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:

\(X: 35C00000 \qquad Y: 34A00000\)

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:

Decode X and Y into sign, exponent, and mantissa form, shift the smaller-exponent number (Y) so both share X's exponent of \(2^{-20}\), add the mantissas as \(1.5 + 0.3125 = 1.8125\), and re-encode; no renormalization is needed since 1.8125 is already less than 2.

Answer 1

The Correct Option is C

Work this out purely in decimal to cross-check the bit-level method.

• Break $X = 35C00000_{16}$ into its bit groups: sign 0, exponent bits $01101011_2 = 107$, fraction bits starting $1\,0000\ldots0$. That gives exponent $107-127=-20$ and fraction $0.5$, so $X = 1.5 \times 2^{-20}$. As a plain decimal number this is $1.5 / 1048576 \approx 1.430511474609375 \times 10^{-6}$.
• Break $Y = 34A00000_{16}$ the same way: sign 0, exponent bits $01101001_2 = 105$, fraction bits starting $01\,000\ldots0$. That gives exponent $105-127=-22$ and fraction $0.25$, so $Y = 1.25 \times 2^{-22}$, which as a decimal number is $1.25/4194304 \approx 2.9802322 \times 10^{-7}$.
• Adding the plain decimal values: $1.430511474609375 \times 10^{-6} + 2.9802322387695312 \times 10^{-7} \approx 1.7285346984863281 \times 10^{-6}$.
• Now test this against each option by converting every option back to a decimal value and comparing. Option (C), $35E80000_{16}$, decodes to exponent field $107$ (same $2^{-20}$ scale) and fraction bits $1101\,0\ldots0$, giving fraction $0.8125$ and mantissa $1.8125$; its decimal value is $1.8125/1048576 \approx 1.7285346984863281 \times 10^{-6}$, which matches the sum exactly.
• Checking the others confirms they are off: option (A) gives about $1.4901 \times 10^{-6}$, option (B) gives about $1.5199 \times 10^{-6}$, and option (D) gives about $1.7583 \times 10^{-6}$, none of which equal the true sum.

So $Z$ equals option (C), $35E80000$.

$$\boxed{Z = 35E80000_{16}\ \text{(option C)}}$$