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Lecture 2

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					INTRODUCTION TO COMPUTER                                                          LECTURE 02


           NUMBER SYSTEM
           As we know every computer stores numbers, letters and other special
           characters in a code form. Before going into the details of these codes, it is
           essential to have a basic understanding of the number system.
           Number systems are basically of two types:
           i)     Non positional number system
           ii)    Positional number system.

      i)    Non-Positional Number System
           In early day, human beings counted on fingers, when ten fingers were not
           adequate, stones, pebbles, or sticks we used to indicate values. This
           method of counting using an additive approach is called non-Positional
           Number System. In this system, we have symbols such as I for 1, II for 2,
           III for 3, IIII for 4, IIIII for 5 etc. Since it is very difficult to perform
           arithmetic operations with such a number system.

ii)        Positional Number System
           In a positional Number system there are only a few symbols called digits,
           and these symbols represents different values, depending on the position
           they occupy in the number. The value of each digit in such a number
           determined by three considerations.
           a)      The digit itself.
           b)      The position of the digit.
           c)      The base of the number system.
           The number system that we use in our daily life is called the decimal number
           system. In this system the base is equal to 10 because there are altogether ten
           digits used in this number system. E.g. the decimal number 2586 written as
           (2586)10 consists of the digit 6 in the units position, 8 in the ten position, 5 in the
           hundreds position and 2 in the thousands position and its value can be written as :
                           (2x1000) + (5x100) + (8x10) + (6x1)
                   OR      2000 + 500 + 80 + +6 = 2586

                 Some of the number systems commonly used in computer design
           and by computer professionals are discussed bellow.
1.         BINARY NUMBER SYSTEM
           The binary number system is exactly like the decimal system except that
           the base is 2 instead of 10. We have only two digits(0 and 1) that can be
           used in this number system. Each position in a binary number represents
           a power of the base (2). E.g. the decimal 21 is equivalent of the binary
           number 10101 written as

           (10101)2 is     (1x 24 ) + (0x 23 ) + (1x22) + (0x21) + (1x20)
                           =16 + 0 + 4 + 0 + 1
                           = 21
                                                                                                 1
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                           LECTURE 02




                         Binary                 Decimal
                                               Equivalent
                           000                     0
                           001                     1
                           010                     2
                           011                     3
                           100                     4
                           101                     5
                           110                     6
                           111                     7


2.   OCTAL NUMBER SYSTEM
     In the octal number system the base is 8. So in this system there are only
     eight symbols or digits: 0,1,2,3,4,5,6 and 7 (8 and 9 do not exist in this
     system) again each position in an octal number represents a power of the
     base(8). The decimal equivalent of the octal number 2057 (written as 20578)
     is as follows:

           (2x83) + (0x82) + (5x81) + (7x80)
           =1024 + 0 + 40 + 7
           = 1071

     So we have 20578=107110
     Observe that since there are only 8 digits in the octal number system, So 3
     bits (23=8) are sufficient to represent any octal number in binary.

3.   HEXADECIMAL NUMBER SYSTEM
     The hexadecimal number system is one with base of 16. The base of 16
     suggests choices of 16 single character digits as follows:
                   0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
     The symbols A, B,C,D,E,F representing the decimal values
     10,11,12,13,14,15 respectively. Each position in a hexadecimal system
     represents a power of the base (16) thus the decimal equivalent of the
     hexadecimal number (1AF)16
                   (1 x 162 ) + (A x 161 ) + (F x 160)
                   =(1x 256) + (10x 16) + (15 x 1)
                   =256 + 160 + 15
                   =431
                                                                                   2
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                          LECTURE 02


                                  Thus (1AF)16 =(431)10
      CONVERTING FROM ONE NUMBER SYSTEM TO
      ANOTHER

      There are many methods or techniques that can be used to convert
      numbers from one base to another. We will see one technique used in
      converting to base 10 from any other base and second technique to be
      used in converting from base 10 to any other base.


