written 6.5 years ago by | modified 5.5 years ago by |
i) (7)10−(15)10
ii) (50)10−(2A)16
Comment on results of (i) and (ii)
written 6.5 years ago by | modified 5.5 years ago by |
i) (7)10−(15)10
ii) (50)10−(2A)16
Comment on results of (i) and (ii)
written 6.4 years ago by | • modified 6.4 years ago |
1) (7)10−(15)10
(7)10=(0111)2
(15)10=(1111)2
1's Complement of (15)10 = (0000)2
2's Complement of (15)10 = (0000)2+(0001)2=(0001)2
Now,
(7)10−(15)10=(0111)2+(0001)2=(1000)2
Since, no carry over is generated, the result of subtraction is Negative.
The result of subtraction is obtained by taking the 2's Complement of (1000)2
The 2's Complement of (1000)2 is -
1's Complement of (1000)2+(0001)2 ----> (0111)2+(0001)2=(1000)2
Therefore, (7)10−(15)10=−(8)10
2) (50)10−(2A)16
(50)10=(110010)2
(2A)16=(101010)2
1's Complement of (2A)16 = (010101)2
2's Complement of (2A)16 = (010101)2+(000001)2=(010110)2
Now,
(50)10−(2A)16=(110010)2+(010110)2=1(001000)2
Since, carry over is generated, the result of subtraction (8)10 is Positive.
Therefore, (50)10−(2A)16=(8)10