# 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

**TOPIC – BINARY SYSTEM**

**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 2Binary 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 EXERCISEConvert the following to base 2:

1. 15

2. 23

3. 78

4. 137

5. 200

SOLUTIONS1. 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 – INTRODUCTIONDecimal 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 11011₂

= 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 21100₂

= 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 31101₂

= 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 411001₂

= 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 51010110₂

= 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 EXERCISEConvert the following to base 10,

1. 1111₂

2. 10111₂

3. 1001110₂

4. 10001001₂

### LESSON 3 – ADDITION AND SUBSTATION OF BINARY NUMBER

ACTIVITY 1 – INTRODUCTIONAddition and subtraction are much like your normal everyday addition (decimal addition), except that it carries on a value of 2 instead of a value of 10.

For example,

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

WORKING EXAMPLE 1¹1 ¹0 ¹1 ¹ 1 1₂

+ 1 1 1 1₂

_____________

1 0 0 1 1 0₂

Therefore,

10111₂ + 1111₂ = 100110₂

WORKING EXAMPLE 210111₂ – 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 EXERCISE1. 11011₂ + 10101₂

2. 1111₂ – 110011₂

3. 11011₂ – 10101₂

4. 11011₂ – 11010₂

ANSWER KEY1. 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 – INTRODUCTIONMultiplication is actually much similar and simpler to calculate than decimal multiplication.

MIND ON ACTIVITIES0 x 0 = 0

0 x 1 = 0

1 x 0 = 0

1 x 1 = 1

WORKING EXAMPLE 110111₂ 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 211101₂ 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 EXERCISE1. 1010₂ x 10₂

2. 1011₂ x 11₂

3. 1111₂ x 101₂

ANSWER KEY1. 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 – INTRODUCTIONThe 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 EXAMPLE11001₂ ÷ 101₂

SOLUTIONStep 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 EXERCISE1. 10010₂ ÷ 10₂

2. 10000₂ ÷ 100₂

3. 1111₂ ÷ 11₂

4. 11010₂ ÷ 1101₂

5. 11110₂ ÷ 1111₂

ANSWER KEY1. 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.

Pupil’s Activities – Follow the teacher’s lead 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.

Next Lesson –

Meaning and Types of Lines, Angles and Bearing Primary 5 – Term 3 Week 2 and Primary 6 Term 3 Week 3 Mathematics

**LESSON EVALUATION**

As stated in the lessons.