed? Thus if m exists it is a countable ordinal c. As f is a well-ordering of the reals that exists in ZFC, m E o+1 = o, where o+1 is the domain of g, and as m is countable, AB is countable. If that is so, then there is a countable set of nested intervals, per the construction of this variation of Cantor's first, that includes a disjoint interval. It's countable. As well, m is the maximal element, which is why I am wondering about the interconversion of infinite sequences, lists, which are countably infinite and have indices with no maximal element, and dom(g), which has a maximal element m else f is not a well-ordering of the reals. They're both countable, thus there exists a bijection between them, if they're both infinite. If m is a countable ordinal, then it must be a limit ordinal, else its predecessor would be an interval with only two points, in the complete ordered field. If the set of sequences is infinite, then it has no maximal element. Why would that be so. Basically I'm trying to figure out why there would be a contradiction in having there be a maximal element as the reals are the complete ordered field, and only the complete ordered field, to show that they are not only the complete ordered field. So I should figure out why m can not be an infinite limit ordinal. I think I may have already stated that. Basically it's the consideration of some eventual predecessor, an existence proof, that were it not to exist then via a reverse of induction g(m) would not be a degenerate interval and thus f would not be a well-ordering of the reals. You see, I think the reals are not just the complete ordered field, but also points not just on a line but in a line, and that's the only way they can be considered at once, right from one to the next. Then, m = 0, it's a limit ordinal, but then again g(m) would contain only two points. In terms of the complete ordered field, that would be a degenerate interval, because there does not exist in the standard reals a distinction between between two real numbers thus that there does not exist a complete continuum between them. In trying to match each real number to an ordinal less than some ordinal that is the set of those ordinals, well-ordering them, when stippling on the line, there is no way to ever find the place again _right next to_ that previous one. Draw a line segment, the pencil goes over the points, the one-sided points in-line on the line, polydimensional points from the directed line perspective. So, my pseudo-reals or R-bar-umlaut are dually complete ordered field and this geometric evolution through all the hypercomplex numbers, R-bar the complete ordered field, and R-umlaut a sequential series of points, the Finlayson reals or I think the real reals. I agree there are obviously many results about the complete ordered field and descriptions of it and basically the axioms of the real numbers and so on and so forth, Cauchy, Dedekind, continued fractions, and other methods of representations of the real numbers. I would want to not-so-subtly hint about extensions to all of those theorems, or restrictions of their application to R-bar, the standard real numbers, except not exactly, what with these considerations of notions like a point at infinity. A definition of the universe in ZF might be there is no definition. There is no universe in ZF, and if you go to classes, there is no universe of classes, there is no universe in any regular theory, the group noun problem, which is the same as Burali-Forti. What's the set of all powersets? How about then with the axiom of infinity? There's talk about these number systems that form a circle. That would have infinity variously being itself, or the same point on the circle as negative one or zero, or sometimes one, which I explain with the confirmator clockwork(s). In physics, negative absolute temperature is hotter than any finite positive temperature. In standard string theory, strings are not mathematical points, yet the machinery there being developed is useful in consideration of that notion. The closer the smaller they appear to be. They're very small. The cosmological constant is probably iota or so. The null axiom theory has all the theorems, its dual specter the universallly axiomatized theory has none, teeing off the infinite ghost chain. They're indistinguishable, and their difference is irrelevant. The universe is infinite, _the_ universe. Ross === Subject: Re: Well Ordering the Reals >> For a set of nested, non-degenerate intervals, none sharing >> both endpoints, they can be put in a 1-1 correspondence >> with disjoint, non-degenerate intervals. >> So, m, the maximal element of dom(g) the domain of a >> function bijecting an ordinal o+1 to the set AB of nested >> intervals with mobile endpoints, with each interval sharing >> no endpoints, generated by induction over a well-ordering f >> of the reals, f a function from an ordinal O' to R, m can not >> be an uncountable ordinal, else there exists a distinct >> rational in each of uncountably many disjoint intervals. Ah, excellent point, Ross. Yes, I agree that these sequences must degenerate at some countable (limit) ordinal. >> The maximal element m of dom(g) exists, else there are >> no degenerate intervals in AB, and thus there exists an >> element of R not in the range of f. Is that agreed? Agreed. >> Thus if m exists it is a countable ordinal c. As f is a >> well-ordering of the reals that exists in ZFC, m E o+1 = o, >> where o+1 is the domain of g, and as m is countable, AB >> is countable. I agree with some of these statements, but some others make no sense to me. I agree that the domain of g is within the countable ordinals, m is countable and AB is a sequence of countable intervals. The line reading m E o+1 = o, where o+1 is the domain of g doesn't seem right, especially the o+1 = o part. And why is the domain of g a successor ordinal? It seems to me that it should be a limit ordinal. >> If that is so, then there is a countable set of nested >> intervals, per the construction of this variation of Cantor's >> first, that includes a disjoint interval. Disjoint from what? This is a sequence of nested intervals, and nowhere are any disjoint. >> It's countable. The it meaning the number of sequences in AB. (Obviously each individual non-degenerate interval is uncountable in size.) >> As well, m is the maximal element, which is why I am >> wondering about the interconversion of infinite sequences, >> lists, which are countably infinite and have indices with no >> maximal element, and dom(g), which has a maximal >> element m else f is not a well-ordering of the reals. I am not sure what you are trying to say here. Could you please restate this? >> They're both countable, thus there exists a bijection >> between them, if they're both infinite. If m is a countable >> ordinal, then it must be a limit ordinal, else its predecessor >> would be an interval with only two points, in the complete >> ordered field. Agreed. >> If the set of sequences is infinite, then it has no maximal >> element. Why would that be so. Why do you say it has no maximal element? Didn't we just prove that it does have a maximal element? >> Basically I'm trying to figure out why there would be a >> contradiction in having there be a maximal element as >> the reals are the complete ordered field, and only the >> complete ordered field, to show that they are not only the >> complete ordered field. I'm not sure why you think there is a contradiction here. A real number can be uniquely determined by a Dedekind cut, which is a set of two countably infinite sequences of rational numbers. This is analogous to your nested intervals, with the difference being that your nested intervals can degenerate to a single point at any countable limit ordinal, whereas Dedekind cuts essentially reach their limit at omega. >> You see, I think the reals are not just the complete ordered >> field, but also points not just on a line but in a line, and >> that's the only way they can be considered at once... Agree so far... >> ...right from one to the next. This is the point at which I disagree. Even in Geometry, points on a line are not sequential. This is easily provable: between any two distinct points, a line segment can be drawn, with points between the original two. You can't have sequential points if there is always a third between any two. >> A definition of the universe in ZF might be there is no >> definition. There is no universe in ZF, and if you go to >> classes, there is no universe of classes, there is no >> universe in any regular theory, the group noun problem, >> which is the same as Burali-Forti. What's the set of all >> powersets? I think you are once again confusing the set theory and the model. You certainly can have a universal collection of everything in your model. It's just that it is not addressable by your theory (ZF), since it is not a ZF-set. If you wish to use a wider theory, such as NBG or others, in which ZF is merely a sub-theory, you are welcome to do that. >> How about then with the axiom of infinity? What about the axiom of infinity? >> There's talk about these number systems that form a circle. The extended real line is useful that way, essentially the real line plus the ideal point of infinity. This is topologically equivalent to the circle. Although oo is not defined for < or > operators, there are certain arithmetic operations in which it is defined. Other differences take place, such as the function 1/x is continuous everywhere on extended R, where it is not on R (the reverse is true of e^x), >> The null axiom theory has all the theorems... A system with no axioms has no theorems. If you tell me of a theorem you think you have proven, I will show you the hidden axiom(s) you used to get there. >> The universe is infinite, _the_ universe. Which universe is the universe? For example, in your universe, is the Parallel Postulate true or false? It seems to me that there are universes in which it is true, and universes in which it is false. No universe has any preference for being more real. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals >> For a set of nested, non-degenerate intervals, none sharing >> both endpoints, they can be put in a 1-1 correspondence >> with disjoint, non-degenerate intervals. >> So, m, the maximal element of dom(g) the domain of a >> function bijecting an ordinal o+1 to the set AB of nested >> intervals with mobile endpoints, with each interval sharing >> no endpoints, generated by induction over a well-ordering f >> of the reals, f a function from an ordinal O' to R, m can not >> be an uncountable ordinal, else there exists a distinct >> rational in each of uncountably many disjoint intervals. Ah, excellent point, Ross. Yes, I agree that these sequences must degenerate at some countable (limit) ordinal. >> The maximal element m of dom(g) exists, else there are >> no degenerate intervals in AB, and thus there exists an >> element of R not in the range of f. æIs that agreed? Agreed. >> Thus if m exists it is a countable ordinal c. æAs f is a >> well-ordering of the reals that exists in ZFC, m E o+1 = o, >> where o+1 is the domain of g, and as m is countable, AB >> is countable. I agree with some of these statements, but some others make no sense to me. I agree that the domain of g is within the countable ordinals, m is countable and AB is a sequence of countable intervals. The line reading m E o+1 = o, where o+1 is the domain of g doesn't seem right, especially the o+1 = o part. And why is the domain of g a successor ordinal? It seems to me that it should be a limit ordinal. >> If that is so, then there is a countable set of nested >> intervals, per the construction of this variation of Cantor's >> first, that includes a disjoint interval. Disjoint from what? This is a sequence of nested intervals, and nowhere are any disjoint. >> It's countable. The it meaning the number of sequences in AB. (Obviously each individual non-degenerate interval is uncountable in size.) >> As well, m is the maximal element, which is why I am >> wondering about the interconversion of infinite sequences, >> lists, which are countably infinite and have indices with no >> maximal element, and dom(g), which has a maximal >> element m else f is not a well-ordering of the reals. I am not sure what you are trying to say here. Could you please restate this? >> They're both countable, thus there exists a bijection >> between them, if they're both infinite. æIf m is a countable >> ordinal, then it must be a limit ordinal, else its predecessor >> would be an interval with only two points, in the complete >> ordered field. Agreed. >> If the set of sequences is infinite, then it has no maximal >> element. æWhy would that be so. Why do you say it has no maximal element? Did we just prove that it does have a maximal element? >> Basically I'm trying to figure out why there would be a >> contradiction in having there be a maximal element as >> the reals are the complete ordered field, and only the >> complete ordered field, to show that they are not only the >> complete ordered field. I'm not sure why you think there is a contradiction here. A real number can be uniquely determined by a Dedekind cut, which is a set of two countably infinite sequences of rational numbers. This is analogous to your nested intervals, with the difference being that your nested intervals can degenerate to a single point at any countable limit ordinal, whereas Dedekind cuts essentially reach their limit at omega. >> You see, I think the reals are not just the complete ordered >> field, but also points not just on a line but in a line, and >> that's the only way they can be considered at once... Agree so far... >> ...right from one to the next. This is the point at which I disagree. Even in Geometry, points on a line are not sequential. This is easily provable: between any two distinct points, a line segment can be drawn, with points between the original two. You can't have sequential points if there is always a third between any two. >> A definition of the universe in ZF might be there is no >> definition. There is no universe in