User:Dfeuer/OR6

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Theorem

Let $\left({R,+,\circ,\le}\right)$ be an ordered ring with zero $0_R$.

Suppose that $x \le y$ and $0_R \le z$.

Then $x \circ z \le y \circ z$ and $z \circ x \le z \circ y$.

Proof

We will show that $x \circ z \le y \circ z$; the other conclusion follows from the same argument.

Since $x \le y$, $0_R \le y - x$.

Since $0_R \le y - x$ and $0_R \le z$, Definition:Ordering Compatible with Ring Structure shows that

\(\ds 0_R\) \(\le\) \(\ds \left({y - x}\right) \circ z\)
\(\ds 0_R\) \(\le\) \(\ds y \circ z - x \circ z\) $\circ$ distributes over $+$.
\(\ds x \circ z\) \(\le\) \(\ds y \circ z\)

$\blacksquare$