Number Systems Types of number systems Number bases Range of possible numbers Conversion between number bases Common powers Arithmetic in different number bases Shifting a number

1. 1
Lecture 2
Number Systems
ITEC 1000 “Introduction to Information Technology”
Prof. Peter Khaiter
Hexadecimal
Decimal Octal
Binary

2. 2
Lecture Outline:
 Types of number systems
 Number bases
 Range of possible numbers
 Conversion between number bases
 Common powers
 Arithmetic in different number bases
 Shifting a number

3. 3
Types of Number Systems
 Additive: Numbers have intrinsic value:
e.g.: Roman Numerals: LVIII = 50 + 5 + 1 + 1
+ 1 = 58
 Positional: Value depends on position: e.g.:
Decimal system: 55 = 5 x 10 + 5 x 1
 Additive Number Systems are not used
much any more:
Awkward to use.
Prone to errors.

4. 4
Definitions
 The Base of a number system – how
many different digits (incl. zero) are
used in the system.
Base 2: 0, 1
Base 5: 0, 1, 2, 3, 4
Base 8: 0, 1, 2, 3, 4, 5, 6, 7
Base 10: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Base 16: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

5. 5
Definitions
 Bit – a cell holding a single binary
number (0 or 1)
 Byte = 8 bits (can hold 28 = 256
different patterns/values)
 Word – a fixed-sized group of bits
that the computer handles together.
Typical word sizes: 4, 8, 16, 32, 64,
128 bits
 1K = 1024 bytes

6. 6
Magnetic Core Memory
http://en.wikipedia.org/wiki/Computer

7. 7
Common Number Systems
System Base Symbols
Used by
humans?
Used in
computers?
Decimal 10 0, 1, … 9 Yes No
Binary 2 0, 1 No Yes
Octal 8 0, 1, … 7 No Yes
Hexa-
decimal
16 0, 1, … 9,
A, B, … F
No Yes

8. 8
Positional decimal system
The number 125 means:
1 group of 100 (100 = 102)
2 groups of 10 (10 = 101)
5 groups of 1 (1 = 100)

9. 9
Place values (1 of 2)
In our usual positional number system, the
meaning of a digit depends on where it is
located in the number
3 groups of 1000
7 groups of 100
3 groups of 10
2 groups of 1
Example: 3 7 3 2

10. 10
Place values (2 of 2)
12510 => 5 x 100 = 5
2 x 101 = 20
1 x 102 = 100
125
Base
Weight

11. 11
Representing in bases: 10, 2, 8, 16
 86510 = 8 x 102 + 6 x 101 + 5 x 100 = 800
+ 60 + 5
 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20
= 8 + 2 + 1 = 1110
 258 = 2 x 81 + 5 x 80 = 16 + 5 = 2110
 A716 = 10 x 161 + 7 x 160 = 160 + 7 =
16710
Note: The subscript naming the base is
itself given in base ten (10), by
convention.
Base

12. 12
Counting in bases (1 of 3)
Decimal Binary Octal
Hexa-
decimal
0 0 0 0
1 1 1 1
2 10 2 2
3 11 3 3
4 100 4 4
5 101 5 5
6 110 6 6
7 111 7 7

13. 13
Counting in bases (2 of 3)
Decimal Binary Octal
Hexa-
decimal
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

14. 14
Counting in bases (3 of 3)
Decimal Binary Octal
Hexa-
decimal
16 10000 20 10
17 10001 21 11
18 10010 22 12
19 10011 23 13
20 10100 24 14
21 10101 25 15
22 10110 26 16
23 10111 27 17

15. 15
Estimating magnitude: Binary
1101 01102 = 21410
1101 01102 > 19210 (128 + 64 + additional bits to the right)
Place 27 26 25 24 23 22 21 20
Value 128 64 32 16 8 4 2 1
Evaluate 1 x 128 1 x 64 0 x 32 1 x16 0 x 8 1 x 4 1 x 2 0 x 1
Sum for
Base 10
128 64 0 16 0 4 2 0

16. 16
Range of possible numbers
 R = BK where
R = range
B = base
K = number of digits
 Example #1: Base 10, 2 digits
R = 102 = 100 different numbers (0…99)
 Example #2: Base 2, 16 digits
R = 216 = 65,536 or 64K
16-bit PC can store 65,536 different number
values

17. 17
Decimal Range for Bit Widths
Bits Digits Range
1 0+ 2 (0 and 1)
4 1+ 16 (0 to 15)
8 2+ 256
10 3+ 1,024 (1K)
16 4+ 65,536 (64K)
20 6+ 1,048,576 (1M)
32 9+ 4,294,967,296 (4G)
64 19+ Approx. 1.6 x 1019
128 38+ Approx. 2.6 x 1038

