,Solution manual for Precalculus Mathematics for
Calculus, 8th Edition James Stewart
Notes
1- The file is chapter after chapter.
2- We have shown you 10 or five pages.
3- The file contains all Appendix and Excel
sheet if it exists.
4- We have all what you need, we make
update at every time. There are many
new editions waiting you.
5- If you think you purchased the wrong file
You can contact us at every time, we can
replace it with true one.
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,PROLOGUE: Principles of Problem Solving
distance 1 1
1. Let r be the rate of the descent. We use the formula time ; the ascent takes h, the descent takes h, and the
rate 15 r
2 1 1 1 1 1
total trip should take h. Thus we have 0, which is impossible. So the car cannot go
30 15 15 r 15 r
fast enough to average 30 mi/h for the 2mile trip.
2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price
decreases to 08P, and after another 20% discount, it becomes 08 08P 064P. Since 06P 064P, a 40% discount
is better.
3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since
the first cut produces 4 pieces, we get the formula f n 4 3 n 1, n 1. Since f 142 4 3 141 427, we
see that 142 parallel cuts produce 427 pieces.
4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two
amoebas into the vessel, it will take 60 3 57 minutes for the vessel to be full of amoebas.
5. The statement is false. Here is one particular counterexample:
Player A Player B
First half 1
1 hit in 99 atbats: average 99 0 hit in 1 atbat: average 01
Second half 1 hit in 1 atbat: average 11 98 hits in 99 atbats: average 98
99
Entire season 2
2 hits in 100 atbats: average 100 99
99 hits in 100 atbats: average 100
6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus,
any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup.
Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream cream
mixture being returned to the pitcher of cream. Suppose it is possible to separate
the cream and the coffee, as shown. Then you can see that the coffee going into the coffee
cream occupies the same volume as the cream that was left in the coffee.
Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream
cream 1 coffee y
is added to the coffee cup, the resulting mixture has the following ratios: and .
mixture y1 mixture y1
1
So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a
y1
y
spoonful of cream and spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is
y 1
1 y
1 of a spoonful. This is the same as the amount of coffee we added to the cream.
y 1 y1
7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius
by 1 foot, the new radius is r 1, so the new circumference is 2 r 1. Thus you need 2 r 1 2r 2 extra
feet of ribbon.
1
,2 Principles of Problem Solving
8. The north pole is such a point. And there are others: Consider a point a1 near the south pole such that the parallel passing
through a1 forms a circle C1 with circumference exactly one mile. Any point P1 exactly one mile north of the circle C1
along a meridian is a point satisfying the conditions in the problem: starting at P1 she walks one mile south to the point a1
on the circle C1 , then one mile east along C1 returning to the point a1 , then north for one mile to P1 . That’s not all. If a
point a2 (or a3 , a4 , a5 , ) is chosen near the south pole so that the parallel passing through it forms a circle C2 (C3 , C4 ,
C5 , ) with a circumference of exactly 12 mile ( 13 mi, 14 mi, 15 mi, ), then the point P2 (P3 , P4 , P5 , ) one mile north
of a2 (a3 , a4 , a5 , ) along a meridian satisfies the conditions of the problem: she walks one mile south from P2 (P3 , P4 ,
P5 , ) arriving at a2 ( a3 , a4 , a5 , ) along the circle C2 (C3 , C4 , C5 , ), walks east along the circle for one mile thus
traversing the circle twice (three times, four times, five times, ) returning to a2 (a3 , a4 , a5 , ), and then walks north one
mile to P2 ( P3 , P4 , P5 , ).
,CHAPTER 1 FUNDAMENTALS 1
1.1 Real Numbers 1
1.2 Exponents and Radicals 6
1.3 Algebraic Expressions 13
1.4 Rational Expressions 19
1.5 Equations 27
1.6 Complex Numbers 36
1.7 Modeling with Equations 39
1.8 Inequalities 50
1.9 The Coordinate Plane; Graphs of Equations; Circles 72
1.10 Lines 91
1.11 Solving Equations and Inequalities Graphically 101
1.12 Making Models Using Variation 111
Chapter 1 Review 115
Chapter 1 Test 132
¥ FOCUS ON MODELING: Fitting Lines to Data 138
,1 FUNDAMENTALS
1.1 REAL NUMBERS
1. (a) The natural numbers are 1 2 3 .
