32 bit single precision IEEE 754 binary floating point number 0 - 1000 0001 - 110 1001 1001 1001 1001 1010 converted to decimal base ten (float)

How to convert 32 bit single precision IEEE 754 binary floating point:
0 - 1000 0001 - 110 1001 1001 1001 1001 1010
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:
1000 0001


The last 23 bits contain the mantissa:
110 1001 1001 1001 1001 1010

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

The exponent is allways a positive integer.

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 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)

110 1001 1001 1001 1001 1010(2) =

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


0.5 + 0.25 + 0 + 0.062 5 + 0 + 0 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0.000 244 140 625 + 0 + 0 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0 + 0 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0 + 0.000 000 238 418 579 101 562 5 + 0 =


0.5 + 0.25 + 0.062 5 + 0.007 812 5 + 0.003 906 25 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 030 517 578 125 + 0.000 015 258 789 062 5 + 0.000 001 907 348 632 812 5 + 0.000 000 953 674 316 406 25 + 0.000 000 238 418 579 101 562 5 =


0.825 000 047 683 715 820 312 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)0 × (1 + 0.825 000 047 683 715 820 312 5) × 22 =


1.825 000 047 683 715 820 312 5 × 22 =


7.300 000 190 734 863 281 25

Conclusion:

0 - 1000 0001 - 110 1001 1001 1001 1001 1010
converted from
32 bit single precision IEEE 754 binary floating point
to
base ten decimal system (float) =

7.300 000 190 734 863 281 25(10)

More operations of this kind:

0 - 1000 0001 - 110 1001 1001 1001 1001 1001 = ?

0 - 1000 0001 - 110 1001 1001 1001 1001 1011 = ?


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 - 1000 0001 - 110 1001 1001 1001 1001 1010 = ? Jan 24 21:34 UTC (GMT)
0 - 1000 0101 - 110 1100 1111 0000 1010 1010 = ? Jan 24 21:34 UTC (GMT)
1 - 1000 0100 - 101 1011 0000 0000 0000 0000 = ? Jan 24 21:33 UTC (GMT)
0 - 0000 0000 - 001 0111 1010 1010 1011 0010 = ? Jan 24 21:33 UTC (GMT)
1 - 1101 0000 - 000 0000 0000 0011 1011 1100 = ? Jan 24 21:33 UTC (GMT)
0 - 0000 0001 - 111 1111 1111 1111 1111 1101 = ? Jan 24 21:32 UTC (GMT)
0 - 0000 0001 - 000 0000 0000 0000 0000 0010 = ? Jan 24 21:32 UTC (GMT)
1 - 0011 0011 - 000 1111 1111 1111 1111 1111 = ? Jan 24 21:32 UTC (GMT)
1 - 0011 0010 - 010 0000 0000 0000 0000 0000 = ? Jan 24 21:31 UTC (GMT)
0 - 1000 0100 - 101 0000 1000 0000 0000 0011 = ? Jan 24 21:30 UTC (GMT)
0 - 0111 1111 - 011 1111 1111 0000 0000 1001 = ? Jan 24 21:30 UTC (GMT)
0 - 1001 0101 - 110 1100 0000 0000 0000 0000 = ? Jan 24 21:30 UTC (GMT)
1 - 1000 0111 - 011 0000 0000 0000 0000 0000 = ? Jan 24 21:30 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)