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

2. The definition of subtraction as “unadding” remains the same. © Math As A Second Language All Rights Reserved next That is when we say that… “f – s means the number that we must add to s to obtain f as the sum” …it does not matter whether f and s are whole numbers or rational numbers.

3. Suppose we want to compute the difference 6 7 / 9 – 4 2 / 9 and express the answer as a mixed number. © Math As A Second Language All Rights Reserved next There are a few ways of finding the answer.

4. The “Direct” Method © Math As A Second Language All Rights Reserved next Since subtraction is “unadding” and since we add mixed numbers by adding the whole numbers and adding the fractions, we want to solve the following two problems… 2 / 9 + _____ = 7 / 9 and… 4 + _____ = 6

5. The “Direct” Method © Math As A Second Language All Rights Reserved next Since 4 + 2 = 6 and 2 / 9 + 5 / 9 = 7 / 9, we see that the answer is 2 5 / 9. 4 2 / 9 + 2 5 / 9 = As a check, we see that… (4 + 2 / 9 ) + (2 + 5 / 9 ) = (4 + 2) + ( 2 / 9 + 5 / 9 ) =67/967/9

6. Converting Mixed Numbers to Improper Fractions © Math As A Second Language All Rights Reserved next A common strategy for solving an unfamiliar problem is to rephrase it in the equivalent form of one or more familiar problems. In this case, since we already know the arithmetic of common fractions, we can convert the mixed numbers to improper fractions.

7. © Math As A Second Language All Rights Reserved next We may rewrite the problem as follows … 6 7 / 9 – 4 2 / 9 = 61 / 9 – 38 / 9 = 23 / 9 =25/925/9 Converting Mixed Numbers to Improper Fractions

8. ► Doing a mixed number problem by converting the mixed numbers to improper fractions is not as transparent as it is when we work directly with the mixed numbers. © Math As A Second Language All Rights Reserved Notes next ► However, it does avoid a pitfall that students often fall into when they subtract one mixed number from another.

9. ► Specifically, students often make a mistake when they borrow or regroup using mixed numbers. © Math As A Second Language All Rights Reserved Notes next ► As we mentioned in the previous presentation the idea of exchanging ten of one denomination for one of the next higher denomination is valid only when the denominator (if there is one) is 10. However, students often forget this when they are borrowing or regrouping by rote.

10. © Math As A Second Language All Rights Reserved next To illustrate the above note, suppose that we want to express the difference 6 2 / 3 – 3 3 / 4 as a mixed number and that we don’t want to rewrite the mixed numbers as equivalent improper fractions. Just as in our previous illustration, in terms of “unadding” the problem is asking us to fill in the blank… 3 3 / 4 + ____ = 6 2 / 3

11. © Math As A Second Language All Rights Reserved next We begin as we did in the previous problem by filling in the blanks in the two statements... and… 3 + _____ = 6 3 / 4 + _____ = 2 / 3

12. © Math As A Second Language All Rights Reserved next We can only add or subtract fractions if they have the same denominator (noun). Therefore, we rewrite 3 / 4 _____ = 2 / 3 in the form… 9 / 12 + _____ = 8 / 12 The fact that 9 / 12 is greater than 8 / 12 means that we will have to “borrow” in order to solve the problem. next

13. © Math As A Second Language All Rights Reserved next This in turn means that we will begin by rewriting 6 2 / 3 in the form 6 8 / 12 and then use the following steps… 6 8 / 12 = (5 + 1) + 8 / 12 = 5 + (1 + 8 / 12 ) = 5 + ( 12 / 12 + 8 / 12 ) = 5 + 20 / 12 We use this information to rewrite the problem in the form… 3 9 / 12 + ____ = 5 20 / 12 next

14. © Math As A Second Language All Rights Reserved next We now proceed just as we did before by filling in the blanks in the two statements.. We fill in the blank in the top equation with 2 and the blank in the bottom equation with 11 / 12. 3 + ____ = 5 9 / 12 + ____ = 20 / 12 2 11 / 12 So altogether, we have to add 2 11 / 12 to 3 3 / 4 to obtain 6 2 / 3 as our sum. next

15. ► The most common mistake that students are prone to make is that when they want to “borrow” 1 from 6 8 / 12, they will write 5 18 / 12 rather than 5 20 / 12. © Math As A Second Language All Rights Reserved Notes next When borrowing using mixed numbers, the new numerator is the sum of the original numerator and the denominator.

16. ► This is analogous to the fact that if you have 8 doughnuts and you buy another dozen, you have 20 doughnuts, not 18! © Math As A Second Language All Rights Reserved Notes next ► Or if you have 3 quarters and you exchange a dollar for quarters, you now have 7 quarters, not 13!

