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3.2 Equivalent Fractions: Simplifying and Building Two fractions are said to be equivalent if they have the same value. Naturally, one approach we could use to determine if two fractions are equivalent is to convert each fraction to a decimal. For 3 15 3 15 example, since = 0.6 and are equivalent, and we could = 0.6 , the fractions and 5 25 5 25 3 15 write = . Alternatively, consider the following forms of the number 1: 5 25 1= 2 3 4 100 n = = = ... = = 2 3 4 100 n Clearly 2 parts out of 2 is equal to 1, as is 100 parts out of 100, or n parts out of n. Now consider the following property (the Fundamental Property of Fractions): If a, b, and c are nonzero: a a•c = b b•c This statement is saying if both the numerator and denominator of a fraction have the same factor (called a common factor), then that factor can be eliminated resulting in an equivalent fraction. a a c a•c a c It is true because = • = , so we are multiplying the fraction by (a form of 1) to b b c b•c b c a•c result in the fraction . Recall that multiplying a number by 1 does not change its value (the b•c 3 15 Identity Property of Multiplication). Using our fractions and , note that: 5 25 15 3 • 5 3 = = 25 5 • 5 5 Thus 15 3 15 3 and are equivalent fractions, or = . 25 5 25 5 145 Example 1 Determine whether the two fractions are equivalent by using the Fundamental Property of Fractions. a. b. c. d. Solution 9 45 , 10 50 5 60 ! ,! 7 84 7 49 , 12 96 4 4xy , 9 9xy a. For the two fractions to be equivalent, there must be a form of 1 (or a common factor) which can be multiplied by one fraction to create the other. Note that: 9 5 45 • = 10 5 50 5 Since is a form of 1, the two fractions are equivalent. 5 b. We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that: 5 12 60 ! • =! 7 12 84 12 Since is a form of 1, the two fractions are equivalent. 12 We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that: 7 7 49 • = 12 8 96 7 Since is not a form of 1, the two fractions are not equivalent. An 8 alternate way to verify this is to convert each fraction to decimal: 7 = 0.583 12 49 = 0.510416 96 Note that these two decimal forms are not the same. c. 146 d. We must find a form of 1 (or common factor) which can be multiplied by one fraction to create the other. Note that: 4 xy 4xy • = 9 xy 9xy xy Since is a form of 1, the two fractions are equivalent. xy Given a fraction, could you find other fractions which are equivalent to it. For example, given 3 the fraction , what would be some other fractions equivalent to it? We could multiply by 7 different forms of 1: 3 2 6 • = 7 2 14 3 5 15 • = 7 5 35 3 a 3a • = 7 a 7a Note that we could list as many equivalent fractions as we can list forms of 1, which is infinite. Example 2 For each fraction, list three equivalent fractions. Use variables in at least one of your fractions. a. 5 8 b. ! c. 7 12 2x 5 147 Solution a. Using the fractions 2 5 ab , , and as forms of 1 (yours will probably be 2 5 ab different): 5 2 10 • = 8 2 16 5 5 25 • = 8 5 40 5 ab 5ab • = 8 ab 8ab 10 25 5ab , , and . 16 40 8ab x2 y 9 100 Using the fractions , , and 2 as forms of 1 (yours will probably be x y 9 100 different): 7 9 63 ! • =! 12 9 108 7 100 700 ! • =! 12 100 1200 2 7 x y 7x 2 y ! • 2 =! 12 x y 12x 2 y Three equivalent fractions are b. 7x 2 y 63 700 ,! , and ! . 12x 2 y 108 1200 xy 8 15 Using the fractions , , and as forms of 1 (again yours will xy 8 15 probably be different): 2x 8 16x • = 5 8 40 2x 15 30x • = 5 15 75 2x xy 2x 2 y • = 5 xy 5xy Three equivalent fractions are ! c. 2x 2 y 16x 30x , , and . Note how we 5xy 40 75 multiplied the numbers, and how the exponent was used to represent x • x in the third fraction. Three equivalent fractions are 148 In addition and subtraction of fractions, it will be necessary to convert a fraction to a specified 5 denominator. For example, given the fraction , how could this fraction be converted to one 6 5 x with a denominator of 72? That is, what numerator x would result in = being equivalent? 