# Real Number to Negative Power/Positive Integer

## Theorem

Let $r \in \R_{> 0}$ be a 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 $P \left({n}\right)$ be the proposition:

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

$P \left({0}\right)$ is the case:

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

### Basis for the Induction

$P \left({1}\right)$ is the case:

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

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:

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

Then we need to show:

$r^{- \left({k + 1}\right)} = \dfrac 1 {r^{k + 1} }$

### Induction Step

This is our induction step:

 $\displaystyle r^{- \left({k + 1}\right)}$ $=$ $\displaystyle \dfrac {r^{-\left({k + 1}\right) + 1} } r$ Definition of Integer Power $\displaystyle$ $=$ $\displaystyle \dfrac {r^{-k} } r$ simplification $\displaystyle$ $=$ $\displaystyle \dfrac 1 {r^k \times r}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \dfrac 1 {r^{k + 1} }$ Definition of Integer Power

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

Therefore:

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

$\blacksquare$