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Mathematics (203)
MATH 157 (19)
Imin Chen (3)
Chapter 1

# chapter 1 solution guide

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School
Department
Mathematics
Course
MATH 157
Professor
Imin Chen
Semester
Spring

Description
CHAPTER 1 EXERCISES 11 page 9 1 The statement is false because 3 is greater than 20 See the number line that follows2 The statement is true because 5 is equal to 5 3 The statement is false because 23 which is equal to 46 is less than 564 The statement is false because 56 which is 1012 is greater than 1112 5 The interval 36 is shown on the number line that follows Note that this is an open interval indicated by and x036 6 The interval 25 is shown on the number line that follows7 The interval 14 is shown on the number line that follows Note that this is a halfopen interval indicated byclosed andopen x14 8 The closed interval 6512 is shown on the number line that follows9 The infinite interval 0 is shown on the number line that follows x01 1Preliminaries 10 The infinite interval 5 is shown in the number line that follows11 First 2x48 Add 4 to each side of the inequality Next 2x4 Multiply each side of the inequality by 12 and x2 We write this in interval notation as 2 12 645x645x105x2x or x2 We write this in interval notation as 2 13 We are given the inequality 4x20Then x5 Multiply both sides of the inequality by 14and reverse the sign of the inequality We write this in interval notation as 53x4x or x4 We write this in interval notation as 4 14 12 15 We are given the inequality 6x24 First 62x42 Add 2 to each member of the inequality and 4x6so the solution set is the open interval 46x14 to 16 We add 1 to each member of the given double inequality 0 obtain 1x3 and the solution set is 13 17 We want to find the values of x that satisfy the inequalitiesx14 or x21 Adding 1 to both sides of the first inequality we obtainx1141 or x3 Similarly adding 2 to both sides of the second inequality we obtainx2212 or x3 Therefore the solution set is 33 18 We want to find the values of x that satisfy the inequalities x12 or x12 Solving these inequalities we find that 1Preliminaries 2 x1 or x1 and the solution set is 11 19 We want to find the values of x that satisfy the inequalitiesx31 and x21 Adding 3 to both sides of the first inequality we obtainx3313 or x2 Similarly adding 2 to each side of the second inequality we obtainx3 and the solution set is 23 20 We want to find the values of x that satisfy the inequalities 1 and x32x4 Solving these inequalities we find that x5 and x1 and the solution set is 15 21 624 22 44448 1248021412 22415231612416240826 122122 3233323353 2512121 28 63633 2729 213221322233342334333730 False If ab then a b abbb and ba0 3132 False Let a2 and b3 Then ab231 2233 False Let a2 and b3 Then a4 and b9 and 49 Note that we only need to provide a counterexample to show that the statement is not always true 34 False Let a2 and b3 Then 1a12 and 1b13 and 12133 1Preliminaries
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