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:
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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