# Real Number to Negative Power/Positive Integer

## Theorem

Let $r \in \R_{> 0}$ be a (strictly) positive real number.

Let $n \in \Z_{\ge 0}$ be a positive integer.

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

Then:

$r^{-n} = \dfrac 1 {r^n}$

## Proof

Proof by induction on $m$:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$r^{-n} = \dfrac 1 {r^n}$

$\map P 0$ is the case:

 $\ds r^{-0}$ $=$ $\ds r^0$ $\ds$ $=$ $\ds 1$ Definition of Integer Power $\ds$ $=$ $\ds \dfrac 1 1$ $\ds$ $=$ $\ds \dfrac 1 {r^0}$ Definition of Integer Power

### Basis for the Induction

$\map P 1$ is the case:

 $\ds r^{-1}$ $=$ $\ds \dfrac {r^{-1 + 1} } r$ Definition of Integer Power $\ds$ $=$ $\ds \dfrac {r^0} r$ $\ds$ $=$ $\ds \dfrac 1 r$ Definition of Integer Power $\ds$ $=$ $\ds \dfrac 1 {r^1}$ Definition of Integer Power

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:

$r^{- k} = \dfrac 1 {r^k}$

Then we need to show:

$r^{- \paren {k + 1}\ } = \dfrac 1 {r^{k + 1} }$

### Induction Step

This is our induction step:

 $\ds r^{- \paren {k + 1} }$ $=$ $\ds \dfrac {r^{-\paren {k + 1} + 1} } r$ Definition of Integer Power $\ds$ $=$ $\ds \dfrac {r^{-k} } r$ simplification $\ds$ $=$ $\ds \dfrac 1 {r^k \times r}$ Induction Hypothesis $\ds$ $=$ $\ds \dfrac 1 {r^{k + 1} }$ Definition of Integer Power

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

Therefore:

$\forall n \in \Z_{\ge 0}: r^{-n} = \dfrac 1 {r^n}$

$\blacksquare$