Data Representation

The bases are:

1. Decimal — Base 10
2. Octal — Base 8
3. Binary — Base 2
4. Hexadecimal — Base 16

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

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

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

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

(a) 1010

Converting to decimal:

Equivalent decimal number = 8 + 2 = 10

Therefore, (1010)₂ = (10)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1010}$

Therefore, (1010)₂ = (A)₁₆

(b) 111010

Converting to decimal:

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

Therefore, (111010)₂ = (58)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{1010}$

Therefore, (111010)₂ = (3A)₁₆

(c) 101011111

Converting to decimal:

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

Therefore, (101011111)₂ = (351)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1111}$

Therefore, (101011111)₂ = (15F)₁₆

(d) 1100

Converting to decimal:

Equivalent decimal number = 8 + 4 = 12

Therefore, (1100)₂ = (12)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1100}$

Therefore, (1100)₂ = (C)₁₆

(e) 10010101

Converting to decimal:

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

Therefore, (10010101)₂ = (149)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1001} \quad \underlinesegment{0101}$

Therefore, (101011111)₂ = (95)₁₆

(f) 11011100

Converting to decimal:

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

Therefore, (11011100)₂ = (220)₁₀

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1101} \quad \underlinesegment{1100}$

Therefore, (11011100)₂ = (DC)₁₆

(a) 23

Converting to binary:

Therefore, (23)₁₀ = (10111)₂

Converting to octal:

Therefore, (23)₁₀ = (27)₈

(b) 100

Converting to binary:

Therefore, (100)₁₀ = (1100100)₂

Converting to octal:

Therefore, (100)₁₀ = (144)₈

(c) 145

Converting to binary:

Therefore, (145)₁₀ = (10010001)₂

Converting to octal:

Therefore, (145)₁₀ = (221)₈

(d) 19

Converting to binary:

Therefore, (19)₁₀ = (10011)₂

Converting to octal:

Therefore, (19)₁₀ = (23)₈

(e) 121

Converting to binary:

Therefore, (121)₁₀ = (1111001)₂

Converting to octal:

Therefore, (121)₁₀ = (171)₈

(f) 161

Converting to binary:

Therefore, (161)₁₀ = (10100001)₂

Converting to octal:

Therefore, (161)₁₀ = (241)₈

(a) A6

(A6)₁₆ = (10100110)₂

(b) A07

(A07)₁₆ = (101000000111)₂

(c) 7AB4

(7AB4)₁₆ = (111101010110100)₂

(d) BE

(BE)₁₆ = (10111110)₂

(e) BC9

(BC9)₁₆ = (101111001001)₂

(f) 9BC8

(9BC8)₁₆ = (1001101111001000)₂

(a) 10011011101

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}$

Therefore, (10011011101)₂ = (4DD)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{011} \quad \underlinesegment{101}$

Therefore, (10011011101)₂ = (2335)₈

(b) 1111011101011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}$

Therefore, (1111011101011011)₂ = (F75B)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{101} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (1111011101011011)₂ = (173533)₈

(c) 11010111010111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}$

Therefore, (11010111010111)₂ = (35D7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{010} \quad \underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (11010111010111)₂ = (32727)₈

(d) 1010110110111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}$

Therefore, (1010110110111)₂ = (15B7)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{111}$

Therefore, (1010110110111)₂ = (12667)₈

(e) 10110111011011

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}$

Therefore, (10110111011011)₂ = (2DDB)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{110} \quad \underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{011}$

Therefore, (10110111011011)₂ = (26733)₈

(f) 1111101110101111

Converting to hexadecimal:

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{1010} \quad \underlinesegment{1111}$

Therefore, (1111101110101111)₂ = (FBAF)₁₆

Converting to Octal:

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{111} \quad \underlinesegment{101} \quad \underlinesegment{110} \quad \underlinesegment{101} \quad \underlinesegment{111}$

Therefore, (1111101110101111)₂ = (175657)₈

The most commonly used number systems are decimal, binary, octal and hexadecimal number systems.

The Hexadecimal number system is used in computers to specify memory addresses (which are 16-bit or 32-bit long). For example, a memory address 1101011010101111 is a big binary address but with hex it is D6AF which is easier to remember. The Hexadecimal number system is also used to represent colour codes. For example, FFFFFF represents White, FF0000 represents Red, etc.

