32 bit single precision IEEE 754 binary floating point number 1 - 1001 0111 (2023)

Follow the steps below to convert a number from 32 bit single precision IEEE 754 binary floating point representation to base 10 decimal system:

• 1. Identify the three elements that make up the binary representation of the number:
  First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
  The next 8 bits contain the exponent.
  The last 23 bits contain the mantissa.
• 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
• 3. Adjust the exponent, subtract the excess bits, 2^{(8 - 1)} - 1 = 127, that is due to the 8 bit excess/bias notation.
• 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
• 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
  (-1)^Sign × (1 + Mantissa) × 2^{(Exponent adjusted)}

Example: convert the number 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 from 32 bit single precision IEEE 754 binary floating point system to base 10 decimal system (float):

• 1. Identify the elements that make up the binary representation of the number:
  First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
  The next 8 bits contain the exponent: 1000 0001
  The last 23 bits contain the mantissa: 100 0001 0000 0010 0000 0000
• 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
  

  1000 0001₍₂₎ =

  1 × 2⁷ + 0 × 2⁶ + 0 × 2⁵ + 0 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰ =

  128 + 0 + 0 + 0 + 0 + 0 + 0 + 1 =

  128 + 1 =

  129₍₁₀₎

• 3. Adjust the exponent, subtract the excess bits, 2^{(8 - 1)} - 1 = 127, that is due to the 8 bit excess/bias notation:
  Exponent adjusted = 129 - 127 = 2
• 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
  

  100 0001 0000 0010 0000 0000₍₂₎ =

  1 × 2^-1 + 0 × 2^-2 + 0 × 2^-3 + 0 × 2^-4 + 0 × 2^-5 + 0 × 2^-6 + 1 × 2^-7 + 0 × 2^-8 + 0 × 2^-9 + 0 × 2^-10 + 0 × 2^-11 + 0 × 2^-12 + 0 × 2^-13 + 1 × 2^-14 + 0 × 2^-15 + 0 × 2^-16 + 0 × 2^-17 + 0 × 2^-18 + 0 × 2^-19 + 0 × 2^-20 + 0 × 2^-21 + 0 × 2^-22 + 0 × 2^-23 =

  0.5 + 0 + 0 + 0 + 0 + 0 + 0.007 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 061 035 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =

  0.5 + 0.007 812 5 + 0.000 061 035 156 25 =

  0.507 873 535 156 25₍₁₀₎

• 5. Put all the numbers into expression to calculate the single precision floating point decimal value:
  

  (-1)^Sign × (1 + Mantissa) × 2^{(Exponent adjusted)} =

  (-1)¹ × (1 + 0.507 873 535 156 25) × 2² =

  -1.507 873 535 156 25 × 2² =

  -6.031 494 140 625

• 1 - 1000 0001 - 100 0001 0000 0010 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point representation to decimal number (float) in decimal system (in base 10) = -6.031 494 140 625₍₁₀₎