1. Mathematics as a Second Language Mathematics as a Second Language Mathematics as a Second Language Developed by Herb I. Gross and Richard A. Medeiros © 2010 Herb I. Gross next Arithmetic Revisited

2. Whole Number Arithmetic Whole Number Arithmetic © 2010 Herb I. Gross next Division or “Unmultiplication” Lesson 2 Part 4

3. Formulas As A Way to “Unite” Multiplication and Division next A rather easy way to see the relationship between multiplication and division is in terms of simple formulas. © 2010 Herb I. Gross In essence, a formula is a recipe that tells us how to find the value of one quantity once we know the value of one or more other quantities.

4. next For example, we all know that there are 12 inches in a foot. Thus, to convert feet to inches, we simply multiply the number of feet by 12. © 2010 Herb I. Gross next As a specific example, to convert 5 feet into an equivalent number of inches, we first notice that 5 feet may be viewed as 5 × 1 foot, and we may then replace 1 foot by 12 inches to obtain that 5 feet = 5 × 12 inches or 60 inches. 9234 5 67 8 1 1011 12

5. next In terms of a more verbal description, the “recipe” (which we usually refer to as a formula) may be stated as follows… © 2010 Herb I. Gross next Step 1: Start with the number of feet. Step 2: Multiply by 12. Step 3: The product is the number of inches. 9234 5 67 8 1 1011 12

6. next Thus, if we start with 5 feet, the recipe tells us that… © 2010 Herb I. Gross next Step 1: Start with 5 feet. Step 2: Multiply by 12 (inches per foot). Step 3: The product is 60 (inches). 9234 5 67 8 1 1011 12

7. next In any case, the verbal process gives us an easy way to see the relationship between multiplication and division. Namely, when we started with 5 feet, we multiplied by 12 to get the number of inches. © 2010 Herb I. Gross next On the other hand, if we had started with 60 inches and wanted to know the number of feet with which we started, we would have had to realize that we obtained 60 inches after we multiplied the number of feet by 12. 9234 5 67 8 1 1011 12

8. next That is, in terms of “fill-in-the-blank”, the computation is… © 2010 Herb I. Gross next 60 inches = 12 × _____ feet Since we obtained 60 after we multiplied by 12, to find out what number we had before we multiplied by 12, we have to “unmultiply” by 12 to determine the number of feet. 9234 5 67 8 1 1011 12

9. next In more familiar terms, to convert 60 inches into feet, we divide 60 by 12. © 2010 Herb I. Gross next 60 inches ÷ 12 = _____ feet In this context, “divide” is the more formal term for “unmultiply”. 1 1 Our own experience seems to indicate that students learn to internalize division better once they connect it with “unmultiplication. For example, they seem to better understand the meaning of 2,600 ÷ 13 = ___, if we paraphrase the problem into the equivalent form 13 × ____ = 2,600. note next 9234 5 67 8 1 1011 12

10. next The verbal description above is easy to understand but cumbersome to write (and it gets even more cumbersome as the “recipe” contains more and more steps). © 2010 Herb I. Gross next The algebraic shortcut is to let a “suggestive” letter of the alphabet represent the number of feet (perhaps the letter “F” because it suggests feet) and similarly let a letter such as I (because it suggests inches) represent the number of inches. The equal sign replaces the word “is”. An Algebraic “Digression”

11. next That is… © 2010 Herb I. Gross next An Algebraic “Digression” Step 3: The product is the number of inches. EnglishAlgebra Step 2: Multiply by 12. Step 1: Start with the number of feet. F 12 × F (or F × 12) 12 × F = I next

12. Because the “times sign” ( × ) and the letter x are easy to confuse (especially when we’re writing by hand), we do not use the “times sign” in algebra. That is, rather than write 12 × F, we would write 12F (even if it were in the form F × 12, we would write 12F probably because of how much more natural it is to say such things as “I have 12 dollars” as opposed to “I have dollars 12”). © 2010 Herb I. Gross An Algebraic “Digression” 12 × F = I