The radix or base of a number system signifies how many unique symbols or digits are used in the number system to represent numbers. For example, the decimal number system has a radix or base of 10 meaning it uses 10 digits from 0 to 9 to represent numbers.

Encoding schemes help Computers represent and recognize letters, numbers and symbols. It provides a predetermined set of codes for each recognized letter, number and symbol. Most popular encoding schemes are ASCI, Unicode, ISCII, etc.

UTF-8 is a variable width encoding that can represent every character in Unicode character set. The code unit of UTF-8 is 8 bits called an octet. It uses 1 to maximum 6 octets to represent code points depending on their size i.e. sometimes it uses 8 bits to store the character, other times 16 or 24 or more bits. It is a type of multi-byte encoding.

UTF-8 is a variable length encoding scheme that uses different number of bytes to represent different characters whereas UTF-32 is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points.

In a binary code, the leftmost bit is called the most significant bit or MSB. It carries the largest weight. The rightmost bit is called the least significant bit or LSB. It carries the smallest weight. For example:

$\begin{matrix} \underset{\bold{MSB}}{1} & 0 & 1 & 1 & 0 & 1 & 1 & \underset{\bold{LSB}}{0} \end{matrix}$

ASCII encoding scheme uses a 7-bit code and it represents 128 characters. Its advantages are simplicity and efficiency. Extended ASCII encoding scheme uses a 8-bit code and it represents 256 characters.

ISCII or Indian Standard Code for Information Interchange can be used to represent Indian languages on the computer. It supports Indian languages that follow both Devanagari script and other scripts like Tamil, Bengali, Oriya, Assamese, etc.

Unicode is a universal character encoding scheme that can represent different sets of characters belonging to different languages by assigning a number to each of the character. It has the following significance:

1. It defines all the characters needed for writing the majority of known languages in use today across the world.
2. It is a superset of all other character sets.
3. It is used to represent characters across different platforms and programs.

Unicode uses UTF-8, UTF-16 and UTF-32 encoding schemes.

ASCII stands for American Standard Code for Information Interchange. It uses a 7-bit code and it can represent 128 characters. ASCII code is mostly used to represent the characters of English language, standard keyboard characters as well as control characters like Carriage Return and Form Feed. ISCII stands for Indian Standard Code for Information Interchange. It uses a 8-bit code and it can represent 256 characters. It retains all ASCII characters and offers coding for Indian scripts also. Majority of the Indian languages can be represented using ISCII.

UTF-8 is a variable length encoding scheme that uses different number of bytes to represent different characters whereas UTF-32 is a fixed length encoding scheme that uses exactly 4 bytes to represent all Unicode code points. UTF-8 is the more popular encoding scheme.

Code point refers to a code from a code space that represents a single character from the character set represented by an encoding scheme. For example, 0x41 is one code point of ASCII that represents character 'A'.

Variable length encoding scheme uses different number of bytes or octets (set of 8 bits) to represent different characters whereas fixed length encoding scheme uses a fixed number of bytes to represent different characters.

Equivalent decimal number = 1 + 4 + 8 = 13

Therefore, (1101)₂ = (13)₁₀

(b) 111010

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

Therefore, (111010)₂ = (58)₁₀

(c) 101011111

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

Therefore, (101011111)₂ = (351)₁₀

Equivalent decimal number = 4 + 8 = 12

Therefore, (1100)₂ = (12)₁₀

(b) 10010101

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

Therefore, (10010101)₂ = (149)₁₀

(c) 11011100

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

Therefore, (11011100)₂ = (220)₁₀

Therefore, (23)₁₀ = (10111)₂

(b) 100

Therefore, (100)₁₀ = (1100100)₂

(c) 145

Therefore, (145)₁₀ = (10010001)₂

(d) 0.25

Therefore, (0.25)₁₀ = (0.01)₂

Therefore, (19)₁₀ = (10011)₂

(b) 122

Therefore, (122)₁₀ = (1111010)₂

(c) 161

Therefore, (161)₁₀ = (10100001)₂

(d) 0.675

(We stop after 5 iterations if fractional part doesn't become 0)

