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

$P \left({0}\right)$ 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, 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\forall n \in \Z_{\ge 0}: r^{n + k} = r^n \times r^k$

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

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