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.

Bruno









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




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