Number System – JSS1 Mathematics Lesson Note (Second Term)
Download Lesson NoteLesson note at a glance
- Subject
- Mathematics
- Class
- JSS1
- Term
- Second Term
- Topic
- Number System
- Order
- Topic 4 of 7 this term
- Reading time
- 2 min
Converting a number from one base to another is a two-stage process. First, change the given number to base ten, then change the result to the required base. Numbers in base two can also be added, subtracted and multiplied using simple rules for the digits 0 and 1.
Behavioural objectives
By the end of the lesson, students should be able to:
- convert a number from one base to another through base ten;
- add numbers in base two;
- subtract numbers in base two;
- multiply numbers in base two.
How do you convert a number from one base to another?
This is a two-stage process. First, convert the given base to base ten. Then convert the result to the required base.
Example 1: Convert 21356 to a number in base two
Step one: Convert to base ten.
21356 = (2 × 63) + (1 × 62) + (3 × 61) + (5 × 60)
= (2 × 216) + (1 × 36) + (3 × 6) + (5 × 1)
= 432 + 36 + 18 + 5
= 49110
Step two: Convert the answer in base ten to a number in base two by dividing repeatedly by 2. Read the remainders from the bottom upwards.
| Divide by | Number | Remainder |
|---|---|---|
| 2 | 491 | |
| 2 | 245 | 1 |
| 2 | 122 | 1 |
| 2 | 61 | 0 |
| 2 | 30 | 1 |
| 2 | 15 | 0 |
| 2 | 7 | 1 |
| 2 | 3 | 1 |
| 2 | 1 | 1 |
| 0 | 1 |
Therefore 21356 = 1111010112
Example 2: Express 101112 as a number in base three
101112 = (1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20)
= 1 × 16 + 0 × 8 + 1 × 4 + 1 × 2 + 1 × 1
= 16 + 0 + 4 + 2 + 1
= 2310
Now divide repeatedly by 3 and read the remainders from the bottom upwards.
| Divide by | Number | Remainder |
|---|---|---|
| 3 | 23 | |
| 3 | 7 | 2 |
| 3 | 2 | 1 |
| 0 | 2 |
Therefore 101112 = 2123
How do you add and subtract in base two?
Use the following rules.
Addition in base two
- 0 + 0 = 0
- 1 + 0 = 1
- 1 + 1 = 10
- 1 + 1 + 1 = 11
Subtraction in base two
- 0 – 0 = 0
- 1 – 0 = 1
- 10 – 1 = 1
- 11 – 1 = 10
Example 3: Add the following
- 11012 and 11112
- 110112, 101012 and 10012
Solution:
(1) Using the addition rule 1 + 1 = 10, write 0 down and carry the 1 to the next column. Carry the 1 in the same way in each column until you reach the last one.
11012
+ 11112
= 111002
(2) Add the three numbers in the same way.
110112
101012
+ 10012
= 1110012
Example 4: Subtract (a) 10112 from 101012 (b) 110112 from 1110102
Solution:
(a) Using the subtraction rule 1 – 1 = 0, write 0 down and move to the next column.
101012
– 10112
= 10102
(b)
1110102
– 110112
= 111112
How do you multiply binary numbers?
Multiply by each digit of the second number, move each row one place to the left, and then add the rows using the rules for addition in base two.
Example 5: Find the product of 10112 and 1012
Solution:
10112
× 1012
1011
0000 (moved one place left)
1011 (moved two places left)
= 1101112
Example 6: Multiply 11112 by 1102
Solution:
11112
× 1102
0000
1111 (moved one place left)
1111 (moved two places left)
= 10110102
Evaluation
- Add the following: (a) 10012 + 11112 (b) 1011102 + 100102
- Subtract 10112 from 11011012.
- Multiply 1101012 by 10112.
- Find the value of the following binary numbers: 10112 + 102 – 1112
This term's Mathematics topics
JSS1 Second Term · 7 topics


