Field Unity Divided by Element equals Multiplicative Inverse

Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $a \in F$.

Then:

$\dfrac {1_F} a = a^{-1}$

where $\dfrac {1_F} a$ denotes division.

Proof

\(\ds \dfrac {1_F} a\)

\(=\)

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

Definition of Division over Field

\(\ds \)

\(=\)

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

Field Axiom $\text M2$: Commutativity of Product

\(\ds \)

\(=\)

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

Field Axiom $\text M3$: Identity for Product

$\blacksquare$

Sources

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