# Real Number to Negative Power/Positive Integer

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## 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:

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

$\blacksquare$