In an 8-bit two's-complement system, what is the largest-magnitude negative number (i.e., the minimum representable value) that can be stored?
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A- 256
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B- 128
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C- 255
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D- 127
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E0
Answer
Correct Answer: - 128
Explanation
Introduction / Context:Two's-complement encoding is the standard for signed integers in digital systems. Understanding its numeric range avoids overflow and sign errors in embedded and computer systems.
Given Data / Assumptions:
- Word size: 8 bits.
- Signed representation: two's complement.
Concept / Approach:
For an n-bit two's-complement integer, the value range is from -2^(n-1) to +2^(n-1) - 1. For n = 8, the range is -128 to +127.
Step-by-Step Solution:
Compute negative limit: -2^(8 - 1) = -2^7 = -128.Compute positive limit: +2^7 - 1 = +127.The “largest negative number” (most negative, minimum) is -128.Verification / Alternative check:
Binary 1000 0000 in two's complement equals -128. No other 8-bit pattern is more negative.
Why Other Options Are Wrong:
- -256 and -255: not representable in 8 bits two's complement.
- -127: not the most negative; it is closer to zero than -128.
- 0: not negative.
Common Pitfalls:
Interpreting “greatest negative” as closest to zero (e.g., -1); in range questions, it means the minimum value (most negative), which is -128 for 8-bit two's complement.
Final Answer:
- 128