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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

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Whole Number Arithmetic Whole Number Arithmetic © 2010 Herb I. Gross next Multiplication Lesson 2 Part 3.2

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The Role of Place Value in the Development of Whole Number Arithmetic --- Multiplication next We ended the first part of this lesson by listing the first nine multiples of 13. By way of review… © 2010 Herb I. Gross 1 × 13 = 13 2 × 13 = 26 3 × 13 = 39 4 × 13 = 52 5 × 13 = 65 6 × 13 = 78 7 × 13 = 91 8 × 13 = 104 9 × 13 = 117

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next Suppose we wanted to use the table to the right to compute the cost of buying 9 items, each of which cost $13. © 2010 Herb I. Gross 1 × 13 = 13 2 × 13 = 26 3 × 13 = 39 4 × 13 = 52 5 × 13 = 65 6 × 13 = 78 7 × 13 = 91 8 × 13 = 104 9 × 13 = 117 The table shows us that 9 × 13 = 117; from which we would conclude that the cost was $117.

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next Suppose, instead, we now wanted to find the price of purchasing 234 items, each costing $13. We could count by 13’s until we got to the 234 th multiple. This would be both tedious and unnecessary! In fact, if we know the “13 table” through 9, the adjective/noun theme takes care of the rest. © 2010 Herb I. Gross next To begin, with we may view 234 in the form… 2 hundreds + 3 tens + 4 ones

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That is, we may think of the 234 items as being arranged in 3 piles; one of which contains 200 of the items; another of which contains 30 of the items and the remaining pile contains 4 of the items. © 2010 Herb I. Gross 2 hundreds + 3 tens + 4 ones 200 + 3 0 + 4

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next 13 × 2 apples = 26 apples © 2010 Herb I. Gross next It is not difficult to find the cost of 200 items. Namely, when we learn that 13 × 2 = 26, our rule for multiplying quantities tells us that… 200 + 30 + 4 13 × 2 people = 26 people 13 × 2 hundreds = 26 hundreds, etc.

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In the language of place value we write… © 2010 Herb I. Gross 13 × 2 hundreds = 26 hundreds as 13 × 200 = 2600 And since 13 × 200 = 200 × 13, we may conclude that at $13 each, 200 items would have cost $2,600. next

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This one step takes the place of our having to count to the 200 th multiple of 13. In other words, if we had continued listing the multiples of 13, and we were “lucky” enough not to have made a computational error, the 200 th line on our list would have read 200 × 13 = 2,600. © 2010 Herb I. Gross next In other words, we already know that at a price of $13 each, 234 items would cost more than $2,600. Note

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next By similar reasoning, the fact that 13 × 3 = 39 tells us that 300 × 13 = 3,900. Hence, we also know that the cost of the 234 items is less than $3,900. © 2010 Herb I. Gross next Note In summary, we have used the adjective/noun theme very efficiently to conclude that at $13 per item, 234 items would cost more than $2,600 but less than $3,900. This is a helpful thing to know once we have computed the exact cost and want to check the plausibility of our answer.

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next In a similar way, we know that… © 2010 Herb I. Gross 13 × 3 = 39 means that 30 × 13 = 390 and 13 × 200 = 2600 next 13 × 4 = 52

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The value of the items in the first pile is $2,600. © 2010 Herb I. Gross Hence… next The value of the items in the second pile is $390. The value of the items in the third pile is $52. Therefore, the answer to our question is $2,600 + $390 + $52 = $3,042 next 200304 $2,600$390$52

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next In this venue, it is relatively easy to see how multiplication is really a special format for organizing rapid, repeated addition. However, in the traditional format in which multiplication is presented, this clarity is either lacking or obscured. © 2010 Herb I. Gross next For example, the most traditional method of finding the sum of 234 “thirteen’s” is to write the multiplication problem in vertical form, making sure that the number with the greater number of digits must be written on top.

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That is, we often write… © 2010 Herb I. Gross …rather than… 2 3 4 next × 1 3 7 0 2 2 3 4 3 0 4 2 × 2 3 4 1 3 5 2 3 9 0 3 0 4 2 2 6 0 0

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next Notice that in this format we are actually finding the cost of 13 items, each of which costs $234. © 2010 Herb I. Gross next This is not the problem we intended to solve, even though it gives us the same answer. 2 3 4 × 1 3 7 0 2 2 3 4 3 0 4 2

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next However, the second form is a more compact version of the method we used above to solve the problem. © 2010 Herb I. Gross next × 2 3 4 1 3 5 2 3 9 3 0 4 2 2 6 9 For example, when we wrote “39” we placed the 9 under the 5, thus putting the 9 in the tens place.

