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 - 13:17:09 PDT