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

Alternative approach - convert to decimal first. Step 1: X = 35C00000 has sign 0, exponent field 107 (unbiased exponent -20), significand \(1.5\). So \(X = 1.5 \times 2^{-20}\). As a plain fraction, this equals \(1.5 / 1048576 \approx 1.430511 \times 10^{-6}\). Step 2: Y = 34A00000 has exponent field 105 (unbiased exponent -22), significand \(1.25\). So \(Y = 1.25 \times 2^{-22} = 1.25/4194304 \approx 2.98023 \times 10^{-7}\). Step 3: Adding the decimals, \(Z \approx 1.430511\times10^{-6} + 0.298023\times10^{-6} = 1.728535\times10^{-6}\). Step 4: Express Z as a power of two: \(1.728535\times10^{-6} \times 2^{20} = 1.8125\), confirming \(Z = 1.8125 \times 2^{-20}\), matching the binary alignment method exactly, and showing no overflow occurs since the significand stays below 2. Step 5: The fraction \(0.8125\) splits into powers of two: \(0.5+0.25+0.0625\), i.e. bits at positions \(2^{-1}, 2^{-2}, 2^{-4}\), giving mantissa bits 1101 followed by zeros. Step 6: Reassembling sign 0, exponent 107 (same as X, written 01101011) and mantissa 11010000000000000000000 gives the hexadecimal word 35E80000. Final answer: Z = 35E80000, which is option C.