GO Classes DPP | GATE CS | Digital Logic | Signed Number Representation

Answer 1

In sign-magnitude representation, the most significant bit (MSB) acts as the sign bit ($0$ for positive, $1$ for negative), and the remaining $n-1$ bits represent the magnitude.

Because zero has a magnitude of zero, the sign bit can be either positive or negative, resulting in two valid representations:

• Positive Zero ($+0$): $000\dots0$

• Negative Zero ($-0$): $100\dots0$

 

In $1$'s complement, positive numbers are represented in standard binary, and negative numbers are formed by flipping every bit (changing $0$s to $1$s and vice versa) of the corresponding positive number.

• Positive Zero ($+0$): $000\dots0$

• Inverting all the bits to get the negative equivalent gives us Negative Zero ($-0$): $111\dots1$

 

In $2$'s complement, negative numbers are formed by taking the $1$'s complement and adding $1$.

• Positive Zero ($+0$): $000\dots0$

• If we take the $1$'s complement of zero ($111\dots1$) and add $1$, it overflows the $n$-bit register. The carry bit is discarded, leaving us right back at $000\dots0$. Therefore, $2$'s complement has only one unique representation for zero.

 

In an excess-$K$ (or biased) representation, a fixed value $K$ (in this case, $K = 2^{n-1}$) is added to the actual number to get the stored integer.

• To represent $0$, the system stores the binary equivalent of $0 + 2^{n-1}$.

• Because every number maps to exactly one unique binary sequence through this simple addition, there is only one representation for zero.