Convert the following pairs of decimal numbers to 5-bit 2's-complement numbers, then add them. State whether or not overflow occurs in each case. (a) 4 and 11 (b) 6 and 14 (c) - 13 and 12 (d) - 4 and 8 (e) - 2 and - 9 (f) - 9 and - 14

Step-by-step explanation:

(a) 4 and 11

binary equivalent of 4 in 5 bit = 00100

binary equivalent of 11 in 5 bit = 01011

decimal number 4 in 2's complement form = 11100

decimal number 11 in 2's complement form = 10101

now,

1 1 1 0 0

+ 1 01 0 1

1 1 000 1

Since, we are doing addition on 5 bit numbers but the result of addition came in 6 digit, so there will be overflow.

(b) 6 and 14

binary equivalent of 6 in 5 bit = 00110

binary equivalent of 14 in 5 bit = 01110

decimal number 6 in 2's complement form = 11010

decimal number 14 in 2's complement form = 10010

now,

1 1 0 1 0

+ 1 00 1 0

1 0 1 1 0 0

Since, we are doing addition on 5 bit numbers but the result of addition came in 6 digit, so there will be overflow.

(c) - 13 and 12

binary equivalent of - 13 in 5 bit = 10011

binary equivalent of 12 in 5 bit = 01100

decimal number - 13 in 2's complement form = 01101

decimal number 12 in 2's complement form = 10100

now,

0 1 1 0 1

+ 1 0 1 0 0

1 0 0 0 0 1

Since, we are doing addition on 5 bit numbers but the result of addition came in 6 digit, so there will be overflow.

(d) - 4 and 8

binary equivalent of - 4 in 5 bit = 11100

binary equivalent of 8 in 5 bit = 01000

decimal number - 4 in 2's complement form = 00100

decimal number 8 in 2's complement form = 11000

now,

0 0 1 0 0

+ 1 1 0 0 0

1 1 1 0 0

Since, we are doing addition on 5 bit numbers and the result of addition also came in 5 digit, so there will not be overflow.

(e) - 2 and - 9

binary equivalent of - 2 in 5 bit = 11110

binary equivalent of - 9 in 5 bit = 10111

decimal number - 2 in 2's complement form = 00010

decimal number - 9 in 2's complement form = 01001

now,

0 0 0 1 0

+ 0 1 0 0 1

0 1 0 1 1

Since, we are doing addition on 5 bit numbers and the result of addition also came in 5 digit, so there will not be overflow.

(f) - 9 and - 14

binary equivalent of - 9 in 5 bit = 10111

binary equivalent of - 14 in 5 bit = 10010

decimal number - 9 in 2's complement form = 01001

decimal number - 10 in 2's complement form = 01110

now,

0 1 0 0 1

+ 0 1 1 1 1

1 1 000

Since, we are doing addition on 5 bit numbers and the result of addition also came in 5 digit, so there will not be overflow.