Question

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?

• A. 01110000
• B. 11100100
• C. 01110001
• D. Result is invalid

Hint:

In two's complement arithmetic, ensure that the sum does not exceed the range of the given number of bits (e.g., -128 to +127 for 8-bit numbers).

Answer 1

The Correct Option is D

We are given the 8-bit two's complement numbers \( A = 10011001 \) and \( B = 11010111 \), and we need to compute their sum.
1. Interpretation of two's complement numbers:
- For \( A = 10011001 \), the most significant bit (MSB) is 1, indicating that it is a negative number. To find the magnitude, we take the two's complement of \( A \):
- Invert the bits: \( 01100110 \)
- Add 1: \( 01100110 + 1 = 01100111 \)
- So, \( A = -55 \).
- For \( B = 11010111 \), the MSB is 1, indicating it is also a negative number. To find the magnitude, we take the two's complement of \( B \):
- Invert the bits: \( 00101000 \)
- Add 1: \( 00101000 + 1 = 00101001 \)
- So, \( B = -41 \).
2. Adding the two numbers:
- \( A + B = -55 + (-41) = -96 \)
3. Converting -96 to 8-bit two's complement:
- First, find the binary representation of 96:
\[ 96 = 01100000 \]
- To convert to two's complement (for negative numbers), invert the bits and add 1:
- Invert the bits: \( 10011111 \)
- Add 1: \( 10011111 + 1 = 10100000 \)
Thus, the result is \( 10100000 \), which is an invalid 8-bit result because it exceeds the range of representable values in an 8-bit two's complement system, which is from -128 to +127.
Thus, the correct answer is \( \boxed{\text{Result is invalid}} \).