(b) The numbers 3 2 1 0 are integers but not natural numbers.
p
(c) Any irreducible fraction with q 1 is rational but is not an integer. Examples: 32 , 12
5 , 1729 .
23
q
p
(d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e.
q
2. (a) ab ba; Commutative Property of Multiplication
(b) a b c a b c; Associative Property of Addition
(c) a b c ab ac; Distributive Property
3. (a) In setbuilder notation: x 3 x 5 (c) As a graph:
_3 5
(b) In interval notation: 3 5
4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.
5. The distance between a and b on the real line is d a b b a. So the distance between 5 and 2 is 2 5 7.
6. (a) If a b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many
ba
numbers—including, for example a n for every positive n. (If an interval extends to infinity in either or both
2
directions, then it obviously contains infinitely many numbers.)
(b) No, because 5 6 does not include 5.
7. (a) No: a b b a b a in general.
(b) No; by the Distributive Property, 2 a 5 2a 2 5 2a 10 2a 10.
8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive.
(b) Yes, b a a b.
9. (a) Natural number: 100 10. (a) Natural numbers: 2, 9 3, 10
(b) Integers: 0, 100, 8 (b) Integers: 2, 100
2 50, 9 3, 10
(c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8 (c) Rational numbers: 45 92 , 13 , 16666 53 ,
(d) Irrational numbers: 7, 2, 100
2 , 9 3, 10
(d) Irrational numbers: 2, 314
11. Commutative Property of addition 12. Commutative Property of multiplication
13. Associative Property of addition 14. Distributive Property
15. Distributive Property 16. Distributive Property
17. Commutative Property of multiplication 18. Distributive Property
19. x 3 3 x 20. 7 3x 7 3 x
21. 4 A B 4A 4B 22. 5x 5y 5 x y
1
,2 CHAPTER 1 Fundamentals
23. 2 x y 2x 2y 24. a b 5 5a 5b
25. 5 2x y 5 2 x y 10x y 26. 43 6y 43 6 y 8y
27. 52 2x 4y 52 2x 52 4y 5x 10y 28. 3a b c 2d 3ab 3ac 6ad
29. (a) 23 57 14 15 29
21 21 21 30. (a) 25 38 16 15 1
40 40 40
5 3 10 9 1
(b) 12 (b) 32 58 16 36 15 4 25
8 24 24 24 24 24 24 24
2
2
31. (a) 23 6 32 23 6 23 32 4 1 3 32. (a) 2 3 2 32 23 12 3 13 93 13 83
2
3
(b) 3 14 1 45 12 4 4
1 5 4 13 1 13
5 5 4 5 20 2 1 2 1 2 1
(b) 15 23 51 21 51 21 10 45 9
10 12 3 3
10 15 10 5 10 5
33. (a) 2 3 6 and 2 72 7, so 3 72 34. (a) 3 23 2 and 3 067 201, so 23 067
(b) 6 7 (b) 23 067
(c) 35 72 (c) 06 06
35. (a) False 36. (a) False: 3 173205 17325.
