Hi Bruno,
                  Yes, this is all clear to me. But when I try to put it into 
practice, confusion reigns. (See bottom of page.) By the way, because I'm 
printing out your lessons and often need to refer back to specific ones, it 
becomes confusing when you append lessons to posts with different subjects.  
Thanks,    marty a.


  ----- Original Message ----- 
  From: Bruno Marchal 
  To: everything-list@googlegroups.com 
  Sent: Thursday, July 23, 2009 6:37 AM
  Subject: Re: Seven Step Series




  On 23 Jul 2009, at 05:44, m.a. wrote:





    >> 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?




  It would help me if you could be more specific on what is it that you don't 
understand. Let me copy the text above, and show how you could perhaps be more 
specific.


  I first said:



    >> 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.


  Do you understand this? Can you compute, with the help of a pocket computer 
the following:


  2^4 = ?       (answer: 16)
  4^3 =           (answer: 64)
  10^5 =         (answer: 100000)


  Do you understand a sentence like, "let us assume that a and b are positive 
integers (= natural numbers) then a^b = a*a* ... *a with b occurrences of a."?


  Then I said that


    >> so a^n times a^m is equal to a^(n+m)


  I assume here that a, n, and m are positive integers. Have you a problem here?


  You can always verify such a statement on examples. You should think like 
this.


  Bruno pretends that a^n times a^m is equal to a^(n+m). This seems weird. I 
guess Bruno means this to be true for all (natural) numbers a, n and m.
  Let us see what this could mean on some little numbers (not too little 
because 0, 1 and even 2 are sometimes more complex than little NUMBERS like 3, 
4, 5 ...).


  So let us see with a =3, n = 4, and m = 5.


  The general statement a^n times a^m = a^(n+m) becomes


      3^4 times 3^5 = 3^(4 + 5)


  Is that true?


  1) brute force verification: (I wrote "*" as a shorthand for "times")


  3^4 = 3*3*3*3 = 81
  3^5 = 3*3*3*3*3 = 243


  3^4 times 3^5 = 81 * 243 = 19,683  (My pocket computer says).


  Now 3^(4+5) = 3^9 OK?
  So 3^(4+5) = 3^9 = 3*3*3*3*3*3*3*3*3*3 = 19,683


  We see that indeed we have that 3^4 times 3^5 = 3^(4 + 5). Both sides are 
equal to 19,683. OK?


  2) verification without computation:


  3^4 * 3^5 = 3*3*3*3    *    3*3*3*3*3
  = 3*3*3*3*3*3*3*3*3
  = 3^9
  = 3^(4+5)


  OK?


  We use the fact that multiplication is associative a*(b*c) = (a*b)*c = a*b*c. 
No need for parenthesis.


  The verification without computation gives an idea how we can convince 
ourself of the truth of the general statement:


  a^n times a^m is equal to a^(n+m)


  a^n = a*a*a* ... *a with n occurences of "a".
  a^n = a*a*a* ... *a with m occurences of "a".


  a^n times a^m = a^n * a^m = (a*a* .. a) * (a*a*...a) with n occurences of a 
in the first parentheses and m occurences of a in the second parentheses. Of 
many "a" appears in the right sides: n+m.


  Tell me if this helps. And if and when this satisfies you, you can read the 
following (but not before!)




  ------
  Then I gave a definition:


    >> This extends to the rationals by defining a^(-n) by 1/a^n.


  I could, like some teacher, just say: accept this as a definition. I could 
also provide motivation for such a definition. You have to separate those two 
things.


  Accepting the definition, you can already deduce that:


   5^(-2) = 1/(5^2) = 1/25 = 0.04


  2^(-4) = ?       (answer: 1/16 = 0.0625)
  4^(-3) =           (answer: 1/64 = 0.015625)
  10^(-5) =         (answer: 1/100000= 0.00001)




  Bu why should we accept this definition? Here are the motivation.


  a/b represent a fraction, and its value is the number obtained by dividing a 
by b. If a and b are natural numbers, most frequently a/b will NOT be a natural 
number. "a/b" is called a fraction. "a" is the numerator, "b" is the 
denominator.
  The value of fraction are called "rational number". In particular all natural 
numbers are rational, but most rational numbers are not natural number. In term 
of set, if N = the natural numbers, and Q = the rational numbers, we have that 
N is included in Q. (the number 2 is the value of the fraction 2/1).


  Examples:


  5/2 = 2.5  (not a natural number!)
  10/5 = 2   (but we were lucky!)
  10/6 = 1.6666... (my pocket computers says). 1.66... is NOT a natural number.


  Do you agree that (10*a)/(6*a) = 10/6. That is, if you have a fraction, and 
if you multiply the numerator by some number, and the denominator by that same 
number, you don't change the value of the fraction. OK?


  So, when we have a fraction, we can multiply or divide the denominator and 
numerator by the same number, and this without changing the value of the 
fraction: 


  So (a*a*a*a)/ (a*a*a) = (by dividing by a both the numerator and the 
denominator) (a*a*a)/(a*a) = (the same) (a*a)/a = a.


  Remember: in case of doubt, verify this on simple number. If the equality 
(a*a)/a = a seems too much esoteric, verify it with a = some simple number, 
like 10 for example. (10*10)/10 = 100/10 = 10, etc.


  OK up to now? I continue if you tell me if it is OK. But to sum up, a bit, 
the definition


  a^(-n) = 1/(a^n)


  is what makes possible to keep the law (a^n * a^m) = a^(n+m) for all n, no 
more just positive integers, but on any integers. (where n + (-m) is defined by 
n-m).


  You don't have to despair. If you don't understand something, it means that 
either you don't know a definition, or you don't recall a definition, or you 
don't have integrate a definition (that is, you are not familiar with the 
definition, so you can recall it, but still cannot use it). But I cannot guess 
where the problem is. I can guess only that you suffer a lack of practice.


  Just a question to proceed later: do you agree that


  (a*a*a*a)/ (a*a*a) = (a*a*a*a) times 1/(a*a*a)


  with this and what I said, you can guess that:


  a^n * a^(-m) = a^(n-m)


  Again, verify this on simple example of your own.

  OK:  If a=10   and n=3 and   m=4  Following the formula above    "a^n * 
a^(-m)"   ,I get as the first half of the equation:

  10^3 * 10/4 =  1000 x 2.5=  2500   but for the second half   "a^(n-m)    I 
get:

  10^(n-m)=  10^ -1=  10/1

  which of course makes no sense at all.     Where did I go wrong?






  Bruno


















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






  

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