1. Dividing Mixed Numbers © Math As A Second Language All Rights Reserved next #7 Taking the Fear out of Math 1313 3 1212 2

2. As in our previous discussions with mixed numbers, let’s begin with a “real world” example… © Math As A Second Language All Rights Reserved next What is the cost per pound if 3 3 / 4 pounds of pears cost $4.50?

3. next © Math As A Second Language All Rights Reserved next 3 3 / 4 pounds is the same amount as 15 fourths of a pound. If 15 fourths of a pound cost $4.50 (that is, 450 cents), then each fourth of a pound costs… 450 cents ÷ 15 (or 30 cents). And since there are 4 fourths of a pound in a pound, each pound of pears costs 4 × 30 cents or $1.20.

4. © Math As A Second Language All Rights Reserved next The cost per pound in dollars is given by the formula… dollars per pound = total cost in dollars ÷ number of pounds. This formula is the same whether we are dealing with whole numbers, common fractions or mixed numbers.

5. cost per pound = 4 1 / 2 dollars ÷ 3 3 / 4 pounds © Math As A Second Language All Rights Reserved next Rewriting $4.50 as $4 1 / 2, our answer takes the form… which can be rewritten as… cost per pound = (4 1 / 2 ÷ 3 3 / 4 ) dollars per pound

6. If you are uncomfortable using mixed numbers to perform the division, you may use improper fractions and rewrite the above computation in the form… © Math As A Second Language All Rights Reserved next cost per pound = ( 9 / 2 ÷ 15 / 4 ) dollars per pound

7. © Math As A Second Language All Rights Reserved next From our study of fractions we know that… 4 1 / 2 ÷ 3 3 / 4 dollars per pound = 9 / 2 ÷ 15 / 4 dollars per pound = 9 / 2 × 4 / 15 dollars per pound = 36 / 30 dollars per pound = 6 / 5 dollars per pound = 1 1 / 5 dollars per pound

8. © Math As A Second Language All Rights Reserved next And since there are 100 cents per dollar… 1 / 5 of a dollar = 100 cents ÷ 5 = 20 cents and therefore… 1 1 / 5 dollars per pound = $1.20 per pound which agrees with our previous answer.

9. © Math As A Second Language All Rights Reserved next As we shall show later in this presentation, it is possible to divide mixed numbers without having to first convert them to improper fractions. However, this can be quite tedious. So while the idea of converting mixed numbers to improper fractions is quite common, it is especially useful when we deal with division.

10. © Math As A Second Language All Rights Reserved next However, converting the division of mixed numbers to division of improper fractions tends to obscure what is actually happening. For example, suppose we want to divide 4 1 / 2 by 2 1 / 2, We could begin by rewriting 4 1 / 2 ÷ 2 1 / 2 as 9 / 2 ÷ 5 / 2. We would then use the “invert and multiply” rule to obtain 9 / 2 × 2 / 5 or 9 / 5. Finally, we divide 9 by 5 and obtain 1 4 / 5 as the answer.

11. © Math As A Second Language All Rights Reserved Notes next Since the definition of division hasn’t changed 4 1 / 2 ÷ 2 1 / 2 means the number we must multiply by 2 1 / 2 to obtain 4 1 / 2 as the product. A quick check shows that this is the case… 2 1 / 2 × 1 4 / 5 = 5 / 2 × 9 / 5 = 9 / 2 = 4 1 / 2

12. © Math As A Second Language All Rights Reserved next As a “reality check”, it is helpful to estimate an answer before proceeding with the actual computation. Noticing that 4 1 / 2 is “around” 4 and 2 1 / 2 is “around” 2, we can estimate that the answer should be “around” 4 ÷ 2 or 2. Actually, since 5 ÷ 2 1 / 2 is exactly 2, 4 1 / 2 ÷ 2 1 / 2 must be a “little less” than 2. Thus, 1 4 / 5 is a plausible answer. Notes

13. © Math As A Second Language All Rights Reserved Notes next As a practical application, suppose we can buy 2 1 / 2 pounds of cheese for $ 4 1 / 2. Then the price per pound (that is, “dollars per pound”) is obtained by dividing $4 1 / 2 by 2 1 / 2 pounds. The quotient tells us that the cost of the cheese is $1 4 / 5 or $1.80 per pound. In terms of relative size, what the quotient tells us is that not only is 4 1 / 2 almost twice as much as 2 1 / 2 but it’s exactly 1 4 / 5 times as much.

