Inverse of Multiplicative Inverse/Proof 2
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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
\(\ds \paren {a^{-1} } \times a\) | \(=\) | \(\ds a \times \paren {a^{-1} }\) | 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 $3 \ \text {(i)}$