CHAPTER 1 PRELIMINARIES
1.1 REAL NUMBERS AND THE REAL LINE
2 3 8 9
1. Executing long division, "
9 œ 0.1, 9 œ 0.2, 9 œ 0.3, 9 œ 0.8, 9 œ 0.9
2 3 9 11
2. Executing long division, "
11 œ 0.09, 11 œ 0.18, 11 œ 0.27, 11 œ 0.81, 11 œ 0.99
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 Ê 2 ! 2 < x ! 2 < 6 ! 2 Ê 0 < x ! 2 < 2.
c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3.
d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2.
e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3.
f) NT. 2 < x < 6 Ê x < 6 Ê (x ! 4) < 2 and 2 < x < 6 Ê x > 2 Ê !x < !2 Ê !x + 4 < 2 Ê !(x ! 4) < 2.
The pair of inequalities (x ! 4) < 2 and !(x ! 4) < 2 Ê | x ! 4 | < 2.
g) NT. 2 < x < 6 Ê !2 > !x > !6 Ê !6 < !x < !2. But !2 < 2. So !6 < !x < !2 < 2 or !6 < !x < 2.
h) NT. 2 < x < 6 Ê !1(2) > !1(x) < !1(6) Ê !6 < !x < !2
4. NT = necessarily true, NNT = Not necessarily true. Given: !1 < y ! 5 < 1.
a) NT. !1 < y ! 5 < 1 Ê !1 + 5 < y ! 5 + 5 < 1 + 5 Ê 4 < y < 6.
b) NNT. y = 5 is a counter example. (Actually, never true given that 4 " y " 6)
c) NT. From a), !1 < y ! 5 < 1, Ê 4 < y < 6 Ê y > 4.
d) NT. From a), !1 < y ! 5 < 1, Ê 4 < y < 6 Ê y < 6.
e) NT. !1 < y ! 5 < 1 Ê !1 + 1 < y ! 5 + 1 < 1 + 1 Ê 0 < y ! 4 < 2.
f) NT. !1 < y ! 5 < 1 Ê (1/2)(!1 + 5) < (1/2)(y ! 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.
g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.
h) NT. !1 < y ! 5 < 1 Ê y ! 5 > !1 Ê y > 4 Ê !y < !4 Ê !y + 5 < 1 Ê !(y ! 5) < 1.
Also, !1 < y ! 5 < 1 Ê y ! 5 < 1. The pair of inequalities !(y ! 5) < 1 and (y ! 5) < 1 Ê | y ! 5 | < 1.
5. !2x # 4 Ê x " !2
6. 8 ! 3x 5 Ê !3x !3 Ê x Ÿ 1 ïïïïïïïïïñqqqqqqqqp x
1
5
7. 5x ! $ Ÿ ( ! 3x Ê 8x Ÿ 10 Ê x Ÿ 4
8. 3(2 ! x) # 2(3 % x) Ê 6 ! 3x # 6 % 2x
Ê 0 # 5x Ê 0 # x ïïïïïïïïïðqqqqqqqqp x
0
7 7
9. 2x ! "
# 7x % 6 Ê ! "# ! 6 5x
ˆ! 10 ‰
Ê "
5 6 x or ! "
3 x
6 !x 3x!4
10. 4 " 2 Ê 12 ! 2x " 12x ! 16
Ê 28 " 14x Ê 2 " x qqqqqqqqqðïïïïïïïïî x
2
,2 Chapter 1 Preliminaries
4
11. 5 (x ! 2) " "
3 (x ! 6) Ê 12(x ! 2) " 5(x ! 6)
Ê 12x ! 24 " 5x ! 30 Ê 7x " !6 or x " ! 67
12. ! x"2 5 Ÿ 12"3x
4 Ê !(4x % 20) Ÿ 24 % 6x
Ê !44 Ÿ 10x Ê ! 22
5 Ÿx qqqqqqqqqñïïïïïïïïî x
!22/5
13. y œ 3 or y œ !3
14. y ! 3 œ 7 or y ! 3 œ !7 Ê y œ 10 or y œ !4
15. 2t % 5 œ 4 or 2t % & œ !4 Ê 2t œ !1 or 2t œ !9 Ê t œ ! "# or t œ ! 9#
