# Divided by Positive Element of Field of Quotients

## Theorem

Let $\struct {K, +, \circ}$ be the field of quotients of a totally ordered integral domain $\struct {D, +, \circ, \le}$.

Then:

$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$

## Proof

By definition of field of quotients:

$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$

Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.

Then $y' < 0$ as $D$ is totally ordered.

Then:

 $\ds x' / y'$ $=$ $\ds x' \circ \paren {y'}^{-1}$ Definition of Division $\ds$ $=$ $\ds \paren {-x'} \circ \paren {-\paren {y'}^{-1} }$ Product of Ring Negatives $\ds$ $=$ $\ds \paren {-x'} \circ \paren {-y'}^{-1}$ Negative of Product Inverse $\ds$ $=$ $\ds \paren {-x'} / \paren {-y'}$ Definition of Division

If $y' < 0$, then $\paren {-y'} > 0$ from Properties of Ordered Ring $(4)$.

So all we need to do is set $x = -x', y = -y'$ and the result follows.

$\blacksquare$