(b) True (b) False
37. (a) True (b) False 38. (a) True (b) True
39. (a) x 0 (b) t 4 40. (a) y 0 (b) z 3
(c) a (d) 5 x 13 (c) b 8 (d) 0 17
(e) 3 p 5 (e) y 2
41. (a) A B 1 2 3 4 5 6 7 8 42. (a) B C 2 4 6 7 8 9 10
(b) A B 2 4 6 (b) B C 8
43. (a) A C 1 2 3 4 5 6 7 8 9 10 44. (a) A B C 1 2 3 4 5 6 7 8 9 10
(b) A C 7 (b) A B C
45. (a) B C x x 5 46. (a) A C x 1 x 5
(b) B C x 1 x 4 (b) A B x 2 x 4
47. 3 0 x 3 x 0 48. 2 8] x 2 x 8
_3 0 2 8
49. [2 8 x 2 x 8 50. 6 12 x 6 x 12
2 8 1
_6 _ _2
, SECTION 1.1 Real Numbers 3
51. [2 x x 2 52. 1 x x 1
2 1
53. x 1 x 1] 54. 1 x 2 x [1 2]
1 1 2
55. 2 x 1 x 2 1] 56. x 5 x [5
_2 1 _5
57. x 1 x 1 58. 5 x 2 x 5 2
_1 _5 2
59. (a) [3 5] (b) 3 5] (c) 3 60. (a) [0 2 (b) 2 0] (c) 0]
61. 2 0 1 1 2 1 62. 2 0 1 1 0
_2 1 _1 0
63. [4 6] [0 8 [0 6] 64. [4 6] [0 8 [4 8
0 6 _4 8
65. 4 4 66. 6] 2 10 2 6]
_4 4 2 6
67. (a) 50 50 68. (a) 2 8 6 6
(b) 13 13 (b) 8 2 8 2 6 6
69. (a) 6 4 6 4 2 2 70. (a) 2 12 2 12 10 10
(b) 1
1 1 1
1 (b) 1 1 1 1 1 1 1 0 1
1 1
71. (a) 2 6 12 12 72. (a) 6
24 4 4
5
(b) 13 15 5 5 (b) 712
127 5 1 1
73. 2 3 5 5 74. 25 15 4 4
,4 CHAPTER 1 Fundamentals
7 1 49 5 54 18 18
75. (a) 17 2 15 76. (a) 15 21 105 105 105 35 35
(b) 21 3 21 3 24 24 (b) 38 57 38 57 19 19.
3 12 55 67 67
(c) 10 11
8 40 40 40 40 (c) 26 18 26 18 08 08.
77. (a) Let x 0777 . So 10x 77777 x 07777 9x 7. Thus, x 79 .
(b) Let x 02888 . So 100x 288888 10x 28888 90x 26. Thus, x 26 13
90 45 .
(c) Let x 0575757 . So 100x 575757 x 05757 99x 57. Thus, x 57 19
99 33 .
78. (a) Let x 52323 . So 100x 5232323 1x 52323 99x 518. Thus, x 518
99 .
(b) Let x 13777 . So 100x 1377777 10x 137777 90x 124. Thus, x 124 62
90 45 .
(c) Let x 213535 . So 1000x 21353535 10x 213535 990x 2114. Thus, x 2114 1057
990 495 .
79. 3, so 3 3. 80. 2 1, so 1 2 2 1.
81. a b, so a b a b b a. 82. a b a b a b b a 2b
83. (a) a is negative because a is positive.
(b) bc is positive because the product of two negative numbers is positive.
(c) a ba b is positive because it is the sum of two positive numbers.
(d) ab ac is negative: each summand is the product of a positive number and a negative number, and the sum of two
negative numbers is negative.
84. (a) b is positive because b is negative.
(b) a bc is positive because it is the sum of two positive numbers.
(c) c a c a is negative because c and a are both negative.
(d) ab2 is positive because both a and b2 are positive.
85. Distributive Property
86. (a) When L 60, x 8, and y 6, we have L 2 x y 60 2 8 6 60 28 88. Because 88 108 the
post office will accept this package.
When L 48, x 24, and y 24, we have L 2 x y 48 2 24 24 48 96 144, and since
144 108, the post office will not accept this package.
(b) If x y 9, then L 2 9 9 108 L 36 108 L 72. So the length can be as long as 72 in. 6 ft.
m1 m2 m1 m m1n2 m2n1
87. Let x and y be rational numbers. Then x y 2 ,
n1 n2 n1 n2 n1 n2
m m m n m 2 n1 m m m m
xy 1 2 1 2 , and x y 1 2 1 2 . This shows that the sum, difference, and product
n1 n2 n1 n2 n1 n2 n1n2
of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily
irrational; for example, 2 2 2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;
for example, 2 2 0 which is rational.
88. 12 2 is irrational. If it were rational, then by Exercise 6(a), the sum 12 2 12 2 would be rational, but
this is not the case.
Similarly, 12 2 is irrational.
