Bruno,
One of my fundamental problems evidently has been a misconception
of the use of exponents (see below in bold).
----- 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. I thought 2^3 meant (2*2)* (4*2)* (8*2)= 16
* >>
>> 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.
From the above misconception you can perhaps get an idea of how utterly
alien these symbols are to me. I have never run across them before in all my
years (and you'd be surprised to learn how many years I'm talking about). When
you say that I "could have found the mistakes by carefully reread the
definitions" it's like saying that given a table of cyrillic letters I should
be able to translate a passage of "Crime and Punishment". A concept like
a^(-n) = 1/a^n is like having to learn a new polysyllabic word. I see it and
the next day I've forgotten it. Having said that, let me reiterate that I do
appreciate your efforts to simplify and explain every step of the way and I
apologize for sometimes needing even more clarification. Your patience is
saint-like and in my case, unfortunately, necessary. m.a.
Which is why when I see you make a simple mistake, I don't feel so bad
because I know how easy it is to do.
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".
http://iridia.ulb.ac.be/~marchal/
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