# Re: Seven Step Series

```Bruno,
See desperate questions below.
marty a.```
```

----- Original Message -----
From: "Bruno Marchal" <marc...@ulb.ac.be>
Sent: Wednesday, July 22, 2009 11:01 AM
Subject: Re: Dreams and Machines

>>
>> 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?  Yes.
>>
>> 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.

Do you really expect us to understand this?

>>
>> 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.

Or this?

>>
>> Take it easy, and ask. Verify the statements a^n/a^m = a^(n-m),
>> with n
>> = 3 and m = 5. What is a*a*a/a*a*a*a*a  "/" = division, and * =
>> times).

Or this?

>>
>> 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, 2 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/
>>
>>>
>>
>
> >

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

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