Transcript Introduction to Number System

Introduction to
Number System
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Number System




When we type some letters or words, the computer translates them
in binary numbers as computers can understand only binary
numbers.
Decimal number system has base 10 as it uses 10 digits from
0 to 9. In decimal number system, the successive positions to the left
of the decimal point represent units, tens, hundreds, thousands and
so on.
A value of each digit in a number can be determined using
The digit
Symbol value (is the digit value 0 to 9)
The position of the digit in the number
Increasing Power of the base (i.e. 10) occupying successive positions
moving to the left
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Example
Decimal number (592):
Number
Symbol
Value
Position from Positional
the right end Value
Decimal
Equivalent
2
0
100
2*100 =
9
1
101
9*101 = 90
5
2
102
5*102 = 500
592
5 9 2
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2
Binary number system
 Uses two digits, 0 and 1.
 Also called base 2 number system
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(110011)2 = (51)10
Number
Symbol
Value
Position from
the right end
Positional
Value
Decimal
Equivalent
1
0
20
1*0 = 1
1
1
21
1*2 = 2
0
2
22
0*4 = 0
0
3
23
0*8 = 0
1
4
24
1*16= 16
1
5
25
1*32= 32
51
1 1 0 0 1 1
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Cont…
A Decimal number can converted into binary
number by the following methods:
 Double-Dabble Method
 Direct Method
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Double-Dabble Method
 Divide the number by 2
 Write the dividend under the number . This
become the new number
 Write the remainder at the right in column
 Repeat these three steps until a ‘0’ is
produced as a new number
 Output (bottom to top).
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Convert decimal 17 into binary
number
Step
Remainder
1
Divide 17 by 2
2 17
8
1
2
Divide 8 by 2
2 8
4
0
3
Divide 4 by 2
2 4
2
0
4
Divide 2 by 2
2 2
1
0
5
Divide 1 by 2
2 1
0
1
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Direct Method
 Write the positional values of the binary number
…. 26 25 24 23 22 21 20
…. 64 32 16 8 4 2 1
 Now compare the decimal number with position value
listed above. The decimal number lies between 32 and
64. Now place 1 at position 32.
45
64 32 16 8 4 2 1
1
 Subtract the positional value to the decimal number
i.e ( 45-32=13)
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Cont..
45
64 32 16 8 4 2 1
1
45-32 =13
1
1
13-8=5
1
1 1
5-4=1
1
1 1
1
1-1=0
Place 0 at the rest of position value
0 1 0 1 1 0 1
(45)10=(101101)2
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Decimal number to fractional
Binary number
 Multiply the decimal fraction by 2
 Write the integer part in a column
 The fraction part become a new fraction
 Repeat step 1 to 3 until the fractional part
become zero.
 Once the required number of digits (say 4)
have been obtained , we can stop.
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Example
 Decimal number is (0.625)
Fractional
decimal
number
Operation
Product
Fractional
part of
product
Integer part
of product
0.625
Multiply
by 2
1.250
.250
1
0.250
-do-
0.500
.500
0
0.500
-do-
1.000
0
1
Ans: (0.625)10= (0.101)2
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Questions
 Convert decimal 89 into equivalent binary
number by using Double-Dabble Method
(89)10= (1011001)2
 Convert decimal 89 into equivalent binary
number by using Direct Method
(89)10= (1011001)2
 Convert decimal 0.8125 into fractional binary
number
(0.8125)10 = (0.1101)2
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Convert Binary to Decimal
 Direct Method
 Double Dabble Method
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Direct Method
Binary Number
Positional
value
operation
1
1*20
1
0
0*21
0
1
1*22
4
0
0*23
0
0
0*24
0
1
1*25
32
1
1*26
64
1
1*27
128 = eITnotes.com
229
11100101
Double Dabble Method
Multiply left most digit by 2 add to
the next digit and so on.
1
1
0
1
2+
1
0
1
3
0
1
6+
0
1
6
1
12+
1
13
(1101)2= (13)10
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Example
 Convert Binary number 10111011 to decimal
(10111011)2 = (187)10
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Convert fractional Binary number to
Fractional Decimal number
 Write out the binary number as (-)ve power of
two. The various digits positions after binary
points are 1,2,3,4…..and so on.
 Convert each power of two into its decimal
equivalent
 Add these to give the decimal number
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Example
. 1
0
1*2-1
0*2-2
0.5
+
0
+
1
1
1*2-3
1*2-4
0.125
+
0.0625
= 0.6875
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Questions
 Convert the fractional binary number to
decimal number
 (0.1101)
ans= 0.8125
 (0.1011)
ans= 0.6875
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Octal number notation
 Octal is base 8 counting system having digit values 0
through 7
 The octal system groups three binary bits together into
one digit symbol.
