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An introduction to number systems

An introduction to number systems

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    Number system Number system Presentation Transcript

    • 011010110011010110011010110011010110101 Data Representation and Number System
    • Data Representation • Data:Data are numbers and other binary-coded information that areoperated on to achieve required computational results. • Control InformationControl information is a bit or a group of bits used to specify thesequence of command signals.
    • Data Representation • Bit:Binary Digit. 0/1 • A group of bits in a computer are used to represent many different things.It can represent a number.It can represent a character.It can represent an instruction. • Byte: A group of 8 bits is a byte. • Nibble A group of 4 bits is a nibble.
    • Number System: Radix / Base • Number System is a code representing quantity. • Radix / Base:A number system of base, or radix, r is a system that uses distinctsymbols for r digits.In this system the number of states each digit has is determined by thebase or radix. • Based on the radix there are four number systems o Decimal o Binary o Octal o Hexadecimal
    • Decimal system Radix / base = 10 • For example :253 means 02*102 + 5*101 + 3*100 1 2 3 4• 1,10,100 (from R to L) are the “weights” 5 6• 10 digits, values 0 through 9 7 8• After 9 comes 10 (double digits) 9
    • Binary system Radix / base = 2 • For example : 01011 means 1*23 + 0*22 + 1*21 + 1*20 1• 2 digits, values 0 and 1• 1, 2, 4, 8 are the weights• After 1 comes 10• Count 0 1 10 11 100 101 110 111 1000
    • Binary system• Used to do calculations in all computers• Used to store values in memory and on disk• Not practical for people• Input-Output done in decimal for user• Software translates in both directions
    • Octal system Radix / base = 8 • For example :253 means : 2*82 + 5*81 + 3*80 0 1• 1, 8, 64 are the weights 2 3• 8 digits, values 0 through 7 4 5• After 7 comes 10 6 7 • Used to display memory addresses in someolder computers
    • Hexadecimal system Radix / base = 16 A B • For example : 0 C3B6 means 3*162 + B*161 + 6*160 1 D 2 E 3 F • 1, 16, 256 are the weights 4 • 16 digits, values 0 - 9 and A-F 5 • After F comes 10 6 • Used to display memory addresses in most 7modern computers e.g., 3C0F 95EA 8 9
    • Categorizing the Conversion Rule• Converting from one number system to the other system can be categorized as• Any radix to Decimal system• Decimal system to any radix• Octal to binary and hexadecimal to binary• Any radix to Any radix (other than binary)
    • Binary to Decimal Conversion For IntegersFor example: (00101010)2 0 0 1 0 1 0 1 0 27 26 25 24 23 22 21 20 Equals: 27 * 0 + 26 * 0 + 2 5 * 1 + 24 * 0 + 23 * 1 + 22 * 0 + 21 * 1 + 20 * 0 = 32 + 8 + 2 = 42 (00101010)2 = 4210 Approach is from LSB to MSB
    • Exercise• 1010002 = ( ??? )10 4010• 10010112 = 7510• 1000112 =• 0110112 = 01001011 Most Significant Bit Least Significant Bit
    • From Decimal to Binary For Integers • Divide by 2 until you reach zero, and then collect theremainders in reverse. • For example: 5610 = ( ???? )22 ) 56 0 Least significant bit2 ) 28 0 2 ) 14 0 2)71 2)31 2)11 0 (0111000)2
    • Exercise• (48)10 = ( ????? )2 (00110000)2
    • Binary to Decimal With FractionBinary point(10110.10101)2 1 0 1 1 0 . 1 0 1 0 1 24 23 22 21 20 . 2-1 2-2 2-3 2-4 2-5= (1 * 16) + (0 * 8) + (1*4) + (1*2) + (0*1) + ( 1 / 2) + (0 / 4)+ ( 1 / 8) + ( 0 / 16) + (1 / 32)= (22.65675) 10
    • From Decimal to Binary With FractionFor example:(56.6875)10 = ( ???? )2 • Convert the integer and fraction part separately(56) 10 = (111000)2 • For fraction part, multiply the fraction part by 2, and each timediscard the integer so obtained. • Collect this discarded integer part as the binary equivalent. • Repeat this process until zero or until the required accuracy.
