Figure 1.1: Positioning numbers on the number line.

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1 1B. INEQUALITIES 13 1b Inequalities In Chapter 1, we introduced the natural numbers N = {1, 2, 3,...}, the whole numbers W = {0, 1, 2, 3,...}, and the integers Z = {..., 3, 2, 1, 0, 1, 2, 3,...}. Later in the chapter, we introduced the rational numbers, numbers of the form p/q, where p and q are integers. We noted that both terminating and repeating decimals are rational numbers. Each of these numbers has a unique position on the number line (see Figure 1.1) Figure 1.1: Positioning numbers on the number line. The natural numbers, whole numbers, and integers are also rational numbers, because each can be expressed in the form p/q, wherep and q are integers. For example, 0=0/12, 4 = 4/1, and 3 = 12/4. Indeed, the rational numbers contain all of the numbers we ve studied up to this point in the course. However, not all numbers are rational numbers. For example, consider the decimal number , which neither terminates nor repeats. The number 2= also equals a decimal number that never terminates and never repeats. A similar statement can be made about the number π = Each of these irrational (not rational) numbers also has a unique position on the number line (see Figure 1.2) π Figure 1.2: Positioning numbers on the number line. The Real Numbers. If we combine all of the rational and irrational numbers into one collection, then we have a set of numbers that is called the set ofreal numbers. The set of real numbers is denoted by the symbol R. Every point on the number line is associated with a unique real number. Conversely, every real number is associated with a unique position on the number line. In lieu of this correspondence, the number line is usually called the real line.

2 14 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Ordering the Real Numbers The real numbers are ordered on the real line in a manner identical to howwe ordered the integers on the number line in Section 1 of Chapter 1. Order on the Real Line. Suppose that a and b are real numbers positioned on the real line as shown below. a b Because a lies to the left of b, wesaythata is less than b, or in mathematical symbols, a < b. The inequality symbol < is read less than. Alternately, b lies to the right of a, so we can also say that b is greater than a, or in mathematical symbols, b>a. The inequality symbol > is read greater than. Here are two more inequality symbols that we will use in this section. Less than or equal to. If we want to say that a lies to the left of b, or shares the same position as b, thenwesaythata is less than or equal to b and write a b. The inequality symbol is pronounced less than or equal to. Greater than or equal to. If we want to say that b lise to the right of a, or shares the same position as a, thenwesaythatb is greater than ore equal to a and write b a. The inequality symbol is pronounced greater than or equal to. Set-Builder Notation Mathematicians use a construct called set-builder notation to describe sets or collections of numbers. The general form of set-builder notation looks as follows: {x :somestatementaboutx} For example, suppose that we want to describe the set of all real numbers that are less than 2. We could use the following notation: A = {x : x<2}

3 1B. INEQUALITIES 15 This is read aloud as follows: A equals the set of all x such that x is less than 2. Some prefer to use a vertical bar instead of a colon. A = {x x<2} In this text we use the colon in set-builder notation, but feel free to usethe vertical bar instead. They mean the same thing. One might still object that the notation {x : x<2} is a bit vague. One objection could be What type of numbers x are you referring to? Do you want the integers that are less than two or do you want the real numbers thatare less than two? As you can see, this is a valid objection. One way of addressing this objection is to write: A = {x R x<2} or A = {x N x<2} The first is read A is the set of all x in R that are less than two, while the second is read A is the set of all x in N that are less than two. Set-builder Assumption. In this text, unless there is a specific refernce to the set of numbers desired, we will assume that the notation {x : x<2} is asking for the set of all real numbers less than 2. In Figure 1.3, we ve shaded the set of real numbers {x : x<2}. Because Figure 1.3: Shading the numbers less than 2. less than is the same as saying left of, we ve shaded (in red) all points on the real line that lie to the left of the number two. Note that there is an empty circle at the number two. The point representing the number two is not shaded because we were only asked to shade the numbers that are strictly less than two. While the shading in Figure 1.3 is perfectly valid, much of the information provided in Figure 1.3 is unnecessary (and perhaps distracting). We only need to label the endpoint and shade the real numbers to the left of two, as we ve done in constructing Figure 1.4. Figure 1.4: You only need to label the endpoint. 2

4 16 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES EXAMPLE 1. Shade the set {x : x 3} on the real line. Solution: The notation {x : x 3} is pronounced the set of all real numbers x such that x is greater than or equal to 3. Thus, we need to shade the number 3 and all real numbers to the right of 3. Shade {x : x 4} on real line. Answer: 3 4 EXAMPLE 2. Use set-builder notation to describe the set of real numbers that are shaded on the number line below. 1 Solution: The number 1 is not shaded. Only the numbers to the left of 1 are shaded. This is the set of all real numbers x such that x is less than 1. Thus, we describe this set with the following set-builder notation: Use set-builder notatio describe the following real numbers: 10. {x : x< 1} Answer: {x : x> 10 Interval Notation In Examples 1 and 2, we used set-builder notation to describe the set of real numbers greater than or equal to 3 and a second set of real numbers less than 1. There is another mathematical symbolism, called interval notation, that can be used to describe these sets of real numbers. Consider the first set of numbers from Example 1, {x : x 3}. 3 Sweeping our eyes from left to right, we use [ 3, ) to describe this set of real numbers. Some comments are in order: 1. The bracket at the left end means that 3 is included in the set. 2. As you move toward the right end of the real line, the numbers grow without bound. Hence, the symbol (positive infinity) is used to indicate that we are including all real numbers to the right of 3. However, is not really a number, so we use a parentheses to indicate we are not including this fictional point.

