The seven step series

[Brian Tenneson]

Comments below.

Bruno Marchal wrote:

Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.

I need to get a better grasp on what a universal machine is, yes.  I am interested in finding out how Tegmark's argument for mathematicalism is invalid, especially since I'm using it to motivate my research.

According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.

I'll have to think more on Jannes' paper.  As I basically resting the motivation of my research on the correctness of "ERH implies MUH," I'm trying to formulate a good refutation to his paper.


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[Bruno Marchal]

On 21 Jul 2009, at 00:22, Brian Tenneson wrote:

Comments below.

Bruno Marchal wrote:

Exercise: criticize the following papers mentioned below in the light of the discovery of the universal machine and its main consequences from incompleteness to first person indeterminacy. Think of the identity thesis. To be sure Tegmark is less "wrong" than Jannes.

Solution: search in the archive of this list where I have already explained this, or use directly UDA, or wait for what will (perhaps) follow.

I should send some of my papers on arXiv, but up to now, only logicians understand the whole "trick", so I have to better appreciated what physicians don't understand in logic, before making a version free of references to mathematical logical baggage. Logicians are not interested in mind, nor really matter, and physicians are still naïve on the link consciousness/reality, I would say.

To be sure Tegmark is closer than most physicists except perhaps Wheeler.

Also, Tegmarks' argument for mathematicalism is invalid (even with strong non-comp axioms). But I prefer to help you to understand this by yourself through the understanding of what a universal machine is, than trying a direct argument.

I need to get a better grasp on what a universal machine is, yes.  I am interested in finding out how Tegmark's argument for mathematicalism is invalid, especially since I'm using it to motivate my research.

At least you are aware that a mathematicalism à-la Tegmark needs a rather sophisticated universal structure, but if we assume even very weak version of comp, the universal machine provides that structure, or that structure has to be reducible as an invariant for a set of effective transformation of that machine. We can come back on this. I may be wrong also.

According of the part of UDA (or perhaps AUDA) you understand, you can already see the weakness of such direct mathematical approach. Note that comp makes physics much more fundamental, and separate it much clearly from possible geograpies. Above all comp does not eliminate the person, which Tegmark is still doing: the frog view is not yet a first person view, in the comp sense.

Interesting stuff, still. Thanks for the references.

I'll have to think more on Jannes' paper.  As I basically resting the motivation of my research on the correctness of "ERH implies MUH," I'm trying to formulate a good refutation to his paper.

OK, nice.

My main critics is that they seem not be aware of the consciousness/reality problem. They are using an identify thesis which is not allowed by comp. The UD argument shows exactly that. It is build to show that if we keep consciousness, eventually, physics is even more fundamental than physicist imagine. The physical world(s) is(are) not just a 'sufficiently rich' part of math, it is somehow the border of the ignorance of any (Löbian) universal machine which introspects itself. This connects in some way all part 'sufficiently rich' part  of math". It explains also the non communicable part of what we can be conscious of, including physical sensations (as modalities related to self-references).

Bruno

http://iridia.ulb.ac.be/~marchal/

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[m.a.-2]

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Hi Bruno,

                   I'm not clear on the sentence in bold below, especially the word "correspondingly". The example of Mister X only confuses me more. Could you please give some simple examples? Thanks,

                                                                                                                                                                                                                                                                                        marty a.

 

 

----- Original Message -----

From: marchal@...

To: everything-list@...

Sent: Monday, July 20, 2009 3:17 PM

Subject: Re: The seven step series

On 20 Jul 2009, at 15:34, m.a. wrote:

And then we have seen that such cardinal was given by 2^n. 

You can see this directly by seeing that adding an element in a set, double the number of subset, due to the dichotomic choice in creating a subset "placing or not placing" the new element in the subset.

 

Likewise with the strings. If you have already all strings of length n, you get all the strings of length n+1, by doubling them and adding zero or one correspondingly.