18. 18
Conversion Among Bases
Hexadecimal
Decimal Octal
Binary

19. 19
Binary to Decimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

20. 20
Binary to Decimal (2 of 3)
 Technique
Multiply each bit by 2n, where n is the
“weight” of the bit
The weight is the position of the bit, starting
from 0 on the right
Add the results

21. 21
Binary to Decimal (3 of 3)
1010112 => 1 x 20 = 1
1 x 21 = 2
0 x 22 = 0
1 x 23 = 8
0 x 24 = 0
1 x 25 = 32
4310
Bit “0”

22. 22
Octal to Decimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

23. 23
Octal to Decimal (2 of 3)
 Technique
Multiply each bit by 8n, where n is the
“weight” of the bit
The weight is the position of the bit,
starting from 0 on the right
Add the results together
Note: 80 = 1, 81 = 8,
82 = 64 83 = 512, Etc.

24. 24
Octal to Decimal (3 of 3)
7248 => 4 x 80 = 4 x 1 = 4
2 x 81 = 2 x 8 = 16
7 x 82 = 7 x 64 = 448
46810

25. 25
Hexadecimal to Decimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

26. 26
Hexadecimal to Decimal (2 of 3)
 Technique
Multiply each bit by 16n, where n is the
“weight” of the bit
The weight is the position of the bit, starting
from 0 on the right
Add the results
Note: 160 = 1, 161 = 16,
162 = 256 163 = 4096, Etc.

27. 27
Hexadecimal to Decimal (3 of 3)
ABC16 => C x 160 = 12 x 1 = 12
B x 161 = 11 x 16 = 176
A x 162 = 10 x 256 = 2560
274810

28. 28
Decimal to Binary (1 of 3)
Hexadecimal
Decimal Octal
Binary

29. 29
Decimal to Binary (2 of 3)
 Technique
Divide by two, keep track of the remainder
First remainder is bit 0 (LSB, least-significant
bit)
Second remainder is bit 1
Etc.

30. 30
Decimal to Binary (3 of 3)
12510 = ?2 2 125
62 12
31 02
15 12
7 12
3 12
1 12
0 1
12510 = 11111012

31. 31
Octal to Binary (1 of 3)
Hexadecimal
Decimal Octal
Binary

32. 32
Octal to Binary (2 of 3)
 Technique
Convert each octal digit to a 3-bit
equivalent binary representation
See Table, Slides 12-14

33. 33
Octal to Binary (3 of 3)
7058 = ?2
7 0 5
111 000 101
7058 = 1110001012

34. 34
Hexadecimal to Binary (1 of 3)
Hexadecimal
Decimal Octal
Binary

35. 35
Hexadecimal to Binary (2 of 3)
 Technique
Convert each hexadecimal digit to a 4-bit
equivalent binary representation
See Table, Slides 12-14

36. 36
Hexadecimal to Binary (3 of 3)
10AF16 = ?2
1 0 A F
0001 0000 1010 1111
10AF16 = 00010000101011112

37. 37
Decimal to Octal (1 of 3)
Hexadecimal
Decimal Octal
Binary

38. 38
Decimal to Octal (2 of 3)
 Technique
Divide by 8
Keep track of the remainder

39. 39
Decimal to Octal (3 of 3)
123410 = ?8
8 1234
154 28
19 28
2 38
0 2
123410 = 23228

40. 40
Decimal to Hexadecimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

41. 41
Decimal to Hexadecimal (2 of 3)
 Technique
Divide by 16
Keep track of the remainder

42. 42
Decimal to Hexadecimal (3 of 3)
123410 = ?16
123410 = 4D216
16 1234
77 216
4 13 = D16
0 4

43. 43
Binary to Octal (1 of 3)
Hexadecimal
Decimal Octal
Binary

44. 44
Binary to Octal (2 of 3)
 Technique
Group bits in threes, starting on right
Convert to octal digits
See Table, Slides 12-14

45. 45
Binary to Octal (3 of 3)
10110101112 = ?8
1 011 010 111
1 3 2 7
10110101112 = 13278

46. 46
Binary to Hexadecimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

47. 47
Binary to Hexadecimal (2 of 3)
 Technique
Group bits in fours, starting on right
Convert to hexadecimal digits
See Table, Slides 12-14

48. 48
Binary to Hexadecimal (3 of 3)
10101110112 = ?16
10 1011 1011
2 B B
10101110112 = 2BB16

49. 49
Octal to Hexadecimal (1 of 3)
Hexadecimal
Decimal Octal
Binary

50. 50
Octal to Hexadecimal (2 of 3)
 Technique
Use binary as an intermediary
See Table, Slides 12-14

