Divided by Positive Element of Quotient Field

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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.

$\blacksquare$