Inverse of Division Product

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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 $a, b \in U_R$.


Then:

$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$

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


Proof

\(\displaystyle \frac {1_R} {\paren {a / b} }\) \(=\) \(\displaystyle 1_R / \paren {a \circ b^{-1} }\) Definition of Division Product
\(\displaystyle \) \(=\) \(\displaystyle 1_R \circ \paren {a \circ b^{-1} }^{-1}\) Definition of Division Product
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ b^{-1} }^{-1}\) Definition of Identity Element of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac a b}^{-1}\) Definition of Division Product
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ b^{-1} }^{-1}\) Definition of Division Product
\(\displaystyle \) \(=\) \(\displaystyle \paren {b^{-1} }^{-1} \circ a^{-1}\) Inverse of Group Product
\(\displaystyle \) \(=\) \(\displaystyle b \circ a^{-1}\) Inverse of Group Inverse
\(\displaystyle \) \(=\) \(\displaystyle \frac b a\) Definition of Division Product

$\blacksquare$


Sources