5. Place Value Multiplication - Single Digits

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Transcript 5. Place Value Multiplication - Single Digits

Taking the Fear
out of Math
next
#5
Multiplying
Whole Numbers
Single Digit Multiplication
9
×9
81
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Single Digit Multiplication
Introduction
Because of your professional
backgrounds, we assume that you all
know the traditional algorithms for
multiplication. However, if your students
memorize the algorithms without
properly understanding them some very
serious errors involving critical thinking
can occur.
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So to begin, we will accept the fact
that most students can perceive that they
have correctly assimilated the various
multiplication algorithms by rote; yet
because of subtleties that they overlook,
they often make serious errors by not
understanding each step in the process.
For example, in computing a product such
as 415 ×101, they often disregard the 0
because it is “nothing”.
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In this context, they might write…
415
× 101
415
415
4565
1
note
1 By
ignoring the zero, the student is computing the product 415 × 11
for which the correct answer is 4,565.
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By understanding the algorithm
415
(and in this context it makes no × 1 0 1
difference whether the logic is
415
“teacher driven” or a form of
415
“student discovery”) students
should realize that 415 × 101 must 4 5 6 5
be greater than 415 × 100, and
since 415 × 100 = 41,500 it is clear
that 415 × 101 > 41,500.
There are many numbers that are greater
than 41,500 but 4,565 isn’t one of them .
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Notes
In terms of our adjective/noun theme, there
are more “concrete” explanations that can
be helpful to the more visually oriented
students.
For example, in terms of money (something
all students can relate to), the person who
has 101 checks each worth $415 has $415
more than the person who has only 100
checks that are worth $415 each.
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Notes
Using a calculator does not make us
immune from making the error
described above.
For example, even with a calculator we can
strike a key too lightly to have it register and
we can also type a number incorrectly. So
even when we use a calculator to compute
415 × 101 we should still be aware of such
“advance information” as
415 × 101 > 41,500.
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Evolution of Multiplication
of Whole Numbers
In order to the understand the traditional
whole number multiplication
algorithm, students should be
nurtured to understand the
concept of rapid repeated addition.
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Suppose you are buying 4 boxes of candy
that cost $7 each. We could think of asking
two questions based on this information.
(1) How many boxes of candy did you buy?”
In this case we can see directly that the
adjective 4 is modifying the noun phrase
“boxes of candy”.
(2) “How much did the 4 boxes of candy cost?”
Explicitly the 4 is still modifying “boxes of
candy”, but to answer the question we see that
it is being used to ask us how many times we
are spending $7.
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In answering these two questions, we
can use the adjective 4 in two ways. Namely,
to answer the first question, we could count
“1 box, 2 boxes, 3 boxes, 4 boxes”, and to
answer the second question, we could
count…
“$7, 1 time;
$7, 2 times ($7 + $7);
$7, 3 times ($7 + $7 + $7);
$7, 4 times ($7 + $7 + $7 + $7)”.
And in this context, we are using 4 to modify
the number of times we are spending $7.
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Notes
The mathematical way of writing “$7 four
times is to write 4 × $7.
4 × $7 is called the 4th multiple of
7(dollars).
No matter what number 7 modifies, 4 × 7
is called the 4th multiple of 7.
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Definition
In the expression 4 × 7, 4 and 7 are
called the factors, and 4×7 is called
the product of 4 and 7.
Notes
We tend to confuse 4 × 7 with 7 × 4.
The fact is that while the product in both
case is 28, the concepts are quite different.
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Notes
For example, we are viewing 4 × 7 as the
4th multiple of 7; that is…
7 + 7 + 7 + 7,
…and we are viewing 7 × 4 as the
7th multiple of 4; that is…
4 + 4 + 4 + 4 + 4 + 4 + 4.
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Notes
Clearly, these two sums look different!
7 + 7 + 7 + 7,
4 + 4 + 4 + 4 + 4 + 4 + 4.
So while students are willing to accept
the fact that 7 × 4 = 4 × 7, there is a
conceptual difference between buying
4 pens at $7 each and buying 7 pens at $4
each (even though the cost is the same in
both cases).
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Internalizing the Multiplication Tables
There are probably many reasons why
some students seem to have difficulty
internalizing the multiplication tables.
Based on our experience, the major
reason is that students tend to
memorize the tables without any
reference to number sense.
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Internalizing the Multiplication Tables
Our hope is that by giving students
insights as to what the various products
mean. They will feel more comfortable
learning the multiplication tables, and
with this new-found comfort,
memorization will occur through
continued use and practice.
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Most students are comfortable learning
to “skip count” by 2’s, 3’s, 4’s and 5’s.
