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 $+$. | |||||||||||
\(\ds x \circ z\) | \(\le\) | \(\ds y \circ z\) |
$\blacksquare$