16. 1 ! t œ 1 or 1 ! t œ !1 Ê !t œ ! or !t œ !2 Ê t œ 0 or t œ 2
9
17. 8 ! 3s œ 2 or 8 ! 3s œ ! #9 Ê !3s œ ! #7 or !3s œ ! 25
# Ê sœ
7
6 or s œ 25
6
s s s s
18. # ! 1 œ 1 or # ! 1 œ !1 Ê # œ 2 or # œ ! Ê s œ 4 or s œ 0
19. !2 " x " 2; solution interval (!2ß 2)
20. !2 Ÿ x Ÿ 2; solution interval [!2ß 2] qqqqñïïïïïïïïñqqqqp x
!2 2
21. !3 Ÿ t ! 1 Ÿ 3 Ê !2 Ÿ t Ÿ 4; solution interval [!2ß 4]
22. !1 " t % 2 " 1 Ê !3 " t " !1;
solution interval (!3ß !1) qqqqðïïïïïïïïðqqqqp t
!3 !1
11
23. !% " 3y ! 7 " 4 Ê 3 " 3y " 11 Ê 1 " y " 3 ;
11 ‰
solution interval ˆ1ß 3
24. !1 " 2y % 5 " " Ê !6 " 2y " !4 Ê !3 " y " !2;
solution interval (!3ß !2) qqqqðïïïïïïïïðqqqqp y
!3 !2
z z
25. !1 Ÿ 5 !1Ÿ1 Ê 0Ÿ 5 Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]
3z 3z
26. !2 Ÿ ! 1 Ÿ 2 Ê !1 Ÿ
# # Ÿ 3 Ê ! 23 Ÿ z Ÿ 2;
solution interval !! 23 ß 2‘ qqqqñïïïïïïïïñqqqqp z
!2/3 2
27. ! "# " 3 ! "
x " "
# Ê ! 7# " ! x" " ! 5# Ê 7
# # "
x # 5
#
2 2
Ê 7 "x" 5 ; solution interval ˆ 27 ß 25 ‰
2 2 x
28. !3 " x !4"3 Ê 1" x "( Ê 1# # # "
7
2 2
Ê 2#x# 7 Ê 7 " x " 2; solution interval ˆ 27 ß 2‰ qqqqðïïïïïïïïðqqqqp x
2/7 2
, Section 1.1 Real Numbers and the Real Line 3
29. 2s 4 or !2s 4 Ê s 2 or s Ÿ !2;
solution intervals (!_ß !2] ' [2ß _)
30. s % 3 "
# or !(s % 3) "
# Ê s ! 5# or !s 7
#
Ê s ! 5# or s Ÿ ! 7# ;
solution intervals ˆ!_ß ! 7# ‘ ' !! 5# ß _‰ ïïïïïïñqqqqqqñïïïïïïî s
!7/2 !5/2
31. 1 ! x # 1 or !(" ! x) # 1 Ê !x # 0 or x # 2
Ê x " 0 or x # 2; solution intervals (!_ß !) ' (2ß _)
32. 2 ! 3x # 5 or !(2 ! 3x) # 5 Ê !3x # 3 or 3x # 7
Ê x " !1 or x # 73 ;
solution intervals (!_ß !1) ' ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî x
!1 7/3
r""
33. # 1 or ! ˆ r"# 1 ‰ 1 Ê r%1 2 or r % 1 Ÿ !2
Ê r 1 or r Ÿ !3; solution intervals (!_ß !3] ' [1ß _)
3r 2
34. 5 !"# 5 or ! ˆ 3r5 ! "‰ # 2
5
3r 7
Ê 5 or ! 3r5 # ! 53 Ê r # 37 or r " 1
# 5
solution intervals (!_ß ") ' ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî r
1 7/3
35. x# " # Ê kxk " È2 Ê !È2 " x " È2 ;
solution interval Š!È2ß È2‹ qqqqqqðïïïïïïðqqqqqqp x
!È # È#
36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ !2;
solution interval (!_ß !2] ' [2ß _) ïïïïïïñqqqqqqñïïïïïïî r
!2 2
37. 4 " x# " 9 Ê 2 " kxk " 3 Ê 2 " x " 3 or 2 " !x " 3
Ê 2 " x " 3 or !3 " x " !2;
solution intervals (!3ß !2) ' (2ß 3) qqqqðïïïïðqqqqðïïïïðqqqp x
