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Number System.pdf
The document provides a comprehensive overview of number systems used by computers, including binary, octal, decimal, and hexadecimal systems, alongside their bases. It explains fundamental concepts such as bits and bytes, ASCII encoding, conversions between these number systems, and methods for binary arithmetic. Additionally, the document covers Binary Coded Decimal (BCD) and various operations like addition, subtraction, multiplication, and division in binary form.
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Explains data representation in computing, emphasizing number systems like binary, octal, decimal, and hexadecimal.
Defines bits and bytes, explaining binary storage, ASCII encoding, and character representation in computers.
Shows conversion methods between decimal and binary. Examples provide clear transformations.
Details conversion processes from decimal to octal and vice versa, offering numerical examples.
Explains conversions between decimal and hexadecimal formats, including numerical instances for clarity.
Explores the conversions between binary and octal, focusing on the relationships and necessary bits for accuracy.
Discusses exercises for converting octal to binary and vice versa, reinforcing conversion understanding.
Demonstrates conversions between hexadecimal and binary, highlighting processes with practical examples.
Continues from previous slide discussing hexadecimal to binary and vice versa with methodical examples.
Calculates the number of bits needed for various symbols, explaining the relation between bit count and representation.
Reiterates conversion between binary and hexadecimal using specific exercises for practice.
Summarizes the principles of number systems and conversions among decimal, binary, octal, and hexadecimal.
Describes binary addition operations with examples of simple calculations to illustrate the process.
Presents complex binary addition and further arithmetic operations showcased through multidigit examples.
Elaborates binary subtraction with demonstrated examples to ensure comprehension and application.
Illustrates binary multiplication and division with examples to solidify understanding of binary computation.
Explains 1’s complement and its applications in representing negative numbers in binary systems.
Outlines the process of finding two's complement, emphasizing practical examples for clarity.
Discusses methods for sign representation in binary, clarifying magnitude and sign determination.
Examines overflow issues in binary addition along with general principles of representation with n bits.
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Number System.pdf
- 1. 1 SWETA KUMARI BARNWAL NUMBERSYSTEM The language we use to communicate with each other is comprised of words and characters. We understand numbers, characters and words. But this type of data is not suitable for computers. Computers only understand the numbers. So, when we enter data, the data is converted into electronic pulse. Each pulse is identified as code and the code is converted into numeric format by ASCII. It gives each number, character and symbol a numeric value (number) that a computer understands. So, to understand the language of computers, one must be familiar with the number systems. The Number Systems used in computers are: o Binary number system o Octal number system o Decimal number system o Hexadecimal number system BASE/RADIX: The base value in a numbering system. For example, • In the decimal numbering system, the radix is 10. • In the hexadecimal numbering system, the radix is 16. • In the octal numbering system, the radix is 8. • In the binary numbering system, the radix is 2. Decimal (10 Symbols) (0, 1, 2, 3…9) Octal (8 ) (0, 1, 2, 3…7) HexaDecimal (16 Symbols) (0, 1, 2, 3…9, A,B,C,D,E,F) Binary (2 Symbols) (0 & 1) Number System
- 2. 2 SWETA KUMARI BARNWAL Bits& Bytes: • a "bit" is atomic: the smallest unit of storage • A bit store just a 0 or 1 • "In the computer it's all 0's and 1's" ... bits • Anything with two separate states can store 1 bit • In a chip: electric charge = 0/1 • In a hard drive: spots of North/South magnetism = 0/1 • A bit is too small to be much use • Group of 8 bits together to make 1 byte • Everything in a computer is 0's and 1's. The bit stores just a 0 or 1: it's the smallest building block of storage. ASCII (American Standard Code for Information Interchange) • ASCII is an encoding representing each typed character by a number • Each number is stored in one byte (so the number is in 0 to 255) • A is 65 • B is 66 • a is 96 • space is 32 ASCII stands for the "American Standard Code for Information Interchange". It was designed in the early 60's, as a standard character set for computers and electronic devices. ASCII is a 7-bit character set containing 128 characters. It contains the numbers from 0-9, the upper- and lower-case English letters from A to Z, and some special characters. The character sets used in modern computers, in HTML, and on the Internet, are all based on ASCII. The following tables list the 128 ASCII characters and their equivalent number.
