Inverse of Multiplicative Inverse

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