!3 !2 2 3
38. "
9 " x# " "
4 Ê "
3 " kxk " "
# Ê "
3 "x" "
# or "
3 " !x " "
#
Ê "
3 "x" or ! "# " x " ! "3 ;
"
#
solution intervals ˆ! "# ß ! "3 ‰ ' ˆ "3 ß "# ‰ qqqqðïïïïðqqqqðïïïïðqqqp x
!1/2 !1/3 1/3 1/2
39. (x ! 1)# " 4 Ê kx ! 1k " 2 Ê !2 " x ! 1 " 2
Ê !1 " x " 3; solution interval (!"ß $) qqqqqqðïïïïïïïïðqqqqp x
!1 3
40. (x % 3)# " # Ê kx % 3k " È2
Ê !È2 " x % 3 " È2 or !3 ! È2 " x " !3 % È2 ;
solution interval Š!3 ! È2ß !3 % È2‹ qqqqqqðïïïïïïïïðqqqqp x
!3 ! È # !3 % È #
, 4 Chapter 1 Preliminaries
1 1 2
41. x# ! x " 0 Ê x# ! x + 4 < 4 Ê ˆx ! 12 ‰ < 1
4 ʹx ! 1
2 ¹< 1
2 Ê ! 12 < x ! 1
2 < 1
2 Ê 0 < x < 1.
So the solution is the interval (0ß 1)
1 9 1 3 1 3
42. x# ! x ! 2 0 Ê x# ! x + 4 4 Ê ¹x ! 2 ¹ 2 Ê x! 2 2 or !ˆx ! 12 ‰ 3
2 Ê x 2 or x Ÿ !1.
The solution interval is (!_ß !1] ' [2ß _)
43. True if a 0; False if a " 0.
44. kx ! 1k œ 1 ! x Í k!(x ! 1)k œ 1 ! x Í 1 ! x 0 Í xŸ1
45. (1) ka % bk œ (a % b) or ka % bk œ !(a % b);
both squared equal (a % b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ !a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka % bk and
y œ kak % kbk so that ka % bk# Ÿ akak % kbkb# Ê ka % bk Ÿ kak % kbk .
46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a " 0 and b " 0, then ab # 0 and kabk œ ab œ (!a)(!b) œ kak kbk .
If a 0 and b " 0, then ab Ÿ 0 and kabk œ !(ab) œ (a)(!b) œ kak kbk .
If a " 0 and b 0, then ab Ÿ 0 and kabk œ !(ab) œ (!a)(b) œ kak kbk .
47. !3 Ÿ x Ÿ 3 and x # ! "# Ê ! "
# " x Ÿ 3.
48. Graph of kxk % kyk Ÿ 1 is the interior
of “diamond-shaped" region.
49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x!1 | < $ . Then | x!1 | < $ Ê 2| x!1 | < 2$ Ê
| 2x ! # | < 2$ Ê | (2x + 1) ! 3 | < 2$ Ê | f(x) ! f(1) | < 2$
50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x ! 0 | < %/2. Then 2| x ! 0 | < % and
| 2x + 3 !3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) ! f(0) | < %.
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê !a < 0. Let !a = b. By definition, | b | œ !b. Since b = !a,
| !a | œ !(!a) œ a and | a | œ | !a | œ a.
ii) For a < 0, | a | œ !a. Now, a < 0 Ê !a > 0. Let !a œ b. By definition, | b | œ b and thus |!a| œ !a. So again
| a | œ |!a|.
iii) By definition | 0 | œ 0 and since !0 œ 0, | !0 | œ 0. Thus, by i), ii), and iii) | a | œ | !a | for any real number.
