Two’s Complement Range — For an 8-bit two’s-complement integer system, the set of negative numbers spans −128 through −1 (order notwithstanding). Evaluate the correctness of the given statement.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction:
The representable range in two’s-complement depends on word width. For 8 bits, the classic range is −128 to +127. This question focuses on the negative side of the range and whether stating “−1 to −128” (reverse ordering) is still a correct characterization of the set of negative values.

Given Data / Assumptions:

  • Word size = 8 bits.
  • Encoding = two’s-complement.
  • We care about the set of negative values, not the ascending order convention.


Concept / Approach:
In two’s-complement, the most negative number is −2^(n−1) and the most positive is +2^(n−1) − 1. For n = 8, negatives are {−128, −127, …, −1}. Writing the set as “−1 to −128” reverses the usual order but references the same elements. Thus, as a truth-value statement about which numbers are negative in 8-bit two’s-complement, it is correct.

Step-by-Step Solution:

Step 1: Compute limits: −2^(8−1) = −128 and +2^(8−1) − 1 = +127.Step 2: Identify the negative subset: {−128 … −1}.Step 3: Conclude that the set description “−1 to −128” names the same members, albeit in descending order.


Verification / Alternative check:

Check specific encodings: 10000000₂ = −128 and 11111111₂ = −1; all intermediate patterns map to intervening negatives.


Why Other Options Are Wrong:

Incorrect: The statement’s content is true; only the ordering is reversed.True only for signed magnitude: That representation has a different range and a negative zero.Depends on endianness: Endianness affects byte ordering in memory, not numeric range.


Common Pitfalls:

Confusing order notation with correctness of the set.Mixing two’s-complement with one’s complement or signed-magnitude ranges.


Final Answer:

Correct

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