# Equality 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:

- $\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$

where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.

## Proof

\(\displaystyle \frac a c\) | \(=\) | \(\displaystyle \frac b d\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a \circ c^{-1}\) | \(=\) | \(\displaystyle b \circ d^{-1}\) | Definition of Division Product | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a \circ c^{-1} \circ c \circ d\) | \(=\) | \(\displaystyle b \circ d^{-1} \circ c \circ d\) | Definition of Cancellable Element of $U_R$ | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {a \circ d} \circ \paren {c^{-1} \circ c}\) | \(=\) | \(\displaystyle \paren {b \circ c} \circ \paren {d^{-1} \circ d}\) | Definition of Commutative Operation | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a \circ d\) | \(=\) | \(\displaystyle b \circ c\) | Definition of Identity Element and Definition of Inverse Element |

$\blacksquare$

Alternatively, a proof can be built using Addition of Division Products.

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 23$: Theorem $23.7 \ (3)$