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
7. 5x $ Ÿ ( 3x Ê 8x Ÿ 10 Ê x Ÿ 5
4
8. 3(2 x) 2(3 x) Ê 6 3x 6 2x
Ê 0 5x Ê 0 x ïïïïïïïïïðqqqqqqqqp x
0
"
9. 2x # 7x 7
6 Ê "# 7
6 5x
Ê "
5
ˆ 10 ‰
6 x or "
3 x
6 x 3x4
10. 4 2 Ê 12 2x 12x 16
Ê 28 14x Ê 2 x qqqqqqqqqðïïïïïïïïî x
2
,2 Chapter 1 Preliminaries
"
11. 4
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. x2 5 Ÿ 123x
4 Ê (4x 20) Ÿ 24 6x
Ê 44 Ÿ 10x Ê 22
5 Ÿ x 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
17. 8 3s œ 9
2 or 8 3s œ #9 Ê 3s œ 7# or 3s œ 25
# Ê sœ
7
6 or s œ 25
6
18. s
# 1 œ 1 or s
# 1 œ 1 Ê s
# œ 2 or s
# 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
23. % 3y 7 4 Ê 3 3y 11 Ê 1 y 11
3 ;
solution interval ˆ1ß 11 ‰
3
24. 1 2y 5 " Ê 6 2y 4 Ê 3 y 2;
solution interval (3ß 2) qqqqðïïïïïïïïðqqqqp y
3 2
25. 1 Ÿ z
5 1Ÿ1 Ê 0Ÿ z
5 Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]
26. 2 Ÿ 1 Ÿ 2 Ê 1 Ÿ
3z
#
3z
# Ÿ 3 Ê 32 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
#
Ê 2
7 x 2
5 ; solution interval ˆ 27 ß 25 ‰
"
28. 3 2
x 43 Ê 1 2
x ( Ê 1 x
# 7
Ê 2x 2
7 Ê 2
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
33. r"
# 1 or ˆ r# 1 ‰ 1 Ê r1 2 or r 1 Ÿ 2
Ê r 1 or r Ÿ 3; solution intervals (_ß 3] [1ß _)
34. 3r
5 " 2
5 or ˆ 3r5 "‰ 2
5
Ê 3r
5
or 3r5 53 Ê r 37 or r 1
7
5
solution intervals (_ß ") ˆ 73 ß _‰ ïïïïïïðqqqqqqðïïïïïïî r
1 7/3
35. x# # Ê kxk È2 Ê È2 x È2 ;
solution interval ŠÈ2ß 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 Š3 È2ß 3 È2‹ qqqqqqðïïïïïïïïðqqqqp x
3 È # 3 È #
, 4 Chapter 1 Preliminaries
Ê ˆx 12 ‰ <
2
41. x# x 0 Ê x# x + 1
4 < 1
4
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)
42. x# x 2 0 Ê x# x + 1
4
9
4 Ê ¹x 1
2 ¹ 3
2 Ê x 1
2
3
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 | x1 | < $ . Then | x1 | < $ Ê 2| x1 | < 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.
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