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:

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

$\blacksquare$