 

This is also illustrated by the iterated self-duplication W, M. Mister X is cut and paste in two rooms containing each a box, in which there is a paper with zero on it, in room W, and 1 on it in room M. After the experience, the 'Mister X' coming out from room W wrote 0 in his diary, and the 'Mister X' coming out from room M wrote 1 in his diary. And then they redo each, the experiment. The Mister-X with-0-in-his-diary redoes it, and gives a Mister-X with-0-in-his-diary coming out from room W, and adding 0 in its diary and a  Mister-X with-0-in-his-diary coming out from room M, adding 1 in its diary: they have the stories 

 

Bruno

http://iridia.ulb.ac.be/~marchal/

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[Brent Meeker-2]

Each binary string of length n has two possible continuations of  length
n+1, one of them by appending a 0 and one of them by appending a 1.  So
to get all binary strings of length n+1 take each string of length n,
make two copies, to one copy append a 0 and to the other copy append a 1.

Brent

m.a. wrote:

> Hi Bruno,
>                    I'm not clear on the sentence in bold below,
> especially the word "correspondingly". The example of Mister X only
> confuses me more. Could you please give some simple examples? Thanks,
>                                                                    
>                                                                    
>                                                                    
>                                                                    
>         marty a.
>  
>  
>
>     ----- Original Message -----
>     *From:* Bruno Marchal <mailto:marchal@...>
>     *To:* everything-list@...
>     <mailto:everything-list@...>
>     *Sent:* Monday, July 20, 2009 3:17 PM
>     *Subject:* Re: The seven step series
>
>
>     On 20 Jul 2009, at 15:34, m.a. wrote:
>
>>     And then we have seen that such cardinal was given by 2^n.
>     You can see this directly by seeing that adding an element in a
>     set, double the number of subset, due to the dichotomic choice in
>     creating a subset "placing or not placing" the new element in the
>     subset.
>      
>     *Likewise with the strings. If you have already all strings of
>     length n, you get all the strings of length n+1, by doubling them
>     and adding zero or one correspondingly.*
>      
>     This is also illustrated by the iterated self-duplication W, M.
>     Mister X is cut and paste in two rooms containing each a box, in
>     which there is a paper with zero on it, in room W, and 1 on it in
>     room M. After the experience, the 'Mister X' coming out from room
>     W wrote 0 in his diary, and the 'Mister X' coming out from room M
>     wrote 1 in his diary. And then they redo each, the experiment. The
>     Mister-X with-0-in-his-diary redoes it, and gives a Mister-X
>     with-0-in-his-diary coming out from room W, and adding 0 in its
>     diary and a  Mister-X with-0-in-his-diary coming out from room M,
>     adding 1 in its diary: they have the stories
>
>      
>     Bruno
>
>
>
>     http://iridia.ulb.ac.be/~marchal/
>     <http://iridia.ulb.ac.be/%7Emarchal/>
>
>
>
>
>
>     >

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[m.a.-2]

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Thanks Brent,

                       Could you supply some illustrative examples?   

                                                                                                     marty a.

 

 

----- Original Message -----

From: "Brent Meeker" <meekerdb@...>

To: <everything-list@...>

Sent: Tuesday, July 21, 2009 3:57 PM

Subject: Re: The seven step series

>

> Each binary string of length n has two possible continuations of  length
> n+1, one of them by appending a 0 and one of them by appending a 1.  So
> to get all binary strings of length n+1 take each string of length n,
> make two copies, to one copy append a 0 and to the other copy append a 1.
>
> Brent
>
> m.a. wrote:
>> Hi Bruno,
>>                    I'm not clear on the sentence in bold below,
>> especially the word "correspondingly". The example of Mister X only
>> confuses me more. Could you please give some simple examples? Thanks,
>>                                                                    
>>                                                                    
>>                                                                    
>>                                                                    
>>         marty a.
>> 
>> 
>>
>>  >>
>>     >
>
>
>

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[Brent Meeker-2]

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.

