Condition for Division by Field Elements to be Unity

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Theorem

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

Let $a, b \in F$.


Then:

$\dfrac a b = 1_F$

if and only if:

$a = b$

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 over Field
\(\ds \) \(=\) \(\ds b \times b^{-1}\) 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 over Field
\(\ds \leadsto \ \ \) \(\ds \paren {a \times b^{-1} } \times b\) \(=\) \(\ds 1_F \times b\) 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

$\blacksquare$


Sources

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