Order of Squares in Totally Ordered Ring without Proper Zero Divisors

Theorem
Let $\left({R, +, \circ, \le}\right)$ be a totally ordered ring without proper zero divisors whose zero is $0_R$.

Let $x, y \in R$ be positive, i.e., $0_R \le x, y$.

Then $x \le y \iff x^2 \le y^2$.

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

Proof
From Order of Squares in Ordered Ring, we have:


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

To prove the opposite implication, we use the Rule of Transposition.

Suppose that $x \not\le y$.

Since $\le$ is a total ordering, this means $y < x$.

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


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

Thus:


 * $0 \le y \circ \left({x - y}\right)$
 * $0 \le \left({x - y}\right) \circ x$

Since $y < x$ and $0_R \le y$, $0_R < x$, so in particular $x \ne 0$.

Since $x - y \ne 0_R$, $x \ne 0_R$, and $R$ has no proper zero divisors, the second inequality must be strict, i.e.:


 * $0_R < \left({x - y}\right) \circ x$

Expanding and rearranging, we get:


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

Hence:


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