Page 43 - Cs_withBlue_J_C11_Flipbook
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c. 1’s complement of 010111.10 is 101000.01 +1
2’s complement of 101000.10 is 101000.10 1 0 0 1 1 0 . 0 1
Adding 100110.01 and 101000.10, we get + 1 0 1 0 0 0 . 1 0
Discarding MSB 1 and ignoring leading 0’s
(1110.11) 1 0 0 1 1 1 0 . 1 1
2
15. Perform binary subtraction (1000.01) - (110111.1) using
2
2
a. 1’s complement method
b. 2’s complement method +1 +1
Ans. a. 1’s complement of 110111.10 is 001000.01 1 0 0 0 . 0 1
Adding 1000.01 and 001000.01, we get + 0 0 1 0 0 0 . 0 1
(-101111.01) 0 1 0 0 0 0 . 1 0
2
1’s complement of 010000.10 is 101111.01
b. 1’s complement of 110111.10 is 001000.01 +1 +1
2’s complement of 001000.01 is 001000.10 1 0 0 0 . 0 1
Adding 1000.01 and 001000.10, we get + 0 0 1 0 0 0 . 1 0
(-101111.01) 2 0 1 0 0 0 0 . 1 1
1’s complement of 010000.11 is 101111.00
2’s complement of 101111.00 is 101111.01
16. Perform the octal addition (1567.2) + (433.4) 8
8
Ans. +1 +1 +1
1 5 6 7 . 2
+ 4 3 3 . 4
2 2 2 2 . 6
(2222.6) 8
17. Perform octal subtraction (3214.6) - (777.7) using
8
8
a. Borrow method
b. 7’s complement method
c. 8’s complement method
Ans. a.
+8 +8 +8 +8
3 2 2 1 1 0 4 3 . 6
7 7 7 . 7
2 2 1 4 . 7
(2214.7) 8
b. 7’s complement of 0777.7 is 7000.0 +1
Adding 3214.6 and 7000.0, we get 3 2 1 4 . 6
(2214.7) + 7 0 0 0 . 0
8
1 2 2 1 4 . 6
1
2 2 1 4 . 7
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