Cancellation Law for Field Product

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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:

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

$\blacksquare$


Sources

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