On 22 July, 16:01, Bruno Marchal <marc....domain.name.hidden> wrote:
> Ma connection at home is again functioning. I am happy to have solved
> the problem rather quickly.
>
> On 22 Jul 2009, at 13:54, David Nyman wrote:
>
>
>
> > 2009/7/22 Bruno Marchal <marc....domain.name.hidden>:
> You thought you could make fun of the poor disconnected one?
Dinna fash yursel laddie, ah was'na makin sport o' ye. It wus a
compliment!
Further lessons available on application.
Hoots mon
David -;
>
> >> Ma connection at home is no functioning.
>
> > 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).
>
> Indeed I am following an intense summer school in Glaswegian.
>
> You thought you could make fun of the poor disconnected one?
>
> > 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?
>
> Once, on a list, someone thought I was using slang from New-York! Now
> Glaswegian!
>
> I am afraid I am just writing to quickly, and then when I read myself
> I concentrate so much on the meaning ...
> Most of the time I see the spelling errors when I read my mail, never
> when I send it.
>
> Sorry sorry sorry ...
>
> Take care of the sense and the spelling will take care of itself. Well
> *that* does not work!
>
> Bruno
>
>
>
> > 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 01 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
> >> = 03 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, 02 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
>
> ...
>
> read more »
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Received on Wed Jul 22 2009 - 09:15:04 PDT