Brent

m.a. wrote:

> *Thanks Brent,*
> *                       Could you supply some illustrative examples?    *
> *                                                                    
>                                 marty a.*
> **
>  
> ----- Original Message -----
> From: "Brent Meeker" <meekerdb@...
> <mailto:meekerdb@...>>
> To: <everything-list@...
> <mailto:everything-list@...>>
> Sent: Tuesday, July 21, 2009 3:57 PM
> Subject: Re: The seven step series
>
> >
> > Each binary string of length n has two possible continuations of  
> length
> > n+1, one of them by appending a 0 and one of them by appending a 1.  So
> > to get all binary strings of length n+1 take each string of length n,
> > make two copies, to one copy append a 0 and to the other copy append
> a 1.
> >
> > Brent
> >
> > m.a. wrote:
> >> Hi Bruno,
> >>                    I'm not clear on the sentence in bold below,
> >> especially the word "correspondingly". The example of Mister X only
> >> confuses me more. Could you please give some simple examples? Thanks,
> >>                                                                    
> >>                                                                    
> >>                                                                    
> >>                                                                    
> >>         marty a.
> >>
> >>
> >>
> >>  >>
> >>     >
> >
> >
> >
> >

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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

Hi Brent,

                I really appreciate the help and I hate to impose on your patience but...(see below)

 

----- Original Message -----

From: "Brent Meeker" <meekerdb@...>

To: <everything-list@...>

Sent: Tuesday, July 21, 2009 5:24 PM

Subject: Re: The seven step series

>
> 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.

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.

 

 

 

 

 

 

>

> Brent
>
> m.a. wrote:
>> *Thanks Brent,*
>> *                       Could you supply some illustrative examples?    *
>> *                                                                    
>>                                 marty a.*
>> **
>> 
>> ----- Original Message -----
>> From: "Brent Meeker" <meekerdb@...
>> <meekerdb@...>>
>> To: <everything-list@...
>> <everything-list@...>>
>> Sent: Tuesday, July 21, 2009 3:57 PM
>> Subject: Re: The seven step series
>>
>> >
>> > Each binary string of length n has two possible continuations of 
>> length
>> > n+1, one of them by appending a 0 and one of them by appending a 1.  So
>> > to get all binary strings of length n+1 take each string of length n,
>> > make two copies, to one copy append a 0 and to the other copy append
>> a 1.
>> >
>> > Brent
>> >
>> > m.a. wrote:
>> >> Hi Bruno,
>> >>                    I'm not clear on the sentence in bold below,
>> >> especially the word "correspondingly". The example of Mister X only
>> >> confuses me more. Could you please give some simple examples? Thanks,
>> >>                                                                   
>> >>                                                                   
>> >>                                                                   
>> >>                                                                   
>> >>         marty a.
>> >>
>> >>
>> >>
>> >>  >>
>> >>     >
>> >
>> >
>> >
>> >
>
>
>

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[Bruno Marchal]

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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[Brent Meeker-2]

m.a. wrote:

> Hi Brent,
>                 I really appreciate the help and I hate to impose on
> your patience but...(see below)
>  
> ----- Original Message -----
> From: "Brent Meeker" <meekerdb@...
> <mailto:meekerdb@...>>
> To: <everything-list@...
> <mailto:everything-list@...>>
> Sent: Tuesday, July 21, 2009 5:24 PM
> Subject: Re: The seven step series
>
>  >
>  > 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.
> *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.*

They aren't permutations.  They're just sticking a 0 or 1 on the end.  One copy
of 01 becomes 010 and the other become 011.

Brent

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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

Going a step further... (see below)

 

----- Original Message -----

From: "Brent Meeker" <meekerdb@...>

To: <everything-list@...>

Sent: Wednesday, July 22, 2009 12:57 PM

Subject: Re: The seven step series

>

> m.a. wrote:
>> Hi Brent,
>>                 I really appreciate the help and I hate to impose on
>> your patience but...(see below)
>> 
>> ----- Original Message -----
>> From: "Brent Meeker" <meekerdb@...
>> <meekerdb@...>>
>> To: <everything-list@...
>> <everything-list@...>>
>> Sent: Tuesday, July 21, 2009 5:24 PM
>> Subject: Re: The seven step series
>>
>>  >
>>  > 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.
>> *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,*
>> **
>>                                                                        
>>                                                                         

They aren't permutations.  They're just sticking a 0 or 1 on the end.  One copy
> of 01 becomes 010 and the other become 011.