Note
“Skip counting” is an informal way of
listing multiples. For example, when we
“skip count” by 3’s and say, “3, 6, 9, 12...”
we are really saying, “1×3 = 3, 2×3 = 6,
3×3 = 9, 4×3 = 12…”.
In terms of our adjective/noun theme we
are saying “1 three, 2 threes, 3 threes,
4 threes…”
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The difficulty lies mostly in internalizing
the multiples of 6, 7, 8 and 9. With this in
mind, let’s look at the multiples of 9 in
more detail.
The trouble with skip counting by 9’s is
that students tend to count on their
fingers, and it is tedious for them to have
to count up to 9.
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For example, if they know that 3×9 = 27,
they tend to compute 4×9 by starting with
27 and then counting on their fingers until
they arrive at the correct answer…
27 28
1
29 30 31 32 33 34
2
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4
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5
6
7
35 36
8
9
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However, most students have no
trouble thinking in terms of money. Thus,
for example, if they were buying items that
cost $9 each, most likely they would
understand that it would be more
convenient to give the clerk a $10- bill and
get back a $1-bill in change than
to tediously count out 9 $1-bills.
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In more mathematical terms, they are
thinking of $9 in the form $10 – $1
rather than in the form of 9 × $1.
This translates into the more abstract
form that to add 9 to a number we can,
instead, add 10 and then subtract 1.
Thus, knowing that 1 × 9 = 9, we can
compute 2 × 9 by starting with 9 and
adding 10 to obtain 19 and then
subtracting 1 to obtain 18.
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Proceeding in this way, we can quickly
write down the “9’s table” by starting with
1 × 9 = 9 and then proceed from multiple
to multiple by adding 1 to the tens place
and then subtracting 1 from the ones
place each time.
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In this
way
we see
that…
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1×9=
09
2×9=
9 + 10 – 1 =
18
3×9=
18 + 10 – 1 =
27
4×9=
27 + 10 – 1 =
36
5×9=
36 + 10 – 1 =
45
6×9=
45 + 10 – 1 =
54
7×9=
54 + 10 – 1 =
63
8×9=
63 + 10 – 1 =
72
9×9=
72 + 10 – 1 =
81
10 × 9 =
81 + 10 – 1 =
90
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The fact that we added 1 to the tens
place and subtracted 1 from the ones place
means that we have not changed the sum
of the digits. Therefore, since 0 + 9 = 9, the
remaining sums will also equal 9.
In other words, if a 2-digit number is divisible
by 9, the sum of its digits is equal to 9.2
note
2 Possibly,
because of how much alike 56 and 54 look, even students who are
relatively fluent with the multiplication facts often confuse 8 × 7 with 9 × 6.
However, based on our above observations, since 5 + 4 = 9, and 5 + 6 ≠ 9, we
would know that 54 is a multiple of 9, but that 56 isn’t.
From this, we would deduce that 9 × 6 = 54.
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Notes
The above discussion can be applied to
any multi-digit number that has
a 9 in the ones place.
For example, if we bought an item for $39
we would most likely pay for it by giving
the clerk 4 $10-bills (or 2 $20-bills) and
then receiving a $1-bill as change. In
essence, to get from one multiple of 39 to
the next we would add 40 (or in terms of
place value, 4 in the tens place) and then
subtract 1 from the ones place.
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In this
way
we see
that…
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1 × 39 =
39
2 × 39 =
39 + 40 – 1 =
78
3 × 39 =
78 + 40 – 1 =
117
4 × 39 =
117 + 40 – 1 =
156
5 × 39 =
146 + 40 – 1 =
195
6 × 39 =
195 + 40 – 1 =
234
7 × 39 =
234 + 40 – 1 =
273
8 × 39 =
273 + 40 – 1 =
312
9 × 39 =
312 + 40 – 1 =
351
10 × 39 =
351 + 40 – 1 =
390
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Generalizing the Multiples 6, 7, and 8
We can compute the multiples of 6, 7,
and 8 in a way that parallels what we did
for the multiples of 9.
For example, suppose we buy an item for
$8 and we give the clerk a $10-bill.
The clerk then gives us back $2.
With this in mind, let’s compute, 7 × 8.
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We may think of it as buying 7 items,
each of which costs $8. Each time we
give the clerk $10, he gives us back $2.
So to pay for the 7 items, we give the
clerk $10 seven times (that is, $70) and
the clerk gives us back $2 seven times
(that is, $14). Thus, all in all, we paid
$70 – $14 or $56.
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In more abstract terms, we may view
8 in the form 10 – 2. In that way, we
essentially replace adding 8 by
adding 10 and subtracting 2, using
the distributive property3.
note
3 We
usually think of the distributive property in the form
a × (b + c) = (a + b) × (a + c).
