Question

The format of a single-precision floating-point number as per the IEEE 754 standard is:} \begin{center} \begin{tabular}{|c|c|c|} \hline Sign (1 bit) & Exponent (8 bits) & Mantissa (23 bits)
\hline \end{tabular} \end{center} Choose the largest floating-point number among the following options:

• A. Sign:} \(0\), Exponent:} \(0111\,1111\), Mantissa:} \(1111\,1111\,1111\,1111\,1111\,111\)
• B. Sign:} \(0\), Exponent:} \(1111\,1110\), Mantissa:} \(1111\,1111\,1111\,1111\,1111\,111\)
• C. Sign:} \(0\), Exponent:} \(1111\,1111\), Mantissa:} \(1111\,1111\,1111\,1111\,1111\,111\)
• D. Sign:} \(0\), Exponent:} \(0111\,1111\), Mantissa:} \(0000\,0000\,0000\,0000\,0000\,000\)

Hint:

For IEEE 754 floating-point numbers, larger exponents and mantissas closer to \(1\) result in larger values. Be cautious about special cases like NaN} and Infinity}.

Answer 1

The Correct Option is B

To determine the largest floating-point number, the following points are considered based on the IEEE 754 standard:
The value of a floating-point number is given by: \[ (-1)^{\text{Sign}} \times (1.\text{Mantissa}) \times 2^{\text{Exponent} - 127}. \]
The larger the exponent and the closer the mantissa is to \(1\), the larger the number.
In the given options:
Option (A): Exponent = \(127\), Mantissa = All \(1\)'s.
Option (B): Exponent = \(254\), Mantissa = All \(1\)'s.
Option (C): Exponent = \(255\), Mantissa = All \(1\)'s (this represents NaN}, not a valid number).
Option (D): Exponent = \(127\), Mantissa = All \(0\)'s.
Analysis:
Option (B) has the largest exponent (\(254\)) among the valid numbers, making it the largest floating-point number.
Option (C) represents NaN} (Not a Number) due to an exponent of \(255\) and a non-zero mantissa.
Option (D) has the smallest value due to its mantissa being all \(0\)'s. Final Answer: \[ \boxed{\text{(B)}} \]