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DATA REPRESENTATION BY-                            Ravi Sharma
Binary number system- [ 0and 1 ]                                                        Radix-2 , e.g.-(101101)2 Decimal number system- [ 0 to 9 ]              Radix-10 , e.g.-(243)10 Octal number system- [ 0 to 7 ]                                       Radix-8 , e.g.-(736.4)8 Hexadecimal - [ 0 to 9 and A to F ]    Radix-16, e.g.-(F3)16 NUMBER SYSTEMS:
Conversion to decimal- A number expressed in base r can be converted to its decimal equivalent by multiplying each coefficient by corresponding power of r and adding . The following is an example of octal to decimal conversion: Conversion
Conversion from decimal to ‘r’ :      Conversion of decimal integer into a base r is done by successive divisions by r and accumulation of the remainders . The conversion of fraction is done by successive multiplication by r and accumulation of integer so obtained.
Conversion from and to binary , octal , hexadecimal- Since 23=8 and 24=16, each octal digits corresponds to three and each hexadecimal corresponds to 4 binary digits . The conversion from binary to octal and hexadecimal is done by partitioning the binary no. into groups of three and four bits respectively .
(r-1)’s  -     - 9’s complement :  It follows that the 9’s complement of a decimal no. is obtained by subtracting each digit from 9.       e.g.-  9’s complement of  546700  is   999999-546700=453299      -1’s complement: The 1’s complement of a binary no. is obtained by subtracting each digit by 1.     e.g.- 1’s complement of 1011001 is 0100110. Complements
( r’s ) – -10’s complement :      10’s complement of a decimal number is obtained by adding 1 to the 9’s complement value.      e.g.- 10’s complement of 2389 is 7610+1=7611. -2’s complement :  2’s complement of binary number is obtained by adding 1 to the 1’s complement. e.g. – 2’s complement of 101100 is 010011+1=010100.
Subtraction of unsigned numbers
Signed Numbers
An overflow condition can be detected by observing the carry into the sign bit position and carry out of the sign bit position . If these two carries are not equal an overflow is occurred . carries:  0  1                              carries: 1  0            +70      0  1000110                  -70       1  0111010 +800  1010000-801  0110000   +150      1  0010110                 -150      0  1101010 Overflow
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Data r epresentation

  • 2. Binary number system- [ 0and 1 ] Radix-2 , e.g.-(101101)2 Decimal number system- [ 0 to 9 ] Radix-10 , e.g.-(243)10 Octal number system- [ 0 to 7 ] Radix-8 , e.g.-(736.4)8 Hexadecimal - [ 0 to 9 and A to F ] Radix-16, e.g.-(F3)16 NUMBER SYSTEMS:
  • 3. Conversion to decimal- A number expressed in base r can be converted to its decimal equivalent by multiplying each coefficient by corresponding power of r and adding . The following is an example of octal to decimal conversion: Conversion
  • 4. Conversion from decimal to ‘r’ : Conversion of decimal integer into a base r is done by successive divisions by r and accumulation of the remainders . The conversion of fraction is done by successive multiplication by r and accumulation of integer so obtained.
  • 5. Conversion from and to binary , octal , hexadecimal- Since 23=8 and 24=16, each octal digits corresponds to three and each hexadecimal corresponds to 4 binary digits . The conversion from binary to octal and hexadecimal is done by partitioning the binary no. into groups of three and four bits respectively .
  • 6. (r-1)’s - - 9’s complement : It follows that the 9’s complement of a decimal no. is obtained by subtracting each digit from 9. e.g.- 9’s complement of 546700 is 999999-546700=453299 -1’s complement: The 1’s complement of a binary no. is obtained by subtracting each digit by 1. e.g.- 1’s complement of 1011001 is 0100110. Complements
  • 7. ( r’s ) – -10’s complement : 10’s complement of a decimal number is obtained by adding 1 to the 9’s complement value. e.g.- 10’s complement of 2389 is 7610+1=7611. -2’s complement : 2’s complement of binary number is obtained by adding 1 to the 1’s complement. e.g. – 2’s complement of 101100 is 010011+1=010100.
  • 10.
  • 11. An overflow condition can be detected by observing the carry into the sign bit position and carry out of the sign bit position . If these two carries are not equal an overflow is occurred . carries: 0 1 carries: 1 0 +70 0 1000110 -70 1 0111010 +800 1010000-801 0110000 +150 1 0010110 -150 0 1101010 Overflow