multiplying whole numbers © math as a second language all rights reserved next #5 taking the fear...

MultiplyingWhole Numbers

MultiplyingWhole Numbers

© Math As A Second Language All Rights Reserved

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#5

Taking the Fearout of Math

9× 9 81

Extending Single Digit Multiplication

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Continuing the Evolution of Single Digit Multiplication

It is not a particularly noteworthy saving of time to write, 4 × 7 in place of 7 + 7 + 7 + 7.

However, with respect to our “boxes of candy” situation that we discussed in single digit multiplication, suppose we wanted to

buy 400 boxes at a cost of $7 per box.

It would indeed be very tedious to write explicitly the sum of four hundred 7’s;

that is… 7 + 7 + 7… + 7

400

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The above problem, when stated as a multiplication problem, should be

written as 400 × 7.

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However, writing the problem as 7 × 400 gives us the equivalent but simpler addition problem…

400 + 400 + 400 + 400 + 400 + 400 + 400.1

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note

1 However, this obscures the fact that we want the sum of four hundred 7’s; not the sum of seven 400’s. While the answer is the same, the “mental image”

is quite different.

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Using the adjective/noun theme there is another way to visualize the “quickway” of multiplying by 400. Once we

know the “number fact” that 4 × 7 = 28, we also know such facts as…

4 × 7 apples = 28 apples

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4 × 7 lawyers = 28 lawyers

4 × 7 hundreds = 28 hundreds

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The latter result, stated in the language of place value (replacing the noun

“hundred” is the same as multiplying by 100 which is the same as annexing two 0’s)

says that 4 × 700 = 2800 (that is, 2,800).

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In other words, once we know that4 × $7 is $28, we also know that

400 × $7 is $2,800.2

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2 Going from 4 ×700 to 400 × 7 might have seemed a bit abrupt. It follows directly from our rules. More specifically,

4 × 700 = 4 × (7 × 100) = 4 × (100 × 7) = (4 × 100) × 7 = 400 × 7.

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This observation gives us an insight to rapid addition. Using our adjective/noun theme, there is no need to know anything beyond the traditional multiplication table

for single digits.

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As an example, let’s make amultiplication table for a number such as

13 which isn’t usually included aspart of the traditional multiplication tables.

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The idea is that we can think of 13 as being an abbreviation for the sum of 1 ten and 3 ones. Thus, a “quick” way to add thirteen is to add 1 in the tens place and

then 3 in the ones place.

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For example, starting with 13,

13+ 10

+ 3

26

1 × 13 =

2 × 13 =

we add 10 to get 23, and then

add 3 ones to get 26.

23

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Starting with 26,

26

+ 10

+ 3

39

1 × 13 = 13

3 × 13 =



36

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2 × 13 =

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Continuing with 39,

39+ 10

+ 3

52

1 × 13 = 13

3 × 13 =



49

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2 × 13 = 26

4 × 13 =

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In this way, our multiplication table for 13 would be…

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

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Suppose you now wanted to find the price of purchasing 234 items, each

costing $13. You could count by 13’s until you got to the 234th multiple.

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This would be both tedious and unnecessary! However, once we know the “13 table” through 9, we can use the

adjective/noun theme to find the answer in a relatively easy way.

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Imagine that the 234 items were stacked into three piles with

200 in the first pile, 30 in the second pile, and

4 in the third pile.

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

next2 × 13 = 26 200 × 13 = 2,600

3 × 13 = 39 30 × 13 = 390

The value of the items in the second pile is $390.

4 × 13 = 52

Finally, the value of the items in the third pile is $52.

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Hence, the answer to our question is $2,600 + $390 + $52 = $3,042.

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With this technique, 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.

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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 greatest number of

digits is written on top.

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7 0 2

2 3 4× 1 3

2 3 4

3, 0 4 2

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5 2

1 3× 2 3 4

3 9

3, 0 4 2

2 6

rather than…

Therefore, we often write…

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Notice that in the this format we are actually finding the cost of 13 items, each of

which costs $234. This is not the problem we intended to

solve, even though it gives us the same answer.

7 0 2

2 3 4× 1 3

2 3 4

3, 0 4 2

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Notice that in this format, we capture what it means to find

the sum of 234 thirteen’s

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In fact, this is precisely how we solved the problem originally, but it is now written in

place value notation.

5 2

1 3× 2 3 4

3 9

3, 0 4 2

2 6

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For example, when we wrote “39” we placed the 9 under

the 5, thus puttingthe 9 in the tens place.

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5 2

1 3× 2 3 4

3 9

3, 0 4 2

2 6

In other words, since the 5 was already holding the tens place there is no need for us to write a 0 next to the 9. However, if we

wish to, we can write the zero.

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This form is a shorter version of…next

5 2

1 3× 2 3 4

3 9

3, 0 4 2

2 6

which is itself a shorter version of…

00 0

= 4 thirteen’s= 30 thirteen’s= 200 thirteen’s

= 234 thirteen’s

…and in this form we see immediately the connection between the traditional

algorithm and rapid repeated arithmetic.

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Sometimes there is a tendency to confuse things foreign to us with what we believe is

illogical.

