# Binary Number System – Conversion, Addition, Subtraction, Multiplication and Division Primary 6 (Basic 6) – Mathematics

Last Updated on September 19, 2021 by Alabi M. S.

### MATHEMATICS

FIRST TERM

WEEK 2

PRIMARY 6

THEME –

PREVIOUS LESSON – Reading and Write Numerals Up to One Billion in Words and Figures Primary 6 (Basic 6) – Mathematics

**TOPIC – BINARY NUMBERS **

**LEARNING AREA**

1. Introductory Activities

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)

### PERFORMANCE 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 is a number system that represent a base 2 number system. This means it only has two numbers: 0 and 1.

**CONVERSION OF BASE 10 TO BASE 2**

Relationship between decimal and binary.

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

LESSON EVALUATIONConvert the following to base 2:

1. 15

2. 23

3. 78

4. 137

5. 200

**LESSON 2 – CONVERSION OF BASE 2 TO BASE 10**

Relationship between binary and decimal

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 EXAMPLES1. 1011 = 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,

1011 = 11

2. 1100 = 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,

1100 = 12

3. 1101 = 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,

1101 = 13

4. 11001 = 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

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

LESSONEVALUATIONConvert the following to base 10,

1. 1111₂

2. 10111₂

3. 1001110₂

4. 10001001₂

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

Addition 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 1 + 1 = 10.

WORKING EXAMPLES1. 10111₂ + 1111₂

2. 10111 – 1111

ADDITION OF BINARY NUMBER1 0 1 1 1₂

+ 1 1 1 1₂

_____________

1 0 0 1 1 0₂

Therefore,

10111₂ + 1111₂ = 100110₂

SUBTRACTION OF BINARY NUMBER

SOLUTION 1 – 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.

LESSON EVALUATIONAdd:

1. 11011₂ + 10101₂

2. 1111₂ – 110011₂

Subtract:

3. 11011₂ – 10101₂

4. 11011₂ – 11010₂

**LESSON 4 – MULTIPLICATION OF BINARY NUMBER **

Multiplication 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 EXAMPLESSimplify:

1. 10111₂ x 1111₂

2. 11101₂ x 1011₂

**LESSON 5 – DIVISION OF BINARY NUMBERS (OPTIONAL) **

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 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

**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. Displays chart showing the relationship between binary and denary numbers.

3. Lets pupils study the relationship.

4. 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.

5. Leads the pupils to add, subtract, multiply and divide binary numbers.

Pupil’s Activities – Pay attention to 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.

CONCLUSIONTo 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.