Inverse of Multiplicative Inverse
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Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Let $a^{-1}$ be the multiplicative inverse of $a$.
Then $\paren {a^{-1} }^{-1} = a$.
Proof 1
The multiplicative inverse is, by definition of a field, the inverse element of $a$ in the multiplicative group $\struct {F^*, \times}$.
The result then follows from Inverse of Group Inverse.
$\blacksquare$
Proof 2
\(\ds \paren {a^{-1} } \times a\) | \(=\) | \(\ds a \times \paren {a^{-1} }\) | Field Axiom $\text M2$: Commutativity of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds 1_F\) | Field Axiom $\text M4$: Inverses for Product | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds \paren {a^{-1} }^{-1}\) | Definition of Multiplicative Inverse in Field |
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties