Properties of Ordered Field

Theorem
Let $(k, +, \cdot)$ be a totally ordered field with unity $1$, zero $0$.

Denote the strict order by $<$ and the weak order by $\leq$.

Let $\operatorname{char}k$ be the characteristic of $k$.

Then the following hold for all $x,y,z \in k$:


 * $(1) \quad x < 0 \iff -x > 0$
 * $(2) \quad x > y \iff x-y > 0$
 * $(3) \quad x < y \iff -x > -y$
 * $(4) \quad (z < 0) \land (x < y) \implies xz > yz$
 * $(5) \quad x \neq 0 \implies x^2 > 0$
 * $(6) \quad 1 > 0$
 * $(7) \quad \operatorname{char}k > 0$
 * $(8) \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$

Proof
By definition of ordering, the relation $\le$ is: and furthermore, every pair of elements is comparable.
 * reflexive
 * transitive
 * antisymmetric

The order is compatible with $k$ in the sense that, for all $x,y,z,c \in k$:
 * $x < y \implies x + z < y + z$
 * $c > 0,\ x < y \implies cx < cy$

The proof is by repeated deduction from these properties.


 * $(1) \quad x < 0 \iff -x > 0$:

Conversely,


 * $(2) \quad x > y \iff x-y > 0$:

Conversely,


 * $(3) \quad x < y \iff -x > -y$:

Conversely,


 * $(4) \quad (z < 0) \land (x < y) \implies xz > yz$:

By parts 1 and 3 above, if $z < 0$, $x < y$ then $-z > 0$ and $-x > -y$.

Then:
 * $ xz = (-x)(-z) > (-y)(-z) = yz$

If $x > 0$, then:
 * $(5) \quad x \neq 0 \implies x^2 > 0$:
 * $x^2 = x \cdot x > x \cdot 0 = 0$

If $x < 0$, then by 1, $-x > 0$, so:
 * $x^2 = (-x) \cdot (-x) > (-x) \cdot 0 = 0$


 * $(6) \quad 1 > 0$:

This is immediate from $(5)$, noting that $1 = 1^2$ is a square.


 * $(7) \quad \operatorname{char}k > 0$:

By $(6)$, we have:
 * $0 < 1 < 1 + 1 < 1 + 1 + 1 < \cdots$

so $n\cdot 1 \neq 0$ for all $n \in \N$.


 * $(8) \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$:

First let $x > 0$, and suppose that $x^{-1} < 0$.

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

The converse follows upon interchanging $x^{-1} \leftrightarrow x$ and $y^{-1} \leftrightarrow y$.