Morris Mano Edition 3 Exercise 1 Question 19 (Page No. 34)

Question

the binary number listed below has a sign in the left most bit position and if the number is negative than in 2’s complement form.

perform the arithmetic operation indicated and verify the answer.

1. 101011 + 111000
2. 001110  + 110010
3. 111001 – 001010
4. 101011 – 100110

Comments

ANS
1- 100011
2- 000000
3- 101111
4-000101

Answer 1

As given, the numbers are in 2’s complement number system.

They asked to perform arithmetic operation and verify the answer –> By verify, I’m assuming the mean us to perform binary arithmetic and then verify whether the answer is same if we’ve first converted the numbers in decimal and then performed decimal arithmetic.

• 101011 + 111000 = 1100011

101011 = –21 , 111000 = –8 and –21 + (–8) = –29

1100011 = –29. 

• ​​​​​​​001110 + 110010 = 1000000

001110 = 14 , 110010 = –14 and 14 + (–14) = 0

1000000 = –64.

Here, if we observe decimal arithmetic results in 0 and binary arithmetic results in –64; also we can se operands are of 6 bits and result is in 7 bits. Now, in result if we ignore the 7^th bit, we get 000000 ie 0.

Similarly, in 1^st example also we can ignore the 7^th bit, we still get 100011 = –29.

TLDR, in 2’s complement addition WE HAVE to discard any extra bit that we get.

• 111001 – 001010 = 101111

Note: 0 – 0 = 0; 1 – 1 = 0; 1 – 0 = 1; 0 – 1 = 1 and 1 Borrow

111001 = –7 , 001010 = 10 and –7 – 10 = –17

101111 = –17

• 101011 – 100110 = 000101

101011 = –21 , 100110 = –26 and –21 – (–26) = 5

000101 = 5