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:
 * $x \le y \implies x \circ x \le y \circ y$

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 $x \le y \implies x^2 \le y^2$.

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 $x \circ x \le y \circ y$.

Also see

 * Order of Squares in Ordered Field
 * Order of Squares in Totally Ordered Ring without Proper Zero Divisors
 * Power Function Strictly Increasing on Positive Elements