Index Laws/Negative Index/Field

Theorem

Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.

Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.

Then:

$\forall a \in \F^* : \forall n \in \Z : \paren{a^n}^{-1} = a^{-n} = \paren{a^{-1}}^n$

Proof

By Definition of Field:

$\struct{F^*, \circ}$ is an Abelian group

By Definition of Power of Field Element:

For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$

From Negative Powers of Group Elements:

$\forall a \in \F^* : \forall n \in \Z : \paren{a^n}^{-1} = a^{-n} = \paren{a^{-1}}^n$

$\blacksquare$