6 72 12 Since 6 • 12 = 72 (we can find 12 by dividing 6 into 72), the form of 1 to use is . Thus: 12 5 12 60 • = 6 12 72 The missing numerator is x = 60 . This idea is often referred to as building fractions. Example 3 Find the variable such that the two given fractions are equivalent. a. b. c. d. Solution a. 9 x = 14 70 3 33 ! =! 4 y a 96 = 15 120 5 200 ! =! b 320 Since 70 ÷ 14 = 5 , the form of 1 to use is 5 . Therefore: 5 9 5 45 • = 14 5 70 The missing numerator is x = 45. b. Since 33 ÷ 3 = 11 , the form of 1 to use is 3 11 33 • =! 4 11 44 The missing denominator is y = 44. ! 149 11 . Therefore: 11 c. Since 120 ÷ 15 = 8 , the form of 1 to use is 8 . Instead of multiplying the 8 8 first fraction by , we can alternatively divide the second fraction: 8 96 8 12 ÷ = 120 8 15 The missing numerator is a = 12. Note how we used the idea that division is the inverse of multiplication to do this problem. d. 40 . Again, we do this problem 40 “backwards” by dividing the second fraction: 200 40 5 ! ÷ =! 320 40 8 The missing denominator is b = 8. Since 200 ÷ 5 = 40 , the form of 1 to use is This last example leads to the idea of simplifying (or reducing) fractions. That is, given a 32 fraction such as , can we apply the Fundamental Property of Fractions to reduce the numbers 40 8 to a “simpler” form? Using the form of 1 as , we can write: 8 32 4 • 8 4 = = 40 5 • 8 5 32 4 4 reduces to . Note that does not reduce further, since there is no other form 40 5 5 8 of 1 we can use in the Fundamental Property of Fractions. But where did come from? Recall 8 from Chapter 1 that the greatest common factor (GCF) of 32 and 40 is the largest number that will divide into both 32 and 40, which is precisely the number 8. In other words, using the GCF of the numerator and denominator as the common factor will always result in the form of 1 to use. In the past, you may have learned to reduce fractions by dividing the numerator and denominator by the same number (this is the same as our form of 1). The big problem, however, is knowing when to stop. We say that 150 For example, we can attempt to reduce 32 as: 40 32 16 • 2 16 = = 40 20 • 2 20 However, the result can be reduced further. Thus the GCF becomes the quickest (and safest, in terms of errors) approach to simplify fractions. Example 4 Use the greatest common factor to simplify each fraction. a. b. c. d. Solution a. 56 80 25 ! 150 48 ! 132 5xy 10x The GCF of 56 and 80 is 8, so the form of 1 to use is 8 . Therefore: 8 56 7 • 8 7 = = 80 10 • 8 10 b. The GCF of 25 and 150 is 25, so the form of 1 to use is 25 . Therefore: 25 25 1 • 25 1 =! =! 150 6 • 25 6 Note how our “invisible” factor of 1 is used in this fraction. ! 151 c. The GCF of 48 and 132 is 12, so the form of 1 to use is ! d. 12 . Therefore: 12 48 4 • 12 4 =! =! 132 11 • 12 11 The GCF of 5xy and 10x is 5x , so the form of 1 to use is 5x . Therefore: 5x 5xy y • 5x y = = 10x 2 • 5x 2 This illustrates how we can simplify fractions with symbols also. Thus far, we have found the GCF by guessing at it, but recall our alternate approach using primes, which works particularly well for larger numbers. For example, to reduce the fraction 168 , it would be difficult to guess at the GCF of 168 and 180. We first factor each number into 180 primes: 168 = 8 • 21 = ( 2 • 4 ) • ( 3 • 7 ) = ( 2 • 2 • 2 ) • ( 3 • 7 ) = 2 • 2 • 2 • 3 • 7 180 = 10 • 18 = ( 2 • 5 ) • ( 3 • 6 ) = ( 2 • 5 ) • ( 3 • 2 • 3) = 2 • 2 • 3 • 3 • 5 Instead of finding the GCF, we will use the primes in our fraction, remembering that common factors of the numerator and denominator will cancel: 168 2 • 2 • 2 • 3 • 7 = 180 2 • 2 • 3 • 3 • 5 2/ • 2/ • 2 • 3/ • 7 = 2/ • 2/ • 3/ • 3 • 5 2•7 = 3• 5 14 = 15 prime factorizations cancelling common factors writing the remaining factors multiplying For fractions with larger numbers, this is usually the most efficient, and more importantly the most accurate, approach. 