Two’s-complement subtraction – does a sign bit of 1 in the difference mean the magnitude is negative and already in true (uncomplemented) binary?

Introduction / Context: In two’s-complement arithmetic, the MSB (sign bit) indicates the sign of a signed result. However, interpretation of the magnitude differs for negative numbers: their bit pattern is not in ordinary (true) binary; it is in two’s-complement form and must be converted to find the positive magnitude.

Given Data / Assumptions:

• Two’s-complement representation.
• Subtraction implemented by adding the two’s complement of the subtrahend.
• We examine the sign and magnitude of the final difference.

Concept / Approach: If the sign bit is 1, the value is negative. The magnitude is not directly in true binary; to get the absolute value, take the two’s complement again (invert bits and add 1). Therefore, saying “negative and in true binary form” is inaccurate; it is negative, but the magnitude is encoded in two’s complement, not plain binary.

Step-by-Step Solution:

Verification / Alternative check: Example (8-bit): 3 – 5 = 3 + (two’s complement of 5 = 11111011) = 11111110. MSB=1 → negative. Magnitude: two’s complement of 11111110 is 00000010 = 2. The stored pattern is not true binary 2 until reconverted.

Why Other Options Are Wrong: End-around carry, saturation, or word-length parity do not change the core rule about two’s-complement magnitude extraction.

Common Pitfalls: Confusing sign indication with magnitude encoding; assuming negative numbers are stored as sign + true magnitude, which is sign-magnitude, not two’s complement.

Final Answer: Incorrect