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
Proof by induction on $m$:

For all $m \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the proposition:
 * $\forall n \in \Z: r^{n + m} = r^n \times r^m$

$P \left({0}\right)$ is true, as this just says:
 * $r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$

Basis for the Induction
$P \left({1}\right)$ is true, as follows:

When $n \ge 0$:
 * $r^{n + 1} = r^n \times r = r^n \times r^1$

by definition of integer power.

When $n < 0$:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ 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:
 * $\forall n \in \Z: r^{n + k + 1} = r^n \times r^{k + 1}$

Induction Step
This is our induction step:

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

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

It remains to be shown that:


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

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

For all $p \in \Z_{> 0}$, let $Q \left({p}\right)$ be the proposition:
 * $\forall n \in \Z: r^{n + \left({- p}\right)} = r^n \times r^{-p}$

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

Basis for the Induction (Negative Index)
$Q \left({1}\right)$ is true, as follows:

When $n > 0$:

When $n \le 0$:

This is our basis for the induction.

Induction Hypothesis (Negative Index)
Now we need to show that, if $Q \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $Q \left({k + 1}\right)$ 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:
 * $\forall n \in \Z: r^{n - \left({k + 1}\right)} = r^n \times r^{- \left({k + 1}\right)}$

Induction Step
This is our induction step:

So $Q \left({k}\right) \implies Q \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

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