Field Unity Divided by Element equals Multiplicative Inverse

From ProofWiki
Jump to navigation Jump to search

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
\(\ds \) \(=\) \(\ds a^{-1} \times 1_F\) Field Axiom $\text M2$: Commutativity of Product
\(\ds \) \(=\) \(\ds a^{-1}\) Field Axiom $\text M3$: Identity for Product

$\blacksquare$


Sources