ZF, and if you go to >> classes, there is no universe of classes, there is no >> universe in any regular theory, the group noun problem, >> which is the same as Burali-Forti. æWhat's the set of all >> powersets? I think you are once again confusing the set theory and the model. You certainly can have a universal collection of everything in your model. It's just that it is not addressable by your theory (ZF), since it is not a ZF-set. If you wish to use a wider theory, such as NBG or others, in which ZF is merely a sub-theory, you are welcome to do that. >> How about then with the axiom of infinity? What about the axiom of infinity? >> There's talk about these number systems that form a circle. The extended real line is useful that way, essentially the real line plus the ideal point of infinity. This is topologically equivalent to the circle. Although oo is not defined for < or > operators, there are certain arithmetic operations in which it is defined. Other differences take place, such as the function 1/x is continuous everywhere on extended R, where it is not on R (the reverse is true of e^x), >> The null axiom theory has all the theorems... A system with no axioms has no theorems. If you tell me of a theorem you think you have proven, I will show you the hidden axiom(s) you used to get there. >> The universe is infinite, the universe. Which universe is the universe? For example, in your universe, is the Parallel Postulate true or false? It seems to me that there are universes in which it is true, and universes in which it is false. No universe has any preference for being more real. === Subject: Re: Well Ordering the Reals Jonathan Hoyle said: >> For a set of nested, non-degenerate intervals, none sharing >> both endpoints, they can be put in a 1-1 correspondence >> with disjoint, non-degenerate intervals. >> So, m, the maximal element of dom(g) the domain of a >> function bijecting an ordinal o+1 to the set AB of nested >> intervals with mobile endpoints, with each interval sharing >> no endpoints, generated by induction over a well-ordering f >> of the reals, f a function from an ordinal O' to R, m can not >> be an uncountable ordinal, else there exists a distinct >> rational in each of uncountably many disjoint intervals. > Ah, excellent point, Ross. Yes, I agree that these sequences must > degenerate at some countable (limit) ordinal. But then the conclusion is that there are countably many reals in the interval. As I said in a different post, while there is undoubtedly a standard rational between any two finitely differing reals, this is not the case for two irrational reals that differ by an infinitesimal amount. >> The maximal element m of dom(g) exists, else there are >> no degenerate intervals in AB, and thus there exists an >> element of R not in the range of f. =C2=A0Is that agreed? > Agreed. There exists an element of R in the range of f which is not resolvable into a degenerate interval in a countable number of subintervals. To get to Cantor's c requires an uncountably infinite number of iterations. What does this mean? =C2=A0Is >> Thus if m exists it is a countable ordinal c. =C2=A0As f is a >> well-ordering of the reals that exists in ZFC, m E o+1 =3D o, >> where o+1 is the domain of g, and as m is countable, AB >> is countable. > I agree with some of these statements, but some others make no sense to > me. I agree that the domain of g is within the countable ordinals, m > is countable and AB is a sequence of countable intervals. The line > reading m E o+1 =3D o, where o+1 is the domain of g doesn't seem right, > especially the o+1 =3D o part. And why is the domain of g a successor > ordinal? It seems to me that it should be a limit ordinal. >> If that is so, then there is a countable set of nested >> intervals, per the construction of this variation of Cantor's >> first, that includes a disjoint interval. > Disjoint from what? This is a sequence of nested intervals, and > nowhere are any disjoint. Each open interval (a_n,a_n+1) is disjoint from every other, no? >> It's countable. > The it meaning the number of sequences in AB. (Obviously each > individual non-degenerate interval is uncountable in size.) Yes, which means the entire interval [a_0,b_0] is also uncountably infinite, and within each of the subsequent intervals which is non-degenerate, there is also a local c which requires an uncountably infinite number of nested intervals to achieve the degenerate interval equivalent to c. >> As well, m is the maximal element, which is why I am >> wondering about the interconversion of infinite sequences, >> lists, which are countably infinite and have indices with no =E2=80=A8>>= > maximal element, and dom(g), which has a maximal >> element m else f is not a well-ordering of the reals. > I am not sure what you are trying to say here. Could you please > restate this? >> They're both countable, thus there exists a bijection >> between them, if they're both infinite. =C2=A0If m is a countable >> ordinal, then it must be a limit ordinal, else its predecessor >> would be an interval with only two points, in the complete =E2=80=A8>> o= > rdered field. > Agreed. >> If the set of sequences is infinite, then it has no maximal >> element. =C2=A0Why would that be so. > Why do you say it has no maximal element? Did we just prove that it > does have a maximal element? Given the particular pair (a_0,b_0) there is a maximal element which is the convergence point of those two. If this is an infinite sequence, it would appear to have no maximal element by virue of that fact, and yet, as the nested intervals close in on the value, at any infinite value of n in the sequence, the difference between a_n and b_n becomes infinitesimal, which in the standard picture of things means they are equal. We may be able to describe this n in terms of an infinite bit string, but as such, is not a very distinct value. >> Basically I'm trying to figure out why there would be a >> contradiction in having there be a maximal element as >> the reals are the complete ordered field, and only the >> complete ordered field, to show that they are not only the >> complete ordered field. > number can be uniquely determined by a Dedekind cut, which is a set of > two countably infinite sequences of rational numbers. This is > analogous to your nested intervals, with the difference being that your > nested intervals can degenerate to a single point at any countable > limit ordinal, whereas Dedekind cuts essentially reach their limit at > omega. Why do either of you think that the interval can degeenrate in a countably infinite number of steps? This seems to rest on the assumption that between any two reals lies a rational, and yet, this assumption really only holds for finitely separated reals, which is not the case when the nesting is continued for an uncountably infinite number of steps. Are either of you particularly sure that between any two infintiesimally different irrational reals lies a rational number? I am rather sure this is not the case, unless one allows infinite numerator and denominator, which means a non-standard rational. Such non-standard rationals are equivalent to the reals, but uncountable. >> You see, I think the reals are not just the complete ordered >> field, but also points not just on a line but in a line, and >> that's the only way they can be considered at once... > Agree so far... >> ...right from one to the next. > line are not sequential. This is easily provable: between any two > distinct points, a line segment can be drawn, with points between the > original two. You can't have sequential points if there is always a > third between any two. Consider the point at which you have divided the line into aleph_0 equal subsegments. What is the difference between the endpoints of any one of those intervals? Are those points the same, or different real numbers? In my view, any infinite subdivision of a finite interval will result in an infinite number of points which are each distinct from their neighbor, and yet indistinguishable in finite location of value. In this sense, Ross's conception of the infinitesimal is correct. Not to mention that it works and plays very well with the inverse function law and leads to notions of near-equivalence between the reals and rationals, which is far more acceptable than an equivalence between the sparse naturals and the dense rationals. >> A definition of the universe in ZF might be there is no >> definition. There is no universe in ZF, and if you go to >> classes, there is no universe of classes, there is no >> universe in any regular theory, the group noun problem, >> which is the same as Burali-Forti. =C2=A0What's the set of all >> powersets? > I think you are once again confusing the set theory and the model. You > certainly can have a universal collection of everything in your model. > It's just that it is not addressable by your theory (ZF), since it is > not a ZF-set. If you wish to use a wider theory, such as NBG or > others, in which ZF is merely a sub-theory, you are welcome to do that. >> How about then with the axiom of infinity? > What about the axiom of infinity? >> There's talk about these number systems that form a circle. > plus the ideal point of infinity. This is topologically equivalent to > the circle. Although oo is not defined for < or > operators, there are > certain arithmetic operations in which it is defined. Other > differences take place, such as the function 1/x is continuous > everywhere on extended R, where it is not on R (the reverse is true of > e^x), Well, 1/x can be considered continuous given the notion that oo=-oo, which is the value at 0 (which is equal to -0). >> The null axiom theory has all the theorems... > you think you have proven, I will show you the hidden axiom(s) you used > to get there. I have to say, that's a valid point. Even if you develop theorems from pure logic, there are axioms which govern how that logic works. There have to be some rules to work with, to get started. Of course, the axioms of set theory do seem, as Ross describes, somewhat non-logical. >> The universe is infinite, _the_ universe. > Which universe is the universe? For example, in your universe, is > the Parallel Postulate true or false? It seems to me that there are > universes in which it is true, and universes in which it is false. No > universe has any preference for being more real. Now, it sounds like you are talking about models, not the universe. The universe includes all models, as well as all theories, all sets, all facts and all rules. -- Smiles, Tony === Subject: Re: Well Ordering the Reals Ah, excellent point, Ross. Yes, I agree that these sequences >> must degenerate at some countable (limit) ordinal. >> But then the conclusion is that there are countably many reals >> in the interval. That would be an incorrect conclusion. There are uncountably many reals in the interval. >> As I said in a different post, while there is undoubtedly a standard >> rational between any two finitely differing reals, this is not the case >> for two irrational reals that differ by an infinitesimal amount. Again untrue. Even in Nonstandard analysis, two standard irrationals must differ by some positive amount. It is true that two hyper-irrationals may be infinitessimally close to each other, but they are separated by hyper-rationals. >> Yes, which means the entire interval [a_0,b_0] is also uncountably >> infinite, and within each of the subsequent intervals which is >> non-degenerate, there is also a local c which requires an >> uncountably infinite number of nested intervals to achieve the >> degenerate interval equivalent to c. Sorry again, but this too is false. You need only countably many nested intervals to degenerate the interval. >> Why do either of you think that the interval can degeenrate in a >> countably infinite number of steps? Consider this example: [3,4], [3.1,3.2], [3.14,3.15], [3.141,3.42], [3.1415,3.1416], ... with each n-th sequence adding another decimal place bounding pi. You never reach pi at any finite n-th place, however at its limit (which is the ordinal omega), the only interval that can nest into the others is [pi,pi]. Thus any one of the uncountably many reals can be approached with only a countable sequence. >> This seems to rest on the assumption that between >> any two reals lies a rational... That assumption does not come into play at all in this construction (although it is true). >> ...and yet, this assumption really only holds for finitely >> separated reals, which is not the case when the nesting >> is continued for an uncountably infinite number of steps. We are dealing only with standard reals (what you are calling finitely separated reals) in this construction; however, even if we were to consider non-standard hyper-reals, you still need only a countable number of steps. >> Are either of you particularly sure that between any two >> infintiesimally different irrational reals lies a rational number? >> I am rather sure this is not the case, unless one allows >> infinite numerator and denominator, which means a >> non-standard rational. Such non-standard rationals are >> equivalent to the reals, but uncountable. You are partially correct. Yes, between any two hyper-irrational there is a hyper-rational, although a hyper-rational is not the same as a standard real number. Hyper-rationals are only in the halo of a standard irrational. And yes, a hyper-rational would be the ratio of two infinite hyper-naturals. >> Consider the point at which you have divided the line into >> aleph_0 equal subsegments. Ah, but exactly *HOW* do you perform this partition? For example, if you were to try by way of sequences, let s_0 = [0,1], s_1 divides it into [0,0.5] and [0.5,1], s_2 divides further into [0,0.25], [0.25,0.5], [0.5,0.75] and [0.75,1], etc. In general, s_n cuts [0,1] into 2^n pieces. Now how do you get to Aleph_0? There is no x (either finite or infinite) such that 2^x = Aleph_0. If you simply let lim s_n = s, then you end up wit s being a set of uncountably many degenerate sequences [r,r] for each real number r. (A total of 2^Aleph_0 sequences.) Let me save you