    • 0.6875 0.5000x2x21.3750 1.00000.3750 .1011x20.7500 (56.6875)10 = (111000.1011)20.7500x21.5000
    • Exercise• (48.3125)10 = ( ????? )2 (00110000.0101)2
    • From Decimal to Radix r• Separate the integer part and fraction part• Convert the integer part and then fraction part separately. • Rule For converting the integer part:Conversion of a decimal to a base r is done by successivedivisions by r and accumulating the remainders.This is repeated until the quotient becomes zero.Collect remainders in the reverse order.
    • • Rule For converting the decimal part:Conversion of a decimal to a base r is done by successivemultiplications by r and accumulating the remainders.• This process is repeated until the fraction parts becomes zero or number of digits gives the required accuracy• Take the integer outputs in the forward direction
    • From Radix r to Decimal• Beginning with the rightmost digit multiply each nth digit by r(n-1), and add all of the results together (considering the position just before the decimal point as the first position.N = AnRn + An-1Rn-1 + …….A2R2+ A1R1 +A0R0. A1R-1+A2R-2 +…….• N - Number• An - Digit in that position (nth Position)• R - Radix or base of the system• - Radix Point
    • Decimal to Octal Conversion • For example: (478.5)10 = ( ?? )8• Convert the fraction and integer part separately.• For Integer part: o The Division Method: Divide by 8 until you reach zero, and then collect the remainders in reverse.8 ) 478 68 ) 59 3 8 ) 7 70 (736)8
    • • For Fraction part: o The Multiplication Method: Multiply the fraction part successively by 8 and accumulate the remainders until you reach zero.0.5x 8 (736.4)84.0
    • Octal to Decimal Conversion • To convert to base 10, beginning with the rightmost digitmultiply each nth digit by 8(n-1), and add all of the resultstogether.For example: (736.4)8 7 3 6 . 4 82 81 80 . 8-1 Equals: 7* 82 + 3 * 81 + 6 * 80 + 4 * 8-1 = 448+24+6+0.5 = (478.5)10
    • Exercise• (0.40)8 = ______10 0.50
    • HexaDecimal to Decimal Conversion• To convert to base 10, beginning with the rightmost digit multiply each nth digit by 16(n-1), and add all of the results together.For example: 1F416 1 F 4 162 161 160 Equals: 1 * 162 + F * 161 + 4 * 160 = 256 + 15*16 + 4 =(500)10
    • Decimal to Hexa Conversion• The Division Method. Divide by 16 until you reach zero, and then collect the remainders in reverse. A 10 For example: 12610 = 7E16 B 1116) 126 14 = E C 12 D 1316) 7 7 E 140 F 15
    • Exercise• (AF)16 = ______10 175
    • Binary to Octal • Group the binary number into groups of 3 bits starting fromthe least significant bit, and convert it into its decimal equivalent. For example: (1 010 101 111)2 Grouping : 1 010 101 111 1257 (1010101111)2 = (1257)8
    • Octal to Binary• Take each digit one by one from the string of digits and convert each digit into its respective binary number, as a group of three bits.(257)8 = ____2 7 is converted as 111 5 is converted as 101 2 is converted as 010(257)8 = (010101 111)2 Binary Triplet Method
    • Binary -Coded Octal NumbersThree-bit Group Decimal Digit Octal Digit000 0 0001 1 1010 2 2011 3 3100 4 4101 5 5110 6 6111 7 7001 000 10010 100 24
    • Binary to Hexadecimal • Group the binary number into groups of 4 bits startingfrom the least significant bit, and convert it into its decimalequivalent. A 10For example: (1010 1111 0110 0011)2 B 11 C 12Grouping : 1010 1111 0110 0011 D 13 E 14AF63 F 15(1010111101100011)2 = (AF63)16
    • Hexadecimal to Binary• Take each digit one by one from the string of digits and convert each digit into its respective binary number, as a group of four bits.(257)16 = ____2 7 is converted as 0111 5 is converted as 0101 2 is converted as 0010(257)16 = (00100101 0111)2
    • Hexadecimal to Binary(BA7)16 = ____2 A 10 B 11 7 is converted as 0111 C 12 A is converted as 1010 D 13 E 14 B is converted as 1011 F 15(BA7)16 = (10111010 0111)2
    • Binary-Coded Hexadecimal Numbers Four-bit Group Decimal Digit Hexadecimal Digit 0000 0 0 0001 1 1 0010 2 2 0011 3 3 0100 4 4 0101 5 5 0110 6 6 0111 7 7 1000 8 8 1001 9 9
    • Binary-Coded Hexadecimal Numbers Four-bit Group Decimal Digit Hexadecimal Digit 1010 10 A 1011 11 B 1100 12 C 1101 13 D 1110 14 E 1111 15 F 0001 0100 14 0011 0010 50
    • Binary to octal and hexadecimal EXERCISE• 1010111101100011 Binary• 1010111101100011 Octal 1275438• 1010111101100011 Hexa AF6316
    • • Note:• The highest digit in octal system is 7 whose binary equivalent is 111.• The highest digit in hexadecimal system in F, whose binary equivalent is 1111.
