The Mathematics 11 Competency Test Adding and Subtracting Fractions

Transcription

1 The Mathematics 11 Competency Test Adding and Subtracting Fractions It is not difficult to visualize what it means to add two fractions, as in 2 + =? 4 As explained before, each of the two fractions represents dividing a whole into a certain number of equal-sized pieces, and including a certain number of those pieces. So, we can represent the two fractions above pictorially as and 4 2 Thus, 4 represents pieces, each of which is 1 4 of the whole, and 2 represents 2 pieces, each 1 of the whole. Then, to form the sum we need to come up with a fraction representing how much of a whole we get when we combine all five shaded pieces in the diagrams above. The difficulty is that we can t just count up the number of shaded pieces to be 5 and the total number of pieces to be 7, and so declare the + ), since the pieces are not all the same size to begin with. In fact, answer to be 5 7 (from you can see that this answer would clearly be wrong, because it represents less than a complete whole, and yet it is obvious from the pictures that the combined shaded areas are much more than a complete whole. There are several errors in the incorrect procedure just described. The most obvious is that the five smaller shaded bits in the diagram are not all the same size, and so simply counting a total of 5 pieces is not a correct reflection of the combined size of the two fractions. However, there is a way around this problem. First, divide each of the quarters in the diagram for 4 into three smaller, equal-sized pieces: = 4 In fact, this diagram is now illustrating both 4 as well as 9 12, because now the whole can also be viewed as being partitioned into 12 equal pieces, of which 9 are shaded. David W. Sabo (200) Adding and Subtracting Fractions Page 1 of 7

2 Secondly, divide each of the thirds in the diagram for 2 into four smaller equal-sized pieces: = 4 Again, this results in the whole being divided into twelve equal pieces of which eight are shaded. But now, both diagrams have shaded pieces which are the same size! Each is 1 12 of the whole. When the total number of these equal-sized shaded pieces is tallied, we get = 17 shaded pieces, each of size 1 12 of the whole. Thus, we apparently get that = + = = You can verify that this is a plausible result by using your calculator to turn the fractions into decimal numbers and checking the addition. What this really shows is that we can add two fractions together only if they have the same denominator (and the same rule will apply to subtraction): a c a + c + = b b b a c a c = b b b because for the sum of the numerators (or the difference of the numerators) to be meaningful, they must refer to pieces of the whole all of which are the same size (that is the same thing as saying that the denominators of the fractions are the same). Thus, before two fractions can be added, they may have to be converted to equivalent fractions so that both have the same denominator, a so-called common denominator. The process for doing this has already been illustrated above, but we will now describe a systematic approach. Finding Lowest Common Denominators We ll be a little bit symbolic here. When attempting a sum such as David W. Sabo (200) Adding and Subtracting Fractions Page 2 of 7

3 we need to first convert both fractions to equivalent fractions which have the same denominator. The method for obtaining equivalent fractions is to multiply the numerator and denominator of the given fraction by the same number: 2 a 2 b + = a b where a and b stand for some whole numbers which have values so that 4a = b = LCD Actually, any values of a and b that satisfy this condition will work, but there are advantages to picking a and b so that the resulting number, LCD, is as small as possible. Since this smallest number will be the new denominator for both fractions, it is called the lowest common denominator or LCD. From the condition above, we see that LCD = 4a, so LCD must be evenly divisible by 4 (since a is supposed to be a whole number). Similarly, LCD = b, so LCD must be divisible by (since b is supposed to be a whole number). The systematic procedure for finding LCD in this case is as follows: step (i): write each of the original denominators as a product of prime factors: Here: 4 = 2 2 = a very simple situation. step (ii): A power of every distinct prime factor that occurs in any of the denominators will be a factor of the LCD. Here: LCD = 2 x y since the only distinct prime factors in the original denominators are 2 and in this case. step (iii): The powers of each factor in the LCD are just the highest power to which that factor is raised in any of the original denominators. So here, x = 2, since 4 contains 2 2 and contains 2 0. Therefore, the highest power of 2 that occurs in either of the original denominators is 2. y = 1, since 4 contains 0 and contains 1. Therefore the highest power of occurring in either of the original denominators is 1. Thus, for this simple example, LCD = 2 2 = 12 (This happens to be equal to the product, 4 x, of the original denominators in this case. However, often the LCD is much smaller than the product of the original denominators because of David W. Sabo (200) Adding and Subtracting Fractions Page of 7

