Bruno, See desperate questions below. marty a.

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----- Original Message ----- From: "Bruno Marchal" <marc...@ulb.ac.be> To: <everything-list@googlegroups.com> Sent: Wednesday, July 22, 2009 11:01 AM Subject: Re: Dreams and Machines >> >> 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? Yes. >> >> 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. Do you really expect us to understand this? >> >> 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. Or this? >> >> 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). Or this? >> >> 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/ >> >>> >> > > > http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---