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1.4: Multiplication of Understanding Elemementary Mathmatics


The teacher said to the student,

"Multiply 3 x 7."

The student replied,

"But I don't know how to multiply."

Then the teacher asked,

"Do you know how to add?"

The student replied,

"Sure, that's easy."

Then the teacher said,

"Well, then, you can multiply. Just add three sevens together!"

The student marveled at this notion. The student simply added 7+7+7, and answered,

"Is 3 x 7 equal to 21?"

The teacher said , "Yes! I knew you could multiply. I'm very proud of you."

The student was proud too...and very happy.

DEFINITION and PROPERTIES

This exercise set presents a different, challenging way for you to look at multiplication. You will be using the Centimeter Strips (C-strips) to explore and discover the commutative, associative and distributive properties of multiplication. Below is a reminder of the different C-Strips we'll be working with along with the abbreviation used for each highlighted in bold:

We will begin by defining a way to "multiply" two C-strips together. Consider multiplying the Purple C-strip by the Yellow C-Strip. Using abbreviations, we will write (P imes Y), which we will read aloud as "P cross Y". The first step in this "multiplication" is to place the first strip (P) down in a horizontal position and then to place the second strip (Y) underneath the first strip so that it is perpendicular to the first strip. The second step is to fill in more yellow C-strips alongside the one already there so that you form a rectangle of yellow strips where the length is the yellow strip and the width is the purple strip. Last, remove the horizontal strip (purple in this case) and form a train of the vertical strip(s) (yellow in this case). This train is (P imes Y) and is shown below. See the illustrations of the three steps.

We say (P imes Y) is the train consisting of 4 yellow C-strips. It is important to notice that the answer is a train that is formed with only yellow strips, which is the second letter in the multiplication.

Exercise 1

1.

a. Make the train (P imes Y) as explained above. Save it until you finish part d.

b. Find (Y imes P) by following the same procedure used to find (P imes Y), by doing the first two steps and showing them below.

c. This train you eventually end up with should consist of only purple C-strips. How many purple strips are there? Draw a picture of the train (Y imes P).

d. Place the two trains, (P imes Y) and (Y imes P), side by side and compare the lengths of the two trains. What do you notice?

To make notation simple, let's agree to say that two trains are equal if they have the same length. In other words, from what you discovered in Exercise 1, (P imes Y = Y imes P).

For the following exercises, describe the trains formed by doing the following multiplications with C-strips. You should physically form each train by going through the same steps used to multiply (P imes Y) on the previous page. For part a and b, fill in the first blank with a numeral (how many) and the second blank with a color. Save each train in order to answer part c. Use your observation to write the equations asked for in Exercise 1d. For instance, since you discovered (P imes Y = Y imes P), this translates to (4 imes 5 = 5 imes 4), since P is 4 cm long and Y is 5 cm long. Use this example to fill in blanks for exercises 2,3 and 4.

Example a

Example a: Find (P imes Y). This train consists of

Example b

Example b: Find (Y imes P). This train consists of

Exercise 2

2.

a. Find (R imes D). This train consists of _____ C-strips

b. Find (D imes R). This train consists of ______ C-strips

c. (R imes D) is made up of ____ white strips and (D imes R) is made up of _____ white strips.

d. What do you notice when you compare the lengths of the trains (R imes D) and (D imes R)?

e. From your work in parts a - d, write an equation including C-strips R and D; then translate the equation to one with numbers.

Exercise 3

3.

a. Find (K imes L). This train consists of _____ C-strips

b. Find (L imes K). This train consists of _____ C-strips

c. (K imes L) is made up of ____ white strips and (L imes K) is made up of _____ white strips.

d. What do you notice when you compare the lengths of the trains (K imes L) and (L imes K)?

e. From your work in parts a - d, write an equation including C-strips K and L; then translate the equation to one with numbers.

Exercise 4

4.

a. Find (H imes W). This train consists of _____ C-strips

b. Find (W imes H). This train consists of _____ C-strips

c. (H imes W) is made up of ____ white strips and (W imes H) is made up of ____ white strips.

d. What do you notice when you compare the lengths of the trains (H imes W) and (W imes H)?

e. From your work in parts a - d, write an equation including C-strips H and W; then translate the equation to one with numbers.

The above problems illustrate the commutative property of multiplication for trains.

The commutative property of multiplication for trains states that if s and t are any two trains, then (s imes t = t imes s)

Note: A train can consist of one or more C-strips. In other words, a single white (or any other color) C-strip is actually a train. There doesn't have to be a caboose and engine!

Exercise 5

5. Fill in the blanks to each of the following problems. To do this, take the color c-strip shown after the multiplication sign and make a train out of that color that is equal in length to the c-strip on the right side of the equal sign. Second, take the train and form it into a rectangle. Then find a c-strip that fits across the width of the rectangle. Write the abbreviation for that c-strip in the blank.

a. ____ ( imes R = H)b. ____ ( imes P = P)c. ____ ( imes L = D)
d. ____ ( imes R = N)e. ____ ( imes L = B)f. ____ ( imes W = K)
g. ____ ( imes R = D)h. ____ ( imes D = H)i. ____ ( imes Y = O)

Exercise 6

6.

a. Take out the Hot Pink (H) C-strip. Use your C-strips (one color at a time) to see if you can form a train made up of c-strips of the same color equal in length to the hot pink c-strip in other words, see if you can do it with all whites (always possible for any train), or all reds, or all light greens, etc. You should be able to make six different trains each train is made up of a single color. Draw a picture of each of these trains under the Hot Pink one shown. I've drawn two trains for you already one is simply the hot pink strip (a train consisting of just one c-strip), and a second is made up of three purple c-strips.

b. Your work from exercise 5 should help complete this exercise. Take each train, one at a time, and put each strip in a vertical position, but side by side to form a rectangle. Then, find a c-strip that fits across the width (or top) of the rectangle. This should tell you from which multiplication problem each train was formed. Write an equation using c-strips and then translate to an equation with numbers. For instance, for train 2, first I would make a rectangle out of the three purple c-strips. Next, I would try to find a c-strip to fit across the top, which would be light green. Therefore, train 2 was formed from the multiplication (L imes P) remember that since the train is formed with purple c-strips, P is the second letter in the multiplication. So, the equation in c-strips is (L imes P = H), and the numerical equivalent is (3 imes 4 = 12). Follow this same procedure for the other four trains you made in part a.

Train 1 illustrates the multiplication (W imes H = H), or (1 imes 12 = 12). (Note that if the hot pink strip is vertical, the white c-strip fits across the top.)

Train 2 illustrates the multiplication (L imes P) = H, or (3 imes 4 = 12)

Train 3 illustrates the multiplication ____ ( imes) ____ = H, or ____ ( imes) ____ = 12

Train 4 illustrates the multiplication ____ ( imes) ____ = H, or ____ ( imes) ____ = _____

Train 5 illustrates the multiplication ____ ( imes) ____ = H, or ____ ( imes) ____ = _____

Train 6 illustrates the multiplication ____ ( imes) ____ = H, or ____ ( imes) ____ = _____

Exercise 7

7. For train 2 in exercise 6, I made a train of purple c-strips equal in length to the hot pink c-strip. Now, make a train of purple c-strips equal in length to the brown (N) c-strip. Take the train of the purple c-strips and make it into a rectangle.

Which color c-strip measures across the width of this rectangle? ______

Put the answer (abbreviation for the c-strip) in the blank to solve the multiplication: ____ ( imes) P = N.

Now, we'll explore more about the equations formed in exercise 5. Consider part a. You should have written (D imes R = H). This is analogous to the statement (6 imes 2 = 12). But, if you actually look at the original train that was formed to find the answer, you would see that there were 6 red c-strips that made a train equal in length to the hot pink c-strip. This gives rise to one of the whole number definitions of multiplication. Multiplication is really just the same thing as Repeated Addition.

Definition: Multiplication of Whole Numbers using the repeated addition approach

If a and b are whole numbers, then a ( imes) b = b + b+ b+ b + ... + b, where there are a addends of b in this sum.

I like to think of using the definition like this:

3 ( imes) 4 means 3 fours added together or 3 ( imes) 4 = 4 + 4 + 4 = 12

4 ( imes) 3 means 4 threes added together, or 4 ( imes) 3 = 3 + 3 + 3 + 3 = 12

5 ( imes) 2 means 5 twos added together, or 5 ( imes) 2 = 2 + 2 + 2 + 2 + 2 = 10

2 ( imes) 5 means 2 fives added together, or 2 ( imes) 5 = 5 + 5 = 10

Note that although 3 ( imes) 4 gives the same answer as 4 ( imes) 3, it does not mean the same thing! One means 4 + 4 + 4 and the other means 3 + 3 + 3 + 3. Before we delve any further into this discussion, I want you to use the repeated addition definition to compute the following:

Exercise 8

8. Use the repeated addition definition of multiplication to compute the following. First, write out the meaning of the multiplication, and then compute the answer.

a. 6 ( imes) 3 =

b. 2 ( imes) 9 =

c. 9 ( imes) 2 =

Now, it may seem obvious to you that 3 ( imes) 4 is the same answer as 4 ( imes) 3. That's because you've no doubt been using the commutative property of addition for years. Whether or not you really thought about why it was true is another thing. But, although 3 ( imes) 4 gives the same answer as 4 ( imes) 3, it does not mean the same thing. One means 4 + 4 + 4 and the other means 3 + 3 + 3 + 3. To a child just learning about addition and multiplication, there is no reason for them to conclude that one sum (4 + 4 + 4) happens to give the same answer as the other sum (3 + 3 + 3 + 3). For instance, if I was asked to compute the following two sums, it wouldn't be evident to me at all that the answer to each would be end up being the same:

17 + 17 + 17 + 17 and 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4

Actually, I find it quite amazing that they are equal. But, indeed they are, because the first sum comes from the multiplication problem 4 ( imes) 17 (since there are 4 seventeens added together), and the second sum comes from the multiplication problem 17 ( imes) 4 (since there are 17 fours added together). Later, when we discuss a more geometrical definition for multiplication, it becomes obvious why the commutative property of multiplication is true.

The commutative property of multiplication for whole numbers states that if a and b are any two numbers, then a ( imes) b = b ( imes) a.

So, the very exciting thing about the commutative property of multiplication is that, given a problem like 100 ( imes) 7, where it would take a long time to write out and compute the answer using the strict definition of multiplication by adding 7 over and over again, we can use our knowledge of the commutative property, that is 100 ( imes) 7 = 7 ( imes) 100, and compute the second multiplication instead. Wouldn't you agree that 100 + 100 + 100 + 100 + 100 + 100 + 100 is quicker and easier to compute than adding together 100 sevens? Thank goodness for the commutative property of multiplication!!!

Before going on to another property, let's practice verifying that the commutative property of multiplication is true for some numbers by using the repeated addition definition of multiplication. The commutative property requires two numbers (they don't have to be two different numbers) and there are two sides of an equation to verify. The difference between the expressions a ( imes) b and b ( imes) a is that the order of the numbers is different. Remember that the commutative property means there has been an order change. To verify the commutative property of multiplication, you need to use the repeated addition definition of multiplication and add from left to right, one number at a time. The next example shows how to verify the commutative property of multiplication, given a set of two numbers.

Example

Example: Verify the commutative property of multiplication is true for the given set of numbers. You'll have to remember how to add in base six to do part b.

a. {5, 3}

Solution: I need to show (5 imes 3 = 3 imes 5). Since (5 imes 3) = 3 + 3 + 3 + 3 + 3 = 6 + 3 + 3 + 3 = 9 + 3 + 3 = 12 + 3 = 15 and since (3 imes 5)= 5 + 5 + 5 = 10 + 5 = 15, then (5 imes 3 = 3 imes 5), because both equal 15.

b. {(3_{ ext{six}}, 4_{ ext{six}})}

Solution: I need to show (3_{ ext{six}} imes 4_{ ext{six}} = 4_{ ext{six}} imes 3_{ ext{six}}) Since (3_{ ext{six}} imes 4_{ ext{six}} = 4_{ ext{six}} + 4_{ ext{six}} + 4_{ ext{six}} = 12_{ ext{six}} + 4_{ ext{six}} = 20_{ ext{six}}) and since (4_{ ext{six}} imes 3_{ ext{six}} = 3_{ ext{six}} + 3_{ ext{six}} + 3_{ ext{six}} + 3_{ ext{six}} = 10_{ ext{six}} + 3_{ ext{six}} + 3_{ ext{six}} = 13_{ ext{six}} + 3_{ ext{six}} = 20_{ ext{six}}), then (3_{ ext{six}} imes 4_{ ext{six}} = 4_{ ext{six}} imes 3_{ ext{six}}), because both equal (20_{ ext{six}}).

Exercise 9

9. Verify the commutative property of multiplication by writing an equation that must be true and using the repeated addition definition of multiplication is true using the given set of numbers. Simplify each expression separately; then show they are equal.

a. {4, 3}

b. {(2_{ ext{five}}) , (3_{ ext{five}})}

c. Make up a set in a base other than 10 and verify the commutative property of multiplication.

We can also use the repeated addition definition of multiplication to multiply in other numeration systems. Study the following examples and then work through the problems in exercise 10.

Example in Tally

Example in Egyptian

Example in Mayan

Notice that all of these examples are done completely in the numeration system given. All of the steps are shown in that numeration system. This is how you should do the problems in exercise 10. If you want, you can check the answer after you complete the problem by converting to base ten. For instance, the Mayan example in base ten is 2 ( imes) 132. So the answer should be 264. If you convert the final answer to base ten, it is also 264.

Exercise 10

10. Use the repeated addition definition of multiplication to compute the following. Make sure you write out the meaning of the multiplication in the system given (see examples on previous page), showing all of the work and exchanges (if necessary), to compute the answer. Do not do the problem in base ten, although you can use base ten to check the answer when you are done.

