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:

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

$\blacksquare$