John Owen, Rockport Fulton HS 1

Computer Science

LESSON ON Number Base

Subtraction and Simple Equations

John Owen, Rockport Fulton HS 2

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 3

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 4

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 5

Subtraction, Base 10

Now look at this problem: 63 -37

In this problem, you need to borrow.

John Owen, Rockport Fulton HS 6

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 7

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 8

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 9

Subtraction, Base 10

513 63 -37 26

Then you subtract the two columns with a result of 26.

John Owen, Rockport Fulton HS 10

Subtraction, Base 8

Now let’s try base eight: 63 -37

Again, in this problem, you need to borrow.

John Owen, Rockport Fulton HS 11

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 12

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 13

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 14

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 15

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 16

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 17

Subtraction, Base 16

Here’s another example in base 16

D6 -3B

How is this one solved? Try it.

John Owen, Rockport Fulton HS 18

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 19

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 20

Subtraction, Base 2

Now base 2: 11 - 1 10 This one is easy…answer is 10

John Owen, Rockport Fulton HS 21

Subtraction, Base 2

Another in base 2: 02 110 - 1 Here we need to borrow from

the twos place…

John Owen, Rockport Fulton HS 22

Subtraction, Base 2

02 110 - 1 101Subtract to get the answer.

John Owen, Rockport Fulton HS 23

Subtraction, Base 2

Still another in base 2: 02 110 - 11 1Now borrow again…

John Owen, Rockport Fulton HS 24

Subtraction, Base 2

2 100 - 11 11Final answer is 11, base 2

John Owen, Rockport Fulton HS 25

Simple Equations

Here an equation to solve (base 10):

x + 6 = 14

John Owen, Rockport Fulton HS 26

Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14 -6 -6x = 8

John Owen, Rockport Fulton HS 27

Simple Equations

Now do it in base 8:

x + 6 = 14

John Owen, Rockport Fulton HS 28

Simple Equations

Solution…subtract 6 from both sides

x + 6 = 14 -6 -6x = ?

John Owen, Rockport Fulton HS 29

Simple Equations

Answer is 6, base 8 12x + 6 = 14 -6 -6x = 6

John Owen, Rockport Fulton HS 30

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 31

Simple Equations

Solution?

x + 2A = F3

John Owen, Rockport Fulton HS 32

Simple Equations

Subtract 2A from both sides: E19x + 2A = F3 - 2A -2Ax = C9

John Owen, Rockport Fulton HS 33

Exercises

Now try these exercises1. 12 - 12 =

2. 78 - 68 =

3. F16 - A16 =

4. 158 - 68 =

5. 4916 - 2B16 =

6. CC16 - AD16 =

John Owen, Rockport Fulton HS 34

Exercises

7. 738 - 348 =

8. 3E16 – 2F16 =

9. 1012 - 102 =

10. 11012 - 112 =

11. 10102 - 1112 =

12. 7168 - 3648 =

13. 7768 + 3378 =

John Owen, Rockport Fulton HS 35

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 36

Exercises

And finally, some equations19. 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 37

ANSWERS (JUMBLED)35

37

69

110

177

332

354

1010

0

1

5

7

11

11

14

19

1335

10000

14037

1E

1F

BF

F

F