# Binary Number System – Conversion of Base 2 to Base 10 | Conversion of Base 10 to Base 2 Primary 5 (Basic 5) Term 3 Week 8 Mathematics

### MATHEMATICS

THIRD TERM

WEEK 8

PRIMARY 6

THEME – NUMBERS AND NUMERATION

PREVIOUS LESSON – Measurement of Height and Distance | Conversion of Units in Height and Distance (Metres and kilometres) Primary 5 (Basic 5) Term 3 Week 6 Mathematics

### LEARNING AREA

1. Introduction

2. Conversion to Base 2 to Base 10

3. Conversion to Base 10 to Base 2

4. Lesson Evaluation and Weekly Assessment (Test)

### LEARNING OBJECTIVES

By the end of the lesson, most of the pupils should have attained the following objectives –

1. Conversion of Base 10 to Base 2.

2. Conversion of Base 2 to Base 10.

3. Solve Quantitative Reasoning.

### ENTRY BEHAVIOURS

The pupils can perform basic mathematics operation such as addition and  subtraction.

### INSTRUCTIONAL MATERIALS

The teacher will teach the lesson with the aid of chart showing relationships between binary (base 2) and decimal (base 10).

### METHOD OF TEACHING

Choose a suitable and appropriate methods for the lessons.

Note – Irrespective of choosing methods of teaching, always introduce an activities that will arouse pupil’s interest or lead them to the lessons.

### REFERENCE MATERIALS

1. Scheme of Work

2. 9 – Years Basic Education Curriculum

3. Course Book

4. All Relevant Material

5. Online Information

### CONTENT OF THE LESSON

LESSON 1 – INTRODUCTION

Binary numbers are composed of only 0 and 1, while decimal numbers are composed of digits from 0 to 9.

ACTIVITY 1 – CONVERSION OF BASE 10 TO BASE 2

Binary is a number system that represent a base 2 number system.

This means it only has two numbers: 0 and 1.

Relationship between decimal and binary is as follows:

0 = 0₂

1 = 1₂

2 = 10₂

3 = 11₂

4 = 100₂

5 = 101₂

6 = 110₂

7 = 111₂

8 = 1000₂

9 = 1001₂

10 = 1010₂

11 = 1011₂

12 = 1100₂

13 = 1101₂

14  = 1110₂

15 = 1111₂

16 = 10000₂

17 = 10001₂

18 = 10010₂

19 = 10011₂

20 = 10100₂

Illustration,

ACTIVITY 3 – CLASS EXERCISE/ASSIGNMENT

Convert the following to base 2:

1. 15

2. 23

3. 78

4. 137

5. 200

### LESSON 2 – CONTINUATION OF LESSON 1 (REVISION)

SOLUTIONS TO WORKING EXERCISE

1. 15

2 | 15

2 | 7 R 1

2 | 3 R 1

2 | 1 R 1

__ | 0 r 1 ↑

15 = 1111₂

2. 23

2 | 23

2 | 11 r 1

2 | 5 r 1

2 | 2 r 1

2 | 1 r 0

_ | 0 r 1 ↑

23 = 10111₂

3. 78

2 | 78

2 | 39 r 0

2 | 19 r 1

2 | 9 r 1

2 | 4 r 1

2 | 2 r 0

2 | 1 r 0

__| 0 r 1 ↑

78 = 1001110₂

4. 137

2 | 137

2 | 68 r 1

2 | 34 r 0

2 | 17 r 0

2 | 8 r 1

2 | 4 r 0

2 | 2 r 0

2 | 1 r 0

__| 0 r 1 ↑

137 = 10001001₂

5. 200

2 | 200

2 | 100 r 0

2 | 50 r 0

2 | 25 r 0

2 | 12 r 1

2 | 6 r 0

2 | 3 r 0

2 | 1 r 1

__| 0 r 1

200 = 11001000₂

### LESSON 3 – CONVERSION OF BASE 2 TO BASE 10

ACTIVITY 1 – INTRODUCTION

Decimal system is a number system that represent a 0 – 9 number system.

Relationship between binary and Decimal is as follows:

0₂ = 0

1₂ = 1

10₂ = 2

11₂ = 3

100₂ = 4

101₂ = 5

110₂ = 6

111₂ = 7

1000₂ = 8

1001₂ = 9

1010₂ = 10

1011₂ = 11

1100₂ = 12

1101₂ = 13

1110₂ = 14

1111₂ = 15

10000₂ = 16

10001₂ = 17

10010₂ = 18

10011₂ = 19

10100₂ = 20

WORKING EXAMPLE 1

1011₂

= 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2º

= 1 x 8 + 0 x 4 + 1 x 2 + 1 x 1

= 8 + 0 + 2 + 1

= 11

Therefore, 1011₂ = 11

WORKING EXAMPLE 2

1100₂

= 1 x 2³ + 1 x 2² + 0 x 2¹ + 0 x 2º

= 1 x 8 + 1 x 4 + 0 x 2 + 0 x 1

= 8 + 4 + 0 + 0

= 12

Therefore, 1100₂ = 12

WORKING EXAMPLE 3

1101₂

= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2º

= 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1

= 8 + 4 + 0 + 1

= 13

Therefore, 1101₂ = 13

WORKING EXAMPLE 4

11001₂

= 1 x 2^4 + 1 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2º

= 1 x 16 + 1 x 8 + 0 x 4 + 0 x 2 + 1 x 1

= 16 + 8 + 0 + 0 + 1

= 25

Therefore, 11001₂ = 25

WORKING EXAMPLE 5

1010110₂

= 1 x 2^6 + 0 x 2^5 + 1 x 2^4 + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2º

= 1 x 64 + 0 x 32 + 1 x 16 + 0 x 8 + 1 x 4 + 1 x 2 + 0 x 1 = 64 + 0 + 16 + 0 + 4 + 2 + 0

= 86

Therefore, 1010110₂ = 86

ACTIVITY 2 – CLASS EXERCISE/ASSIGNMENT

Convert the following to base 10,

1. 1111₂

2. 10111₂

3. 1001110₂

4. 10001001₂

### LESSON 3 – CONTINUATION OF LESSON 2 (REVISION)

LESSON 5 – WEEKLY ASSESSMENT (TEST)

### PRESENTATION

To deliver the lesson, the teacher adopts the following steps:

1. To introduce the lesson, the teacher revises the previous lesson. Based on this, he/she asks the pupils some questions;

2. Teacher organizes pupils in groups or pairs depending on the size of the class.

3. Teacher displays chart showing the relationship between binary and denary numbers.

4. Teacher lets pupils study the relationship between the two forms of number.

5. Teacher uses the chart and pupil’s responses to introduce the lesson and discuss the concept of binary system.

6. Teacher guides pupils to convert numbers in binary system to denary system and vice versa.

Pupil’s Activities – Convert numbers in binary system to other bases and vice versa.

7. Guides pupils to solve quantitative aptitude involving binary numbers system.

Pupil’s Activities – Solve quantitative aptitude involving binary number system.

8. Teacher summarizes each of the lesson on the board with appropriate lesson evaluation.

Pupil’s Activities – Participate actively in the summary of the lesson by responding correctly to the questions and write as instructed.

### CONCLUSION

To conclude the lesson for the week, the teacher revises the entire lesson and links it to the following week’s lesson.

### LESSON EVALUATION

As stated in the lessons.