    • ComplementsThere are two types of complements for each base r system:• r’s complement• ( r-1)’s complement
    • (r-1)’s complement• Given a number N in base r having n digit, the (r-1)’s complement of N is (rn –1) –N.• For decimal numbers, there exist 9’s complement.• For binary numbers, there exist 1’s complement.
    • 9’s Complement • For example:For decimal number N= 546700, n= 6 and r =109’s complement equals:= (rn –1) – N= (106 –1) - 546700= (1000000 –1) - 546700= 999999 – 546700 = 453299• That is, 9’s complement of a number would be same as subtracting each digit from 9.
    • 1’s Complement • For example:For binary number N= 1011, n= 4 and r =21’s complement equals:= (rn –1) – N= (24 –1) - 1011= (10000 –1) – 1011 (24 in binary)= 1111 – 1011= 0100• That is, 1’s complement of a number would be same as subtracting each digit from 1.
    • 1’s Complement• For a binary number 1011001, 1’s complement can be obtained by 1111111 If you look at the result, you can see, the 1’s 1011001 complement of a binary number can be obtained by _______ reversing the bits. 0100110 _______
    • r’s complement• Given a number N in base r having n digit, the r’s complement of N is rn –N for N < > 0 and 0 for N=0.• Also, r’s complement is equal to: = rn –N = rn –N – 1 + 1 (Add and subtract 1) = [(rn –1) –N] +1 (Rearranging the terms) = (r-1)’s complement + 1• For decimal numbers, there exist 10’s complement.• For binary numbers, there exist 2’s complement.
    • 10’s Complement• For decimal numbers, 10’s complement of a number is equal to its 9’s complement +1. • For example:10’s complement of 546700 == 9’s complement of 546700 + 1= 453299 + 1= 453300
    • 2’s Complement • Given a number in binary say N, having ‘n’ digits, then 2’s complement of N is defined as (2n-N), if N < > 0 else 0, when N=0 • For binary numbers, 2’s complement of a number is equal to its 1’s complement +1. • For example:2’s complement of 1011 == 1’s complement of 1011 + 1= 0100 +1 = 0101
    • Exercise• Find the 2’s complement of 10101011 01010101• Find the 2’ complement of 01010101 10101011
    • Integer Representations• Two different representations exists for integers• The signed representation: in that case the most significant bit (MSB) represents the sign o Positive number (or zero) if MSB = 0 o Negative number if MSB = 1• The unsigned representation: in that case all the bits are used to represent a magnitude o It is thus always a positive number or zero
    • Signed and Unsigned Interpretation • To obtain the value of a integer in memory we need to chose an interpretation • For example: a byte of memory containing 1111 1111 can represent either one of these numbers: o -1 if a signed interpretation (2’s complement) is used o 255 if an unsigned interpretation is used
    • Subtraction of Unsigned Numbers • The subtraction of two n-digit unsigned numbers M – N (N < > 0) in base r can be done as follows: 1. Add the minuend M to the r’s complement of the subtrahend N. This performs M + (rn – N) = M – N + rn.Case 1 : If M >= N, the sum will produce an end carry rn which is discarded,and what is left is the result M – N.Case 2 : If M < N, the sum does not produce an end carry and is equal torn – (N – M), which is the r’s complement of (N – M). To obtain the answerin a familiar form, take the r’s complement of the sum and place a negativesign in front.This will equate to : rn – (rn – ( N – M)) = M - N
    • Subtraction of Unsigned NumbersCase 1: Minuend > Subtrahend • Take the r’s complement of the subtrahend. • Add this to the minuend. • Discard the end carry. 3456 10’s complement of 2234 = 7766 _______ 3456 - 2234 11222 radix 10 Discard the end carry 10000 1222
