32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 12.225 468 52 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 12.225 468 52(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: 12.
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;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.


12(10) =


1100(2)


3. Convert to binary (base 2) the fractional part: 0.225 468 52.

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) 0.225 468 52 × 2 = 0 + 0.450 937 04;
  • 2) 0.450 937 04 × 2 = 0 + 0.901 874 08;
  • 3) 0.901 874 08 × 2 = 1 + 0.803 748 16;
  • 4) 0.803 748 16 × 2 = 1 + 0.607 496 32;
  • 5) 0.607 496 32 × 2 = 1 + 0.214 992 64;
  • 6) 0.214 992 64 × 2 = 0 + 0.429 985 28;
  • 7) 0.429 985 28 × 2 = 0 + 0.859 970 56;
  • 8) 0.859 970 56 × 2 = 1 + 0.719 941 12;
  • 9) 0.719 941 12 × 2 = 1 + 0.439 882 24;
  • 10) 0.439 882 24 × 2 = 0 + 0.879 764 48;
  • 11) 0.879 764 48 × 2 = 1 + 0.759 528 96;
  • 12) 0.759 528 96 × 2 = 1 + 0.519 057 92;
  • 13) 0.519 057 92 × 2 = 1 + 0.038 115 84;
  • 14) 0.038 115 84 × 2 = 0 + 0.076 231 68;
  • 15) 0.076 231 68 × 2 = 0 + 0.152 463 36;
  • 16) 0.152 463 36 × 2 = 0 + 0.304 926 72;
  • 17) 0.304 926 72 × 2 = 0 + 0.609 853 44;
  • 18) 0.609 853 44 × 2 = 1 + 0.219 706 88;
  • 19) 0.219 706 88 × 2 = 0 + 0.439 413 76;
  • 20) 0.439 413 76 × 2 = 0 + 0.878 827 52;
  • 21) 0.878 827 52 × 2 = 1 + 0.757 655 04;
  • 22) 0.757 655 04 × 2 = 1 + 0.515 310 08;
  • 23) 0.515 310 08 × 2 = 1 + 0.030 620 16;
  • 24) 0.030 620 16 × 2 = 0 + 0.061 240 32;

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:


0.225 468 52(10) =


0.0011 1001 1011 1000 0100 1110(2)


5. Positive number before normalization:

12.225 468 52(10) =


1100.0011 1001 1011 1000 0100 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


12.225 468 52(10) =


1100.0011 1001 1011 1000 0100 1110(2) =


1100.0011 1001 1011 1000 0100 1110(2) × 20 =


1.1000 0111 0011 0111 0000 1001 110(2) × 23


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): 3


Mantissa (not normalized):
1.1000 0111 0011 0111 0000 1001 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(8-1) - 1 =


(3 + 127)(10) =


130(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;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


130(10) =


1000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 100 0011 1001 1011 1000 0100 1110 =


100 0011 1001 1011 1000 0100


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) =
1000 0010


Mantissa (23 bits) =
100 0011 1001 1011 1000 0100


The base ten decimal number 12.225 468 52 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0010 - 100 0011 1001 1011 1000 0100

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