Binary Number System – Conversion, Addition, Subtraction, Multiplication and Division Primary 6 (Basic 6) Term 3 Week 2 Mathematics

MATHEMATICS

FIRST TERM

WEEK 2

PRIMARY 6

THEME –

PREVIOUS LESSON – Ratio and Proportion Primary 6 (Basic 6) Second Term Week 2 and Third Term Week 1 (Revision) Mathematics

LEARNING AREA

1. Introduction

2. Conversion to Base 2

3. Conversion to Base 10

4. Addition and Subtraction of Binary Numbers

5. Multiplication and Division of Binary Numbers

6. 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. Perform Basic Addition and Subtraction of Binary Number.

4. Perform Basic Multiplication of Binary Number.

5. Solve Quantitative Reasoning.

ENTRY BEHAVIOURS

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

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

Convert the following to base 2:

1. 15

2. 23

3. 78

4. 137

5. 200

SOLUTIONS

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 2 – 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

Convert the following to base 10,

1. 1111₂

2. 10111₂

3. 1001110₂

4. 10001001₂

LESSON 3 – ADDITION AND SUBSTATION OF BINARY NUMBER

ACTIVITY 1 – INTRODUCTION

For example,

8 + 2 = 10 (decimal number), corresponding value in binary is 1₂ + 1₂ = 10.

WORKING EXAMPLE 1

¹1  ¹0  ¹1 ¹ 1  1₂

+  1  1  1  1₂

_____________

1  0  0  1  1  0₂

Therefore, 10111₂ + 1111₂ = 100110₂

WORKING EXAMPLE 2

10111₂ – 1111₂

1º  0²  1  1  1₂

–  1  1  1  1₂

_____________

1  0  0  0₂

Therefore, 10111₂ – 1111₂ = 1000₂

Note – 1 borrow from the next digit is equal to 2 just as in decimal, the one you borrow from the next digit is equal to 10.

ACTIVITY 2 – CLASS EXERCISE

1. 11011₂ + 10101₂

2. 1111₂ – 110011₂

3. 11011₂ – 10101₂

4. 11011₂ – 11010₂

1. 11011₂ + 10101₂

1 1 0 1 1₂

+ 1 0 1 0 1₂

_____________

1 1 0 0 0 0₂

2. 1111₂ – 1011₂

1 1 1 1₂

– 1 0 1 1₂

_____________

1 0 0₂

3. 11011₂ – 10101₂

1 1 0 1 1₂

– 1 0 1 0 1₂

_____________

1 1 0₂

4. 11011₂ – 11010₂

11011₂

– 11010₂

_____________

1₂

LESSON 4 – MULTIPLICATION OF BINARY NUMBER

ACTIVITY 1 – INTRODUCTION

Multiplication is actually much similar and simpler to calculate than decimal multiplication.

MIND ON ACTIVITIES

0 x 0 = 0

0 x 1 = 0

1 x 0 = 0

1 x 1 = 1

WORKING EXAMPLE 1

10111₂ x 11₂

1 0 1 1 1₂

x           1 1₂

_____________

1 0 1 1 1₂

+ 1 0 1 1 1₂…

_____________

1 0 0 0 1 0 1₂

WORKING EXAMPLE 2

11101₂ x 101₂

1 1 1 0 1₂

x 1 0 1₂

_____________

1 1 1 0 1₂

0 0 0 0 0₂..

+ 1 1 1 0 1₂….

_____________

1 0 0 1 0 0 0 1₂

ACTIVITY 2 – CLASS EXERCISE

1. 1010₂ x 10₂

2. 1011₂ x 11₂

3. 1111₂ x 101₂

1. 1010₂ x 10₂

1 0 1 0₂

x 1 0₂

_____________

0 0 0 0₂

+ 1 0 1 0₂..

_____________

1 0 1 0 0₂

2. 1011₂ x 11₂

1011₂

x 11₂

_____________

1 0 1 1₂

+ 1 0 1 1₂..

_____________

1 0 0 0 0 1₂

3. 1111₂ x 101₂

1 1 1 1₂

x 1 0 1₂

_____________

1 1 1 1₂

0 0 0 0₂..

+ 1 1 1 1₂….

_____________

1 0 0 1 0 1 1₂

LESSON 5 – DIVISION OF BINARY NUMBERS (OPTIONAL)

ACTIVITY 1 – INTRODUCTION

The easy way to solve division of binary numbers is to,

• Step 1 – convert the two binary numbers to base 10.
• Step 2 – Divide the dividend by the divisor.
• Step 3 – Convert the quotient back to binary number.

WORKING EXAMPLE

11001₂ ÷ 101₂

SOLUTION

Step 1 – Convert to base 10

• 11001₂ = 25
• 101₂ = 5

Step 2 – Divide dividend by divisor

25 ÷ 5 = 5

Step 3 – Convert 5 to binary.

5 = 101₂

ACTIVITY 2 – CLASS EXERCISE

1. 10010₂ ÷ 10₂

2. 10000₂ ÷ 100₂

3. 1111₂ ÷ 11₂

4. 11010₂ ÷ 1101₂

5. 11110₂ ÷ 1111₂

1. 10010₂ ÷ 10₂

10010₂ = 18

10₂ = 2

18 ÷ 2 = 9

Therefore, 10010₂ ÷ 10₂ = 1001₂

2. 10000₂ ÷ 100₂

10000₂ = 16

100₂ = 4

16 ÷ 4 = 4

10000₂ ÷ 100₂ = 100₂

3. 1111₂ ÷ 11₂

1111₂ = 15

11₂ = 3

15 ÷ 3 = 5

Therefore, 1111₂ ÷ 11₂ = 101₂

4. 11010₂ ÷ 1101₂

11010₂ = 26

1101₂ = 13

26 ÷ 13 = 2

Therefore, 11010₂ ÷ 1101₂ = 10₂

5. 11110₂ ÷ 1111₂

11110₂ = 30

1111₂ = 15

30 ÷ 15 = 2

Therefore, 11110₂ ÷ 1111₂ = 10₂

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.

6. Teacher uses the basic conversion between binary and decimal numbers to lead the groups or pairs to add, subtract, multiply and divide binary numbers.

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

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

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

Pupil’s Activities –

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.