Order of Squares in 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 $$x, y \in \left({R, +, \circ; \le}\right)$$ such that $$0_R \le x, y$$.

Then $$x \le y \iff x \circ x \le y \circ y$$.

When $$R$$ is one of the standard sets of numbers, i.e. $$\mathbb{Z}, \mathbb{Q}, \mathbb{R}$$, then this translates into:

If $$x, y$$ are positive then $$x \le y \iff x^2 \le y^2$$.

Note it does not hold for the complex numbers $$\mathbb{C}$$, as $$\mathbb{C}$$ is not an ordered ring.

Proof

 * Assume $$x \le y$$.

As $$\le$$ is compatible with the ring structure of $$\left({R, +, \circ; \le}\right)$$, we have:


 * $$x \ge 0 \Longrightarrow x \circ x \le x \circ y$$
 * $$y \ge 0 \Longrightarrow x \circ y \le y \circ y$$

and thus as $$\le$$ is transitive, it follows that $$x \circ x \le y \circ y$$.


 * Now assume that $$x \circ x \le y \circ y$$.

Thus:

$$ $$ $$ $$

As $$0_R \le x, y$$ we have $$0_R \le x + y$$.

Hence from Properties of an Ordered Ring we have $$0_R \le \left({x + y}\right)^{-1}$$.

So as $$0_R \le \left({y + \left({-x}\right)}\right) \circ \left({y + x}\right)$$ we can multiply both sides by $$\left({x + y}\right)^{-1}$$ and get $$0_R \le \left({y + \left({-x}\right)}\right)$$.

Adding $$-x$$ to both sides gives us $$x \le y$$.