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next In other words, since the 5 was already holding the tens place there was no need for us to write the 0. However, if we wanted to, we could have. © 2010 Herb I. Gross next × 2 3 4 1 3 5 2 3 9 0 3 0 4 2 2 6 6 Similarly when we wrote “26” we placed the 6 under the 3, thus putting the 6 in the hundreds place, 0 and annexing the two zeroes we obtain… next

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© 2010 Herb I. Gross …and this in turn is a shorter version of… × 2 3 4 1 3 5 2 3 9 0 3 0 4 2 2 6 0 0 (4 thirteen’s) (30 thirteen’s) (200 thirteen’s) (234 thirteen’s) In this form, we see immediately the connection between the traditional algorithm and rapid, repeated addition. next

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Comparing 234 × 13… © 2010 Herb I. Gross next 2 3 4 × 1 3 7 0 2 2 3 4 3 0 4 2 × 2 3 4 1 3 5 2 3 9 0 3 0 4 2 2 6 0 0 (4 thirteen’s) (30 thirteen’s) (200 thirteen’s) (234 thirteen’s) (3 “234’s”) (10 “234’s”) and 13 × 234, we see that… (13 “234’s”) Note

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next This format results in finding the 13 th multiple of 234. © 2010 Herb I. Gross next Note 2 3 4 × 1 3 × 2 3 4 1 3 On the other hand, this format results in finding the 234 th multiple of 13…which was our goal in this problem.

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next However, both formats have the same property. Namely each digit in one factor multiplies each digit in the other factor. © 2010 Herb I. Gross next Note This is known more formally as the Distributive Property of Multiplication over Addition (or more simply, the Distributive Property) which will now be discussed in more detail.

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The Distributive Property next Many of us in our high school (middle school?) algebra course learned the so called rule of FOIL which was a rote device for remembering (First, Outer, Inner, Last). © 2010 Herb I. Gross What it meant was that if we were multiplying two numbers each of which was the sum of two terms, we could find the product by adding the following four terms ---- the product of the first terms in each factor; the product of the two outer terms; the product of the two inner terms; and the product of the two last terms in each factor.

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next Foil is a special case of when we multiply a sum of numbers by another sum of numbers we multiply each number in one grouping by one number in the other grouping. Thus, for example, to find the product of (3 + 4 + 5) and (8 + 9), we could form the sum… © 2010 Herb I. Gross next (3 × 8) + (3 × 9) + (4 × 8) + (4 × 9) + (5× 8) + (5 × 9) Note that we might have found it more convenient to rewrite 3 + 4 + 5 as 12 and 8 + 9 as 17; after which we would simply compute the product 12 × 17.

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next However, while we can simplify 3 + 4 + 5, it is not possible to simplify a + b + c in a similar manner. Thus, the Distributive Property is essential if we wish to rewrite an expression in which letters are used to represent numbers (such as we do in algebra). © 2010 Herb I. Gross next (a × d) + (a × e) + (b × d) + (b × e) + (c × d) + (c × e) Thus, for example, to form the product of a + b + c and d + e, we would use the Distributive Property to write the product in the form…

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next To see why the Distributive Property is plausible, it might be helpful to think in terms of the area of a rectangle. © 2010 Herb I. Gross next For example, in the diagram below, a, b, c, d, and e represent lengths. ab c d e Hence, the length (base) of the rectangle is given by a + b + c ab c, and the width (height) is given by d + e. d e

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Since the area of a rectangle is the product of its length and width (base and height), on the one hand the area of the rectangle below is given by (a + b + c) × (d + e). © 2010 Herb I. Gross next On the other hand, it is also the sum of the areas of the 6 smaller rectangles. abc dddd aa eeee bbcc × × × ×× ×

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We have just demonstrated the Distributive Property in terms of the area model. Now we will demonstrate it in terms of the adjective/noun theme. © 2010 Herb I. Gross For example, when we multiply 30 by 20, we are really multiplying 3 tens by 2 tens, and according to our adjective/noun theme 3 tens × 2 tens = 6 “ten tens” or 6 hundred. next