Exercise 11

11. Next, we will form the trains for ((R imes L) imes P) and for (R imes (L imes P)). We can only work on one at a time. In both cases, you must do what's inside the parentheses first.

a. First, we'll do ((R imes L) imes P). Begin by forming the train (R imes L) to use in part b.

b. Next, we need to form the train ((R imes L) imes P). The first part, (R imes L), which you formed in part a, should be in a horizontal position, and the purple c-strip goes underneath it, perpendicular to the train (R imes L). Note that although (R imes L) is made up of more than one c-strip, it still forms one train across the top. In the space to the right, draw a picture of what this looks like so far.

c. Do the multiplication by filling in the rest of the purple c-strips in the vertical position under the train (R imes L); then form a train out of the rectangle of purple c-strips. Keep this train until you have finished all of exercise 11. How many purple c-strips are in the train formed by doing the multiplication ((R imes L) imes P)?

d. Now, we'll do (R imes (L imes P)). Begin by forming the train (L imes P) to use in part e.

e. Next, form the train (R imes (L imes P)). Put the first part, R, in the horizontal position. The second part, (L imes P), formed in part d, should go in the vertical position as one long train, underneath and perpendicular to the train R. Don't separate the purple strips in (L imes P) line up as one long vertical train (made up of more than one purple C-strip) under the red c-strip. To the right, draw a picture of what this looks like so far.

f. Do the multiplication by filling in the rest of the purple c-strips to form the rectangle; then make a train out of the rectangle of purple c-strips. How many purple c-strips are in the train formed by doing the multiplication (R imes (L imes P))?

g. Compare the length of train ((R imes L) imes P), from part c with length of train (R imes (L imes P)), from part f. What do you notice?

h. From your observation, write an equation using the c-strips:

i. Translate the equation from part h to one with numbers:

j. Write the left side of the equation you wrote down in part i (which has numbers, not c-strips), and simplify by following the order of operations do what is in parentheses first. Then, do the same for the right side of the equation.

Left side:

Right side:

k. Are the two expressions you simplified in part (j) equal? If so, you have verified the equation from part (i) is true

Exercise 12

12. Next, we will form the trains for ((Y imes R) imes L) and for (Y imes (R imes L)).

a. First, we'll do ((Y imes R) imes L). Begin by forming the train (Y imes R).

b. Next, we need to form the train ((Y imes R) imes L). The first part, (Y imes R), which you formed in part a, should be in a horizontal position, and the light green c-strip should go underneath it, perpendicular to the train (Y imes R). Note that although (Y imes R) is made up of more than one c-strip, it still forms one train that is across the top. That's okay. Below, draw a picture of what this looks like so far.

c. Do the multiplication by filling in the rest of the light green c-strips and then making a train out of the rectangle of light green c-strips. Keep this train until you have finished all of exercise 12. How many light green c-strips are in the train formed by doing the multiplication ((Y imes R) imes L) ?

d. Now, we'll do (Y imes (R imes L)). Begin by forming the train (R imes L).

e. Next, form the train (Y imes (R imes L)). Put the first part, Y, in the horizontal position. The second part, (R imes L), formed in part d, should go in the vertical position as one long train, underneath and perpendicular to the train Y. Don't separate the light green strips in (R imes L) line up as one long vertical train (made up of more than one light green C-strip) under the yellow c-strip. To the right, draw a picture of what this looks like so far.

f. Do the multiplication by filling in the rest of the light green c-strips to form the rectangle, and then making a train out of the rectangle of light green c-strips. How many light green c-strips are in the train formed by the multiplication (Y imes (R imes L))?

g. Compare the length of train ((Y imes R) imes L), from part c with the length of train (Y imes (R imes L)), from part f. What do you notice?

h. From your observation, write an equation using the c-strips:

i. Translate the equation from part h to one with numbers:

j. Write the left side of the equation you wrote down in part i (which has numbers, not c-strips), and simplify by following the order of operations do what is in parentheses first. Then, do the same for the right side of the equation.

Left side:

Right side:

k. Are the two expressions you simplified in part (j) equal? If so, you have verified the equation from part (i) is true.

Exercises 11 and 12 illustrate the associative property of multiplication for trains.

The associative property of multiplication for trains states that if r, s and t are any three trains, then ((r imes s) imes t = r imes (s imes t)).

Before going on to any more properties, we'll practice verifying that the associative property of multiplication is true for some numbers in base ten later on, we'll do it in other bases, too. The associative property requires three numbers (they don't have to be three different numbers) and there are different expressions on each side of the equation. The difference between the left and right sides is that the parentheses are around a different pair of numbers. Remember that the associative property means there is a parentheses change, whereas the commutative property means there has been an order change. To verify, you need to use the order of operations when simplifying each side. If each expression simplifies to same thing, the equation is true. In the order of operations, you always simplify what is in parentheses first. The following are some examples of how to verify the associative property of multiplication, given a set of three numbers.

The associative property of multiplication for whole numbers states that if a, b and c are any three numbers, then (( a imes b ) imes c = a imes ( b imes c )).

Example 1

Example 1: Verify the associative property of multiplication is true using the three numbers {4, 5, 7} in your example.

Note: Since the associative property states ((a imes b) imes c = a imes (b imes c)), one of the numbers given is put in for a, the other for b and the other for c. The choice is yours. For example, we can show any of the following six equations are true:

a. ((4 imes 5) imes 7 = 4 imes (5 imes 7)) or d. ((4 imes 7) imes 5 = 4 imes (7 imes 5))

b. ((5 imes 4) imes 7 = 5 imes (4 imes 7)) or e. ((7 imes 4) imes 5 = 7 imes (4 imes 5))

c. ((7 imes 5) imes 4 = 7 imes (5 imes 4)) or f. ((5 imes 7) imes 4 = 5 imes (7 imes 4)).

If this was a question on a test, you would only have to verify one of the six equations shown — there are always six possibilities one might choose, depending on which numbers you put in for a, b or c. I will show you how to do it if you decided to verify the equation (e). I'll also verify it for another equation (f). Then, for practice, you get to verify the other four equations.

One Solution to Example 1: I'll show: ((7 imes 4) imes 5 = 7 imes (4 imes 5))

Left side: ((7 imes 4) imes 5 = 28 imes 5 = 140) and Right side: (7 imes (4 imes 5) = 7 imes 20 = 140)

Since I used the order of operations to simplify each expression, and both expressions are equal, the associative property is true using the numbers 4, 5 and 7.

Here is a different way to verify it using a different equation.

Another Solution to Example 1: I'll show: ((5 imes 7) imes 4 = 5 imes (7 imes 4))

Left side: ((5 imes 7) imes 4 = 35 imes 4 = 140) and Right side: (5 imes (7 imes 4) = 5 imes 28 = 140)

Since I used the order of operations to simplify each expression, and both expressions are equal, the associative property is true using the numbers 4, 5 and 7.

Exercise 13

13. Verify the associative property of multiplication is true using the numbers 4, 5 and 7, by doing the order of operations on each side. Do this four times, using the equations a, b, c and then d, as shown on the previous page.

a. Verify: ((4 imes 5) imes 7 = 4 imes (5 imes 7))

b. Verify: ((5 imes 4) imes 7 = 5 imes (4 imes 7))

c. Verify: ((7 imes 5) imes 4 = 7 imes (5 imes 4))

d. Verify: ((4 imes 7) imes 5 = 4 imes (7 imes 5))

To verify the associative property of multiplication is true by example, first you must write an equation that must be true using any three numbers (or by using three particular numbers given to you.) Then, by doing the order of operations on each side (each individual expression), you must show each side of the equation gives the same result.

Example 2

Example 2: Verify the associative property is true using the numbers 10, 8 and 3.

a. I will show this equation is true: ((10 imes 8) imes 3 = 10 imes (8 imes 3))

Since ((10 imes 8) imes 3 = 80 imes 3 = 240 and (10 imes (8 imes 3) = 10 imes 24 = 240), then the equation ((10 imes 8) imes 3 = 10 imes (8 imes 3)) is true.

Exercise 14

14. Verify the associative property is true using the numbers given.

a. {9, 7, 6}

b. any three numbers of your choice.

Time to work on another property...

STEP 1: Form the train (P imes (R + L)). To do this, the purple strip should be place in a horizontal position, and the train R + L should be placed in a vertical position underneath the purple c-strip. Then, the rectangle needs to filled in. Lastly, make a long train out of the rectangle. Note that this train is made up of 8 c-strips (4 reds and 4 light greens). Save this train.

STEP 2: Form two trains: (P imes R) and (P imes L). Then add the two trains together (in other words, (P imes R + P imes L)) to form one long train. Note that this train is made up of 8 c-strips (4 reds and 4 light greens). Compare the length of the train formed in Step 1 with this train. Are they are the same length? Are they made up of the same exact c-strips?

We have just shown and verified that (P imes (R + L) = P imes R + P imes L).

Exercise 15

15.

a. Form the train (L imes (P + Y)) by first forming the train P + Y. Then do the multiplication (L imes (P + Y)). Exactly what c-strips make up this train?

b. Form the train (L imes P) and (L imes Y). Now add them together to form one long train. Exactly what c-strips make up this train?

c. Compare the trains from part a and b, and write an equation (with c-strips).

d. Translate the equation from part c by writing an equation using numbers.

e. Verify the equation you wrote in part d by first simplifying the left hand side of the equation using the order of operations and then by simplifying the right hand side as well. If the answer to each expression is the same, you have verified the equation.

left side:

right side:

Exercise 16

16.

a. Form the train (R imes (W + D)) by first forming the train W + D. Then do the multiplication (R imes (W + D)). Exactly what c-strips make up this train?

b. Form the train (R imes W) and (R imes D). Now add them together to form one long train. Exactly what c-strips make up this train?

c. Compare the trains from part a and b, and write an equation (with c-strips).

d. Translate the equation from part c by writing an equation using numbers.

e. Verify the equation you wrote in part d by first simplifying the left hand side of the equation using the order of operations and then by simplifying the right hand side as well. If the answer to each expression is the same, you have verified the equation.

left side:

right side:

The above problems illustrate the distributive property of multiplication over addition for trains.

The distributive property of multiplication over addition for trains states that if r, s and t are any three trains (each train may be made up of one or more c-strips), then (r imes (s + t) = (r imes s) + (r imes t)).

Now, it's time to practice verifying the distributive property of multiplication over addition is true for some numbers in base ten — later on, we'll do it in other bases, too. The distributive property requires three numbers (they don't have to be three different numbers) and there are two sides of an equation to verify. Note the difference between the left and right sides. The following are some examples of how to verify the distributive property of multiplication, given a set of three numbers.

The distributive property of multiplication over addition for whole numbers states that if a, b and c are any three numbers, then (a imes (b + c) = (a imes b) + (a imes c)).

Example 1

Example 1: Verify the distributive property of multiplication is true using the three numbers {4, 5, 7} in your example.

Note: Since the distributive property states (a imes (b + c) = (a imes b) + (a imes c)), one of the numbers given is put in for a, the other for b and the other for c. For example, we can show any of the following six equations are true:

a. (4 imes (5 + 7) = (4 imes 5) + (4 imes 7) or d. (5 imes (7 + 4) = (5 imes 7) + (5 imes 4))

b. (4 imes (7 + 5) = (4 imes 7) + (4 imes 5) or e. (7 imes (4 + 5) = (7 imes 4) + (7 imes 5))

c. (5 imes (4 + 7) = (5 imes 4) + (5 imes 7) or f. (7 imes (5 + 4) = (7 imes 5) + (7 imes 4))

If this was a question on a test, you would only have to verify one of the six equations shown. There are always six possibilities one might choose, depending on which numbers you put in for a, b or c. I will show you how to do it if you decided to verify the equation (c). Then, for practice, you get to verify the other four equations.

One Solution to Example 1: I'll show: (5 imes (4 + 7) = (5 imes 4) + (5 imes 7))

Left side: (5 imes (4 + 7) = 5 imes (11) = 55) and Right side: ((5 imes 4) + (5 imes 7) = 20 + 35 = 55)

Since I used the order of operations to simplify each expression, and both expressions are equal, the distributive property is true using the numbers 4, 5 and 7.

Here is a different way to verify it using a different equation.

Another Solution to Example 1: I'll show: (7 imes (5 + 4) = (7 imes 5) + (7 imes 4))

Left side: (7 imes (5 + 4) = 7 imes 9 = 63) and Right side: ((7 imes 5) + (7 imes 4) = 35 + 28 = 63)

Since I used the order of operations to simplify each expression, and both expressions are equal, the distributive property is true using the numbers 4, 5 and 7.

Exercise 17

17. Verify the distributive property of multiplication is true using the numbers 4, 5 and 7, by doing the order of operations on each side. Do this four times, using the equations a, b, d and then e, as shown on the previous page.

a. Verify: (4 imes (5 + 7) = (4 imes 5) + (4 imes 7))

b. Verify: (4 imes (7 + 5) = (4 imes 7) + (4 imes 5))

c. Verify: (5 imes (7 + 4) = (5 imes 7) + (5 imes 4))

d. Verify: (7 imes (4 + 5) = (7 imes 4) + (7 imes 5))

To verify the distributive property of multiplication is true by example, first you must write an equation that must be true using any three numbers (or by using three particular numbers given to you.) Then, by doing the order of operations on each side (each individual expression), you must show each side of the equation gives the same result.

Example 2

Example 2: Verify the distributive property is true using the numbers 10, 8 and 5.

a. I will show this equation is true: (10 imes (8 + 5) = (10 imes 8) + (10 imes 5))

Since (10 imes (8 + 5) = 10 imes 13) and ((10 imes 8) + (10 imes 5) = 80 + 50), then the equation (10 imes (8 + 5) = (10 imes 8) + (10 imes 5)) is true.

Exercise 18

18. Verify the distributive property of multiplication over addition is true for the three numbers specified:

a. {7, 8, 12}

b. {28, 65, 35}

c. any three numbers of your choice.

Exercise 19

19. Use the definition of multiplication for trains to compute the following. Then translate to make an equation using numbers.

a. (W imes R) = _____ translates to _______________

b. (W imes B) = _____ translates to _______________

c. (W imes D) = _____ translates to _______________

d. (W imes S) = _____ translates to _______________

e. (N imes W) = _____ translates to _______________

f. (H imes W) = _____ translates to _______________

Exercise 19 illustrates that 1 is the identity element for multiplication. In other words, for any number m, (m imes 1 = m) and (1 imes m = m).

Exercise 20

20. The identity element for multiplication if we are referring to trains is what you fill in the blank to make the following statement true. Fill in the blanks to find the identity.

For any train t, (t imes _____ = t) and _____ imes t = t).

The distributive property of multiplication over addition is used quite often to simplify arithmetic problems. Consider the following examples:

Someone who needed to multiply (57 imes 102) might find it easier to think of the problem as (57 imes (100 + 2) = (57 imes 100) + (57 imes 2) = 5700 + 114 = 5814).