(a) Following the hint, suppose that r t q, a rational number. Then by Exercise 6(a), the sum of the two rational
numbers r t and r is rational. But r t r t, which we know to be irrational. This is a contradiction, and
hence our original premise—that r t is rational—was false.
, SECTION 1.1 Real Numbers 5
a
(b) r is a nonzero rational number, so r for some nonzero integers a and b. Let us assume that rt q, a rational
b
c a c bc
number. Then by definition, q for some integers c and d. But then r t q t , whence t , implying
d b d ad
that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number
and an irrational number is irrational.
89.
x 1 2 10 100 1000
1 1 1 1 1 1
x 2 10 100 1000
As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero.
x 1 05 01 001 0001
1 1 1 1 1 1
x 05 2 01 10 001 100 0001 1000
As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.
90. We can construct the number 2 on the number line by
Ï2
transferring the length of the hypotenuse of a right triangle 1
with legs of length 1 and 1.
_1 0 1 Ï2 2 3
Similarly, to locate 5, we construct a right triangle with legs
of length 1 and 2. By the Pythagorean Theorem, the length Ï5
1
of the hypotenuse is 12 22 5. Then transfer the
length of the hypotenuse to the number line. _1 0 1 2 Ï5 3
The square root of any rational number can be located on a
number line in this fashion.
The circle in the second figure in the text has circumference , so if we roll it along a number line one full rotation, we have
found on the number line. Similarly, any rational multiple of can be found this way.
a b a b abab
91. (a) Suppose that a b, so max a b a and a b a b. Then a.
2 2
On the other hand, if b a, then max a b b and a b a b b a. In this case,
a b a b abba
b.
2 2
If a b, then a b 0 and the result is trivial.
a b a b a b b a
(b) If a b, then min a b a and a b b a. In this case a.
2 2
a b a b
Similarly, if b a, then b; and if a b, the result is trivial.
2
92. Answers will vary.
93. (a) Subtraction is not commutative. For example, 5 1 1 5.
(b) Division is not commutative. For example, 5 1 1 5.
(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes,
the result is not the same as if you proceed the other way around.
(d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result.
(e) Washing laundry and drying it are not commutative.
Calculus, 8th Edition James Stewart
Notes
1- The file is chapter after chapter.
2- We have shown you 10 or five pages.
3- The file contains all Appendix and Excel
sheet if it exists.
4- We have all what you need, we make
update at every time. There are many
new editions waiting you.
5- If you think you purchased the wrong file
You can contact us at every time, we can
replace it with true one.
Our email:
,PROLOGUE: Principles of Problem Solving
distance 1 1
1. Let r be the rate of the descent. We use the formula time ; the ascent takes h, the descent takes h, and the
rate 15 r
2 1 1 1 1 1
total trip should take h. Thus we have 0, which is impossible. So the car cannot go
30 15 15 r 15 r
fast enough to average 30 mi/h for the 2mile trip.
2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price
decreases to 08P, and after another 20% discount, it becomes 08 08P 064P. Since 06P 064P, a 40% discount
is better.
3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since
the first cut produces 4 pieces, we get the formula f n 4 3 n 1, n 1. Since f 142 4 3 141 427, we
see that 142 parallel cuts produce 427 pieces.
4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two
amoebas into the vessel, it will take 60 3 57 minutes for the vessel to be full of amoebas.
5. The statement is false. Here is one particular counterexample:
Player A Player B
First half 1
1 hit in 99 atbats: average 99 0 hit in 1 atbat: average 01
Second half 1 hit in 1 atbat: average 11 98 hits in 99 atbats: average 98
99
Entire season 2
2 hits in 100 atbats: average 100 99
99 hits in 100 atbats: average 100
6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus,
any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup.
Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream cream
mixture being returned to the pitcher of cream. Suppose it is possible to separate
the cream and the coffee, as shown. Then you can see that the coffee going into the coffee
cream occupies the same volume as the cream that was left in the coffee.
Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream
cream 1 coffee y
is added to the coffee cup, the resulting mixture has the following ratios: and .
mixture y1 mixture y1
1
So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a
y1
y
spoonful of cream and spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is
y 1
1 y
1 of a spoonful. This is the same as the amount of coffee we added to the cream.
y 1 y1
7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius
by 1 foot, the new radius is r 1, so the new circumference is 2 r 1. Thus you need 2 r 1 2r 2 extra
feet of ribbon.
