2009/7/22 Bruno Marchal <marc...@ulb.ac.be>: > Ma connection at home is no functioning.

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As a linguistic aside, Bruno has cleverly expressed the above statement in perfect Glaswegian (i.e. the spoken tongue of Glasgow, Scotland - my home town). Other well-known examples are: "Is'arra marra on yer barra Clarra?" (Is that large vegetable on your barrow a marrow, Clara?); and "Gie's'a sook on yer soor ploom" (Let me taste the "sour plum" (a globular sweet-sour confection) that you are presently sucking). Perhaps he intends to continue further in this vein? David ;-) > Hi, > > Ma connection at home is no functioning. So I am temporarily > disconnected. I hope I will be able to solve that problem. I am sending > here some little comments from my office. > > I include some more material for Kim and Marty, and others, just to > think about, in case I remain disconnected for some time. Sorry. > > Bruno > > Le 22-juil.-09, à 10:27, Torgny Tholerus a écrit : > >> >> Rex Allen skrev: >>> Brent, >>> >>> So my first draft addressed many of the points you made, but it that >>> email got too big and sprawling I thought. >>> >>> So I've focused on what seems to me like the key passage from your >>> post. If you think there was some other point that I should have >>> addressed, let me know. >>> >>> So, key passage: >>> >>> >>>> Do these mathematical objects "really" exist? I'd say they have >>>> logico-mathematical existence, not the same existence as tables and >>>> chairs, or quarks and electrons. >>>> >>> >>> So which kind of existence do you believe is more fundamental? Which >>> is primary? Logico-mathematical existence, or quark existence? Or >>> are they separate but equal kinds of existence? >>> >>> >> >> The most general form of existence is: All mathematical possible >> universes exist. Our universe is one of those mathematical possible >> existing universes. > > This is non sense. Proof: see UDA. Or interrupt me when you have an > objection in the current explanation. I have explained this many times, > but the notion of universe or mathematical universe just makes no > sense. The notion of "our universe" is too far ambiguous for just > making even non sense. > > I could say the same to Brent. First I don't think it makes sense to > say that epistemology comes before ontology, given that the ontology, > by definition, in concerned with what we agree exist independently of > the observer/knower ... Then what you say contradict the results in the > computationalist theory, where the appearances of universe emerges from > the collection of all computations > > BTW, thanks to Brent for helping Marty. > > Rex, when you say: > >> I would say that most people PERCEIVE logico-mathematical objects >> differently than they perceive tables and chairs, or quarks and >> electrons. But this doesn't tell us anything about whether these >> things really have different kinds of existence. That we perceive >> them differently is just an accident of fate. > > We perceive them differently because "observation" is a different > modality of self-reference than "proving". It has nothing to do with > accident or fate. The comp physics is defined by what is invariant, > from the "observation" point of view of universal machine. Later this > will shown to be given by the 3th, 4th, and 5th hypostases. > > ==== math lesson ==== (2 posts): > > Hi, > > I wrote: > << > The cardinal of { } = 0. All singletons have cardinal one. All pairs, > or doubletons, have cardinal two. > > Problem 1 has been solved. They have the same cardinal, or if you > prefer, they have the same number of elements. The set of all subsets > of a set with n elements has the same number of elements than the set > of all strings of length n. > > Let us write B_n for the sets of binary strings of length n. So, > > B_0 = { } > B_1 = {0, 1} > B_2 = {00, 01, 10, 11} > B_3 = {000, 010, 100, 110, 001, 011, 101, 111} > > We have seen, without counting, that the cardinal of the powerset of a > set with cardinal n is the same as the cardinal of B_n. > >> > > > And now the killing question by the sadistic math teacher: > > What is the cardinal, that is, the number of element, of B_0, that is > the set of strings of length 0. > > The student: let see, you wrote above B_0 = { },, and you were kind > enough to recall that the cardinal of { } is zero (of course, there is > zero element in the empty set). So the cardinal of B_0 is zero. 