152 Example 5 Use prime numbers to simplify each fraction. a. b. c. d. Solution a. 21 112 90 ! 198 168 ! 224 5x 2 y 3 4x 4 y First find the prime factorizations of 21 and 112: 21 = 3 • 7 112 = 4 • 28 = ( 2 • 2 ) • ( 2 • 2 • 7 ) = 2 • 2 • 2 • 2 • 7 Now rewrite the fraction using prime numbers, and simplify: 21 3• 7 = prime factorizations 112 2 • 2 • 2 • 2 • 7 3 • 7/ = cancelling common factors 2 • 2 • 2 • 2 • 7/ 3 = writing the remaining factors 2•2•2•2 3 = multiplying 16 b. First find the prime factorizations of 90 and 198: 90 = 9 • 10 = ( 3 • 3) • ( 2 • 5 ) = 2 • 3 • 3 • 5 198 = 2 • 99 = 2 • ( 9 • 11) = 2 • 3 • 3 • 11 Now rewrite the fraction using prime numbers, and simplify: 90 2 • 3• 3• 5 ! =! prime factorizations 198 2 • 3 • 3 • 11 2/ • 3/ • 3/ • 5 =! cancelling common factors 2/ • 3/ • 3/ • 11 5 =! writing the remaining factors 11 153 c. First find the prime factorizations of 168 and 224: 168 = 4 • 42 = ( 2 • 2 ) • ( 2 • 21) = ( 2 • 2 ) • ( 2 • 3 • 7 ) = 2 • 2 • 2 • 3 • 7 224 = 4 • 56 = ( 2 • 2 ) • ( 4 • 14 ) = ( 2 • 2 ) • ( 2 • 2 • 2 • 7 ) = 2 • 2 • 2 • 2 • 2 • 7 Now rewrite the fraction using prime numbers, and simplify: 168 2 • 2 • 2 • 3• 7 ! =! prime factorizations 224 2•2•2•2•2•7 2/ • 2/ • 2/ • 3 • 7/ =! cancelling common factors 2/ • 2/ • 2/ • 2 • 2 • 7/ 3 =! writing the remaining factors 2•2 3 =! multiplying 4 d. This may not seem to “fit”, but look carefully at the fraction. Treating the variables as prime numbers, we simplify: 5x 2 y 3 5•x•x•y•y•y = prime factorizations 4 4x y 2 • 2 • x • x • x • x • y 5 • x/ • x/ • y • y • y/ = cancelling common factors 2 • 2 • x/ • x/ • x • x • y/ 5•y•y = writing the remaining factors 2•2•x•x 5y 2 = 2 multiplying 4x This example is a typical algebra problem, and it is included here to illustrate that the ideas we are developing extend directly to algebra. In the last section, we found that some fractions result in a terminating decimal, while others result in a repeating decimal. Is there a way to tell in advance which will occur? Consider the two fractions and their decimal forms: 17 = 0.425 40 19 = 0.63 30 154 Now look at the prime factorizations of the denominators: 17 17 = = 0.425 40 2 • 2 • 2 • 5 19 19 = = 0.63 30 2 • 3 • 5 Recall that the decimal system has denominators which are powers of 10 = 2 • 5. Suppose we 17 want to build the first fraction up to a power of 10 in the denominator. Since each 10 = 2 • 5, 40 we will need to multiply by two additional factors of 5: 17 17 5•5 17 • 25 425 = • = = = 0.425 40 2 • 2 • 2 • 5 5 • 5 10 • 10 • 10 1000 19 , the prime factor of 3 will always be in the prime 30 factorization of the denominator. Thus we can never build its denominator to a power of 10, and thus it can’t be represented as a terminating decimal. In summary, only fractions whose denominators have prime factors of 2 and 5 can be converted to terminating decimals. If a denominator of a fraction has prime factors other that 2 or 5, it will result in a repeating decimal (assuming the fraction is simplified). Thus the vast majority of fractions have repeating, rather than terminating, decimal forms. However, with the second fraction Example 6 Determine whether the decimal form of each rational number will be terminating or repeating. Do not actually convert the fraction to decimal! a. b. c. d. 29 85 23 ! 