# Properties of Ordered Field

## 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 $x, y, z \in k$:

- $(1): \quad x < 0 \iff -x > 0$
- $(2): \quad x > y \iff x-y > 0$
- $(3): \quad x < y \iff -x > -y$
- $(4): \quad \paren {z < 0} \land \paren {x < y} \implies x z > y z$
- $(5): \quad x \ne 0 \implies x^2 > 0$
- $(6): \quad 1 > 0$
- $(7): \quad \Char k = 0$
- $(8): \quad x > y > 0 \iff y^{-1} > x^{-1} > 0$

## Proof

By definition of ordering, the relation $\le$ is:

and furthermore, every pair of elements is comparable.

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.

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

\(\displaystyle x\) | \(<\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x - x\) | \(<\) | \(\displaystyle 0 - x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -x\) | \(>\) | \(\displaystyle 0\) |

Conversely:

\(\displaystyle x\) | \(>\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x - x\) | \(>\) | \(\displaystyle 0 - x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -x\) | \(<\) | \(\displaystyle 0\) |

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

\(\displaystyle x\) | \(>\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x - y\) | \(>\) | \(\displaystyle y - y = 0\) |

Conversely:

\(\displaystyle x - y > 0\) | \(>\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x - y + y\) | \(>\) | \(\displaystyle y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(>\) | \(\displaystyle y\) |

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

\(\displaystyle x\) | \(<\) | \(\displaystyle y\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x - x - y\) | \(<\) | \(\displaystyle y - x - y\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -y\) | \(<\) | \(\displaystyle -x\) |

Conversely:

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

\(\displaystyle \leadsto \ \ \) | \(\displaystyle -y + y + x\) | \(<\) | \(\displaystyle -x + y + x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(<\) | \(\displaystyle y\) |

- $(4): \quad (z < 0) \land (x < y) \implies x z > y z$:

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.

- $(7): \quad \Char k = 0$:

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 $x > 0$, 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:

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

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

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

$\blacksquare$