User:Dfeuer/OR6

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 shows that $0_R \le \left({y - x}\right) \circ z$.

{{eqn |l= 0_R \le |o= |r= |c=

{{eqn |l= \left({x - y}\right) \circ z     |o= \le |r= 0_R }} {{eqn |l= x \circ z - y \circ z     |o= \le |r= 0_R }} {{eqn |l= x \circ z     |o= \le |r= y \circ z }} {{end-eqn}}

{{qed}}