In binary fixed-point arithmetic, compute the 2’s complement of the binary number 010111.1100 (invert all bits and add 1 across the fixed-point boundary). Which option matches the correct 2’s complement representation?

Introduction / Context:
Two’s complement is the most common representation for signed binary numbers. For fixed-point values (with a binary point), the method is the same as for integers: invert every bit and then add 1 to the least significant position, carrying across the fractional boundary if necessary. This question reinforces precise bitwise manipulation for fractional binaries.

Given Data / Assumptions:

• Original number: 010111.1100 (six integer bits and four fractional bits).
• We apply two’s complement over the entire fixed-width pattern, including the fractional part.
• Binary point location remains fixed.

Concept / Approach:
Compute two’s complement in two steps: (1) one’s complement (bitwise invert) and (2) add 1 at the least significant fractional bit. Care must be taken to preserve the bit positions on both sides of the binary point. The procedure mirrors integer two’s complement but with the fixed-point layout retained.

Step-by-Step Solution:
Start with 010111.1100.Invert all bits (one’s complement): 101000.0011.Add 1 to the least significant bit position: 101000.0011 + 0.0001 = 101000.0100.Keep the binary point fixed; no overflow occurs beyond the fixed width.Match with options: 101000.0100.

Verification / Alternative check:
Reverse the process: two’s complement again of 101000.0100 should return the original. Invert to 010111.1011 and add 1 → 010111.1100, confirming correctness.

Why Other Options Are Wrong:
101001.1100: incorrect bit pattern; not the add-1 result after inversion.010111.0011: this is not a two’s complement; looks like partial inversion.101000.0011: only one’s complement, missing the +1 step.None of the above: incorrect since a correct option exists.

Common Pitfalls:
Forgetting to add 1 after inversion; shifting the binary point; performing addition only to integer bits and ignoring fractional carries.

Final Answer:
101000.0100