32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.527 64 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.527 64(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: 0.527 64.

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) 0.527 64 × 2 = 1 + 0.055 28;
  • 2) 0.055 28 × 2 = 0 + 0.110 56;
  • 3) 0.110 56 × 2 = 0 + 0.221 12;
  • 4) 0.221 12 × 2 = 0 + 0.442 24;
  • 5) 0.442 24 × 2 = 0 + 0.884 48;
  • 6) 0.884 48 × 2 = 1 + 0.768 96;
  • 7) 0.768 96 × 2 = 1 + 0.537 92;
  • 8) 0.537 92 × 2 = 1 + 0.075 84;
  • 9) 0.075 84 × 2 = 0 + 0.151 68;
  • 10) 0.151 68 × 2 = 0 + 0.303 36;
  • 11) 0.303 36 × 2 = 0 + 0.606 72;
  • 12) 0.606 72 × 2 = 1 + 0.213 44;
  • 13) 0.213 44 × 2 = 0 + 0.426 88;
  • 14) 0.426 88 × 2 = 0 + 0.853 76;
  • 15) 0.853 76 × 2 = 1 + 0.707 52;
  • 16) 0.707 52 × 2 = 1 + 0.415 04;
  • 17) 0.415 04 × 2 = 0 + 0.830 08;
  • 18) 0.830 08 × 2 = 1 + 0.660 16;
  • 19) 0.660 16 × 2 = 1 + 0.320 32;
  • 20) 0.320 32 × 2 = 0 + 0.640 64;
  • 21) 0.640 64 × 2 = 1 + 0.281 28;
  • 22) 0.281 28 × 2 = 0 + 0.562 56;
  • 23) 0.562 56 × 2 = 1 + 0.125 12;
  • 24) 0.125 12 × 2 = 0 + 0.250 24;

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:


0.527 64(10) =


0.1000 0111 0001 0011 0110 1010(2)


5. Positive number before normalization:

0.527 64(10) =


0.1000 0111 0001 0011 0110 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.527 64(10) =


0.1000 0111 0001 0011 0110 1010(2) =


0.1000 0111 0001 0011 0110 1010(2) × 20 =


1.0000 1110 0010 0110 1101 010(2) × 2-1


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): -1


Mantissa (not normalized):
1.0000 1110 0010 0110 1101 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


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


(-1 + 127)(10) =


126(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;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 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) =


126(10) =


0111 1110(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. 000 0111 0001 0011 0110 1010 =


000 0111 0001 0011 0110 1010


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 1110


Mantissa (23 bits) =
000 0111 0001 0011 0110 1010


The base ten decimal number 0.527 64 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1110 - 000 0111 0001 0011 0110 1010

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation