Addition of Division Products

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.

Let $\left({U_R, \circ}\right)$ be the group of units of $\left({R, +, \circ}\right)$.

Let $a, c \in R, b, d \in U_R$. Then:


 * $\displaystyle \frac a b + \frac c d = \frac {a \circ d + b \circ c} {b \circ d}$

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

The operation $+$ is well-defined.

That is:
 * $\displaystyle \frac a b = \frac {a'} {b'}, \frac c d = \frac {c\,'} {d\,'} \implies \frac a b + \frac c d = \frac {a'} {b'} + \frac {c\,'} {d\,'}$

Proof

 * First we demonstrate the operation has the specified property:

Notice that this works only if $\left({R, +, \circ}\right)$ 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:

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

Hence:

Thus:

showing that $+$ is indeed well-defined.