Someone who needed to compute (47 imes 38 + 47 imes 62) might realize this is really just the same as (47 imes (38 + 62) = 47 imes 100 = 4700).

This next example illustrates that multiplication also distributes over subtraction:

Someone who needed to multiply 38 x 99 might find it easier to think of the problem as (38 imes (100 - 1) = (38 imes 100) - (38 imes 1) = 3800 - 38 = 3762).

Exercise 21

21. Use the distributive property of multiplication over addition (or subtraction) to rewrite the following problems. Then, simplify. For d, e and f, make up three of your own problems where it would be easier to use the distributive property first before simplifying. Show how to use the property and simplify.

a. (764 imes 999)

b. (324 imes 102)

c. (83 imes 74 + 83 imes 26)

d.

e.

f.

Exercise 22

22. So far, you have worked with three operations; addition, subtraction and multiplication. You have learned many properties. Take a little time to reflect on and review these by writing the properties or answering the questions as stated below:

a. State the associative property of multiplication and provide an example.

b. State the commutative property of addition and provide an example.

c. State the identity element for addition and what it means and provide an example.

d. State the commutative property of multiplication and provide an example.

e. State the distributive property of multiplication over addition and provide an example.

f. State the identity element for multiplication and provide an example.

g. Provide a counterexample to show that subtraction is not commutative

h. State the associative property of addition.

i. Provide a counterexample to show that subtraction is not associative.

There is another way to define multiplication using Set Theory. In order to do this, we have to remember how to take the Cartesian product of two sets, A and B.

The Cartesian product of set A with set B, which is written and is read as "A cross B" is the set of all possible ordered pairs (a,b), where (a imes A) and (b imes B).

If you are finding the Cartesian product, the answer is a set that contains ordered pairs.

Example 1

Example 1: If A = {x, y, z} and B = {a, b}, find (A imes B).

Solution

Solution: (A imes B) = {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}

Example 2

Example 2: If E = {1, 2} and F = {2, 3}, find (E imes F).

Solution

Solution: (E imes F) = {(1, 2), (1, 3), (2, 2), (2, 3)}

Exercise 23

23. Write the Cartesian product. Each answer is a set containing ordered pairs!

a. {3, 4} ( imes) {2, 6} = __________________________________________________

b. {6, 7, 8, 9} ( imes) {5} = _________________________________________________

c. {r, s, t} ( imes) { } = __________________________________________________

d. {a} ( imes) {a} = __________________________________________________

e. {x, y} ( imes) {x, y} = __________________________________________________

f. {1, 3, 5} ( imes) {1, 3, 5} = __________________________________________________

Set Theory Definition for Multiplying two whole numbers:

For any two sets A and B, (n(A) imes n(B) = n(A imes B)).

In other words, to multiply two numbers, a and b, write one set A which has a elements in it and another set B that has b elements in it. To find the product, write the Cartesian
product of A and B and count the elements in the set.
Note: There are no restrictions on what elements you choose for sets A and B. They can be disjoint or they can have elements in common. It’s up to you.

Example 1

Example 1: Use the set theory definition of multiplication to show that (3 imes 2 = 6)

Solution

Let A = {v, w, x} and B = {1, 2}. Since n(A) = 3 and n(B) = 2, then
(3 imes 2 = n(A) imes n(B)) by substituting n(A)for 3 and n(B)for 2
= (n(A imes B)) by the set theory definition of multiplication
= n({(v,1), (v, 2), (w, 1), (w, 2), (x, 1), (x, 2)}) by computing (A imes B)
=6 by counting the elements in (A imes B)
Therefore, (3 imes 2 = 6).

Example 2

Example 2: Use the set theory definition of multiplication to show that (1 imes 3 = 3)

Solution

Let A = {v} and B = {1, 2, 3}. Since n(A) = 1 and n(B) = 3, then
(1 imes 3 = n(A) imes n(B)) by substituting n(A)for 1 and n(B)for 3
= (n(A imes B)) by the set theory definition of multiplication
= n({(v,1), (v, 2), (v, 3)}) by computing (A imes B)
= 3 by counting the elements in (A imes B)
Therefore, (1 imes 3 = 3)

Example 3

Example 3: Use the set theory definition of multiplication to show that (2 imes 0 = 0)

Solution

Let A = {v, w} and B = { }. Since n(A) = 2 and n(B) = 0, then

(2 imes 0 = n(A) imes n(B)) by substituting n(A)for 2 and n(B)for 0
= (n(A imes B)) by the set theory definition of multiplication
= n({ }) by computing (A imes B)
= 0 by counting the elements in (A imes B)

Therefore, (2 imes 0 = 0)

Example 4

Example 4: Use the set theory definition of multiplication to show that (4 imes 3 = 12)

Solution
Solution:
Let A = {e, f, g, h} and B = {g, h, i }. Since n(A) = 4 and n(B) = 3, then


(4 imes 3 = n(A) imes n(B)) by substituting n(A)for 2 and n(B)for 0
= (n(A imes B)) by the set theory definition of multiplication
= n({(e, g), (e, h), (e, i), (f, g), (f, h), (f, i), (g, g), (g, h), (g, i), (h, g), (h, h), (h, i) }) by computing (A imes B)
= 12 by counting the elements in (A imes B)


Therefore, (4 imes 3 = 12)

Exercise 24

24. Use the set theory definition of multiplication to verify each multiplication. Show and justify each step. Only the solution to 24a is shown in the solutions.

a. (5 imes 2 = 10)

b. (3 imes 1 = 10)

c. (0 imes 2)

Here's a more geometrical approach to defining multiplication of two whole numbers.

Definition:

To find the product of any two whole numbers, m and n, make a rectangular array of objects having m rows and n columns. The product of m and n, written (m imes n), equals the number of objects in the array.

The cool thing about using this geometrical approach is that any objects can be used, and one only needs to be able to count to get the answer. You don't need to use addition in order to multiply! I personally advocate having lots of graph paper around for anyone just learning to multiply. To compute (4 imes 7), you or the learner can mark off a rectangle on the graph paper that has 4 rows and 7 columns of boxes. The number of boxes in the rectangle is the answer to the problem. This is a useful tool for first learning a particular multiplication fact, or for anyone (adults, too) who forgot a fact and needs to figure it out again. The "objects" used on a graph paper are the individual boxes on the graph paper.

Here is how six different students might geometrically show how to multiply (4 imes 7).

There is no point to making anyone stress out if they forgot, didn't memorize or couldn't recall the answer to (4 imes 7). The point is to KNOW WHAT MULTIPLICATION MEANS so you can figure a problem out again if you have to. Of course, there is merit to using flashcards and eventually memorizing basic multiplication facts like your multiplication tables. If not, it takes longer to do problems, you aren't able to focus on more advanced issues, and you might not trust your own mind. You shouldn't have to think about multiplication every time you have to multiply two numbers together, but you should be able to think about it. Knowing how to get the answer is better than having memorized a bunch of facts if you don't know where they came from, what multiplication really means, or how to figure it out again in case you forget the answer to some fact. And, as far as I'm concerned, nothing's wrong with using your fingers, either! Let people do what is comfortable for them, if they understand and can get the right answer, that's GREAT!

Exercise 25

25. Write the multiplication problem represented by each geometrical representation:

a. ____

b. ____

c. ____

d. ____

e. ____

f. ____

Exercise 26

26. Show two different geometrical representations for (4 imes 3)

Exercise 27

27. For each multiplication, do the following to represent the multiplication: For part a, draw a rectangular array of objects, for part b use the graph paper, and for part c use the dotted paper. There are two ways you can choose to use the dotted paper.

a. (2 imes 6)

b. (5 imes 3)

c. (3 imes 7)

Exercise 28

28.

a. Draw a rectangular array of boxes (a rectangle) on the graph paper provided to do the multiplication (4 imes 6):

b. Turn your page sideways and look at the rectangle. Draw what the picture looks like:

c. What multiplication problem is now represented? ________________

Exercise 28 should convince you that multiplication is commutative. In fact, the commutative property of multiplication is easiest to see using the geometrical approach. A rectangular array of objects having m rows and n columns represents the multiplication (m imes n). Once the rectangle is rotated 90˚ (or looked at sideways), there are now n rows and m columns, which represents the multiplication (n imes m). But of course, it's really the same rectangle of objects, so the number of objects in m n equals the number of objects in (n imes m). Therefore, (m imes n = n imes m). Okay, just for fun, one more time:

The commutative property of multiplication for whole numbers states that if a and b are any two numbers, then (a imes b = b imes a).

Wow! Isn't this exciting? There are several ways to think about and define multiplication, and we can show the commutative property still holds each time. Does it get any better than this? Well,...YES...but that's another story.

It's a little bit trickier to show on a two-dimensional page that the associative property of multiplication holds. With your work with the C-strips, you know multiplication is associative. Here's one way to geometrically model the multiplication of three numbers.

Definition:

To find the product of any three whole numbers, ((a imes b) imes c), first make a rectangular array of cubes (dice, sugar cubes, etc.) having a rows and b columns. Then, fill in c layers of the bottom rectangle to obtain a rectangular box, where the base of the box has width a and length b, and the height of the box is c.

Basically, you need to make (or visualize) a 3-dimensional box to do multiplication on three numbers. As it turns out, no matter how you construct it, if at the end, one dimension is a, one is b and the other is c, you've got the same box, and it holds the same number of cubes in it. Therefore, multiplication is associative. We won't be going through the motions of showing it in 3-dimensions, but elementary school children should have the benefit of working through it with cubes. I hope you'll model it this way if you become an elementary school teacher.

Since multiplication is both commutative and associative, any time more than two numbers are being multiplied together, the parentheses do not have to be shown, and the numbers can be multiplied together in any order.

For instance, notice how differently one might compute (4 imes 10 imes 3 imes 5)

Student 1: This person just goes left to right. First, (4 imes 10) is 40, then (40 imes 3) is 120, then (120 imes 5 = 600).

Student 2: This person first multiplies (4 imes 5) to get 20, then multiplies (10 imes 3) to get 30, then multiplies the two answers together, (20 imes 30) to get 600.

Student 3: This person does (4 imes 5) first to get 20, then multiplies by 3 to get 60, then multiplies by 10 to get 600.

Student 4: This person multiplies (3 imes 5) first to get 15, then multiplies that by 4 to get 60, and then multiplies by 10 to get 600.

And there are many more possibilities. The point is that you have the option of multiplying in any order and any combination you want. The following show how this is very convenient for certain computations:

Example

Example: (5 imes 87 imes 2): Well, it's easier to multiply the 5 and 2 first to get 10 and then multiply by 87 to get 870. Otherwise, you could do (5 imes 87), which isn't so easy to do in your head.

Example

Example: (8 imes 22 imes 5): Well, I'd do 8 times 5 first to get 40 and then multiply by 22 to get 880.

Exercise 29

29. Make up a multiplication problem, where it would be easier to multiply by doing some rearranging first. Explain your steps and how to get the answer.

Exercise 30

30. We're going to show that (3 imes (4 + 2) = (3 imes 4) + (3 imes 2)). Let's first work on the right side of the equation, which has two parts.

a. Draw a rectangular array of dots or stars to represent 3 4.

b. Now, draw a rectangular array of dots or stars to represent 3 2.

c. Since the right side of the equation says to join the two parts together (addition), then draw the array from part a below, and right next to it, draw the array from part b.

d. Now, let's do the left hand side of the equation, (3 imes (4 + 2)). We first have to compute in the parentheses to get (3 imes 6). So, make a rectangular array of dots or stars to represent (3 imes 6).

e. Is the number of dots or starts in part c equal to the number of dots or starts in part d?

Exercise 31

31. What property did you just illustrate in exercise 30?

Well, I hope you realize you just showed that multiplication distributes over addition. In fact, it is also true that multiplication distributes over subtraction. In other words,

For any three whole numbers, a, b and c, where b > c, (a imes (b + c) = (a imes b) + (a imes c)) and (a imes (b – c) = (a imes b) – (a imes c)).

Exercise 32

32. Show that (2 imes (3 + 4) = (2 imes 3) + (2 imes 4)) using a geometrical approach, like you did in exercise 30. Explain the steps.

Exercise 33

33.

a. Form the train (P imes (Y – L)) by first forming the train Y – L. Find out what C-strip Y – L is. Then do the multiplication (P imes (Y – L)). Exactly how many white C-strips make up this train?

b. Form the train (P imes Y) and (P imes L). Now subtract the second train from the first train. Exactly how many white C-strips make up this train?

c. Compare the lengths of the trains from part a and b.

Exercise 33 showed that (P imes (Y – L) = P imes Y – P imes L). This is one illustration that multiplication distributes over subtraction.

Multiplication Algorithms

You will need: Base Blocks (Material Cards 4-15)

There are many algorithms for multiplication. Personally, I think the way most of us learned it in school is one of the hardest algorithms. It's an uphill battle getting new methods into the school system. But, it's worth pursuing. It's a good thing to know what multiplication means, and how to multiply any two numbers together. However, there is no reason that everyone should have to use the same method to obtain the answer. In this exercise set, you'll learn some new algorithms for multiplication. You'll also learn how to multiply in different bases.

Many people use repeated addition all the time to multiply. Consider the multiplication problem 26 times 17. Basically, this means add 26 seventeens added together. This could be done by adding them together one at a time. That would take a long time.

Another approach would be to add 10 seventeens together first to get 170, then add 10 more seventeens together to get 170 again, then add 5 more seventeens together which is 85, and then add one more seventeen. Adding 10 seventeens (170), 10 seventeens (170), 5 seventeens (85) and 1 seventeen (17) gives 170 + 170 + 85 + 17 = 442, which is exactly the answer to 26 times 17. That should be no surprise since we added 26 (10+10+5+1) seventeens together.

Someone else might add 10 seventeens (170), 10 seventeens (170), 3 seventeens (51) and 3 more seventeens (51) to get 170 + 170 + 51 + 51 = 340 + 102 =442.

There are several other ways people might break up the number 26 and still get the correct answer. This approach may be one you frequently or maybe this is the first time you've thought of it this way. In any case, it's different from just doing it the same old way.

Because of the commutative property, you could have done 17 times 26 instead. Maybe you'd add 10 twenty-sixes (260), then 5 more twenty-sixes (130) and 2 more twenty-sixes (52) to get 260 + 130 + 52 = 442.