time and give you the answer now: There is *NO WAY* to partition [0,1] into Aleph_0 equal intervals (if by equal you mean have equal measure). It is a result from Measure Theory that proves this. As it turns out, you can (with the Axiom of Choice) create Aleph_0 subsets of [0,1], but they would be non-contiguous and they are non-measurable (and thus impossible to equate). You can alternatively (of course) create Aleph_0 intervals of differing lengths: [0,1/2], [1/2,3/4], [3/4,7/8], ... but I suspect that is not what you had in mind. Irrespective, you must drop either the requirement of Aleph_0 subsets or equal measure. Hope that helps, Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals Jonathan Hoyle said: >> Ah, excellent point, Ross. Yes, I agree that these sequences >> must degenerate at some countable (limit) ordinal. >> But then the conclusion is that there are countably many reals >> in the interval. > That would be an incorrect conclusion. There are uncountably many > reals in the interval. I agree. >> As I said in a different post, while there is undoubtedly a standard >> rational between any two finitely differing reals, this is not the case >> for two irrational reals that differ by an infinitesimal amount. > Again untrue. Even in Nonstandard analysis, two standard irrationals > must differ by some positive amount. It is true that two > hyper-irrationals may be infinitessimally close to each other, but they > are separated by hyper-rationals. Well, yes, that is what I said. The only way to have two infinitesimally close irrationals with a rational in between is to allow infinite numerator and denominator, which is a hyper-rational, no? When we speak of infinitesimals we are already talking about nonstandard analysis, aren't we? And, in that case, when I speak of two irrationals that are infintesimally close, then I am talking about hyper-irrationals in that context. So, I don't think we disagree here. Do you? Surely, you aren't claiming that there are a countably infinite number of hyper-rationals, are you? >> Yes, which means the entire interval [a_0,b_0] is also uncountably >> infinite, and within each of the subsequent intervals which is >> non-degenerate, there is also a local c which requires an >> uncountably infinite number of nested intervals to achieve the >> degenerate interval equivalent to c. > Sorry again, but this too is false. You need only countably many > nested intervals to degenerate the interval. Then you are saying that there are countably many reals in the finite interval, since any can be pinpointed with a countable number of nestings. But, that's not correct, or in agreement with Cantor's first, which proves that an uncountably infinite number of subintervals is required to get to c. Otherwise, what do you think the degeneration of the interval signifies? >> Why do either of you think that the interval can degeenrate in a >> countably infinite number of steps? > Consider this example: [3,4], [3.1,3.2], [3.14,3.15], [3.141,3.42], > [3.1415,3.1416], ... with each n-th sequence adding another decimal > place bounding pi. You never reach pi at any finite n-th place, > however at its limit (which is the ordinal omega), the only interval > that can nest into the others is [pi,pi]. Thus any one of the > uncountably many reals can be approached with only a countable > sequence. Okay, where this is where my understanding diverges with the norm, but I guess you are saying something like, if the subdivisions each eliminate an uncountably infinite number of reals, then a countably infinite number of them suffices to pinpoint one of the uncountably infinite number of reals, much like you consider 2^aleph_0 to be uncountable. Is that about right? So, as long as you have performed more than a finite number of iterations, you have less than a finite subinterval, and the endpoints are the same. Okay. >> This seems to rest on the assumption that between >> any two reals lies a rational... > That assumption does not come into play at all in this construction > (although it is true). >> ...and yet, this assumption really only holds for finitely >> separated reals, which is not the case when the nesting >> is continued for an uncountably infinite number of steps. > We are dealing only with standard reals (what you are calling finitely > separated reals) in this construction; however, even if we were to > consider non-standard hyper-reals, you still need only a countable > number of steps. I'll accept that statement, despite my own perspective, in the context of the standard understanding. >> Are either of you particularly sure that between any two >> infintiesimally different irrational reals lies a rational number? >> I am rather sure this is not the case, unless one allows >> infinite numerator and denominator, which means a >> non-standard rational. Such non-standard rationals are >> equivalent to the reals, but uncountable. > You are partially correct. Yes, between any two hyper-irrational there > is a hyper-rational, although a hyper-rational is not the same as a > standard real number. Hyper-rationals are only in the halo of a > standard irrational. And yes, a hyper-rational would be the ratio of > two infinite hyper-naturals. What do you mean by the halo of a standard irrational? >> Consider the point at which you have divided the line into >> aleph_0 equal subsegments. > you were to try by way of sequences, let s_0 = [0,1], s_1 divides it > into [0,0.5] and [0.5,1], s_2 divides further into [0,0.25], > [0.25,0.5], [0.5,0.75] and [0.75,1], etc. In general, s_n cuts [0,1] > into 2^n pieces. Now how do you get to Aleph_0? There is no x (either > finite or infinite) such that 2^x = Aleph_0. If you simply let lim s_n > = s, then you end up wit s being a set of uncountably many degenerate > sequences [r,r] for each real number r. (A total of 2^Aleph_0 > sequences.) Yes, which is why you say the countable number of nestings produces uncountably many intervals, as I said above. I see it differently, but in the standard world you are correct. > Let me save you time and give you the answer now: > There is *NO WAY* to partition [0,1] into Aleph_0 equal intervals (if > by equal you mean have equal measure). It is a result from Measure > Theory that proves this. As it turns out, you can (with the Axiom of > Choice) create Aleph_0 subsets of [0,1], but they would be > non-contiguous and they are non-measurable (and thus impossible to > equate). You can alternatively (of course) create Aleph_0 intervals of > differing lengths: [0,1/2], [1/2,3/4], [3/4,7/8], ... but I suspect > that is not what you had in mind. Irrespective, you must drop either > the requirement of Aleph_0 subsets or equal measure. Well, this is part of the problem with the standard treatment of infinity, in my opinion. It's good to see Ross struggling it out with the standard notions, because I think he has to do that for his own edification. I am afraid your statement here will turn him away from that endeavor, which is ultimately what has to happen. His is a system of infinitesimals which are equinumerous with the naturals, within the unit interval. There may be no standard way to construct aleph_0 even subintervals in the standard world, but in fact his notion meets with my inverse function notion to provide a system that begins to address these issues. You see, if we have a number aleph_0, then we should be able to perform log2(aleph_0) divisions, and get aleph_0 subintervals. So, in standard theory, yes, I must drop the idea of aleph_0 infinitesimal subintervals of equal measure, but outside of it, there is no requirement that I do so. > Hope that helps, > Jonathan Hoyle > Eastman Kodak -- Smiles, Tony === Subject: Re: Well Ordering the Reals Well, yes, that is what I said. The only way to have two infinitesimally >> close irrationals with a rational in between is to allow infinite >> numerator and denominator, which is a hyper-rational, no? When we >> speak of infinitesimals we are already talking about nonstandard >> analysis, aren't we? And, in that case, when I speak of two >> irrationals that are infintesimally close, then I am talking about >> hyper-irrationals in that context. So, I don't think we disagree here. >> Do you? Surely, you aren't claiming that there are a countably >> infinite number of hyper-rationals, are you? No, I am not. I misunderstood your posts here, sorry. I think it's a terminology issue. When I see rational and irrational without the hyper- prefix, I assumed you were referring to the standard versions of these things. So yes, we agree. >> Sorry again, but this too is false. You need only countably >> many nested intervals to degenerate the interval. >> Then you are saying that there are countably many reals in >> the finite interval, since any can be pinpointed with a >> countable number of nestings. No, that is not correct. Any of the uncountably many reals can be uniquely specified by a countable sequence. >> But, that's not correct, or in agreement with Cantor's first, >> which proves that an uncountably infinite number of >> subintervals is required to get to c. That is not what Cantor's first says. In this theorem, he proves that you can't do it in a *finite* number of steps. You can indeed do it in a countably infinite number of steps. >> Otherwise, what do you think the degeneration of the >> interval signifies? The degeneration of intervals takes place after all the finite steps take place, but not after all the countably infinite steps. >> Okay, where this is where my understanding diverges >> with the norm, but I guess you are saying something >> like, if the subdivisions each eliminate an uncountably >> infinite number of reals, then a countably infinite >> number of them suffices to pinpoint one of the >> uncountably infinite number of reals, much like you >> consider 2^aleph_0 to be uncountable. Is that about >> right? Sort of, but not exactly. It has more to do with the fact that you have 10 choices in Aleph_0 positions, and then we get |10^Aleph_0| = |2^Aleph_0| which is uncountable. Consider the Dedekind cut { <3,3.1,3.14,3.141,3.1415,...>, <4,3.5,3.15,3.142,3.1416,...> }. At each step n, we add the n-th decimal place. There are Aleph_0 elements in the sequence, with a 10 choices of digits at each place, yielding |10^Aleph_0| = c, which is uncountable. >> What do you mean by the halo of a standard irrational? The halo of x is defined to be the set of all hyper-reals infinitely of x, denoted gal(x) is the set of all hyper-reals finitely close to x. Thus hal(0) is simply the set of all infinitessimals, and gal(0) is simply the set of finite hyper-reals. There is an interesting fact about halos: for every finite x, hal(x) contains exactly one standard real, called the shadow of x (or the standard part of x). Although it's obvious that hal(x) could not contain more than one standard real (because any two standard numbers are more than an infinitessimal apart), it was not obvious (at least to me) that there had to be at least one. I thought that maybe there would be halos of fluff between the standard reals, but it turns out that the standard reals are so dense, you can't have a gap without a even a standard real in it. >> There is *NO WAY* to partition [0,1] into Aleph_0 equal >> intervals >> Well, this is part of the problem with the standard treatment >> of infinity, in my opinion. It's good to see Ross struggling it >> out with the standard notions, because I think he has to do >> that for his own edification. I am afraid your statement here >> will turn him away from that endeavor, which is ultimately >> what has to happen. His is a system of infinitesimals which >> are equinumerous with the naturals, within the unit interval. >> There may be no standard way to construct aleph_0 even >> subintervals in the standard world, but in fact his notion >> meets with my inverse function notion to provide a system >> that begins to address these issues. You see, if we have >> a number aleph_0, then we should be able to perform >> log2(aleph_0) divisions, and get aleph_0 subintervals. So, >> in standard theory, yes, I must drop the idea of aleph_0 >> infinitesimal subintervals of equal measure, but outside >> of it, there is no requirement that I do so. I do not see any problem with creating an axiomatic system with quanta like the way you describe (essentially, that what the natural numbers are). I am sure there are models you can construct to give you this behavior. It's just that this is not the model of the real numbers. The main problem with the quanta approach to reals is this: you have to drop the notion that between any two reals is another between them. If you have a sequence of quanta, what is their granularity? If the granularity is some large finite number N, then you can only drill down as far as 1/N, and 1/(N+1) has no meaning. (Setting N=1 gives you just the integers, which is interesting.) Suppose the granularity is set to some infinite number, say omega (denoted w). Then 1/w is your granularity, and w is your largest number. But is it? Is the inverse of w equal to 0 or 1/w? Presumably it is equal to 1/w. Then what pray tell is 1/0 equal to? Then there are other arithmetic problems, but they would have to be solved by your axiomatic system. fascinating to ignore, so i am happy to be involved. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals Ross A. Finlayson said: < snippogram For a set of nested, non-degenerate intervals, none sharing both > endpoints, they can be put in a 1-1 correspondence with disjoint, > non-degenerate intervals. > So, m, the maximal element of dom(g) the domain of a function bijecting > an ordinal o+1 to the set AB of nested intervals with mobile endpoints, > with each interval sharing no endpoints, generated by induction over a > well-ordering f of the reals, f a function from an ordinal O' to R, m > can not be an uncountable ordinal, else there exists a distinct > rational in each of uncountably many disjoint intervals. > The maximal element m of dom(g) exists, else there are no degenerate > intervals in AB, and thus there exists an element of R not in the range > of f. Is that agreed? > Thus if m exists it is a countable ordinal c. As f is a well-ordering > of the reals that exists in ZFC, m E o+1 = o, where o+1 is the domain > of g, and as m is countable, AB is countable. I cannot see how the reals in AB can be enumerated in a countably infinite number of nestings. Your logic here is good, it seemed at first, but I think it suffers from the general notion that a rational exists between any two reals. When the difference between two irrational reals is infinitesimal, then there is no guarantee that a rational exists between them, and in fact, a rational CANNOT exist between them, without allowing infinite numerator and denominator. Now, I agree that the number of rationals is much more similar to the number of reals than to the number of naturals, countable or not, so your point is not lost on me. Yet, I hope this objection is not lost on you. What do you think about this notion? > If that is so, then there is a countable set of nested intervals, per > the construction of this variation of Cantor's first, that includes a > disjoint interval. It's countable. As well, m is the maximal element, > which is why I am wondering about the interconversion of infinite > sequences, lists, which are countably infinite and have indices with no > maximal element, and dom(g), which has a maximal element m else f is > not a well-ordering of the reals. They're both countable, thus there > exists a bijection between them, if they're both infinite. If m is a > countable ordinal, then it must be a limit ordinal, else its > predecessor would be an interval with only two points, in the complete > ordered field. If the set of sequences is infinite, then it has no > maximal element. Why would that be so. Because there are an uncountably infinite number of disjoint subintervals in any finite interval of the reals? > Basically I'm trying to figure out why there would be a contradiction > in having there be a maximal element as the reals are the complete > ordered field, and only the complete ordered field, to show that they > are not only the complete ordered field. So I should figure out why m > can not be an infinite limit ordinal. I think I may have already > stated that. Basically it's the consideration of some eventual > predecessor, an existence proof, that were it not to exist then via a > reverse of induction g(m) would not be a degenerate interval and thus f > would not be a well-ordering of the reals. That's just it. The well ordering is sequential, in that every element has a successor. In order to well-order the reals, this must be true of the well- ordering. BUT, in order for the order to be a well ordering, every subset must have a minimal element, and since this includes subsets defined using the term finite, and since the set of finite values is considered infinite, we can identify some element at some uncountable index, and call its set of finite predecessors infinite, thus defining an infinite descending sequence. This is why any well-ordering of the reals requires predecessor discontinuities separating countably infinite subsequences with no predecessor, and why the only way to do this would be to form a bijection with the countably infinite ordinals. Unfortunately, my friend, I don't think this is really possible. One would have to partition the set of reals into countably infinite subsequences, and also make sure there are only a countably infinite sequences of such subsequences, or else we can have the infinite descending sequence of the first elements of those subsequences. Unless the union of a countably infinite set of countably infinite sets can produce an uncountably infinite set, which seems to depend on one's choice of axioms, it seems, the goose has already headed south. > You see, I think the reals are not just the complete ordered field, but > also points not just on a line but in a line, and that's the only way > they can be considered at once, right from one to the next. I do think your concept of infinitesimals is essentially correct, but doesn't fit in ZFC. > Then, m = 0, it's a limit ordinal, but then again g(m) would contain > only two points. In terms of the complete ordered field, that would be > a degenerate interval, because there does not exist in the standard > reals a distinction between between two real numbers thus that there > does not exist a complete continuum between them. In trying to match > each real number to an ordinal less than some ordinal that is the set > of those ordinals, well-ordering them, when stippling on the line, > there is no way to ever find the place again _right next to_ that > previous one. Draw a line segment, the pencil goes over the points, > the one-sided points in-line on the line, polydimensional points from > the directed line perspective. > So, my pseudo-reals or R-bar-umlaut are dually complete ordered field > and this geometric evolution through all the hypercomplex numbers, > R-bar the complete ordered field, and R-umlaut a sequential series of > points, the Finlayson reals or I think the real reals. > I agree there are obviously many results about the complete ordered > field and descriptions of it and basically the axioms of the real > numbers and so on and so forth, Cauchy, Dedekind, continued fractions, > and other methods of representations of the real numbers. I would want > to not-so-subtly hint about extensions to all of those theorems, or > restrictions of their application to R-bar, the standard real numbers, > except not exactly, what with these considerations of notions like a > point at infinity. You know, of course, that one can also number the number circle in terms of finite fractions of oo, in which case the one-point compactification occurs around 0, with all finite values at that point. Two sides of the same coin, dontcha know? > A definition of the universe in ZF might be there is no definition. > There is no universe in ZF, and if you go to classes, there is no > universe of classes, there is no universe in any regular theory, the > group noun problem, which is the same as Burali-Forti. What's the set > of all powersets? How about then with the axiom of infinity? > There's talk about these number systems that form a circle. That would > have infinity variously being itself, or the same point on the circle > as negative one or zero, or sometimes one, which I explain with the > confirmator clockwork(s). In physics, negative absolute temperature is > hotter than any finite positive temperature. In standard string > theory, strings are not mathematical points, yet the machinery there > being developed is useful in consideration of that notion. The closer > the smaller they appear to be. They're very small. The cosmological > constant is probably iota or so. Give or take iota^2. :) > The null axiom theory has all the theorems, its dual specter the > universallly axiomatized theory has none, teeing off the infinite ghost > chain. They're indistinguishable, and their difference is irrelevant. > The universe is infinite, _the_ universe. > Ross -- Smiles, Tony === Subject: Re: Well Ordering the Reals >> An infinite sequence doesn't have a maximal element. > I have given you an example of an infinite sequence that does: [3,4], > [3.1,3.2], [3.14,3.15], ... This sequence is infinite but degenerates > at ordinal omega with degenerate interval [pi,pi]. Except that it never reaches ordinal omega. No infinite sequence, in the standard sense of that phrase, ever has an element corresponding to any but finite ordinals. Infinite sequences, in any standard sense, are functions whose arguments are restricted to members of the set of finite ordinals. The sequence of right endpoints has a maximal element, the first, but has no minimal one. the sequence of left endpoints has no maximal but does have a minimal element, and the sequence of interval lengths has a maximal but no minimal value. === Subject: Re: Well Ordering the Reals Except that it never reaches ordinal omega. No infinite sequence, >> in the standard sense of that phrase, ever has an element >> corresponding to any but finite ordinals. That is true, however, the context was in Ross creating a mapping of these intervals against ordinal numbers. He was apparently attempting to extend Cantor's first proof of the uncountability of the reals [see: http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof ] from the finite integers to ordinals and show that the same argument can be used to show that the reals are not 1-1 with |P(N)|. His error was in his handling of limit ordinals, where his proof breaks down. Jonathan Hoyle Eastman Kodak === Subject: Re: Well Ordering the Reals >> Except that it never reaches ordinal omega. No infinite sequence, >> in the standard sense of that phrase, ever has an element >> corresponding to any but finite ordinals. > That is true, however, the context was in Ross creating a mapping of > these intervals against ordinal numbers. He was apparently attempting > to extend Cantor's first proof of the uncountability of the reals > [see: > http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof ] > from the finite integers to ordinals and show that the same argument > can be used to show that the reals are not 1-1 with |P(N)|. His error > was in his handling of limit ordinals, where his proof breaks down. > Jonathan Hoyle > Eastman Kodak Ross' errors are too numerous to ennumerate. === Subject: hypergeometric functions in Gradshteyn Hi all! What das Gradshteyn (Tables of Integrals Series and Products) mean (i.e. F_4(alpha,beta ; gamma,delta ; x,y) and what is F(0,1/2;1,1;x,y) ? Has it realy two arguments or is it an abrevetion? Please help me. The corresponding pages in my copy are missing! Andreas === Subject: Re: Upper case cursive Greek letters Are you sure they are Greek? They could be Gothic script which looks quite ornate. === Subject: Hi I don't really know that much about science and math but i want to learn more. === Subject: Re: Hi > I don't really know that much about science and math but i want to > learn more. For science: Start reading Scientific American. If you want to learn more, , there should be recomended reading, or ask around (e.g., in Usenet) for suggested background material. For math: I recommend taking a look at one of the following: Courant and Robbins, What is Mathematics? Newman, J.R., ed., The World of Mathematics The latter is a four volulme set of essays, most not very technical. You will find that the mathematics is a very wide area. Look at the parts that interest you. If the book gives no references for backup reading, ask here, and people will surely give you suggestions. As a guess, you might also want to start working through a book like Lehman, C., College Algebra or any book with a similar title from the 1960's. The subject got dumbed down in later years. Importantly, be careful of what you get from the Usenet newsgroups. Crackpots lurk here. Tony Orlow is one of them. I am not. -- Stephen J. Herschkorn sjherschko@netscape.net Math Tutor on the Internet and in Central New Jersey and Manhattan === Subject: Re: Hi Hannah said: > I don't really know that much about science and math but i want to > learn more. Well, if you want to learn about infinity, by all means avoid mathematicians standard analysis. It's wrong. ;) A particular question might be a good place to start. Why do you want to learn more, anyway? -- Smiles, Tony === Subject: Re: Hi mathworld.wolfram.com planetmath.org en.wikipedia.org These three are brilliant for mathematical encyclopedias. http://www.maths.mq.edu.au/~wchen/ln.html This site has some great free lecture notes for pure degree courses. metamath.org A derivation of the whole of mathematics from the axioms of logic and set theory (hard). === Subject: Re: Seven Cards >order - what would be considered statistically significant ie if I got 4 out >of 7 right would this be ok? Is this down to the tail in the PDF or some >such? What would be considered to be beyond chance? It depends on the rules of the game. So does the answer to your original question, actually. The people here who answered it made assumptions which weren't clear from your original post. If you get a card wrong, are you told it is wrong before your next guess? Are you told what the right card is? The more information you have, the easier it is to guess. If you are told nothing about whether you got the card right or wrong, and you don't repeat any guesses, your original answer of 1/5040 is the correct probability of getting all 7 right. There are 70 ways to get exactly 4 right, so the chances of getting 4 out of 7 right are 1/72. P(7) = 1/5040 P(6) = 0 P(5) = 21/5040 = 1/240 P(4) = 70/5040 = 1/72 --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: Continuum hypothesis <43adfa6a$24$fuzhry+tra$mr2ice@news.patriot.net> <43b2e169$43$fuzhry+tra$mr2ice@news.patriot.net> <43b44431$6$fuzhry+tra$mr2ice@news.patriot.net> <43b55e8c$13$fuzhry+tra$mr2ice@news.patriot.net> 01/01/2006 at 01:01 PM, Han.deBruijn@DTO.TUDelft.NL said: >Allright. Would you also object if I replace the word set by >sequence Of course; they're very different. >and say that any sequence is finite, If you said that then I'd have to ask you what set theory you were working with. In, e.g., ZF, there are infinite sequences. >Can you accept a sequence like 1,2,3,4,5, ... ,N and the idea that >such a sequence can still exist for the limit case N -> oo ? That would depend on how you defined the limit case N -> oo. >If not, what's your problem? My problem is that you refuse to either use standard definitions or to stipulate your nonstandard definitions. My problem is that you simultaneously deny the existence of infinite sets and make claims as to their existence. Both of which I've repeatedly pointed out. >Would you please explain what this basic problem has to do with >it? Because as a consequence of the Axiom of Archimedes, you cannot have an infinite sequence with identical nonzero terms whose partial sums are bounded, and thus you cannot have such a sequence whose partial sums converge to 1. >I think I've found on the Internet what the meaning is of >Archimedian: there exists no infinitesimals in R? Right? Wrong? That would depend on how you defined infinitesimal. >What does it mean? For any x>0 and y>0, there exists a natural n such that n*x>y. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Continuum hypothesis <43adfa6a$24$fuzhry+tra$mr2ice@news.patriot.net> <43b2e169$43$fuzhry+tra$mr2ice@news.patriot.net> <43b44431$6$fuzhry+tra$mr2ice@news.patriot.net> <43b55e8c$13$fuzhry+tra$mr2ice@news.patriot.net> <43ba6b08$19$fuzhry+tra$mr2ice@news.patriot.net Because as a consequence of the Axiom of Archimedes, you cannot have > an infinite sequence with identical nonzero terms whose partial sums > are bounded, and thus you cannot have such a sequence whose partial > sums converge to 1. Is it so bad? Consider instead the integral (0..1) 1.dx , which is equal to 1. It can be calculated as limit(N -> oo) N.1/N = 1, starting with a subdivision of the (0..1) interval in N pieces. This is an _finite_ sequence with identical nonzero terms 1/N whose sum is bounded=1. The point is that N becomes infinite while 1/N becomes zero. Both at the same time ! Hence the limit. Hence the Axiom of Archimedes is simply not applicable for such a case. Very much the same thing happens with probabilities as I define them on the naturals, resulting in a theorem like P(a natural is divisible by a) = 1/a . The point is, again, that such a probability for each individual natural goes to zero, yes. But the number of naturals goes to infinity at the same time ! Exactly like with the above integral, these two compensate each other. So I still cannot comprehend what the flaw is in my reasoning. Han de Bruijn === Subject: Re: Continuum hypothesis <43adfa6a$24$fuzhry+tra$mr2ice@news.patriot.net> <43b2e169$43$fuzhry+tra$mr2ice@news.patriot.net> <43b44431$6$fuzhry+tra$mr2ice@news.patriot.net> <43b55e8c$13$fuzhry+tra$mr2ice@news.patriot.net 12/30/2005 >Any set of events is finite. > This isn't sci.engineering. Terms have well defined meanings, and your > statement is demonstrably false in, e.g., ZF. Now, you can specify > your own set theory and discuss what is true in it, but to pretend > that others are talking about your private theory is dishonest. > Allright. Would you also object if I replace the word set by > sequence > and say that any sequence is finite, I mean: from the start? (Sequences > may be better things to work with, after all) Can you accept a sequence > like 1,2,3,4,5, ... ,N and the idea that such a sequence can still > exist for the limit case N -> oo ? So you're saying that there exists an infinite sequence (i.e., it makes sense to talk about it as a mathematical object)? For example, does the sequence (1, 2, 3, 4, ..., n, ...) exist in the limit? Isn't that just the same as the naturals? > If not, what's your problem? Consistency? > [ .. snip .. snip .. ] To take a limit as N->oo there has > to be a function of N that converges. Now, you could define P(E_N_j) > and discuss the limit N->oo for a specific j, but that still wouldn't > do what you want. The basic problem is that R is an Archimedean field. > Would you please explain what this basic problem has to do with it? > I think I've found on the Internet what the meaning is of > Archimedian there exists no infinitesimals in R? Right? Wrong? What does it mean? R has the Archimedean property: http://en.wikipedia.org/wiki/Archimedean_property By asking for a uniform distribution P on the naturals, we require that there be some /particular/ p in R such that for /all/ natural numbers i, P(E_i) = p; and that the sum of P(E_i) over /all/ naturals be 1; i.e., lim n->oo sum(i = 1 to n, P(E_i)) = lim n->oo sum(i = 1 to n, p) = 1. The basic problem is that since R is Archimedean, then given p in R with p <> 0, we have that lim n->oo (sum(i = 1 to n, p)) fails to converge. Proof: Assume p > 0. Exists integer m > 0 such that m*p > 1 by the Archimedean property. Therefore, for all a > 0, exists k = m*a such that for all n > k, sum(1 to n, p) = n*p > k*p = (m*a)*p = (m*p)*a > a; thus the sum is unbounded. A similar argument follows for p < 0. And if p = 0, then lim n->oo (sum(i = 1 to n, p)) = 0 by any reasonable meaning of limit. In none of these cases (p<0, p=0, p>0) is the sum equal to 1; by the trichotomy law, p must fall into one of those three cases; therefore no P exists meeting our needs. The thing that seems to elude you is that we then don't need to worry about missing such a creature, possibly constructed through some strange and elusive limit process no one has ever thought up. Its existence is logically inconsistent with the /definition/ of a uniform distribution on the naturals. If you think such a thing still /could/ exist, you are using a different definition of uniform distribution on the naturals; thus the general penchant for clear definitions in math. One might wish to consider why the above argument does /not/ hold when proposing, say, a uniform distribution on the reals [0,1]; but that's not particularly germane to why it /does/ hold when trying to propose a uniform distribution on the naturals. === Subject: Re: Continuum hypothesis <43adfa6a$24$fuzhry+tra$mr2ice@news.patriot.net> <43b2e169$43$fuzhry+tra$mr2ice@news.patriot.net> <43b44431$6$fuzhry+tra$mr2ice@news.patriot.net> <43b55e8c$13$fuzhry+tra$mr2ice@news.patriot.net So you're saying that there exists an infinite sequence (i.e., it makes > sense to talk about it as a mathematical object)? For example, does the > sequence (1, 2, 3, 4, ..., n, ...) exist in the limit? Isn't that > just the same as the naturals? That's what I hoped for. But the standard naturals seem to have additional properties. And these do not automatically show up as the above sequence becomes infinitely large (whatever that means according to you). > If not, what's your problem? > Consistency? Consistency indeed, but within mathematics AND beyond. [ ... reasoning with Archmedian property involved skipped for now; should be here: ] Take the finite sequence of naturals (1 ... n) . When I count the even and odd numbers (cumulative) _within_ that sequence, I find the following: If n = even then the number of even integers in (1..n) is n/2 and the number of odd integers in (1..n) is n/2 If n = odd then the number of even integers in (1..n) is (n-1)/2 and the number of odd integers in (1..n) is (n+1)/2 Now suppose that the number of elements in (1..n) is very large. Then we can define the probability P that a natural in (1..n) will be even/odd: If n = even then P(even) = (n/2) / n = 1/2 and P(odd) = (n/2) / n = 1/2 If n = odd then P(even) = ((n-1)/2) / n -> 1/2 and P(odd) = ((n+1)/2) / n -> 1/2 In the limiting case, of an infinitely large interval (1..n), quite as expected, we find: P(even) = P(odd) = 1/2 I'm still at lost what precisely would be the flaw in this reasoning. A generalization could be the theorem that P(a natural n is divisible by a) = 1/a . I _know_ that mainstream mathematics finds it wrong to say this, but I can not comprehend why. Han de Bruijn === Subject: Re: Continuum hypothesis > 12/30/2005 >Any set of events is finite. > This isn't sci.engineering. Terms have well defined meanings, and your > statement is demonstrably false in, e.g., ZF. Now, you can specify > your own set theory and discuss what is true in it, but to pretend > that others are talking about your private theory is dishonest. > Allright. Would you also object if I replace the word set by > sequence > and say that any sequence is finite, I mean: from the start? (Sequences > may be better things to work with, after all) Can you accept a sequence > like 1,2,3,4,5, ... ,N and the idea that such a sequence can still > exist for the limit case N -> oo ? > So you're saying that there exists an infinite sequence (i.e., it makes > sense to talk about it as a mathematical object)? For example, does the > sequence (1, 2, 3, 4, ..., n, ...) exist in the limit? Isn't that > just the same as the naturals? > If not, what's your problem? > Consistency? > [ .. snip .. snip .. ] To take a limit as N->oo there has > to be a function of N that converges. Now, you could define P(E_N_j) > and discuss the limit N->oo for a specific j, but that still wouldn't > do what you want. The basic problem is that R is an Archimedean field. > Would you please explain what this basic problem has to do with it? > I think I've found on the Internet what the meaning is of > Archimedian there exists no infinitesimals in R? Right? Wrong? What does > it mean? > R has the Archimedean property: > http://en.wikipedia.org/wiki/Archimedean_property > By asking for a uniform distribution P on the naturals, we require that > there be some /particular/ p in R such that for /all/ natural numbers > i, P(E_i) = p; and that the sum of P(E_i) over /all/ naturals be 1; > i.e., lim n->oo sum(i = 1 to n, P(E_i)) = lim n->oo sum(i = 1 to n, p) > = 1. > The basic problem is that since R is Archimedean, then given p in R > with p <> 0, we have that lim n->oo (sum(i = 1 to n, p)) fails to > converge. > Proof: Assume p > 0. Exists integer m > 0 such that m*p > 1 by the > Archimedean property. Therefore, for all a > 0, exists k = m*a such > that for all n > k, > sum(1 to n, p) = n*p > k*p = (m*a)*p = (m*p)*a > a; > thus the sum is unbounded. A similar argument follows for p < 0. > And if p = 0, then lim n->oo (sum(i = 1 to n, p)) = 0 by any reasonable > meaning of limit. > In none of these cases (p<0, p=0, p>0) is the sum equal to 1; by the > trichotomy law, p must fall into one of those three cases; therefore no > P exists meeting our needs. > The thing that seems to elude you is that we then don't need to worry > about missing such a creature, possibly constructed through some > strange and elusive limit process no one has ever thought up. Its > existence is logically inconsistent with the /definition/ of a uniform > distribution on the naturals. > If you think such a thing still /could/ exist, you are using a > different definition of uniform distribution on the naturals; thus > the general penchant for clear definitions in math. > One might wish to consider why the above argument does /not/ hold when > proposing, say, a uniform distribution on the reals [0,1]; but that's > not particularly germane to why it /does/ hold when trying to propose a > uniform distribution on the naturals. Nice! === Subject: Re: Continuum hypothesis <43adfa6a$24$fuzhry+tra$mr2ice@news.patriot.net> <43b2e169$43$fuzhry+tra$mr2ice@news.patriot.net> <43b44431$6$fuzhry+tra$mr2ice@news.patriot.net> <43b55e8c$13$fuzhry+tra$mr2ice@news.patriot.net> The basic problem is that R is an Archimedean field. with p <> 0, we have that lim n->oo (sum(i = 1 to n, p)) fails to > converge. 01/01/2006 >maybe it can display mathematical formulae with LaTeX codes input, >like what some other maths fora do. Rendering is done by software. Since Usenet is defined as text, news clients can all handle text but can't all handle extensions. There are many different news clients. That's very different from a forum where all readers use the same software. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Two Questions on Probability and Measure >> 1. Let Y_n and Z_n be two independent sequences of iidrvs. >> E[(logY_1)+]minus infinity. Show that the >> series sum_{n=1..infinity}(Y_n Z_1 ... Z_n) converges to a finite >> number almost surely. >This is false. There are trivial counterexamples. Perhaps you left >out some assumptions. The Z condition needed is E(log|Z_1|) < 0. It can fail to exist and be minus infinity; this is adequate. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: Two Questions on Probability and Measure >Oh yes, you are right. I am so sorry. The conditions I missed out are >Y_1>0 and 0 1. Let Y_n and Z_n be two independent sequences of iidrvs. > E[(logY_1)+]minus infinity. Show that the > series sum_{n=1..infinity}(Y_n Z_1 ... Z_n) converges to a finite > number almost surely. Hints: (a) If a > 0, E[ln(Y_1)+] > a sum_{j=1}^infty Prob{ln(Y_n) > an}, implying that almost surely ln(Y_n) <= an for all but finitely many n. (b) Find b > 0 such that for all but finitely many n, sum_{j=1}^n ln(Z_n) <= -bn. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Two Questions on Probability and Measure === Subject: Re: Degree of a map S^1 -> S^1 12/31/2005 at 11:56 PM, Rufus.Zee@gmail.com said: >So the basic technique that is being employed is to view first f not >as a map from the circle to itself, but rather as a path on the >circle by making the identification of the circle with the unit >interval. The fact the endpoints must be identified makes these >special paths; in fact loops. (However not with a fixed base >point). Homotopy works equally well whether you use f:S^1->X or f:[0,1]->X with f(0)=f(1). >We would expect there to be something like the lift of a >composition is the composition of lifts but of course this mode of >thought is shot down by the fact that we cannot compose the lifts >since they are functions [0,1] to R. That's not an obstacle when you have the constraint f(0)=f(1). -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: proving orthonormal basis, struggling with completeness Hello im trying to prove in L^2 ([-pi, pi]) that E_n is a orthonormal basis where E_n let e_0 (t) =1/sqrt(2*Pi) E_n (t) =(sin (nt))/sqrt(Pi) n>0 E_n (t) =(cos (nt))/ sqrt(Pi) n <0 i have shown it is orthnormal, i just cannot seem to show that it is a complete basis (eg every function is L2 ([-Pi,Pi]) can be expressed in terms of a summation of a_n *E_n 's where a_n is a sequence in the reals..) === Subject: Re: proving orthonormal basis, struggling with completeness >Hello >im trying to prove in L^2 ([-pi, pi]) that E_n is a orthonormal basis >where E_n let >e_0 (t) =1/sqrt(2*Pi) >E_n (t) =(sin (nt))/sqrt(Pi) n>0 >E_n (t) =(cos (nt))/ sqrt(Pi) n <0 >i have shown it is orthnormal, i just cannot seem to show that it is a >complete basis (eg every function is L2 ([-Pi,Pi]) can be expressed in >terms of a summation of a_n *E_n 's where a_n is a sequence in the >reals..) What theorems do you know about convergence of Fourier series? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: proving orthonormal basis, struggling with completeness I havent come across fourier series in this course, so i would like to use a method that doesnt use fourier series, I do know a little about fourier series though, so do you mind me asking what theorems i could use? === Subject: Re: proving orthonormal basis, struggling with completeness >I havent come across fourier series in this course, so i would like to >use a method that doesnt use fourier series, >I do know a little about fourier series though, so do you mind me >asking what theorems i could use? Well, the point is that the completeness of this orthonormal set _is_ a statement about Fourier series. One common approach is to first use the fact that functions in some suitable subspace of L^2 can be approximated by linear combinations of the basis elements, and then show that this subspace is dense in L^2. You could use Dirichlet's convergence theorem for differentiable functions, or the convergence of the Cesaro means for continuous functions, or the Stone-Weierstrass Theorem, for example. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: GCD(0,0) >> >>[...] > I think the following is the standard: > > a|b iff a<>0 and b=a*x for some x. >>I always thought it was standard to leave out the part a<>0 . >> >> That's just the point. For some of these concepts there is a clear >> standard, nearly universal, while for others there are competing >> alternatives. >> >> For a|b, the choice as to whether or not to bar 0|0 is not so >> universal, but it is barred in most of the texts I've looked at. >> >> So if you think it's standard to leave out the restriction a<>0, how >> about some references? I'm sure there are references that allow 0|0 as >> well as many that bar it. But here too, the power of a reference >> depends to some extent on whether the author's text is itself viewed >> as a standard. >> >> If it's clear that there is no real standard, then you can't insist >> one is right -- it becomes author's choice. However, in such cases, >> it's only fair for the author to warn the reader of competing >> definitions or at least clearly declare the choice being made. >You are right here. Certainly local clarity is valued higher >than global consistency (after all this is mathematics, not engineering). >For references see below. >> However if there is an overwhelming standard, then in that case, using >> a nonstandard definition is surely confusing. It's still author's >> choice, but if an author does too much of that without some very good >> reasons explained in advance, the resulting confusion will typically >> cause the text to be rejected by the target audience, Who wants to >> constantly be doing mental translations between nonstandard and >> standard? >> > The above definition works in any commutative ring, and while it > doesn't require x to be unique, it does bar 0|b. > > But even though I prefer 0 not to be regarded as a divisor of > anything, I can see how the case can be made for 0|0. >>By not imposing artificial conditions. While it is at least >>conceivable to bar 0 altogether from the natural numbers, >> >> In fact, for many books (not all) N = Z+, that is, {natural numbers} = >> {positive integers}. However this definition is far from universal, >> the alternative being N = Z+ union {0}, that is, {natural numbers} = >> {nonnegative integers}. >> >> I think (but I'm not sure about this) that most number theory books >> use N = Z+, while I know some computer science and discrete math texts >> use N = Z+ union {0}. >> >>once it is accepted there, it should not be barred from divisibility. >> >> Not necessarily. Many (and perhaps most?) abstract algebra and ring >> theory texts define a|b in a ring theory context (where positive vs >> nonnegative is not applicable), but still bar a=0 in the notation a|b. >> > After all 0=x*0 would seem to suggest that 0 is both a multiple of 0 > and a factor of 0. > > I have a feeling there's less agreement by texts as to whether 0|0 > than as to whether gcd(0,0)=0. >>Some people like maximal elements. >> >> I think we all do, depending on what's being maximized. >> > Certainly any text which bars 0|0 would also make gcd(0,0) undefined. > > But conceivably, there are some texts where 0|0 is accepted but for > which gcd(0,0) is still undefined (because of the g in gcd). >>A common misconception, see <41levcF1fatb8U1@news.dfncis.de>. >> >[...] >| If you unfold the definition if GCD you will see, that GCD(m,n) >| is just the infimum of m and n with respect to the order |. >| (also LCM(m,n) is the supremum of m an n). >| Hence GCD(m,m)=m and in particular GCD(0,0)=0. >| >| If you consider all integers, the relation | is not antisymmetric >| any more. But you still have the notion of a greatest element, >| and _a_ GCD(i,k) is _a_ greatest element of the set of common divisors >| of i and k. >| >| That the _usual_ order on the naturals is not really important for >| the concept of GCD becomes clear when more general Rings (i.e. k[X]) >| are considered. >However, gcd is of course not the infimum with respect to the usual order >on the naturals, even if 0 is excluded from consideration. >So, using | instead of <= produces the notions of gcd and lcm as >instances of simple and well known general concepts. This is IMO a plain >down to earth _mathematical_ reason, why | should be used. >A more didactical reason lies in the above observation that one cannot >expect a counterpart of <= in other rings, but still has |; so one >might as well start right away with the more sustainable choice. >Note, that so far I do not insist on introducing gcd(m,n) for >the case mn=0; the main point is the sensible choice of greatest. >It makes sense to chose some submonoid of (R,*) and the corresponding >restriction of |. >[...] > For the integers, the traditional definition of gcd use the word > greatest in the normal way. However, the only way to consistently > generalize gcd to rings is to base the definition of gcd on the > divisibility preorder. That doesn't mean you have to redefine it for > the integers since it's sufficient to observe that the evaluation of > gcd would stay the same. > Of course you could redefine gcd for the ordinary integers so as > have exactly the same definition as for general rings, and it's hard > to argue against that -- it really does makes sense, > So I'll concede that point -- that for consistency across rings in > general, the definition of gcd for the ordinary integers should be > based on the divisibility partial order on the positive integers. But > just because I think it should be defined in a certain way, doesn't > mean I can single-handedly change the standard. The question is, whether it is actually standard, even for the integers. Of course the phrase greatest common divisor of a and b may well be applied to (nonzero) natural numbers a and b without ever specifying whether greatest refers to <= or |. But already for the integers (even if restricted to nonzero arguments) using <= makes the reason for possible choice among associated elements appear quite unmotivated whereas it appears obvious with |. The basic requirements (i) gcd(a,b) divides both a and b (ii) every d that divides both a and b also divides gcd(a,b) are always used, even for the integers; they are part of the standard definition in textbooks. So it is not necessarily a matter of redefining but just of clearly stating what the definition says. > However this post was mainly focused on gcd(0,0), with some posters > insisting that gcd(0,0)=0. For me, that went against the grain for > several reasons: > (1) Various posters were asserting the validity of gcd(0,0)=0 as if it > were a known fact. In fact, after reading those claims, I suddenly > wasn't sure if I had it right. But after doing some checks, it became > clear that the overwhelming majority of reputable texts define gcd in > such a way as to preclude gcd(0,0). As I mentioned, most of the time this happens because they only consider gcd(a,b) for nonzero a and b. I do not argue against such a restriction. Those texts most often introduce gcd etc. in the context of PID's where this restriction works fine. > (2) In my understanding, 0 is not a divisor of anything, not even > itself. In other words, 0|b is barred for all b, including b=0. But if > I were to accept gcd(0,0)=0, that would mean I would also have to > accept 0|0. With so many posts asserting gcd(0,0)=0, I also began to > worry about whether barring 0|0 was standard or not. It appears that > most reputable texts do bar 0|0, but this is less universal than the > barring of gcd(0,0). Again, I would explain this as in (1) above. > (3) There's really no need to allow 0|0 or gcd(0,0)=0. In practice, it > simply doesn't apply. Since the conflict with the traditional > definition has the potential to cause confusion, and since there's not > enough benefit, why bother? Whether there is need to allow 0|0 or gcd(0,0)=0 certainly depends on the particular situation. Again, I have no reason to argue against leaving gcd(0,0) undefined if it suits the particular situation. And I do not think those posters arguing for gcd(0,0)=0 had in mind to have it defined compulsory. Instead their (as well as my) point was, that _if_ you want to define gcd(0,0) then gcd(0,0)=0 is the only sensible choice, that is consistent with the usual definition of the gcd for nonzero arguments. >[...] >> To recap -- we appear to disagree on what's standard. >> >> I claim: >> >> (1) Most (but not all) reputable texts define the notation a|b as: a|b >> iff a<>0 and b=a*x for some x. >> >> (2) The overwhelming majority of reputable texts define gcd in a way >> that precludes giving a value to gcd(0,0). >> >> If you think otherwise, let's see some references. >I claim: >(a) they do not put any such funny condition as in (1) >on the relation |. Instead they restrict | to a suitable submonoid >of the multiplicative monoid of the ring in question, for which they >define the gcd. > Well, that sounds nice -- restrict it to a submonoid, but that's > simply not consistent with standard usage. The standard does allow a|0 > if a<>0, but bars 0|b for any b. Thus, there's no submonoid since the > relation doesn't allow 0 on the left but does allow it on the right > (provided the left is nonzero). I was only referring to the use of | as far as it concerns the definition of gcd. But I even doubt that your (1) above is standard usage. I have never seen that. Either those texts only consider this relation among nonzero a and b (in which case the extra condition is redundant), or they do not restrict | at all. There are even respectable texts which demolishing both our claims: Atiyah, Macdonald: Introduction to commutative algebra (add-wesl, 1969) Chap. 1 , (Section on zero-divisors, nilpotent elements and units), p2: | A zero-divisor in a ring A is an element x which divides 0, i.e., an | element for which there exists y =/= 0 in A such that xy = 0. | A ring with no zero-divisors =/= 0 (and in which 1 =/= 0) is called | an integral domain. So for Atiyah/Macdonald 0|0 but a|0 only if a is a zero divisor. > Of course, I did point out that while most reputable texts bar 0|0, > some do allow it. Still, the standard is clear. Now if you want to > allow 0|0, fine, but it has to be specified since it's not the > default. >(b) for a, b in this submonoid, the gcd(a,b) is defined as a greates >common lower bound of a and b w.r.t. the restriction of |. > As pointed out above, the submonoid idea doesn't work if we want > gcd(a,b) be defined when exactly one of a,b is zero. I wonder if this is so widespread. >(c) that submonoid may or may not include 0. If it does not, then gcd(0,0) >is obviously not defined. If it does, then gcd(0,0)=0, and this follows >automatically from the definition in (b) via gcd(a,0) = a = gcd(0,a). >There is absolutely no need to regard gcd(0,0)=0 as a special case. >My first reputable choice is >Saunders Mac Lane, Garret Birkhoff: Algebra (Chelsea, 1967, 3rd ed 1993) >[...] >My second reputable choice is >[ unfortunately I do not have it with me right now, so it is from memory ] It is at the bottom of page 111 in that edition. >He considers only an entire ring, and defines the gcd as follows: >suppose ab =/= 0 [ so both a and b are nonzero ] >then d is a greatest common divisor if d divides a and b and every >other common divisor of a and b divides d. >It may surprise you, that I claim him as witness for me, although he >obviously does define gcd only for nonzero arguments. But in fact, this >is consistent with my claim: he _first_ chooses a suitable submonoid >of (A,*), in this case A{0}, then uses the restriction of | to this >submonoid (without any artificial conditions) and then defines gcd(a,b) >exactly as a greates common lower bound of a and b w.r.t. |, like in >the first example. > Well, I also cited Lang as a reference since he does bar gcd(0,0), > and that's the _main_ focus of this thread. So you're back to even. > One text in favor of gcd(0,0)=0 (MacLane & Birkhoff), one text that > bars it (Lang). > As far as your other point, whether you conceptualize it in terms of a > submonoid or not, the effect is still to allow or bar certain values > from being used. There's no real difference, provided the acceptable > inputs are the same. However, as I pointed out, for Lang's definition, > you can't really use the submonoid concept anyway (without making > exceptions to that) since Lang does allow gcd(a,0) if a<>0. Not at all. Lang introduces gcd(a,b) _only_for_entire_rings_, where the condition ab =/= 0 and a=0 & b=0 are equivalent. >My third reputable choice is >Heinz L.9fneburg: Vorlesungen .9fber lineare Algebra (B.I. Mannheim, 1993) >Kap. 3 (Ringe und K.9arper), s.73: >| Sind a,b,c in Z, so hei¤t c gr.9a¤ter gemeinsamer Teiler von a und b, >| falls c ein Teiler von a und b ist, und falls jeder gemeinsame Teiler >| von a und b auch ein Teiler von c ist. >At this stage, divisibility seems to be taken for granted for integers. >Later there follows a definition of | and the gcd: >Kap. 4 (Polynomringe), s.91/92: >| Es sei R ein kommutativer Ring und a und b seinen Elemente von R. >| Man nennt b Teiler von a, falls es ein c in R gibt mit a=bc. >| >| Es