 

Then I assume the next step would be making two copies of each of those:

 

000    000       001     001      010      010       011     011     100      100       101         101         110           110             111          111

 

...and sticking a 0 or 1 at the end:

 

0000   0001    0010    0011     0100    0101    0110    0111    1000     1001     1010       1011         1100         1101           1110      1111

 

and this is the binary sequence of length 4.

 

How do these translate into ordinary numerals? 1,2,3,4...

>
> Brent
>
>
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[Brent Meeker-2]

m.a. wrote:

> *Going a step further... (see below)*
> **
> ----- Original Message -----
> From: "Brent Meeker" <meekerdb@...
> <mailto:meekerdb@...>>
> To: <everything-list@...
> <mailto:everything-list@...>>
> Sent: Wednesday, July 22, 2009 12:57 PM
> Subject: Re: The seven step series
>
>  >
>  > m.a. wrote:
>  >> Hi Brent,
>  >>                 I really appreciate the help and I hate to impose on
>  >> your patience but...(see below)
>  >>
>  >> ----- Original Message -----
>  >> From: "Brent Meeker" <meekerdb@...
> <mailto:meekerdb@...>
>  >> <mailto:meekerdb@...>>
>  >> To: <everything-list@...
> <mailto:everything-list@...>
>  >> <mailto:everything-list@...>>
>  >> Sent: Tuesday, July 21, 2009 5:24 PM
>  >> Subject: Re: The seven step series
>  >>
>  >>  >
>  >>  > 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.
>  >> *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,*
>  >> **
>  >>                                                                        
>  >>                                                                        
> They aren't permutations.  They're just sticking a 0 or 1 on the end.  
> One copy
>  > of 01 becomes 010 and the other become 011.
>  
> *Then I assume the next step would be making two copies of each of those:*
> **
> *000    **000       001     001      010      010       011     011    
> 100      100       101         101         110           110            
> 111          111*
> **
> *...and sticking a 0 or 1 at the end:*
> **
> *0000   0001    0010    0011     0100    0101    0110    0111    
> 1000     1001     1010       1011         1100         1101          
> 1110      1111*
> **
> *and this is the binary sequence of length 4.*

Right, it's all the binary strings of length 4

> **
> *How do these translate into ordinary numerals? 1,2,3,4...*

Bruno's using them to represent sets and subsets.  So if we have a set {a b c}
we can represent the subset {a c} by 101 and {a b} by 110, etc.  That's quite
different from using a binary string to represent a number in positional
notation.  I'll leave it to Bruno whether he wants to go into that.

Brent

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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

Hi Bruno,

                I asked Brent Meeker a question which he referred back to you. Will you be covering it? (see para in bold below)

 

----- Original Message -----

From: "Brent Meeker" <meekerdb@...>

To: <everything-list@...>

Sent: Wednesday, July 22, 2009 11:49 PM

Subject: Re: The seven step series

>> *0000   0001    0010    0011     0100    0101    0110    0111   

>> 1000     1001     1010       1011         1100         1101          
>> 1110      1111*
>> **
>> *and this is the binary sequence of length 4.*
>
> Right, it's all the binary strings of length 4
>
>> **
>> *How do these translate into ordinary numerals? 1,2,3,4...*
>
> Bruno's using them to represent sets and subsets.  So if we have a set {a b c}
> we can represent the subset {a c} by 101 and {a b} by 110, etc.  That's quite
> different from using a binary string to represent a number in positional
> notation.  I'll leave it to Bruno whether he wants to go into that.

 

 

 

 

 

 

>
> Brent
>
>
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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

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: marchal@...

To: everything-list@...

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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[Bruno Marchal]

Hi Marty,

I can if you really want it, but it is out of topic and could introduced some confusion.  I suggest we could come back on this later perhaps. But if you insist, I can do it. have you get my last post?

Note that I have also already explained how binary strings can represent number in some older post. Honestly we will not need this, so it is better not to accumulate too many "new" materials, especially when I can fear some confusion. It is good to be familiar with the object "binary strings" seen as an object by itself.

