Order of Squares in Totally Ordered Ring without Proper Zero Divisors
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Theorem
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring without proper zero divisors whose zero is $0_R$.
Let $x, y \in R$ be positive, that is, $0_R \le x, y$.
Then $x \le y \iff x \circ x \le y \circ y$.
That is, the square mapping is an order embedding of $\struct {R_{\ge 0}, \le}$ into itself.
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.
Aiming for a contradiction, suppose $x \circ x \le y \circ y$ but $x \not\le y$.
Since $\le$ is a total ordering, this means $y < x$.
Since $0_R \le y$, Extended Transitivity shows that $0_R < x$, so in particular $x ≠ 0_R$.
As $\le$ is compatible with the ring structure $\struct {R, +, \circ}$ and $x$ and $y$ are both positive, we have:
- $y \circ y \le y \circ x \le x \circ x$
Since we assume $x \circ x \le y \circ y$, Ordering Cycle implies Equality shows that:
- $y \circ y = x \circ x = y \circ x$
Therefore:
\(\ds 0_R\) | \(=\) | \(\ds x \circ x + \paren {-\paren {y \circ x} }\) | Definition of Ring Negative | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ x + \paren {-y} \circ x\) | Product with Ring Negative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x + \paren {-y} } \circ x\) | Ring Axiom $\text D$: Distributivity of Product over Addition |
Since $x \ne y$, we have that $x + \paren {-y} \ne 0_R$; otherwise, by Ring Axiom $\text A1$: Associativity of Addition, we would have $x = \paren {x + \paren {-y} } + y = y$.
We have already shown that $x \ne 0_R$.
But then $x$ and $x + \paren {-y}$ are proper zero divisors of $R$, contradicting the hypothesis that $R$ has no proper zero divisors.
$\blacksquare$