Equality of Division Products
Jump to navigation
Jump to search
Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
- $\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
Proof
\(\ds \frac a c\) | \(=\) | \(\ds \frac b d\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a \circ c^{-1}\) | \(=\) | \(\ds b \circ d^{-1}\) | Definition of Division Product | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a \circ c^{-1} \circ c \circ d\) | \(=\) | \(\ds b \circ d^{-1} \circ c \circ d\) | Definition of Cancellable Element of $U_R$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {a \circ d} \circ \paren {c^{-1} \circ c}\) | \(=\) | \(\ds \paren {b \circ c} \circ \paren {d^{-1} \circ d}\) | Definition of Commutative Operation and Definition of Associative Operation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a \circ d\) | \(=\) | \(\ds b \circ c\) | Definition of Identity Element and Definition of Inverse Element |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (3)$