Addition 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 $a, c \in R, b, d \in U_R$.

Then:

$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$

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


The operation $+$ is well-defined.

That is:

$\dfrac a b = \dfrac {a'} {b'}, \dfrac c d = \dfrac {c'} {d'} \implies \dfrac a b + \dfrac c d = \dfrac {a'} {b'} + \dfrac {c'} {d'}$

Proof

First we demonstrate the operation has the specified property:

\(\displaystyle \frac a b + \frac c d\) \(=\) \(\displaystyle a \circ b^{-1} + c \circ d^{-1}\) Definition of Division Product
\(\displaystyle \) \(=\) \(\displaystyle a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1}\) Definition of Inverse Element and Definition of Identity Element under $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ d} \circ \paren {d^{-1} \circ b^{-1} } + \paren {b \circ c} \circ \paren {d^{-1} \circ b^{-1} }\) Definition of Commutative Operation
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ d + b \circ c} \circ \paren {b \circ d}^{-1}\) Definition of Distributive Operation $\circ$ over $+$
\(\displaystyle \) \(=\) \(\displaystyle \frac {a \circ d + b \circ c} {b \circ d}\) Definition of Division Product


Notice that this works only if $\struct {R, +, \circ}$ is commutative.


Now we show that $+$ is well-defined.

Let $a, c, a', c' \in D, b, d, b', d' \in D^*$ such that $\dfrac a b = \dfrac {a'} {b'}$ and $\dfrac c d = \dfrac {c'} {d'}$.


Then:

\(\displaystyle a \circ b'\) \(=\) \(\displaystyle \paren {a \circ b^{-1} } \circ b \circ b'\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a' \circ \paren {b'}^{-1} } \circ b \circ b'\) as $a / b = a' / b'$
\(\displaystyle \) \(=\) \(\displaystyle a' \circ b\)


Similarly, $c \circ d' = c' \circ d$.


Hence:

\(\displaystyle \paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d\) \(=\) \(\displaystyle a' \circ b \circ d' \circ d + c' \circ d \circ b' \circ b\)
\(\displaystyle \) \(=\) \(\displaystyle a \circ b' \circ d' \circ d + c \circ d' \circ b' \circ b\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'\)


Thus:

\(\displaystyle \frac {\paren {a' \circ d'} + \paren {c' \circ b'} } {b' \circ d'}\) \(=\) \(\displaystyle \paren {\paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b^{-1} \circ d^{-1}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {a \circ d} + \paren {c \circ b} } {b \circ d}\)


showing that $+$ is indeed well-defined.

$\blacksquare$


Sources