Chapter 2: Data Representation

Introduction

As we know that computer system stores any data in binary form that’s why we use to add the word DIGITAL with all the work related to computer. The data stored in computer is knows as Digital Data.

In this chapter we will see various techniques to represent data in a computer system.

Human beings have adopted Decimal Number System in their day to day life. In the same manner, computer system has adopted Binary Number System, Octal Number System and Hexadecimal Number System which are combinedly known as Digital Number System.

These Number Systems are -

Decimal Number System

Decimal system is consists of 10 digits which are as under

• 0,1,2,3,4,5,6,7,8,9

Base of this system is 10 and it is to be shown as-

• (5479)₁₀

This technique is based on positional value where the weightage of a digit is as per its position.

For ex- in number 526, the value of 5 is 500, value of 2 is 20 and value of 6 is 6. (it is as per the method of hundreds , tens and ones).

We can write above given example as-

486 = 4 X 102 + 8 X 101 + 6 X 100

The left most digit is called MSD (Most Significant Digit ).
The right most digit is called LSD (Least Significant Digit )

Binary Number System

Binary system consists of 2 digits 0,1 known as bit.
Base of this system is 2 and it is to be shown as (1001010101)₂

In Digital systems, use of decimal system is impossible therefore use of binary system for a computer system is meaningful. Use of circuit to maintain two voltage level is very easy.

See the examples of binary number method-

1010 = 1 X 23 + 0 X 22 + 1 X 21 + 0 X 20

10.11 = 1 X 21 + 0 X 20 + 1 X 2-1 + 1 X 2-2

The left most digit is called MSB (Most Significant Bit ).
The right most digit is called LSB (Least Significant Bit ).

Octal Number System

Octal system consists of 8 digits which are as under.

0,1,2,3,4,5,6,7

Base of this system is 8 and it is to be shown as (564)₈

See the examples of Octal number method-

2256 =2 X 83 + 2 X 82 + 5 X 81 + 6 X 80

Hexadecimal Number System

Octal system consists of 16 digits which are as under.

0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

Here A=10, B=11, C=12, D=13, E=14, F=15

Base of this system is 16 and it is to be shown as (BCD54)₁₆

See the examples of Hexadecimal number method-

AC23 =10 X 163 + 12 X 162 + 2 X 161 + 3 X 160

Relation between different Number Systems

Number Conversion

Follow the following diagram to convert a number from one system to another.

QUESTION /ANSWER

Question 1

What are the bases of decimal, octal, binary and hexadecimal systems ?

Answer

The bases are:

Decimal — Base 10
Octal — Base 8
Binary — Base 2
Hexadecimal — Base 16

Question 2

What is the common property of decimal, octal, binary and hexadecimal number systems ?

Answer

Decimal, octal, binary and hexadecimal number systems are all positional-value system.

Question 3

Complete the sequence of following binary numbers : 100, 101, 110, ............... , ............... , ............... .

Answer

100, 101, 110, 111 , 1000 , 1001 .

Question 4

Complete the sequence of following octal numbers : 525, 526, 527, ............... , ............... , ............... .

Answer

525, 526, 527, 530 , 531 , 532 .

Question 5

Complete the sequence of following hexadecimal numbers : 17, 18, 19, ............... , ............... , ............... .

Answer

17, 18, 19, 1A , 1B , 1C .

Question 6

Convert the following binary numbers to decimal and hexadecimal:

(a) 1010

(b) 111010

(d) 1100

(e) 10010101

(f) 11011100

Answer

(a) 1010

Converting to decimal:

Binary No	Power	Value	Result
0 (LSB)	2⁰	1	0x1=0
1	2¹	2	1x2=2
0	2²	4	0x4=0
1 (MSB)	2³	8	1x8=8

Equivalent decimal number = 8 + 2 = 10

Therefore, (1010)₂ = (10)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{1010}$

Binary Number	Equivalent Hexadecimal
1010	A (10)

Therefore, (1010)₂ = (A)₁₆

(b) 111010

Converting to decimal:

Binary No	Power	Value	Result
0 (LSB)	2⁰	1	0x1=0
1	2¹	2	1x2=2
0	2²	4	0x4=0
1	2³	8	1x8=8
1	2⁴	16	1x16=16
1 (MSB)	2⁵	32	1x32=32

Equivalent decimal number = 32 + 16 + 8 + 2 = 58

Therefore, (111010)₂ = (58)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{0011}$

Binary Number	Equivalent Hexadecimal
1010	A (10)
0011	3

Therefore, (111010)₂ = (3A)₁₆

Converting to decimal:

Binary No	Power	Value	Result
1 (LSB)	2⁰	1	1x1=1
1	2¹	2	1x2=2
1	2²	4	1x4=4
1	2³	8	1x8=8
1	2⁴	16	1x16=16
0	2⁵	32	0x32=0
1	2⁶	64	1x64=64
0	2⁷	128	0x128=0
1 (MSB)	2⁸	256	1x256=256

Equivalent decimal number = 256 + 64 + 16 + 8 + 4 + 2 + 1 = 351

Therefore, (101011111)₂ = (351)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{0001}$

Binary Number	Equivalent Hexadecimal
1111	F (15)
0101	5
0001	1

Therefore, (101011111)₂ = (15F)₁₆

(d) 1100

Converting to decimal:

Binary No	Power	Value	Result
0 (LSB)	2⁰	1	0x1=0
0	2¹	2	0x2=0
1	2²	4	1x4=4
1 (MSB)	2³	8	1x8=8

Equivalent decimal number = 8 + 4 = 12

Therefore, (1100)₂ = (12)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{1100}$

Binary Number	Equivalent Hexadecimal
1100	C (12)

Therefore, (1100)₂ = (C)₁₆

(e) 10010101

Converting to decimal:

Binary No	Power	Value	Result
1 (LSB)	2⁰	1	1x1=1
0	2¹	2	0x2=0
1	2²	4	1x4=4
0	2³	8	0x8=0
1	2⁴	16	1x16=16
0	2⁵	32	0x32=0
0	2⁶	64	0x64=0
1 (MSB)	2⁷	128	1x128=128

Equivalent decimal number = 1 + 4 + 16 + 128 = 149

Therefore, (10010101)₂ = (149)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{1001}$

Binary Number	Equivalent Hexadecimal
0101	5
1001	9

Therefore, (101011111)₂ = (95)₁₆

(f) 11011100

Converting to decimal:

Binary No	Power	Value	Result
0 (LSB)	2⁰	1	0x1=0
0	2¹	2	0x2=0
1	2²	4	1x4=4
1	2³	8	1x8=8
1	2⁴	16	1x16=16
0	2⁵	32	0x32=0
1	2⁶	64	1x64=64
1 (MSB)	2⁷	128	1x128=128

Equivalent decimal number = 4 + 8 + 16 + 64 + 128 = 220

Therefore, (11011100)₂ = (220)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underset{undefined}{1101}$

Binary Number	Equivalent Hexadecimal
1100	C (12)
1101	D (13)

Therefore, (11011100)₂ = (DC)₁₆

Question 7

Convert the following decimal numbers to binary and octal :

(a) 23

(b) 100

(d) 19

(e) 121

(f) 161

Answer

(a) 23

Converting to binary:

2	Quotient	Remainder
2	23	1 (LSB)
2	11	1
2	5	1
2	2	0
2	1	1 (MSB)
	0

Therefore, (23)₁₀ = (10111)₂

Converting to octal:

8	Quotient	Remainder
8	23	7 (LSB)
8	2	2 (MSB)
	0

Therefore, (23)₁₀ = (27)₈

(b) 100

Converting to binary:

2	Quotient	Remainder
2	100	0 (LSB)
2	50	0
2	25	1
2	12	0
2	6	0
2	3	1
2	1	1 (MSB)
	0

Therefore, (100)₁₀ = (1100100)₂

Converting to octal:

8	Quotient	Remainder
8	100	4 (LSB)
8	12	4
8	1	1 (MSB)
	0

Therefore, (100)₁₀ = (144)₈

Converting to binary:

2	Quotient	Remainder
2	145	1 (LSB)
2	72	0
2	36	0
2	18	0
2	9	1
2	4	0
2	2	0
2	1	1 (MSB)
	0

Therefore, (145)₁₀ = (10010001)₂

Converting to octal:

8	Quotient	Remainder
8	145	1 (LSB)
8	18	2
8	2	2 (MSB)
	0

Therefore, (145)₁₀ = (221)₈

(d) 19

Converting to binary:

2	Quotient	Remainder
2	19	1 (LSB)
2	9	1
2	4	0
2	2	0
2	1	1 (MSB)
	0

Therefore, (19)₁₀ = (10011)₂

Converting to octal:

8	Quotient	Remainder
8	19	3 (LSB)
8	2	2 (MSB)
	0

Therefore, (19)₁₀ = (23)₈

(e) 121

Converting to binary:

2	Quotient	Remainder
2	121	1 (LSB)
2	60	0
2	30	0
2	15	1
2	7	1
2	3	1
2	1	1 (MSB)
	0

Therefore, (121)₁₀ = (1111001)₂

Converting to octal:

8	Quotient	Remainder
8	121	1 (LSB)
8	15	7
8	1	1 (MSB)
	0

Therefore, (121)₁₀ = (171)₈

(f) 161

Converting to binary:

2	Quotient	Remainder
2	161	1 (LSB)
2	80	0
2	40	0
2	20	0
2	10	0
2	5	1
2	2	0
2	1	1 (MSB)
	0

Therefore, (161)₁₀ = (10100001)₂

Converting to octal:

8	Quotient	Remainder
8	161	1 (LSB)
8	20	4
8	2	2 (MSB)
	0

Therefore, (161)₁₀ = (241)₈

EXCERCISE

A. Convert the following Hexadecimal numbers to Binary Number:

(a) A6

(b) A07

(d) BE

(e) BC9

(f) 9BC8

B. Convert the following Decimal numbers to Binary Number:

(a) 543

(b) 796

(d) 7654

(e) 7432

(f) 7453

C. Convert the following Binary numbers to Decimal Number:

(a) 1000111

(b) 111101

(d) 11011001

(e) 101010101

(f) 10001001

D. Convert the following Octal numbers to Decimal Number:

(a) 524

(b) 354

(d) 7532

(e) 1234

(f) 6524

Class XI Computer Science(083) Chapter 2: Data Representation (Number System)

Chapter 2: Data Representation

Introduction

Decimal Number System

Binary Number System

Octal Number System

Hexadecimal Number System

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

EXCERCISE

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