1
,2 Principles of Problem Solving
8. The north pole is such a point. And there are others: Consider a point a1 near the south pole such that the parallel passing
through a1 forms a circle C1 with circumference exactly one mile. Any point P1 exactly one mile north of the circle C1
along a meridian is a point satisfying the conditions in the problem: starting at P1 she walks one mile south to the point a1
on the circle C1 , then one mile east along C1 returning to the point a1 , then north for one mile to P1 . That’s not all. If a
point a2 (or a3 , a4 , a5 , ) is chosen near the south pole so that the parallel passing through it forms a circle C2 (C3 , C4 ,
C5 , ) with a circumference of exactly 12 mile ( 13 mi, 14 mi, 15 mi, ), then the point P2 (P3 , P4 , P5 , ) one mile north
of a2 (a3 , a4 , a5 , ) along a meridian satisfies the conditions of the problem: she walks one mile south from P2 (P3 , P4 ,
P5 , ) arriving at a2 ( a3 , a4 , a5 , ) along the circle C2 (C3 , C4 , C5 , ), walks east along the circle for one mile thus
traversing the circle twice (three times, four times, five times, ) returning to a2 (a3 , a4 , a5 , ), and then walks north one
mile to P2 ( P3 , P4 , P5 , ).
,CHAPTER 1 FUNDAMENTALS 1
1.1 Real Numbers 1
1.2 Exponents and Radicals 6
1.3 Algebraic Expressions 13
1.4 Rational Expressions 19
1.5 Equations 27
1.6 Complex Numbers 36
1.7 Modeling with Equations 39
1.8 Inequalities 50
1.9 The Coordinate Plane; Graphs of Equations; Circles 72
1.10 Lines 91
1.11 Solving Equations and Inequalities Graphically 101
1.12 Making Models Using Variation 111
Chapter 1 Review 115
Chapter 1 Test 132
¥ FOCUS ON MODELING: Fitting Lines to Data 138
,1 FUNDAMENTALS
1.1 REAL NUMBERS
1. (a) The natural numbers are 1 2 3 .
(b) The numbers 3 2 1 0 are integers but not natural numbers.
p
(c) Any irreducible fraction with q 1 is rational but is not an integer. Examples: 32 , 12
5 , 1729 .
23
q
p
(d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e.
q
2. (a) ab ba; Commutative Property of Multiplication
(b) a b c a b c; Associative Property of Addition
(c) a b c ab ac; Distributive Property
3. (a) In setbuilder notation: x 3 x 5 (c) As a graph:
_3 5
(b) In interval notation: 3 5
4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.
5. The distance between a and b on the real line is d a b b a. So the distance between 5 and 2 is 2 5 7.
6. (a) If a b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many
ba
numbers—including, for example a n for every positive n. (If an interval extends to infinity in either or both
2
directions, then it obviously contains infinitely many numbers.)
(b) No, because 5 6 does not include 5.
7. (a) No: a b b a b a in general.
(b) No; by the Distributive Property, 2 a 5 2a 2 5 2a 10 2a 10.
8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive.
(b) Yes, b a a b.
