32 bit single precision IEEE 754 binary floating point number 0 - 0111 0010 - 101 0011 0110 0100 1001 0111 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 0 - 0111 0010 - 101 0011 0110 0100 1001 0111 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 0010


The last 23 bits contain the mantissa:
101 0011 0110 0100 1001 0111

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

The exponent is allways a positive integer.

0111 0010(2) =


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


0 + 64 + 32 + 16 + 0 + 0 + 2 + 0 =


64 + 32 + 16 + 2 =


114(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 = 114 - 127 = -13


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 0011 0110 0100 1001 0111(2) =

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


0.5 + 0 + 0.125 + 0 + 0 + 0.015 625 + 0.007 812 5 + 0 + 0.001 953 125 + 0.000 976 562 5 + 0 + 0 + 0.000 122 070 312 5 + 0 + 0 + 0.000 015 258 789 062 5 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.5 + 0.125 + 0.015 625 + 0.007 812 5 + 0.001 953 125 + 0.000 976 562 5 + 0.000 122 070 312 5 + 0.000 015 258 789 062 5 + 0.000 001 907 348 632 812 5 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.651 507 258 415 222 167 968 75(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)0 × (1 + 0.651 507 258 415 222 167 968 75) × 2-13 =


1.651 507 258 415 222 167 968 75 × 2-13 =


0.000 201 600 007 130 764 424 800 872 802 734 375

0 - 0111 0010 - 101 0011 0110 0100 1001 0111 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
0.000 201 600 007 130 764 424 800 872 802 734 375(10)

More operations of this kind:

0 - 0111 0010 - 101 0011 0110 0100 1001 0110 = ?

0 - 0111 0010 - 101 0011 0110 0100 1001 1000 = ?


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)

0 - 0111 0010 - 101 0011 0110 0100 1001 0111 = ? Mar 08 12:04 UTC (GMT)
0 - 0000 0000 - 000 0000 1000 0000 1011 1111 = ? Mar 08 12:04 UTC (GMT)
1 - 1010 1000 - 010 0101 0000 1010 1010 0011 = ? Mar 08 12:03 UTC (GMT)
0 - 0000 0000 - 000 0000 0000 0000 0000 1100 = ? Mar 08 12:03 UTC (GMT)
0 - 0111 1111 - 000 1101 0001 1110 1011 1000 = ? Mar 08 12:03 UTC (GMT)
0 - 0000 0000 - 000 0000 0000 0000 0000 1011 = ? Mar 08 12:03 UTC (GMT)
0 - 0111 1010 - 000 1100 1110 0110 1001 1110 = ? Mar 08 12:03 UTC (GMT)
1 - 1001 0110 - 111 1010 1011 1000 1001 1111 = ? Mar 08 12:03 UTC (GMT)
0 - 1011 1000 - 000 0000 0000 0000 0000 0000 = ? Mar 08 12:02 UTC (GMT)
0 - 1000 0010 - 001 0010 1111 1111 1111 0110 = ? Mar 08 12:01 UTC (GMT)
1 - 1001 0000 - 100 0001 1111 0001 0011 0111 = ? Mar 08 12:01 UTC (GMT)
0 - 0111 1000 - 011 1011 1111 0101 1000 1000 = ? Mar 08 12:01 UTC (GMT)
0 - 1000 0100 - 001 0100 0111 1111 1111 1011 = ? Mar 08 12:00 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)