17. ► While converting mixed numbers to improper fractions is usually a more tedious way to subtract mixed numbers, it does avoid the “borrowing” problem. © Math As A Second Language All Rights Reserved Notes next

18. ► For example, we may rewrite 6 2 / 3 and 3 3 / 4 as… © Math As A Second Language All Rights Reserved Notes next 6 2 / 3 = (6 + 2 / 3 ) = ( 6 / 1 + 2 / 3 ) = 20 / 3 = ( 18 / 3 + 2 / 3 ) = 80 / 12 3 3 / 4 = (3 + 3 / 4 ) = ( 3 / 1 + 3 / 4 ) = 15 / 4 = ( 12 / 4 + 3 / 4 ) = 45 / 12 next

19. ► Hence, 6 2 / 3 – 3 3 / 4 © Math As A Second Language All Rights Reserved Notes next = 80 / 12 – 45 / 12 next = 35 / 12 = 2 11 / 12 However, as you can see, this method, while giving us the correct answer, gives us no hint as to what is actually happening.

20. next © Math As A Second Language All Rights Reserved next As usual, the abstract approach becomes more transparent when we supply nouns (units) for the numbers to modify. Let’s return to the problem we were discussing in our previous presentation. Recall that we had assumed that you rented a particular machine and used it for 6 hours and 40 minutes one day and for 3 hours and 45 minutes the second day; and the question involved having to find the total time you used the machine.

21. © Math As A Second Language All Rights Reserved This time, however, let’s suppose that we knew that we had used the machine for a total of 10 hours and 25 minutes during the two days and that we also knew that we used it for 3 hours and 45 minutes the second day, and now we want to know how long we used it on the first day. A Subtraction Problem

22. 10 hours + 25 minutes = © Math As A Second Language All Rights Reserved next Probably, we would subtract 3 hours 45 minutes from 10 hours 25 minutes and in order to subtract in an easier way we would have written… (9 hours + 1 hour) + 25 minutes = 9 hours + (1 hour + 25 minutes) = 9 hours + 85 minutes = 9 hours 85 minutes 9 hours + (60 minutes + 25 minutes) 1 = note 1 Notice that when we exchanged an hour for the equivalent number of minutes, we replaced the hour by 60 minutes, not 10 minutes. next

23. © Math As A Second Language All Rights Reserved next The above subtraction would now look like… 10 hours 25 minutes – 3 hours 45 minutes 9 hours 85 minutes – 3 hours 45 minutes 6 hours 40 minutes next

24. 6 hours = © Math As A Second Language All Rights Reserved next If we wanted to use only one noun, we could have either, -- used the fact that 60 minutes = 1 hour and written… 6 × 60 minutes = 360 minutes 6 hours 40 minutes = 400 minutes 360 minutes + 40 minutes =

25. 40 minutes = © Math As A Second Language All Rights Reserved next or, -- again using the fact that 1 hour = 60 minutes and write… 40 / 60 of an hour = 2 / 3 of an hour 6 hours 40 minutes = 6 hours + 40 minutes = 6 hours + 2 / 3 of an hour = 6 2 / 3 hours

26. © Math As A Second Language All Rights Reserved However, notice how quickly the transparency of what we did above disappears as soon as we elect to omit the noun “hours” (and/or “minutes”). Suppose rather than writing 10 hours and 25 minutes, we simply wrote 10 5 / 12, and instead of writing 3 hours and 45 minutes, we wrote 3 3 / 4. We would then be faced with the equivalent problem of computing the difference 10 5 / 12 – 3 3 / 4 and expressing the answer as a mixed number. next

27. © Math As A Second Language All Rights Reserved Most likely we would have expressed the two fractions in an equivalent form in which they have the same denominator and written… The plausible error is that we might have rewritten 10 5 / 12 as 9 15 / 12 forgetting that 1 = 12 twelfths, not 10 twelfths. next 10 5 / 12 – 3 3 / 4 = 10 5 / 12 – 3 9 / 12

28. Or we might have rewritten the problem in the form of equivalent improper fractions and obtained… © Math As A Second Language All Rights Reserved next = 125 / 12 – 45 / 12 = (125 – 45) / 12 = 80 / 12 10 5 / 12 – 3 3 / 4 = 125 / 12 – 15 / 4 = 6 8 / 12 = 6 2 / 3

29. © Math As A Second Language All Rights Reserved Notice that if we now reinsert the noun “hours”, we see that 6 2 / 3 hours is the same as 6 hours and 40 minutes, which agrees with the answer we obtained previously. However, notice how much more transparent the problem becomes if we insert units for the mixed numbers to modify. next

30. © Math As A Second Language All Rights Reserved Notice also that any common denominator can be used. In fact, if we had been used to supplying a unit based solely on the above steps, the chances are we would have chosen “dozens” or “feet” instead of “hours” because there are 12 in a dozen and 12 inches in 1 foot.

31. next In our next section we will discuss the process of multiplying one mixed number by another. © Math As A Second Language All Rights Reserved 6 2 / 3 × 3 3 / 4 = ?