GATE CS 2024 Set 2 — Question 14

In the IEEE 754 single-precision floating-point standard, the value of a normalized number is represented as:$$V = (-1)^S \times 2^{E - 127} \times (1.M)$$where $S$ is the sign bit, $E$ is the biased exponent, and $M$ is the fractional part of the mantissa.Let's evaluate the options:

• Option (A): $S = 0$, $E = (01111111)_2 = 127$, $M = (111...1)_2 = 1 - 2^{-23}$.

Value $\approx 2^{127-127} \times (1 + 1) = 2^0 \times 2 = 2$.

• Option (B): $S = 0$, $E = (11111110)_2 = 254$, $M = (111...1)_2 = 1 - 2^{-23}$.

Value $\approx 2^{254-127} \times (1 + 1) = 2^{127} \times 2 = 2^{128}$. This is the largest finite representable number in single precision.

• Option (C): $S = 0$, $E = (11111111)_2 = 255$, $M \neq 0$.

According to the standard, an exponent of all 1s with a non-zero mantissa represents NaN (Not a Number). If the mantissa were zero, it would represent $+\infty$.

• Option (D): $S = 0$, $E = (01111111)_2 = 127$, $M = 0$.

Value $= 2^{127-127} \times (1.0) = 2^0 \times 1 = 1$.Comparing the finite values, $2^{128}$ (Option B) is significantly larger than $2$ (Option A) or $1$ (Option D). NaN is not a comparable value. Therefore, Option (B) is the correct answer.