Binary ↘ Float: The 32 Bit Single Precision IEEE 754 Binary Floating Point Standard Representation Number 0 - 0000 0000 - 101 0000 0000 0000 0000 0000 Converted and Written as a Base Ten Decimal System Number (as a Float)

0 - 0000 0000 - 101 0000 0000 0000 0000 0000: 32 bit single precision IEEE 754 binary floating point standard representation number converted to decimal system (base ten)

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 0000


The last 23 bits contain the mantissa:
101 0000 0000 0000 0000 0000


2. Reserved bitpattern.

We notice that all the bits that make up the exponent are on 0 (clear) and at least one bit of the mantissa is set on 1 (set).

This is one of the reserved bitpatterns of the special values of: Denormalized.


Denormalized numbers are too small to be correctly represented so they approximate to zero.

Depending on the sign bit, -0 and +0 are two distinct values though they both compare as equal (0).


3. Convert the exponent from binary (from base 2) to decimal (in base 10).

The exponent is allways a positive integer.

0000 0000(2) =


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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0(10)

4. 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 = 0 - 127 = -127


5. 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).


101 0000 0000 0000 0000 0000(2) =

1 × 2-1 + 0 × 2-2 + 1 × 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 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 =


0.5 + 0 + 0.125 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.5 + 0.125 =


0.625(10)

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

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =


(-1)0 × (1 + 0.625) × 2-127 =


1.625 × 2-127 =


0

0 - 0000 0000 - 101 0000 0000 0000 0000 0000 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(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

The latest 32 bit single precision IEEE 754 floating point binary standard numbers converted and written as decimal system numbers (in base ten, float)

The number 0 - 1000 0110 - 001 1111 0001 0100 1101 0101 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:31 UTC (GMT)
The number 0 - 0000 0001 - 001 0011 0010 0100 0000 0010 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:31 UTC (GMT)
The number 1 - 0111 1101 - 011 1000 0000 0000 0000 0000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:31 UTC (GMT)
The number 1 - 1010 0000 - 100 0000 1000 0000 0010 0001 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:31 UTC (GMT)
The number 1 - 0111 1101 - 100 1000 0000 0000 0000 1100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
The number 1 - 0001 0000 - 010 1010 0000 0000 0000 1100 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
The number 0 - 1100 1100 - 011 1111 1111 1111 1110 1000 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
The number 1 - 1011 1001 - 110 0111 1111 1111 1111 0111 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
The number 0 - 1100 0000 - 010 0011 1111 1111 1100 1101 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
The number 1 - 1100 0010 - 110 0001 0011 1011 1000 0101 converted from 32 bit single precision IEEE 754 binary floating point system and written as a decimal number (float) written in base ten = ? Jul 13 12:30 UTC (GMT)
All 32 bit single precision IEEE 754 binary floating point representation numbers converted to base ten decimal numbers (float)

How to convert numbers from 32 bit single precision IEEE 754 binary floating point standard to decimal system in base 10

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)