Official PYQ · 2003Q43Number Representations (Fixed and Floating Point)MCQ1 mark

GATE CS 2003 Q43 — Number Representations (Fixed and Floating Point)

The following is a scheme for floating point number representation using $16$ bits.

Let $s, e,$ and $m$ be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:

$\begin{cases}(-1)^s \left(1+m \times 2^{-9}\right) 2^{e-31}, & \text{ if the exponent } \neq 111111 \\ 0, & \text{ otherwise} \end{cases}$

What is the maximum difference between two successive real numbers representable in this system?

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Answer: C

For a fixed exponent $e$ , the value of the floating-point number is given by:

V = (-1)^s \left(1 + m \times 2^{-9}\right) 2^{e-31}

Two successive representable real numbers with the same exponent $e$ differ by a single unit in the least significant bit of the mantissa $m$ (i.e., $\Delta m = 1$ ).

The difference $\Delta$ between these two successive numbers is:

\Delta = \left(1 + (m+1) \times 2^{-9}\right) 2^{e-31} - \left(1 + m \times 2^{-9}\right) 2^{e-31}

\Delta = 2^{-9} \times 2^{e-31} = 2^{e-40}

To maximize this difference, we must choose the maximum possible value for the exponent $e$ . Since $e$ is a $6$ -bit field, its values range from $0$ to $2^6 - 1 = 63$ .

According to the given condition, the exponent cannot be $111111_2 = 63$ . Thus, the maximum valid exponent is $e = 62$ .

Substituting $e = 62$ into the difference formula:

\Delta_{\text{max}} = 2^{62 - 40} = 2^{22}

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Published 10 Oct 2026

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