14. © Math As A Second Language All Rights Reserved next For additional practice, let’s use the above method to express 6 1 / 3 ÷ 1 3 / 4 as a mixed number. 6 1 / 3 ÷ 1 3 / 4 = 19 / 3 ÷ 7 / 4 = 19 / 3 × 4 / 7 = 76 / 21 = 76 ÷ 21 = 3 13 / 21

15. © Math As A Second Language All Rights Reserved next To check that our answer is at least reasonable, we observe that rounded off to the nearest whole number 6 1 / 3 = 6 and 1 3 / 4 = 2. Hence, our answer should be “reasonably close to” 6 ÷ 2 or 3. However, once we obtain our answer, we can check to see if it’s correct by remembering that 6 1 / 3 ÷ 1 3 / 4 = 3 13 / 21 means that 3 13 / 21 × 1 3 / 4 = 6 1 / 3. Notes

16. next © Math As A Second Language All Rights Reserved next Thus, to check our answer we compute the value of 3 13 / 21 × 1 3 / 4 to verify that it is equal to 6 1 / 3. 3 13 / 21 × 1 3 / 4 = 76 / 21 × 7 / 4 = (19 × 2 × 2 × 7) / (3 × 7 × 2 × 2) = 19 / 3 = 6 1 / 3 = (76 × 7) / (21 ×4)

17. © Math As A Second Language All Rights Reserved next As we mentioned earlier, while converting mixed numbers to improper fractions gives us the desired result when we divide two mixed numbers, the fact is that it doesn’t give us much insight as to what is actually happening. From a mathematical perspective, it would be nice to know that the arithmetic of mixed numbers is self-contained (at least in the sense that we aren’t forced to rewrite mixed numbers as improper fraction in order to do the arithmetic).

18. © Math As A Second Language All Rights Reserved next From a teaching point of view, we might want to illustrate how we can divide mixed numbers in ways that may be more intuitive to students. To divide 4 1 / 2 by 2 1 / 2, think of both mixed numbers as modifying, say, “a carton of books” where each carton contains 2 books 1 Using Mixed Numbers as Adjectives Modifying the Same Noun. note 1 More generally, we would use the (least) common multiple of both denominators. For example, had the problem been 4 1 / 3 ÷ 2 5 / 7, we would have assumed that each carton contained 21 books and then multiplied both the dividend and the divisor by 21. next

19. © Math As A Second Language All Rights Reserved next In this case, 4 1 / 2 cartons is another name for 9 books and 2 1 / 2 cartons is another name for 5 books. Therefore, we may visualize the problem in terms of the following steps… 4 1 / 2 ÷ 2 1 / 2 = 9 ÷ 5 = 1 4 / 5 = 4 1 / 2 cartons ÷ 2 1 / 2 cartons = 9 books ÷ 5 books

20. © Math As A Second Language All Rights Reserved next Students might find it interesting to see that we can divide mixed numbers by “translating” the problem into an equivalent whole number problem without having to refer to such things as books and cartons. The key is that it is still a fact that we do not change a quotient when we multiply both the dividend and the divisor by the same (non zero) whole number. Multiplying Both Numbers in a Ratio by the Same Number

21. © Math As A Second Language All Rights Reserved next To this end, if we are given the division problem 4 1 / 2 ÷ 2 1 / 2, we simply multiply both numbers by 2… 2 and obtain… 4 1 / 2 ÷ 2 1 / 2 = (4 1 / 2 × 2) ÷ (2 1 / 2 × 2) = 1 4 / 5 = 9 ÷ 5 note 2 When we multiply a mixed number by the denominator of its fractional part we always obtain a whole number. For example, 5 × 4 2 / 5 = 5 × (4 + 2 2 / 5 ) = (5 × 4) + (5 × 2 / 5 ) = 20 + 2 = 22. next