13. next It is also traditional to write the letter that stands by itself (namely, the “answer”), on the left of the equal sign, but this is not really vital. © 2010 Herb I. Gross An Algebraic “Digression” In any case, when we write I = 12F, it is “shorthand” for the verbal recipe. next 12F = I

14. Again, we assume that you know the standard algorithm for performing long division. What is not always as clear to students is that division is really a form of repeated, rapid subtraction. © 2010 Herb I. Gross Revisiting the Division Algorithm for Whole Numbers Let’s look at a typical kind of problem that one uses to illustrate a division problem. next

15. A certain book company ships its books in cartons, each of which contains 13 books. © 2010 Herb I. Gross The Problem If a customer makes an order for 2,821 books, how many cartons will it take to ship the books? next

16. To answer this question, we divide 2,821 by 13 to obtain 217 as the quotient. The usual algorithm is illustrated below… © 2010 Herb I. Gross next 1 32 8 2 1 2 2 2 1 1 9 0 7 2 6 1 3 × 2 2 6– – – 1 3 × 1 1 3 9 1 1 3 × 7 9 1 next

17. However, even if it is successfully memorized by students, this algorithm is rarely understood by them. However, with a little help from our “adjective/noun” theme the “mysticism” is quickly evaporated! © 2010 Herb I. Gross More specifically, the “fill in the blank” problem we are being asked to solve is… next 13 × _____ = 2821

18. next Since 13 × 2 = 26, our adjective/noun theme tells us that… © 2010 Herb I. Gross … or in the language of place value… next 13 × 2 hundred = 26 hundred 13 × 200 = 2600 Hence, after we have packed 200 cartons, we have packed 2,600 books. 2 2 And since 13 × 300 = 3,900 and there are only 2,821 books, we know that we will need fewer than 300 cartons. Thus, we already know that we will need more than 200 but less than 300 cartons. note next

19. Since we started with 2,821 books and 2,600 books have now been packed, there are now 2,821 – 2,600 or 221 books still left to pack. © 2010 Herb I. Gross Since 13 × 1 = 13, we know that 13 ×10 = 130. next Hence, after we pack an additional 10 cartons, there are still 91 (that is, 221 – 130) books still left to pack.

20. Finally, since 13 × 7 = 91, we see that it takes 7 more cartons to pack the remaining books. © 2010 Herb I. Gross Thus, in all, we had to use 217 (that is, 200 + 10 + 7) cartons. next 217

21. next We can now write the result in tabulated form as shown below. © 2010 Herb I. Gross 2 8 2 1 books next – 2 6 0 0 books 2 2 1 books left 1 3 0 books 9 1 books left 9 1 books 0 books left 200 cartons 10 cartons 7 cartons 217 cartons + – –

22. In our opinion, repeated, rapid subtraction is easier to understand. However, the “traditional” recipe appears to be more familiar. © 2010 Herb I. Gross next 1 32 8 2 1 2 2 2 1 1 9 1 0 7 2 6– – – 1 3 9 1

23. next However, if we simply insert the “missing digits” into the above format (which are emphasized in color), © 2010 Herb I. Gross next 1 32 8 2 1 2 2 2 1 1 9 1 0 7 2 6– – – 1 3 9 1 2 8 2 1 –2 6 0 0 2 2 1 1 3 0 9 1 0 – – 0 1 0 we see that the numbers in the emended diagrams look exactly the same. next

24. In terms of having students see the above transition more gradually, we might want to introduce the intermediate representation… © 2010 Herb I. Gross next 2 0 0 1 0 7 next Note 1 32 8 2 12 1 2 2 1 9 1 0 2 6 0 0– – – 1 3 0 9 1 next

25. In checking a division problem, we multiply the quotient by the divisor, and if we divided correctly, the product we obtain should be equal to the dividend. In terms of the present illustration, the usual check is to show that 217 × 13 = 2,821. © 2010 Herb I. Gross next Note More specifically… 2 1 7 × 1 3 6 5 1 2 1 7 2 8 2 1

