Properties of Ordered Ring

Theorem
Let $\left({R, +, \circ, \le}\right)$ be an ordered ring whose zero is $0_R$ and whose unity is $1_R$.

Let $U_R$ be the group of units of $R$.

Let $x, y, z \in \left({R, +, \circ, \le}\right)$.

Then the following properties hold:


 * $(1): \quad x < y \iff x + z < y + z$. Hence $x \le y \iff x + z \le y + z$ (because $\left({R, +, \le}\right)$ is an ordered group).


 * $(2): \quad x < y \iff 0 < y + \left({-x}\right)$. Hence $x \le y \iff 0 \le y + \left({-x}\right)$


 * $(3): \quad 0 < x \iff \left({-x}\right) < 0$. Hence $0 \le x \iff \left({-x}\right) \le 0$


 * $(4): \quad x < 0 \iff 0 < \left({-x}\right)$. Hence $x \le 0 \iff 0 \le \left({-x}\right)$


 * $(5): \quad \forall n \in \Z_{>0}: x > 0 \implies n \cdot x > 0$


 * $(6): \quad x \le y, 0 \le z: x \circ z \le y \circ z, z \circ x \le z \circ y$


 * $(7): \quad x \le y, z \le 0: y \circ z \le x \circ z, z \circ y \le z \circ x$

Total Ordering
If, in addition, $\left({R, +, \circ, \le}\right)$ is totally ordered, the following properties also hold:


 * $(8): \quad 0 < x \circ y \implies \left({0 < x \land 0 < y}\right) \lor \left({x < 0 \land y < 0}\right)$


 * $(9): \quad x \circ y < 0 \implies \left({0 < x \land y < 0}\right) \lor \left({x < 0 \land 0 < y}\right)$


 * $(10): \quad 0 \le x \circ x$. In particular, if $R$ is non-null and has a unity, $0_R < 1_R$


 * $(11): \quad x \in U_R \implies 0 < x \iff 0 < x^{-1}, x \le 0 \iff x^{-1} \le 0$