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:
- $\dfrac a b = 1_F$
- $a = b$
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
- 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(ii)}$