Ordered Fields | Real Analysis

Sort the fields that admit an order from those that cannot. One positive cone yields every inequality rule.

An ordered field is a field plus a set of positives, the positive cone. Positives must stay positive under addition and multiplication, and every element is positive, zero, or its negative is positive. The rationals and reals obey this; a cyclic clock-like world cannot.

For example, on a 🕐 clock, seven steps of one hour return to the start. In a world where positive steps reach 0, larger and smaller cannot be defined. Lengths differ: no sum of positive lengths is ever 0.

Examples & Exercises

Prove completely, from the positive cone axioms, that in an ordered field $a^2 \ge 0$ for every element $a$ , and $1 > 0$ .

a \ne 0 \ \Longrightarrow \ a^2 > 0, \qquad 1 = 1^2 > 0

A square can never be negative.

Show proof

Case $a > 0$ : the cone is closed under multiplication, so $a^2 = a \cdot a > 0$ . Case $a = 0$ : $a^2 = 0$ .
Case $a < 0$ : trichotomy gives $-a > 0$ , so $(-a)^2 > 0$ . But $(-a)^2 = a^2$ , hence $a^2 > 0$ . Combining the cases: $a^2 > 0$ whenever $a \ne 0$ , and $a^2 \ge 0$ always.
$1 > 0$ : in a field $1 \ne 0$ , and $1 = 1^2$ , so the result above gives $1 > 0$ .

Prove completely, from the positive cone axioms, that $a < b$ implies $a + c < b + c$ for every $c$ , and $ac < bc$ when $c > 0$ .

a < b \iff b - a \in P

Order is judged by a single difference.

Show proof

By definition $a < b$ means $b - a \in P$ . And $(b + c) - (a + c) = b - a \in P$ , so $a + c < b + c$ .
When $c > 0$ , cone closure under multiplication gives $bc - ac = c(b - a) \in P$ . Hence $ac < bc$ .

Prove completely that the complex field $\mathbb{C}$ cannot be made into an ordered field.

i^2 = -1, \qquad a \ne 0 \ \Longrightarrow \ a^2 > 0

The complexes are not the only field that refuses an order.

Show proof

Suppose $\mathbb{C}$ carried an order satisfying the positive cone axioms. Since $i \ne 0$ , the square result forces $i^2 > 0$ .
But $i^2 = -1$ . And $1 > 0$ , so adding $-1$ to both sides gives $0 > -1$ . Thus $-1 > 0$ and $-1 < 0$ at once, contradicting trichotomy.
Hence no positive cone can order $\mathbb{C}$ .