Question

Consider the real valued variables X, Y and Z represented using the IEEE 754 single-
precision floating-point format. The binary representations of X and Y in hexadecimal
notation are as follows:
X: 35C00000 Y: 34A00000
Let 𝑍 = 𝑋+ 𝑌.
Which one of the following is the binary representation of 𝑍, in hexadecimal
notation?

• A. 35C80000
• B. 35CC0000
• C. 35E80000
• D. 35EC0000

Hint:

Align the smaller exponent (Y) to match the larger exponent (X) by shifting its significand right, then add the significands and check if renormalization is needed before reassembling the IEEE-754 word.

Answer 1

The Correct Option is C

Step 1: Decode X in IEEE-754 single precision. X = 35C00000 in hex expands to the bit pattern 0 01101011 10000000000000000000000. Sign = 0 (positive). Exponent field = 01101011 = 107 in decimal, so the unbiased exponent is \(107-127=-20\). The mantissa field is 1000...0, so the significand is \(1.1_2 = 1.5\). Hence \(X = 1.5 \times 2^{-20}\).

Step 2: Decode Y. Y = 34A00000 expands to 0 01101001 01000000000000000000000. Exponent field = 01101001 = 105, unbiased exponent = \(105-127=-22\). Mantissa field is 01000...0, so significand = \(1.01_2 = 1.25\). Hence \(Y = 1.25 \times 2^{-22}\).

Step 3: Align exponents. Since X has the larger exponent (-20), rewrite Y with exponent -20: \(Y = 1.25 \times 2^{-22} = 0.3125 \times 2^{-20}\) (dividing the significand by \(2^{2}\)).

Step 4: Add the significands. \(1.5 + 0.3125 = 1.8125\). So \(Z = 1.8125 \times 2^{-20}\). This is already normalized since the significand 1.8125 lies between 1 and 2, so the exponent field stays 107 = 01101011.

Step 5: Convert the fractional part to binary. \(0.8125 = 0.5+0.25+0.0625 = (0.1101)_2\). So the significand is 1.1101 and the 23-bit mantissa field is 11010000000000000000000.

Step 6: Assemble Z. Sign = 0, exponent = 01101011, mantissa = 11010000000000000000000, giving 0 01101011 11010000000000000000000, which regrouped into hex nibbles is 35E80000.

Final Answer: \(\boxed{Z = 35E80000 \text{ (Option C)}}\)