GATE CS 2014 Set 2 Q45 — Number Representations (Fixed and Floating Point)

To represent the decimal value $-14.25$ in IEEE-754 single-precision floating-point format, we follow these steps:

1. Sign Bit ($S$): Since the number is negative, the sign bit is $S = 1$.
2. Binary Conversion: Convert the magnitude $14.25$ to binary: $$14_{10} = 1110_2$$ $$0.25_{10} = 0.01_2$$ Thus, $14.25_{10} = 1110.01_2$.
3. Normalization: Normalize the binary representation to the form $1.M \times 2^E$: $$1110.01_2 = 1.11001_2 \times 2^3$$ Here, the actual exponent is $3$, and the mantissa bits ($M$) are $11001$ (padded with zeros to 23 bits).
4. Biased Exponent ($E$): The bias for single-precision is $127$. $$\text{Biased Exponent} = 3 + 127 = 130_{10} = 10000010_2$$

Combining the fields ($S$, $E$, and $M$):

Grouping the 32 bits into groups of 4 to convert to hexadecimal:

Thus, the hexadecimal representation is $\text{C1640000H}$.