Inverse of Multiplicative Inverse/Proof 2

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

\(\ds \paren {a^{-1} } \times a\)

\(=\)

\(\ds a \times \paren {a^{-1} }\)

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

• 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(i)}$