# 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:

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)

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.

Bruno

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

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