One reason it's advantageous to do multiplication using repeated addition is that many problems can be done in your head this way. But, of course, every new method takes practice and getting used to, and it's common for people to revert back to using their old, familiar methods even if it takes longer and requires paper. That's okay, too. It is, however, very important that a teacher knows several ways so that he or she can provide more opportunities for students. What seems obvious to one student is foreign to another, so you give alternatives and present different ways of looking at things.

Exercise 1

1. Use some form of repeated addition to do the following multiplications. Explain how you do each one.

a. (23 imes 13)

b. (14 imes 14)

One new algorithm (for most of you reading this) is called Duplation. This is actually a very old algorithm used in ancient times by the Egyptians. It also uses the principle that multiplication is simply repeated addition. It is called duplation because it uses the idea of doubling numbers to get the answer. Let's look again at the first multiplication we did in this section. It was (26 imes 17). Since 26 seventeens must be added together, the number 26 is broken down into powers of two. It's easy to get powers of two: simply start with 1, then double it to get 2, then double it to get 4, then double it to get 8, then double it to get 16, etc. To figure out how far you have to go in the doubling process, stop if doubling gives a number bigger than the number you are trying to break down (26, in this case). Now, all this sounds more complicated than it really is. Before explaining too much more, notice how you'll write it. Write 26 at the top and underline it. Underneath, start with 1, then, double continuously, stopping at 16 since if you doubled another time, you'd get a number bigger than 26. To the right of 26, write the number 17 and underline it. Underneath it, first write 17, then double 17 to get 34, then double 34 to get 68, double 68 to get 136, etc., until you have the same amount of numbers under the 17 as you do under the 26.

(26 imes 17)

1 17

2 34

4 68

8 136

16 272

Do you see the corresponding numbers? 1 corresponds with 17, because 17 is (1 imes 17), 2 corresponds with 34, because 34 is (2 imes 17), 4 corresponds with 68, because 68 is (4 imes 17), 8 corresponds with 136, because 136 is (8 imes 17), and 16 corresponds with 272, because 272 is (8 imes 17). Now, you may need to think about this for a few minutes. The left side keeps track of how many of some number you are adding together. So, the (1 imes 17) should be clear. So, (2 imes 17) is simply 17 doubled. Now if (2 imes 17) is 34, then (4 imes 17) is twice as many as (2 imes 17), so double 34 to get 68. This continues on and on. So, if (8 imes 17) is 136, then if we double 136, we should have (16 imes 17). Isn't it neat how we know that (16 imes 17 = 272) and we just double a few numbers to get there?

Okay, now we only need to add 26 seventeens together. Simply start at the bottom of the first column, and check off numbers that add up to 26 (this is like doing it in base two). After you check off the numbers in the left column, circle or point to their corresponding numbers in the right column. It's those numbers in the right column that you add together to get the answer. Watch how the rest of the problem is done:

Let's do the same problem over again, but use the commutative property of multiplication. In other words, use the duplation method to compute: (17 imes 26). This time, in the left column, you check off numbers that add up to 17 it's coincidental that I had to double to 16 again and stop. In the right hand column, start with 26 and double away.

Here are two more examples for you to study. I'll do (15 imes 35) and (35 imes 15).

Of course, because of the commutative property of multiplication, the answer is 525, no matter which way you do it. Are you ready to try a couple of duplation problems of your own? Well, ready or not, here they come!

Exercise 2

2. Multiply (13 imes 29) using duplation. Then, multiply (29 imes 13) using duplation. Show all of your work. Make it clear what is added together for each problem.

Exercise 3

3. Use duplation to multiply 27 by 14, and then 14 by 27. Show your work.

Here is an example of how the Egyptians multiplied using duplation.

Exercise 4

4. Use duplation to multiply | | | | | |. Do the problem entirely in Egyptian.

You are going to learn a cool and easy way to multiply numbers together. But first, I’ll show you how to multiply 2 single digits together. Make a square with a diagonal in it. Put one number on top and the other on the right of the square. Multiply the two digits together and put the answer in the box with the diagonal. Look at the examples to the right.

The method you are now going to learn is called the LATTICE METHOD and it could be used for multiplying 2 digit or 3 digit or even bigger numbers together. The example well do is (47 imes 32).

STEP 1: To begin, put one of the numbers at the top (47) of a rectangle (that has a space for each digit in the number) and the other number along the side (32) of the same rectangle (that has a space for each digit). In this case, we need a rectangle that has two spaces across the top and 2 spaces along the side. Then, you make diagonals. Look to the right.

STEP 2: Multiply each of the digits on the top by each of the digits along the side, and put the answer where they would meet. For instance, look where the 12 goes when you multiply 4 times 3.

STEP 3: There are three other multiplications to do: 7 times 3, 4 times 2 and 7 times 2. Do you see how all the answers are filled in? It doesn’t matter which ones you multiply and put in first.

STEP 4: There are four diagonals (made up of little triangles) slanting to the right inside the rectangle. Starting with the one in the right bottom corner, add the digits inside each diagonal and fill put the sum in the “cup” under its diagonal. In the first diagonal is 4, so 4 is put in the “cup” under its diagonal (bottom right of rectangle). In the next diagonal are three digits a 1, 1 and 8 which adds up to 10, so put a 0 in the cup under that diagonal (at the bottom to the left) and a 1 (for the carry) in one of the triangles to the left. The next diagonal has a 2, 2, 0 and 1, which adds up to 5, and goes in the “cup” under its diagonal (left, bottom). In the last diagonal, there is just a 1, so a 1 is put in the “cup”under its diagonal (left, top). Look at how I filled in the numbers one at a time.

STEP 5: The numbers you filled in under each diagonal in the “cups” (outside the rectangle) form the answer. It is read by starting at the left top and reading down and then across the bottom. In this case, the answer is 1504. Isn’t that cool? That’s all there is to it!

Most of my students love this very old Lattice method of multiplying. It works as easily for small numbers as it does for large numbers. First, you simply do a series of simple single digit multiplications, and then you add, with carrying. Actually you could use the left to right (with underlining for carrying) method of addition for the last step. The main disadvantage is that you have to draw a lattice, which is a small price to pay to make multiplication easier and less complicated. Most likely, it takes less time than using the standard algorithm with carrying, because there's less chance of error and confusion with all that multiplying and carrying, etc. I propose all students learn multiplication this way first, and let them stay with it if they like it. Who needs to shorten it up and make it more complicated than it really is? Ah, well, I take it you get my point.

Exercise 5

5. Use the lattices provided in a - d to multiply the given numbers. Then write the answer in the space provided. For e, draw your own lattice to compute the multiplication. By the way, you don't have to draw the little cups. That was my own little invention so it would be easy to identify the answer. There are two more problems (f and g) on the next page. Check your answers with a calculator

a. (63 imes 57) = ____

b. (342 imes 85) = ____

c. Make up a problem:

d. Make up a problem:

e. (58 imes 67) = ____ Draw your own lattice.

Exercise 6

6.

a. Get out your Base Four Blocks. Take 41 units, and trade them in using your Base Four blocks. You should have some flats, longs and units which you can write as a numeral in Base Four.

Write the Base Four numeral here:

b. If you wanted to multiply 2 times that number, you would need to make two piles representing that number, join the two equal piles together (since multiplication is repeated addition) and then take the new pile and make trade-ins. Make two equal piles with the blocks, where each pile is the same as the one you formed from trade-ins in part a.Draw a picture of what the two piles look like.

c. Join the two piles together and make trade-ins to get a new base four numeral. Write this Base Four numeral here:

d. Fill in the first blank with what you got for part a and the second blank with what you got for part c. You just did a multiplication problem in Base Four using repeated addition!

(2_{ ext{four}} imes _____ _{ ext{four}} = _____ _{ ext{four}})

Exercise (PageIndex{1})

7.

a. Write the Base Four numeral from 6a here:

b. If you wanted to multiply 3 times that number, you would need to make three piles representing that number, join the three equal piles together (since multiplication is repeated addition) and then take the new pile and make trade ins. Make three equal piles with the blocks, where each pile is the same as the one you formed from trade-ins in part a. Draw a picture of what the three piles look like.

c. Join the three piles together and make trade-ins to get a new base four numeral. What Base Four numeral did you get?

d. Fill in the first blank with what you got for part a and the second blank with what you got for part c. You just did another multiplication problem in Base Four using repeated addition!

(3_{ ext{four}} imes _______ _{ ext{four}} = _____ _{ ext{four}})

Exercise 8

8. Using the same procedure as in 6 and 7, take out your base three blocks and do the following multiplication: (2_{ ext{three}} imes 221_{ ext{three}}). Explain your work, and show pictures.

Exercise 9

9. If you wanted to do (10_{ ext{three}} imes 221_{ ext{three}}), you would need to trade the long ((10_{ ext{three}})) in for three units, make three piles using Base three blocks representing (221_{ ext{three}}), and do the same thing you did in exercise 7. Explain how you do this and show pictures.

This method of repeated addition with the blocks would be too cumbersome if we were working with higher numbers. For instance, if you were trying to use this method to multiply (321_{ ext{four}} imes 312_{ ext{four}}), the first number when broken down into single units would consist of more than 50 units, and that would be way too many piles to make! Instead, we'll think about and figure out what it means to multiply using larger blocks than units. For instance, we'll figure out what a flat times a long is, or what a long times a flat is in any base! We'll be using these abbreviations: U for unit, L for long, F for flat, B for block, LB for long block, FB for flat block, and BB for block block.

Exercise 10

10.

a. Get out your base three blocks. You're going to find the product of a long and a long similar to how you did products with the C-strips in Exercise Set 1. (L imes L) is obtained by putting one long in the horizontal position and the second long in the vertical position, underneath and perpendicular to the first long. Then form a rectangle of longs underneath. Take the rectangle (don't include the top horizontal long) and trade up in base three, if possible. What is the product of long and a long in base three?

b. Get out your base four blocks. Find the product of long and a long in base four using the method explained in part a. What is the product of long and a long in base four?

c. Get out your base five blocks. Find the product of long and a long in base five using the method explained in part a. What is the product of long and a long in base five?

d. In any base, what is the product of a long and a long? (L imes L =)

Exercise 11

11.

a. You're going to find the product of a flat and a long now. First, we need to convert the flat to longs so that we can make a horizontal train across the top of the T. Do this. Then, put a long in the vertical position, underneath and perpendicular to the long train of longs making the top of the T. Then form a rectangle of longs underneath. Take the rectangle (don't include any longs from the top horizontal part of the T) and trade up in base three, if possible. What is the product of flat and a long in base three?

b. Get out your base four blocks. Find the product of flat and a long in base four using the method explained in part a. What is the product of flat and a long in base four?

c. Get out your base six blocks. Find the product of flat and a long in base six using the method explained in part a. What is the product of flat and a long in base six?

d. In any base, what is the product of a flat and a long? (F imes L =)

Exercise 12

12.

a. You're going to find the product of a block and a long now. First, we need to convert the block to flats, and then the flat to longs so that we can make a horizontal train across the top of the T. Do this. Then, put a long in the vertical position, underneath and perpendicular to the long train of longs making the top of the T. Then form a rectangle of longs underneath. Take the rectangle (don't include any longs from the top horizontal part of the T) and trade up in base three, if possible. What is the product of block and a long in base three?

b. Get out your base four blocks. Find the product of block and a long in base four using the method explained in part a. What is the product of block and a long in base four?

c. Get out your base two blocks. Find the product of block and a long in base two using the method explained in part a. What is the product of block and a long in base two?

d. In any base, what is the product of a block and a long? (B imes L =)

Exercise 13

13. Summarize the results from part d of exercises 10, 11 and 12.

a. (L imes L = _______) b. (L imes F = _______) c. (L imes B = ________)

Use the commutative property of multiplication to compute the following:

d. (F imes L = ________) e. (B imes L = ________)

Exercise 14

14. In your own words, what happens when something is multiplied by a long?

Did you notice that when something is multiplied by a long, it bumps up to the next place value. This is analogous to multiplying by 10 in base 10, since 10 is a long in base ten. For instance, in base 10, 347 represents 3 hundreds, 4 tens and 7 ones; when you multiply (347 imes 10), you get 3470, 3 thousands, 4 hundreds and 7 tens, the place value for each digit moved up one place value.

Exercise 15

15. Consider multiplying (312_{ ext{four}} imes 10_{ ext{four}}). This means you are multiplying a number in base four by a long in base four. Get out your base four blocks and represent the number (312_{ ext{four}}).

a. You have ____ flats, _____ longs and _____ units.

b. If you multiply each of these by the long, now how many of each type of place value block do you have in base four?

To multiply a two-digit numeral, X, in a base by some number, Y, in that base, the idea of repeated addition can be used. For instance, if we represent the number with base blocks, we'll want to create an "answer pile" where we put the blocks representing the product as we multiply by the two-digit numeral. Every unit in X multiplied by Y gives us one set of base blocks contained in Y in the answer pile. Every long in X multiplied by Y gives us the next place value up for each of the base blocks contained in Y. This is much more complicated to explain than to do with the blocks. So, let's do an example.

Get out your Base Four Blocks. We'll multiply (23_{ ext{four}} imes 32_{ ext{four}}). With your blocks, represent this as a multiplication problem. It should look something like this:

Recall that each long in the bottom multiplied by the top makes each top piece move up to the next higher place value block. So, when the first bottom long is multiplied by the top number, 2 flats and 3 longs go into the answer pile. When the second bottom long is multiplied by the top number, another 2 flats and 3 longs go into the answer pile. When the third bottom long is multiplied by the top number, yet another 2 flats and 3 longs go into the answer pile. When I do this with the blocks, I usually move each part of the multiplier (bottom number) I've done so far to the right or out of the way. In any case, in your answer pile, you should now have 6 flats and 9 longs in the answer pile. We still have to multiply by each of the two units. Each unit multiplied by the top number gives a set of the top number; it's like multiplying by 1. So, when the first bottom unit is multiplied by the top number, 2 longs and 3 units go into the answer pile. When the second bottom unit is multiplied by the top number, another 2 longs and 3 units go into the answer pile. Create a place for the answer pile and do this with your blocks. Try it yourself, then check on the next page.

I'll show the answer pile below: each row shows what I got as I multiplied each bottom piece (3 longs and 2 units) by the top. By the way, you could have multiplied by the units first and then by the longs I'll do it that way for the next example. It all just eventually ends up in the answer pile and all exchanges have to made in the end, so you can write it in Base Four. This is (23_{ ext{four}} imes 32_{ ext{four}}):

THE ANSWER PILE IS SHOWN BELOW:

Now, make all your trade-ins with these pieces. You should end up with 2 blocks, 1 flat, 2 longs and 2 units, which is the number (2122_{ ext{four}}), written in base four.