5 1B. INEQUALITIES 17 The set of numbers from Example 2 is {x : x< 1}. 1 Sweeping our eyes from left to right, this set of real numbers is described with (, 1). Again, comments are in order: 1. The number 1 is not included in this set. To indicate that it is not included, we use a parenthesis. 2. As you move toward the left end of the real line, the numbers decrease without bound. Hence, the symbol (negative infinity) is used to indicate that we are including all real numbers to the left of 1. Again, is not an actual number, so we use a parenthesis to indicate that we are not including this fictional point. Sweep your eyes from left to right. If you would like to insure that you correctly use interval notation, place the numbers in your interval notation in the same order as they are encountered as you sweep your eyes from left to right on the real line. Here are a few more examples of sets of numbers described using both setbuilder and interval notation. Shading on the real line Set-builder Interval {x : x> 5} ( 5, ) {x : x 5} [ 5, ) {x : x< 5} (, 5) {x : x 5} (, 5] Table 1.1: Examples of set-builder and interval notation. Equivalent Inequalities Like equations, two inequalities are equivalent if they have the same solutions.

6 18 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Adding the Same Quantity to Both Sides of an Inequality. Adding the same quantity to both sides of an inequality does not change the solution set. That is, if a<b, then adding c to both sides of the inequality produces the equivalent inequality a + c<b+ c. EXAMPLE 3. Solve for x: x 2 7. Sketch the solution on the real line, Solve for x: then use set-builder and interval notation to describe your solution. x 7 < 8 Solution: To undo subtracting 2, we add 2 to both sides of the inequality. x 2 7 x x 9 Original inequality. Add 2 to both sides. To shade the real numbers less than or equal to 9, we shade the number 9 and all real numbers to the left of 9. 9 Using set-builder notation, the solution is {x : x 9}. Using interval notation, the solution is (, 9]. Check: Note that the solution tells us that any number less than or equal to 9 is a solution of the original inequality x 2 7. For example, 100, 3, and 9 are all less than or equal to 9, and when substituted for x in the original inequality, true statements are obtained. x 2 7 x 2 7 x On the other hand, 10,23, and 200 are not less than or equal to 9, and when substituted for x in the original inequality, false statements are obtained. x 2 7 x 2 7 x

7 1B. INEQUALITIES 19 {x : x< 1} Therefore, 10, 23, and 200 are not solutions. convince readers that x 9 is a valid solution. Hopefully, this is enough to Subtracting the Same Quantity from Both Sides of an Inequality. Subtracting the same quantity to both sides of an inequality does not change the solution set. That is, if a<b, then subtracting c from both sides of the inequality produces the equivalent inequality a c<b c. r x: x EXAMPLE 4. Solve for x: x +2 > 7. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To undo adding 2, we subtract 2 from both sides of the inequality. x + 2 > 7 x +2 2 > 7 2 x> 9 Original inequality. Subtract 2 from both sides. Shade the real numbers greater than 9. Note that we do not shade the point associated with 9 because it is not greater than 9. [ 15, ) 9 Using set-builder notation, the solution is {x : x > 9}. Using interval notation, the solution is ( 9, ). If we multiply or divide both sides of an inequality by a positive number, we have an equivalent inequality. Multiplying or Dividing by a Positive Number. numbers with a<b.ifcis a real positive number, then Let a and b be real ac < bc and a c < b c.

8 20 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES EXAMPLE 5. Solve for x: 3x 9 Sketch the solution on the real line, Solve for x: then use set-builder and interval notation to describe your solution. 2x > 8 Solution: To undo multiplying by 3, divide both sides of the inequality by 3. Because we are dividing both sides by a positive number, we do not reverse the inequality sign. 3x 9 3x x 3 Original inequality. Divide both sides by 3. Shade the real numbers less than or equal to 3. 3 Using set-builder notation, the solution is {x : x 3}. Using interval notation, the solution is (, 3]. Answer: ( 4, ) Reversing the Inequality Sign Up to this point, it seems that the technique for solving inequalities is pretty much identical to the technique used to solve equations. And that s pretty much the case. However, in this section we are going to encounter one exception. Suppose we start with the valid inequality 2 < 5, then we multiply both sides by 2, 3, and 4. 2 < 5 2 < 5 2 < 5 2( 2) < 2(5) 3( 2) < 3(5) 4( 2) < 4(5) 4 < 10 6 < 15 8 < 20 Note that in each case, the resulting inequality is still valid. Caution! We re about to make an error! Start again with 2 < 5, but this time multiply both sides by 2, 3, and 4. 2 < 5 2 < 5 2 < 5 2( 2) < 2(5) 3( 2) < 3(5) 4( 2) < 4(5) 4 < 10 6 < 15 8 < 20