22. © Math As A Second Language All Rights Reserved next We can now use the above method to write 6 1 / 3 ÷ 2 3 / 5 as a mixed number. Since the denominators are 3 and 5, we can eliminate them by multiplying both numbers by 15. 15 × 6 1 / 3 = (15 × 6) + (15 × 1 / 3 ) = 95 = 90 + 5 15 × 2 3 / 5 = (15 × 2) + (15 × 3 / 5 ) = 39 = 30 + 9

23. © Math As A Second Language All Rights Reserved next Since… = 95 ÷ 39 15 × 6 1 / 3 = 95and 15 × 2 3 / 5 = 39 = 2 17 / 39 …then, 6 1 / 3 ÷ 2 3 / 5

24. © Math As A Second Language All Rights Reserved next As a check we see that… = 95 / 39 × 13 / 5 = 19 / 3 2 17 / 39 × 2 3 / 5 = (19 × 5 × 13) / (3 × 13 × 5) = 6 1 / 3 = (95 ×13) / (39 × 5)

25. © Math As A Second Language All Rights Reserved next In terms of relative size, the above result tells us that 6 1 / 3 is approximately 2 1 / 2 times as great as 2 3 / 5. (It is actually 2 17 / 39 times as great.) We don’t usually think about using common denominators when we want to divide two fractions, but we can. The denominator of a fraction is the noun and when we divide two numbers that modify the same noun, the nouns “cancel”. Notes

26. © Math As A Second Language All Rights Reserved next Thus, if we decide to rewrite the mixed numbers as improper fractions, we obtain… = 19 / 3 ÷ 13 / 5 = 95 / 15 ÷ 39 / 15 6 1 / 3 ÷ 2 3 / 5 = (19 × 5) / (3 × 5) ÷ (13 × 3) / (5 × 3) = 95 fifteenths ÷ 39 fifteenths = 2 17 / 39 = 95 ÷ 39

27. © Math As A Second Language All Rights Reserved next While the topic might be beyond the scope of the elementary school student, it is interesting to note that the long division algorithm for whole numbers, as a form of rapid subtractions, also applies to the division of mixed number. For example, 6 1 / 3 ÷ 2 3 / 5 = ( ) means the same thing 2 3 / 5 × ( ) = 6 1 / 3.

28. © Math As A Second Language All Rights Reserved next Using trial-and error to solve this equation we see that… 2 3 / 5 × 1 = 2 3 / 5 2 3 / 5 × 2 = 4 6 / 5 = 5 1 / 5 2 3 / 5 × 3 = 6 9 / 5 = 7 4 / 5 less than 6 1 / 3 greater than 6 1 / 3 In other words, 6 1 / 3 is more than twice as big as 2 3 / 5 but less than three times as big. next

29. © Math As A Second Language All Rights Reserved next The difference between 6 1 / 3 and 5 1 / 5 is 6 1 / 3 – 5 1 / 5 = 6 5 / 15 – 5 3 / 15 = 1 2 / 15. Hence, if we use the long division algorithm we see that… 2 5 1 / 5 R 23/523/5 61/361/3 …and if we now write the remainder over the divisor, we see that… 1 2 / 15 1 2 / 15 6 1 / 3 ÷ 2 3 / 5 = 2 + 1 2 / 15 23/523/5

30. © Math As A Second Language All Rights Reserved next We can simplify the complex fraction above by multiplying numerator and denominator by 15. = 15 × 2 1 2 / 15 23/523/5 15 × 1 2 / 15 = (15 × 1) + (15 × 2 / 15 ) = 17 15 × 2 3 / 5 = (15 × 2) + (15 × 3 / 5 ) = 39 = 30 + 9

31. © Math As A Second Language All Rights Reserved next Therefore… 6 1 / 3 ÷ 2 3 / 5 = 2 + 1 2 / 15 23/523/5 = 17 / 39 = 2 17 / 39 23/523/5 61/361/3 next 1 2 / 15 23/523/5 = 2 + 17 / 39

32. next This completes our discussion of the four basic operations of arithmetic using mixed numbers, and in our next presentation we shall introduce the notion of percents. © Math As A Second Language All Rights Reserved 4 / 5 × 100 = 80%