## Learning Math: Number and Operations

# Meanings and Models for Operations Part C: Colored-Chip Models (35 minutes)

**In This Part: Addition**

In this section, we will use manipulatives to model addition, subtraction, multiplication, and division with positive and negative integers. By working with these models, we will see why the laws for signed numbers (positive and negative) work the way they do. See Note 9 below.

We’ll begin with models for addition in which we will represent positive numbers with black chips and negative numbers with red chips:

Using red and black chips, every number can be represented in a variety of ways. For example, here are some ways to represent positive two (+2):

**•**Two black chips:

**•**Three black chips and one red chip:

In this case, one black chip and one red chip together sum to zero: +1 + (-1) = 0. What remains are two black chips, or +2.

**•**Five black chips and three red chips:

And so on! Now let’s explore some addition problems using these chips.

**•**To find the sum of +3 + (+1), join three black chips (three positives) with one black chip (one positive), for a total, or sum, of four black chips (four positives, or +4):

**•**To find the sum of -5 + (-2), join five red chips (five negatives) with two red chips (two negatives), for a sum of seven red chips (seven negatives, or -7):

**•**To find the sum of -4 + (+6), join four red chips (four negatives) with six black chips (six positives):

Because one black chip and one red chip together sum to zero, four pairs of black and red chips zero out. What remains are two black chips (two positives, or +2):

**•**To find the sum of +2 + (-5), join two black chips (two positives) with five red chips (five negatives):

Because one black chip and one red chip together sum to zero, two pairs of black and red chips zero out. What remains are three red chips (three negatives). The sum is negative three (-3):

Draw or make a diagram to show the colored-chip model for each of the following:

**Problem C1
**+4 + (-6)

**Problem C2**

-4 + (-6)

**In This Part: Subtraction
**When doing subtraction using this model, you cannot take away things that are not present. Adding pairs of zeros gives us the actual pieces we need to take away. This can be shown symbolically as follows: +6 – (-4) = (6 + 4) – (4 + -4) = 10 – 0 = 10. Notice that +4 was added to both numbers, which is the same as adding and subtracting 4. See Note 10 below.

**• **To find the difference of +3 – (+1), begin with three black chips (three positives). Subtract one black chip (one positive). Record what is left (two positives, or +2):

**• **To find the difference of -5 – (-2), begin with five red chips (five negatives). Subtract two red chips (two negatives). Record what is left (three negatives, or -3):

**• **To find the difference of +4 – (+6), begin with four black chips (four positives):

Your next step would be to subtract six black chips (six positives), but there are not enough black chips to subtract six. So put in zeros (pairs of positives and negatives) until there are enough black chips to subtract six:

(Note that the collection is still worth four black chips.) Now subtract six black chips, and record what is left (two negatives, or -2):

**• **To find the difference of +2 – (-3), begin with two black chips (two positives):

Since there are no red chips to subtract, put in zeros (pairs of positives and negatives) until there are enough red chips to subtract three:

(Note that the collection is still worth two black chips.) Now subtract three red chips, and record what is left (five positives, or +5):

Draw or make a diagram to show the colored-chip model for each of the following:

**Problem C3
**+5 – (-6)

**Problem C4
**-5 – (-6)

**In This Part: Multiplication
**In this section, multiplication is defined as “putting in” to a circle, or “taking out” from a circle. Remember, in Part A we described 3 • 4 as three groups of four. Three could be thought of as the operator — it tells us what to do with the second number. In this section, we need to know the sign of the operator. If the sign is positive, you put things into the circle. If the sign is negative, you take things out of the circle. In this case, you must put in zeros until you have enough to follow instructions.

**•**To find the product of +2 • (+3), start with an empty circle:

Since the 2 is positive, +2 • (+3) means that you should “put in” “two” sets of “three black chips”:

Record what is left; i.e., six positives (+6):

+2 • (+3) = +6

**•**To find the product of +2 • (-3), start with an empty circle. Since the 2 is positive, +2 • (-3) means that you should “put in” “two” sets of “three red chips”:

Record what is left; i.e., six negatives (-6):

+2 • (-3) = -6

**•**To find the product of -2 • (+3), start with an empty circle. Since the 2 is negative, -2 • (+3) means that you should “take out” “two” sets of “three black chips”:

But there are no black chips. So put in zeros (pairs of positive and negative chips) until there are enough black chips for you to take out two sets of three. Then take them out:

Record what is left; i.e., six negatives (-6):

-2 • (+3) = -6

**•**To find the product of -2 • (-3), start with an empty circle. Since the 2 is negative, -2 • (-3) means that you should “take out” “two” sets of “three red chips.”

But there are no red chips. So put in zeros (pairs of positive and negative chips) until there are enough red chips for you to take out two sets of three. Then take them out:

Record what is left; i.e., six positives (+6):

-2 • (-3) = +6

Draw or make a diagram to show the colored-chip model for each of the following:

**Problem C5
**-3 • (-3)

**Problem C6
**+3 • (-3)

**In This Part: Division
**So far we have used this model for addition, subtraction, and multiplication. Let’s see what happens when we use it for division.

