Subtraction by 1’s Complement

In subtraction by 1’s complement we subtract two binary numbers using carried by 1’s complement.

The steps to be followed in subtraction by 1’s complement are:

i) To write down 1’s complement of the subtrahend.

ii) To add this with the minuend.

iii) If the result of addition has a carry over then it is dropped and an 1 is added in the last bit.

iv) If there is no carry over, then 1’s complement of the result of addition is obtained to get the final result and it is negative.

Evaluate:

(i) 110101 – 100101

Solution:

1’s complement of 10011 is 011010. Hence

                                     Minued -              1 1 0 1 0 1

      1’s complement of subtrahend -              0 1 1 0 1 0

                                Carry over -     1       0 0 1 1 1 1

                                                                           1

                                                              0 1 0 0 0 0

The required difference is 10000

(ii) 101011 – 111001

Solution:

1’s complement of 111001 is 000110. Hence

                                    Minued -              1 0 1 0 1 1

                           1’s complement -            0 0 0 1 1 0

                                                              1 1 0 0 0 1

Hence the difference is – 1 1 1 0

(iii) 1011.001 – 110.10

Solution:

1’s complement of 0110.100 is 1001.011 Hence

                                     Minued -              1 0 1 1 . 0 0 1

      1’s complement of subtrahend -              1 0 0 1 . 0 1 1

                                Carry over -     1       0 1 0 0 . 1 0 0

                                                                                1

                                                              0 1 0 0 . 1 0 1

Hence the required difference is 100.101

(iv) 10110.01 – 11010.10

Solution:

1’s complement of 11010.10 is 00101.01

                                                  1 0 1 1 0 . 0 1

                                                  0 0 1 0 1 . 0 1

                                                  1 1 0 1 1 . 1 0

Hence the required difference is – 00100.01 i.e. – 100.01

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