1.    Converting To Decimal From Another Base
      The following three steps are used to convert to a base 10 value from any
      other number system.
      Step1. Determine the column (positional) value of each digit ( this
             depends on the position of the digit and the base of the number
             system)
      Step2 Multiply the obtained column values (in Step1 ) by the digit in
             corresponding columns.
      Step3. Sum the products calculated in Step2. The total is the equivalent
             value in decimal.

      Example1:    110012 = ?10

      Solution:
      Step1. Determine column value.

      Column No. from right       Column Value
             1                       20=1
              2                      21=2
              3                      22=4
              4                      23=8
               5                     24=16

Step2. Multiply column values by corresponding column digit.

                   16    8     4     2     1
                     x   x     x     x     x
                    1    1     0     0     1
                   __________________________
                   16     8    0     0     1

Step3. Sum the product
                   16 + 8 + 0 + 0 + 1=25
                                                                                  3
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                        LECTURE 02


      Hence           110012=2510


Example 2.    (4706)8=?10

Solution:                     4x8 + 7x8 + 0x8 + 6x8
                             =4x512 + 7x64 +0x8 + 6x1
                             =2048 + 448 + 0 + 6
                             =2502

      Hence (4706)8=(2502)10


Example 3.    (1AC)16 =?10

Solution:                    (1AC)16=1x162 + 10x161 +12x160
                                    =256 + 160 + 12
              Hence                 =42810

Example 4.    40527=?10

Solution:              4x73 + 0x72 + 5x71 + 2x70
                      =4x343 + 0x49 + 5x7 + 2x1
                      =1372 + 0 + 35 + 2
      Hence           =(1409)10

Example 5.    (4052)6=?10

Solution:              4x6 + 0x6 + 5x6 + 2x6
                      =4x216 + 0 + 5x6 + 2x1
                      =864 + 0 + 30 + 2
                      =89610
      Hence           (4052)6 =(896)10

2.    Converting From Base 10 To A New Base

      The following four steps are used to convert a number from base 10 to
      another base.
      Step1. Divide the decimal number to be converted by the value of the new
      base.
      Step2. Record the remainder from Step1 as the rightmost digit ( least
             significant digit) of the new bas number.
      Step3. Divide the quotient of the previous divide by the new base.

                                                                             4
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                            LECTURE 02


     Step4. Record the remainder from Step3 as the next digit (to the left) of the
            new base number.
            Repeat Step3 and 4 recording remainders from right to left, until
            the quotient becomes zero in Step3. Note that the last remainder
            thus obtained will be the most significant digit(MSD) of the new
            base number.
Example:    2510=?2

             Step 1&2:     25/2= 12 and remainder 1
             Step 3&4:     12/2= 6 and remainder 0
             Step 3&4:     6/2 = 3 and remainder 0
             Step 3&4:     3/2 = 1 and remainder 1
             Step 3&4:     1/2 = 0 and remainder 1

            As mentioned in step 2&4, the remainders have to be arranged in
            the reverse order so that the first remainder becomes the least
            significant (LSD) and the last remainder becomes the most
            significant digit (MSD).
      Hence         (25)10 = (11001)2

Example 1.   4210 = ?2

                                  2      42

                                   2     21     0

                                   2    10      1

                                   2     5      0

                                   2     2      1

                                         1      0

             Hence:               (42)10= (101010)2




                                                                                 5
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                           LECTURE 02


Example 2.      (952)10 =?8
                                         8            952

                                          8          119      0

                                             8       14       7

                                          2              1     6

        Hence          (952)10 = (1670)8

Example 3.      (428)10 = ?16
                                          16         428

                                          16         26       12

                                                     1        10

Hence           (428)10 = (1AC)16

3.      Converting From A Base Other Than 10 To A Base Other
        Than 10
     The following two steps are used to convert one number system to
     another number system.
     Step 1.       Convert the original number to a decimal number (Base 10)
     Step 2.       Convert the decimal so obtained to the new base.
Example.    (545)6 = ?4