1.1 REAL NUMBERS AND THE REAL LINE
2 3 8 9
1. Executing long division, "
9 œ 0.1, 9 œ 0.2, 9 œ 0.3, 9 œ 0.8, 9 œ 0.9
2 3 9 11
2. Executing long division, "
11 œ 0.09, 11 œ 0.18, 11 œ 0.27, 11 œ 0.81, 11 œ 0.99
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 Ê 2 ! 2 < x ! 2 < 6 ! 2 Ê 0 < x ! 2 < 2.
c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3.
d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2.
e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3.
f) NT. 2 < x < 6 Ê x < 6 Ê (x ! 4) < 2 and 2 < x < 6 Ê x > 2 Ê !x < !2 Ê !x + 4 < 2 Ê !(x ! 4) < 2.
The pair of inequalities (x ! 4) < 2 and !(x ! 4) < 2 Ê | x ! 4 | < 2.
g) NT. 2 < x < 6 Ê !2 > !x > !6 Ê !6 < !x < !2. But !2 < 2. So !6 < !x < !2 < 2 or !6 < !x < 2.
h) NT. 2 < x < 6 Ê !1(2) > !1(x) < !1(6) Ê !6 < !x < !2
4. NT = necessarily true, NNT = Not necessarily true. Given: !1 < y ! 5 < 1.
a) NT. !1 < y ! 5 < 1 Ê !1 + 5 < y ! 5 + 5 < 1 + 5 Ê 4 < y < 6.
b) NNT. y = 5 is a counter example. (Actually, never true given that 4 " y " 6)
c) NT. From a), !1 < y ! 5 < 1, Ê 4 < y < 6 Ê y > 4.
d) NT. From a), !1 < y ! 5 < 1, Ê 4 < y < 6 Ê y < 6.
e) NT. !1 < y ! 5 < 1 Ê !1 + 1 < y ! 5 + 1 < 1 + 1 Ê 0 < y ! 4 < 2.
f) NT. !1 < y ! 5 < 1 Ê (1/2)(!1 + 5) < (1/2)(y ! 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.
g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.
h) NT. !1 < y ! 5 < 1 Ê y ! 5 > !1 Ê y > 4 Ê !y < !4 Ê !y + 5 < 1 Ê !(y ! 5) < 1.
Also, !1 < y ! 5 < 1 Ê y ! 5 < 1. The pair of inequalities !(y ! 5) < 1 and (y ! 5) < 1 Ê | y ! 5 | < 1.
5. !2x # 4 Ê x " !2
6. 8 ! 3x 5 Ê !3x !3 Ê x Ÿ 1 ïïïïïïïïïñqqqqqqqqp x
1
5
7. 5x ! $ Ÿ ( ! 3x Ê 8x Ÿ 10 Ê x Ÿ 4
8. 3(2 ! x) # 2(3 % x) Ê 6 ! 3x # 6 % 2x
Ê 0 # 5x Ê 0 # x ïïïïïïïïïðqqqqqqqqp x
0
7 7
9. 2x ! "
# 7x % 6 Ê ! "# ! 6 5x
ˆ! 10 ‰
Ê "
5 6 x or ! "
3 x
6 !x 3x!4
10. 4 " 2 Ê 12 ! 2x " 12x ! 16
Ê 28 " 14x Ê 2 " x qqqqqqqqqðïïïïïïïïî x
2
,2 Chapter 1 Preliminaries
4
11. 5 (x ! 2) " "
3 (x ! 6) Ê 12(x ! 2) " 5(x ! 6)
Ê 12x ! 24 " 5x ! 30 Ê 7x " !6 or x " ! 67
12. ! x"2 5 Ÿ 12"3x
4 Ê !(4x % 20) Ÿ 24 % 6x
Ê !44 Ÿ 10x Ê ! 22
5 Ÿx qqqqqqqqqñïïïïïïïïî x
!22/5
13. y œ 3 or y œ !3
14. y ! 3 œ 7 or y ! 3 œ !7 Ê y œ 10 or y œ !4
15. 2t % 5 œ 4 or 2t % & œ !4 Ê 2t œ !1 or 2t œ !9 Ê t œ ! "# or t œ ! 9#
