Index Laws/Negative Index/Field
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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:
- $\forall a \in \F^* : \forall n \in \Z : \paren{a^n}^{-1} = a^{-n} = \paren{a^{-1}}^n$
$\blacksquare$