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In this sense, given a problem such as 437 x 28, we can view the multiplication algorithm in the form… © 2010 Herb I. Gross next Ten ThousandsThousandsHundredsTensOnes 28 × 437 5 5 6 14 7 24 6 3 3 2 + 8 4 0 0 0 0 0 0 0 0 0

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next For students who are visual learners, the above algorithm can be explained in terms of an area model. Imagine that there is a rectangle whose dimensions are, say, 28 feet by 437 feet. © 2010 Herb I. Gross next Note on Area model 437 28

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next On the one hand, the area of the rectangle is 437 feet × 28 feet or 12,236 square feet (that is, 12,236 “feet feet” or 12,236 ft 2 ) © 2010 Herb I. Gross next Note on Area model 437 28 437 feet × 28 feet = 12,236 ft 2

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next On the other hand, we can compute the same area by subdividing the rectangle as shown below. © 2010 Herb I. Gross next Note on Area model 730400 20 8 next 11,200 840196++= 12,236 next 7 × 8 56 30 × 8 240 7 × 20 140 30 × 20 600400 × 20 8000 400 × 8 3200 next

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On the other hand, we can compute the same area by subdividing the rectangle as shown below. © 2010 Herb I. Gross next Note on Area model 730400 20 8 next 11,200 840196++= 12,236 next 7 × 8 56 30 × 8 240 7 × 20 140 30 × 20 600400 × 20 8000 400 × 8 3200 next

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In this sense, we can rewrite the bottom row in the chart below to obtain… © 2010 Herb I. Gross next Ten ThousandsThousandsHundredsTensOnes 28 × 437 5 5 6 14 24 6 3 2 + 8 11 136 11 12 36 2 36 1 2 2 36

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next The 3 models are summarized below. © 2010 Herb I. Gross next 400 × 20 400 × 8 30 × 20 30 × 8 7 × 20 7 × 8 730400 20 8 Area Model 8000 600140 Adjective /Noun Traditional 1 1,2 0 0 8 4 0 1 9 6 ++ = 1 2,2 3 6 56 240 3200 4 3 7 × 2 8 8 7 4 0 3 4 9 6 + 2 8 × 4 3 7 1 9 68 4 01 1,2 0 0 3 4 9 6 8 7 4 0 1 2,2 3 6 next

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Remember that when we use the multiplication algorithm to multiply two whole numbers, we have to remember that each digit (including 0) in one number has to multiply each digit in the other number. © 2010 Herb I. Gross Beware of the Missing Zero next

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Thus, the correct way to compute a product such as… © 2010 Herb I. Gross is to write… × 1 0 3 2 4 6 × 1 0 3 2 4 6 7 3 8 0 0 0 2,5 3 3 8 2 4 6 013 next

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However students are often tempted to “ignore” the 0 and instead compute the product as follows… © 2010 Herb I. Gross × 1 0 3 2 4 6 7 3 8 0 0 0 2,5 3 3 8 2 4 6 next × 1 0 3 2 4 6 7 3 8 3,1 9 8 2 4 6 013

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next By placing the 6 in 246 under the 3 in 738, they were computing the value of ten 246’s rather than of a hundred 246’s. © 2010 Herb I. Gross × 1 0 3 2 4 6 7 3 8 0 0 0 2,5 3 3 8 2 4 6 next × 1 0 3 2 4 6 7 3 8 3,1 9 8 2 4 6 013 In other words, they found the correct answer to the problem 246 × 13.

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next In terms of the adjective/noun theme, the fact that 1 × 246 = 246 means that 1 hundred × 246 = 246 hundreds. In the language of place value this says that… © 2010 Herb I. Gross next And since 103 × 246 is greater than 100 × 246, it means that 103 × 246 has to be greater than 24,698. Thus, 2,698 is too small to be the correct answer. 100 × 246 = 2,400 next

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© 2010 Herb I. Gross On the other hand, using the adjective/noun theme we see that… × 1 0 3 2 4 6 7 3 8 2 4 6 0 0 2 5,3 3 8 (three 246’s) (one hundred 246’s) (one hundred three 246’s) next

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The ancient Egyptians anticipated the binary number system long before the invention of either place value or computers. More specifically, they realized that every non-zero whole number could be expressed as a sum of powers of 2. © 2010 Herb I. Gross The Ancient Egyptian Method of Duplation (Enrichment)

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next © 2010 Herb I. Gross Classroom Activity next To appreciate the Method of Duplation, you might want to have students work on the following project. Have them pretend that place value was based on trading in by two’s rather than by ten’s.

