In 8-bit two's complement arithmetic, compute the result of the following addition: A = 10011001, B = 11010111. What is the resulting 8-bit binary value?

Let’s first convert both binary numbers into decimal (two’s complement form):

• $A = 10011001_2$ → negative because MSB = 1 → invert bits: $01100110_2 = 102_{10}$ → add 1 → $103_{10}$ So, $A = -103$.
• $B = 11010111_2$ → MSB = 1 (negative) → invert bits: $00101000_2 = 40_{10}$ → add 1 → $41_{10}$ So, $B = -41$.

Add them:

$A + B = -103 + (-41) = -144$

Now, in 8-bit two’s complement, the range is $-128$ to $+127$. Since $-144$ is out of range, overflow occurs.

But let’s compute the 8-bit result (ignoring overflow):

$10011001 + 11010111 =$ 10011001
+ 11010111
= 101110000 (9 bits)

Drop the carry beyond 8 bits → 01110000.

Therefore, the resulting 8-bit binary value is:

✅ Result = 01110000₂

(Overflow occurred, actual signed result would have been −144, but the 8-bit stored value is +112.)