## 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:

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

 $\ds a \circ b'$ $=$ $\ds \paren {a \circ b^{-1} } \circ b \circ b'$ $\ds$ $=$ $\ds \paren {a' \circ \paren {b'}^{-1} } \circ b \circ b'$ as $a / b = a' / b'$ $\ds$ $=$ $\ds a' \circ b$

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

Hence:

 $\ds \paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d$ $=$ $\ds a' \circ b \circ d' \circ d + c' \circ d \circ b' \circ b$ $\ds$ $=$ $\ds a \circ b' \circ d' \circ d + c \circ d' \circ b' \circ b$ $\ds$ $=$ $\ds \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'$

Thus:

 $\ds \frac {\paren {a' \circ d'} + \paren {c' \circ b'} } {b' \circ d'}$ $=$ $\ds \paren {\paren {\paren {a' \circ d'} + \paren {c' \circ b'} } \circ b \circ d} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }$ $\ds$ $=$ $\ds \paren {\paren {\paren {a \circ d} + \paren {c \circ b} } \circ b' \circ d'} \circ \paren {b^{-1} \circ d^{-1} \circ \paren {b'}^{-1} \circ \paren {d'}^{-1} }$ $\ds$ $=$ $\ds \paren {\paren {a \circ d} + \paren {c \circ b} } \circ b^{-1} \circ d^{-1}$ $\ds$ $=$ $\ds \frac {\paren {a \circ d} + \paren {c \circ b} } {b \circ d}$

showing that $+$ is indeed well-defined.

$\blacksquare$