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n^0 Power = 1: Defined or Proved?

Date: 04/18/2003 at 08:28:14
From: Tam
Subject: Number

I have been wondering if a^0 is DEFINED to be 1 or is PROVED to be 1. 

I think that we define a^0 = 1, but one of my friends said that a^0 
is proved to be 1. 

I think that we have to define it first, because as we define a^n = 
a.a.a....a (n times) with n a whole number, this definition will not 
work for n=0. Therefore, we must define a^0 in another way, which is 
a^0 = 1 (with a not equal to 0). As for the reason why we define 
a^0 = 1 but not any other number, it is just to keep the fomula 
a^m.a^n = a^(m+n), and (a^m)^n = a^(m.n)

But as my friend said, we define a^n first in the way I used above, 
but n can be any number including 0. As we can't figure out what a^0 
is from this definition, (in other words, it is meaningless to say 
a^0 is defined to be a.a...a [0 times]) we have to calculate it, and 
he proved to me that 1 = a^n/a^n = a^(n-n) = a^0. Therefore he said 
a^0 is PROVED to be 1.

I think before we talk about something, we must first define it, and 
the definition must show what that something is (and we don't need 
to go through some theorems to figure out or prove what we have 
already defined, as with my friend's argument). I think this is just 
normal mathematical logic that we have to define something clearly 
first before using it. 

Am I right?

Date: 04/18/2003 at 09:27:27
From: Doctor Peterson
Subject: Re: Number

Hi, Tam.

You are correct, but your friend is not too far off. Since we have no 
preexisting definition of a^0, we have to choose a definition for it; 
but we can prove that this particular definition is the only one that 
retains all the important properties of exponents. The only thing 
wrong with your friend's approach is that you can't make a definition 
that supposedly includes the case n=0, but then say that it can't be 
applied in that case. However, if he said that exponentiation was 
defined in general by its properties, rather than as repeated 
multiplication, then he could prove that a^0 = 1 for all a other than 
0 strictly from those properties (which would be taken as axioms).

A definition with a reason given (as in your presentation) is not too 
far from a proof; and a proof based on a non-standard definition (as 
in my version of your friend's) is not far from a definition.

We discuss these ideas in the Dr. Math FAQ:

   N to 0 power 

But it is actually possible to start with a definition that includes 
0 (but not negative exponents). Just say

   a^n = 1 * a * ... * a
         n multiplications

and it makes sense to say that a^0 is 1, not multiplied by anything. 
You still have to extend this definition to negative, and then 
rational, and then real, exponents.

So ultimately, which of you is right depends on what axiomatic system 
you start with, and how you define exponents initially. I myself 
would say that we are making an extended definition, and proving it 
to be consistent.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Exponents
Middle School Exponents

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