Therefore, (0.675)₁₀ = (0.10101)₂

Therefore, (19)₁₀ = (23)₈

(b) 122

Therefore, (122)₁₀ = (172)₈

(c) 161

Therefore, (161)₁₀ = (241)₈

(d) 0.675

Therefore, (0.675)₁₀ = (0.53146)₈

(A6)₁₆ = (10100110)₂

(b) A07

(A07)₁₆ = (101000000111)₂

(c) 7AB4

(7AB4)₁₆ = (111101010110100)₂

(23D)₁₆ = (1000111101)₂

(b) BC9

(BC9)₁₆ = (101111001001)₂

(c) 9BC8

(9BC8)₁₆ = (1001101111001000)₂

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{1101} \quad \underlinesegment{1101}$

Therefore, (10011011101)₂ = (4DD)₁₆

(b) 1111011101011011

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{0111} \quad \underlinesegment{0101} \quad \underlinesegment{1011}$

Therefore, (1111011101011011)₂ = (F75B)₁₆

(c) 11010111010111

Grouping in bits of 4:

$\underlinesegment{0011} \quad \underlinesegment{0101} \quad \underlinesegment{1101} \quad \underlinesegment{0111}$

Therefore, (11010111010111)₂ = (35D7)₁₆

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1011} \quad \underlinesegment{0111}$

Therefore, (1010110110111)₂ = (15B7)₁₆

(b) 10110111011011

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{1101} \quad \underlinesegment{1101} \quad \underlinesegment{1011}$

Therefore, (10110111011011)₂ = (2DDB)₁₆

(c) 0110101100

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{1010} \quad \underlinesegment{1100}$

Therefore, (0110101100)₂ = (1AC)₁₆

Equivalent decimal number = 7 + 40 + 128 = 175

Therefore, (257)₈ = (175)₁₀

(b) 3527

Equivalent decimal number = 7 + 16 + 320 + 1536 = 1879

Therefore, (3527)₈ = (1879)₁₀

(c) 123

Equivalent decimal number = 3 + 16 + 64 = 83

Therefore, (123)₈ = (83)₁₀

(d) 605.12

Integral part

Fractional part

Equivalent decimal number = 5 + 384 + 0.125 + 0.0312 = 389.1562

Therefore, (605.12)₈ = (389.1562)₁₀

Equivalent decimal number = 6 + 160 = 166

Therefore, (A6)₁₆ = (166)₁₀

(b) A13B

Equivalent decimal number = 11 + 48 + 256 + 40960 = 41275

Therefore, (A13B)₁₆ = (41275)₁₀

(c) 3A5

Equivalent decimal number = 5 + 160 + 768 = 933

Therefore, (3A5)₁₆ = (933)₁₀

Equivalent decimal number = 9 + 224 = 233

Therefore, (E9)₁₆ = (233)₁₀

(b) 7CA3

Equivalent decimal number = 3 + 160 + 3072 + 28672 = 31907

Therefore, (7CA3)₁₆ = (31907)₁₀

Therefore, (132)₁₀ = (84)₁₆

(b) 2352

Therefore, (2352)₁₀ = (930)₁₆

(c) 122

Therefore, (122)₁₀ = (7A)₁₆

(d) 0.675

(We stop after 5 iterations if fractional part doesn't become 0)

Therefore, (0.675)₁₀ = (0.ACCCC)₁₆

Therefore, (206)₁₀ = (CE)₁₆

(b) 3619

Therefore, (3619)₁₀ = (E23)₁₆

(38AC)₁₆ = (11100010101100)₂

Grouping in bits of 3:

$\underlinesegment{011}\medspace\underlinesegment{100}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{100}$

(38AC)₁₆ = (34254)₈

(b) 7FD6

(7FD6)₁₆ = (111111111010110)₂

Grouping in bits of 3:

$\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{111}\medspace\underlinesegment{010}\medspace\underlinesegment{110}$

(7FD6)₁₆ = (77726)₈

(c) ABCD

(ABCD)₁₆ = (1010101111001101)₂

Grouping in bits of 3:

$\underlinesegment{001}\medspace\underlinesegment{010}\medspace\underlinesegment{101}\medspace\underlinesegment{111}\medspace\underlinesegment{001}\medspace\underlinesegment{101}$

(ABCD)₁₆ = (125715)₈

Therefore, (123)₈ = ($\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{011}}$)₂

(b) 3527

Therefore, (3527)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{111}}$)₂

(c) 705

Therefore, (705)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}$)₂

Therefore, (7642)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}$)₂

(b) 7015

Therefore, (7015)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{101}}$)₂

(c) 3576

Therefore, (3576)₈ = ($\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{110}}$)₂

(d) 705

Therefore, (705)₈ = ($\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{101}}$)₂

