Properties of Ordered Field

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

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

Proof
Recall that the order is, by definition, reflexive, transitive, antisymmetric and further more, every pair of elements is comparable.

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 repeated deduction from these properties.

1.

Conversely,

2.

Conversely,

3.

Conversely,

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


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

5. If $x > 0$, then


 * $x^2 = x \cdot x > x \cdot 0 = 0$

If $x < 0$, then by 1, $-x > 0$, so


 * $x^2 = (-x) \cdot (-x) > (-x) \cdot 0 = 0$

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

7. By 6, we have


 * $0 < 1 < 1 + 1 < 1 + 1 + 1 < \cdots$

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

8. 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 $y^{-1} \leftrightarrow y$.