Computer Science
LESSON ON
Number Base
Subtraction and Simple
Equations
John Owen, Rockport Fulton HS
1
Objective


In this lesson you’ll learn how to do
subtraction and how to solve
simple equations involving Base 2,
8, and 16.
Again, it is essentially the same
concept as Base 10, just in a
different base!
John Owen, Rockport Fulton HS
2
Review Base Ten Subtraction


In Base 10 subtraction, you use a
very simple process.
Look at this problem:
48
-37
= 11
John Owen, Rockport Fulton HS
3
Review Base Ten Subtraction

48
-37
= 11
Each column is subtracted to get
an answer of 11…pretty easy, huh?
John Owen, Rockport Fulton HS
4
Subtraction, Base 10


Now look at this problem:
63
-37
In this problem, you need to
borrow.
John Owen, Rockport Fulton HS
5
Subtraction, Base 10
513
63
-37

Borrowing means taking a value
from the next column and adding it
to the column you need.
John Owen, Rockport Fulton HS
6
Subtraction, Base 10
513
63
-37

In this case, borrow from the 6,
which becomes five, and add 10 to
the 3, making 13.
John Owen, Rockport Fulton HS
7
Subtraction, Base 10
513
63
-37

When you borrow 1 from one column, it
becomes the value of the base in the
next column, or 10 in this case.
John Owen, Rockport Fulton HS
8
Subtraction, Base 10
513
63
-37
26

Then you subtract the two columns
with a result of 26.
John Owen, Rockport Fulton HS
9
Subtraction, Base 8


Now let’s try base eight:
63
-37
Again, in this problem, you need to
borrow.
John Owen, Rockport Fulton HS
10
Subtraction, Base 8
511
63
-37

Borrow from the 6, which becomes
five, and add 8 to the 3, making
11!
John Owen, Rockport Fulton HS
11
Subtraction, Base 8
511
63
-37

When you borrow 1 from a column,
it becomes the value of the base in
the next column, or 8 in this case.
John Owen, Rockport Fulton HS
12
Subtraction, Base 8
511
63
-37
24

Then you subtract the two columns
with a result of 24, base 8.
John Owen, Rockport Fulton HS
13
Subtraction, Base 16
Now base 16:
519
63
-37

Again, we borrow from the 6, which
becomes five, and add 16 to the 3,
making 19!
John Owen, Rockport Fulton HS
14
Subtraction, Base 16
519
63
-37

When you borrow 1 from a column, it
becomes the value of the base in the
next column, or 16 in this case.
John Owen, Rockport Fulton HS
15
Subtraction, Base 16

519
63
-37
2C
In the ones column, 19 minus 7 is
12, which is C in base sixteen, with
2 in the second column.
John Owen, Rockport Fulton HS
16
Subtraction, Base 16
Here’s another example in base 16
D6
-3B

How is this one solved? Try it.
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17
Subtraction, Base 16
C22
D6
-3B

We must borrow from D, which
becomes C, then add 16 to 6,
which makes 22.
John Owen, Rockport Fulton HS
18
Subtraction, Base 16
C22
D6
-3B
9B



22 minus B (11) is B.
C minus 3 is 9.
Answer is 9B
John Owen, Rockport Fulton HS
19
Subtraction, Base 2
Now base 2:
11
- 1
10

This one is easy…answer is 10
John Owen, Rockport Fulton HS
20
Subtraction, Base 2
Another in base 2:
02
110
- 1

Here we need to borrow from the
twos place…
John Owen, Rockport Fulton HS
21
Subtraction, Base 2
02
110
- 1
101
Subtract to get the answer.
John Owen, Rockport Fulton HS
22
Subtraction, Base 2
Still another in base 2:
02
110
- 11
1
Now borrow again…
John Owen, Rockport Fulton HS
23
Subtraction, Base 2
2
100
- 11
11
Final answer is 11, base 2
John Owen, Rockport Fulton HS
24
Simple Equations
Here an equation to solve (base 10):
x + 6 = 14
John Owen, Rockport Fulton HS
25
Simple Equations
Solution…subtract 6 from both sides
x + 6 = 14
-6
-6
x
= 8
John Owen, Rockport Fulton HS
26
Simple Equations
Now do it in base 8:
x + 6 = 14
John Owen, Rockport Fulton HS
27
Simple Equations
Solution…subtract 6 from both sides
x + 6 = 14
-6
-6
x
= ?
John Owen, Rockport Fulton HS
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Simple Equations
Answer is 6, base 8
12
x + 6 = 14
-6
-6
x
= 6
John Owen, Rockport Fulton HS
29
Simple Equations
Here’s an equation in base sixteen
(remember, A and F are NOT variables,
but base sixteen values):
x + 2A = F3
John Owen, Rockport Fulton HS
30
Simple Equations
Solution?
x + 2A = F3
John Owen, Rockport Fulton HS
31
Simple Equations
Subtract 2A from both sides:
E19
x + 2A = F3
- 2A -2A
x
= C9
John Owen, Rockport Fulton HS
32
Exercises

1.
2.
3.
4.
5.
6.
Now try these exercises
12 - 12 =
78 - 68 =
F16 - A16 =
158 - 68 =
4916 - 2B16 =
CC16 - AD16 =
John Owen, Rockport Fulton HS
33
Exercises
7.
8.
9.
10.
11.
12.
13.
738 - 348 =
3E16 – 2F16 =
1012 - 102 =
11012 - 112 =
10102 - 1112 =
7168 - 3648 =
7768 + 3378 =
John Owen, Rockport Fulton HS
34
Exercises
Now let’s mix it up a bit!
14. AE16 + 768 = _________8
15. 2348 + 110110112 = _________16
16. 101102 - F16 + 768 = _________10
17. 38 + 3910 - 1101012 = _________16
18. 11112 - F16 + 1510 = _________16
John Owen, Rockport Fulton HS
35
Exercises
And finally, some equations
19. x16 + 7616 = AB16
20. x2 - 10112 = 1012
21. x8 + 568 = 728
22. x2 + 2510 = 1F16
23. x8 + 3748 - 65568 = BAD16
24. 378 + X16 = 110111102
John Owen, Rockport Fulton HS
36
ANSWERS (JUMBLED)
0
1
5
7
11
11
14
19
35
37
69
110
177
332
354
1010
1335
10000
14037
1E
1F
BF
F
F
John Owen, Rockport Fulton HS
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