Sum of Indices of Real Number/Positive Integers

Theorem
Let $r \in \R_{> 0}$ be a positive real number. Let $n, m \in \Z_{\ge 0}$ be positive 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 $\map P m$ be the proposition:
 * $\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$

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

Basis for the Induction
$\map P 1$ is true, by definition of power to an integer:
 * $r^{n + 1} = r^n \times r = r^n \times r^1$

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

Then we need to show:
 * $\forall n \in \Z_{\ge 0}: r^{n + k + 1} = r^n \times r^{k + 1}$

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