9. (a) Natural number: 100 10. (a) Natural numbers: 2, 9 3, 10
(b) Integers: 0, 100, 8 (b) Integers: 2, 100
2 50, 9 3, 10
(c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8 (c) Rational numbers: 45 92 , 13 , 16666 53 ,
(d) Irrational numbers: 7, 2, 100
2 , 9 3, 10
(d) Irrational numbers: 2, 314
11. Commutative Property of addition 12. Commutative Property of multiplication
13. Associative Property of addition 14. Distributive Property
15. Distributive Property 16. Distributive Property
17. Commutative Property of multiplication 18. Distributive Property
19. x 3 3 x 20. 7 3x 7 3 x
21. 4 A B 4A 4B 22. 5x 5y 5 x y
1
,2 CHAPTER 1 Fundamentals
23. 2 x y 2x 2y 24. a b 5 5a 5b
25. 5 2x y 5 2 x y 10x y 26. 43 6y 43 6 y 8y
27. 52 2x 4y 52 2x 52 4y 5x 10y 28. 3a b c 2d 3ab 3ac 6ad
29. (a) 23 57 14 15 29
21 21 21 30. (a) 25 38 16 15 1
40 40 40
5 3 10 9 1
(b) 12 (b) 32 58 16 36 15 4 25
8 24 24 24 24 24 24 24
2
2
31. (a) 23 6 32 23 6 23 32 4 1 3 32. (a) 2 3 2 32 23 12 3 13 93 13 83
2
3
(b) 3 14 1 45 12 4 4
1 5 4 13 1 13
5 5 4 5 20 2 1 2 1 2 1
(b) 15 23 51 21 51 21 10 45 9
10 12 3 3
10 15 10 5 10 5
33. (a) 2 3 6 and 2 72 7, so 3 72 34. (a) 3 23 2 and 3 067 201, so 23 067
(b) 6 7 (b) 23 067
(c) 35 72 (c) 06 06
35. (a) False 36. (a) False: 3 173205 17325.
(b) True (b) False
37. (a) True (b) False 38. (a) True (b) True
39. (a) x 0 (b) t 4 40. (a) y 0 (b) z 3
(c) a (d) 5 x 13 (c) b 8 (d) 0 17
(e) 3 p 5 (e) y 2
41. (a) A B 1 2 3 4 5 6 7 8 42. (a) B C 2 4 6 7 8 9 10
(b) A B 2 4 6 (b) B C 8
43. (a) A C 1 2 3 4 5 6 7 8 9 10 44. (a) A B C 1 2 3 4 5 6 7 8 9 10
(b) A C 7 (b) A B C
45. (a) B C x x 5 46. (a) A C x 1 x 5
(b) B C x 1 x 4 (b) A B x 2 x 4
47. 3 0 x 3 x 0 48. 2 8] x 2 x 8
_3 0 2 8
49. [2 8 x 2 x 8 50. 6 12 x 6 x 12
2 8 1
_6 _ _2
, SECTION 1.1 Real Numbers 3
51. [2 x x 2 52. 1 x x 1
2 1
53. x 1 x 1] 54. 1 x 2 x [1 2]
1 1 2
55. 2 x 1 x 2 1] 56. x 5 x [5
_2 1 _5
57. x 1 x 1 58. 5 x 2 x 5 2
_1 _5 2
59. (a) [3 5] (b) 3 5] (c) 3 60. (a) [0 2 (b) 2 0] (c) 0]
61. 2 0 1 1 2 1 62. 2 0 1 1 0
_2 1 _1 0
63. [4 6] [0 8 [0 6] 64. [4 6] [0 8 [4 8
0 6 _4 8
65. 4 4 66. 6] 2 10 2 6]
_4 4 2 6
67. (a) 50 50 68. (a) 2 8 6 6
(b) 13 13 (b) 8 2 8 2 6 6
69. (a) 6 4 6 4 2 2 70. (a) 2 12 2 12 10 10
(b) 1
1 1 1
1 (b) 1 1 1 1 1 1 1 0 1
1 1
71. (a) 2 6 12 12 72. (a) 6
24 4 4
5
(b) 13 15 5 5 (b) 712
127 5 1 1
73. 2 3 5 5 74. 25 15 4 4
,4 CHAPTER 1 Fundamentals
7 1 49 5 54 18 18
75. (a) 17 2 15 76. (a) 15 21 105 105 105 35 35
(b) 21 3 21 3 24 24 (b) 38 57 38 57 19 19.
3 12 55 67 67
(c) 10 11
8 40 40 40 40 (c) 26 18 26 18 08 08.
77. (a) Let x 0777 . So 10x 77777 x 07777 9x 7. Thus, x 79 .
(b) Let x 02888 . So 100x 288888 10x 28888 90x 26. Thus, x 26 13
90 45 .
(c) Let x 0575757 . So 100x 575757 x 05757 99x 57. Thus, x 57 19
99 33 .
78. (a) Let x 52323 . So 100x 5232323 1x 52323 99x 518. Thus, x 518
99 .
