How to prove $a^ma^n = a^{m+n}$ using field axioms?

Question

I came across this while doing questions on field axioms. I need this proved for where $m,n \in \mathbb Z$ and $a \neq 0$. Please do it in complete steps. I looked this up on the site but I couldn't understand what's been done.

Answer 1

If you are thinking of a real number a and integers m and n it could be easily proven by definition.

So what is a^m: a * a * ... * a m times. (1)

So what is a^n: a * a * ... * a n times. (2)

The same is with a^(m + n): a * a * ... * a m + n times. (3)

When we multiply (1) and (2) we got a * a * ... * a m + n times which is equal to (3).

It means that we proved the identity.

Answer 2

The definition of $a^m$ is by induction on $m$. So the best way to prove what you want to prove is by induction. As $a^0= 1$by definition, what you want to prove is immediate for $n=0$. The step $n=1$ is interesting for the step $n\to n+1$. $a^{m+1} = a^m a$ by definition and so $a^{m+1} = a^m a^1$. Now the step $n\to n+1$ follows easily from the definitions

Answer 3

Okay I found the answer. You have to consider all the values for m and n such that $m,n\in\mathbb Z^+$, $m,n\in\mathbb Z^-$, $m,n = 0$ and $m\in\mathbb Z^+ and, n\in\mathbb Z^- s.t. m>n$.