sei R ein Integrit.8atsbereich und a, b und c seien Elemente von R. >| Genau dann hei¤t c gr.9a¤ter gemeinsamer Teiler von a und b, wenn gilt >| >| 1) c ist Teiler von a und auch von b, dh., c ist gemeinsamer Teiler >| von a und b. >| >| 2) Jeder gemeinsame Teiler von a und b teilt c. >| >| Da a und b in aller Regel viele gr.9a¤te gemeinsame Teiler haben, bezeichnen >| wir die Menge ihrer gr.9a¤ten gemeinsamen Teiler mit ggT(a,b). Im Ring Z der >| ganzen Zahlen gilt ggT(4,-6) = ggT(2,-2) = {2,-2} und ggT(0,0) = {0}. > While I can't read German, I get the sense of the last line -- the > author is explicitly giving gcd(0,0)=0 as an example. So now you have > 2 texts in favor of gcd(0,0)=0, 1 against. Oh, actually I thought you even were german (hard to guess with pseudonyms), but I must have confused you with somebody else. Essentially he uses unrestricted | for general rings. He introduces the gcd in the context of integral domains, but (unlike Lang) puts no restriction on the arguments of the gcd. One of my reasons for giving this example was to show that the issue comes up already in basic texts (linear algebra is a first year course at our universities) because polynomial rings are considered early. > I'm not knowledgeable about German authors, but I'll assume that Heinz > L.9fneburg is a reputable author. However, I do know of one famous > German author -- Van Der Warden. I wonder how Van Der Warden would > choose on this issue. I do not know. Perhaps similar to Lang. >My fourth reputable choice is >Hideyuki Matsumura: Commutative Ring Theory (CUP 1986) >Section 20 (UFDs), p163/4: >| Let A be an integral domain; for any two non-zero elements a,b in A. >| The notion of greatest common divisor (g.c.d) and >| least common multiple (l.c.m) are defined as in the ring of integers. >| That is, d is a g.c.d. of a and b if d divides both a and b, and any >| element x dividing both a and b divides d;[...] >which is the same as in Lang. > Not exactly the same. > Matsumura won't allow gcd(a,b) unless a,b are both nonzero, while Lang > allows gcd(a,b) unless a,b are both zero. As pointed out above, both allow only nonzero a and b. > Take another look at the title of this thread. Matsumura bars > gcd(0,0). Thus, you're back to break even on this issue since 2 of > your 4 references bar gcd(0,0). Well at least all four examples support my interpretation (a)-(c). I admit that I did read the title differently, perhaps because this question comes up with every new generation of students. As already mentioned above, I have no problem with restricting the gcd, resulting in gcd(0,0) not being defined. I have also no problem with restricting [x -> x^2] resulting in 0^2 not to be defined. I only fail to see any compelling mathematical reason for doing either. That some textbooks exclude (0,0) as possible arguments for the gcd _without_ any proper reason or explanation just adds to the confusion of students, who may well think that there is an inherent _mathemathical_ difficulty which might prevent considering gcd(0,0). >[...] > I'll admit though, the arguments against gcd(0,0)=0 are not that > strong. At this point, I could take either side. It's just a question > of what's standard. We already agreed, that it is more difficult to insist on a fixed standard in mathematics than in other fields. I also agree with your remark, that authors should at least warn their readers about that, whenever variants of a particular definition are in use. However, I think the question of this perceived standard usage should not prevent us from looking at mathematical considerations when it comes to the choice of a definition. Marc === Subject: Re: GCD(0,0) It's not defined. But there is no great problem with this. (0,0) just isnt a point in the domain of GCD (the domain is ZxN). It's just like saying sin(My face) or log(the Eiffel Tower) aren't defined. Adam. === Subject: Re: GCD(0,0) <7580088.1136308285802.JavaMail.jakarta@nitrogen.mathforum.org It's not defined. But there is no great problem with this. > (0,0) just isnt a point in the domain of GCD (the domain is ZxN). > It's just like saying sin(My face) or log(the Eiffel Tower) aren't defined. > Adam. If only the real world was as black and white as you wish it were: Excel ----- GCD(0,0) 0 === Subject: Re: GCD(0,0) <9egcr1lilaq1j8g0td38ugpj6lc6f7qaf7@4ax.com> <41oahgF1f73caU2@news.dfncis.de> <41tbruF1f44ddU1@news.dfncis.de> <6vuir1ts9h2kjj6vjh3bo9b1di42jvqrec@4ax.com> I'll admit though, the arguments against gcd(0,0)=0 are not that strong. At this point, I could take either side. It's just a question of what's standard. ------------------------- Hi Quasi, With all due respect, I expect you'd find that most mathematicians would replace your It's just a question of what's standard with It's just a question of what's useful. And I don't really understand the idea of taking sides for or against a definition. In an elementary number theory text that doesn't go beyond the integers (as in my book), there may be no particular reason to assign a value to gcd(0,0). In other contexts, such as when working with Dedekind domains and the like, it's often more convenient to think of the gcd as being an ideal, and with the appropriate definition in that situation, one gets gcd(0,0)=0. To try to illustrate the difference between accepting what's standard and accepting what's useful, let me describe two definitions of generalized greatest common divisors, one from a recent paper of Corvaja and Zannier, one from a recent paper of mine. My point here is that people seem to spend a lot of time arguing over which definition is right, but as others have pointed out in this and other threads, a definition is neither right nor wrong. How generally you want to define something depends on what you're trying to accomplish. Def [C-Z]: Let a and b be algebraic numbers, say both in the number field K. The generalized gcd of a and b is the following product over all (normalized) absolute values v of K: product of max { |a|_v, |b|_v }. Notice that this interesting definition assigns a gcd to rational numbers, not just to integers. And in number fields, it tends to give something rather different than using ideals. Corvaja and Zannier then proceed to prove a very deep and beautiful theorem giving a highly nontrivial upper bound for gcd(a-1,b-1). Okay, now here's another definition that I realize may include a bunch of unfamiliar terminology. Def [JS]: Let X be the variety P^1 x P^1 blown up at the point (0,0) and let E be the exceptional divisor. Then for any point (a,b) not equal to (0,0), the generalized (logarithmic) gcd of a and b is the Weil height of (a,b), considered as a point of X, with respect to the divisor E. This definition does have your preferred restriction that (a,b) is not (0,0). OTOH, it's also only well-defined up to a bounded amount (that's independent of a and b). What's the point of such a complicated definition? Well, this definition can be readily generalized to define gcds in very general situations in the context of modern algebraic geometry. And then one can relate it to a wonderful conjecture of Paul Vojta, which is then seen to imply something interesting about gcds over the ordinary integers. So would I advocate explaining either of these definitions of gcd in an introductory number theory class. Of course not. But would I expect the graduate students working with me to be able to understand and work with these definitions. Yes, absolutely. Anyway, sorry this got a bit long. But really, definitions evolve by generalization and, sometimes they even get changed because it turns out that the original definition wasn't the one that turned out to be most useful. (The definition of scheme comes to mind here. I believe the original definition of scheme included an extra condition that is no longer used. And schemes lie at the heart of modern algebraic geometry.) === Subject: Re: GCD(0,0) >I'll admit though, the arguments against gcd(0,0)=0 are not that >strong. At this point, I could take either side. It's just a question >of what's standard. >------------------------- >Hi Quasi, >With all due respect, I expect you'd find that most mathematicians >would replace your It's just a question of what's standard with It's >just a question of what's useful. When this thread was born, the original question was whether or not gcd(0,0) was defined and if so, defined to be what? Generally a student doesn't have the luxury of choosing their own preferred definition. The professor can, and a textbook can, but a student is stuck with whatever definition is being used in the context of their course. Often a course will use some terminology and definitions from prior courses without explicitly defining them. This puts the burden on the student to find the appropriate interpretation. There are terms that are so absolutely standard that decoding is easy -- just look up the term in any reputable text or web site. For other terms, there are competing alternatives and it's often a dilemma as to which choice is the right one in the context. Of course, asking the professor is one way to resolve the issue, but that's not always an option. My main point is that unless you have the authority to select your own definition, you really don't have a choice. Thus for students, in a given context, some definitions are right, others are wrong. But there's another point here that's also important. If you've browsed and/or participated in some of the sci.math crank threads, it should be obvious that many of the arguments are based on lack of adherence to standard definitions. If everyone rolls their own and each applies their own definition when interpreting the arguments of others, the result is an unbridgeable communication gap.In these threads, the non-cranks often try to force the cranks to use the terminology consistent with the standard definitions (often to no avail). It's a little hypocritical to say, say that it's not ok for the cranks to reuse a standard term in some way of their own, but it is ok if the non-cranks do it. On the other hand, if you're saying, go ahead, no definition is wrong, choose whichever one is more useful, then the resulting chaos will make you regret you ever said that. I pointed out earlier that students generally don't have the right to choose their own definitions for standard terms. But students are not the only ones who don't have full freedom of choice in this respect. In fact, textbook authors and professors really can't innovate that much either, except in the more advanced courses. For all the core math subjects, if a text strays from the standard too many times, the text will probably get rejected. Similarly, if a professor chooses too many custom definitions while teaching a standard course, the professor may end up getting rejected by students and rebuked by the department. So the luxury of choosing a definition based on what's useful is just that -- it's a luxury reserved for those who are in a position to choose. For the rest of us, it's a question of what definition is correct in a given context. >And I don't really understand the idea of taking sides for or against a definition. No, I'm sure you do understand -- you've been there before, we all have. There are some definitions that are so bad that anyone with any mathematical common sense would agree that the definition is seriously flawed -- that is everyone else except the author or professor who, having the power to choose, actually made those horrendous choices. >In an elementary number theory text that doesn't go beyond the integers >(as in my book), there may be no particular reason to assign a value to >gcd(0,0). In other contexts, such as when working with Dedekind domains >and the like, it's often more convenient to think of the gcd as being >an ideal, and with the appropriate definition in that situation, one >gets gcd(0,0)=0. >To try to illustrate the difference between accepting what's standard >and accepting what's useful, let me describe two definitions of >generalized greatest common divisors, one from a recent paper of >Corvaja and Zannier, one from a recent paper of mine. My point here is >that people seem to spend a lot of time arguing over which definition >is right, but as others have pointed out in this and other threads, a >definition is neither right nor wrong. How generally you want to define >something depends on what you're trying to accomplish. >Def [C-Z]: Let a and b be algebraic numbers, say both in the number >field K. The generalized gcd of a and b is the following product over >all (normalized) absolute values v of K: > product of max { |a|_v, |b|_v }. >Notice that this interesting definition assigns a gcd to rational >numbers, not just to integers. And in number fields, it tends to give >something rather different than using ideals. Corvaja and Zannier then >proceed to prove a very deep and beautiful theorem giving a highly >nontrivial upper bound for gcd(a-1,b-1). >Okay, now here's another definition that I realize may include a bunch >of unfamiliar terminology. >Def [JS]: Let X be the variety P^1 x P^1 blown up at the point (0,0) >and let E be the exceptional divisor. Then for any point (a,b) not >equal to (0,0), the generalized (logarithmic) gcd of a and b is the >Weil height of (a,b), considered as a point of X, with respect to the >divisor E. >This definition does have your preferred restriction that (a,b) is not >(0,0). OTOH, it's also only well-defined up to a bounded amount (that's >independent of a and b). What's the point of such a complicated >definition? Well, this definition can be readily generalized to define >gcds in very general situations in the context of modern algebraic >geometry. And then one can relate it to a wonderful conjecture of Paul >Vojta, which is then seen to imply something interesting about gcds >over the ordinary integers. >So would I advocate explaining either of these definitions of gcd in an >introductory number theory class. Of course not. But would I expect the >graduate students working with me to be able to understand and work >with these definitions. Yes, absolutely. >Anyway, sorry this