Equal Elements of Field of Quotients

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain whose zero is $0_D$.

Let $\left({K, +, \circ}\right)$ be the quotient field of $\left({D, +, \circ}\right)$.

Let $x = \dfrac p q \in K$.

Then:
 * $\forall k \in D^*: x = \dfrac {p \circ k} {q \circ k}$

where:
 * $D^* := D \setminus \left\{{0_D}\right\}$

that is, $D$ with its zero removed.

Proof
We have that the quotient field $\left({K, +, \circ}\right)$ of an integral domain is its inverse completion.

Thus we have:
 * $\forall x_1, x_2 \in D, y_1, y_2 \in D^*: \frac {x_1} {y_1} = \dfrac {x_2} {y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$

So:

Hence the result.

Note that in order for $\dfrac {p \circ k} {q \circ k}$ to be defined, $q \circ k \ne 0_D$, that is, $k \ne 0_D$.