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- 4. 4 SWETA KUMARI BARNWAL Decimalto Binary (951.205)10 → (951.205)10 → (1110110111.0011)2 Binary to Decimal (110101.110)2 → ( )10
- 5. 5 SWETA KUMARI BARNWAL (110101.110)2→ (53.75)10 Decimal to Octal (98.56)10 → (142.4365)8 Octal to Decimal (41.56)8 → ( 33.7187)10
- 6. 6 SWETA KUMARI BARNWAL Decimalto Hexadecimal (935.126)10 → (3A7.204F)16 Hexadecimal to Decimal (3A7.204F)16 → (935.1261)10 (2) (1) (0) (-1) (-2) (-3) (-4) Position 3 A 7 . 2 0 4 F x x x x x x x 162 161 160 16-1 16-2 16-3 16-4 768 + 160 + 7 . (.125) + (0) + (0.0009)+(0.0002)
- 7. 7 SWETA KUMARI BARNWAL Binaryto/from Octal a) Binary → Decimal → Octal (1010101)2 → (85)10 → (125)8 (110011.110) → (51.75) → (63.6) b) Octal → Decimal → Binary (125) → (85) → (1010101) (1010101) (721.25) → (465.3281) → (111010001.0101) If no of bits = n Number represented by that bits (N) = 2n – 1 Ex: Octal System (8 symbols: 0 to 7) N = 7 N = 2n – 1 = 7 n = 3 When we will convert Octal to Binary, we have required 3 bits for 1 digit. Decimal to Binary Binary to Decimal 2(n-1) ……… 22 21 20 32 16 8 4 2 1 No ABC 4 2 1 0 0 0 0
- 8. 8 SWETA KUMARI BARNWAL 10 0 1 2 0 1 0 3 0 1 1 4 1 0 0 5 1 0 1 6 1 1 0 7 1 1 1 (125)8 = (001 010 101) (1010101) 721.25 → (111 010 001. 010 101) (110 011 . 110) → 63.6 (001 110 010 101 . 110 100)2 → (1625.64) (56.54) → (000056.540000) (5600.00054) Exercise: Octal to Binary (128.777) (765.154) (5564.721) (254.631) (435.61) Exercise: Binary to Octal (110111011.111101) (101110100.011101) (11100011.010101) (1010101010.1101001)
- 9. 9 SWETA KUMARI BARNWAL Hexadecimalto/from Binary a) Hexadecimal → Decimal → Binary (A5.9D) → (165.6132) → (10100101.1001) (FB1.C4) → (4017.7656) → (111110110001.1100) b) Binary → Decimal → Hexadecimal (11110011010.10110) → (1946.6875) → (79A.B0) Hexadecimal System (16 symbols: 0 to 9 & A B C D E F) N = 15 N = 2n – 1 = 15 n = 4 When we will convert Octal to Binary, we have required 3 bits for 1 digit. Decimal to Binary Binary to Decimal 2(n-1) ……… 22 21 20 32 16 8 4 2 1
- 10. 10 SWETA KUMARI BARNWAL NoABCD 8 4 2 1 0 0 0 0 0 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 4 0 1 0 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1 A 1 0 1 0 B 1 0 1 1 C 1 1 0 0 D 1 1 0 1 E 1 1 1 0 F 1 1 1 1 a) Hexadecimal → Binary (A5.9D) → (1010 0101 . 1001 1101) → (1010 0101.1001) (FB1.C4) → (1111 1011 0001 . 1100 0100) → (1111 1011 0001.1100) (1FB1.C4) → (0001 1111 1011 0001 . 1100 0100) b) Binary → Hexadecimal (0111 1001 1010 . 1011 0) → (79A.B0) → (79A.B0) Q. How many bits are required to represent a number system having 128 symbols? (N 0 to 127) N = 127 N = 2n -1 n = 7
- 11. 11 SWETA KUMARI BARNWAL Q.How many bits are required to represent a number system having 156 symbols? N = 155 2n -1 > = N n >= 7 n = 8 Q. How many bits are required to represent a number system having 394 symbols? 2n -1 > = N = 394 Let n = 8, condition is not true Let n = 9, condition is true n = 9 Ans.: 9 Q. How many symbols can be represented by using 10 bits? Ans.: 1024 Q. Find the Maximum Number that can be represented by using 10 bits? Ans.: 1023 Q. How many symbols can be represented by using 12 bits? Ans.: 4096 Q. Find the Maximum Number that can be represented by using 9 bits? Ans.: 511 Q. Hexadecimal to Binary (BC5.982) (A712.0056) (5D8.234) (98FE.BAF)