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To have this seem more relevant, have them consider a monetary system in which the only denominations are bills of denomination $1, $2, $4, $8, $16, etc. © 2010 Herb I. Gross Then have them “discover” how any whole number of dollars can be expressed, using no more than one of any given denomination. For example, $19 = $16 + $2 + $1. 1 1 2 2 4 4 16 8 8 2 2 1 1 ++ next

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The Method of Duplation is a rather elegant way of performing rapid addition by knowing only how to multiply by 2 and adding. © 2010 Herb I. Gross 1 × 67 = 67 For example, to use the Method of Duplation to find the product 19 × 67 the ancient Egyptians would first notice that 19 = 16 + 2 + 1, after which would make the following table just by knowing how to double a number (hence, the term, duplation)… 2 × 67 = 134 4 × 67 = 268 8 × 67 = 536 16 × 67 = 1072

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next Then they would check the powers of 2 that were used in arriving at 19 as the sum as well as the corresponding products. That is… © 2010 Herb I. Gross 1 × 67 = 67 2 × 67 = 134 4 × 67 = 268 8 × 67 = 536 16 × 67 = 1072 Finally, they would add the checked products, in this case obtaining as the sum 67 + 134 + 1072 = 1273. 67 134 1072

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next We can check the duplation with our traditional approach… © 2010 Herb I. Gross 4 × 6 7 = 2 6 8 8 × 6 7 = 5 3 6 1 6 × 6 7 = 1 0 7 2 1 9 × 6 7 1 3 3 1 1 4 1 2 7 3 6 7 × 1 9 6 0 3 6 7 1 2 7 3 2 × 6 7 = 1 3 4 1 × 6 7 = 6 7 1 9× 6 7 = next

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Here we have another subtle application of the “adjective/noun” theme. © 2010 Herb I. Gross next Note sixteen 67’s + two 67’s + one 67 = nineteen 67’s Namely, since 16 + 2 + 1 = 19… next

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© 2010 Herb I. Gross Classroom Activity Have the students do several problems using the Method of Duplation, and then have them check each answer by using the traditional algorithm.

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next © 2010 Herb I. Gross Classroom Activity Such an activity not only helps them learn several things (including an introduction to binary numbers and an application of the Distributive Property), but it gives them a “painless” motivation for practicing with the traditional algorithm in order to check the answers obtained by Duplation.

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next © 2010 Herb I. Gross In the next part of this lesson, we will discuss “unmultiplying” or as it is better known, division. 6 7 8 ÷ 2 4

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next The Distributive Property helps to explain the logic that is involved in the traditional multiplication algorithm. © 2010 Herb I. Gross × 4 3 7 2 8 1 9 6 8 4 1 2 2 3 6 1 1 2 For example, the multiplication algorithm for finding the sum of 437 twenty eight’s is… 7 3 4 next

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In the format to the right, the nouns have been omitted. However, if we put them in, it becomes easy to see what is happening. © 2010 Herb I. Gross × 4 3 7 2 8 1 9 6 8 4 1 2 2 3 6 1 1 2 For example, when we multiplied 3 by 2, we were really multiplying 3 tens by 2 tens; and according to our adjective/noun theme 3 tens × 2 tens = 6 “ten tens” or 6 hundred. next

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In fact, the “carrying” process may obscure the fact that what we really did was use the Distributive Property to obtain… © 2010 Herb I. Gross × 4 3 7 2 8 14 56 6 24 8 38 38 56 8 32 next

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That is, the product of 28 and 437 can be represented by the sum of 8 thousands, 38 hundreds, 38 tens and 56 ones; after which we simply kept “trading in” 10 of any power of ten for 1 of the next greater power of ten to obtain… © 2010 Herb I. Gross × 4 3 7 2 8 14 56 6 24 8 38 38 56 8 32 next 8 38 43 6 8 42 3 6 12 2 3 6

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next Notice how the areas of each piece match the set of partial sums we obtained using the algorithm. That is… © 2010 Herb I. Gross next Ten ThousandsThousandsHundredsTensOnes 28 × 437 5 5 6 140 240 600 3 200 8000 400 × 20 8000 400 × 8 3200 30 × 20 600 30 × 8 240 7 × 20 140 7 × 8 56

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