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Transcript
Section 5.3 The Rational Numbers Defining the Rational Numbers Objectives 1. Define the rational numbers. 2. Reduce rational numbers. 3. Convert between mixed numbers and improper fractions. 4. Express rational numbers as decimals. 5. Express decimals in the form a / b. 6. Multiply and divide rational numbers. 7. Add and subtract rational numbers. 8. Apply the density property of rational numbers. 9. Solve problems involving rational numbers. 8/7/2009 Section 5.3 • The set of rational numbers is the set of all numbers which can be expressed in the form a , where a and b are integers and b is not equal to 0. b • The integer a is called the numerator. • The integer b is called the denominator. Examples: The following are examples of rational numbers: ¼, -½, ¾, 5, 0 1 8/7/2009 Reducing a Rational Number Section 5.3 Reducing a Rational Number a • If b is a rational number and c is any number other than 0, a⋅c a = b⋅c b Section 5.3 3 8/7/2009 Section 5.3 • A mixed number consists of the sum of an integer and a rational number, expressed without the use of an addition sign. Example: 130 130 ÷ 65 2 = = 455 455 ÷ 65 7 There are no common divisors of 2 and 7 other than 1. Thus, the rational number 2 is in its lowest terms. • An improper fraction is a rational number whose numerator is greater than its denominator. Example: 7 19 5 8/7/2009 Section 5.3 4 Mixed Numbers and Improper Fractions Divide the numerator and the denominator of the given rational number by 5 · 13 or 65. or 130 to lowest terms. 455 Thus, 130 = 2 · 5 · 13, and 455 = 5 · 7 · 13. The greatest common divisor is 5 · 13 or 65. Reducing a Rational Number Example Continued 130 2 ⋅ 5 ⋅ 13 2 = = 455 5 ⋅ 7 ⋅ 13 7 Example: Reduce Solution: Begin by finding the greatest common divisor of 130 and 455. • The rational numbers a and a ⋅ c are called equivalent b b⋅c fractions. • To reduce a rational number to its lowest terms, divide both the numerator and denominator by their greatest common divisor. 8/7/2009 2 5 8/7/2009 19 is larger than 5 Section 5.3 6 1 Mixed Numbers and Improper Fractions Mixed Numbers and Improper Fractions Converting from Mixed Number to an Improper Fraction Converting from an Improper Fraction to a Mixed Number 1. Multiply the denominator of the rational number by the integer and add the numerator to this product. 2. Place the sum in step 1 over the denominator of the mixed number. Example: Convert to an improper fraction. 1. Divide the denominator into the numerator. Record the quotient and the remainder. 2. Write the mixed number using the following form: Solution: 8/7/2009 Section 5.3 7 8/7/2009 Mixed Numbers and Improper Fractions 42 to a mixed number. 5 • Any rational number can be expressed as a decimal by dividing the denominator into the numerator. Example: Express each rational number as a decimal. a. 5 b. 7 Solution: Step 1. Divide the denominator into the numerator. 8 Section 5.3 9 Rational Numbers and Decimals Example Continued a. 0.625 8 5.000 b. 0.6363 11 7.000 48 66 20 16 40 33 40 0 Notice the decimal stops. This is called a terminating decimal. 8/7/2009 Section 5.3 10 Expressing Decimals as a Quotient of Two Integers Notice the digits 63 repeat over and over indefinitely. This is called a repeating decimal. 70 66 40 11 Solution: In each case, divide the denominator into the numerator. Step 2. Write the mixed number with the Thus, 8/7/2009 8 Rational Numbers and Decimals Converting from an Improper Fraction to a Mixed Number Example: Convert Section 5.3 • Terminating decimals can be expressed with denominators of 10, 100, 1000, 10,000, and so on. • Using the chart, the digits to the right of the decimal point are the numerator of the rational number. 40 Example: Express each terminating decimal as a quotient of integers: 33 70 a. 0.7 b. 0.49 c. 0.048. M 8/7/2009 Section 5.3 11 8/7/2009 Section 5.3 12 2 Expressing Decimals as a Quotient of Two Integers Expressing Decimals as a Quotient of Two Integers Example Continued Repeating Decimals Example: Express 0.6 as a quotient of integers. Solution: 7 because the 7 is in the tenths position. a. 0.7 = 10 Solution: Step1. Let n equal the repeating decimal such that n = 0 .6 , or 0.6666… Step 2. If there is one repeating digit, multiply both sides of the equation in step 1 by 10. 49 b. 0.49 = because the digit on the right, 9, is in the 100 hundredths position. 