32 bit single precision IEEE 754 binary floating point number 1 - 0111 0111 - 101 0110 1101 1010 1110 0010 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 1 - 0111 0111 - 101 0110 1101 1010 1110 0010 to decimal system (base ten) = ?

1. Identify the elements that make up the binary representation of the number:

First bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.


The next 8 bits contain the exponent:
0111 0111


The last 23 bits contain the mantissa:
101 0110 1101 1010 1110 0010

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

The exponent is allways a positive integer.

0111 0111(2) =


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


0 + 64 + 32 + 16 + 0 + 4 + 2 + 1 =


64 + 32 + 16 + 4 + 2 + 1 =


119(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 = 119 - 127 = -8


4. Convert the mantissa from binary (base 2) to decimal (base 10):

Mantissa represents the number's fractional part (the excess beyond the number's integer part, comma delimited)

101 0110 1101 1010 1110 0010(2) =

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


0.5 + 0 + 0.125 + 0 + 0.031 25 + 0.015 625 + 0 + 0.003 906 25 + 0.001 953 125 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0.000 061 035 156 25 + 0 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0 =


0.5 + 0.125 + 0.031 25 + 0.015 625 + 0.003 906 25 + 0.001 953 125 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 061 035 156 25 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0.000 000 238 418 579 101 562 5 =


0.678 554 773 330 688 476 562 5(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 + 0.678 554 773 330 688 476 562 5) × 2-8 =


-1.678 554 773 330 688 476 562 5 × 2-8 =


-0.006 556 854 583 323 001 861 572 265 625

1 - 0111 0111 - 101 0110 1101 1010 1110 0010 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
-0.006 556 854 583 323 001 861 572 265 625(10)

More operations of this kind:

1 - 0111 0111 - 101 0110 1101 1010 1110 0001 = ?

1 - 0111 0111 - 101 0110 1101 1010 1110 0011 = ?


Convert 32 bit single precision IEEE 754 floating point standard binary numbers to base ten decimal system (float)

32 bit single precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

Latest 32 bit single precision IEEE 754 floating point binary standard numbers converted to decimal base ten (float)

1 - 0111 0111 - 101 0110 1101 1010 1110 0010 = ? Apr 18 09:17 UTC (GMT)
0 - 1000 0011 - 010 0010 0000 0000 0000 0000 = ? Apr 18 09:17 UTC (GMT)
0 - 0100 0011 - 111 1000 0010 0000 0000 0000 = ? Apr 18 09:16 UTC (GMT)
0 - 1001 1000 - 000 0001 0101 1010 1110 1100 = ? Apr 18 09:16 UTC (GMT)
1 - 0101 1011 - 011 1111 1111 1111 1111 1111 = ? Apr 18 09:16 UTC (GMT)
0 - 0000 0000 - 000 0000 0000 0000 0000 0111 = ? Apr 18 09:16 UTC (GMT)
1 - 1000 1011 - 000 0010 1011 1110 1111 1110 = ? Apr 18 09:15 UTC (GMT)
1 - 1001 0110 - 111 1010 1011 1000 1001 1110 = ? Apr 18 09:15 UTC (GMT)
1 - 0001 1000 - 000 0101 0000 0000 0000 0000 = ? Apr 18 09:15 UTC (GMT)
1 - 1000 0111 - 000 1110 0000 0000 0000 0000 = ? Apr 18 09:14 UTC (GMT)
0 - 1000 0000 - 011 1001 0000 0000 0000 0000 = ? Apr 18 09:13 UTC (GMT)
0 - 0111 0011 - 010 0100 1100 0100 0110 1011 = ? Apr 18 09:12 UTC (GMT)
1 - 1000 0111 - 110 0000 0000 0000 0000 0101 = ? Apr 18 09:11 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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)