Signed Magnitude Representation Range Explained

The question asks for the range of numbers that can be represented using the n-bit signed magnitude format.

Understanding Signed Magnitude Format

In the signed magnitude representation, one bit is dedicated to representing the sign of the number, and the remaining bits represent the magnitude (absolute value) of the number.

• Sign Bit: Typically, the most significant bit (MSB) is used as the sign bit. A '0' usually indicates a positive number, and a '1' indicates a negative number.
• Magnitude Bits: The remaining n-1 bits are used to represent the magnitude of the number.

Calculating the Range

With n bits in total and 1 bit used for the sign:

• There are n-1 bits available for the magnitude.
• These n-1 bits can represent magnitudes from 0 up to $2^{(n-1)} - 1$. For example, if n-1 is 3, the magnitudes can be 000 (0), 001 (1), 010 (2), ..., 111 (7), where $2^3 - 1 = 7$.

Considering the sign bit:

• Positive Numbers: The range starts from positive zero ($+0$) and goes up to the largest possible magnitude using n-1 bits, which is $+(2^{(n-1)} - 1)$.
• Negative Numbers: The range starts from the smallest possible magnitude using n-1 bits, which is $-(2^{(n-1)} - 1)$, and goes down to negative zero ($-0$).

Therefore, the complete range for an n-bit signed magnitude representation spans from $-(2^{(n-1)} - 1)$ to $+(2^{(n-1)} - 1)$. Note that this system has two representations for zero: $+0$ and $-0$.

Analyzing the Options

Let's compare our calculated range with the given options:

1. 0 to $2^n - 1$: This is the range for unsigned integers.
2. $ -(2^{n-1} - 1) $ to $ (2^{n-1} - 1) $: This matches our calculated range for signed magnitude representation.
3. $ -(2^n - 1) $ to $ (2^n - 1) $: This range incorrectly uses all n bits for magnitude, ignoring the sign bit's role.
4. 0 to $2^{n-1} - 1$: This only covers the positive values and zero, omitting the negative values.

Conclusion

Based on the analysis, the correct range for an n-bit signed magnitude representation is $ -(2^{n-1} - 1) $ to $ (2^{n-1} - 1) $. This includes both positive and negative numbers, with zero having two representations.