We have discussed about addition of integers and its representation as chips. Recall in our previous discussion that a single positive chip added up to a single negative chip is equal to 0. The addition below as we have discussed will give us a notion of how to add signed numbers.

Now let us discuss on how use signed chips can be used as a strategy to subtract integers. Let us represent the following using chips.

^{+}7 –^{+}4–^{+}3^{+}**5**– (^{+}5)^{–}3– (^{–}3)^{–}4**-5**–^{+}2

Before discussing subtraction of integers using signed chips, let us recall some concepts. First, we know that any number added to 0 is equal to that number. That is, if *n* + 0 = *n* for any number *n*. Second, if we add the same number of positive and negative chips to *n*, then we are just adding 0; therefore, it does not change the value of *n*.

We will also agree on the following. First, we can think of subtraction as “taking away,” and second, when we say positive chip, we mean^{+}1 chip, unless stated otherwise. This means that we when we say 3 positive chips, we mean three pieces of ^{+}1 chips.

**Case 1:** ^{+}7 – ^{+}3

The solution for case 1 is very obvious. ^{+}7 – ^{+}4 is equal to ^{+}3. But if we represent it using the chips, we have 7 positive chips and we want to take away 3 positive chips. So, the answer is 3 chips. Again, ^{+}7 – ^{+}3 = ^{+}4.

**Case 2: ^{+}3 – ^{+}5 **

In the second problem, we have three positive chips, and we want to take away 5 positive chips. We have a slight problem; we don’t have enough chips to take away!

Hmm…. but we can use the second concept that we have recalled above. . In *A* (in the second figure), we need 5 chips so we can add 2 more positive chips, but that will change our original number of chips to ^{+}5. To remedy this problem, we add another two negative chips to cancel out the ^{+}2 that we have added. Note that we added ^{+}2 and added ^{–}2, which means that we added 0. In *B*, we represented ^{+}3 with ^{+}5 and ^{–}2 chips. Now, we can take away ^{+}5 and we are left with -2. Therefore, ^{+}3 – ^{+}5 = ^{–}2.

**Case 3: ^{+}5 – (^{–}3) **

In the third case, we have 5 positive chips, and we want to take away 3 negative chips. Again we don’t have negative chips.** ** Just like in case 2, we add -3 and +3 chips which mean that just added 0. Therefore ^{+}5 – (^{–}3) = ^{+}8

**Case 4: ^{–}3** – (

**)**

^{–}4In A, we have 3 negative chips and we take away 4 negative chips, so we add 1 negative chip and 1 positive chip which is equivalent to adding 0. In B, we already have four negative chips and 1 positive chip; we take away four negative chips and we are left with 1 positive chip. Therefore,** ^{ –}3** – (

**) =**

^{–}4

^{+}1

**Case 5: ^{–}5** –

^{+}2In A, we have 5 negative chips and we want to take away 2 negative chips. So we add 0 by adding 2 positive chips and 2 negative chips. In B, we have 7 negative chips and 2 positive chips. We take away ^{+}2. We are left with 7 negative chips. Therefore, ** ^{–}5** –

^{+}2 =^{–}7

The five cases above also review us of what we have learned in addition of integers. In case 2, we have ^{+}3 = ^{+}5 + (^{–}2); in case 3, we have ^{+}8 + (^{–}3) = ^{+}5; in case 4, we have ^{–}3 = ^{–}4 + ^{+}1; and in case 5, we have ^{–}5 = ^{–}7 + ^{+}2.