Proving properties of exponents.

Question

For positive integer $n$,

$$a^n:=a\cdot a\cdot a...a \ \ \ (n\ \text{times})$$ $$a^{-n}:=\frac{1}{a^n}$$ $$a^0:=1$$

I want to prove the properties, $$a^n\cdot a^m=a^{n+m}$$ $$\frac{a^n}{a^m}=a^{n-m}$$ $$(a^n)^m=a^{nm}$$ for all integer values of $n,m$. It's easy to prove them for positive integers using the 3 definitions. Is there any way I could extend the proof to any integer value (including 0) directly (perhaps by mathematical induction), or do I have consider each possible combination of positive, negative and 0 separately?

This might be a very basic question but I'm trying to improve by proof writing skills and would appreciate any help.

Edit:

My text book says that 0 and negative exponents are defined the way they are as that is the only way to make the properties hold for all values of $n$ & $m$. Wouldn't that mean that I'll have to prove the 0 and negative cases separately, usingg 2 extra defintions?

Answer 1

Yes, a complete proof will cover all the cases. But you can simplify the work a lot: note that proving $a^{m+n}=a^m\cdot a^n~~~(\dagger)$ for integers $m,n$ imply the two other properties:

$$\frac{a^m}{a^n}=\begin{cases}a^m\cdot\dfrac 1{a^n}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m+(-n)}=a^{m-n}&,n\gt 0\\ \dfrac{a^m}{a^0}=a^m=a^m\cdot a^0\overset{\dagger}{=}a^{m-0}&,n=0\\ \dfrac{a^m}{a^{-(-n)}}=a^m\cdot\dfrac 1{a^{-(-n)}}=a^m\cdot a^{-n}\overset{\dagger}{=}a^{m-n}&,n\lt 0\end{cases}$$

$$(a^m)^0:=1=a^{0}=a^{m\cdot 0}$$

$$(a^m)^n=\begin{cases}\underbrace{a^m\cdot a^m\cdots a^m}_{n\text{ times}}\overset{\dagger}{=}a^{\underbrace{m+m+\cdots+m}_{n\text{ times}}}=a^{mn}&,n\gt 0\\ \dfrac 1{\underbrace{a^m\cdot a^m\cdots a^m}_{-n\text{ times}}}\overset{\dagger}{=}\dfrac 1{a^{\underbrace{m+m+\cdots+m}_{-n\text{ times}}}}=\dfrac 1{a^{m(-n)}}=\dfrac 1{a^{-(mn)}}=a^{mn}&,n\lt 0\end{cases}$$

So, all you need to do is to prove $(\dagger)$ for all integers $m,n$

Once you have proved it for positive integers $m,n$, you can wlog consider only one of $m,n$ to be negative and by symmetry, you prove the cases where exactly one of $m,n$ is negative. For the case of both $m,n$ negative, we note that $$a^{m+n}=a^{-(-m-n)}=\frac 1{a^{(-m)+(-n)}}\overset{\dagger}{=}\frac 1{a^{-m}\cdot a^{-n}}=\frac 1{a^{-m}}\cdot\frac 1{a^{-n}}=a^m\cdot a^n$$

where we apply $(\dagger)$ with positive integers $-m,-n$

which leaves us with the case of one or both of $m,n$ be $0$ which is easy to finish up. By symmetry, wlog prove with just one of $m,n$ being $0$ and for the case of $m=n=0$, we have $a^{0+0}=a^{0}=1=1\cdot 1=a^0\cdot a^0$. $_\square$