Visual Arithmetic
Many students struggle with adding and subtracting integers – particularly when negatives are involved.
Unfortunately, most textbooks have a very procedure-oriented approach, and if you’ve struggled
memorizing procedures in the past, seeing those same procedures again might not help.
One alternative is to take a visual approach. This can be very helpful, particularly for visual learners.
To help visualize adding and subtracting integers, we’ll use two different symbols – one to represent a
positive 1 and one to represent a negative 1. (You can actually do this physically as well. Two different
types of beans work great!)
+
positive 1
–
negative 1
Signs
Before we try a few problems, it’s important to remember that every number has a sign. When we see
the symbols 6+5, what we really mean is “positive 6 plus positive 5”. Symbolically, that’d be (+6)+(+5).
You can see how cumbersome that becomes, which is why we abbreviate it with 6+5.
But that’s for positive numbers. Negative numbers can’t be treated the same way, so they always have
the negative sign. Otherwise, we wouldn’t know if the number we were dealing with was positive or
negative!
Zero
One of the key principles in the visual arithmetic of integers is the concept of zero. In principle, we could
think of a blank canvas as zero, but we can also treat a pair of one positive 1 and one negative 1 as zero.
+ –
zero
zero
This will be important later, so hold on to that idea of each pair being equivalent to zero.
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Adding
The first operation we’ll look at is addition. These first two examples illustrate the idea when we add
numbers that have the same sign.
Example 1: 6 + 4
Visually, we should start with 6 blue circles and add 4 blue circles.
++
++
++
+
+6
++
++
=
+++
++++
+++
+10
+4
In this example, we can see that since both 6 and 4 are positive, we’re just adding a bunch of
pieces of the same color to get a total of 10 blues – or positive 10.
Answer: 6 + 4 = 10. Easy stuff!
Example 2: (−6) + (−3)
Visually, we’ll add 6 red circles and 3 red circles.
– –
– –
– –
-6
+
–
–
–
-3
=
– – –
– – –
– – –
-9
Again, since we’re adding pieces of the same color, we just combine them to get 9 reds – or
negative 9.
Answer: (−6) + (−3) = −9.
Hopefully these are straight-forward so far. The next couple of examples show what happens when we
add integers that have opposite signs.
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Example 3: 11 + (−8)
Visually, this means we add 11 blue circles and 7 red circles.
+++
+
+++
++++
+
– –
–
– –
– – –
+11
-8
Well, now we have a problem – they’re not the same color! Here’s where the pairs adding to
zero come in. We can see clearly that there are fewer reds than blues, so pair up each red with
a blue – each pair is equivalent to zero – and leave the rest of the blues by themselves.
+++
+
+++
++++
+
– –
–
– –
– – –
+11
-8
=
+
+
+
+– +–
+–
+– +–
+– +– +–
+
=
8 “zeros”
+3
+
+
+
+3
If we look closely at what’s happening here, we can see that the 7 red negatives group with 7
blue positives to equal zero, leaving only the four blue positives.
Answer: 11 + (−8) = 3.
Let’s see what happens when the negative is larger.
Example 4: (−15) + 6
Visually, this means adding 15 red circles and 6 blue circles.
– – – – –
– – – – –
– – – – –
+
-15
+ +
+ +
+ +
+6
=
– – –
– – –
– – –
+
-9
–+ –+
–+ –+
–+ –+
6 “zeros”
=
– – –
– – –
– – –
-9
This is basically the same pattern as the previous example, except here there are more red
circles (the -15 is larger). In this example, we’re left with 9 reds – or negative nine.
Answer: (−15) + 6 = −9.
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Subtracting
Subtracting can be a little trickier, primarily in the cases when there isn’t enough to subtract. Confused?
Just wait!
Example 5: 9 − 3
Visually, this means starting with 9 blue circles and then taking 3 away.
+ + +
+ + +
+ + +
=
+ +
+ +
+ +
+6
+9
Answer: 9 − 3 = 6. Easy!
Example 6: (−8) − (−2)
For this example, we’ll start with 8 red circles and then take 2 red circles away.
– –
–
– –
– – –
=
– –
– –
– –
-6
-8
Answer: (−8) − (−2) = −6.
Example 7: 4 − 8
Visually, 4 − 8 means starting with 4 blue circles and then taking 8 blue circles away. ??? Wait,
how is that possible?
+ +
++
+4
There are only four blues – how can we take away 8? We need to somehow get more blues in,
without changing the value. Here’s where those handy zeros come in again.
+ +
++
+4
+ ++ ++ +++
– – – – – – – –
some “zeros”
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Now we have enough blues to take 8 away.
++ + ++ ++ +++
+ + – – – – – – – –
=
+ ++ +
– – – – – – – –
-4
+4
extra “zeros”
After we add the “zeros”, we now have 8 blues to take away, leaving us with 4 reds (and some
extra “zeros”) – or negative four.
Answer: 4 − 8 = −4.
One question you might have is how to decide the number of zeros to add. Or, you might be wondering
why I added all the extra zeros (and not just 4). These are good questions. To answer the first: there are
infinitely many zeros at your disposal, so feel free to add as many as you like. And second, you may
have noticed that we really only need to add enough to make up the “difference” between what we
have and what we need to take away. If you noticed that pattern – good for you! If not, take a minute
and look back over the previous example again before moving on.
Example 8: (−5) − (−12)
For this example, we should start with 5 reds and take away 12 reds. Hopefully you can
recognize the previous problem here as well – we don’t have 12 reds to take away! We’ll have
to add in some “zeros” again until we have enough.
–
+++++ ++
– –
– – – – – – –
– –
-5
=
++++ ++ +
+7
some “zeros”
(You’ll notice here that I only added 7 “zeros”, since that was really all I needed to get to 12 total
reds.)
Answer: (−5) − (−12) = 7.
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Example 9: −3 − 2
Concept Check: Take a quick look at this problem – if your first inclination is -1, you still need
practice!
For this one, we start with 3 reds and are supposed to take away 2 blues. ??? We don’t even
have any blues! Wait….we can always add some, right?
–
–
–
– –
+ +
-3
2 “zeros”
–
=
– –
– –
-5
Answer: −3 − 2 = −5.
You may be looking at this all right now and think “Isn’t there any easier way? Why can’t I just….”
And you know what, that’s great. Hopefully, after seeing these examples, you have some of your own
procedures in mind for doing arithmetic with positive and negative integers. There are some standard
procedures I could give you, but they’re probably the same ones you had a problem remembering
before, so what good is that?
If you don’t have any general strategies in mind for these problems, go back and look at the examples
again. They weren’t randomly chosen – they all illustrate key properties and relationships between
integers. The hope here is that after seeing a visual representation, you’ll be able to develop your own
strategies. You’ll be much more likely to remember the procedures you develop yourself.
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Now you try!
OK, now it’s time for you to try some. These problems are a mixture of problems from the various types
shown above. Answers are at the end of this handout.
1 −8 + 15
2 −12 + (−6)
3 −22 + 4
4 −14 − (−3)
5 −9 − 6
6 8 − (−12)
7 7 − 16
8 −26 − 14
9 −7 + 9
10 −4 − 1
Answers: 1. 7 2. −18 3. −18 4. −11 5. −15 6. 20 7. −9 8. −42 9. 2 10. −5
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