Properties of Ordered Ring

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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 $x \le y \iff x + z \le y + z$
$(2): \quad x < y \iff 0 < y + \left({-x}\right)$. 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 $x \le 0 \iff 0 \le \left({-x}\right)$
$(5): \quad \forall n \in \N^*: x > 0 \implies n \cdot x > 0$
$(6): \quad x \le y, 0 \le z: x \circ z \le y \circ z, z \circ x \le z \circ y$
$(7): \quad x \le y, z \le 0: y \circ z \le x \circ z, z \circ y \le z \circ x$


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 $R$ has a unity, $0_R < 1_R$
$(11): \quad x \in U_R \implies 0 < x \iff 0 < x^{-1}, x \le 0 \iff x^{-1} \le 0$


Proof


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