definition of exponents

Question

if you wanted to define $a^x$ for all integers would this be ok:

For non negative integers:

$a^0=1$

$a^{(n+1)}=a^n\cdot a$

For negative integers:

$a^{-n} = \frac{1}{a^n}$ an identity which holds for $n$ a positive integer, by definition.

So what my question really is: is the purpose of this definition to give $a^x$ meaning and so to say i wanted to prove this definition would just not be valid? But Then would it be valid to prove the identity $a^n*a^m=a^{(n+m)}$ From this definition for all integers? And lastly could you prove that the identity $a^{-m}=\frac{1}{a^m}$ (which is i think by definition true for positive m) is true for all integers by the defintion given above.

Sorry if I'm way off the mark here, thanks.

Answer 1

You propose the following as a definition of $a^n$ for $n\in\mathbb Z$:

1. $a^0=1$
2. $a^{n+1}=aa^n, n>0$
3. $a^{n}=\frac1{a^{-n}},n<0$

This is indeed a valid definition. And you're correct that it's never meaningful in mathematics to talk about "proving" a definition. You can't prove a definition, a definition just is.

You have two mathematical questions:

1. Can you prove $a^{n+m}=a^na^m$ for $n,m\in\mathbb Z$ from this definition?
2. Can you prove $a^{-n}=\frac{1}{a^n}$ for $n\in\mathbb Z$?

The answer to both questions is yes, although the details are kind of annoying and you have to break the proof into lots of cases. For the first property, it's probably simplest to fix $n$ and prove the result by induction for $m$. For the second, you're correct that for $n>0$, this is true by definition. For $n<0$, $-n$ is positive, so by point (3) of your definition, $a^{-n}=\frac1{a^n}$, which is equivalent to $a^n=\frac1{a^n}$. For $n=0$ the property is obvious since by definition $a^0=1$.

Answer 2

By definition

$$\overbrace{a\cdot \dots \cdot a}^{\text{n terms}} =a^n$$

thus

$$\overbrace{a\cdot \dots \cdot a}^{\text{n terms}} \cdot \overbrace{a\cdot \dots \cdot a}^{\text{m terms}} =a^{n+m}$$

and you can check that the definition is consistent also with the other rules.