Next, we'll multiply (132_{ ext{four}} imes 23_{ ext{four}}). It should look something like this:

Each unit multiplied by the top number gives a set of the top number it's like multiplying by 1. So, when the first bottom unit is multiplied by the top number, a flat, 3 longs and 2 units go into the answer pile. When the second bottom unit is multiplied by the top number, another flat, 3 longs and 2 units go into the answer pile. When the third bottom unit is multiplied by the top number, yet another flat, 3 longs and 2 units go into the answer pile. When I do this with the blocks, I usually move each part of the multiplier I've done so far to the right or out of the way. In any case, in your answer pile, you should now have 3 flats, 9 longs and 6 units in the answer pile. We still have to multiply by each of the two longs. Recall that each long multiplied by the top makes each top piece move up to the next higher place value block. So, when the first bottom long is multiplied by the top number, a block, 3 flats and 2 longs go into the answer pile. When the second bottom long is multiplied by the top number, another block, 3 flats and 2 longs go into the answer pile. I'll show the answer pile below: each row shows what I got as I multiplied each bottom piece (3 units and 2 longs) by the top. By the way, you could have multiplied by the longs first and then by the units. This is (132_{ ext{four}} imes 23_{ ext{four}}):

THE ANSWER PILE IS SHOWN BELOW:

Now, make all your trade-ins with these pieces. You should end up with 1 long block, 1 block, 2 longs and 2 units, which is the number (11022_{ ext{four}}), written in base four. Make sure you do and understand these last two problems before going on.

Exercise 16

16. Multiply each of the following using base blocks. Then explain and draw the steps in the space below as done in the previous two examples.

a. Use Base Four blocks to multiply (12_{ ext{four}} imes 31_{ ext{four}})

b. Use Base Three blocks to multiply (112_{ ext{three}} imes 21_{ ext{three}})

Exercise 17

17.

a. Get out your base three blocks. You're going to find the product of a flat and a flat now. We need to convert the first flat to longs, make a train, and put this horizontal train across the top of the T. Then, convert the second flat to longs, make a train, put this train in the vertical position, underneath and perpendicular to the long train of longs making the top of the T. Then form a rectangle of a bunch of longs underneath. Take the rectangle (don't include any longs from the top horizontal part of the T) and trade up as much as possible in base three, if possible. What is the product of flat and a flat in base three?

b. Here's another way to find the product of a flat and a flat in base three. For each unit in the first flat, put a flat in an answer pile. Take all those flats in the answer pile and trade them in to see what you get. What is the product of flat and a flat in base three?

c. Get out your base two blocks. Find the product of flat and a flat in base two using the method explained in part a or b. What is the product of flat and a flat in base two?

d. Complete the following: In any base, (F imes F = _____)

Another way to think about what happens when you multiply by a flat is to think of a flat as a long times a long. Let's say you are multiplying something by a flat. First, multiply it by a long, which means it moves up one place value, and then multiply that answer by a long once more, which means it moves up one more place value. So, anything multiplied by a long moves up one place value, and anything multiplied by a flat moves up two place values. Multiplying by a flat is analogous to multiplying by 100 in base ten. 100b in any base b represents a flat. Think about what happens when a block is multiplied by anything.

Exercise 18

18. Complete the following:

a. (U imes U = ____)f. (F imes U = _______)
b. (U imes L = _____)g. (B imes U = _______)
c. (U imes F = _____)h. (L imes L = _______)
d. (U imes B = _____)i. (L imes F = _______)
e. (L imes U = _____)j. (L imes B = _______)

Exercise 19

19. Use base six blocks to multiply the following. (105_{ ext{six}} imes 123_{ ext{six}}). If you have trouble, go on to the next page, where it is explained and then try it again. Show the steps and diagram.

If you had trouble with #19, below is an abbreviated picture of of the steps. The multiplication shown on the left is (105_{ ext{six}} imes 123_{ ext{six}}) and the multiplication shown on the right is (123_{ ext{six}} imes 105_{ ext{six}}) In both cases, I started multiplying with the biggest block at the bottom and moved left to right. In both cases, the final result is the same, and after all exchanges have been made, there is 1 long block, 3 blocks, 4 flats and 3 units, which is (13403_{ ext{six}}), written in base six. It takes a lot of practice to really do this with the blocks. But it's fun once you get it!

Exercise 20

20. Use base three blocks to multiply (102_{ ext{three}} imes 120_{ ext{three}}). Show steps like done in the example shown above.

We can shorten up this process of multiplying with the blocks even further. In exercise 18, you showed the products obtained when multiplying two base blocks together. The interesting thing is that a flat times a long is a block, no matter what base you are in. You can actually make up a multiplication chart for using blocks.

Exercise 21

21. Complete the rest of the multiplication chart below.

In our first example with the blocks, we computed (23_{ ext{four}} imes 32_{ ext{four}}). Since we are multiplying 2 longs and 3 units by 3 longs and 2 units, the multiplication can be written as ((2L + 3U) imes (3L + 2U)). Applying the distributive property (or FOIL), we get ((2L + 3U) imes (3L + 2U))

= (2L imes 3L + 2L imes 2U + 3U imes 3L + 3U imes 2U)

= ((2 imes 3)(L imes L) + (2 imes 2)(L imes U) + (3 imes 3)(L imes U) + (3 imes 2)(U imes U))

= 6F + 4L + 9L + 6U

= 6F + 13L + 6U **

At this point exchanges have to be made. In Base Four, 6F = 1B + 2F, 13L = 3F + 1L and 6U = 1L + 2U

= 1B + 2F + 3F + 1L + 1L + 2U

= 1B + 5F + 2L + 2U

The 5 flats are now exchanged for a block and a flat

= 1B + 1B + 1F + 2L + 2U

= 2B + 1F + 2L + 2U

which is written as (2122_{ ext{four}}) in base four. This is the same answer we got before.

**If I were doing (23 imes 32) in a different base, all of steps up to this point are exactly the same. For instance, on the next page, the same basic problem is done in base six. All of the steps are the same, but notice that once exchanges are made, they are done with base six blocks in mind.

Also, it isn't necessary to write out the step that is in bold type. That's an application of the commutative and associative properties, and I wanted to show that step for clarification.

Here is the same basic problem done in Base Six.

(23_{ ext{six}} imes 32_{ ext{six}} = (2L + 3U) imes (3L + 2U))

= (2L imes 3L + 2L imes 2U + 3U imes 3L + 3U imes 2U)

= ((2 imes 3)(L imes L) + (2 imes 2)(L imes U) + (3 imes 3)(L imes U) + (3 imes 2)(U imes U))

= 6F + 4L + 9L + 6U

= 6F + 13L + 6U

= 1B + 2F + 1L + 1L

= 1B + 2F + 2L

= (1220_{ ext{six}})

Here is the same basic problem done in Base Nine.

(23_{ ext{nine}} imes 32_{ ext{nine}} = (2L + 3U) imes (3L + 2U))

= (2L imes 3L + 2L imes 2U + 3U imes 3L + 3U imes 2U)

= 6F + 4L + 9L + 6U

= 6F + 13L + 6U

= 6F + 1F + 4L + 6U

= 7F + 4L + 6U

= (746_{ ext{nine}})

Exercise 22

22. Show the steps using the blocks as shown in the previous examples to multiply the following.

a. (23_{ ext{five}} imes 32_{ ext{five}})

b. (42_{ ext{eight}} imes 53_{ ext{eight}})

Here is an example of using this method for numerals with more than two digits each. You still use the distributive property. Each term in the first parentheses is multiplied by each term in the second parentheses.

(212_{ ext{five}} imes 102_{ ext{five}} = (2F + 1L + 2U) imes (1F + 2U))

= (2F imes 1F + 2F imes 2U + 1L imes 1F + 1L imes 2U + 2U imes 1F + 2U imes 2U)

= 2LB + 4F + 1B + 2L + 2F + 4U

= 2LB + 1B + 6F + 2L + 4U

= 2LB + 1B + 1B + 1F + 2L + 4U

= 2LB + 2B + 1F + 2L + 4U

= (22124_{ ext{five}})

(212_{ ext{eight}} imes 102_{ ext{eight}} = (2F + 1L + 2U) imes (1F + 2U))

= (2F imes 1F + 2F imes 2U + 1L imes 1F + 1L imes 2U + 2U imes 1F + 2U imes 2U)

= 2LB + 4F + 1B + 2L + 2F + 4U

= 2LB + 1B + 6F + 2L + 4U

= (21624_{ ext{eight}})

In this last example, we got away without having to make any exchanges. We aren't so lucky in this next example, which is in base three. Base Three is harder to work with since it only takes three of something before you have to make an exchange. Sometimes it

seems that all you are doing is exchanging!

(212_{ ext{three}} imes 102_{ ext{three}} = (2F + 1L + 2U) imes (1F + 2U))

= (2F imes 1F + 2F imes 2U + 1L imes 1F + 1L imes 2U + 2U imes 1F + 2U imes 2U)

= 2LB + 4F + 1B + 2L + 2F + 4U

= 2LB + 1B + 6F + 2L + 4U

= 2LB + 1B + 2B + 2L + 1L + 1U

= 2LB + 3B + 3L + 1U

= 2LB + 1LB + 1F + 1U

= 3LB + 1F + 1U

= 1FB + 1F + 1U

= (100101_{ ext{three}})

Exercise 23

23. Show the steps using the blocks as shown in the previous examples to multiply:

(212_{ ext{four}} imes 102_{ ext{four}})

Exercise 24

24. Show the steps using the blocks as shown in the previous examples to multiply:

a. (361_{ ext{nine}} imes 15_{ ext{nine}})

b. (111_{ ext{two}} imes 11_{ ext{two}})

Another way to multiply with the blocks is by using a chart. This is similar to how we did it for addition. It's actually very similar to doing it using the regular multiplication algorithm, except we don't have to write the carry as we go. You simply multiply each digit of one numeral by each digit in the other numeral, and put it in the proper column. Then, you add the like terms, and still do exchanges. Here is an example writing it out with the distributive property. Below, contrast with the method using a chart.

(23_{ ext{six}} imes 32_{ ext{six}} = (2L + 3U) imes (3L + 2U))

= (2L imes 3L + 2L imes 2U + 3U imes 3L + 3U imes 2U)

= 6F + 4L + 9L + 6U

= 6F + 13L + 6U

= 1B + 2F + 1L + 1L

= 1B + 2F + 2L

= (1220_{ ext{six}})

In step one, the individual multiplications are done. If you get a 2-digit numeral, just write it in. No need to carry. In step 2, the like terms are added. Then, exchanges are made. I did the exchanges by exchanging the units first. In step 3, When 6 units were exchanged, I had 0 units and 1 more long, so I crossed off the 13 and put 14 in the next column to the left. In step 4, I exchanged 14 longs for 2 flats and 2 longs, so I crossed off 6 in the flat column, and added 2 more to get 8, and crossed off 14 in the long column since there were only 2 left. In the last step, 8 flats are exchanged for 1 block and 2 flats. I've shown the individual steps, all the work is shown in Step 5. That's all the space it takes.

Here are three more examples of using the charts to multiply with the blocks.

Exercise 25

25. Use the base block charts, writing the numbers in terms of actual base blocks at first, as shown in the above examples, to compute the following. Show all your work and steps below.

Eventually, it would be nice to be able to do the multiplications without having to think in terms of the actual blocks. The traditional algorithm could be used. To do this, you have to make sure you immediately convert to the given base before writing the answer. For instance, (4_{ ext{five}} imes 4_{ ext{five}} = 31_{ ext{five}}) because 4 units times 4 units is 16 units, which in base five converts to 3 longs and 1 unit, which is written as (31_{ ext{five}}) in base five. A good start is to make up some multiplication tables in other bases. A base seven multiplication table is shown on the next page. Keep in mind that there are seven digits in base seven, so you'll have a table to fill in that is 7 rows by 7 columns. Because of the commutative property, there is symmetry in the table, which means many answers are duplicated. Furthermore, multiplying by 0 or 1 is trivial. At the top of the table, we'll indicate it is in base seven, and then leave off writing the base to the right and below each numeral.

Make sure you don't read these as base ten numerals. 26 means "two six, base seven", which is 2 longs and 6 units, or 20 base ten units! An easy way to get started is to do the two rows and two columns where you multiply by 0 and 1, then, go down the diagonal, from upper left to bottom right. Then, fill in the numbers above the diagonal. Using the commutative property, fill in the bottom of the diagonal.

Exercise 26

26. Make up a multiplication table for each base indicated.

a. Base Two Multiplication Tableb. Base Three Multiplication Table
c. Base Six Multiplication Tabled. Base Eight Multiplication Table
e. Base Twelve Multiplication Table

If you can quickly figure out the answer to multiplying two single digits in any given base, you can now use a lattice or the standard algorithm to do multiplication in different bases. Here are some examples of multiplying numbers in different bases using the lattice method.

(54_{ ext{six}} imes 23_{ ext{six}} = 2210_{ ext{six}})

(42_{ ext{twelve}} imes 58_{ ext{twelve}} = 1E74_{ ext{twelve}})

(202_{ ext{three}} imes 21_{ ext{three}} = 12012_{ ext{three}})

(111_{ ext{two}} imes 11_{ ext{two}} = 10101_{ ext{two}})

Exercise 27

27. Use a lattice to multiply each of the numbers in the base given.

Finally, you could use the standard multiplication algorithm with carrying. This is the shortest algorithm in terms of space on the paper, but you have to be fairly adept at mental calculations, which doesn't come naturally to most of us when working in other bases.

Before going on to the standard multiplication algorithm, we need to notice a few more things. First of all, remember when we were working with the blocks to figure out a (L imes L), etc.? When something was multiplied by a long, everything moved up one place value. In base ten, when you multiply a whole number by 10, the effect is that the answer is the original number with a zero tacked on to the end. That is, every digit moved up a place value. This is true in all bases. For example, (524_{ ext{seven}} imes 10_{ ext{seven}} = 5240_{ ext{seven}}) and (1T6_{ ext{twelve}} imes 10_{ ext{twelve}} = 1T60_{ ext{twelve}}). If you are multiplying by 100 in any base, two zeroes are tacked on to the end, and so on. This principle can be used to multiply (2000 imes 400) by taking advantage of the commutative and associative properties of multiplication as shown below:

(2000 imes 400 = 2 imes 1000 imes 4 imes 100 = 2 imes 4 imes 1000 imes 100 = 8 imes 100000 = 800000).

