Unsigned Binary Division


Unsigned Binary Division is more complex than multiplication to implement however, like Unsigned Binary Multiplication it can be done using two approaches:

  • Paper Approach
  • Hardware Approach


    • Paper approach

    The paper method consists of a dividend, divisor and partial remainders. It is done by first setting quotient to 0 then you align leftmost digits in dividend and divisor. If that portion of the dividend above the divisor is greater than or equal to the divisor then subtract divisor from that portion of the dividend and concatentate 1 to the right hand end of the quotient.Else concatentate 0 to the right hand end of the quotient and shift the divisor one place right. These steps are done Until dividend is less than the divisor quotient is correct, dividend is remainder.

    Example 1

    Divisor = 1011(11 decimal) and Dividend = 10010011 (147 decimal)

             00001101    Quotient
    1011 10010011
                1011
              001110    Partial remainder
                  1011
                  001111   Partial remainder
                       1011
                          100    Remainder

    Answer(decimal) = 13 remainder 4

    Example 2

    Divisor = 100(4 decimal) and Dividend = 11111 (31 decimal)

             0111    Quotient
    100 11111
            100
            0111    Partial remainder
               100
             00111   Partial remainder
                  100
              00011    Remainder

    Answer(decimal) = 7 remainder 3


    Hardware Approach

    For the hardware approach we le Q (n-bit) be the dividend and M (n-bit) be the divisor. We assume that Q < M thus we find the quotient and remainder of Q.2n divided by M.

    Operation

    Instead of shifting to the right as in Unsigned Binary Multiplication we shift the quotient to the left and the subtraction operation is used instead of adding.

    Algorithm:
  • A (n + 1-bit) = 0;
    • Counter = n; Repeat the following steps until Count = 0:
  • Shift AQ together one bit to the left.
  • A = A - M
  • If An (MSB of A) = 1, then Q0 (LSB of Q) = 0 and A = A +M
  • Count = Count - 1
  • Stop Quotient is in Q and remainder is in A.
    •

    Example 1

    M = 11 and Q = 1000

             10   
    11 1000
           11
             10   
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