        Step 1.        Conversion from base 6 to base 10

                  5x6 + 4x6 + 5x6
                = 5x36 + 4x6 + 5x1
                =180 + 24 + 5
                = (209)10
Step 2.         Convert (209)10 to base 4

                                    4         209

                                    4          52        1
                                     4        13         0

                                                 3        1

        Hence          (545)6 = (3101)4
                                                                               6
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                            LECTURE 02


4.    Shortcut Method For Binary To Octal Conversion
     The following steps are used in this method.
     Step 1.      Divide the binary digits into groups of three (starting from
                  right).
     Step 2.      Convert each group of three binary digits into one octal digit
                  by using weight 1,2 and 4 from right side.
Example.    (101110)2 = ?8
     Step 1.      divide the binary digits into groups of 3 starting from right
                  (LSD)
                           101 110

      Step 2.       Convert each group into one digit of octal.
                    (101)2 = 1x22 + 0x21 + 1x20
                           =4 + 0 + 1
                           =58
                    (110)2 = 1x22 + 1x21 + 0x20
                           =4 + 2 + 0
                           =68
      Hence         (101110)2 = (56)8
5.    Shortcut Method For Octal To Binary Conversion
      The following steps are used in this method.
      Step 1. Convert each octal digit to a 3 digit binary number.
      Step 2. Combine all the resulting binary groups into a single binary
      number.
      Example.     (562)8 = ?2

      Solution.
      Step 1.       Convert each octal digit to 3 binary digits.
                         58 = 1012
                         68 = 1102
                         28 = 0102

      Step 2.       Combine the binary groups.
      Hence:             5628 = (101110010)2

6.    Shortcut Method For Binary To Hexadecimal Conversion
      The following steps are used in this method.

      Step 1.       Divide the binary digits into groups of four.
      Step 2.       Convert each group of four binary digits to one hexadecimal
                    digit.



                                                                               7
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                                          LECTURE 02


     Example.     (11010011)2 = ?16
     Solution.
     Step 1.      Divide the binary digits into group of 4.

     Step 2.             1101 0011
            Convert each group of 4 binary digits to 1 hexadecimal digit.

           11012 = 1x23 + 1x22 + 0x21 + 1x20
                 =8+4+0+1
                 = 1310
                 = D16
           00112 = 0x23 + 0x22 + 1x21 + 1x20
                 =0+0+2+1
                 = 316
     Hence       (1010011)2 = D316

7.   Shortcut Method For Hexadecimal To Binary Conversion

     The following steps are used in this method

     Step 1.      Convert the decimal equivalent of each hexadecimal digit to
                  4 binary digits.
     Step 2.      Combine all the resulting binary groups into a single binary
                  number.
     Example.     (2AB)16 = ?2

     Solution.
     Step 1.      Convert the decimal equivalent of each hexadecimal digit to
                  4 binary digits.
                         216 = 210 = (0010)2
                         A16 = (10)10 = (1010)2
                         B16 = (11)10 = (1011)2
     Step 2.      Combine the binary groups.
                  2AB16 = (001010101011)2
8.   Fractional Numbers
     In a binary number system, fractional numbers are formed in the same
     general way as in the decimal system.

           0.235 = (2x10-1) + (3x10-2) + (5x10-3)
           and

           68.53 = (6x101) +(8x100) + (5x10-1) + (3x10-2)


                                                                             8
PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.
INTRODUCTION TO COMPUTER                               LECTURE 02


         Similarly in binary system

         0.101 = (1x2-1) + (0x2-2) + (1x2-3)
         and
         10.01 = (1x21) + (0x20) + (0x2-1) + (1x2-2)




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PREPARED BY SADIQUE A. BUGTI LECTURER BUITMS QUETTA.

				
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