Transcript 7. Place Value Extended Multiplication

Taking the Fear
out of Math
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#5
Multiplying
Whole Numbers
Extending Single Digit
Multiplication
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9
×9
81
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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
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400
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Note
The above problem, when stated as a
multiplication problem, should be
written as 400 × 7.
However, writing the problem as 7 × 400
gives us the equivalent but simpler addition
problem…
400 + 400 + 400 + 400 + 400 + 400 + 400.1
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 “quick
way” 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
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).
In other words, once we know that
4 × $7 is $28, we also know that
400 × $7 is $2,800.2
note
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.
2 Going
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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.
As an example, let’s make a
multiplication table for a number such as
13 which isn’t usually included as
part 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.
For example,
starting with 13,
we add 10 to
get 23, and then
add 3 ones to
get 26.
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1 × 13 =
13
+ 10
23
+ 3
2 × 13 =
26
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Starting with 26,
we add 10 to
get 36, and then
add 3 ones to
get 39.
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1 × 13 =
2 × 13 =
13
26
+ 10
36
+ 3
3 × 13 =
39
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Continuing with
39, we add 10 to
get 49, and then
add 3 ones to
get 52.
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1 × 13 =
2 × 13 =
3 × 13 =
13
26
39
+ 10
49
+ 3
4 × 13 =
52
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In this way, our multiplication table
for 13 would be…
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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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.
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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2 × 13 = 26
200 × 13 = 2,600
Thus, the value of the items in the first pile is
$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.
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.
Therefore, we often write…
13
234
×234
× 13
52
702
rather than…
39
234
26
3, 0 4 2
3, 0 4 2
next
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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.
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234
× 13
702
234
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
13
×234
52
39
26
3, 0 4 2
In fact, this is precisely how we solved the
problem originally, but it is now written in
place value notation.
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For example, when we wrote
“39” we placed the 9 under
the 5, thus putting
the 9 in the tens place.
13
×234
52
39
26
3, 0 4 2
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…
which is itself a shorter version of…
13
×234
=
4 thirteen’s
52
39 0
= 30 thirteen’s
26 00
= 200 thirteen’s
3, 0 4 2
= 234 thirteen’s
…and in this form we see immediately the
connection between the traditional
algorithm and rapid repeated arithmetic.
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Note
Sometimes there is a tendency to confuse
things foreign to us with what we believe is
illogical.
traditional
234
× 13
702
234
3, 0 4 2
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non- traditional
13
×234
52
39
26
3, 0 4 2
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Note
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.
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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Note
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.
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Note
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.
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…
×
28
437
196
84
112
1 2, 2 3 6
3
note
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.
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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.
×
28
437
196
84
112
1 2, 2 3 6
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.
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In this sense, we can view the above
algorithm in the form…
ten thousands
thousands
hundreds
4
1
3
8
11
11
12
2
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1
2
6
2
tens
ones
2
3
5
4
4
8
7
6
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
11
12
2
2
13
3
3
3
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6
6
6
6
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Note on the Area Model
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.
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.
400
20
8
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30
7
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Computing the area of each smaller
rectangle, we obtain…
400
30
7
20
8,000
600
140
8
3,200
240
56
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Notice how
the areas of
each piece
match the set
of partial sums
we obtained
using the
algorithm.
400
30
20
8,000
600 140
8
3,200
240
ten thousands
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hundreds
4
1
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thousands
3
8
2
1
2
6
2
0
2
7
56
tens
ones
2
3
5
4
4
0
0
0
3
8
7
6
0
0
0
0
0
6
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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.
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.
To have this seem more relevant, consider
a monetary system in which the only
denominations are
$1, $2, $4, $8, $16, etc.
$128 $64
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$32
$16
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$8
$4
$2
$1
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In this form, you would never have to
give someone more than one bill of each
denomination.
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.
$128 $64
$32
$16
$8
$4
*
16
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$2
$1
**
+
2 + 1 = 19
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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 = 67
2 × 67 = 134
4 × 67 = 268
8 × 67 = 536
16 × 67 = 1,074
32 × 67 = 2,148
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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…
1 × 67 = 67
67
2 × 67 = 134
134
4 × 67 = 268
8 × 67 = 536
16 × 67 = 1,074
1,074
32 × 67 = 2,148
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1,273
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Note
Here we have another subtle application
of the “adjective/noun” theme.
Namely, since 19 = 16 + 2 + 1,
16 sixty-sevens
2 sixty-sevens
+ 1 sixty-sevens
19 sixty-sevens
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multiplication
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19 × 67
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In our next
presentation, we
introduce the concept
of unmultiplying ( our
name for division).