Order of Squares in Ordered Field

Theorem
Let $\left({R, +, \circ, \le}\right)$ be an ordered field 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 $x \le y \iff x^2 \le y^2$.

$x \le y$ implies $x \circ x \le y \circ y$
Suppose that $x \le y$.

By Order of Squares in Ordered Ring:


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

$x \circ x \le y \circ y$ implies $x \le y$
Suppose that $x \circ x \le y \circ y$.

Thus:

As $0_R \le x, y$ we have $0_R \le x + y$.

Hence from the definition of an Ordered Field we have $0_R \le \left({x + y}\right)^{-1}$.

So as $0_R \le \left({y + \left({-x}\right)}\right) \circ \left({y + x}\right)$ we can multiply both sides by $\left({x + y}\right)^{-1}$ and get $0_R \le \left({y + \left({-x}\right)}\right)$.

Adding $-x$ to both sides gives us $x \le y$.

Also see

 * Order of Squares in Totally Ordered Ring without Zero Divisors
 * Order of Squares in Ordered Ring