This document discusses data representation and number systems in computing. It covers the following key points in 3 sentences: Data such as numbers and coded information are represented using bits and bytes which can represent values, characters, or instructions. Common number systems used in computing include binary, decimal, octal, and hexadecimal, which use different radixes or bases to represent quantities with distinct symbols. Methods for converting between number systems involve grouping bits or digits into the appropriate radix and determining the place value of each position to arrive at the value in the target base.

1. 011010110011010110011010110011010110101
Data Representation
and
Number System

2. Data Representation
• Data:
Data are numbers and other binary-coded information that are
operated on to achieve required computational results.
• Control Information
Control information is a bit or a group of bits used to specify the
sequence of command signals.

3. 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.

4. 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 distinct
symbols for r digits.
In this system the number of states each digit has is determined by the
base or radix.
• Based on the radix there are four number systems
o Decimal
o Binary
o Octal
o Hexadecimal

5. Decimal system
Radix / base = 10
• For example :
253 means 0
2*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

6. Binary system
Radix / base = 2
• For example :
0
1011 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

7. 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

8. 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 some
older computers

9. Hexadecimal system
Radix / base = 16
A
B
• For example : 0 C
3B6 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 7
modern computers e.g., 3C0F 95EA 8
9

10. 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)

11. Binary to Decimal Conversion
For Integers
For 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

12. Exercise
• 1010002 = ( ??? )10 4010
• 10010112 =
7510
• 1000112 =
• 0110112 =
01001011
Most Significant Bit Least Significant Bit

13. From Decimal to Binary
For Integers
• Divide by 2 until you reach zero, and then collect the
remainders in reverse.
• For example: 5610 = ( ???? )2
2 ) 56 0 Least significant bit
2 ) 28 0
2 ) 14 0
2)71
2)31
2)11
0 (0111000)2

14. Exercise
• (48)10 = ( ????? )2
(00110000)2

15. Binary to Decimal
With Fraction
Binary 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

16. From Decimal to Binary
With Fraction
For 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 time
discard the integer so obtained.
• Collect this discarded integer part as the binary equivalent.
• Repeat this process until zero or until the required accuracy.

17. 0.6875 0.5000
x2x2
1.3750 1.0000
0.3750 .1011
x2
0.7500
(56.6875)10 = (111000.1011)2
0.7500
x2
1.5000

18. Exercise
• (48.3125)10 = ( ????? )2
(00110000.0101)2

19. 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 successive
divisions by r and accumulating the remainders.
This is repeated until the quotient becomes zero.
Collect remainders in the reverse order.

20. • Rule For converting the decimal part:
Conversion of a decimal to a base r is done by successive
multiplications 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

21. 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

22. 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 6
8 ) 59 3 8 ) 7 7
0
(736)8

23. • For Fraction part:
o The Multiplication Method: Multiply the fraction part
successively by 8 and accumulate the remainders until
you reach zero.
0.5
x 8 (736.4)8
4.0

24. Octal to Decimal Conversion
• To convert to base 10, beginning with the rightmost digit
multiply each nth digit by 8(n-1), and add all of the results
together.
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

25. Exercise
• (0.40)8 = ______10 0.50

26. 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

27. 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 11
16) 126 14 = E C 12
D 13
16) 7 7 E 14
0 F 15

28. Exercise
• (AF)16 = ______10
175

29. Binary to Octal
• Group the binary number into groups of 3 bits starting from
the 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

30. 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

31. Binary -Coded Octal Numbers
Three-bit Group Decimal Digit Octal Digit
000 0 0
001 1 1
010 2 2
011 3 3
100 4 4
101 5 5
110 6 6
111 7 7
001 000 10
010 100 24

32. Binary to Hexadecimal
• Group the binary number into groups of 4 bits starting
from the least significant bit, and convert it into its decimal
equivalent.
A 10
For example: (1010 1111 0110 0011)2 B 11
C 12
Grouping : 1010 1111 0110 0011 D 13
E 14
AF63 F 15
(1010111101100011)2 = (AF63)16

33. 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

34. 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

35. 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

36. 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

37. Binary to octal and hexadecimal
EXERCISE
• 1010111101100011 Binary
• 1010111101100011
Octal 1275438
• 1010111101100011
Hexa AF6316

38. • 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.

39. Complements
There are two types of complements for each base r system:
• r’s complement
• ( r-1)’s complement

40. (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.

41. 9’s Complement
• For example:
For decimal number N= 546700, n= 6 and r =10
9’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.

42. 1’s Complement
• For example:
For binary number N= 1011, n= 4 and r =2
1’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.

43. 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
_______

44. 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.

45. 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

46. 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

47. Exercise
• Find the 2’s complement of 10101011
01010101
• Find the 2’ complement of 01010101
10101011

48. 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

49. 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

50. 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 to
rn – (N – M), which is the r’s complement of (N – M). To obtain the answer
in a familiar form, take the r’s complement of the sum and place a negative
sign in front.
This will equate to : rn – (rn – ( N – M)) = M - N

51. Subtraction of Unsigned Numbers
Case 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

52. Subtraction of Unsigned Numbers
Case 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

53. 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.

54. Subtraction of Unsigned Numbers
• In a similar manner, the subtraction with complements is done with
binary numbers.
• For example:
X: 1010100
Y: 1000011
• To perform X – Y :
X = 1010100
2’s complement of Y = 0111101
Sum = 10010001
Discard the end carry 10000000
0010001

55. Exercise
Y: 1000011
X: 1010100
radix 2
Perform Y – X = ????
- 0010001

56. 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= 11100100
00011101- 00011011 _________
radix 2 100000001
1
RESULT 00000010

57. 1’s Complement Subtraction
Unsigned Representation
Case 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 11111011
00011001 - 00011101
radix 2
RESULT -00000100

58. Exercise (using 1’s complement)
X: 00110011
Y: 00101101 Perform X - Y
radix 2
00000110

59. 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 representation
2. Signed - 1’s complement representation
3. Signed - 2’s complement representation.

60. Example for Negative number
Representation
• To represent -14
1. Signed magnitude representation
1 0001110
Note : This representation of – 14 is obtained from +14 by
complementing only the sign bit.
2. Signed - 1’s complement representation
1 1110001
Note : This representation of – 14 is obtained by complementing all
the bits of + 14, including the sign bit.

61. Example for Negative number
Representation
3. Signed - 2’s complement representation.
1 1110010
Note : This representation of – 14 is obtained by taking the 2’s
complement of +14, including the sign bit.

62. 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

63. NOTE
1’s complement form
• + 0 in binary 00000000
• - 0 in 1’s complement form 11111111
Two representations of –0 and +0, which is absurd.
2’s complement form
• + 0 in binary 00000000
• - 0 in 2’s complement form 00000000
Same representation of +0 and –0.

64. END