9.2 ORDERED FIELDS 305(c) Let x be a nonzero element of the ordered field F. Then either
x E 9 or -x E 8. If x E 9, then x2 = (x)(x) E 9 by (b) of Definition 1.
If - x E 9 then, for the same reason, ( - x)( - x) E 9. But (- x)( - x) =
x2, by the corollary to Theorem 3, Article 9.1, so that, again, x2 E 8, as
desired.Only slight knowledge of algebraic properties of C (which we won't cover
in detail until Article 9.4) is required to observe now that no ordering of
C is possible. For if C were ordered, with positive part 9, we would neces-
sarily have i2 E 9, by (c) of Theorem 1. But i2 = - 1 and - 1 E 9 contradicts
(a) of Theorem 1. The following result is established.
COROLLARY
The field C of complex numbers does not admit an ordering.Other examples of fields from Article 9.1 also prove to be nonorder-
able. It is impossible, for instance, to order Z,, because if 9 is the posi-
tive part of Z,, then 1 E 9 SO that 1 + 1 +.. + 1 (7 times) = 0 E 9, as
well. But 0 E 9 contradicts (c) of Definition 1 [see Exercise l(c)].
On the other hand, in addition to the ordered fields R and Q, the
field described in Example 6, Article 9.1, is ordered by the ordering in-
herited from R. Also, we may order the field in Example 7, Article 9.1,
by an ordering based on the sign of the lead coefficient of a polynomial
(see Example 1, Article 10.3 for details).
We next show that the ordering in an ordered field gives rise to a total
ordering of the elements of that field, in the sense of Definition 4, Article
7.4.
DEFINITION 2
Let F be an ordered field with positive part 9. We define a relation "less
than," denoted < , on F by the rule x < y (x is less than y, or xis strlctfy less than y)
if and only if y - x E 8, for any x, y E F. We also write y > x in case x c y.THEOREM 2
In an ordered field (F, 9):(a) x < x is false for any x E F.
(b) If x < y ahd y < z, then x < z, for any x, y, z E F.
(c) x > 0 if and only if x E 9, whereas x < 0 if and only if -x E 9.
(d) Given any x E F, precisely one of the three possibilities, x > 0, x < 0, or x = 0
obtains.
(e) Given any x, YE F, precisely one of the three possibilities, x > y, x < y, or
x = y obtains.Partial proof We leave (a) and (e) as exercises (see Exercise 2). For (b), as-
sume x < y and y < z; to show x < z, we must prove z - x E 9. Now,
by our assumptions, we have y - x E 9 and z - y E 9. By (a) of