# Divided by Positive Element of Quotient Field

## Theorem

Let $\left({K, +, \circ}\right)$ be the quotient field of a totally ordered integral domain $\left({D, +, \circ, \le}\right)$.

Then:

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

## Proof

By definition:

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

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

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

Then:

 $\displaystyle x' / y'$ $=$ $\displaystyle x' \circ \left({y'}\right)^{-1}$ Definition of Division of $x$ by $y$ $\displaystyle$ $=$ $\displaystyle \left({- x'}\right) \circ \left({- y'}\right)^{-1}$ Product of Ring Negatives $\displaystyle$ $=$ $\displaystyle \left({- x'}\right) / \left({- y'}\right)$ Definition of Division of $x$ by $y$

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

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

$\blacksquare$