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5 2

1 3× 2 3 4

3 9

3, 0 4 2

2 6

7 0 2

2 3 4× 1 3

2 3 4

3, 0 4 2

traditional non- traditional

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While both forms give the same answer, they represent different questions.

The traditional format shows us the sum of thirteen 234’s; while the non-traditional format shows the sum of two hundred

thirty-four 13’s.

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Notice, however, that in both formats each digit in the top number multiplies each digit

in the bottom number.

This lead us to…

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The Generalized Distributive Property

We learned the distributive property in the form b(c + d) = bc + bd.

The Generalized Distributive Property To multiply a sum of numbers by another sum of numbers, we form the sum of all possible

products where one factor comes from the first sum, and the second factor comes from the second sum. For example, to compute the

product of 3 + 4 + 5 and 8 + 9, we could form sum…

(3 × 8) + (3 × 9) + (4 × 8) + (4 × 9) + (5 × 8) + (5 × 9)

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In real life, 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. However, while we can

simplify 3 + 4 + 5 when we are doing arithmetic, in algebra it is not possible to

simplify b + c + d in a similar manner.

Note

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Aside from any other of its practical uses, it is essential to know the generalized

distributive property, it if we want to rewrite an expression in which letters are used to

represent numbers.

nextNote

For example, to form the product of b + c + d and e + f, we might use the

distributive property to write the product in the form…

(b × e) + (b × f) + (c × e) + (c × f) + (d × e) + (d × f)

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The Adjective/Noun Theme And the Multiplication Algorithm

To see how the adjective/noun theme

was used in the multiplication

algorithm, let’s revisit a typical computation

such as…

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2 84 3 7×

1 9 68 4

1 1 2

1 2, 2 3 6note

3 We have deliberately written the number with the few digits on top to help you internalize the fact that this is just as logical as doing it the traditional way. Students might like to do the problem the traditional way as well to see that

the answer is the same in both cases.

3

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In this format, the nouns have been omitted. However, if we put them in, it becomes

easy to see what is happening.

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For example, when we multiplied the 3 in 437 by the 2 in 28 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.

2 84 3 7×

1 9 68 4

1 1 2

1 2, 2 3 6

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In this sense, we can view the above algorithm in the form…

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thousands hundreds tens onesten thousands

2 8

4 3 7 5 6 1 4 2 4 6

3 2 8 11 11 13 6 11 12 3 6 12 2 3 6 2 2 3 6 1

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7 ones × 8 ones = 56 ones

7 ones × 2 tens = 14 tens

3 tens × 8 ones = 24 tens

3 tens × 2 tens = 6 hundreds

4 hundreds × 8 ones = 32 hundreds

4 hundreds × 2 tens = 8 thousands

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For students who are visual learners, we can explain the above algorithm in terms of an area model. Imagine that there is a rectangle whose dimensions are 28 feet by 437 feet.

nextNote on the Area Model

437 ft

28 ft

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On the one hand, the area of the rectangle is 28 feet × 437 feet or 12,236 square feet (that is, 12,236 “feet feet” or 12,236 ft2). On the other hand, we can

compute the same area by subdividing the rectangle as shown below.

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400

20

8

30 7

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Computing the area of each smaller rectangle, we obtain…

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400

20

8

30 7

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8,000 600 140

3,200 240 56

400 30 7

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Notice how the areas of each piece

match the set of partial sums

we obtained using the algorithm.

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5 6

thousands hundreds tens onesten thousands

2 8

4 3 7

1 4 0 2 4 0 6 0 0

3 2 0 0 8 0 0 0 2 2 3 6 1

20

8

30 7400

8,000 600 140

3,200 240 56

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The Ancient Method of Duplation

The ancient Egyptians anticipated the binary number system long before the invention of

either place value or computers.

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They realized that every natural number could be expressed as a sum

of powers of 2.

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

how to multiply by 2 and adding.

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To appreciate the Method of Duplation you might 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, consider a monetary system in which the only

denominations are$1, $2, $4, $8, $16, etc.

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$1$2$4$8$16$32$64$128

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For example, to give Rick $19, you could give him a $16 bill, leaving a balance of $3,

and then give him a $2 bill and a $1 bill.

$1$2$4$8$16$32$64$128

16 2 1+ + = 19

* * *

nextnext In this form, you would never have to give someone more than one bill of each

denomination.

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next So for example, to compute the sum of

nineteen 67’s (that is: 19 × 67), the Egyptians would make the following table just by

knowing how to double a number (the term, duplation).

1 × 67 = 672 × 67 = 1344 × 67 = 268

8 × 67 = 536

16 × 67 = 1,07432 × 67 = 2,148

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next And knowing that 19 sixty-sevens was the sum of 16 sixty-sevens, 2 sixty-sevens,

and 1 sixty-seven, they would simply perform the addition as shown below…

4 × 67 = 268

8 × 67 = 536

32 × 67 = 2,148

67

134

1,07416 × 67 = 1,074

2 × 67 = 134

1 × 67 = 67

1,273

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Here we have another subtle application of the “adjective/noun” theme.Namely, since 19 = 16 + 2 + 1,

nextNote

19 sixty-sevens

16 sixty-sevens 2 sixty-sevens + 1 sixty-sevens

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In our next presentation, we

introduce the concept of unmultiplying ( our

name for division).


19 × 67multiplication

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