GO Classes DPP | GATE CS | Digital Logic | Binary Arithmetic

Answer 1

To find the sum of the two $8$-bit $2$'s complement numbers $A$ and $B$, we can perform standard binary addition and discard any carry that exceeds the $8$th bit (since we are restricted to an 8-bit system).

Here are the numbers:

$A = 10000001$

$B = 11111110$

Let's align them and add them bit by bit:

$$\begin{array}{@{}c@{\quad}cccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ + & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \mathbf{1} & \mathbf{0} & \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} & \mathbf{1} \end{array}$$

The result of the binary addition is $101111111$, which is $9$ bits long. Because this is an 8-bit system, we discard the most significant bit (the carry-out bit).

The remaining $8$ bits give us:

$01111111$

(Note: If you look at their decimal values, $A = -127$ and $B = -2$. Their true sum is $-129$, which falls outside the valid 8-bit 2's complement range of $-128$ to $+127$. This means an overflow occurred, which is why the addition of two negative numbers resulted in a positive binary representation. However, the hardware simply performs the binary addition and truncates the carry, giving the 8-bit result).

The correct option is A. $01111111$.

Answer 2

Option A

Answer 3

Here is the Details Explanation and Don't Forget to read last 4,5 line given by Deepak Sir