Compute –11 + (–2) using 8-bit two’s complement. Which 8-bit two’s-complement result is obtained?

Difficulty: Medium

Correct Answer: 1111 0011

Explanation:


Introduction / Context:
Binary arithmetic with signed numbers relies on two’s complement encoding. This question checks your ability to encode negatives and perform binary addition correctly within a fixed word size (8 bits).


Given Data / Assumptions:

  • Operands: –11 and –2.
  • Word length: 8 bits.
  • Two’s complement arithmetic; discard carry out of MSB.


Concept / Approach:
To represent –N in two’s complement: write N in binary, invert all bits, then add 1. Add using binary rules; the carry beyond bit 7 is dropped. Interpret the final 8 bits as two’s complement.


Step-by-Step Solution:

+11 = 0000 1011 → invert 1111 0100 → add 1 → 1111 0101 (this is –11).+2 = 0000 0010 → invert 1111 1101 → add 1 → 1111 1110 (this is –2).Add: 1111 0101 + 1111 1110 = 1 1111 0011 (9-bit). Discard carry → 1111 0011.Interpret: MSB = 1 → negative. Magnitude check: invert 0000 1100; add 1 → 0000 1101 = 13 → result = –13, which matches –11 + –2 = –13.


Verification / Alternative check:
Decimal check: –11 + –2 = –13; result encoding 1111 0011 corresponds to –13 in 8-bit two’s complement. Range –128…+127 supports this result without overflow.


Why Other Options Are Wrong:

  • 1110 1101 / 1110 1001 / 1111 1001: These bit patterns decode to other negative values (not –13) and do not match the arithmetic.


Common Pitfalls:
Not discarding the end carry; forgetting the invert+1 rule; mixing sign-magnitude with two’s complement decoding.


Final Answer:
1111 0011

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