Order of Squares in Totally Ordered Ring without Proper Zero Divisors

Theorem
Let $\left({R, +, \circ, \le}\right)$ be a totally ordered ring without zero divisors 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 $\C$ cannot be given a total order compatible with its ring structure.

$x \le y \implies x^2 \le y^2$
Order of Squares in Ordered Ring

$x \not\le y \implies x^2 \not\le y^2$
Suppose that $x \not\le y$.

Since $R$ is totally ordered, $y < x$.

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


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

Thus
 * $0 \le y \circ (x-y)$
 * $0 \le (x-y) \circ x$

Since $y < x$, $y ≠ x$, so $x - y \ne 0$.

Since $R$ has no zero divisors and $y ≠ x$, at least one of the inequalities must be strict.

Suppose without loss of generality that
 * $0 < y \circ (x-y)$
 * $0 \le (x-y) \circ x$

Expanding and rearranging, we get


 * $y \circ y < y \circ x$
 * $y \circ x \le x \circ x$

Thus $y \circ y < x \circ x$, so

$x \circ x \not\le y \circ y$