GATE CS 2004 Q66 — Number Representations (Fixed and Floating Point)

To find the product of two $8$-bit $2$'s complement numbers $A$ and $B$, we convert them to decimal, multiply, and convert the result back to $8$-bit $2$'s complement:

1. Convert $A = 1111 1010$ to decimal: Since the MSB is $1$, $A$ is negative. Its magnitude is the $2$'s complement of $1111 1010$, which is $0000 0101 + 1 = 0000 0110_2 = 6_{10}$. Thus, $A = -6$.
2. Convert $B = 0000 1010$ to decimal: Since the MSB is $0$, $B$ is positive. Thus, $B = 2^3 + 2^1 = 10_{10}$.
3. Calculate the product: $\text{Product} = A \times B = -6 \times 10 = -60_{10}$.
4. Convert $-60$ back to $8$-bit $2$'s complement: The binary representation of $+60$ is $0011 1100_2$. Taking its $2$'s complement gives $1100 0011 + 1 = 1100 0100_2$.

Therefore, the product in $2$'s complement is $1100 0100$.