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.

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-- 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/ --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---