Of course, there is no need to write all the steps out. Simply multiply 2 times 4 and tack on five zeroes. Here's one in base eight: (3000_{ ext{eight}} imes 500_{ ext{eight}} = 1700000_{ ext{eight}}).

The Partial Products Algorithm uses the above fact. Doing the partial products algorithm is similar to how we used the distributive property to multiply with the blocks (back in exercise 22). For instance, to multiply, rewrite the problem like this:

(346 imes 72 = (300 + 40 + 6) imes (70 + 2))

= ((300 + 40 + 6) imes 70 + (300 + 40 + 6) imes 2)

= (300 imes 70 + 40 imes 70 + 6 imes 70 + 300 imes 2 + 40 imes 2 + 6 imes 2)

= 21000 + 2800 + 420 + 600 + 80 + 12

= 24912

The partial products are shown in the third line in the example above. They are:

(300 imes 70, 40 imes 70, 6 imes 70, 300 imes 2, 40 imes 2 ext{and} 6 imes 2)

It's easier to write this in a vertical format. I'll show this problem two ways. It doesn't matter which partial products you multiply first. The second way it is shown is more similar to our standard algorithm, where we start with the unit's value of the bottom number and eventually work up to the larger place values.

On the left the first three partial products are (346 imes 70) and the second three partial products are (346 imes 2). On the right, the first three partial products are (2 imes 346) and the second three partial products are (70 imes 346).

Our standard algorithm is simply a shortening up of the partial products algorithm. We don't write all the zeroes and we doing the carrying involved with adding more than one partial product at a time in our head. Sure, it's shorter, but it is easier to make mistakes.

Usually we just move over one place to the left to account for the zero in the one's place and don't write the zero in 24220. We "think" 7 times 346.

Exercise 28

28. For each problem, write out the partial products in a vertical format to multiply the numbers together. Show your work, identifying the multiplication for each partial product as shown in the example on the bottom of the previous page.

a. 538

( imes) 34

_________

b. 257

( imes) 941

_________

Finally, I'll provide some examples of what the multiplication looks like when the standard algorithm is used to multiply numbers in different bases. You have to remember to write down the number and all carries in the base you are working in. We tend to always put the numeral with the most digits on the top, but this is not necessary. Here are three ways to multiply (563_{ ext{eight}} imes 24_{ ext{eight}}).

IMPORTANT NOTE:

Each numeral below is in base eight, even though the "eight" isn't written in each intermediate step. E.g., (4 imes 3) in the first multiplication below, is done in base 8; therefore (4 imes 3 = 14).

(563_{ ext{eight}})

( imes24_{ ext{eight}})

2714

13460

(16374_{ ext{eight}})

(24_{ ext{eight}})

( imes 563_{ ext{eight}})

74

1700

14400

(16374_{ ext{eight}})

(563_{ ext{eight}})

( imes 24_{ ext{eight}})

14 ((4 imes 3))

300 ((4 imes 60))

2400 ((4 imes 500))

60 ((20 imes 3))

1400 ((20 imes 60))

12000 ((20 imes 500))

(16374_{ ext{eight}})

Multiply the following numbers together using two different algorithms (or switching the order of the numbers around as shown above). Show your work. Do not do it by converting each number to base ten, multiplying them together, and then convert back to the given base. Use each algorithm at least one time; lattice, partial products, standard algorithm, using the blocks with charts, using the distributive property with or without the blocks. You have lots of choices, but always keep in mind what base you are working in. You need to know all the algorithms for the exam so be sure to try them all. Okay, have fun!

Exercise 29

29. (23_{ ext{five}} imes 43_{ ext{five}})

Exercise 30

30. (612_{ ext{seven} imes 35_{ ext{seven}})

Exercise 31

31. (1011_{ ext{two}} imes 1101_{ ext{two}})

Exercise 32

32. (2121_{ ext{three}} imes 202_{ ext{three}})

Exercise 33

33. (4T1_{ ext{tweleve}} imes 27_{ ext{tweleve}})

Exercise 34

34. (423_{text{six}} imes 145_{ ext{six}})

Exercise 35

35. (2212_{ ext{three}} imes 210_{ ext{three}})

Exercise 36

36. (5421_{ ext{six}} imes 44_{ ext{six}})

Exercise 37

37. (10110_{ ext{two}} imes 1111_{ ext{two}})

Exercise 38

38. (E4T_{ ext{twelve}} imes E2_{ ext{twelve}})

Exercise 39

39. (6661_{ ext{eight}} imes 72_{ ext{eight}})

Exercise 40

40. (1212_{ ext{four}} imes 302_{ ext{four}})

Exercise 41

41. (4234_{ ext{five}} imes 244_{ ext{five}})

· Submit homework separately from this workbook and staple all pages together. (One staple for the entire submission of all the unit homework)

· Start a new module on the front side of a new page and write the module number on the top center of the page.

· Answers without supporting work will receive no credit.

· Some solutions are given in the solutions manual.

· You may work with classmates but do your own work.

HW #1

1. Use the repeated addition definition of multiplication to compute the following. First, write out the meaning of the multiplication, and then compute the answer.

a. (8 imes 4)b. (4 imes 11)

HW #2

2. Use the repeated addition definition of multiplication to compute the following. Make sure you write out the meaning of the multiplication in the system given showing all of the work and exchanges. Do not do the problem in base ten.

HW #3

3. Use the definition of multiplication for trains to compute the following. Then translate to make an equation using numbers.

a. (P imes W = _____) translates to _______________
b. (K imes W = _____) translates to _______________

HW #4

4. Write the Cartesian product

a. {3, x} ( imes) {0, 1, 6}b. {a, b, 0} ( imes) {1, 2}

HW #5

5. Use the set theory definition of multiplication to verify (3 imes 2 = 6)

HW #6

6. Complete the following using your base blocks. Show all work.

a. (F imes L)b. (B imes L)c. (F imes F)d. (F imes B)e. (B imes F)

HW #7

7. Write the base four multiplication table

HW #8

8. Compute the following (2506_{ ext{seven}} imes 451_{ ext{seven}}) using the lattice method.

a. (2506 imes 451)b. (2506_{ ext{seven}} imes 451_{ ext{seven}})

HW #9

9. Compute (19 imes 24) and (24 imes 19) using the Duplation method. Show all steps.

HW #10

10. For each set of three numbers given, illustrate an example of the associative property of multiplication, and then illustrate an example of the distributive property of multiplication over addition. Follow procedures outlined in this module

a. (2_{ ext{five}}, 3_{ ext{five}}, 4_{ ext{five}})b. (3_{ ext{six}}, 4_{ ext{six}}, 5_{ ext{six}})

Multiplication Number Talks Using Models

In my previous post I discussed the importance of planning number talks with the four stages of using models. I used a 1st grade example in that post, and almost immediately my colleague Heidi Fessenden shared this wondering.

Great minds think alike, because this is exactly what I’ve been thinking about lately!

In Round Rock ISD, we want our students to learn thinking strategies for multiplication, rather than attempting to memorize facts in isolation. Thinking strategies have the following benefits for our students:

  • There’s less to memorize because there are 5 thinking strategies to learn instead of 121 isolated facts.
  • They create consistent language across grade levels.
  • They afford a strategic mindset around how we think about computation facts.
  • Their utility extends beyond basic facts to computation with larger numbers.

The thinking strategies we want our students to learn are from ORIGO’s Book of Facts series. (Each strategy is linked to a one-minute video if you’d like to learn more.)

One-Minute Overview Videos

In our curriculum, students learn about these thinking strategies in their core instruction. We have two units in 3rd grade that focus on building conceptual understanding of multiplication and division across a total of 51 instructional days. In between those units, students practice these thinking strategies during daily numeracy time so they can build procedural fluency from their conceptual understanding. My hope is that planning number talks with the four stages of using models will facilitate this rigorous work.

I also hope it supports students in maintaining their fluency at the start of both 4th and 5th grade. Our daily numeracy time at the beginning of both of those grade levels focuses on multiplication and division. Even if every 3rd grade student ended the year fluent, it’s naive to think that fluency will continue into perpetuity without any sort of maintenance.

To help teachers envision what a number talk might look like at different stages of using models, I’ve designed a bank of sample number talks for each thinking strategy.

Each bank includes a variety of examples from the different stages of using models:

  • Stage 2 Referring to a complete model (Number Talks 1-4)
  • Stage 3 Referring to a partial model (Number Talks 5-8)
  • Stage 4 Solving the problem mentally (Number Talks 9-10)

You’ll notice some “Ask Yourself” questions on many slides. You’re welcome to delete them if you don’t want them visible to students. Ever since reading Routines for Reasoning by Grace Kelemanik and Amy Lucenta, I’ve been utilizing the same pedagogical strategies they baked into their routines to support emergent bilingual students and students with learning disabilities:

  • Think-Pair-Share
  • Ask Yourself Questions
  • Annotation
  • Sentence Stems and Sentence Starters
  • The 4Rs: Repeat, Rephrase, Reword, Record

Since not all of the teachers in our district might be aware of “Ask Yourself” questions, I embedded them on the slides to increase the likelihood they’ll be used by any given teacher utilizing these slides.

Caveat

These sample banks are not designed to be followed in order from Number Talk 1 through Number Talk 10. Student thinking should guide the planning of your number talks. As Kathy Richardson shared in a tweet responding to my previous post, the four stages of using models are about levels of student thinking, not levels of instruction.

What these number talks afford is different ways of thinking about computation. A traditional number talk that presents a symbolic expression allows students to think and share about the quantities and operations the symbols represent. The teacher supports the students by representing their thinking using pictures, objects, language, and/or symbols.

A number talk that presents models, on the other hand, allows students to think and share about the the quantities shown and the operation(s) implied. The teacher supports the students by representing their thinking with language and/or symbols.

Trying It Out in the Classroom

For example, I led a number talk in a 5th grade class today, and I started with this image:

A student said she saw 10 boxes with 3 dots in each box. I wrote that language down verbatim, and then asked her how we could represent what she said with symbols. She responded with 10 × 3.

I asked the 5th graders to turn and talk about why we can use multiplication to represent this model. This was challenging for them! They’ve been multiplying since 3rd grade, but they haven’t necessarily revisited the meaning of multiplication in a while.

They were able to use the model to anchor their understanding. They said it’s because the number 3 repeats. This led us into talking about how there are 10 groups of 3 and how multiplication is a way that we can represent counting equal groups of things.

The number talk continued with this second image:

The first student I called on to defend their answer said, “I know 10 times 3 is 30, so I just took away 3.”

I recorded (10 × 3) – 3 = 27, but I didn’t let the students get away with that. I reminded them that multiplication is about equal groups. If we had 10 groups of 3, then we didn’t just take 3 away, we took away something else.

One of the students responded, “You took away a group.”

We continued talking which led to me recording (10 groups of 3) – (1 group of 3) = 9 groups of 3 under the original equation and then (10 × 3) – (1 × 3) = 27 under that.

I have to admit I screwed up in that last equation because I should have written 9 × 3 instead of 27. Thankfully number talks are an ongoing conversation. Students’ number sense is not dependent on any given day’s number talk, which means they’re forgiving of the occasional mistake.

What we did today is hopefully the start of a series of number talks to get students thinking about how taking away groups is one thinking strategy to help them derive facts they don’t know. Students don’t own that strategy right now, but our conversation today using the model was an excellent start.

Final Thoughts

I’m hoping these samples might inspire you to create number talks of your own based on the kinds of conversations you’re having with your students. Here is a document with dot images you can copy and paste from to create your own number talk images.

If you try out these number talks in your classroom, I’d love to hear how it went. Either tag me in a tweet (@EMathRRISD) or share your experience in the comments.


Prospective elementary teachers’ knowledge of fraction division

Prospective elementary teachers must understand fraction division deeply in order to meaningfully teach this topic to their future students. This paper explores the nature of the subject content knowledge of fraction division possessed by a group of Taiwanese prospective elementary teachers at the beginning of their mathematics methods course. The findings provide preliminary evidence that many prospective Taiwanese elementary teachers have developed the knowledge package of fraction division as described by Ma (Knowing and teaching elementary mathematics: Teachers’ understanding of fundamental mathematics in China and the United States. Lawrence Erlbaum Associates, Mahwah, 1999). The nature of various strategies used by these teachers provides further illustration of a secure common content knowledge that can serve as a benchmark for the development of mathematics courses for prospective teachers. However, the findings also show that the tasks of representing fraction division, through either word problems or pictorial diagrams, are challenging even for those highly proficient in elementary and middle school mathematics. The broader implications of this research for the international community are discussed, and recommendations for elementary teacher education programs are presented.

This is a preview of subscription content, access via your institution.


Even with a deep focus on the progression towards fluency with the standard algorithms, IM K–5 Math continues to offer students opportunities to practice their facts and demonstrate flexibility with multiplication and division through instructional routines.

In this video clip, Savannah Sanders, a fourth grader in Saugus, CA explains how she efficiently reasoned through 15 x 40, the last expression in the Number Talk shown.

As a fourth grader, Savannah can use diagrams and written methods to solve problems like 15 x 40, but these Number Talk routines help students to think flexibly about the facts they know to compute mentally using their conceptual understanding of place value, the properties of operations, and the relationships between operations.

As students become comfortable with using mental strategies and the standard algorithms, they are able to determine whether it is appropriate to use the standard algorithm to find the value of a given product or quotient, or if a different strategy is more efficient.


Series Math Routines with Kristin Gray: Fourth Grade: Multiplication Number String: Fourth Grade

  • Math: Math
  • 4: Grade 4
  • NF: Number & Operations--Fractions
  • B: Build fractions from unit fractions
  • 4a:
    Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.

a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).

b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.)

Common core State Standards

  • Math: Math
  • 4: Grade 4
  • NF: Number & Operations--Fractions
  • B: Build fractions from unit fractions
  • 4b:
    Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.

a. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 × (1/4), recording the conclusion by the equation 5/4 = 5 × (1/4).

b. Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 × (2/5) as 6 × (1/5), recognizing this product as 6/5. (In general, n × (a/b) = (n × a)/b.)


