# Cancellation Law for Field Product

## Theorem

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

Let $a, b, c \in F$.

Then:

$a \times b = a \times c \implies a = 0_F \text { or } b = c$

## Proof

Let $a \times b = a \times c$.

Then:

 $\ds a$ $\ne$ $\ds 0_F$ $\ds \leadsto \ \$ $\ds \exists a^{-1} \in F: \,$ $\ds a^{-1} \times a$ $=$ $\ds 1_F$ $\ds \leadsto \ \$ $\ds a^{-1} \times \paren {a \times b}$ $=$ $\ds a^{-1} \times \paren {a \times c}$ $\ds \leadsto \ \$ $\ds \paren {a^{-1} \times a} \times b$ $=$ $\ds \paren {a^{-1} \times a} \times c$ Field Axiom $\text M1$: Associativity of Product $\ds \leadsto \ \$ $\ds 1_F \times b$ $=$ $\ds 1_F \times c$ Field Axiom $\text M4$: Inverses for Product $\ds \leadsto \ \$ $\ds b$ $=$ $\ds c$ Field Axiom $\text M3$: Identity for Product

That is:

$a \ne 0 \implies b = c$

$\Box$

Suppose $b \ne c$.

Then by Rule of Transposition:

$\map \neg {a \ne 0}$

that is:

$a = 0_F$

and we note that in this case:

$a \times b = 0_F = a \times c$

$\blacksquare$