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

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

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


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

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:


 * $(1) \quad y \circ y \le y \circ x \quad $
 * $y \circ x \le x \circ x$

Because $\le$ is compatible with $+$, the second inequality can be rewritten thus:


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

By the definition of additive inverse, the left side is $0_R$.

By Product with Ring Negative, $-(y \circ x) = (-y)\circ x$, so we can write:


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

Since $\circ$ distributes over $+$, the equation becomes:


 * $(2) \quad 0_R \le \left({x + (-y)}\right) \circ x$

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

Since $x ≠ y$ and additive inverses are unique, $x + (-y) \ne 0_R$.

Since we also know that $x \ne 0_R$, and $R$ has no proper zero divisors, the inequality in $(2)$ must be strict, i.e.:


 * $(3) \quad 0_R < \left({x + (-y)}\right) \circ x$

Expanding and rearranging $(3)$, we get:


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

Combining this transitively with $(1)$:


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

Hence, since $\le$ is an ordering:


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

Also see

 * Order of Squares in Ordered Ring, a weaker result in a more general context.
 * Order of Squares in Ordered Field, an equivalent result in a different context.