    • Subtraction of Unsigned NumbersCase 2: Minuend < Subtrahend • Take the r’s complement of the subtrahend. • Add this to the minuend. • Find the r’s complement of the result and append a negative sign in front of it. 2234 10’s complement of 3456 = 6544 _______ 2234 - 3456 8778 radix 10 -1222 10’s complement of 8778
    • Subtraction of Unsigned Numbers• In case 2, after the 10’s complement of 8778, we get 1222 only and not -1222.• When working manually it can be noticed that the subtrahend is > minuend and so it needs a -ve sign for the result.• When subtracting with complements it is found that the answer where there is no end carry and a negative sign should be added.
    • Subtraction of Unsigned Numbers • In a similar manner, the subtraction with complements is done with binary numbers. • For example:X: 1010100Y: 1000011 • To perform X – Y :X = 10101002’s complement of Y = 0111101Sum = 10010001 Discard the end carry 10000000 0010001
    • ExerciseY: 1000011X: 1010100radix 2Perform Y – X = ???? - 0010001
    • 1’s Complement Subtraction Unsigned representation Case 1: Minuend > Subtrahend (M – N) • Take the 1’s complement of the subtrahend. • Add this to the minuend. • Remove the carry and add it to the result. This is called END AROUND CARRY. 00011101 1’s complement of 00011011= 1110010000011101- 00011011 _________radix 2 100000001 1 RESULT 00000010
    • 1’s Complement Subtraction Unsigned RepresentationCase 2: Minuend < Subtrahend• Take the 1’s complement of the subtrahend.• Add this to the minuend.• Find the 1’s complement of the result and append a negative sign in front of it. 00011001 1’s complement of 00011101 = 11100010 1111101100011001 - 00011101radix 2 RESULT -00000100
    • Exercise (using 1’s complement)X: 00110011Y: 00101101 Perform X - Yradix 2 00000110
    • Signed Representation • In signed representation, the most significant bit (MSB) represents the sign. • When a binary number is positive, the sign is represented by 0 and the magnitude by a positive binary number. • When the number is negative, the sign is represented by 1 but the rest of the number may be represented in three possible ways.1. Signed magnitude representation2. Signed - 1’s complement representation3. Signed - 2’s complement representation.
    • Example for Negative number Representation • To represent -141. Signed magnitude representation1 0001110Note : This representation of – 14 is obtained from +14 bycomplementing only the sign bit.2. Signed - 1’s complement representation1 1110001Note : This representation of – 14 is obtained by complementing allthe bits of + 14, including the sign bit.
    • Example for Negative number Representation3. Signed - 2’s complement representation.1 1110010Note : This representation of – 14 is obtained by taking the 2’scomplement of +14, including the sign bit.
    • Advantage of 2’s Complement System• Representing in 2’s complement is preferred over 1’s complement as well as signed magnitude system.• Representing in signed magnitude is easy for manual arithmetic processing and not for the computer.• The reason is 1’s complement takes two representation for +0 and -0 which is absurd.• In 2’s complement system both -0 and +0 will have the same representation
    • NOTE1’s complement form • + 0 in binary 00000000 • - 0 in 1’s complement form 11111111Two representations of –0 and +0, which is absurd.2’s complement form • + 0 in binary 00000000 • - 0 in 2’s complement form 00000000Same representation of +0 and –0.
    • END