Proof. (a) Exercise l.
(b) By Definition 1.2.12,I -x f = { -x if
-(-x) if1.2 The Order Properties 17- x20
- x<O
{- x if - ( -x) :::: 0
-(-x) if - (-x) > 0
{-x if x :::; 0 = {-x if x < 0
x if x > 0 x if x 2 0= fxf.
(c) Let x E F. Then either x 2 0 or x < 0.
Case 1 (x 2 0): Then 0 :::; x = fxf, so -fxf :::; 0 :::; x :::; fxf. Thus,
-fxf:::; x:::; fx f.
Case 2 (x < 0): Then fxl = -x, so -fxf = x < 0 :::; fxf. Thus,
-lxf:::; x:::; fxf.
In either case,-fxl:::; x:::; fxf.
( d) Exercise l.
( e) Exercise l. •Theorem 1.2.14 (Absolute Value Inequalities) Let a 2 0 be a fixed non-
negative element in an ordered field F. Then \:/x, y E F,
(a) fxl < a¢=> -a < x < a.
(b) fxl > a ¢=> x > a or x < -a.
(c) fx-yf<a¢=>y-a<x<y+a.
Proof. (a) Exercise 2.
(b) First, the "::::}" part. Suppose fxf >a.
Case 1 (x 2 0) : Then fxf = x, so x > a.
Case 2 (x < 0): Then fx f = -x, so - x > a, which is equivalent to
x <-a.
Since either Case 1 or Case 2 must be true, either x > a or x < -a.
Now, for the "<=" part. Suppose x >a or x < -a.
Case 1 (x >a): Then x > 0, so fxl = x, so fxl >a.
Cas.e 2 (x < -a): Then - x > a. But in this case, x < 0, so fxl = -x,
from which it follows that f x f > a.
In either case, fx l > a.
( c) Exercise 2. •