This is such a beautiful representation of the progression of multiplication! It makes me so sad I, or rather, my teachers, didn’t know about this when I was learning multiplication in elementary school. But don’t worry, memorizing facts and steps to algorithms that made no sense still didn’t stop me from becoming a huge math dork! I’m super excited to share this with teachers in my district so we can support students’ understanding and discovery of multiplication strategies and algorithms. I have very high hopes for the next generation of math dorks!!

Yep…the struggle is real. But hopefully as the ratio of math dorks to students increases we’ll begin to see some change.

Thank you! I’m fighting the battle of appropriate times to teach the US standard algorithm for both multiplication and division. They are pushing for it to be done in 4th grade so any and every explanation and resource I can use of why this is detrimental is like gold to me right now.

Glad to be of service my Elizabeth and thanks for fighting the good fight. I have to keep reminding myself and the teachers I work with that “the turtle won the race.”

I love how you connect the concrete tools to the visual representation and finally to the numbers so that teachers will be able to mindful of how to implement as their students are learning about these concepts. Thank you for sharing

Grant, this is fantastic! I can see using this w/ teachers and families in my work. Thanks!

err… Graham… not Grant… Was evidently thinking of my nephew…

Thank you for the video Graham. I liked it a lot.

After Descartes began his work that led to the Cartesian Plane and Isaac Newton, (who said if an affirmative number goes to the right a negative number goes to the left) we might have evolved a model of multiplication that spanned both the discrete (integers) and continuous (reals).

I’ve put up a video about a simple multiplication model the world may have missed. And yes, it also reveals why -2 x -3 = +6.

I definitely appreciate this Jonathon and the more we can flesh out conceptual understanding of content before we engage with students, the better off we’ll all be.

This is cool…and needed. Have you considered making one for middle and high school, eg the evolution from ratios to promotions to linear functions to quadratics and beyond?

If not, want to? I think it’d be fun to noodle over.

Thanks for the invite Karim…I’ve already began a R&P through linear functions. Love to get your perspective on it.
I know Robert Kaplinsky has also talked about the importance of developing these up through high school. Definitely not my lane of expertise but love to learn from you all. We’ll be in touch.

There are a lot of amazing things on the internet such as pandas playing in snow and babies giggling non-stop and directions on how to fix the improperly wired 3-way switch in my hallway that was driving me crazy, but this might even top all of those. Thank you!

Hi Graham! First time caller. I’ve shared your blogs on my pinterest page too many times to count. I worked with a team of 3rd grade teachers just yesterday who developed their own “progression” for multiplication by deconstructing the standard and doing a little digging. During lunch this video made its way into my inbox, echoing the work these teachers had done (with some added detail, of course)! It was fantastic and so well-timed! As with everyone else, I also have a request. How about a progression in multiplication as it pertains to a number line? (This could include the idea of MIRA using money, just to really get Simon Gregg worked up.) And I agree with others that I wish the last model continued to show how the “area model”/”box model”/”partial products” can be adapted to demonstrate multiplication with decimals. Suddenly the hundred flat represents a whole, and BOOM, decimals into the hundredths. Thank you for constantly pushing my thinking!!

I really appreciate the comment Nova and thanks for posting. I like the idea of using a number line as a progression tool and connecting how it aligns vertically. Adding this to my list now.
Still chewing on how to incorporate decimals. I’m thinking about including them with division but still trying to figure out how to make things align. Division is definitely messier than multiplication.

I’ve finally gotten around to publishing the wonderful lesson I was a part of onto my blog. Enjoy! http://mathiemomma.com/blog/multiplying-decimals-with-area-model

Graham, this is amazing! I hope to share it with all of our elementary teachers. If we could break through and help students conceptually understand multiplication and fractions, all our mathematics woes would end!! Can you just imagine a culture of students really getting this? Awesome! Thank you…

I feel hesitant to inject an issue here that hasn’t been touched on at all: what is multiplication? In particular, is it accurate to claim “multiplication IS repeated addition” (MIRA for short)? The models looked at for 2nd/3rd grade seem to be focused on that notion. There is some question as to what the models for 4th/5th grade are doing (are we no longer teaching multiplication of decimal numbers in 5th grade, getting there via a review of integer multiplication? That seemed to be the case a decade ago, but what you said here makes me wonder if something actually got pushed to LATER in K-12 in the Common Core world, or if you just don’t touch on decimals here for some other reason). Are we still doing repeated addition when we start with arrays, manipulatives, drawings, or what-have-you for multi-digit multiplication? It’s not crystal-clear to me, though it still feels like we’re defining all the partial products as repeated addition and then doing a grand sum at the end to get a product that feels like a sum to me. Isn’t that odd?

Natural analogues to my question are going to be: “Is division just repeated subtraction?” and “Is exponentiation just repeated multiplication? That last one promises to get really hairy when negative exponents, rational exponents, and, eventually, irrational exponents rear their heads, but that will be after K-5 so no need to worry about that for elementary teachers, right? :^)

Getting back to multiplication and MIRA: if multiplication IS repeated addition, why is it that when we multiply numbers with units of measure, we need to have the same units to add/subtract but not to multiply or divide? How come I can’t add workers and hours, but I can multiply them and get worker-hours? That feels a bit puzzling. Or even before units of measure show up, what about fraction arithmetic? Why can’t I add or subtract fractions with different denominators but I can multiply (and divide) them with impunity? (I’m not even going to ask why we have associative and commutative laws for addition and multiplication but not for subtraction and division. That’s just crazy).

I hate to be a troublemaker, but I’m also wondering why if MIRA is true, then why does addition of positive numbers always result in larger and larger sums, but multiplying positive numbers doesn’t always seem to work like that (ignoring for the moment the question of multiplying by 1).

Is anyone getting the sinking feeling that maybe multiplication ISN’T repeated addition?

And if it isn’t, what is it? Is there something that multiplication is that would account for all or at least SOME of the above? Or does it turn out that math is broken? (I tried doing some trig problems involving the law of sines (using Desmos on my iPhone) on Dec. 1 and the answers I got made no sense. I was convinced that math was broken (again) until it struck me that I hadn’t used Desmos for trig problems before and was assuming that the default setting for angle measurements was the same as on TI calculators – degrees. But it turned out that the default in Desmos is radians. Problem solved! Math wasn’t broken after all. Whew!)

But now I’m feeling like someone may have broken math again. Any thoughts?

I don’t think you’re being a troublemaker Michael. This seems like a right place to discuss this kind of thing.

To me as an elementary (or what we in Europe call primary) teacher, it makes good sense that one of our approaches to multiplication is as repeated addition. It seems like a great way to first understand it. And arrays and areas seem to give a really good intuitive grasp of what multiplication means. (You must have seen Ben Orlin’s post on arrays: http://mathwithbaddrawings.com/2014/12/03/the-sixth-sense-for-multiplication/ ) Also, once we have an area, it is different from two lengths in quality or dimmensionality if that’s the right word. Maybe that reflects the situation you talk about with units?

To me, it’s a good idea to approach concepts via multiple routes. And although he’s outlined a main route into multiplication, that will give students a really good grasp of it, I’ve got a feeling that Graham would approach it in other ways that just the ones he shows here. I certainly do. With my students, for instance, we look at factors and prime numbers a lot:
http://pinkmathematics.blogspot.fr/2013/02/important-factors.html
Which you could say is division, but gives insight into multiplication.
Last year we did a series of Grade 3 lessons on scaling:
http://followinglearning.blogspot.fr/2015/05/scaling.html

I’d really like to know how you would supplement (or replace??) these kinds of lessons for elementary children. What kind of pictures come into mind for you? I myself like the idea of stretching a number line – although I’m not sure how this could be a manipulative that the students could experiment with.

What you say in your last sentence about stretching the number line begins to get at a perspective I believe can give kids a sense that multiplication isn’t JUST repeated addition and perhaps isn’t really repeated addition at all. Of course, since we have little choice in the matter as long as we start early grade mathematics with the notion of counting than to focus on the counting numbers and work our way towards integers and rational numbers towards then end of K-5, it’s very difficult for teachers and students to do anything BUT counting ==> adding ==> repeated adding (multiplication) ==> repeated multiplying (exponentiation). And the questions I posed previously begin to rise when you go to irrational numbers and try to make the counting metaphor/model work. I’ve seen folks twist themselves into knots trying to defend MIRA even when discussing the real numbers, but I think it’s a losing approach that leaves students fairly hamstrung when they get to calculus.

But there is another way (probably more than one, in fact). I recommend looking at some of what the Soviet Union was doing with K-3 mathematics back in the 󈨊s, specifically via the work of V. V. Davydov and his colleagues. They approached mathematics teaching to young children from the perspective of measurement, not counting. While it is very difficult to find much of their work in print in English (I have a set of textbooks by some of Davydov’s colleagues ostensibly still in use in some parts of Russia someone kindly sent these to me a few years ago, but they are in Russian, and while I have a smattering of knowledge of that language, it is not even close to being adequate for understanding those books well enough to judge what the authors were up to). I do have one book by Davydov in pdf format I can send via email and an interesting 2013 article by Yuri Karpov – “A Way to Implement the Neo-Vygotskian Theoretical Learning Approach in the Schools” that I can send.

There was research done in upstate New York by the late Jean Schmittau on the Davydov approach. She has articles in print about Davydov that you can likely download free, but she was extremely reticent about providing information about the actual materials she used in her work with local schoolchildren. Basically, she would not respond to email or phone requests from me about the materials when I pursued information for several years. A former graduate assistant of hers who wrote a dissertation on the work now teaches at a university not too far from me, but as of a couple of years ago, she said that only Dr. Schmittau could help me see the actual textbooks. So I gave up.

There are a couple of Americans who have created their own measurement-based materials. One is Susan Addington and you can see about her work here: http://www.quadrivium.info/MtWindex.html. I have materials from her but I am not at liberty to share those without her permission.

Another is Peter Moxhay, who used materials based on Davydov in a Maine school district. He sent me copies and I can share those via email.

I hope this gets people thinking about measurement vs. counting (not mutually exclusive approaches, of course) and the implications of the former for how we think about, model, and teach arithmetic operations. Do we really HAVE to start with repeated addition as THE model for multiplication? Would we do better to start with measurement, leading to SCALING as an effective way to think about multiplication (note that scaling has both stretching and shrinking built into it and allows us to avoid some of the paradoxes I mentioned in my previous comment)? Is it simply the case that repeated addition yields the right result for multiplication within some sets of numbers, but that is a deceptive coincidence that, if extended, gets kids into difficulties when they hit real numbers and issues with limits, continuitiy, etc., in calculus? At least one mathematician in addition to Susan Addington thinks so: Keith Devin of Stanford, the fellow whose articles on MIRA got me thinking about all this around 2008 or so. And his writing led me to the work of English researchers Terezinha Nunes and Peter Bryant, whose book CHILDREN DOING MATHEMATICS is worth looking at on these issues.

If you send me your email address ([email protected]) I will send you what I can should you be interested.

Wow! Wow again! I don’t know which of my colleagues and admins to send this to first!! This is so powerful for so many reasons. The Progressions are critical for teachers to read and discuss however soooooo dry to read. This really brings the Progression for Multiplication to LIFE! I’m curious if you purposely focused on multiplication, leaving out division except for one mention? Division seems the hardest operation for students, teachers, and parents to teach and learn through a progression. I hope you are planning another video for that since most 4th grade teachers jump right to the long division algorithm with NO concrete-representational approach. Thanks for sharing this!

Thanks for the kind words Lynn! There’s definitely more on the way and division is up next.

Wow. Great visual model to model the progression. I think it deepened my understanding of the models and how they are connected to the algorithm. Thank you.

This is great Graham. This really reminded me of reading Chris Danielson’s book Common Core math for dummies, because you and him break it down so well and show the progressions. The guys over at University of Arizona with the progression documents would be proud!

Plus that poster you made is perfect to go up in a faculty lounge. I want all elementary teachers in our district to see this.

Thanks Martin! The poster in itself is a powerful visual! I’ll be sure take a panoramic pic and share that as well.
And YES….Christopher’s book is awesome.

Where can I find the panoramic picture? I would love to have it as a constant visual reference for my teachers! Love, love, love this- swoon….!

I’m curious as to why they’re doing the “representational” rows and columns in second grade — and then use manipulatives in third??
I fervently agree with comments about the importance of words and connecting language to what’s happening (especially for students who might be stronger visual thinkers & be able to imitate what looks right without making that bridge to understanding).
Hooray for the turtles! (This makes plenty room for the more efficient thinkers, too.)

In 2nd grade it is actually listed as a geometry standard:
2.G.A.2 Partition a rectangle into rows and columns of same-size squares and count to find the total number of them.
Then when they begin with the manipulative in 3rd grade it transitions over to an Operations and Algebraic Thinking standard.
Thanks for the comment!

This is wonderful! So clear and concise in such a short amount of time. I’m totally going to share it with my colleagues and even with families! Can’t wait for an addition/subtraction one… : )

This is fantastic Graham. I love everything from the content to how you laid it out. I would love to see you extend this into middle and high school. For example connecting this method to the distributive property and then multiplying polynomials.

This was so well produced! You might also consider taking screen shots of each key slide so that after watching the video you have something to reference quickly as a PDF.

Thanks Roberts! The MS and HS ones are on the way for sure and I really like the idea of attaching PDFs for friends to read and digest as a quick reference. Thanks as always.

You’ve laid it all on the table, Definitely one for all our primary / elementary colleagues to see!

“And the big piece here is that we don’t push an understanding on students we let students explain what their model represents. The big piece is that the context will explain the equation, not the teacher.”

Perhaps too, you’ve inspired the #MTBoS to do similar things…?

Thanks Simon and from the recent comments I’ve seen on Twitter there’s definitely some more of these on the horizon…from myself and others.

A man, some markers, and some chart paper. Brilliant work here Graham. I see this video becoming a PD staple.

Thank you so much for this Graham! This link is definitely being shared with everyone.

Well done, Graham! You nailed this in under 6 minutes. It’s just like the original, but better! I think I want to take this progression through middle school. Everything in your video is applicable for multiplication in 6-8 and beyond! Again, well done.

Thanks Mike. It’s on my radar and I’ll probably be recruiting your help!

Yes, I see the sequel. And all the spin offs. Everything is connected which is the hard part. Y could go off in so many directions. Well done in getting to the core.

Yes, this is great! I had been wanting to either find something or create something like this. This would be great to extend up to 6-8 so they can see what to build on from elementary. I also would love similar videos for the other operations. Any plans for this? THANK YOU. I already tweeted this!