However, it is often expressed in the form a × (b – c) = (a – b) × (a – c).
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In this way we obtain…
1 × 8 = 1 × (10 – 2) = (1 × 10) – (1 × 2) = (10 – 2) =
8
2 × 8 = 2 × (10 – 2) = (2 × 10) – (2 × 2) = (20 – 4) =
16
3 × 8 = 3 × (10 – 2) = (3 × 10) – (3 × 2) = (30 – 6) = 24
4×8=
4 × (10 – 2) = (4 × 10) – (4 × 2) = (40 – 8) = 32
5×8=
5 × (10 – 2) = (5 × 10) – (5 × 2) = (50 – 10) = 40
6×8=
6 × (10 – 2) = (6 × 10) – (6 × 2) = (60 – 12) = 48
7×8=
7 × (10 – 2) = (7 × 10) – (7 × 2) = (70 – 14) = 56
8×8=
8 × (10 – 2) = (8 × 10) – (8 × 2) = (80 – 16) = 64
9×8=
9 × (10 – 2) = (9 × 10) – (9 × 2) = (90 – 18) = 72
10 × 8 = 10 × (10 – 2) = (10 × 10) – (10 × 2) = (100 – 20) = 80
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We may obtain another version of the
above table by recognizing that
we can skip count by 8’s simply by adding
10 and then subtracting 2.
So starting with 8 × 1 = 8, we might say
“8 plus 10 is 18 and 18 minus 2 is 16,
16 plus 10 is 26 and 26 minus 2 is 24,
24 plus 10 is 34 and 34 minus 2 is 32,
32 plus 10 is 42 and 42 minus 2 is 40,
40 plus 10 is 50 and 50 minus 2 is 48,
48 plus 10 is 58 and 58 minus 2 is 56,
56 plus 10 is 66 and 66 minus 2 is 64;
and 64 plus 10 is 74 and 74 minus 2 is 72”.
next
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And if we wanted to skip count by 7’s we
could instead add 10 and then subtract 3.
So starting with 7 × 1 = 7, we might say
“7 plus 10 is 17 and 17 minus 3 is 14,
14 plus 10 is 24 and 24 minus 3 is 21,
21 plus 10 is 31 and 31 minus 3 is 28,
28 plus 10 is 38 and 38 minus 3 is 35,
35 plus 10 is 45 and 45 minus 3 is 42,
42 plus 10 is 52 and 52 minus 3 is 49,
49 plus 10 is 59 and 59 minus 3 is 56,
and 56 plus 10 is 66 and 66 minus 3 is 63”.
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Other Ways to Internalize the
Multiplication Tables
Most students prefer to add rather than
to subtract.
With that in mind, our adjective/noun theme
allows us to construct the rest of the single
digit multiplication tables once we know
how to “skip count” by 2’s, 3’s, and 4’s.
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For example, suppose we want to find
the sum of 7 sixes (that is, 7 × 6).
Using the fact that 7 = 3 + 4, we may
proceed as follows…
7 × 6 = 7 sixes = 3 sixes + 4 sixes
18
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+
24
= 42
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Viewing a product as the sum of
smaller products is a good way to help
students begin to develop a number sense.
For example, if a student recognizes that
6 × 7 is twice 3 × 7 and that 3 × 7 = 21, the
student will then know that 6 ×7 = 2 × 21 =
21 + 21 = 42.4
note
4
More formally, 6 × 7 = (2 × 3) × 7 = 2 × (3 × 7) = 2 × 21 = 21 + 21 = 42.
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For further practice, let’s use the fact
that 5 = 3 + 2 to construct the 5’s table.
5×1=
5 ones =
3 ones + 2 ones =
3
5×2=
5 twos =
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2
3 twos + 2 twos =
6
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+
5
4
10
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5 × 3 = 5 threes =
5×4=
5 fours =
3 threes + 2 threes = 15
9
+
6
3 fours + 2 fours =
12
5×5=
5 fives =
5 sixes =
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+
25
10
3 sixes + 2 sixes = 30
18
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3 fives + 2 fives =
15
5×6=
+
20
+
12
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5 × 7 = 5 sevens = 3 sevens + 2 sevens =
21
5 × 8 = 5 eights =
5 fives =
5 tens =
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+
+
40
16
45
18
3 tens + 2 tens =
30
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+
3 nines + 2 nines =
27
5 × 10 =
14
3 eights + 2 eights =
24
5×9=
+
35
20
50
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There are many exercises such as the
ones shown previously that should help
students develop a better number sense.
multiplication
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9×9
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In our next
presentation, we shall
show how by using the
adjective/noun theme,
we can multiply any
number of multi-digit
numbers just by
knowing the
multiplication tables
through 9.