Subtract 36 trom 83 usin
Using l’s complement method, solve 54(10)-87(10)
Answers
Explanation:
) 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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