1. LESSON ON Number Base Subtraction and Simple Equations
Computer Science
LESSON ON
Number Base Subtraction and Simple Equations
John Owen, Rockport Fulton HS

2. John Owen, Rockport Fulton HS
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 Subtraction48
-37
= 11
Each column is subtracted to get an answer of 11…pretty easy, huh?
John Owen, Rockport Fulton HS

5. John Owen, Rockport Fulton HS
Subtraction, Base 10
Now look at this problem: 63
-37
In this problem, you need to borrow.
John Owen, Rockport Fulton HS

6. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
Subtraction, Base 10
513 63
-37 26
Then you subtract the two columns with a result of 26.
John Owen, Rockport Fulton HS

10. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
Subtraction, Base 2
Now base 2: 11
- 1 10
This one is easy…answer is 10
John Owen, Rockport Fulton HS

21. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
Subtraction, Base 2 02
110
- 1
101
Subtract to get the answer.
John Owen, Rockport Fulton HS

23. John Owen, Rockport Fulton HS
Subtraction, Base 2
Still another in base 2: 02
110
- 11 1
Now borrow again…
John Owen, Rockport Fulton HS

24. John Owen, Rockport Fulton HS
Subtraction, Base 2 2
100
- 11 11
Final answer is 11, base 2
John Owen, Rockport Fulton HS

25. John Owen, Rockport Fulton HS
Simple Equations
Here an equation to solve (base 10):
x + 6 = 14
John Owen, Rockport Fulton HS

26. John Owen, Rockport Fulton HS
Simple Equations
Solution…subtract 6 from both sides
x + 6 = 14
x = 8
John Owen, Rockport Fulton HS

27. John Owen, Rockport Fulton HS
Simple Equations
Now do it in base 8:
x + 6 = 14
John Owen, Rockport Fulton HS

28. John Owen, Rockport Fulton HS
Simple Equations
Solution…subtract 6 from both sides
x + 6 = 14
x = ?
John Owen, Rockport Fulton HS

29. John Owen, Rockport Fulton HS
Simple Equations
Answer is 6, base 8 12
x + 6 = 14
x = 6
John Owen, Rockport Fulton HS

30. John Owen, Rockport Fulton HS
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. John Owen, Rockport Fulton HS
Simple Equations
Solution?
x + 2A = F3
John Owen, Rockport Fulton HS

32. John Owen, Rockport Fulton HS
Simple Equations
Subtract 2A from both sides:
E19
x + 2A = F3
- 2A -2A
x = C9
John Owen, Rockport Fulton HS

33. John Owen, Rockport Fulton HS
Exercises
Now try these exercises = =
F16 - A16 = =
B16 =
CC16 - AD16 =
John Owen, Rockport Fulton HS

34. John Owen, Rockport Fulton HS
Exercises =
3E16 – 2F16 = = = = = =
John Owen, Rockport Fulton HS

35. John Owen, Rockport Fulton HS
Exercises
Now let’s mix it up a bit!
AE = _________8
= _________16
F = _________10
= _________16
F = _________16
John Owen, Rockport Fulton HS

36. John Owen, Rockport Fulton HS
Exercises
And finally, some equations
x = AB16
x = 1012
x = 728
x = 1F16
x = BAD16
378 + X16 =
John Owen, Rockport Fulton HS

37. John Owen, Rockport Fulton HS
ANSWERS (JUMBLED) 1 5 7 11 14 19 35 37 69
110
177
332
354
1010
1335
10000
14037 1E 1F BF F
John Owen, Rockport Fulton HS