Order of Squares in Totally 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. $\Z, \Q, \R$, 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 $\C$, as $\C$ 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:

• $x \ge 0 \implies x \circ x \le x \circ y$
• $y \ge 0 \implies x \circ y \le y \circ y$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \circ x\)	\(\le\)	\(\displaystyle y \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \circ x + \left({- \left({x \circ x}\right)}\right)\)	\(\le\)	\(\displaystyle y \circ y + \left({- \left({x \circ x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0_R\)	\(\le\)	\(\displaystyle y \circ y + \left({- \left({x \circ x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0_R\)	\(\le\)	\(\displaystyle \left({y + \left({-x}\right)}\right) \circ \left({y + x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Difference of Two Squares

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 $x \le y$.

$\blacksquare$