Order of Squares in Ordered Field

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Theorem

Let $\struct {R, +, \circ, \le}$ 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 $x, y \in \struct {R, +, \circ, \le}$ 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 $\struct {R_{\ge 0}, \le}$ 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:

\(\displaystyle x \circ x\) \(\le\) \(\displaystyle y \circ y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x \circ x + \paren {-\paren {x \circ x} }\) \(\le\) \(\displaystyle y \circ y + \paren {-\paren {x \circ x} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0_R\) \(\le\) \(\displaystyle y \circ y + \paren {-\paren {x \circ x} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0_R\) \(\le\) \(\displaystyle \paren {y + \paren {-x} } \circ \paren {y + x}\) 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 $0_R \le \paren {x + y}^{-1}$.

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

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

$\blacksquare$


Also see