# Ordered ring

An **ordered ring** is a ring \(R=(X,\oplus,\otimes)\) with a total order \(\leq\) compatible with the ring structure. Specifically, it must satisfy these axioms for any \(a,b,c \in X\):

If \(a \leq b\), then \(a \oplus c \leq b \oplus c\).

If \(0 \leq a\) and \(0 \leq b\), then \(0 \leq a \otimes b\)

An element \(a\) of the ring is called “positive” if \(0<\) and “negative” if \(a<\). The second axiom, then, says that the product of nonnegative elements is nonnegative.

An ordered ring that is also a field is an ordered field.

# Basic Properties

For any element \(a\), \(a \leq 0\) if and only if \(0 \leq -a\).

The product of nonpositive elements is nonnegative.

The square of any element is nonnegative.

The additive identity \(1 \geq 0\). (Unless the ring is trivial, \(1>0\).)

# Examples

The real numbers are an ordered ring (in fact, an ordered field), as is any subring of \(\mathbb R\), such as \(\mathbb Q\).

The complex numbers are not an ordered ring, because there is no way to define the order between \(0\) and \(i\). Suppose that \(0 \le i\), then, we have \(0 \le i \times i = -1\), which is false. Suppose that \(i \le 0\), then \(0 = i + (-i) \le 0 + (-i)\), but then we have \(0 \le (-i) \times (-i) = -1\), which is again false. Alternatively, \(i^2=-1\) is a square, so it must be nonnegative; that is, \(0 \leq -1\), which is a contradiction.

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