Properties of Totally Ordered Fields

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:

• reflexive
• transitive
• antisymmetric

and furthermore, 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 by repeated deduction from these properties.

• $(1): \quad x < 0 \iff -x > 0$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x - x\)	\(<\)	\(\displaystyle 0 - x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle -x\)	\(>\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Conversely,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(>\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x - x\)	\(>\)	\(\displaystyle 0 - x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle -x\)	\(<\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• $(2): \quad x > y \iff x-y > 0$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(>\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x - y\)	\(>\)	\(\displaystyle y - y = 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Conversely,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x - y > 0\)	\(>\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x - y + y\)	\(>\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(>\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

• $(3): \quad x < y \iff -x > -y$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x - x - y\)	\(<\)	\(\displaystyle y - x - y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle - y\)	\(<\)	\(\displaystyle - x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Conversely,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle - y\)	\(<\)	\(\displaystyle -x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle - y + y + x\)	\(<\)	\(\displaystyle -x + y + x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(<\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

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

Then:

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

• $(5): \quad x \neq 0 \implies x^2 > 0$:

If $x > 0$, then:

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

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

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

• $(6): \quad 1 > 0$:

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x^{-1}y^{-1}x\)	\(>\)	\(\displaystyle x^{-1}y^{-1}y > 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y^{-1}\)	\(>\)	\(\displaystyle x^{-1} > 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

$\blacksquare$