32 bit single precision IEEE 754 binary floating point number 1 - 1011 1010 - 011 0101 1000 0000 0000 0011 converted to decimal base ten (float)

32 bit single precision IEEE 754 binary floating point 1 - 1011 1010 - 011 0101 1000 0000 0000 0011 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:
1011 1010


The last 23 bits contain the mantissa:
011 0101 1000 0000 0000 0011

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

The exponent is allways a positive integer.

1011 1010(2) =


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


128 + 0 + 32 + 16 + 8 + 0 + 2 + 0 =


128 + 32 + 16 + 8 + 2 =


186(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 = 186 - 127 = 59


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)

011 0101 1000 0000 0000 0011(2) =

0 × 2-1 + 1 × 2-2 + 1 × 2-3 + 0 × 2-4 + 1 × 2-5 + 0 × 2-6 + 1 × 2-7 + 1 × 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 + 1 × 2-22 + 1 × 2-23 =


0 + 0.25 + 0.125 + 0 + 0.031 25 + 0 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.25 + 0.125 + 0.031 25 + 0.007 812 5 + 0.003 906 25 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 =


0.417 969 107 627 868 652 343 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)1 × (1 + 0.417 969 107 627 868 652 343 75) × 259 =


-1.417 969 107 627 868 652 343 75 × 259 =


-817 403 538 526 175 232

1 - 1011 1010 - 011 0101 1000 0000 0000 0011 converted from 32 bit single precision IEEE 754 binary floating point to base ten decimal system (float) =
-817 403 538 526 175 232(10)

More operations of this kind:

1 - 1011 1010 - 011 0101 1000 0000 0000 0010 = ?

1 - 1011 1010 - 011 0101 1000 0000 0000 0100 = ?


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 - 1011 1010 - 011 0101 1000 0000 0000 0011 = ? Mar 08 11:51 UTC (GMT)
0 - 0110 1111 - 100 0000 1111 0000 0010 0000 = ? Mar 08 11:51 UTC (GMT)
1 - 0111 1111 - 010 0111 1111 1111 1111 1000 = ? Mar 08 11:50 UTC (GMT)
0 - 0011 0000 - 110 1000 0000 0000 0000 0001 = ? Mar 08 11:50 UTC (GMT)
1 - 1000 0001 - 011 0011 0010 0001 1111 1110 = ? Mar 08 11:50 UTC (GMT)
0 - 1000 1001 - 010 1101 0111 0101 0110 0111 = ? Mar 08 11:49 UTC (GMT)
0 - 0111 1110 - 110 0101 1101 0001 1100 1010 = ? Mar 08 11:49 UTC (GMT)
1 - 1000 1000 - 100 1100 1111 1000 0011 0111 = ? Mar 08 11:49 UTC (GMT)
0 - 0111 1010 - 010 0000 0000 0000 0000 0000 = ? Mar 08 11:48 UTC (GMT)
1 - 1100 0110 - 100 1001 1111 1111 1111 1100 = ? Mar 08 11:48 UTC (GMT)
0 - 1100 0010 - 001 0001 1111 1111 1111 1100 = ? Mar 08 11:48 UTC (GMT)
1 - 1000 0001 - 110 1100 0010 0000 1100 1001 = ? Mar 08 11:47 UTC (GMT)
1 - 1010 1111 - 111 1011 1111 1111 1111 1100 = ? Mar 08 11:47 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)