Great! I selfishly hope it’s a start of a #progression series…

My Christmas present to my third and fourth grade teachers especially. Beautifully done. Thank you!

AWESOME. Thanks for sharing!

Somewhere along the way they need to figure out, with or without help, that twentythree can be seen as twenty and three, and get a grasp of the place value notation. Words are important, and often overlooked in the rush to use the symbolic notation. And whose idea (CCSS) and others) that we have to stuff their heads with extra jargon – commutative, distributive, etcetera.

@howardat58: at some point (and it’s hardly an absolute one), the issue of communicating with others arises. We can have every student invent his/her own terminology for “it doesn’t matter in what order we add” and a different term for “it doesn’t matter in what order we multiply” and so on. Now, let’s have students in a class write to one another to explain what they’re talking about, using solely or primarily their own terminology (and notation if they so choose). What will be the result of this experiment, do you think? How about writing to a student across the hall? In another school? In another. . . ?

Similarly, at what point would we hope that K-12 teachers stop referring to things like “the top number” and the “bottom number”? Is there a point at which we’d like students to know the difference among “dividend, divisor, and quotient”? Is that point due to pedantry, or is there some other reason or reasons?

@howardat58 & Michael,
I think the important piece is that the formal language of mathematics usually proceeds the understanding, or at least in my world. Once students have began to develop a conceptual understanding of an idea, formal words needs to be introduce for the purpose of communication (like you mentioned Michael).
In my mind I’m picturing a scenario like this, “Alright, today we are going to use the distributive property, let me show you how”. This is usually followed by a series of steps and practiced with rote thinking taking place. The conceptual understanding of place value and what’s happening when we decompose a number needs to take place before the introduction of the term “distributive property”. The introduction of formal vocabulary needs to be in context.
With that being said, the practice standards and mathematics in general, ask that we be precise when writing and communicating mathematically. If the formal language supersedes the conceptual understanding, I can quickly see how students would take the understanding that follows as a series of steps and procedures. I think we need to be strategic as to when we introduce formal language. It all comes back to slowing down the process.

When I work with K-5 teachers, I urge them to refrain from early introduction of official terminology. Why is it important for a primary grade kid to know the term “commutative property” before knowing that order for addition or multiplication doesn’t change the outcome, but it does for subtraction and (later) division? The answer is that it isn’t. However, kids should be thinking about whether or not order matters for a given operation, for other operations, for all operations, (later, of course, the set AND the operation must be considered: matrix multiplication is not, in general, commutative). And they should also be asked to think about why order doesn’t matter for addition and multiplication but does for subtraction and division (and this raises the issue of why it may be vital to teach operations as inverses rather than “opposites” since the former carries an implication of undoing. There are also the fundamental algebraic notions of inverse elements and identity elements, how those two relate, and why it might be problematic if subtraction and division were commutative. None of these things require formal language/terminology to explore.

And no, I don’t think this is way over kids’ heads. It’s been looked at successfully in K-5 by people such as Robert A. Davis in his Madison Project. Materials from that project can be downloaded free. It’s a tragedy of 20th-century math education that the “New Math” became associated in the public mind with the Dolciani textbooks rather than the materials Davis developed for the Madison Project. His teaching and materials were effective where they were employed, but they didn’t receive wide distribution.

It’s the thinking, not the words, that matters. But eventually, terminology and notation become important. Kids can and should have the chance to play and work with the ideas long before the terminology and/or notation is shoved down their throats. But that chance can only be allowed to last up to a point. Then it is important to ask them how they will communicate with other kids or other classrooms. And once the notion of standard/common terminology and notation is on the table, it should be easier in future to get students to want to know the “right” way of talking about or writing down some mathematics. No need to introduce distributivity by stating the term. Rather, I’d recommend asking, “Hey, can we figure out 3(2 +5)? And what if we did it this way? [whichever they didn’t use the first time].”

This is amazing! Thank you so much for sharing this! I’ll be passing this on to the teachers in my district! We have been working on understanding the CRA sequence and how it fits with developing understanding for the traditional algorithm. Very well done! Thank you! Thank you!

Thanks Megan and Christine! I’d love to hear how it goes and I’ll take any feedback you colleagues can offer.

Genius! Seriously!! I can’t wait to share this one.

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    - […] The Progression of Multiplication (5 min 56 sec) Here are some terms to ‘google’ to learn more after watching&hellip - […] as well as Graham Fletcher’s videos about the progression of how we teach division and multiplication through the grades. While&hellip - […] Today I am going to post a link to another blog where the blogger has made a great video&hellip - […] I owe Graham Fletcher a big thanks for his progression videos. Great work that led to the story idea.&hellip - […] The Progression of Multiplication […] - […] The Progression of Multiplication by Graham Fletcher. […] - […] Graham Fletcher deserves credit for inspiration with his amazing progression videos. I certainly won’t be able to match those, but&hellip - […] of Addition and Subtraction. Jenise Sexton went in depth into the first episode about the Progression of Multiplication, here.&hellip - […] The Progression of Multiplication by Graham Fletcher. […] - […] Multiplication: http://gfletchy.com/2015/12/18/the-progression-of-multiplication/ […] - […] the multiplication and division videos which were introduced separately, this video is an addition/subtraction mash […] - […] month I posted The Progression of Multiplication hoping that a couple of friends and parents would find it helpful. Well, here’s&hellip - […] do so from a middle school perspective. This week, the elementary teacher in me screamed at The Progression of&hellip

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3 Answers 3

Although the duplicate question has an excellent top-voted answer, it may be a bit too dense for beginners. Here's a simple but instructive example. Fibonacci numbers! We all know them, don't we? =) $def n>$

Let $F_0 = 0$ and $F_1 = 1$ and $F_ = F_ + F_n$ for any $n in n$ .

$F$ is of course the Fibonacci sequence. To capture the underlying structure of the sequence, we would like to 'factor out' the " $n$ " from the above defining recurrence relation. How do we do that? Matrices are a natural solution, as we shall see.

A matrix can be used to encode a linear transformation. How do we view the recurrence relation as a transformation? Our objective is to pass along enough information so that we can generate the sequence by iterating some transformation. Clearly then we need to maintain a pair of consecutive terms, and the transformation is to go from that pair of terms to the next pair:

That immediately gives us the transformation:

Why have I written it like this? Because now we can 'factor out those " $n$ "s by fiat! To do so, define:

Note that the notation on the left is purely notation (for now), and has absolutely nothing to do with matrices or multiplication or anything at all. It just means what is on the right!

Then we can write our transformation as:

Great! We seem to have 'factored out' the recurrence structure from the terms in the sequence. But as mentioned this is merely notation so far. For this representation to be useful, we need to interpret the notation better.

Firstly, we can treat $pmatrix$ as a function on $2$ -vectors (which are those things like $pmatrix$ that consists of a single column of $2$ numbers), specifically:

Now we can fully capture the iteration of a transformation, which is just repeated application of a function. As usual, for any function $f$ we use " $f^k$ " to mean its $k$ -fold composition.

The Fibonacci sequence is now completely expressed by:

Notice two important things. Firstly, this representation cleanly divides into the recurrence and the initial values, and works for any choice of $(F_0,F_1)$ . Secondly, all we need to do now is to be able to compute $pmatrix<1&11&0>^n$ directly before applying the resulting function to $pmatrixF_<0>>$ . For this we must find out what is the composition of two functions $pmatrix$ and $pmatrix$ . Completely by our above definition we will get:

By the definition of equality of functions we get:

Conventionally, each of these three functions is called a matrix, and we shall now define matrix multiplication to be exactly composition! This is the reason for matrix multiplication.

Finally, one should notice that the most efficient way to compute they $n$ -fold composition of $pmatrix$ is to use the following facts for any function $f$ and $n in n$ :

This actually yields one of the fastest algorithms possible to compute the $n$ -th Fibonacci number with arbitrary precision. (The closed form formula is of course faster for fixed precision.)

See my other answer that explains the motivation and interpretation of matrix multiplication. In this follow-up answer I give a brief explanation of elementary row operations, which is a rather different topic actually.

Notice that an elementary row operation does not change the solution set of the represented system of linear equations. What exactly is happening, in terms of the linear transformation viewpoint?

Well, firstly the elementary matrices are invertible transformations, because we can write down explicitly the inverse elementary matrices, which are of course nothing more than the inverse functions. But what really are these functions?

The scale-row operation is a scaling along a coordinate axis.

The swap-row operation is a reflection about a coordinate axis.

The add-multiple-to-another-row operation is a shear along a coordinate axis that moves another coordinate axis but keeps all other coordinate axes in their places.

A lot of properties of matrices fall out from these three facts. For one, the determinant can be easily shown to be changed by scale-row operations (by exactly the same factor), negated by the swap-row operation, but unchanged by the third operation. Notice also that the signed volume is changed in exactly the same way, so this immediately tells us that the determinant of a matrix is the signed volume of the unit cube after it has been transformed by the matrix! This can be used to explain the change of variables theorem for multi-variable integration.

Secondly, those row operations are often performed on the so-called augmented matrix, simply because it corresponds to applying the transformations to both sides of the system of equations. The system starts of as:

where $A$ is a matrix and $v,w$ are vectors. If you succeed in applying elementary row operations until $A$ becomes the identity matrix, then the system has a unique solution because you would end up with:

If that is impossible, it means that $A$ is non-invertible. You can then ask in general about the reduced row-echelon form. That corresponds to breaking $A$ up into $E circ R$ where $E$ is a composition of elementary matrices and $R$ is in reduced row-echelon form. Then you have:

and after the row operations what you would have gotten is:

Depending on the form of $R$ and the vector on the right, you either get no solution or at least one solution. In the latter case, the solution set is obvious from the form of $R$ , and corresponds to a subspace (point / line / plane / . ) of the domain of $R$ .

In some sense therefore the reduced row-echelon form is the core of the matrix transformation, in that the rest of the transformation is invertible. If $A$ is square, and $R$ is not the identity, then $R$ is geometrically a projection.


4th Grade Math

In Grade 4, instructional time should focus on three critical areas: (1) developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends (2) developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers (3) understanding that geometric figures can be analyzed and classified based on their properties, such as having parallel sides, perpendicular sides, particular angle measures, and symmetry.

  1. Students generalize their understanding of place value to 1,000,000, understanding the relative sizes of numbers in each place. They apply their understanding of models for multiplication (equal-sized groups, arrays, area models), place value, and properties of operations, in particular the distributive property, as they develop, discuss, and use efficient, accurate, and generalizable methods to compute products of multi-digit whole numbers. Depending on the numbers and the context, they select and accurately apply appropriate methods to estimate or mentally calculate products. They develop fluency with efficient procedures for multiplying whole numbers understand and explain why the procedures work based on place value and properties of operations and use them to solve problems. Students apply their understanding of models for division, place value, properties of operations, and the relationship of division to multiplication as they develop, discuss, and use efficient, accurate, and generalizable procedures to find quotients involving multi-digit dividends. They select and accurately apply appropriate methods to estimate and mentally calculate quotients, and interpret remainders based upon the context.
  2. Students develop understanding of fraction equivalence and operations with fractions. They recognize that two different fractions can be equal (e.g., 15/9 = 5/3), and they develop methods for generating and recognizing equivalent fractions. Students extend previous understandings about how fractions are built from unit fractions, composing fractions from unit fractions, decomposing fractions into unit fractions, and using the meaning of fractions and the meaning of multiplication to multiply a fraction by a whole number.

3. Students describe, analyze, compare, and classify two-dimensional shapes. Through building, drawing, and analyzing two-dimensional shapes, students deepen their understanding of properties of two-dimensional objects and the use of them to solve problems involving symmetry.


Understanding the Common Core State Standards for Mathematics

Since 2010, states have been adopting the Common Core in both English Language Arts and Mathematics across the United States. Thus far, 45 states have adopted the Common Core State Standards (CCSS) in their entirety and are implementing them throughout each state. As parents, it is important to understand these new standards and the changes being made in the classroom. As a mother of a third grader and kindergartener, I have asked many questions and read the standards to familiarize myself with them. And as a math consultant, I have spent the last two years working with teachers and districts across my state to help build understanding of the Common Core and how it should be implemented in classrooms.

Change can sometimes not be easy, especially when it is not how we learned as children ourselves. The CCSS were created to ensure college readiness and career preparation. The standards are extremely specific and focus on concepts of mastery and fluency. In mathematics, there are fewer standards, which allow for more focus time of concepts and procedures. You may be noticing your child's homework looking different or strategies you are not familiar with. I have many parents asking, "How can I help my child when I don't understand it myself?" Don't be afraid to ask questions and read the standards to feel more comfortable with what your children are learning. You are not alone! The CCSS in Mathematics is broken into 3 sections:

1. 8 Mathematical Practices: There are 8 Math Practices that begin in kindergarten and are taught all the way until high school. These practices don't change depending on grade level but will look different as students grow and become older. For example, how a 1st grader uses math tools will not look exactly the same as an 8th grader. However, they are carried over from year to year so students can build upon these practices and become more proficient in their skills. The math practices are in place to develop the habits of mind of an excellent math student. For example: perseverance – teachers will be creating situations for students to persevere in the math class developing this skill to extend a child's thinking past the first possible answer and so as to not give up. Below are the 8 Mathematical Practices (You can read more in-depth information about these here.)

1) Make sense of problems and persevere in solving them.
2) Reason abstractly and quantitatively.
3) Construct viable arguments and critique the reasoning of others.
4) Model with mathematics.
5) Use appropriate tools strategically.
6) Attend to precision.
7) Look for and make use of structure.
8) Look for and express regularity in repeated reasoning.

2. Critical Areas: Each grade level has several critical areas for instruction in the introduction of the CCSS. These critical areas outline the important topics and concepts to be taught throughout the year. It is important for parents to understand what their children will be learning throughout the school year. These critical areas highlight the main topics and give understanding to the concepts being taught. (The critical areas for each grade are in the introduction section of the standards) For example, the three critical areas in Grade 4 are:
In Grade 4, instructional time should focus on three critical areas: (1) developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit dividends (2) developing an understanding of fraction equivalence, addition and subtraction of fractions with like denominators, and multiplication of fractions by whole numbers (3) understanding that geometric figures can be analyzed and classified based on their properties, such as having parallel sides, perpendicular sides, particular angle measures, and symmetry.


Watch the video: Gange med decimaltal - forståelse (October 2021).