Number Base Mathematics JSS 1 Second Term Lesson Note for Week Five

Second Term Lesson Note for Week Five

Class : JSS 1

Subject : Mathematics

Topic : Number Bases

Duration : 80 Minutes

Period : Double Period

Reference Book :

• Essential Mathematics for Junior Secondary School, JSS 1.
• Lagos State Unified schemes of work for Junior Secondary Schools, JSS 1 – 3.
• Online Resources

Instructional Material : Chart

Learning Objectives : By the end of the lesson learners will be able to :

Content :

NUMBER SYSTEM

5.1 Converting Numbers Base From One Base to Other Bases
This is a 2-stage conversion process. The steps to do this are simple.

First convert the given base to base ten and then convert the result to the required base.

Example 5.1: Convert 21356 to a number in base two

Solution :
Step one: Converting to base ten

3  2  1  0

2  1  3  5 ₆ = (2×6³) + (1×6²) + (3×6¹) + (5×6°)

= (2 x 216) + (1 x 36) + (3 x 6) + (5 x 1)
= 432 + 36 + 18 + 5

= 491 ten

Step two: Convert the answer in base ten to a number in base two
2 |491
2 | 245 R 1
2 |122 R 1
2 |  61 R 0
2  |30 R 1
2 | 15 R 0
2 |  7 R 1
2 |  3 R 1
2 |  1 R 1
|  0 R 1  ⇑
Therefore 2135₆ = 111101011₂ 

Examples 5.2 : Express 10111₂ to a number in base three

Solution :

First convert the binary to base ten, from base ten change to base three.

4  3  2  1  0
1.  1  0  1  1  1 = (1 X 2⁴) + (0 X 2³) + (1 X 2²) + (1 X 2¹) + (1 X 2°)
= (1 x 16) + (0 x 8) + (1 x 4) + (1 x 2) + (1 x 1)
= 16 + 0 + 4 + 2 + 1
= 23₁₀

Then divide through by three and record the remainder as shown below:

3   |23
3   | 7 R 2
3   | 2 R 1
| 0 R 2
Therefore, 10111₂ = 212 ₃

5.2 Addition and Subtraction of Number Base

Note the following STEPS in adding and subtracting.
To add in base two
0 + 0 = 0
1 + 0 = 1
1 + 1 = 10
1 + 1 + 1 = 11

To subtract in base two
0 – 0 = 0
1 – 0 = 1
10 – 1 = 1
11 – 1 = 10

Example 5.3 : Add the following

1. 11012 and 11112

2. 110112, 101012, and 10012

Solution:
1. 1101 + 1111 =  1 1 1 0 0 [using the addition steps 1 + 1 = 10, write 0 down and move the 1 to the next number]
+ 1 1 1 1 [also take the 1 to next, and move them till you get to last one].

1 1 0 1

 +  1 1 1 1

 1 1 1 0  0 ₂

2. 11011 + 10101 + 1001 =

1 1 0 1 1₂
1 0 1 0 1₂
+   1 0 0 1₂
1 1 1 0 0 1₂

Example 5.4: Subtract

(a) 1011₂ from 10101₂

(b) 11011₂ from 111010₂

Solution:
1. 1 0 1 0 1[using the subtraction steps

1 – 1 = 0, write 0 down and move to the next number]

1 0 1 0 1
–  1 0 1 12
 1 0 1 02

2. 1 1 1 0 1 02
-1 1 0 1 12
1 1 1 1 12

5.3 Multiplication of Binary
Example 5.5: Find the product of 10112 and 1012
Solution:
1 0 1 12
x 1 0 12
1 0 1 1
0 0 0 0
1 0 1 1
1 1 0 1 1 12

Example 5.6: multiply 1111 by 110

Solution:

		1	1	1	1
		x	1	1	0
		0	0	0	0
	1	1	1	1
1	1	1	1
10	1	1	0	1	0 2

 

EVALUATION :

DO THESE
1. Add the following :

(a) 1001₂ + 1111₂

(b) 101110₂ + 10010₂

2. Subtract 1011₂ from 1101101₂

3. Multiply 110101₂ by 1011₂

4. find the value of the following binary numbers 1011_² + 10_²  – 111_²

ASSIGNMENT:

EXERCISE 9.5 : page 107, NO. 1(B,C,D), NO. 2 A and B, NO. 3 (E,F,G,H)

EXERCISE 9.6 : Page 108,  NO. 2, 5, 8 and 11.