mathematics as a second language mathematics as a second language mathematics as a second language...

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

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Arithmetic RevisitedArithmetic Revisited

Whole Number Arithmetic

Whole Number Arithmetic

© 2010 Herb I. Gross

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Multiplication

Lesson 2 Part 3.2

The Role of Place Value in theDevelopment of Whole Number

Arithmetic --- Multiplication

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We ended the first part of this lesson by listing the first nine multiples of 13. By way of review…


1 × 13 = 132 × 13 = 26

3 × 13 = 39

4 × 13 = 52

5 × 13 = 656 × 13 = 787 × 13 = 91

8 × 13 = 1049 × 13 = 117

nextnext

Suppose we wanted to use the table to the right to compute the

cost of buying 9 items, each of which cost $13.


1 × 13 = 132 × 13 = 263 × 13 = 394 × 13 = 525 × 13 = 656 × 13 = 787 × 13 = 918 × 13 = 1049 × 13 = 117

The table shows us that 9 × 13 = 117; from which we would

conclude that the cost was $117.

9 × 13 = 117

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 234th 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.


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.


2 hundreds + 3 tens + 4 ones

200 + 3 0 + 4

next

13 × 2 apples = 26 apples


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.

nextnext

next

In the language of place value we write…


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

next

This one step takes the place of our having to count

to the 200th 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 200th line on our list would have

read 200 × 13 = 2,600.


next

In other words, we already know that at a price of $13 each, 234 items would cost

more than $2,600.

Note

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.


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.

next

In a similar way, we know that…


13 × 3 = 39 means that 30 × 13 = 390

and

13 × 200 = 2600next

13 × 4 = 52

next

next

The value of the items in the first pile is $2,600.


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

nextnext200 30 4

$2,600 $390 $52

next

In this venue, it is relatively easy to see how multiplication is really a specialformat for organizing rapid, repeated addition. However, in the traditional

format in which multiplication is presented, this clarity is either lacking or obscured.


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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.

next

That is, we often write…


…rather than…

2 3 4next

× 1 3 7 0 2

2 3 43 0 4 2

× 2 3 4 1 3

5 23 9 0

3 0 4 2

2 6 0 0

next

Notice that in this format we are

actually finding the cost of 13 items,

each of which costs $234.


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 43 0 4 2

next

However, the second form is a more compact

version of the method we used above to solve the

problem.


next

× 2 3 4 1 3

5 23 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.

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.


next

× 2 3 4 1 3

5 23 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 0

and annexing the twozeroes we obtain…

next

next


…and this in turn is a shorter version of…

× 2 3 4 1 3

5 23 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.

nextnext

next

Comparing 234 × 13…


next

2 3 4× 1 3 7 0 2

2 3 43 0 4 2

× 2 3 4 1 3

5 23 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

next

This format results in

finding the 13th multiple of 234.


next Note

2 3 4

× 1 3

× 2 3 41 3

On the other hand, this format results in

finding the 234th multiple of 13…which was our goal in this

problem.

next

However, both formats have the same property.

Namely each digit in one factor multiplies each digit in the other factor.


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.

The Distributive Propertynextnext

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).


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.

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…


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.

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).


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…

next To see why the Distributive Property is plausible, it might be helpful to think in terms of the area of a rectangle.


next

For example, in the diagram below, a, b, c, d, and e

represent lengths.

a b c

d

e

Hence, the length (base) of the rectangle is given by a + b + c

a b c

, and the width (height) is given by d + e.

d

e

nextnext

next 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).


next

On the other hand, it is also the sum of the areas of the 6 smaller rectangles .

a b c

dddd

aa

eeee

bb cc

× × ×

× × ×

next

next

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.


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

next In this sense, given a problem such as437 x 28, we can view the multiplication

algorithm in the form…


nextnextnextnextnext

Ten Thousands Thousands Hundreds Tens Ones

2 8 × 4 3 7 55 6 1 4

7

2 4 6

3

3 2 + 8

4

00

0 00 0

0 0 0

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.


next Note on Area model

437

28

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 ft2)



437

28 437 feet × 28 feet = 12,236 ft2

next

On the other hand, we can compute the same area by

subdividing the rectangle as shown below.