16. 1 ! t œ 1 or 1 ! t œ !1 Ê !t œ ! or !t œ !2 Ê t œ 0 or t œ 2
9
17. 8 ! 3s œ 2 or 8 ! 3s œ ! #9 Ê !3s œ ! #7 or !3s œ ! 25
# Ê sœ
7
6 or s œ 25
6
s s s s
18. # ! 1 œ 1 or # ! 1 œ !1 Ê # œ 2 or # œ ! Ê s œ 4 or s œ 0
19. !2 " x " 2; solution interval (!2ß 2)
20. !2 Ÿ x Ÿ 2; solution interval [!2ß 2] qqqqñïïïïïïïïñqqqqp x
!2 2
21. !3 Ÿ t ! 1 Ÿ 3 Ê !2 Ÿ t Ÿ 4; solution interval [!2ß 4]
22. !1 " t % 2 " 1 Ê !3 " t " !1;
solution interval (!3ß !1) qqqqðïïïïïïïïðqqqqp t
!3 !1
11
23. !% " 3y ! 7 " 4 Ê 3 " 3y " 11 Ê 1 " y " 3 ;
11 ‰
solution interval ˆ1ß 3
24. !1 " 2y % 5 " " Ê !6 " 2y " !4 Ê !3 " y " !2;
solution interval (!3ß !2) qqqqðïïïïïïïïðqqqqp y
!3 !2
z z
25. !1 Ÿ 5 !1Ÿ1 Ê 0Ÿ 5 Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]
3z 3z
26. !2 Ÿ ! 1 Ÿ 2 Ê !1 Ÿ
# # Ÿ 3 Ê ! 23 Ÿ z Ÿ 2;
solution interval !! 23 ß 2‘ qqqqñïïïïïïïïñqqqqp z
!2/3 2
27. ! "# " 3 ! "
x " "
# Ê ! 7# " ! x" " ! 5# Ê 7
# # "
x # 5
#
2 2
Ê 7 "x" 5 ; solution interval ˆ 27 ß 25 ‰
2 2 x
28. !3 " x !4"3 Ê 1" x "( Ê 1# # # "
7
2 2
Ê 2#x# 7 Ê 7 " x " 2; solution interval ˆ 27 ß 2‰ qqqqðïïïïïïïïðqqqqp x
2/7 2
, Section 1.1 Real Numbers and the Real Line 3
29. 2s 4 or !2s 4 Ê s 2 or s Ÿ !2;
solution intervals (!_ß !2] ' [2ß _)
30. s % 3 "
# or !(s % 3) "
# Ê s ! 5# or !s 7
#
Ê s ! 5# or s Ÿ ! 7# ;
solution intervals ˆ!_ß ! 7# ‘ ' !! 5# ß _‰ ïïïïïïñqqqqqqñïïïïïïî s
!7/2 !5/2
31. 1 ! x # 1 or !(" ! x) # 1 Ê !x # 0 or x # 2
Ê x " 0 or x # 2; solution intervals (!_ß !) ' (2ß _)
32. 2 ! 3x # 5 or !(2 ! 3x) # 5 Ê !3x # 3 or 3x # 7
Ê x " !1 or x # 73 ;
solution intervals (!_ß !1) ' ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî x
!1 7/3
r""
33. # 1 or ! ˆ r"# 1 ‰ 1 Ê r%1 2 or r % 1 Ÿ !2
Ê r 1 or r Ÿ !3; solution intervals (!_ß !3] ' [1ß _)
3r 2
34. 5 !"# 5 or ! ˆ 3r5 ! "‰ # 2
5
3r 7
Ê 5 or ! 3r5 # ! 53 Ê r # 37 or r " 1
# 5
solution intervals (!_ß ") ' ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî r
1 7/3
35. x# " # Ê kxk " È2 Ê !È2 " x " È2 ;
solution interval Š!È2ß È2‹ qqqqqqðïïïïïïðqqqqqqp x
!È # È#
36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ !2;
solution interval (!_ß !2] ' [2ß _) ïïïïïïñqqqqqqñïïïïïïî r
!2 2
37. 4 " x# " 9 Ê 2 " kxk " 3 Ê 2 " x " 3 or 2 " !x " 3
Ê 2 " x " 3 or !3 " x " !2;
solution intervals (!3ß !2) ' (2ß 3) qqqqðïïïïðqqqqðïïïïðqqqp x
!3 !2 2 3