Bruno

On 23 Jul 2009, at 14:02, m.a. wrote:

Hi Bruno,

                I asked Brent Meeker a question which he referred back to you. Will you be covering it? (see para in bold below)

 

----- Original Message -----

From: "Brent Meeker" <meekerdb@...>

To: <everything-list@...>

Sent: Wednesday, July 22, 2009 11:49 PM

Subject: Re: The seven step series

>> *0000   0001    0010    0011     0100    0101    0110    0111    
>> 1000     1001     1010       1011         1100         1101           
>> 1110      1111*
>> ** 
>> *and this is the binary sequence of length 4.*
> 
> Right, it's all the binary strings of length 4
> 
>> ** 
>> *How do these translate into ordinary numerals? 1,2,3,4...*
> 
> Bruno's using them to represent sets and subsets.  So if we have a set {a b c} 
> we can represent the subset {a c} by 101 and {a b} by 110, etc.  That's quite 
> different from using a binary string to represent a number in positional 
> notation.  I'll leave it to Bruno whether he wants to go into that.

 

 

 

 

 

 

> 
> Brent
> 
> 

http://iridia.ulb.ac.be/~marchal/

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[Bruno Marchal]

On 23 Jul 2009, at 15:09, m.a. wrote:

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.

Don't worry, I understand that very well. And this illustrates also that your "despair" is more psychological than anything else. I have also abandoned the study of a mathematical book until I realize that the difficulty was more my bad eyesight than any conceptual difficulties. With good spectacles I realize the subject was not too difficult, but agglomeration of little symbols can give a bad impression, even for a mathematician.

I will make some effort, tell me if my last post, on the relation

     (a^n) * (a^m) = a^(n + m)

did help you.

You are lucky to have an infinitely patient teacher. You can ask any question, like "Bruno,

is (a^n) * (a^m) the same as a^n times a^m?" 

 Answer: yes, I use often "*", "x", as shorthand for "times", and I use "(" and ")" as delimiters in case I fear some ambiguity.

Bruno

 

 

 

 

-- Original Message -----

From: marchal@...

To: everything-list@...

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/

http://iridia.ulb.ac.be/~marchal/

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[ronaldheld]

Bruno:
   I am following, but have not commented, because there is nothing
controversal.
   When you are done, can your posts be consolidated into a paper or a
document that can be read staright through?
                                                         Ronald

On Jul 23, 9:28 am, Bruno Marchal <marc...@...> wrote:

> On 23 Jul 2009, at 15:09, m.a. wrote:
>
> > 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.
>
> Don't worry, I understand that very well. And this illustrates also  
> that your "despair" is more psychological than anything else. I have  
> also abandoned the study of a mathematical book until I realize that  
> the difficulty was more my bad eyesight than any conceptual  
> difficulties. With good spectacles I realize the subject was not too  
> difficult, but agglomeration of little symbols can give a bad  
> impression, even for a mathematician.
>
> I will make some effort, tell me if my last post, on the relation
>
>       (a^n) * (a^m) = a^(n + m)
>
> did help you.
>
> You are lucky to have an infinitely patient teacher. You can ask any  
> question, like "Bruno,
>
> is (a^n) * (a^m) the same as a^n times a^m?"
>   Answer: yes, I use often "*", "x", as shorthand for "times", and I  
> use "(" and ")" as delimiters in case I fear some ambiguity.
>
> Bruno
>
>
>
>
>
>
>
> > -- Original Message -----
> > From: Bruno Marchal
> > To: everything-list@...
> > 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/
>
> http://iridia.ulb.ac.be/~marchal/- Hide quoted text -
>
> - Show quoted text -

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[Bruno Marchal]

On 27 Jul 2009, at 16:07, ronaldheld wrote:
>
>   I am following, but have not commented, because there is nothing
> controversal.

Cool. Even the sixth first steps of UDA?


>
>   When you are done, can your posts be consolidated into a paper or a
> document that can be read staright through?

I should do that.

