2009/7/22 Bruno Marchal <marchal.domain.name.hidden>:
> 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). 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 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
> 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/
>
> >
>
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Received on Wed Jul 22 2009 - 12:54:39 PDT