1. Identify the elements that make up the binary representation of the number:
The first bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.
0
The next 8 bits contain the exponent:
0000 0011
The last 23 bits contain the mantissa:
110 0000 0000 0000 0000 0100
1. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
0000 0011(2) =
0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 1 × 20 =
0 + 0 + 0 + 0 + 0 + 0 + 2 + 1 =
2 + 1 =
3(10)
2. Adjust the exponent.
Subtract the excess bits: 2(8 - 1) - 1 = 127,
that is due to the 8 bit excess/bias notation.
The exponent, adjusted = 3 - 127 = -124
2. Convert the mantissa from binary (from base 2) to decimal (in base 10).
The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).
110 0000 0000 0000 0000 0100(2) =
1 × 2-1 + 1 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 =
0.5 + 0.25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 =
0.5 + 0.25 + 0.000 000 476 837 158 203 125 =
0.750 000 476 837 158 203 125(10)
= 0.000 000 000 000 000 000 000 000 000 000 000 000 07
0 - 0000 0011 - 110 0000 0000 0000 0000 0100 converted from a 32 bit single precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (float) = 0.000 000 000 000 000 000 000 000 000 000 000 000 07(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.