Question

4) Show how each of the following floating-point values would be stored using IEEE-754 single precision (be sure to

indicate the sign bit, the exponent, and the significand fields):

a) 8.25

Answer 1

8.25 in binary => 1000.01
so, 8.25 in normal binary is 1000.01 => 1.00001 * 2^3
single precision:
--------------------
sign bit is 0(+ve)
exp bits are (127+3=130) => 10000010
frac bits are 00001000000000000000000
so, 8.25 in single-precision format is 0 10000010 00001000000000000000000
in hexadecimal it is 0x41040000
Answer: 0x41040000