Sum of Indices of Real Number/Integers

Theorem
Let $r \in \R_{> 0}$ be a positive real number. Let $n, m \in \Z$ be integers.

Let $r^n$ be defined as $r$ to the power of $n$.

Then:


 * $r^{n + m} = r^n \times r^m$

Proof
From Sum of Indices of Real Number: Positive Integers, we have that:


 * $m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$

It remains to be shown that:


 * $\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$

The proof will proceed by induction on $m$.

As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.

For all $p \in \Z_{>0}$, let $\map P p$ be the proposition:
 * $\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$

that is:
 * $\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$

Basis for the Induction
$\map P 1$ is true, as follows:

When $n > 0$:

When $n \le 0$:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\forall n \in \Z: r^{n - k} = r^n \times r^{- k}$

Then we need to show:
 * $\forall n \in \Z: r^{n - \paren {k + 1} } = r^n \times r^{-\paren {k + 1} }$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n, m \in \Z: r^{n + m} = r^n \times r^m$