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{010}$

Therefore, (111010)₂ = (72)₈

(b) 110110101

Grouping in bits of 3:

$\underlinesegment{110} \quad \underlinesegment{110} \quad \underlinesegment{101}$

Therefore, (110110101)₂ = (665)₈

(c) 1101100001

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \underlinesegment{001}$

Therefore, (1101100001)₂ = (1541)₈

Grouping in bits of 3:

$\underlinesegment{011} \quad \underlinesegment{001}$

Therefore, (11001)₂ = (31)₈

(b) 10101100

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100}$

Therefore, (10101100)₂ = (254)₈

(c) 111010111

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (111010111)₂ = (727)₈

$\begin{matrix} & & \overset{1}{1} & \overset{1}{0} & 1 & 1 & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 \\ + & & & 1 & 1 & 0 & 0 & 1 & 0 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{1} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (10110111)₂ + (1100101)₂ = (100011100)₂

(ii) 110101 and 101111

$\begin{matrix} & & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{0} & 1 \\ + & & 1 & 0 & 1 & 1 & 1 & 1 \\ \hline & \bold{1} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (110101)₂ + (101111)₂ = (1100100)₂

(iii) 110111.110 and 11011101.010

$\begin{matrix} & & \overset{1}{0} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & \overset{1}{1} & . & \overset{1}{1} & 1 & 0 \\ + & & 1 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & . & 0 & 1 & 0 \\ \hline & \bold{1} & \bold{0} & \bold{0} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{0} \end{matrix}$

Therefore, (110111.110)₂ + (11011101.010)₂ = (100010101)₂

(iv) 1110.110 and 11010.011

$\begin{matrix} & & \overset{1}{0} & \overset{1}{1} & \overset{1}{1} & 1 & \overset{1}{0} & . & \overset{1}{1} & 1 & 0 \\ + & & 1 & 1 & 0 & 1 & 0 & . & 0 & 1 & 1 \\ \hline & \bold{1} & \bold{0} & \bold{1} & \bold{0} & \bold{0} & \bold{1} & \bold{.} & \bold{0} & \bold{0} & \bold{1} \end{matrix}$

Therefore, (1110.110)₂ + (11010.011)₂ = (101001.001)₂

Binary representation of 'A' in ASCII will be binary representation of its code point 65.

Converting 65 to binary:

Therefore, binary representation of 'A' in ASCII is 1000001.

Converting 65 to Hexadecimal:

Therefore, hexadecimal representation of 'A' in ASCII is (41)₁₆.

Similarly, converting 97 to binary:

Therefore, binary representation of 'a' in ASCII is 1100001.

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 1 + 4 + 32 + 0.5 + 0.125 = 37.625

Therefore, (100101.101)₂ = (37.625)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{100} \quad \underlinesegment{101} \quad \bold{.} \quad \underlinesegment{101}$

Therefore, (100101.101)₂ = (45.5)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{0010} \quad \underlinesegment{0101} \medspace . \medspace \underlinesegment{1010}$

Therefore, (100101.101)₂ = (25.A)₁₆

(ii) 10101100.01011

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 = 172.34375

Therefore, (10101100.01011)₂ = (172.34375)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{110}$

Therefore, (10101100.01011)₂ = (254.26)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1000}$

Therefore, (10101100.01011)₂ = (AC.58)₁₆

(iii) 1010

Decimal Conversion:

Equivalent decimal number = 2 + 8 = 10

Therefore, (1010)₂ = (10)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010}$

Therefore, (1010)₂ = (12)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010}$

Therefore, (1010)₂ = (A)₁₆

(iv) 10101100.010111

Decimal Conversion of integral part:

Decimal Conversion of fractional part:

Equivalent decimal number = 4 + 8 + 32 + 128 + 0.25 + 0.0625 + 0.03125 + 0.015625 = 172.359375

Therefore, (10101100.010111)₂ = (172.359375)₁₀

Octal Conversion

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{101} \quad \underlinesegment{100} \quad \bold{.} \quad \underlinesegment{010} \quad \underlinesegment{111}$

Therefore, (10101100.010111)₂ = (254.27)₈

Hexadecimal Conversion

Grouping in bits of 4:

$\underlinesegment{1010} \quad \underlinesegment{1100} \medspace . \medspace \underlinesegment{0101} \medspace \underlinesegment{1100}$

Therefore, (10101100.010111)₂ = (AC.5C)₁₆