# 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 $n > 0$:

 $\ds r^{n - 1} \times r$ $=$ $\ds r^n$ Definition of Integer Power $\ds \leadsto \ \$ $\ds r^{n - 1}$ $=$ $\ds r^n \dfrac 1 r$ multiplying both sides by $\dfrac 1 r$ $\ds$ $=$ $\ds r^n \times r^{-1}$ Real Number to Negative Power: Integer

When $n \le 0$:

 $\ds r^{n - 1}$ $=$ $\ds \dfrac {r^n} r$ Definition of Integer Power $\ds$ $=$ $\ds r^n \times r^{-1}$ Real Number to Negative Power: Integer

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:

 $\ds r^n \times r^{-\paren {k + 1} }$ $=$ $\ds r^n \times \dfrac {r^{-k} } r$ Definition of Integer Power $\ds$ $=$ $\ds \paren {r^n \times r^{-k} } \times \dfrac 1 r$ Real Multiplication is Associative $\ds$ $=$ $\ds r^{n - k} \times \dfrac 1 r$ Induction Hypothesis $\ds$ $=$ $\ds r^{n - k} \times r^{-1}$ Real Number to Negative Power: Integer $\ds$ $=$ $\ds r^{n - k - 1}$ Basis for the Induction $\ds$ $=$ $\ds r^{n - \paren {k + 1} }$

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

Therefore:

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

$\blacksquare$