Properties of Ordered Field
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Theorem
Let $\struct {F, +, \cdot}$ be an ordered field with unity $1$, zero $0$.
Denote the strict order by $<$ and the weak order by $\le$.
Let $\Char F$ denote the characteristic of $F$.
Then the following hold for all $x, y, z \in F$:
- $(1): \quad x < 0 \iff -x > 0$
- $(2): \quad x > y \iff x - y > 0$
- $(3): \quad x < y \iff -x > -y$
- $(4): \quad \paren {z < 0} \land \paren {x < y} \implies x z > y z$
- $(5): \quad x \ne 0 \implies x^2 > 0$
- $(6): \quad 1 > 0$
- $(7): \quad \Char F = 0$
- $(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$
Proof
By definition of ordering, the relation $\le$ is:
and furthermore, every pair of elements is comparable.
The order is compatible with $F$ in the sense that, for all $x, y, z, c \in F$:
- $x < y \implies x + z < y + z$
- $c > 0,\ x < y \implies c x < c y$
The proof is by repeated deduction from these properties.
- $(1): \quad x < 0 \iff -x > 0$:
| \(\ds x\) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - x\) | \(<\) | \(\ds 0 - x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -x\) | \(>\) | \(\ds 0\) |
Conversely:
| \(\ds x\) | \(>\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - x\) | \(>\) | \(\ds 0 - x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -x\) | \(<\) | \(\ds 0\) |
- $(2): \quad x > y \iff x - y > 0$:
| \(\ds x\) | \(>\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - y\) | \(>\) | \(\ds y - y = 0\) |
Conversely:
| \(\ds x - y > 0\) | \(>\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - y + y\) | \(>\) | \(\ds y\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(>\) | \(\ds y\) |
- $(3): \quad x < y \iff -x > -y$:
| \(\ds x\) | \(<\) | \(\ds y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x - x - y\) | \(<\) | \(\ds y - x - y\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -y\) | \(<\) | \(\ds -x\) |
Conversely:
| \(\ds -y\) | \(<\) | \(\ds -x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -y + y + x\) | \(<\) | \(\ds -x + y + x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(<\) | \(\ds y\) |
- $(4): \quad (z < 0) \land (x < y) \implies x z > y z$:
By parts 1 and 3 above, if $z < 0$, $x < y$ then $-z > 0$ and $-x > -y$.
Then:
- $x z = \paren {-x} \paren {-z} > \paren {-y} \paren {-z} = y z$
- $(5): \quad x \ne 0 \implies x^2 > 0$:
If $x > 0$, then:
- $x^2 = x \cdot x > x \cdot 0 = 0$
If $x < 0$, then by 1, $-x > 0$, so:
- $x^2 = \paren {-x} \cdot \paren {-x} > \paren {-x} \cdot 0 = 0$
- $(6): \quad 1 > 0$:
This is immediate from $(5)$, noting that $1 = 1^2$ is a square.
- $(7): \quad \Char F = 0$:
By $(6)$, we have:
- $0 < 1 < 1 + 1 < 1 + 1 + 1 < \cdots$
so $n \cdot 1 \ne 0$ for all $n \in \N$.
- $(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$:
First let $x > 0$.
Aiming for a contradiction, suppose $x^{-1} < 0$.
Then by $(4)$:
- $0 = 0 \cdot x^{-1} > x \cdot x^{-1} = 1$
which contradicts $(6)$, so $x^{-1} > 0$.
Now let $x > y > 0$.
Then:
| \(\ds x^{-1} y^{-1} x\) | \(>\) | \(\ds x^{-1} y^{-1} y > 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y^{-1}\) | \(>\) | \(\ds x^{-1} > 0\) |
The converse follows upon interchanging $x^{-1} \leftrightarrow x$ and $y^{-1} \leftrightarrow y$.
$\blacksquare$