32 bit single precision IEEE 754 binary floating point number 1 - 0110 0111 - 010 1111 1111 1111 1111 1101 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 1 - 0110 0111 - 010 1111 1111 1111 1111 1101 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:
0110 0111


The last 23 bits contain the mantissa:
010 1111 1111 1111 1111 1101

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

The exponent is allways a positive integer.

0110 0111(2) =


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


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


64 + 32 + 4 + 2 + 1 =


103(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 = 103 - 127 = -24


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)

010 1111 1111 1111 1111 1101(2) =

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


0 + 0.25 + 0 + 0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0 + 0.000 000 119 209 289 550 781 25 =


0.25 + 0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0.000 000 119 209 289 550 781 25 =


0.374 999 642 372 131 347 656 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 + 0.374 999 642 372 131 347 656 25) × 2-24 =


-1.374 999 642 372 131 347 656 25 × 2-24 =


-0.000 000 081 956 365 249 880 036 571 994 423 866 27

1 - 0110 0111 - 010 1111 1111 1111 1111 1101 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
-0.000 000 081 956 365 249 880 036 571 994 423 866 27(10)

More operations of this kind:

1 - 0110 0111 - 010 1111 1111 1111 1111 1100 = ?

1 - 0110 0111 - 010 1111 1111 1111 1111 1110 = ?


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 - 0110 0111 - 010 1111 1111 1111 1111 1101 = ? Mar 02 13:39 UTC (GMT)
1 - 1111 1010 - 111 1001 1001 1001 1001 1001 = ? Mar 02 13:39 UTC (GMT)
1 - 0111 1011 - 111 1001 0001 0001 1100 1011 = ? Mar 02 13:39 UTC (GMT)
0 - 1111 1111 - 101 0000 0000 0000 0000 0000 = ? Mar 02 13:39 UTC (GMT)
0 - 1100 1101 - 100 0011 0101 0111 1111 1100 = ? Mar 02 13:39 UTC (GMT)
0 - 0110 0100 - 101 1100 0110 0111 1101 0110 = ? Mar 02 13:38 UTC (GMT)
1 - 0000 0001 - 111 0110 1010 0011 1101 0100 = ? Mar 02 13:38 UTC (GMT)
1 - 1000 0011 - 000 0101 0000 0000 0000 0000 = ? Mar 02 13:38 UTC (GMT)
1 - 0111 1000 - 011 0000 1000 0000 0000 0010 = ? Mar 02 13:38 UTC (GMT)
0 - 0101 0000 - 010 1111 1111 1111 1111 1001 = ? Mar 02 13:37 UTC (GMT)
1 - 1000 0010 - 110 0011 0000 0000 0000 0000 = ? Mar 02 13:36 UTC (GMT)
1 - 1000 0011 - 010 1100 0000 0000 0000 0010 = ? Mar 02 13:36 UTC (GMT)
0 - 0000 0011 - 111 0111 1111 1111 1111 1011 = ? Mar 02 13:36 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)