- 12. 12 SWETA KUMARI BARNWAL Exercise:Binary to Hexadecimal (110111011.111101) (101110100.011101) (11100011.010101) (1010101010.1101001) We have already Covered: Number System & Conversion between a) Decimal to/from (Binary, Octal & Hexadecimal) b) Binary to/from Octal & Hexadecimal OCTAL TO/FROM HEXADECIMAL Octal → Binary → Hexadecimal (27.65)8 → (00010 111 . 110 10100) → (17.D4)16 (7765.6271)8 → (111 111 110 101 . 110 010 111 001) → (F F 5. C B 9) B (11) (8421) (1011) Hexadecimal → Binary → Octal (FD90.5C7) → (001111 1101 1001 0000 . 0101 1100 0111) → (176620.2707)8 Ex: (12FD.97C) (9532.4471) (10DE4.10B) (7AB.F9EB) (710.554) (5517.346) (7712.354) (1275.2606) BCD (Binary Coded Decimal) Another process for converting decimal numbers into their binary equivalents. • It is a form of binary encoding where each digit in a decimal number is represented in the form of bits. • This encoding can be done in either 4-bit or 8-bit (usually 4-bit is preferred).
- 13. 13 SWETA KUMARI BARNWAL •It is a fast and efficient system that converts the decimal numbers into binary numbers as compared to the existing binary system. • These are generally used in digital displays where is the manipulation of data is quite a task. • Thus, BCD plays an important role here because the manipulation is done treating each digit as a separate single sub-circuit. In the BCD numbering system, the given decimal number is segregated into chunks of four bits for each decimal digit within the number. Each decimal digit is converted into its direct binary form (usually represented in 4-bits). DECIMAL NUMBER BCD 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 (852.371)10 → (1000 0101 0010 . 0011 0111 0001)BCD 8 → 1000 5 → 0101 2 → 0010 3 → 0011 7 → 0111 1 → 0001
- 14. 14 SWETA KUMARI BARNWAL ArithmeticSystem a) Addition 1+0 = 1 0+0 = 0 1+1 = 10 → {(1*21 ) + (0*20 ) = 2} 1+1+1 = 11 → {(1*21 ) + (1*20 ) = 3} 1+1+1+1 = 100 → {(1*22 ) + (0*21 ) + (0*20 ) = 4} 1100 +1011 10111 11111 +11111 111110 12 12 24 99 99 198
- 15. 15 SWETA KUMARI BARNWAL Ex: 1011+1111+1100+1001 1111+1111+1111+1111= 111100 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 95 98 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1100+0011 10101111+10001111 = 111111110, 100111110 10101111 10001111 11000011 + 1111100 8 4 2 1 0 1 0→ 2, 011/11 → 3, 100
- 16. 16 SWETA KUMARI BARNWAL 1+1 = 2, 1+1+1 = 3, 1+1+1+1=4 1+1 = 10, 1+1+1 = 11, 1+1+1+1=100 Ex: a) 1011101110 +1111000110 b) 1111100001+11000110010+10001110101+1011011111= 1000101100111 01111100001 11000110010 10001110101 01011011111 b) Subtraction: (9-5 = 4, 5-9 = -4) A B Difference Borrow 0 0 0 0 0 1 1 0 1 0 1 1 0 0 4 +5 9 -5 4 1+1 = 10 -1= 1 1+1=2-1=1 1+1=10-1= 1
- 17. 17 SWETA KUMARI BARNWAL 10 -1 11 1 1 1 0 0 1 5000 1925 100 → 4 2n-1 ………..21 , 20 11111111100000 → 16352 10000111000011 → 08643 07709 (213 *0)+ (212 *1)+ (211 *1)+ (210 *1)+ (29 *1)+ (28 *0)+ (27 *0)+ (26 *0)+ (25 *0)+ (24 *1)+ (23 *1)+ (22 *1)+ (21 *0)+ (20 *1) = 0+4096+2048+1024+512+0+0+0+16+8+4+0+1=7709 Ex: a) 110011101 – 100101011 b) 101010101-010101010 c) 111111000-111000111
