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
\(\ds \frac {1_R} {\paren {a / b} }\) | \(=\) | \(\ds 1_R / \paren {a \circ b^{-1} }\) | Definition of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds 1_R \circ \paren {a \circ b^{-1} }^{-1}\) | Definition of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b^{-1} }^{-1}\) | Definition of Identity Element of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac a b}^{-1}\) | Definition of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a \circ b^{-1} }^{-1}\) | Definition of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b^{-1} }^{-1} \circ a^{-1}\) | Inverse of Group Product | |||||||||||
\(\ds \) | \(=\) | \(\ds b \circ a^{-1}\) | Inverse of Group Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac b a\) | 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 $23.7 \ (5)$