(CIE A2 Compsci) Data representation - floating point binary

Representing floating point numbers using binary

  • Fixed-point representations of numbers using unsigned/two’s-complement/BCD binary limit number ranges in a way that’d forbid fractional value representation

  • As a result the denary standard-form floating point system (where larger numbers can be written as something like 2.5Ă—1024) can be adapted for usage in binary notation as MĂ—2E (where M = the mantissa and E = the exponent)

Converting denary numbers into floating point binary representations and vice versa

an example of a denary number being represented in the mantissa-exponent formatthe columns that can be used to represent denary numbers like the one above in floating point binary

Examples from textbook

  1. Firstly add up all the 1-values in the mantissa (1/2 + 1/8 + 1/16 + 1/64 = 0.703 = 45/64)

  2. Then add up all the 1-values in the exponent and use the M Ă— 2E formula (E = 4 so M Ă— 2E = 45/64 Ă— 24 = 45/64 Ă— 16 = 11.25)

  1. Firstly add up all the 1-values in the mantissa (-1 + 1/2 + 1/16 + 1/32 = 0.40625)

  2. Then add up all the 1-values in the exponent and use the M Ă— 2E formula (E = 8+4 = 12 so M Ă— 2E = 0.40625 Ă— 12 = 4.875)

  1. Firstly check if the fraction’s in the correct form (numerator < denominator if fraction < 1) - in this case it is indeed since 0.171875 is approx. equal to 11/64

  2. Then split the fraction (11/64) up into the individual values it’s summed up from (1/8 + 1/32 + 1/64) which gives 0.0010110 as the mantissa and 0 as the exponent

Potential rounding errors/approximations

  • Stuff like this often comes up with fractional numbers whose values can only be approximated (ie. have odd denominators - eg. 1/3) - reducing an error like this is possible when the mantissa’s increased in terms of bit size (eg. a 16-bit mantissa’s larger than an 8-bit one)

  • To increase the mantissa’s size the binary floating point has to be moved as far to the left as possible (eg. if 5.88 = 0101.11100001 then moving the binary point three places to the left causes that number to be stored as 5.75)

Normalization example

  1. Firstly the bits in the mantissa get shifted by two places to the left to get 1.0110000

  2. Then the exponent gets reduced by two to 00000011 and recombined with the mantissa to provide the normalized result (1.011000 00001000)

Possible issues with using floating point binary

  • Certain numbers can only be approximated using floating point binary due to mantissa size limitations (an issue that can be minimized when using double/quadruple precision-permitting programming languages)

  • Overflow errors also risk being produced if a floating point binary calculation leads to a value that exceeds the maximum possible storable value

  • Underflow errors also risk being produced if division by a very large number leads to any value lower than the smallest storable value

  • The mantissa also doesn’t allow for a zero value as it must be valued at either 0.1 or 1.0