38. "
9 " x# " "
4 Ê "
3 " kxk " "
# Ê "
3 "x" "
# or "
3 " !x " "
#
Ê "
3 "x" or ! "# " x " ! "3 ;
"
#
solution intervals ˆ! "# ß ! "3 ‰ ' ˆ "3 ß "# ‰ qqqqðïïïïðqqqqðïïïïðqqqp x
!1/2 !1/3 1/3 1/2
39. (x ! 1)# " 4 Ê kx ! 1k " 2 Ê !2 " x ! 1 " 2
Ê !1 " x " 3; solution interval (!"ß $) qqqqqqðïïïïïïïïðqqqqp x
!1 3
40. (x % 3)# " # Ê kx % 3k " È2
Ê !È2 " x % 3 " È2 or !3 ! È2 " x " !3 % È2 ;
solution interval Š!3 ! È2ß !3 % È2‹ qqqqqqðïïïïïïïïðqqqqp x
!3 ! È # !3 % È #
, 4 Chapter 1 Preliminaries
1 1 2
41. x# ! x " 0 Ê x# ! x + 4 < 4 Ê ˆx ! 12 ‰ < 1
4 ʹx ! 1
2 ¹< 1
2 Ê ! 12 < x ! 1
2 < 1
2 Ê 0 < x < 1.
So the solution is the interval (0ß 1)
1 9 1 3 1 3
42. x# ! x ! 2 0 Ê x# ! x + 4 4 Ê ¹x ! 2 ¹ 2 Ê x! 2 2 or !ˆx ! 12 ‰ 3
2 Ê x 2 or x Ÿ !1.
The solution interval is (!_ß !1] ' [2ß _)
43. True if a 0; False if a " 0.
44. kx ! 1k œ 1 ! x Í k!(x ! 1)k œ 1 ! x Í 1 ! x 0 Í xŸ1
45. (1) ka % bk œ (a % b) or ka % bk œ !(a % b);
both squared equal (a % b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ !a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka % bk and
y œ kak % kbk so that ka % bk# Ÿ akak % kbkb# Ê ka % bk Ÿ kak % kbk .
46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a " 0 and b " 0, then ab # 0 and kabk œ ab œ (!a)(!b) œ kak kbk .
If a 0 and b " 0, then ab Ÿ 0 and kabk œ !(ab) œ (a)(!b) œ kak kbk .
If a " 0 and b 0, then ab Ÿ 0 and kabk œ !(ab) œ (!a)(b) œ kak kbk .
47. !3 Ÿ x Ÿ 3 and x # ! "# Ê ! "
# " x Ÿ 3.
48. Graph of kxk % kyk Ÿ 1 is the interior
of “diamond-shaped" region.
49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x!1 | < $ . Then | x!1 | < $ Ê 2| x!1 | < 2$ Ê
| 2x ! # | < 2$ Ê | (2x + 1) ! 3 | < 2$ Ê | f(x) ! f(1) | < 2$
50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x ! 0 | < %/2. Then 2| x ! 0 | < % and
| 2x + 3 !3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) ! f(0) | < %.
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê !a < 0. Let !a = b. By definition, | b | œ !b. Since b = !a,
| !a | œ !(!a) œ a and | a | œ | !a | œ a.
ii) For a < 0, | a | œ !a. Now, a < 0 Ê !a > 0. Let !a œ b. By definition, | b | œ b and thus |!a| œ !a. So again
| a | œ |!a|.
iii) By definition | 0 | œ 0 and since !0 œ 0, | !0 | œ 0. Thus, by i), ii), and iii) | a | œ | !a | for any real number.