9 1B. INEQUALITIES 21 Note that in each of the resulting inequalities, the inequality symbol is pointing the wrong way! aders might prefer a rmal reason as to reversethe ty when we multiply es by a negative. Suppose that a<b. btracting b from es gives the result 0. This means that a negative number. c is a negative, then the product is positive. Then: (a b)c >0 ac bc > 0 bc + bc > 0+bc ac > bc you start with a<b 0, then ac > bc. When you multiply both sides of an inequality by a negative number, you must reverse the inequality sign. Starting with 2 < 5, multiply both sides by 2, 3, and 4, but reverse the inequality symbol. 2 < 5 2 < 5 2 < 5 2( 2) > 2(5) 3( 2) > 3(5) 4( 2) > 4(5) 4 > 10 6 > 15 8 > 20 Note that each of the resulting inequalities is now a valid inquality. Multiplying or Dividing by a Negative Number. Let a and b be real numbers with a<b.ifc is a real negative number, then ac > bc and a c > b c. That is, when multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign. r x: 3x 6 EXAMPLE 6. Solve for x: 2x <4. Sketch the solution on the real line, then use set-builder and interval notation to describe your solution. Solution: To undo multiplying by 2, divide both sides by 2. Because we are dividing both sides by a negative number, we reverse the inequality sign. 2x <4 2x 2 > 4 2 x> 2 Original inequality. Divide both sides by 2. Reverse the inequality sign. Shade the real numbers greater than 2. 2

10 22 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES Using set-builder notation, the solution is {x : x > 2}. Using interval notation, the solution is ( 2, ). Answer: (, 2] Multiple Steps Sometimes you need to perform a sequence of steps to arrive at the solution. EXAMPLE 7. Solve for x: 2x +5> 7. Sketch the solution on the real Solve for x: line, then use set-builder and interval notation to describe your solution. 3x 2 4 Solution: To undo adding 5, subtract 5 from both sides of the inequality. 2x + 5 > 7 2x +5 5 > 7 5 2x > 12 Original inequality. Subtract 5 from both sides. To undo multiplying by 2, divide both sides by 2. Because we are dividing both sides by a positive number, we do not reverse the inequality sign. 2x 2 > 12 2 x> 6 Divide both sides by 2. Shade the real numbers greater than 6. 6 Using set-builder notation, the solution is {x : x > 6}. Using interval notation, the solution is ( 6, ). Answer: (, 2] EXAMPLE 8. Solve for x: 3 5x 2x Sketch the solution on the Solve for x: real line, then use set-builder and interval notation to describe your solution. 4 x>2x +1 Solution: We need to isolate terms containing x on one side of the inequality. Start by subtracting 2x from both sides of the inequality. 3 5x 2x +17 Original inequality. 3 5x 2x 2x +17 2x Subtract 2x from both sides. 3 7x 17

11 1B. INEQUALITIES 23 We continue to isolate terms containing x on one side of the inequality. Subtract 3 from both sides. 3 7x Subtract 3 from both sides. 7x 14 To undo multiplying by 7, divide both sides by 7. Because we are dividing both sides by a negative number, we reverse the inequality sign. 7x x 2 Divide both sides by 7. Shade the real numbers greater than or equal to 2. 2 (, 1) Using set-builder notation, the solution is {x : x 2}. Using interval notation, the solution is [ 2, ). We clear fractions from an inequality in the usual manner, by multiplying both sides by the least common denominator. EXAMPLE 9. Solve for x: 3 4 x 12 > 1 3. Solution: First, clear the fractions from the inequality by multiplying both sides by the least common denominator x 12 > 1 Original inequality. [ x ] [ ] 1 > 12 Multiply both sides by [ ] 3 [ [ ] x 1 12 > 12 Distribute the ] 3 9 x>4 Cancel and multiply. To undo adding 9, subtract 9 from both sides. 9 x 9 > 4 9 Subtract 9 from both sides. x > 5

12 24 MODULE 1. LINEAR EQUATIONS AND INEQUALITIES We could divide both sides by 1, but multiplying both sides by 1 will also do the job. Because we are multiplying both sides by a negative number, we reverse the inequality sign. ( 1)( x) < ( 5)( 1) Multiply both sides by 1. Reverse the inequality sign. Shade the real numbers less than 5. x<5 5 Using set-builder notation, the solution is {x : x < 5}. Using interval notation, the solution is (, 5). We clear decimals from an inequality in the usual manner, by multiplying both sides by the appropriate power of ten. EXAMPLE 10. Solve for x: x >4.6. Solution: First, clear the decimals from the inequality by multiplying both sides by 100, which moves each decimal point two places to the right x >4.6 Original inequality x >460 Multiply both sides by x 325 > Subtract 325 from both sides. 120x> x 120 < x< Shade the real numbers less than 27/24. Divide both sides by 120. Reverse the inequality sign. Reduce to lowest terms. 27/24 Using set-builder notation, the solution is {x : x < 27/24}. Using interval notation, the solution is (, 27/24).