(b) Let x 13777 . So 100x 1377777 10x 137777 90x 124. Thus, x 124 62
90 45 .
(c) Let x 213535 . So 1000x 21353535 10x 213535 990x 2114. Thus, x 2114 1057
990 495 .
79. 3, so 3 3. 80. 2 1, so 1 2 2 1.
81. a b, so a b a b b a. 82. a b a b a b b a 2b
83. (a) a is negative because a is positive.
(b) bc is positive because the product of two negative numbers is positive.
(c) a ba b is positive because it is the sum of two positive numbers.
(d) ab ac is negative: each summand is the product of a positive number and a negative number, and the sum of two
negative numbers is negative.
84. (a) b is positive because b is negative.
(b) a bc is positive because it is the sum of two positive numbers.
(c) c a c a is negative because c and a are both negative.
(d) ab2 is positive because both a and b2 are positive.
85. Distributive Property
86. (a) When L 60, x 8, and y 6, we have L 2 x y 60 2 8 6 60 28 88. Because 88 108 the
post office will accept this package.
When L 48, x 24, and y 24, we have L 2 x y 48 2 24 24 48 96 144, and since
144 108, the post office will not accept this package.
(b) If x y 9, then L 2 9 9 108 L 36 108 L 72. So the length can be as long as 72 in. 6 ft.
m1 m2 m1 m m1n2 m2n1
87. Let x and y be rational numbers. Then x y 2 ,
n1 n2 n1 n2 n1 n2
m m m n m 2 n1 m m m m
xy 1 2 1 2 , and x y 1 2 1 2 . This shows that the sum, difference, and product
n1 n2 n1 n2 n1 n2 n1n2
of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily
irrational; for example, 2 2 2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;
for example, 2 2 0 which is rational.
88. 12 2 is irrational. If it were rational, then by Exercise 6(a), the sum 12 2 12 2 would be rational, but
this is not the case.
Similarly, 12 2 is irrational.
(a) Following the hint, suppose that r t q, a rational number. Then by Exercise 6(a), the sum of the two rational
numbers r t and r is rational. But r t r t, which we know to be irrational. This is a contradiction, and
hence our original premise—that r t is rational—was false.
, SECTION 1.1 Real Numbers 5
a
(b) r is a nonzero rational number, so r for some nonzero integers a and b. Let us assume that rt q, a rational
b
c a c bc
number. Then by definition, q for some integers c and d. But then r t q t , whence t , implying
d b d ad
that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number
and an irrational number is irrational.
89.
x 1 2 10 100 1000
1 1 1 1 1 1
x 2 10 100 1000
As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero.
x 1 05 01 001 0001
1 1 1 1 1 1
x 05 2 01 10 001 100 0001 1000
As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.
90. We can construct the number 2 on the number line by
Ï2
transferring the length of the hypotenuse of a right triangle 1
with legs of length 1 and 1.
_1 0 1 Ï2 2 3
Similarly, to locate 5, we construct a right triangle with legs
of length 1 and 2. By the Pythagorean Theorem, the length Ï5
1
of the hypotenuse is 12 22 5. Then transfer the
length of the hypotenuse to the number line. _1 0 1 2 Ï5 3
The square root of any rational number can be located on a
number line in this fashion.
The circle in the second figure in the text has circumference , so if we roll it along a number line one full rotation, we have
found on the number line. Similarly, any rational multiple of can be found this way.
a b a b abab
91. (a) Suppose that a b, so max a b a and a b a b. Then a.
2 2
On the other hand, if b a, then max a b b and a b a b b a. In this case,
a b a b abba
b.
2 2
If a b, then a b 0 and the result is trivial.
a b a b a b b a
(b) If a b, then min a b a and a b b a. In this case a.
2 2
a b a b
Similarly, if b a, then b; and if a b, the result is trivial.
2
92. Answers will vary.
93. (a) Subtraction is not commutative. For example, 5 1 1 5.
(b) Division is not commutative. For example, 5 1 1 5.
(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes,
the result is not the same as if you proceed the other way around.
(d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result.
(e) Washing laundry and drying it are not commutative.
