# Condition for Division by Field Elements to be Unity

## Theorem

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

Let $a, b \in F$.

Then:

$\dfrac a b = 1_F$
$a = b$

where $\dfrac a b$ denotes division.

## Proof

### Necessary Condition

Let $a = b$.

Then:

 $\ds \dfrac a b$ $=$ $\ds a \times b^{-1}$ Definition of Division $\ds$ $=$ $\ds b \times b^{-1}$ as $a = b$ $\ds$ $=$ $\ds 1_F$ Field Axiom $\text M4$: Inverses for Product

$\Box$

### Sufficient Condition

Let $\dfrac a b = 1_F$.

Then:

 $\ds a \times b^{-1}$ $=$ $\ds 1_F$ Definition of Division $\ds \leadsto \ \$ $\ds \paren {a \times b^{-1} } \times b$ $=$ $\ds 1_F \times b$ multiplying both sides by $b$ $\ds \leadsto \ \$ $\ds a \times \paren {b \times b^{-1} }$ $=$ $\ds 1_F \times b$ Field Axiom $\text M1$: Associativity of Product $\ds \leadsto \ \$ $\ds a \times 1_F$ $=$ $\ds 1_F \times b$ Field Axiom $\text M4$: Inverses for Product $\ds \leadsto \ \$ $\ds a$ $=$ $\ds b$ Field Axiom $\text M3$: Identity for Product

$\blacksquare$