26. next Notice, however, that when we perform the multiplication this way, the rows in the above multiplication (namely, 651 and 2,170) have no relationship to the three rows in the traditional division algorithm (namely, 2,600, 130 and 91). © 2010 Herb I. Gross Note 2 1 7 × 1 3 6 5 1 2 1 7 0 2 8 2 1 1 32 8 2 1 2 2 2 1 1 9 1 0 7 2 6– – – 1 3 9 1 0 1 0

27. next The reason is that the division problem showed us that if there were 13 books in each carton, then 217 cartons would have to be used. © 2010 Herb I. Gross Note On the other hand, the above check by multiplication showed that if there were 217 books in a carton, then 13 cartons would have to be used. 3 next 3 In other words, the multiplication problem we did was 13 × 217. However, to check the division problem, the multiplication problem should have been 217 × 13. note next

28. However, a more enlightening way to perform the check is to compute the product in the less traditional but more accurate form 217 × 13. Indeed, when we do the problem that way we see an “amazing” connection between multiplication and division. © 2010 Herb I. Gross next Note Namely… 1 3 × 2 1 7 9 1 1 3 0 2 6 0 0 2 8 2 1

29. next Notice that when written in this form we see that the same three numbers we are adding to get the product (namely 91, 130 and 2,600 are the same three numbers (in the reverse order) we subtracted to find the quotient. © 2010 Herb I. Gross next 1 3 × 2 1 7 9 1 1 3 0 2 6 0 0 2 8 2 1 1 32 8 2 1 2 2 2 1 1 9 1 0 7 2 6– – – 1 3 9 1 0 1 0

30. next This illustrates quite vividly the connection between division and multiplication as inverse processes. 4 © 2010 Herb I. Gross next 4 The reason for using the traditional check is that the fewer digits in the bottom number, the fewer rows we will need to compute the product, thus saving paper. Yet this format, inspired by economic frugality, obscures the connection that illustrates why we often refer to division as being the inverse of multiplication. note × ÷

31. next When students memorize the traditional long division recipe, they often have trouble with what we might call the missing 0. © 2010 Herb I. Gross Beware of the Missing Zero For example, suppose instead of 2,821 books, there were 2,652 books. In this case, the number of cartons would have been 2,652 ÷ 13. However, in performing the division the following problem arises. next

32. Namely, the division proceeds normally until they get to… © 2010 Herb I. Gross next 1 32 6 5 2 2 5 2 0 4 2 6– –5 2 The tendency is then to observe that since 13 doesn’t “go into” 5, the student brings down the next digit (2) and then finishes the problem as follows… next

33. © 2010 Herb I. Gross next If students understood the concept of “unmultiplying”, they would see that it is impossible that 2652 ÷ 13 = 24 (or equivalently that it is impossible that 13 × 24 = 2652). 13 × 24 < 1300 < 2652 Namely, 100 thirteen’s is 1300; therefore, 24 thirteen’s is less than 1300. More symbolically,

34. © 2010 Herb I. Gross next 1 32 6 5 2 2 5 2 5 0 2 6– –0 The “trick” is that every time we “bring down” a digit from the dividend, we have to place a digit above it in the quotient. Hence, the correct computation would be… next 4 0 –5 2 Since 200 × 13 = 2600, we see that 204 is a “reasonable” answer. next

35. © 2010 Herb I. Gross next This problem wouldn’t occur if students used our adjective/noun theme. More specifically… 2 0 0 thirteen’s 2 0 00 – 4 4 thirteen’s 2 0 4 thirteen’s + 5 2 1 32 6 5 2 2 6 0 0 5 2– 0 next

36. © 2010 Herb I. Gross next In any event, this concludes our discussion of the role of place value in whole number arithmetic. Our next endeavor will be to show that a knowledge of whole number arithmetic, together with our adjective/noun theme, is all we need in order to do the arithmetic associated with fractions.