730400

20

8

nextnextnextnextnextnext

11,200 840 196

+

+

= 12,236

next

7 × 8 56 30 × 8 240

7 × 20140 30 × 20 600 400 × 20 8000

400 × 8 3200

nextnext

next In this sense, we can rewrite the bottom row in the chart below to obtain…


next

Ten Thousands Thousands Hundreds Tens Ones

2 8 × 4 3 7 55 6 1 4 2 4 6 3 2

+ 8 11 11 13 6 11 12 3 6 12 2 3 6

1 2 2 3 6

next The 3 models are summarized below.


nextnext

400 × 20

400 × 8

30 × 20

30 × 8

7 × 20

7 × 8

730400

20

8

Area Model

8000 600 140

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 6

8 4 0

1 1,2 0 0

3 4 9 6

8 7 4 0

1 2,2 3 6

1 2,2 3 6

nextnextnextnextnextnext

next

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.


Beware of the Missing Zeronext

next

Thus, the correct way to compute

a product such as…


is to write…

× 1 0 3 2 4 6

× 1 0 3 2 4 6

7 3 80 0 0

2,5 3 3 8

2 4 6

0

1 3

nextnextnextnext

next

However students are often tempted to “ignore”

the 0 and instead compute the product

as follows…


× 1 0 3 2 4 6

7 3 80 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

0

1 3

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.


× 1 0 3 2 4 6

7 3 80 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

0

1 3

In other words, they found the correct answer to the problem 246 × 13.

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…


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

next


On the other hand, using the adjective/noun theme we see that…

× 1 0 3 2 4 6

7 3 82 4 6 0 0

2 5,3 3 8

(three 246’s)(one hundred 246’s)

(one hundred three 246’s)

nextnext

next

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.


The Ancient Egyptian Method of Duplation (Enrichment)

next


Classroom Activitynext

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.

nextnext To have this seem more relevant, have them consider a monetary system in which

the only denominations are bills ofdenomination $1, $2, $4, $8, $16, etc.


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.

11

22

44

1616

88

1616

22

11

+ +

nextnextnext

nextnext The Method of Duplation is a rather elegant way of performing rapid addition by knowing

only how to multiply by 2 and adding.


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

nextnext 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…


1 × 67 = 672 × 67 = 1344 × 67 = 2688 × 67 = 536

16 × 67 = 1072

Finally, they would add the checked products, in this case obtaining as the

sum 67 + 134 + 1072 = 1273.

67134

1072

nextnext We can check the duplation with our traditional approach…


4 × 6 7 = 2 6 88 × 6 7 = 5 3 6

1 6 × 6 7 = 1 0 7 2

1 9× 6 7 1 3 3

1 1 41 2 7 3

6 7× 1 9 6 0 36 7

1 2 7 31 2 7 3

2 × 6 7 = 1 3 4 1 × 6 7 = 6 7

1 9 × 6 7 =

nextnextnext

next

Here we have another subtle application of the “adjective/noun” theme.


next Note

sixteen 67’s + two 67’s + one 67 = nineteen 67’s

Namely, since 16 + 2 + 1 = 19…

next

next


Classroom Activity

Have the students do several problems using the Method of Duplation, and then have them check each answer by

using the traditionalalgorithm.

next


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.

next


In the next part of this lesson, we will discuss

“unmultiplying” or as it isbetter known, division.

6 7 8 ÷ 2 4

next The Distributive Property helps to explain the logic that is involved in the

traditional multiplication algorithm.


× 4 3 7 2 8

1 9 68 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

nextnextnextnextnext

next

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.


× 4 3 7 2 8

1 9 68 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

next In fact, the “carrying” process may

obscure the fact that what we really didwas use the Distributive Property to obtain…


× 4 3 7

2 8

14 56

6 24

8 38 38 56

8 32

next

next

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…


× 4 3 7 2 8

14 566 24

8 38 38 56

8 32

next

8 38 43 6

8 42 3 612 2 3 6

1 2 2 3 6

next Notice how the areas of each piece match the set of partial sums we obtained using the

algorithm. That is…


nextnext

Ten Thousands Thousands Hundreds Tens Ones 2 8 × 4 3 7

55 6 1 4 0 2 4 0 6 0 0 3 2 0 0

8 0 0 0

400 × 20 8000

400 × 8 3200

30 × 20 600

30 × 8 240

7 × 20140

7 × 8 56

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