48 c. 0.048 = because the digit on the right, 8, is in the 1000 thousandths position. Reducing to lowest terms, n = 0.66666… 10n = 10(0.66666…) 10n = 6.66666… 48 48 ÷ 8 6 = = 1000 1000 ÷ 8 125 8/7/2009 Section 5.3 13 8/7/2009 Multiplying by 10 moves the decimal point one place to the right. Section 5.3 14 Expressing Decimals as a Quotient of Two Integers Repeating Decimals Example Continued Multiplying Rational Numbers Step 3. Subtract the equation in step 1 from the equation in step 2. • The product of two rational numbers is the product of their numerators divided by the product of their denominators. a c a⋅c • If a and c are rational numbers, then ⋅ = . b d b⋅d b d Step 4. Divide both sides of the equation in step 3 by the number in front of n and solve for n. We solve 9n = 6 for n: 9n = 6 2 9n 6 Thus, 0.6 = . = Example: Multiply. If possible, reduce the product to its lowest terms: 3 9 9 6 2 n= = 9 3 8/7/2009 2 9 − − 3 4 Section 5.3 15 8/7/2009 Section 5.3 Multiplying Rational Numbers Example Continued Dividing Rational Numbers Solution: • The quotient of two rational numbers is a product of the first number and the reciprocal of the second number. If a and c are rational numbers, then 1 2 9 (− 2 )(− 9 ) 18 3 ⋅ 6 3 = = = or 1 − − = 3⋅ 4 12 2 ⋅ 6 2 2 3 4 Multiply across. b d a c a d a⋅d ÷ = ⋅ = b d b c b⋅c Simplify to lowest terms. Example: Divide. If possible, reduce the quotient to its lowest terms: − 8/7/2009 Section 5.3 16 17 8/7/2009 3 7 ÷ 5 11 Section 5.3 18 3 Dividing Rational Numbers Example Continued Adding and Subtracting Rational Numbers Identical Denominators Solution: The sum or difference of two rational numbers with identical denominators is the sum or difference of their numerators over the common denominator. a c If and are rational numbers, then 3 7 3 11 3 ⋅ 11 33 =− − ÷ =− ⋅ =− 5 11 5 7 5⋅7 35 Change to multiplication by using the reciprocal. b b a c a+c + = b b b Multiply across. and a c a−c − = b b b 8/7/2009 Section 5.3 19 Example: Perform the indicated operations: a. 3 + 2 b. 11 − 5 c. − 5 1 − − 2 3 7 4 12 12 7 7 4 a a c a⋅c = ⋅ = b b c b⋅c 7 b. 11 − 5 = 11 − 5 = 6 = 1 ⋅ 6 1 = 12 12 12 12 2⋅6 20 • If the rational numbers to be added or subtracted have different denominators, we use the least common multiple of their denominators to rewrite the rational numbers. • The least common multiple of their denominators is called the least common denominator or LCD. Solution: a. 3 + 2 = 3 + 2 = 5 7 Section 5.3 Adding and Subtracting Rational Numbers Unlike Denominators Adding and Subtracting Rational Numbers Identical Denominators 7 8/7/2009 2 c. − 5 1 − − 2 3 = − 21 − − 11 = − 21 + 11 = − 21 + 11 = − 10 = − 5 or − 2 1 2 4 4 4 4 8/7/2009 4 4 4 4 2 Section 5.3 21 8/7/2009 Adding and Subtracting Rational Numbers Unlike Denominators Example: Find the sum of 3 1 + . 4 6 We multiply the first rational 3 1 3 3 1 2 + = ⋅ + ⋅ number by 3/3 and the second one 4 6 4 3 6 2 by 2/2 to obtain 12 in the denominator for each number. 9 2 = + Notice, we have 12 in the 12 12 denominator for each number. 11 = Thus, we add across to obtain the answer. 12 Section 5.3 22 Density of Rational Numbers Solution: Find the least common multiple of 4 and 6 so that the denominators will be identical. LCM of 4 and 6 is 12. Hence, 12 is the LCD. 8/7/2009 Section 5.3 23 If r and t represent rational numbers, with r < t, then there is a rational number s such that s is between r and t: r < s < t. Example: Find a rational number halfway between ½ and ¾. Solution: First add ½ and ¾. 1 3 1 2 3 2 3 5 + = ⋅ + = + = 2 4 2 2 4 4 4 4 8/7/2009 Section 5.3 24 4 Density of Rational Numbers Example Continued Problem Solving with Rational Numbers • A common application of rational numbers involves preparing food for a different number of servings than what the recipe gives. • The amount of each ingredient can be found as follows: Next, divide this sum by 2. 5 5 2 5 1 5 ⋅1 5 = ÷2 = ÷ = ⋅ = 4 4 1 4 2 4⋅2 8 The number 5 is halfway between ½ and ¾. Thus, 8 amount of ingredient needed = desired serving size × ingredient amount recipe serving size in the recipe 1 5 3 < < . 2 8 4 8/7/2009 Section 5.3 25 Problem Solving with Rational Numbers Changing the Size of a Recipe 26 The amount of sugar, in cups, needed is determined by multiplying the rational numbers: Solution: desired serving size amount of sugar needed = × ingredient amount recipe serving size in the recipe 8/7/2009 Section 5.3 Problem Solving with Rational Numbers Changing the Size of a Recipe A chocolate-chip recipe for five dozen cookies requires ¾ cup of sugar. If you want to make eight dozen cookies, how much sugar is needed? = 8/7/2009 8 3 8 ⋅ 3 24 6 1 × = = = =1 5 4 5 ⋅ 4 20 5 5 1 Thus, 1 5 cups of sugar is needed. 8 dozen 3 × cup 5 dozen 4 Section 5.3 27 8/7/2009 Section 5.3 28 5