**•**To find the quotient of (+6)(+2), start with six black chips.

The quotient of (+6)(+2) could be found in one of two ways:

Method 1: Finding the number of sets of positive two that are contained in positive six (measurement, repeated subtraction, or quotative division)

Method 2: Finding the number of items in each set when positive six is split into two equal sets (equal groups or partitive division)

Using Method 1, we group the black chips into sets of two:

There are three sets of two black chips contained within six black chips: (+6)(+2) = +3.

Using Method 2, we divide the chips into two equal sets:

There are three black chips in each of the two equal sets: (+6)(+2) = +3.

**• **To find the quotient of (-6)(+2), start with six red chips.

The quotient of (-6)(+2) cannot be determined by Method 1 since there are no sets of positive two contained within negative six (i.e., within your six red chips, there are no black chips).

But…

Using Method 2, we can find the number of items in each set when negative six is divided into two equal sets:

There are three red chips in each of the sets: (-6)(+2) = -3.

**• **To find the quotient of (-6)(-2), start with six red chips.

The quotient of (-6)(-2) cannot be determined by Method 2 since we cannot divide a set into negative two equal groups.

But…

Using Method 1, we can find the number of sets of negative two contained within negative six:

There are three sets of two red chips within the six red chips: (-6)(-2) = +3.

**• **To find the quotient of (+6)(-2), start with six black chips.

The quotient of (+6)(-2) cannot be determined by Method 1 since there are no sets of negative two in positive six. And the quotient of (+6)(-2) cannot be determined by Method 2 since we cannot divide a set into negative two equal groups.

So…

Our model falls apart at this point! See Note 11 below. Therefore, division in this case must be computed by writing a multiplication problem with a missing number in the following way:

(+6)(-2) | = ? means (-2) • ? | = +6 |

**Video Segment
**Watch this video segment for a quick demonstration of why this model falls apart when dividing a positive number by a negative one.

Note that the participants in this segment use yellow chips instead of black ones to represent positive values.

You can find this segment on the session video approximately 19 minutes and 25 seconds after the Annenberg Media logo.

One of the interesting things about using models in mathematics is that no one model works for everything. In this case, the colored-chip model worked for addition, subtraction, multiplication, and most division problems. However, when we tried to divide a positive number by a negative one, the model fell apart. No model is going to carry over to all computations.

**Problem C7
**Draw or make a diagram to show the colored-chip model for each of the following:

**a.**(+8)(+4)

**b.**(-4)(+2)

**c.**(-8)(-2)

**d.**(+4)(-2)

### Notes

**Note 9
**This model provides a visual image of the operations with signed numbers. You can literally see that the pairs of red and black chips cancel each other out, which gives you a sense of the magnitude of the answer. This works for all of the operations except division (but relating division to multiplication solves this problem).

**Note 10
**This process can help you do subtraction of whole numbers using mental math. For example, to do 94 – 37 in your head, you might add 3 to both numbers. That gives you 97 – 40, which is easy to do in your head.

**Note 11**

No model works for all computations, which is why we must not rely on models to do everything. Models provide a visual and kinesthetic feeling about the process, but eventually you must learn how to do the computation with symbols alone. It is important, however, to see the relationships between what you do with the models and what you do with the symbols.

### Solutions

**Problem C1**

Here, we add six red chips to four black chips. After doing this, four zero pairs can be removed, leaving two red chips. So +4 + (-6) = -2.

**Problem C2
**

Here, we add six red chips to four red chips. After doing this, we have 10 red chips. So -4 + (-6) = -10.

**Problem C3
**Start with five black chips. We want to subtract (take away) six red chips. Since there are no red chips to take away, we must add a number of zero pairs (one black and one red chip each). Add six of these pairs so that there are six red chips to take away. Eleven black chips remain, so +5 – (- 6) = +11.

**Problem C4
**Start with five red chips. We want to subtract (take away) six red chips. Since there are not enough red chips to take away, we must add one zero pair (one black and one red chip each). Now there are six red chips to take away. Removing them leaves one black chip, so -5 – (-6) = +1.

**Problem C5
**Start with an empty circle. We want to take away three groups (the first -3) of three red chips (the second -3). Since there are no red chips to take away, we must add three zero pairs for each group, or nine zero pairs in all. Taking away three red chips from the group leaves three black chips in each group. Since there are three groups, a total of nine black chips remain. So -3 • (-3) = +9.

**Problem C6
**Start with an empty circle. We want to add three groups (+3) of three red chips (-3). We can do this directly. After doing this, there are nine red chips in the circle, so +3 • (-3) = -9.

**Problem C7
**

**a.**Using the partitive method, there are two black chips in each of four groups: (+8) / (+4) = +2.

**b. **Using the partitive method, there are two red chips in each of the two groups: (-4) / (+2) = -2.

**c. **Using the quotative method, there are four sets of two red chips: (-8) / (-2) = +4.

**d. **This division problem cannot be modeled using the colored-chip model.