Product of Division Products
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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:
- $\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
Proof
\(\ds \frac a c \circ \frac b d\) | \(=\) | \(\ds a \circ c^{-1} \circ b \circ d^{-1}\) | Definition of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} }\) | Definition of Commutative Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b} \circ \paren {c \circ d}^{-1}\) | Inverse of Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a \circ b} {c \circ d}\) | Definition of Division Product |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (4)$