51. 51
Octal to Hexadecimal (3 of 3)
10768 = ?16
1 0 7 6
001 000 111 110
2 3 E
10768 = 23E16

52. 52
Hexadecimal to Octal (1 of 3)
Hexadecimal
Decimal Octal
Binary

53. 53
Hexadecimal to Octal (2 of 3)
 Technique
Use binary as an intermediary
See Table, Slides 12-14

54. 54
Hexadecimal to Octal (3 of 3)
1F0C16 = ?8
1 F 0 C
0001 1111 0000 1100
1 7 4 1 4
1F0C16 = 174148

55. 55
Exercise – Convert ...
Don’t use a calculator!
Decimal Binary Octal
Hexa-
decimal
33
1110101
703
1AF
Skip answer Answer

56. 56
Exercise – Convert (answers)
Decimal Binary Octal
Hexa-
decimal
33 100001 41 21
117 1110101 165 75
451 111000011 703 1C3
431 110101111 657 1AF

57. 57
Common Powers (1 of 2)
 Base 10
Power Preface Symbol
10-12 pico p
10-9 nano n
10-6 micro 
10-3 milli m
103 kilo k
106 mega M
109 giga G
1012 tera T
Value
.000000000001
.000000001
.000001
.001
1000
1000000
1000000000
1000000000000

58. 58
Common Powers (2 of 2)
 Base 2 Power Preface Symbol
210 kilo k
220 mega M
230 Giga G
Value
1024
1048576
1073741824
What is the value of “k”, “M”, and “G”?
In computing, particularly w.r.t. memory,
the base-2 interpretation generally applies

59. 59
Example
/ 230 =
1. Double click on My Computer
2. Right click on C:
3. Click on Properties

60. 60
Multiplying powers
 For common bases, add powers
26  210 = 216 = 65,536
or…
26  210 = 64  210 = 64k
ab  ac = ab+c

61. 61
Table of powers
Power
Base 8 7 6 5 4 3 2 1 0
2
256 128 64 32 16 8 4 2 1
8
32,768 4,096 512 64 8 1
16
65,536 4,096 256 16 1

62. 62
Fractions
 Number point or radix point
Decimal point in base 10
Binary point in base 2
 No exact relationship between
fractional numbers in different
number bases
Exact conversion may be impossible

63. 63
Decimal fractions
 Move the number point one place to the
right
Effect: multiplies the number by the base
number
Example: 139.010 139010
 Move the number point one place to the
left
Effect: divides the number by the base
number
Example: 139.010 13.910

64. 64
Fractions: Base 10
Place 10-1 10-2 10-3 10-4
Value 1/10 1/100 1/1000 1/10000
Evaluate 2 x 1/10 5 x 1/100 8 x 1/1000 9 x1/10000
Sum .2 .05 .008 .0009
.258910
10-1 10-2 10-3 10-4

65. 65
Fractions: Base 2
.1010112 = 0.67187510
Place 2-1 2-2 2-3 2-4 2-5 2-6
Value 1/2 1/4 1/8 1/16 1/32 1/64
Evaluate 1 x 1/2 0 x 1/4 1x 1/8 0 x 1/16 1 x 1/32 1 x 1/64
Sum .5 0.125 0.03125 0.015625
2-1 2-2 2-3 2-4 2-5 2-6

66. 66
Fractions: Base 10 and Base 2
 No general relationship between fractions
of types 1/10k and 1/2k
Therefore a number representable in base 10
may not be representable in base 2
But: the converse is true: all fractions of the
form 1/2k can be represented in base 10
 Fractional conversions from one base to
another are stopped
If there is a rational solution, or
When the desired accuracy is attained

67. 67
Fractions: From Base B To Base 10
 Determine the appropriate weight for
each fractional digit as a negative power
of the base
 Multiply each digit by its weight
 Add the values
 Example:
.A116 = 10x16-1 + 1x16-2 = 10x0.0625 +
1x0.00390625 = 0.6289062510

68. 68
Fractions: From Base 10 To Base B
 Multiply the fraction by the base
value B (i.e., 2, 8 or 16)
 Record the values that move to the
left of the radix point and drop them
 Repeat the process until
the value being multiplied is zero, or
the desired number of digits of
accuracy is attained

69. 69
Fractions: From Base 10 To Base B
0.67312510 = ?2 0.673125
x 2
1.346250
x 2
0.692500
x 2
1.385000
x 2
0.770000
x 2
0.1010112 1.540000
x 2
1.080000

70. 70
Fractions: From Base 2 To Base 8, 16
 Group digits from left to right in
groups of 3 (base 8) or 4 (base 16)
 Supplement the right-most group
with 0’s, if necessary
 Convert each group to the desired
base.
See Table, Slides 12-14

