Re: Seven Step Series

From: m.a. <>
Date: Sat, 25 Jul 2009 09:35:22 -0400

             One of my fundamental problems evidently has been a misconception of the use of exponents (see below in bold).
  ----- Original Message -----
  From: Bruno Marchal
  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 = 01 (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".


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Received on Sat Jul 25 2009 - 09:35:22 PDT

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