$a^{m+n} = a^m*a^n$ where $m,n \in \mathbb{Z}$

Question

I can use the exponents laws only $m,n \in \mathbb{N}$, and need to prove them for $m,n \in \mathbb{Z}$.

note that $0 \neq a \in \mathbb{R}$

I proved some cases (mainly the trivial ones) and I'm having a hard time proving this case:

Assuming $m<0 \wedge n>0$ and $\left | n \right | > \left | m \right |$ we know that $(m+n) \in \mathbb{N}$

I was thinking about expressing $n+m$ as $k\in\mathbb{N}$ but then I need to handle the cases where $k = 1$ and $k > 1$

I started with $a^{m+n} = \frac{1}{a^{-m-n}}$ but then again I get that $-m \in\mathbb{N}$ and that $-n<0$ so I can't seem to apply any exponent rules with the naturals.

How should I approach this?

edit:

Would it be legit performing:

$a^{m+n} = \frac{1}{a^{-m-n}} = \frac{1}{a^{-m}*\frac{1}{a^{n}}} = \frac{1}{\frac{1}{a^m}*\frac{1}{a^n}} = a^m*a^n$

Answer 1

Defining $a^{-k}:=\frac{1}{a^k}$ for $k>0$, we first prove $a^{m+1}=a^m a$ for all $m\in\Bbb Z$: the desired result is the definition of $a^{m+1}$ for $m\ge 0$, is trivial if $m=-1$, and extends to $m=-k$ for all $k>0$ viz. $$a^{-k+1}=\frac{1}{a^{k-1}}=\frac{a}{a^k}=a^{-k}a.$$Now your original problem is solved for $n=1$, all $n\ge 1$ follow from the inductive step $$a^{m+l+1}=a^ma^l\cdot a=a^m a^{l+1},$$and $n=0$ is trivial. Finally, if $n=-k$ with $k>0$ then $$a^m a^n=\frac{a^m}{a^k}=a^{m-k},$$with the last $=$ following from $a^sa^k=a^{s+k}$ with $s:=m-k$. This is valid by a previous step because $k>0$.

Answer 2

Important note: I consider always $m,n,k\in\mathbb N$, and use $-n,-m,-k$ for negative exponents.

If both negative exponents then $$a^{-m-n}=(\frac 1a)^{m+n}\color{red}=(\frac 1a)^n(\frac 1a)^m=a^{-n}a^{-m}$$

$\color{red}=\quad$ base $\frac 1a\neq 0$ and $m,n$ positive exponents.

If one is negative and the other positive, since the expression is symmetric we don't lose in generality assuming $m\le n$.

Then consider $n-m=k\ge 0$.

$$\require{cancel}a^{-n}=a^{0-n}=a^{k-k-n}=a^{\cancel{n}-m-k-\cancel{n}}=a^{-m-k}\color{blue}=a^{-m}a^{-k}$$

$\color{blue}=\quad$ base $a\neq 0$ and $-m,-k$ both negative exponents.

Rearranging gives $a^k=a^na^{-m}$ which is the desired result.