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 $\Z, \Q, \R$, 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 a proof by contradiction.

Suppose that $x \circ x \le y \circ y$ but $x \not\le y$.

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

As $\le$ is compatible with the ring structure $\left({R, +, \circ}\right)$ and $x$ and $y$ are both positive, we have:


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

By the transitivity and antisymmetry of $\le$, it follows that:
 * $y \circ y = x \circ x$

Applying the antisymmetry of $\le$ again, this implies:
 * $x \circ x = y \circ x$

Therefore:

Since $x \ne y$, we have that $x + \left({-y}\right) \ne 0_R$; otherwise, by the associativity of $+$, we would have $x = \left({x + \left({-y}\right)}\right) + y = y$.

We have that $x \ne 0_R$; otherwise, by the antisymmetry of $\le$, we would have $y = 0_R = x$.

But then $x + \left({-y}\right)$ is a proper zero divisor of $R$, contradicting the hypothesis that $R$ has no proper zero divisors.

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.