32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.311 9 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.311 9(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =


0(2)


3. Convert to binary (base 2) the fractional part: 0.311 9.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.311 9 × 2 = 0 + 0.623 8;
  • 2) 0.623 8 × 2 = 1 + 0.247 6;
  • 3) 0.247 6 × 2 = 0 + 0.495 2;
  • 4) 0.495 2 × 2 = 0 + 0.990 4;
  • 5) 0.990 4 × 2 = 1 + 0.980 8;
  • 6) 0.980 8 × 2 = 1 + 0.961 6;
  • 7) 0.961 6 × 2 = 1 + 0.923 2;
  • 8) 0.923 2 × 2 = 1 + 0.846 4;
  • 9) 0.846 4 × 2 = 1 + 0.692 8;
  • 10) 0.692 8 × 2 = 1 + 0.385 6;
  • 11) 0.385 6 × 2 = 0 + 0.771 2;
  • 12) 0.771 2 × 2 = 1 + 0.542 4;
  • 13) 0.542 4 × 2 = 1 + 0.084 8;
  • 14) 0.084 8 × 2 = 0 + 0.169 6;
  • 15) 0.169 6 × 2 = 0 + 0.339 2;
  • 16) 0.339 2 × 2 = 0 + 0.678 4;
  • 17) 0.678 4 × 2 = 1 + 0.356 8;
  • 18) 0.356 8 × 2 = 0 + 0.713 6;
  • 19) 0.713 6 × 2 = 1 + 0.427 2;
  • 20) 0.427 2 × 2 = 0 + 0.854 4;
  • 21) 0.854 4 × 2 = 1 + 0.708 8;
  • 22) 0.708 8 × 2 = 1 + 0.417 6;
  • 23) 0.417 6 × 2 = 0 + 0.835 2;
  • 24) 0.835 2 × 2 = 1 + 0.670 4;
  • 25) 0.670 4 × 2 = 1 + 0.340 8;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.311 9(10) =


0.0100 1111 1101 1000 1010 1101 1(2)


5. Positive number before normalization:

0.311 9(10) =


0.0100 1111 1101 1000 1010 1101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.311 9(10) =


0.0100 1111 1101 1000 1010 1101 1(2) =


0.0100 1111 1101 1000 1010 1101 1(2) × 20 =


1.0011 1111 0110 0010 1011 011(2) × 2-2


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): -2


Mantissa (not normalized):
1.0011 1111 0110 0010 1011 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-2 + 2(8-1) - 1 =


(-2 + 127)(10) =


125(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


125(10) =


0111 1101(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =


1. 001 1111 1011 0001 0101 1011 =


001 1111 1011 0001 0101 1011


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
0111 1101


Mantissa (23 bits) =
001 1111 1011 0001 0101 1011


The base ten decimal number 0.311 9 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1101 - 001 1111 1011 0001 0101 1011

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