GATE CS 2025 Set 1 Q15 — Number Representations (Fixed and Floating Point)

To represent a signed integer with more bits in $2$'s complement representation, we use sign extension. This involves replicating the most significant bit (MSB), which represents the sign, into all the newly added higher-order bit positions.

Given:

• The $4$-bit $2$'s complement representation of $-6$ is $1010$.
• The MSB (sign bit) is $1$ (since the number is negative).

1. Representation in $8$-bits:

To extend from $4$ bits to $8$ bits, we replicate the sign bit ($1$) four times to the left:

$$\underline{1 \: 1 \: 1 \: 1} \: 1 \: 0 \: 1 \: 0_2 = 1111 \: 1010_2$$

This matches Option B.

2. Representation in $16$-bits:

To extend from $4$ bits to $16$ bits, we replicate the sign bit ($1$) twelve times to the left:

$$\underline{1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1 \: 1} \: 1 \: 0 \: 1 \: 0_2 = 1111 \: 1111 \: 1111 \: 1010_2$$

This matches Option D.

Why Options A and C are incorrect:

• Option A ($1000 \: 1010$): In $8$-bit $2$'s complement, this represents $-128 + 8 + 2 = -118 \neq -6$.
• Option C ($1000 \: 0000 \: 0000 \: 1010$): In $16$-bit $2$'s complement, this represents $-32768 + 8 + 2 = -32758 \neq -6$.

Thus, the correct options are B and D.