Divided by Positive Element of Field of Quotients

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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
\(\ds \) \(=\) \(\ds \paren {-x'} \circ \paren {-y'}^{-1}\) Product of Ring Negatives
\(\ds \) \(=\) \(\ds \paren {-x'} / \paren {-y'}\) {Defof

}}



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.

$\blacksquare$