Bruno

> On Jul 23, 9:28 am, Bruno Marchal <marc...@...> wrote:
>> On 23 Jul 2009, at 15:09, m.a. wrote:
>>
>>> 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.
>>
>> Don't worry, I understand that very well. And this illustrates also
>> that your "despair" is more psychological than anything else. I have
>> also abandoned the study of a mathematical book until I realize that
>> the difficulty was more my bad eyesight than any conceptual
>> difficulties. With good spectacles I realize the subject was not too
>> difficult, but agglomeration of little symbols can give a bad
>> impression, even for a mathematician.
>>
>> I will make some effort, tell me if my last post, on the relation
>>
>>       (a^n) * (a^m) = a^(n + m)
>>
>> did help you.
>>
>> You are lucky to have an infinitely patient teacher. You can ask any
>> question, like "Bruno,
>>
>> is (a^n) * (a^m) the same as a^n times a^m?"
>>   Answer: yes, I use often "*", "x", as shorthand for "times", and I
>> use "(" and ")" as delimiters in case I fear some ambiguity.
>>
>> Bruno
>>
>>
>>
>>
>>
>>
>>
>>> -- Original Message -----
>>> From: Bruno Marchal
>>> To: everything-list@...
>>> 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/
>>
>> http://iridia.ulb.ac.be/~marchal/- Hide quoted text -
>>
>> - Show quoted text -
> >

http://iridia.ulb.ac.be/~marchal/




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[Bruno Marchal]

Hi,

OK, I will come back on the square root of 2 later.

We have talked on sets.

Sets have elements, and elements of a set define completely the set,  
and a set is completely defined by its elements.

Example: here is a set of numbers {1, 2, 3}
and a set of sets of numbers {{1, 2}, {3}, { }}.

We can do some operations, like their union, or their intersection.
Examples:
{1,2,3} union {3,4,5} = {1,2,3,4,5},
{1,2,3} intersection {3,4,5} = { }.

We can verify if some relation hold for them, like equality, or  
inclusion.
{1, 2} = {2, 1, 1}   (yes!)
{1, 2} included-in {3, 2, 1}
{1, 2} not-included-in {1, 3}

We can compute their powerset.
Powerset {1, 2} = {{ }, {1}, {2}, {1, 2}}

We have discovered SBIJECTION between powersets of a set with cardinal  
n, and the set of binary strings of length n.
And we have presented reasons for the existence of a bijection between  
the powerset of N = {0, 1, 2, ...} and the set of infinite binary  
strings.

OK?

Today, I suggest we look at two new operations on sets. The product of  
sets, and the exponentiation of sets. Well, I will probably do only  
the product today.

First I have to introduce a new, well actually *very* well known,  and  
absolutely important, notion: the couple.

A couple is when there is two things, but with some order. It looks  
like a pair, but the order counts.

Usually a couple of things a , b is designated, in math, like this:

(a, b).

It looks like a pair {a, b}, but it is not. Indeed, {a, b} = {b, a},  
but the couple (a, b) is NOT equal to the couple (b, a).

When are two couples (a, b) and (c, d) equal? Only when a = c and b = d.

Examples. the couple of number (2, 3) is not equal to the couple (3,  
2), but the couple (0, 666) is equal to the couple (0, 666).

OK?

APARTE: Are couples sets? No. Nor are numbers. But yes, you can easily  
represent them by sets, so we could work only with sets, but we will  
not do that. Much later we will work only with numbers, in fact. The  
very notion of representation will be important, though.

Now we are ready to define the so called "cartesian" product of sets.  
It is indeed a cousin of Descartes' discovery that you can represent a  
point of the plane by a couple of (real) numbers. I read somewhere  
that Descartes discovered this by trying to describe a spider walking  
on a window with squared little piece of glass. But such a  
localization works also for cities like Los Angeles where you address  
is something like 15th avenue 61th street. The whole field of  
analytical geometry is founded on this idea.

That cartesian idea generalises on sets A and B. It is written A X B,  
and it is defined by
the set of couples (x, y) such that x belongs to A, and y belongs to B.

AXB = {(x, y) such-that x belongs-to A, and y belongs to B}.        
(compare with the preceding definitions).

