Ordered field

The usual rules for inequalities follow from the axioms. Trichotomy holds (for each pair \(a,b\) exactly one of \(a<b\), \(a=b\), \(a>b\) is true), the same element can be added to both sides, and one can multiply by a positive element. Multiplying by a negative element reverses the direction: from \(-3<2\) we get \(12>-8\) after multiplying by \(-4\).

A basic result is that \(a^2\ge0\) for every \(a\), hence \(1=1^2>0\). This has strong consequences: \(\mathbb C\) cannot be ordered because \(i^2=-1<0\), and no finite field \(\mathbb Z/p\) can be, since the sum \(1+\cdots+1\) of \(p\) terms would be positive yet equals \(0\).

The rationals \(\mathbb Q\) and the reals \(\mathbb R\) are ordered fields, and so is \(\mathbb Q(\sqrt2)\). So order properties do not distinguish \(\mathbb R\) from \(\mathbb Q\): what separates them is completeness, introduced through the supremum and the completeness axiom.

A common mistake is to use \(a<b\Rightarrow a^2<b^2\) carelessly; it only holds if \(0\le a<b\). With \(-3<2\) it fails, since \(9>4\).