'zero" > said the student. > > 'zero' indeed, said the teacher, but it is your note. You are wrong. > > B_0 is not empty! It *looks* empty, but beware the appearance, it looks > empty because it contains the empty string, which, if you remember some > preceding post is invisible (even under the microscope, telescope, > radioscope, ..). > > A solution could have been to notate the empty string by a symbol like > "_", and write all sequences "0111000100" starting from "_": > _0111000100, with rules __ = _, etc. Then B_0 = { _ }, B_1 = {_0, > _1}, etc. But this is too much notation. > > > And now the time has come for contrition when the teacher feels guilty! > > Ah..., I should have written directly something like > > B_0 = { _ }, with _ representing the empty sequence. > B_1 = {0, 1} > B_2 = {00, 01, 10, 11} > B_3 = {000, 010, 100, 110, 001, 011, 101, 111} > > OK? > > Remember we have seen that the cardinal of the powerset of a set with n > elements is equal to the cardinal of B_n, is equal to 2^n. > > The cardinal of B_0 has to be equal to to 2^0, which is equal to one. > Why? > > if a is a number, usually, a^n is the result of effectuating (a times a > times a time a ... times a), with n occurences of a. For example: 2^3 = > 2x2x2 = 8. > > so a^n times a^m is equal to a^(n+m) > > This extends to the rational by defining a^(-n) by 1/a^n. In that case > a^(m-n) = a^m/a^n. In particular a^m/a^m = 1 (x/x = 1 always), and > a^m/a^m = a^(m-m) = a^0. So a^0 = 1. So in particular 2^0 = 1. > > But we will see soon a deeper reason to be encouraged to guess that a^0 > = 1, but for this we need to define the product and the exponentiation > of sets. if A is a set, and B is a set: the exponential B^A is a very > important object, it is where the functions live. > > Take it easy, and ask. Verify the statements a^n/a^m = a^(n-m), with n > = 3 and m = 5. What is a*a*a/a*a*a*a*a "/" = division, and * = times). > > Bruno > > > > http://iridia.ulb.ac.be/~marchal/ > > ----------------- > > Hi, > > I am thinking aloud, for the sequel. > > > There will be a need for a geometrical and number theoretical interlude. > > Do you know what is a periodic decimal? > > Do you know that a is periodic decimal if and only if it exists n and > m, integers, such that a = n/m. And that for all n m, n/m is a > periodic decimal? > > Could you find n and m, such that > 12.95213213213213213213213213213213213213 ... (= 12.95 213 213 ...) > > Solution: > > Let k be a name for 12.95213213213213213213213213213213213213213 ... > > Let us multiply k by 100 000. > > 100 000k = 1295213.213213213213213213213213213213213213 ... = 1295213 > + 0.213213213 ... > > Let us multiply k by 100 > > 100k = 1295.213213213213... = 1295 + 0.213213213213213.. > > > We have 100000k - 100k = 1295213 + 0.213213213... - 1295 > - 0.213213213... = 1295213 - 1295 = 1293918 > > So 99900k = 1293918 > > Dividing by 99900 the two sides of the egality we get: > > k = 1293918/99900 > > We have n and m such that k = n/m = 12.95213213213213213... > n = 1293918, and m = 99900. > > This should convince you that all periodic decimal are fractions. > > Exercice: find two numbers n and m such that n/m = > 31,2454545454545454545... = 31, 2 45 45 45 45 ... > > > Convince yourself that for all n and m, n/m gives always a periodic > decimal.(hint: when n is divided by m, m bounds the number of possible > remainders). > > And now geometry (without picture, do them). > > Do you know that the length of the circle divided by its diameter is > PI? (PI = 3.141592...) > Do you know that the length of the square divided by its diagonal is > the square root of 2? (sqrt(2)= 1,414213562...) > - can you show this? > - can you show this without Pythagorus theorem? (like in Plato!) > > Do you know if it exists n and m such that n/m = the square root of 2 > (relation with incommensurability) > Do you know if the Diophantine equation x^2 = 2y^2 has a solution? > > No. > I think I will prove this someday, if only to have an example of > simple, yet non trivial, proof. > > This entails that the sqaure root of 2 cannot be equal to any fraction > n/m. > And it means the square root of 2 is a non periodic decimal. (its > decimal will provide a good example of a non trivial computable > function). > > Bruno > > http://iridia.ulb.ac.be/~marchal/ > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. 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