got a bit long. But really, definitions evolve by >generalization and, sometimes they even get changed because it turns >out that the original definition wasn't the one that turned out to be >most useful. (The definition of scheme comes to mind here. I believe >the original definition of scheme included an extra condition that is >no longer used. And schemes lie at the heart of modern algebraic >geometry.) The examples you cite are very interesting, but I'll point out that in each case, the generalized gcd was defined explicitly so as not to cause confusion. This thread started with the OP asking for the value, if any, of gcd(0,0). What surprised me was that so many people immediately jumped in and asserted gcd(0,0)=0, leaving the clear impression that was the standard definition. I felt obligated to challenge that since I was pretty sure it was not the standard. I gave some references -- Sierpinski, Hardy, Gilmer, Lang, and I'm sure you would agree that those authors are all highly reputable. To the extent that I took sides, it was from the point of view of elementary number theory. In that context, both definitions are acceptable, and both are equally useful, but I felt barring gcd(0,0) makes the concepts of gcd clearer for the beginning student. But I think there are issues here that go beyond just gcd(0,0). Especially now that we have wikipedia and MathWorld (where I've seen some horrible definitions), both being viewed as official sources, the question as to what's the right choice for a standard definition will increasingly be a subject of heated debate. quasi if that was the standard. === Subject: Re: GCD(0,0) <9egcr1lilaq1j8g0td38ugpj6lc6f7qaf7@4ax.com> <41oahgF1f73caU2@news.dfncis.de> <41tbruF1f44ddU1@news.dfncis.de> <6vuir1ts9h2kjj6vjh3bo9b1di42jvqrec@4ax.com> <1rgjr11cedl16jmbnuu23evrc8opi35u3v@4ax.com> gcd(0,0). What surprised me was that so many people immediately jumped > in and asserted gcd(0,0)=0, In my case, I made no such assertion. I merely pointed out that in two different programming environments, gcd(0,0) returns 0. > leaving the clear impression that was the standard definition. And it may very well be, where software is concerned, since as David Ullrich pointed out, it is a consequence of the Euclidean Algorithm. > I felt obligated to challenge that since I was > pretty sure it was not the standard. And it may very well be that what is standard in programming differs from what is standard in text books. In which case, the student and researcher should at least be aware of the discrepency. > I gave some references -- Sierpinski, Hardy, Gilmer, Lang, and I'm > sure you would agree that those authors are all highly reputable. > To the extent that I took sides, it was from the point of view of > elementary number theory. In that context, both definitions are > acceptable, and both are equally useful, but I felt barring gcd(0,0) > makes the concepts of gcd clearer for the beginning student. As long as they don't get their hands on Excel or Python. Your goal, although admirable, is a lost cause. > But I think there are issues here that go beyond just gcd(0,0). > Especially now that we have wikipedia and MathWorld (where I've seen > some horrible definitions), both being viewed as official sources, > the question as to what's the right choice for a standard definition > will increasingly be a subject of heated debate. > quasi > if that was the standard. === Subject: Re: a Looks like a plug for search engine company MIVA [ http://www.miva.com/uk/content/about/symbol.asp ] Infinity plus growth = infinite growth Don'cha just love those marketing mottos ;-) === Subject: Re: a > a This remembers me Dada's philosophy, from Tristan Tzara. Fernando. === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> <4cKdnVLIfr64ii7eRVn-rw@wideopenwest.com> Go back on your meds, jinny. You have ceased to be amusing. James Powell Of course, Hy'mie, you wouldn't DARE challenge the polls, surveys, studies, and ELECTION RESULTS which prove beyond the shadow of all doubt that sodomy will NEVER be legalized in this putative Christian nation, because you know what this will simply confirm what we already KNOW, which is that you're a pistol packing faggot who should NEVER be permitted to spend ONE SECOND around our White Christian Israelite children. Yes, jews and niggers DO think sodomy is a snivel right, and YES, you deserve nothing less than to spend eternity with both of these amoral, despicable, vile, murderous, hateful, worthless, expensive, and now GONE, races. http://christianparty.net/pollblacks.htm proves that this is NOT what America wants, deserves, NOR SHALL HAVE. John Knight === Subject: Re: Standard Deviation of PISA >Of course, Hy'mie, you wouldn't DARE challenge the polls, surveys, >studies, and ELECTION RESULTS which prove beyond the shadow of all >doubt that sodomy will NEVER be legalized in this putative Christian >nation, http://www.sodomylaws.org/usa/usa.htm Sodomy had been made legal in 36 of the 50 states, before the USSC ruling struck down the other laws as unconstitutional. http://www.sodomylaws.org/ Note that most of the remaining sodomy laws that exist in the world are in black countries, including Liberia (but not Madagascar). The rest of the laws are in countries populated by what you have called mud people. NOT ONE European country has laws against sodomy. http://www.sodomylaws.org/world/world.htm#Europe So much for your visions of white European superiority - maybe it is YOU who should move to Liberia. >Yes, jews and niggers DO think sodomy is a snivel right, The map cited above contradicts you. >and YES, you >deserve nothing less than to spend eternity with both of these amoral, >despicable, vile, murderous, hateful, worthless, expensive, and now >GONE, races. They are still right here, every last one of them. All of your strutting has not caused one person to go into exile. >http://christianparty.net/pollblacks.htm proves that this is NOT what >America wants, deserves, NOR SHALL HAVE. You just said that it already had it, nincompoop. Make up your mindlessness. I suspect that America would like you to sodomize yourself. lojbab === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> <4cKdnVLIfr64ii7eRVn-rw@wideopenwest.com> the nation to deliberately approve same-sex marriages and handing a > political hot potato to an already beleaguered Gov. Arnold Schwarzenegger > (R). > You ignore Proposition 22, which CANNOT be undone, because we the > people PASSED that proposition with flying colors > http://christianparty.net/sodomycalifornia.htm [read: TWO THIRDS OF > CALIFORNIANS OPPOSE LEGALIZED FAGGOTRY] This is the same state which elected the Terminator governor. Your point? > We waste time-- oh so much time!-- fighting one another over symbolic > issues like same-sex marriage and flag-burning while the truly practical > problems, the ones that require compromise and sacrifice and creativity, are > left for later. We fiddle with these water-cooler issues while America burns > with the need for pragmatic ideas and courageous action. Why? Because these > so-called moral issues do not leave room for compromise and thus by > definition do not allow us to come together as a nation and accept one > another's values; instead, they encourage us to be polarized, a division > which more and more cynical politicians are taking advantage of. Every > second wasted tilting at the windmills of the silly flag-burning debate is a > second not spent trying to make our schools better, our streets safer, and > our borders and future more secure. Every minute Congress spends on symbolic > legislation is a minute it does not spend trying to figure out how rapacious > corporations (and evil and greedy individuals) were able to so quickly take > advantage of public funding at Ground Zero, in Iraq, and New Orleans.Andrew > Cohen (http://cbs4denver.com/gavel/local story 362231046.html) > Who gives a rat's a.. what a STUPID faggot jew lawyer thinks [though I > repeat myself]? Yes, jews ARE faggots, so 87% of them think faggotry > is a snivel right, but jews are less than 2% of the population (which > is a grand total of only 1.74%), faggotry is STILL a SIN and a CRIME, > and no matter how much you fudge packing jew faggots think sodomy is > simply a snivel right, THIRTY FIVE STATES have passed laws OUTLAWING > faggot marriages, and it's only a matter of time before ALL states will > have to: Jews are faggots? That might explain the whole circumcision thing, but how are little Jews made then? > Supported by most members of the Liberals, the Bloc Qu.8eb.8ecois and the NDP, > the legislation passed easily, making Canada only the third country in the > world, after the Netherlands and Belgium, to officially recognize same-sex > marriage. > SO WHAT? WE DON'T HAVE TO DO WHAT CANUCKS DO. AND WE WON'T. Very true. The U.S. doesn't have to do that. Which is why in the U.S. water freezes at 32 and boils at 212. And why the U.S. goes on little wars to find weapons which they know don't exist. BTW, Degrassi pwned 90210. === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> <4cKdnVLIfr64ii7eRVn-rw@wideopenwest.com> Jews are faggots? That might explain the whole circumcision thing, but how are little Jews made then? 87% of jews think sodomy should be legalized, so I suppose it's the other 13% who keep the jew race from going the way of the albatross: http://christianparty.net/jewpollsodomy.htm You can bet that, just like faggot Hy'mie, they're too self centered, arrogant, and ignorant to be demanding sodomite rights on behalf of OTHER faggots, particularly honkey faggots [if there is any such thing], so their only possible objective would be to attempt to legalize their own pathetic, amoral, anti-social, destructive, unhealthy, and CRIMINAL behavior. Even though these attempts FAILED miserably, faggots like pistol packing Hy'mie still hold out hope that we'll simply stab God in the Eye and totally abandon His Word. But you can see from the following url that this will NEVER happen in this putative Christian nation (which makes Madagascar, Liberia, and perhaps even Israel the perfect faggot retreats for such folks). http://christianparty.net/sodomy.htm John Knight === Subject: Re: Standard Deviation of PISA >Jews are faggots? That might explain the whole circumcision >thing, but how are little Jews made then? >87% of jews think sodomy should be legalized, It has been. >But you can see from the following >url that this will NEVER happen in this putative Christian nation >(which makes Madagascar, Liberia, and perhaps even Israel the perfect >faggot retreats for such folks). Sodomy is illegal in Liberia, but legal everywhere in the United States. You lose, loser. lojbab === Subject: Re: Standard Deviation of PISA <1-6dnaaBov7SXjPeRVn-vA@wideopenwest.com> <4cKdnVLIfr64ii7eRVn-rw@wideopenwest.com> We waste time-- oh so much time!-- fighting one another over symbolic issues like same-sex marriage and flag-burning while the truly practical problems, the ones that require compromise and sacrifice and creativity, are left for later. We fiddle with these water-cooler issues while America burns with the need for pragmatic ideas and courageous action. Why? Because these so-called moral issues do not leave room for compromise and thus by definition do not allow us to come together as a nation and accept one another's values; instead, they encourage us to be polarized, a division which more and more cynical politicians are taking advantage of. Every second wasted tilting at the windmills of the silly flag-burning debate is a second not spent trying to make our schools better, our streets safer, and our borders and future more secure. Every minute Congress spends on symbolic legislation is a minute it does not spend trying to figure out how rapacious corporations (and evil and greedy individuals) were able to so quickly take advantage of public funding at Ground Zero, in Iraq, and New Orleans.Andrew Cohen (http://cbs4denver.com/gavel/local_story_362231046.html) These IDIOTIC comments from this STUPID faggot fudge packing jew [though I repeat myself] deserve a bit more scrutiny. It's precisely JEWS who encourage[ed] us to be polarized, a division which more and more cynical politicians are taking advantage of, which is why jews must LEAVE, along with the niggers http://blackexile.com 87% of jews think sodomy is a snivel right, which is why they trivialize this WORD OF GOD as a mere water cooler issue: http://christianparty.net/jewpollsodomy.htm This is proof enough that at least 87% of jews are faggots, just like Alan Dirshowitz is a faggot who claimed [ON TAPE] that we have no right never deny us free exercise of religion: What a twisted brain this jew moron [though I repeat myself] has: http://christianparty.net/dirshowitz.htm What did Jesus say about the jews?: Matthew 7:6 Give not that which is holy unto the dogs, neither cast ye your pearls before swine, lest they trample them under their feet, and turn again and rend you. John Knight === Subject: Continuous of integrable I'm not sure about the answer of this problem. Is it true that Let f be an (Lebesgue)integrable on the set of real numbers R. Then the function F(x) = (Integrate from - infinity to x of) f(x) is continuous? If it's true how to prove it and if it's not true what's a === Subject: Re: Continuous of integrable It's more than continuous, it's absolutely continuous. -- Julien Santini === Subject: Re: Continuous of integrable > I'm not sure about the answer of this problem. Is it true that Let f > be an (Lebesgue)integrable on the set of real numbers R. Then the > function F(x) = (Integrate from - infinity to x of) f(x) is > continuous? If it's true how to prove it and if it's not true what's a Yes, it's true. Let a_n --> a monotonely. Now mu(A) = int_A f(x) dx is a measure, and mu((-infinity,a_n]) --> mu((-infinity,a]) either because a_n decreasing to a implies (-inf,a] = intersection of (-inf,a_n], or because (-inf,a] = union of (-inf,a_n] (in case the a_n are eventually constant) or (-inf,a)=union of (-inf,a_n] otherwise. As for why mu is a measure, use the dominated convergence theorem. --Ron Bruck