Answers & Hints for Selected Exer cises 637Case 2: a = b-/2 => a = b = 0, since otherwise % = -/2, which would tell
us that -/2 is rational. So, in this case a + b-/2 = 0.
Case 3: a< b-/2 => (-a) > (-b)-/2 => (-a)+(-b)-/2 E P' => -(a+b-/2) E
P'.- i^2 = -1 < 0 by Cor. 1.2.7, which would contradict Thm. 1.2.6 (c).
EXERCISE SET 1.2-B- (a) By (03), x 2: 0 or x < 0. In the former case lxl = x 2: 0 and in the
latter, lxl = - x 2: 0.
(d) Ix - YI= I - (y - x)I = IY - xi by (b).
(e) We have four cases:
(1) x 2: 0, y 2: 0. Then xy 2: 0 and lxyl = xy = lxllYI·(2) x 2: 0 ,y < 0. Then xy:::; 0 and lxyl = -xy = x(-y) = lxllYI·
(3) x < 0 ,y 2: 0. Then xy:::; 0 and lxyl = -xy = (-x)y = lxllYI·
(4) x < 0, y < 0. Then xy > 0 and lxyl = xy = (-x)(-y) = lxllYI·
- (c) By (b), lx l - IYI:::; Ix -yl and IYI - lx l:::; Ix -yl. Since llx l - IYll = either
lx l - IYI or IYI - lx l, the desired result follows. - Let A= u{[y, z] : y , z EI}. Show A= I.
x E A=> x E [y, z] for some y, z EI=> x E I since I is an interval. Thus,
A~I.
x E I => [x, x] ~ I => x E A. Thus, I ~ A. - x:::; y => min{x, y} = x = -(-x) = - max{-x, -y} since -y:::; -x.
- Multiply both sides of the given inequality by the lowest common denomi-
nator. Prove the resulting inequality and then divide both sides by the LCD.
EXERCISE SET 1.3- 1 - 1 tj. Np; 1 ...;- 2 tj. Np.
In Exercises 3-19 we show only the induction step, P(k) => P(k + 1). Begin by
assuming P(k). Then,
- 1+2+3+·. ·+k+(k+l) = k(k;-l) +(k+l) = (k+l) [~ + 1] = (k+l~k+^2 l.
5. 13 + 23 + ... + k^3 +(k+1)^3 = k
2
(k:l)2
+(k+1)^3 =(k+1)^2 [~2
+(k+1))
= (k + 1)2 k2+:k+4 = (k+l)2Jk+2)2.- 1+4 + 7 + ... + [3(k + 1) - 2] = 1+4 + 7 + ... + (3k - 2) + [3k + 1]