# Binary Number System

### LESSON – BINARY SYSTEM

###### TOPIC
• Introduction
• Conversion of Base 2 to Base 10
• Conversion of Base 10 to Base 2
• Addition and Subtraction of Binary Numbers
• Multiplication and Division Binary System

###### INTRODUCTION

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

###### 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,

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

###### CONVERSION OF BASE 2 TO BASE 10

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

1. 1111₂

2. 10111₂

3. 1001110₂

4. 10001001₂

###### ADDITION AND SUBSTATION OF BINARY NUMBERS

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.

###### CLASS EXERCISE 3 – Addition and Subtraction of Binary

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

###### MULTIPLICATION OF BINARY NUMBER

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

CLASS EXERCISE – Multiplication of Binary Numbers

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

###### 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

###### CLASS EXERCISE – Division of Binary Numbers

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