12 Chapter 1 • The Real Number System( c) x :::; y iff x 1 y; x ::::: y iff x </. y.
(d) Ifx:s;yandy:s;x,thenx=y.
Proof. (a) Exercise 2.
(b) Exercise 3.
( c) Exercise 4.(Anti-symmetric property)(d) Suppose x:::; y and y:::; x. For contradiction, suppose x # y. Then our
hypotheses become x < y and y < x. But this contradicts the alternate law of
trichotomy, above. Therefore, x = y. •
Theorem 1.2.6 (Combinations of Positive and Negative Elements) In any or-
dered field F ,
(a) The sum of two negative elements is negative.(b) The product of two negative elements is positive.(c) The square of any nonzero element is positive.(d) The prod'uct of a positive element and a negative element is negative.(e) Vx, y E F, if xy > 0, then x and y have the same sign.
(f) Vx, y E F , if x y < 0, then x and y have opposite signs.
Proof. (a) Suppose x and y are negative elements. By definition, this means
that -x E P and -y E P. Thus, by Axiom (01), (- x) + (-y) E P. That is,
-(x+y) E P. But by definition of "negative,'' this means that x+y is negative.
(b) Exercise 5.
(c) Suppose x # 0. By the law of trichotomy, either x E P or -x E P.
Case 1 (x E P): By Axiom (02), x^2 E P. That is, x^2 is positive.
Case 2 (-x E P): By Axiom (02), (-x )^2 E P. But (-x)^2 = x^2.
In either case, x^2 is positive.
( d) Exercise 6.
( e) Exercise 7.
(f) Exercise 8. •
Corollary 1.2. 7 1 > 0. [Thus, -l < O.]
Proof. Exercise 9. •