- 18. 18 SWETA KUMARI BARNWAL d)11001100-10101011 c) Multiplication: 1001 1000 1111 1111 11000001 (11100001) Ex: i) 11010*11100 ii) 11101*11111 iii) 111111*101010 iv) 101010*1010101 v) 111111*101 d) Division: 1001/1000 1000
- 19. 19 SWETA KUMARI BARNWAL Ex: i)11111/1100 ii) 101010/111 iii) 1110011/1010 iv) 1111010/1100 1876→ 10 762 → 8 1011 → 2 98A→ 16 1’s Complement:- A binary no system used to write the -ve no. Method to write this no.- transforming the 0 bit to 1 and the 1 bit to 0 Ex:- 1010111110 → 0101000001 = 1111111111 9 → 1001 (-9) → 0110 (in 1’s complement) 129 → 10000001 (2048 1024 512 256 128 64 32 16 8 4 2 1) Write (-129) → using 1’s complement → 01111110 Write 148 in 1’s complement → (10010100) → 01101011 (-148)
- 20. 20 SWETA KUMARI BARNWAL NOTE:If we take complement of any no in two times, then we will get the original no. Write in 1’s Complement: (HW) a) 1110001010 → 0001110101 b) 11111000011 c) 10101010101 d) 1001111000 e) 256 → 011111111 → (100000000) f) -412 → (412→110011100) → (-412 → 001100011) g) 98 h) -567 2’s Complement 101010111 → (1’s Complement) 010101000 +1 →(2’s Complement) 010101001 010101001 → 101010110+1 → 101010111 10110110 → 01001001 → 10110110 256 → (100000000) (Binary) -256 → (011111111) (in 1’s Complement) -256 → (011111111+1 → 100000000) (in 2’s Complement) 011111111 1 -412 (412→110011100 → Binary) → (-412 → 001100011 → 1’s Complement) → (-412 → 001100100 → 2’s complement) 001100100 → 110011011+1 → 110011100
- 21. 21 SWETA KUMARI BARNWAL Ex: a)298 → 100101010 → (-298 → 011010101 → 1’s) (-298 → 011010110 → 2’s) b) -298 → 011010110 (011011000) c) 97 d) -492 e) 187 f) -299 g) 100011101 h) 111001111 i) 100000010 Sign Representation -355 → (355) +355 → (355) (28 → 000011100) (-28 → 100011100) 28 0 Sign 00011100 (Magnitude -28 1 Sign 00011100 (Magnitude (No. of bits: 1, 2, 4, 8, 16…….) 5 (+5) → 0101 → 0 0101 -5 → 0101 (5) → 1 0101 37 → 00100101 (2n no of bits) 32 16 8 4 2 1
- 22. 22 SWETA KUMARI BARNWAL 54→ 110110 (00110110) (1101 0110) 129 → 10000001 37 → 0 00100101 000100101 -37 → 1 00100101 100100101 453 → 111000101 → 0000000111000101 0 0000000111000101 00000000111000101 -453 1 0000000111000101 10000000111000101 Ex: (HW) Write these numbers in 1’s, 2’s Compliment & Sign Representation a) 137 & -137 b) 98 & -98 c) 266 & -266 d) 32 & -32 e) 78 & -78 The given binary no is in sign representation. Find the magnitude & Sign. a) 1 1001 = (-9)
- 23. 23 SWETA KUMARI BARNWAL b)0 1101 (+13) c) 0 00111010 (+58) d) 1 00111111 (-63) e) 1 11111111 (-255) The given binary no (10010) is in sign representation. Find the magnitude & Sign. Ans: (-2) Overflow: 1111 +1111 11110 10111 0111 11110 2n → No of symbols 2n -1 → Maximum no represented by n bits. n → bits number n = 1 No of symbols = 21 = 2 ( 0, 1) Maximum no represented by 1 bits = 21 -1 = 1 n = 2 No of symbols = 22 = 4 ( 00, 01,10, 11) Maximum no represented by 2 bits = 22 -1 = 3 n = 5 No of symbols = 25 = 32 ( 00000, 00001, 00010……… 11111) Maximum no represented by 5 bits = 25 -1 = 31