Question

Two numbers, a and b, are stored in one byte floating point notation using the least significant (rightmost) 3 bits for the exponent. both parts of each number are in 2's complement. if the a and b are 00110111 and 11011000, then a+b in the same notation as the operands is (select all correct answers)

Answer 1

$a=00110111 = 00110_2 \times 2^{111_2} = 6 \times 2^{-1} = 3$

$b=11011000 = 11011_2 \times 2^{000_2} = -(0100_2 + 1) =-(4+1)=-5$

$a+b=-2 = -1 \times 2^{001} = -2 \times 2^{0} = -4 \times 2^{-1}$ etc.

$a+b=-2 = 11111001 = 11110000 = 11100111 = ...$

Hard to select the correct answers without seeing the choices. Let's check a couple,

$11111001 = -(00000+1) \times 2^{1} = -2 \quad\checkmark$

$11110000 = -(00001 + 1) \times 2^{0} = -2 \quad\checkmark$

$11100111 = -(00011 + 1) \times 2^{-1} = -4/2= -2 \quad\checkmark$