Order of Squares in Ordered Field

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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 $x \le y$.

$\blacksquare$


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