# Order of Squares in Ordered Field It has been suggested that this article or section be renamed. One may discuss this suggestion on the talk page.

## Theorem

Let $\left({R, +, \circ, \le}\right)$ be an ordered field whose zero is $0_R$ and whose unity is $1_R$.

Suppose that $\forall a \in R: 0 < a \implies 0 < a^{-1}$.

Let $x, y \in \left({R, +, \circ, \le}\right)$ such that $0_R \le x, y$.

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

That is, the square function is an order embedding of $\left({R_{\ge 0}, \le}\right)$ into itself.

When $R$ is one of the standard fields of numbers $\Q$ and $\R$, 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 reverse implication, suppose that $x \circ x \le y \circ y$.

Thus:

 $\displaystyle x \circ x$ $\le$ $\displaystyle y \circ y$ $\displaystyle \implies \ \$ $\displaystyle x \circ x + \left({- \left({x \circ x}\right)}\right)$ $\le$ $\displaystyle y \circ y + \left({- \left({x \circ x}\right)}\right)$ $\displaystyle \implies \ \$ $\displaystyle 0_R$ $\le$ $\displaystyle y \circ y + \left({- \left({x \circ x}\right)}\right)$ $\displaystyle \implies \ \$ $\displaystyle 0_R$ $\le$ $\displaystyle \left({y + \left({-x}\right)}\right) \circ \left({y + x}\right)$ Difference of Two Squares, which applies because a field is a commutative ring.

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

Hence by the premise 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 $x \le y$.

$\blacksquare$