4 common prime factors between them. The fact that the LCD can be smaller or simpler than the product of the original denominators is why this procedure for finding the LCD is worthwhile. Using the LCD is particularly important when working with algebraic fractions.) step (iv): Convert the original fractions to equivalent fractions with this LCD as denominator and do the arithmetic. The numbers a and b involved in this conversion, as illustrated above, are just LCD divided by the original denominators of each fraction. So, in our simple example, the fraction 4 must be multiplied top and bottom by LCD = 12 =, and the fraction 2 must be multiplied top and bottom by 4 4 LCD = 12 = 4 : = + = + = = Sometimes people write this conversion to equivalent fractions as multiplication by fractions with identical numerators and denominators, as in: = + = + = = but because of the rule for multiplication of fractions, the result is the same. We ve described this process in considerable detail. Now we ll illustrate it more concisely with a few examples. Example: Perform the addition 7 +, and express your final answer in simplest form solution: The prime factorizations of the two denominators are 10 = 2 1 x = 1 x 5 1 So, the prime factors 2,, and 5 occur, each to at most the first power. Thus LCD = 2 x x 5 = 0 (Notice that this is smaller than the product, 10 x 15 = 150, of the original denominators. The factor 5 occurs in both of the original denominators, but need appear only once in the LCD.) Now, to convert 10 to an equivalent fraction with a denominator of 0, we need to multiply top and bottom by 0 10 =. To convert 7 15 to an equivalent fraction with denominator of 0, we need to multiply top and bottom by =. So David W. Sabo (200) Adding and Subtracting Fractions Page 4 of 7

5 = + = + = = Since 2 is a prime number, no simplification of this result is possible, and so our final answer is = Example: Perform the addition and express your final answer in simplest form solution: First we write the two denominators as products of prime factors: Thus, 48 = 2 4 x 1 18 = 2 1 x 2 LCD = 2 x y since the prime factorizations of 48 and 18 contain only 2 and as factors. Then and So, x = 4, because the highest power of 2 is 4, occurring in the factorization of 48, y = 2, because the highest power of is 2, occurring in the factorization of 18. LCD = = 144. Now, to convert to an equivalent fraction with a denominator of 144, we must multiply top and bottom by =. To convert 7 18 to an equivalent fraction with a denominator of 144, we must multiply top and bottom by 144 = 8. So, our problem becomes = + = + = = To check for the possibility of simplification, we need to express the numerator and denominator of this result as a product of prime factors. We already know that 144 = It takes just a minute to verify that 11 is not divisible by either 2,, 5, 7, or 11, and so 11 must be a prime number already. Therefore no further simplification is possible and our final answer is David W. Sabo (200) Adding and Subtracting Fractions Page 5 of 7

6 = (In case you re wondering why we had to check only that none of 2,, 5, 7, and 11 divided evening into 11 to conclude that 11 is a prime number the reason is this. We don t have to check any divisors which are not prime numbers themselves, or which are larger than the square root of the number being factored. Since the square root of 11 is less than 12, we only have to check potential prime divisors which are less than 12, and this is the list 2,, 5, 7, and 11.) Example: Perform the subtraction 7 8 and express the final answer in simplest form solution: 9 = 2 and 21 = 1 x 7 1 so LCD = 2 x 7 1 = 6 Thus = = = = 6 = 7 The factorization shown in brackets indicates that the numerator and denominator of this answer have no common factors, so no further simplification is possible. Thus, in simplest form = This procedure works when more than two fractions are involved. In such cases, the LCD must be formed for all fractions present. We illustrate with an example. Example: Perform the arithmetic and express the final answer in simplest form solution: First write each of the three denominators as products of prime factors: 7 = = 1 x 7 1 David W. Sabo (200) Adding and Subtracting Fractions Page 6 of 7

7 5 = 5 1 x 7 1 Thus, for these three fractions, LCD = 1 x 5 1 x 7 1 = 105. Thus, = = + = = Now, 122 = 2 x 61 (and 61 is prime) 105 = x 5 x 7 Therefore, the numerator and denominator of this answer have no prime factors in common and so no further simplification is possible. So our final answer here is: = Converting Mixed Numbers to Pure Fractions We now have the tools to justify the method used previously for converting mixed numbers to pure fractions. For example, 5 = 5 + (by definition) = + (because = ) 5 8 = (LCD is 8 here) = + = = This example also illustrates how we can add or subtract whole numbers and fractions. David W. Sabo (200) Adding and Subtracting Fractions Page 7 of 7