16 Chapter^1
We recall that by a linearly ordered field (briefly, an ordered field) is
meant a field F that contains a nonempty subset Fp having the following
properties:
1. If a, b E Fp, then a+ b E Fp-
2. If a, b E Fp, then ab E Fp.
- For each a E F exactly one of the following holds:
a=O, (trichotomy law)The elements of Fp are called the positive elements of F. If there exists
in F such a subset Fp, a linear order can be introduced in F by defining
a > b to mean a - b E Fp. It follows that a > 0 iff a E Fp.
We are now ready to prove the followingTheorem 1.2 The field C of the complex numbers is not a linearly
ordered field.
Proof Suppose that C contains a subset Cp with properties 1, 2, 3, and
that a linear order is introduced in C as indicated above. If z -:/= O, we must
have z^2 > 0. In fact, if z E Cp, then z^2 E Cp or z^2 > 0 by property 2, and
if -z E <Cp, then (-z)^2 = (-1)^2 z^2 = z^2 E Cp or z^2 > 0. Hence 12 = 1
is positive, as well as i^2 = -1. But 1 E Cp implies that -1 rt Cp, by
(3). This contradiction shows that C does not contain a subset Cp with
the required properties.
The question arises as to whether some kind of order (not a linear one)
can be introduced in C. The answer is in the affirmative. In fact, it can be
done in several ways; for instance, if (a, b) -:/= ( c, d) and a -:/= c, we may define
(a, b) > ( c, d) if a > c, and for the case a = c, b -:/= d, we set (a, b) > ( c, d)
if b > d.
However, a more significant ordering of C is obtained .by introduc-
ing the notion of a direct two-order of three elements as follows. Let
S = {x, y, z, ... } be any set with more than two elements, and suppose,
that there is a function f defined on the classes of equivalent ternary per-
mutations of the elements of S (two permutations being equivalent if they
consist of the same elements and are both of the same parity). In addition,
suppose that the range off is the set {-1, O, 1} and that f satisfies the
following properties:
1. f(x1,x2,xa) = f(x2,xa,x1) = -f(x2,xi,xa)
2. If f(y1,x2,xa)f(x1,Y2,Ya) 2 O, f(y2,x2,xa)f(yi,x1,ya) > 0, and
f(ya,x2,xa)f(y1,Y2,x1) 2 O, then
f(x1,x2,xa)f(y1,y2,Ya);?: 0