400 49 320 97 ! 440 155 Solution a. Finding the prime factorization of 85: 85 = 5 • 17 Since the denominator has a prime factor other than 2 or 5 (17), the decimal form is a repeating decimal. b. Finding the prime factorization of 400: 400 = 4 • 100 = ( 2 • 2 ) • (10 • 10 ) = (2 • 2) • (2 • 5 • 2 • 5) = 2•2•2•2•5•5 Since the denominator has only 2 and 5 prime factors, the decimal form is a terminating decimal. c. Finding the prime factorization of 320: 320 = 10 • 32 = (2 • 5) • ( 4 • 8) = (2 • 5) • (2 • 2 • 2 • 2 • 2) = 2•2•2•2•2•2•5 Since the denominator has only 2 and 5 prime factors, the decimal form is a terminating decimal. d. Finding the prime factorization of 440: 440 = 10 • 44 = ( 2 • 5 ) • ( 4 • 11) = ( 2 • 5 ) • ( 2 • 2 • 11) = 2 • 2 • 2 • 5 • 11 Since the denominator has a prime factor other than 2 or 5 (11), the decimal form is a repeating decimal. Terminology equivalent fractions common factor simplifying (or reducing) fractions Fundamental Property of Fractions building fractions greatest common factor (GCF) 156 Exercise Set 3.2 Determine whether the two fractions are equivalent by using the Fundamental Property of Fractions. 1. 3. 5. 7. 9. 5 40 , 7 56 13 52 ! ,! 15 75 23 368 ! ,! 25 400 5x 25ax , 4y 20ay 3a 6ab ! ,! 7b 14ab 6 48 , 7 56 5 20 4. ! , ! 16 80 15 525 6. ! , ! 28 980 7y 35y 2 , 8. 9a 45ay 9w 27awz 10. ! ,! 10z 30az 2. For each fraction, list three equivalent fractions. Use variables in at least one of your fractions. 6 13 5 13. ! 12 3x 15. ! 5 6a 17. 7b 7 11 8 14. ! 15 4a 16. ! 7 5s 18. 9t 11. 12. Find the variable such that the two given fractions are equivalent. 5 x = 7 56 14 x 21. ! =! 25 500 13 195 = 23. 17 y 7 112 25. ! = ! 15 y 6 x = 13 91 9 x 22. ! = ! 15 210 12 216 = 24. 19 y 9 162 26. ! = ! 16 y 19. 20. 157 a 60 = 7 84 a 187 29. ! = ! 15 255 9 117 31. = b 143 12 300 33. ! = ! b 425 a 121 = 12 132 a 198 30. ! = ! 13 286 6 108 32. = b 198 16 432 34. ! = ! b 567 27. 28. Use the greatest common factor to simplify each fraction. 35. 37. 39. 41. 43. 45. 47. 49. 12 20 70 ! 75 35 72 60 ! 100 35 77 36 ! 60 12xy 15ax 16abx ! 30axy 36. 38. 40. 42. 44. 46. 48. 50. 158 25 40 32 ! 84 49 93 45 ! 150 26 143 56 ! 78 25ab 40ax 45axy ! 75aby Use prime numbers to simplify each fraction. 51. 53. 55. 57. 59. 61. 63. 65. 67. 69. 71. 48 72 36 ! 40 216 234 270 ! 320 81 192 155 ! 248 384 432 261 ! 290 45x 2 y 5 15x 3 y 3 8a 7b 6 6a 5b 2 16x 8 y12 ! 20x 9 y 7 52. 54. 56. 58. 60. 62. 64. 66. 68. 70. 72. 159 70 170 45 ! 81 220 242 275 ! 286 245 246 153 ! 272 364 468 265 ! 424 30x 3 y 7 15x 6 y 4 12a 6b 5 18a 4 b 3 36x 6 y11 ! 24x 9 y 8 Determine whether the decimal form of each rational number will be terminating or repeating. Do not actually convert the fraction to decimal! 73. 75. 77. 79. 81. 14 75 19 ! 64 67 ! 448 21 (Hint: Simplify first) 120 9 ! 480 74. 76. 78. 80. 82. 19 90 38 ! 125 53 ! 480 49 (Hint: Simplify first) 175 21 ! 448 Answer each question as true or false. If it is false, give a specific example to show that it is false. If it is true, explain why. 83. Every fraction has either a terminating or repeating decimal form. 84. Every repeating decimal can be written as a fraction. a 85. If the numbers a and b are relatively prime, then the fraction is already simplified. b a 86. If the numbers a and b are not relatively prime, then the fraction is not simplified. b 87. There is only one decimal form for every fraction. (Hint: Refer back to Exercises 95 and 96 of the previous section.) 88. There is only one fraction for every terminating decimal. 160