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.
1
The next 8 bits contain the exponent:
1101 1011
The last 23 bits contain the mantissa:
100 0010 0001 0001 1000 0001
1. Convert the exponent from binary (from base 2) to decimal (in base 10).
The exponent is allways a positive integer.
1101 1011(2) =
1 × 27 + 1 × 26 + 0 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 =
128 + 64 + 0 + 16 + 8 + 0 + 2 + 1 =
128 + 64 + 16 + 8 + 2 + 1 =
219(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 = 219 - 127 = 92
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).
100 0010 0001 0001 1000 0001(2) =
1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 1 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 1 × 2-15 + 1 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 1 × 2-23 =
0.5 + 0 + 0 + 0 + 0 + 0.015 625 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 119 209 289 550 781 25 =
0.5 + 0.015 625 + 0.000 488 281 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 000 119 209 289 550 781 25 =
0.516 159 176 826 477 050 781 25(10)
= -7 507 656 603 693 835 276 601 262 080
1 - 1101 1011 - 100 0010 0001 0001 1000 0001 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) = -7 507 656 603 693 835 276 601 262 080(10)
Spaces were used to group digits: for binary, by 4, for decimal, by 3.