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next Mathematics as a Second Language Arithmetic Revisited © 2010 Herb I. Gross Developed by Herb I. Gross and Richard A. Medeiros next Lesson 2 Part 4 Whole Number Arithmetic Division or “Unmultiplication” © 2010 Herb I. Gross next Formulas As A Way to “Unite” Multiplication and Division A rather easy way to see the relationship between multiplication and division is in terms of simple formulas. 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. © 2010 Herb I. Gross next 1 2 3 4 5 6 7 8 9 10 11 12 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. 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. © 2010 Herb I. Gross next next 1 2 3 4 5 6 7 8 9 10 11 12 In terms of a more verbal description, the “recipe” (which we usually refer to as a formula) may be stated as follows… Step 1: Start with the number of feet. Step 2: Multiply by 12. Step 3: The product is the number of inches. © 2010 Herb I. Gross next next 1 2 3 4 5 6 7 8 9 10 11 12 Thus, if we start with 5 feet, the recipe tells us that… Step 1: Start with 5 feet. Step 2: Multiply by 12 (inches per foot). Step 3: The product is 60 (inches). © 2010 Herb I. Gross next 1 2 3 4 5 6 7 8 9 10 11 12 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. 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. © 2010 Herb I. Gross next 1 2 3 4 5 6 7 8 9 10 11 12 That is, in terms of “fill-in-the-blank”, the computation is… 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. © 2010 Herb I. Gross next 1 2 3 4 5 6 7 8 9 10 11 12 In more familiar terms, to convert 60 inches into feet, we divide 60 by 12. 60 inches ÷ 12 = _____ feet In this context, “divide” is the more formal term for “unmultiply”.1 note 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. © 2010 Herb I. Gross next An Algebraic “Digression” 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). 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”. © 2010 Herb I. Gross next next An Algebraic “Digression” That is… English Step 1: Start with the number of feet. Step 2: Multiply by 12. Step 3: The product is the number of inches. © 2010 Herb I. Gross Algebra F 12 × F (or F × 12) 12 × F = I next An Algebraic “Digression” 12 × F = I 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 next An Algebraic “Digression” 12F = I 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. In any case, when we write I = 12F, it is “shorthand” for the verbal recipe. © 2010 Herb I. Gross next Revisiting the Division Algorithm for Whole Numbers 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. Let’s look at a typical kind of problem that one uses to illustrate a division problem. © 2010 Herb I. Gross next The Problem A certain book company ships its books in cartons, each of which contains 13 books. If a customer makes an order for 2,821 books, how many cartons will it take to ship the books? © 2010 Herb I. Gross next To answer this question, we divide 2,821 by 13 to obtain 217 as the quotient. The usual algorithm is illustrated below… 1 3 – 2 1 7 2 8 2 1 – 2 – 9 0 © 2010 Herb I. Gross 1 3 × 2 2 6 1 3 × 1 1 3 1 3 × 7 9 1 next 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! More specifically, the “fill in the blank” problem we are being asked to solve is… 13 × _____ = 2821 © 2010 Herb I. Gross next Since 13 × 2 = 26, our adjective/noun theme tells us that… 13 × 2 hundred = 26 hundred … or in the language of place value… 13 × 200 = 2600 Hence, after we have packed 200 cartons, we have packed 2,600 books.2 note 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. 2And © 2010 Herb I. Gross next 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. Since 13 × 1 = 13, we know that 13 ×10 = 130. Hence, after we pack an additional 10 cartons, there are still 91 (that is, 221 – 130) books still left to pack. © 2010 Herb I. Gross next Finally, since 13 × 7 = 91, we see that it takes 7 more cartons to pack the remaining books. Thus, in all, we had to use 217 (that is, 200 + 10 + 7) cartons. © 2010 Herb I. Gross next We can now write the result in tabulated form as shown below. 2 8 2 – 2 6 0 2 2 – 1 3 9 – 9 1 books 0 books 1 books left 0 books 1 books left 1 books 0 books left © 2010 Herb I. Gross 200 cartons 10 cartons + 7 cartons 217 cartons next In our opinion, repeated, rapid subtraction is easier to understand. However, the “traditional” recipe appears to be more familiar. 2 1 3 2 8 –2 6 2 – 1 1 7 2 1 2 3 9 1 – 9 1 0 © 2010 Herb I. Gross next However, if we simply insert the “missing digits” into the above format (which are emphasized in color), 2 8 2 – 2 6 0 2 2 – 1 3 9 – 9 7 1 0 1 0 1 1 0 0 we see that the numbers in the emended diagrams look exactly the same. © 2010 Herb I. Gross 1 0 1 0 1 1 2 1 3 2 8 –2 6 2 – 1 1 2 0 2 3 9 – 9 next Note In terms of having students see the above transition more gradually, we might want to introduce the intermediate representation… © 2010 Herb I. Gross 7 1 0 2 0 0 1 3 2 8 2 1 – 2 6 0 0 2 2 1 – 1 3 0 9 1 – 9 1 0 next In checking a division problem, Note 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. More specifically… 2 1 7 × 1 3 6 5 1 2 1 7 2 8 2 1 © 2010 Herb I. Gross Notice, however, that when we Note 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). 2 1 7 2 1 7 1 3 2 8 2 1 × 1 3 –2 6 0 0 1 2 2 6 5 1 – 1 3 0 2 1 7 0 9 1 – 9 1 2 8 2 1 0 next © 2010 Herb I. Gross next The reason is that the division Note problem showed us that if there were 13 books in each carton, then 217 cartons would have to be used. 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 note 3In other words, the multiplication problem we did was 13 × 217. However, to check the division problem, the multiplication problem should have been 217 × 13. © 2010 Herb I. Gross next However, a more enlightening Note 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 1 3 and division. × 2 1 7 Namely… © 2010 Herb I. Gross 9 1 3 2 6 0 2 8 2 1 0 0 1 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. 2 1 7 1 3 1 3 2 8 2 1 × 2 1 7 –2 6 0 0 2 2 1 9 1 – 1 3 0 1 3 0 9 1 2 6 0 0 – 9 1 2 8 2 1 0 © 2010 Herb I. Gross next × ÷ This illustrates quite vividly the connection between division and multiplication as inverse processes.4 note 4The 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. © 2010 Herb I. Gross next Beware of the Missing Zero When students memorize the traditional long division recipe, they often have trouble with what we might call the missing 0. 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. © 2010 Herb I. Gross next Namely, the division proceeds normally until they get to… 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… © 2010 Herb I. Gross 2 4 1 3 2 6 5 2 –2 6 – 5 2 0 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). Namely, 100 thirteen’s is 1300; therefore, 24 thirteen’s is less than 1300. More symbolically, 13 × 24 < 1300 < 2652 © 2010 Herb I. Gross next 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… 2 0 4 1 3 2 6 5 2 –2 6 – 0 5 – 5 2 0 Since 200 × 13 = 2600, we see that 204 is a “reasonable” answer. © 2010 Herb I. Gross next next This problem wouldn’t occur if students used our adjective/noun theme. More specifically… 2 0 4 0 1 3 © 2010 Herb I. Gross 2 6 5 –2 6 0 5 – 5 2 0 2 2 0 2 0 0 thirteen’s + 4 thirteen’s 2 0 4 thirteen’s 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. © 2010 Herb I. Gross