Show that if < is a relation on a ring R satisfying the properties of trichotomy, transitivity, and isotonicity stated in Theorem 25.5, then there exists a subset P of R satisfying the conditions for a set of positive elements in Definition 25.1, and such that the relation < p defined by a < p b if and only if (b - a) ∈ P is the same as the relation <.

Data from 25.5 Theorem

Let R be an ordered ring with set P of positive elements. Let <, read "is less than," be the relation on R defined by

a < b if and only if (b - a) ∈ P 

for a, b ∈ R. The relation < has these properties for all a, b, c ∈ R. 

Trichotomy One and only one of the following holds: 

a < b, a = b, b < a. 

Transitivity If a < band b < c, then a < c. 

Isotonicity If b < c, then a+ b  a + c

If b < c, and 0 < a then ab < ac and ba < ca.

Conversely, given a relation < on a nonzero ring R satisfying these three conditions, the set P = {x ∈ R |0 < x} satisfies the two criteria for a set of positive elements in Definition 25 .1, and the relation <_p defined as in Condition ( 1) with this P is the given relation<.

Proof Let R be an ordered ring with set P of positive elements, and let a < b mean (b - a) ∈ P. We prove the three properties for <.

Trichotomy Let a, b ∈ R. By the trichotomy property of Pin Definition 25.1 applied to b - a, exactly one of 

(b - a) ∈ P, b - a = 0, (a - b) ∈ P holds. 

These translate in terms of <  to 

a < b, a = b, b < a respectively.

Transitivity Let a < band b < c. Then (b - a) ∈ P and (c - b) ∈ P. By closure of P under addition, we have 

(b - a) + (c - b) = (c - a) ∈ P 

so a < C.

Isotonicity Let b < c, so (c - b) ∈ P. Then (a + c) - (a+ b) = (c - b) ∈ P so a + b < a + c. Also if a > 0, then by closure of P both a(c - b) = ac - ab and (c - b)a = ca - ba are in P, so ab < ac and ba < ca.