Binary Addition Using 2's Complement

To perform binary addition of a positive and a negative number using 2's complement, we follow these steps:

1. Convert the decimal numbers to binary.
2. Represent both numbers with the same number of bits, typically using 2's complement representation for the negative number.
3. Add the binary numbers using standard binary addition.
4. Interpret the result based on the sign bit.

Converting Decimal to Binary

First, convert the magnitudes of the decimal numbers (16 and 83) into their binary equivalents.

• \( 16_{10} = 10000_2 \)
• \( 83_{10} = 64 + 16 + 2 + 1 = 2^6 + 2^4 + 2^1 + 2^0 = 1010011_2 \)

2's Complement Representation

We need to represent both 16 and -83 using 2's complement. The magnitude of 83 requires 7 bits (1010011). To work with signed numbers, we typically use a fixed number of bits, often 8 bits for simplicity in examples like this. We will use 8 bits, where the leftmost bit is the sign bit (0 for positive, 1 for negative).

• For 16:

  Positive 16 in 8 bits is \( 00010000_2 \).

• For -83:

  First, represent positive 83 in 8 bits: \( 01010011_2 \).

  To find the 2's complement of -83:

  1. Find the 1's complement by inverting all bits:

     \( 01010011_2 \) becomes \( 10101100_2 \).

  2. Add 1 to the 1's complement:

     \( 10101100_2 + 1_2 = 10101101_2 \).

  So, -83 in 8-bit 2's complement is \( 10101101_2 \).

Performing Binary Addition

Now, add the 8-bit 2's complement representations of 16 and -83:

\[ \begin{array}{@{}c@{\,}c@{}c@{}c@{}c@{}c@{}c@{}c} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \quad (16_{10}) \\ + & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 & \quad (-83_{10}) \\ \hline & 1 & 0 & 1 & 1 & 1 & 1 & 0 & 1 & \quad (\text{Result in 2's complement}) \\ \end{array} \]

The addition results in \( 10111101_2 \). There is no carry-out from the most significant bit, which is expected when the result is negative.

Interpreting the Result

The result \( 10111101_2 \) is in 2's complement form. The most significant bit is 1, indicating that the result is a negative number.

To find the decimal value represented by \( 10111101_2 \), we can take its 2's complement and find the magnitude:

• Original result: \( 10111101_2 \)
• 1's complement: \( 01000010_2 \)
• Add 1: \( 01000010_2 + 1_2 = 01000011_2 \)

The magnitude is \( 01000011_2 \), which in decimal is:

\( 2^6 + 2^1 + 2^0 = 64 + 2 + 1 = 67_{10} \)

Since the sign bit of the 2's complement result \( 10111101_2 \) was 1, the decimal result is -67.

Let's check this with the original decimal calculation: \( 16 + (-83) = 16 - 83 = -67 \). The binary addition result matches the decimal calculation.

Relating Result to Options

The result of the 2's complement addition is a value representing -67. The options provided are in a signed binary format, showing a minus sign followed by the binary magnitude.

The magnitude of 67 in binary is \( 1000011_2 \). Therefore, -67 represented in this format is \( -1000011_2 \).

This matches option 1.