Example:  what is the product {0, 1} X {a, b}?  Well it is the set of  
all the couples made from elements of A in company of elements of  B,  
and in that order, with A = {0, 1}, and B = {a, b}.
So (0, a) is in there, and there are others. The product of {0, 1)  
with {a, b} is equal to

{(0,a), (0, b), (1, a), (1, b)}

The convenient usual cartesian drawing is, for AXB, with A = {0, 1},  
and B = {a, b} :


a     (0, a)   (1, a)

b     (0, b)  (1, b)

         0          1

A product of numbers a and b,  ab, can be conceived as the area of a  
rectangle of sides a and b. Here you can see that the product of sets  
AXB can fit in a rectangle when you dispose horizontally the elements  
of A, and vertically the elements of B. By convention, usually A is  
put horizontally, and B vertically.
But note that if the number ab is equal to the number ba, it is not  
the case that the set AXB is equal to the set BXA. (0, a) does not  
belong to BXA, for example.

Exercise: to the cartesian drawing for BXA.

1)
Compute
{a, b, c} X {d, e} =
{d, e} X {a, b, c} =
{a, b} X {a, b} =
{a, b} X { } =

2)
Convince yourself that the cardinal of AXB is the product of the  
cardinal of A and the cardinal of B.
A and B are finite sets here. Hint: meditate on their cartesian drawing.

3) Draw a piece of NXN.

Solution and sequel tomorrow.

Any question?

Bruno

http://iridia.ulb.ac.be/~marchal/




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[ronaldheld]

Bruno:
 I meant the mathematical formalism you are teaching us. When we
eventually get to the UDA steps, I wil be better able to do that
assessment.
                                                      Ronald

On Jul 27, 1:27 pm, Bruno Marchal <marc...@...> wrote:

> On 27 Jul 2009, at 16:07, ronaldheld wrote:
>
>
>
> >   I am following, but have not commented, because there is nothing
> > controversal.
>
> Cool. Even the sixth first steps of UDA?
>
>
>
> >   When you are done, can your posts be consolidated into a paper or a
> > document that can be read staright through?
>
> I should do that.
>
> Bruno
>
>
>
>
>
> > On Jul 23, 9:28 am, Bruno Marchal <marc...@...> wrote:
> >> On 23 Jul 2009, at 15:09, m.a. wrote:
>
> >>> 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.
>
> >> Don't worry, I understand that very well. And this illustrates also
> >> that your "despair" is more psychological than anything else. I have
> >> also abandoned the study of a mathematical book until I realize that
> >> the difficulty was more my bad eyesight than any conceptual
> >> difficulties. With good spectacles I realize the subject was not too
> >> difficult, but agglomeration of little symbols can give a bad
> >> impression, even for a mathematician.
>
> >> I will make some effort, tell me if my last post, on the relation
>
> >>       (a^n) * (a^m) = a^(n + m)
>
> >> did help you.
>
> >> You are lucky to have an infinitely patient teacher. You can ask any
> >> question, like "Bruno,
>
> >> is (a^n) * (a^m) the same as a^n times a^m?"
> >>   Answer: yes, I use often "*", "x", as shorthand for "times", and I
> >> use "(" and ")" as delimiters in case I fear some ambiguity.
>
> >> Bruno
>
> >>> -- Original Message -----
> >>> From: Bruno Marchal
> >>> To: everything-list@...
> >>> 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/
>
> >>http://iridia.ulb.ac.be/~marchal/-Hide quoted text -
>
> >> - Show quoted text -
>
> http://iridia.ulb.ac.be/~marchal/- Hide quoted text -
>
> - Show quoted text -

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[m.a.-2]

Some parts of this message have been removed. Learn more about Nabble's security policy.

Bruno,

             I have searched my notes for an exposition of BIJECTION and found only one mention in an early email which promises to define it in a later lesson. Do you have a reference to that lesson or perhaps an instant explanation of it? Thanks,

                                                                              Chief Ignoramus

 

 

 

             

----- Original Message -----

From: "Bruno Marchal" <marchal@...>

To: <everything-list@...>

Sent: Monday, July 27, 2009 4:54 PM

Subject: Re: The seven step series

>

> We have discovered SBIJECTION between powersets of a set with cardinal 
> n, and the set of binary strings of length n.
> And we have presented reasons for the existence of a bijection between 
> the powerset of N = {0, 1, 2, ...} and the set of infinite binary  
> strings.
>
> OK?

 

 

 

 

 

 

 

 

 

>
>
>

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