Octal
Binary
0
000
1
001
2
010
3
011
4
100
5
101
6
110
7
111
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Convert binary number into
octal
 Divide the given binary number into group of
three bits (from right to left)
 Replace each group by its octal equivalent
 Examples:
11001
101010001110
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Convert decimal to octal
 Divide the number by 8
 Write the dividend under the number. This
become the new number
 Write the remainder at the right in a column
 Repeat steps 1 to 3 until a ‘0’ is produced as a
new number
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Question
 Convert decimal 17 to octal number
Ans= (17)10 = (21)8
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Convert octal to decimal number
 Write out the octal digits as power of 8
 Convert each power of 8 into its decimal
equivalent term
 Add these terms to produce the required
decimal number
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Example
7
2
1
=7*82
2*81
1*80
=448
16
1
465
(721)8= (465)10
Ques: Convert the octal 131 to its equivalent
decimal number
ans: 89
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Hexadecimal
 Hexadecimal number system is a base 16
counting system
 It uses 16 Symbols: 0 to 9 and the capital
letter A,B…F.
 Each Hexadecimal is equivalent to a group of
4 binary bits.
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Hexadecimal Binary
Hexadecima
l
Binary
0
0000
8
1000
1
0001
9
1001
2
0010
A
1010
3
0011
B
1011
4
0100
C
1100
5
0101
D
1101
6
0110
E
1110
7
0111
F
1111
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Convert binary to
Hexadecimal
 Divide the given binary number into groups of 4 bits
each(from right to left).
 Replace each group by its hexadecimal Equivalent.
Questions:
1. Convert (101111100001)2 into its hexadecimal.
Ans: (BEI)16.
2. Convert (10101111.0010111)2 into its hexadecimal.
Ans: (AF.2E)16
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Convert Decimal to
Hexadecimal
 Divide the number by 16.
 Write the dividend under the number. This
become the new number.
 Write the remainder at the right in a column.
 Repeat steps 1 to 3 until a ‘0’ is produced as a
new number.
Question: Convert the Decimal 87 to
hexadecimal number.
(87)10= (57)16
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Convert hexadecimal to
Decimal
 Write out the Hexadecimal digits as power of
16.
 Convert each power of 16 into its decimal
equivalent term.
 Add these terms to produce the required
decimal number.
Question: (A2D)16=(2605)10
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Data Representation
 We known that computer work with binary
numbers and therefore the numbers, letters,
and other symbols have to be converted into
their binary equivalents.
 However, this is not enough in the sense that
still we do not know how to store this binary
information so that it become suitable for
computer processing.
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Cont..
 The Representation of a positive integer
number is quite straight forward but we are
interested to represent positive as well as
negative numbers.
 For a Positive number , the sign bit set to 0
and for negative number the sign bit is set to
1.
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Integer Representation
 An integer can be represented by fixed point
representation
 The left most bit is considered as sign bit.
 The magnitude of the number can be
represented in following three ways:
Signed magnitude representation.
2. Signed 1’s complement representation.
3. Signed 2’s complement representation.
1.
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Signed Magnitude
 In this representation , if n bit of storage is available then
1 bit is reserved for sign and n-1 bits for the magnitude.
Sig
n
bit
magnitude
 The Disadvantage of this representation is that during
addition and Subtraction, the sign bit has to be
considered along with the magnitude.
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0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
Signed 1’s Compliment
 The 1’s Compliment of a binary integer can be
obtained by simply replacing the digit 0 by 1 and
digit 1 by 0
 Example: 00001100 is 11100111
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
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(+0)10
(-0)10
Signed 2’s Compliment
 The 2’s Compliment of a binary number is obtained by
adding 1 to 1’s Compliment.
 Example: (+12)10= 1100
0
0
0
0
1
1
0
0
 11110011
1
1
1’s Compliment
1
1
0
0
1
1
1
11110100
2,s Compliment
1’s
integer
2’s compliment
is the
1Therefore,
1 Positive
1
1
0
1
0 negative
0 integer(-12)10
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Question
 Express the following in signed magnitude
form, 1’s Compliment, 2’s Compliment:
 (35)10 = 100011
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Floating point representation
 We can represent a floating point binary number in the following
form:
±M * 2±e
 Where M : is the mantissa or significant
e : is the exponent
 Example: 101.11
10111 * 2-2
101.11 * 20
10.111 *21
1.0111 *22
.10111 * 23
.010111 * 24
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Cont..
.10111 * 23
M e
 The Mantissa part of the number is suitably shifted (left
or right) to obtain a non zero digit at a most significant
position. The activity is known as normalization.
 In a 16 bit representation, let us assume that 10 bits are
reserved for mantissa and 6 for exponent.
Sign
Sign
Mantissa
0
1
0
1
1
1
exponent
0
0
0
0
0
0
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0
0
1
1
Question
 Represent floating point binary number in 16
bit representation (1110.001)
The normalization number is = .1110001 * 24
16 bit representation:
Sign
Sign
0 111000100 0 00100
M
e
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