34 Chapter 1 l'!I The Real Number System
Proof. (a) Suppose that Ve: > 0, x:::::; c:. For contradiction, suppose x 1:. 0.
Then x > 0, so ~x > 0. Then, taking c: = ~x, x:::::; ~x. Contradiction. Therefore,
x:::::; o.
(b) Exercise 9.
( c) Exercise 10.
( d) Exercise 11. •
Although it may be hard to imagine, there are non-Archimedean ordered
fields. Such fields are not often encountered, but can be constructed without
great difficulty. See Exercise 13 below.
EXERCISE SET 1.5l. Prove that the ordered field of rational numbers is Archimedean.
- Prove Theorem 1.5.2, (a) ==> Archimedean Property.
- Prove Theorem 1.5.2, Archimedean Property==> (b).
- Prove Theorem 1.5.2, (c) ==> (a).
- Prove Corollary 1.5.4.
- Prove Theorem 1.5.7 (b).
- Prove that any ordered field F, whether or not it is Archimedean, is
"dense in itself;"that is, between any two elements of F, there exists
another element of F. - Prove Theorem 1.5.8.
- Prove Part (b) of the "forcing principle" (Theorem 1.5.9).
- Prove Part (c) of the "forcing principle" (Theorem 1.5.9).
- Prove Part (d) of the "forcing principle" (Theorem 1.5.9).
- Prove the following extension of the "forcing principle" (Theorem 1.5.9):
(a) If Ve:> 0, x ::'.:: -c:, then x ::'.:: 0.
(b) Ve:> 0, x ::'.::a - c:, then x ::'.::a.- (Project) A non-Archimedean ordered field: Recall that a polyno-
mial (in one variable) is a function of the form
p(x) = anxn + an-1Xn-l + · · · + a1x + ao