Bruno,
Yes, yours and Brent's explanations seem very clear. I hate to ask
you to spell things out step by step all the way, but I can tell you that when
I'm confronted by a dense hedge or clump of math symbols, my mind refuses to
even try to disentangle them and reels back in terror. So I beg you to always
advance in baby steps with lots of space between statements. I want to assure
you that I'm printing out all of your 7-step lessons and using them for study
and reference. Thanks for your patience, m.a.
-- Original Message -----
From: Bruno Marchal
To: everything-list@googlegroups.com
Sent: Wednesday, July 22, 2009 12:20 PM
Subject: Re: The seven step series
Marty,
Brent wrote:
On 21 Jul 2009, at 23:24, Brent Meeker wrote:
Take all strings of length 2
00 01 10 11
Make two copies of each
00 00 01 01 10 10 11 11
Add a 0 to the first and a 1 to the second
000 001 010 011 100 101 110 111
and you have all strings of length 3.
Then you wrote
I can see where adding 0 to the first and 1 to the second gives 000 and 001
and I think I see how you get 010 but the rest of the permutations don't seem
obvious to me. P-l-e-a-s-e explain, Best,
m.
(mathematically hopeless) a.
Let me rewrite Brent's explanation, with a tiny tiny tiny improvement:
Take all strings of length 2
00
01
10
11
Make two copies of each
first copy:
00
01
10
11
second copy
00
01
10
11
add a 0 to the end of the strings in the first copy, and then add a 1 to the
end of the strings in the second copy:
first copy:
000
010
100
110
second copy
001
011
101
111
You get all 8 elements of B_3.
You can do the same reasoning with the subsets. Adding an element to a set
multiplies by 2 the number of elements of the powerset:
Exemple. take a set with two elements {a, b}. Its powerset is {{ } {a} {b}
{a, b}}. How to get all the subset of {a, b, c} that is the set coming from
adding c to {a, b}.
Write two copies of the powerset of {a, b}
{ }
{a}
{b}
{a, b}
{ }
{a}
{b}
{a, b}
Don't add c to the set in the first copy, and add c to the sets in the second
copies. This gives
{ }
{a}
{b}
{a, b}
{c}
{a, c}
{b, c}
{a, b, c}
and that gives all subsets of {a, b, c}.
This is coherent with interpreting a subset {a, b} of a set {a, b, c}, by a
string like 110, which can be conceived as a shortand for
Is a in the subset? YES, thus 1
Is b in the subset? YES thus 1
Is c in the subset? NO thus 0.
OK?
You say also:
The example of Mister X only confuses me more.
Once you understand well the present post, I suggest you reread the Mister X
examples, because it is a key in the UDA reasoning. If you still have problem
with it, I suggest you quote it, line by line, and ask question. I will answer
(or perhaps someone else).
Don't be afraid to ask any question. You are not mathematically hopeless. You
are just not familiarized with reasoning in math. It is normal to go slowly. As
far as you can say "I don't understand", there is hope you will understand.
Indeed, concerning the UDA I suspect many in the list cannot say "I don't
understand", they believe it is philosophy, so they feel like they could object
on philosophical ground, when the whole point is to present a deductive
argument in a theory. So it is false, or you have to accept the theorem in the
theory. It is a bit complex, because it is an "applied theory". The mystery are
in the axioms of the theory, as always.
So please ask *any* question. I ask this to everyone. I am intrigued by the
difficulty some people can have with such reasoning (I mean the whole UDA
here). (I can understand the shock when you get the point, but that is always
the case with new results: I completely share Tegmark's idea that our brain
have not been prepared to have any intuition when our mind try to figure out
what is behind our local neighborhood).
Bruno
http://iridia.ulb.ac.be/~marchal/
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