71. 71
Fractions: From Base 8, 16 To Base 2
 Convert each octal (base 8) or
hexadecimal (base 16) digit to its 3-
bit or 4-bit representation
See Table, Slides 12-14

72. 72
Fractions: between Base 8 and
Base 16
 Use binary conversion as an
intermediary
 Example:
0.C816 = ?8
1100 10002
6 2 0 0.C816 = 0.628

73. 73
Mixed number conversion
 Convert whole part and fraction part
separately
See Table, Slides 12-14

74. 74
Arithmetic operations (1 of 14)
 Binary Addition
Two 1-bit values
A B A + B
0 0 0
0 1 1
1 0 1
1 1 10
0 and carry 1 to the
next more
significant bit, i.e.
“two”

75. 75
Arithmetic operations (2 of 14)
 Two n-bit values
Add individual bits (see Table)
Propagate carries
10101 21
+ 11001 + 25
101110 46
11
Note: superscripts are carried amounts.

76. 76
Addition (different bases) (3 of 14)
Base Problem Largest Single Digit
Decimal
6
+3
9
Octal
6
+1
7
Hexadecimal
6
+9
F
Binary
1
+0
1

77. 77
Addition (different bases) (4 of 14)
Base Problem Carry Answer
Decimal
6
+4
Carry the 10 10
Octal
6
+2
Carry the 8 10
Hexadecimal
6
+A
Carry the 16 10
Binary
1
+1
Carry the 2 10

78. 78
Addition Table: Base 10 (5 of 14)
+ 0 1 2 3 4 5 6 7 8 9
0 0 1 2 3 4 5 6 7 8 9
1 1 2 3 4 5 6 7 8 9 10
2 2 3 4 5 6 7 8 9 10 11
3 3 4 5 6 7 8 9 10 11 12
4 4 5 6 7 8
etc
9 10 11 12 13
310 + 610 = 910

79. 79
Addition Table: Base 8 (6 of 14)
+ 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7
1 1 2 3 4 5 6 7 10
2 2 3 4 5 6 7 10 11
3 3 4 5 6 7 10 11 12
4 4 5 6 7 10 11 12 13
5 5 6 7 10 11 12 13 14
6 6 7 10 11 12 13 14 15
7 7 10 11 12 13 14 15 16
38 + 68 = 118

80. 80
Arithmetic operations (7 of 14)
 Binary Subtraction
Two 1-bit values
A B A - B
0 0 0
0 1 1
1 0 1
1 1 0
and borrow 1 from
the next more
significant bit

81. 81
Arithmetic operations (8 of 14)
 Two n-bit values
Subtract individual bits (see Table)
Keep track of the borrowings
00100101 – 00010001 = 00010100
00100101 3710
- 00010001 - 1710
00010100 2010
0 10

82. 82
Arithmetic operations (9 of 14)
 Multiplication
Decimal (just for fun)
35
x 105
175
000
35
3675

83. 83
Arithmetic operations (10 of 14)
 Binary multiplication, two 1-bit values
A B A  B
0 0 0
0 1 0
1 0 0
1 1 1

84. 84
Arithmetic operations (11 of 14)
 Binary multiplication, two n-bit values
As with decimal values
1110
x 1011
1110
1110
0000
1110
10011010

85. 85
Multiplication Table: Base 10 (12 of
14)
x 0 1 2 3 4 5 6 7 8 9
0 0
1 1 2 3 4 5 6 7 8 9
2 2 4 6 8 10 12 14 16 18
3 3 6 9 12 15 18 21 24 27
4 0 4 8 12 16 20 24 28 32 36
5 5 10 15 20 25 30 35 40 45
6 6 12 18 24 30 36 42 48 54
7 7 14 21 28 35 42 49 56 63
etc.
310 x 610 = 1810

86. 86
Multiplication Table: Base 8 (13 of
14)
x 0 1 2 3 4 5 6 7
0 0
1 1 2 3 4 5 6 7
2 2 4 6 10 12 14 16
3 0 3 6 11 14 17 22 25
4 4 10 14 20 24 30 34
5 5 12 17 24 31 36 43
6 6 14 22 30 36 44 52
7 7 16 25 34 43 52 61
38 x 68 = 228

87. 87
Arithmetic operations (14 of 14)
 Binary division, two n-bit values
As with decimal values
100001/11 = 1011 11 ) 100001
1011
11
00100
11
11
11
0
divisor
dividend
quotient

88. 88
Shifting a number
 Shifting a decimal number to the left
by one position is equivalent to
multiplying by 10
 Shifting a binary number to the left by
one position is equivalent to multiplying
by 2
 General rule: shifting a number in any
base left one digit multiplies its value
by the base; shifting one digit right
divides its value by the base

89. 89
Thank you!
Reading: Lecture slides and notes, Chapter 3