32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 66.149 993 896 484 5 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 66.149 993 896 484 5(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: 66.
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;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.


66(10) =


100 0010(2)


3. Convert to binary (base 2) the fractional part: 0.149 993 896 484 5.

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) 0.149 993 896 484 5 × 2 = 0 + 0.299 987 792 969;
  • 2) 0.299 987 792 969 × 2 = 0 + 0.599 975 585 938;
  • 3) 0.599 975 585 938 × 2 = 1 + 0.199 951 171 876;
  • 4) 0.199 951 171 876 × 2 = 0 + 0.399 902 343 752;
  • 5) 0.399 902 343 752 × 2 = 0 + 0.799 804 687 504;
  • 6) 0.799 804 687 504 × 2 = 1 + 0.599 609 375 008;
  • 7) 0.599 609 375 008 × 2 = 1 + 0.199 218 750 016;
  • 8) 0.199 218 750 016 × 2 = 0 + 0.398 437 500 032;
  • 9) 0.398 437 500 032 × 2 = 0 + 0.796 875 000 064;
  • 10) 0.796 875 000 064 × 2 = 1 + 0.593 750 000 128;
  • 11) 0.593 750 000 128 × 2 = 1 + 0.187 500 000 256;
  • 12) 0.187 500 000 256 × 2 = 0 + 0.375 000 000 512;
  • 13) 0.375 000 000 512 × 2 = 0 + 0.750 000 001 024;
  • 14) 0.750 000 001 024 × 2 = 1 + 0.500 000 002 048;
  • 15) 0.500 000 002 048 × 2 = 1 + 0.000 000 004 096;
  • 16) 0.000 000 004 096 × 2 = 0 + 0.000 000 008 192;
  • 17) 0.000 000 008 192 × 2 = 0 + 0.000 000 016 384;
  • 18) 0.000 000 016 384 × 2 = 0 + 0.000 000 032 768;
  • 19) 0.000 000 032 768 × 2 = 0 + 0.000 000 065 536;
  • 20) 0.000 000 065 536 × 2 = 0 + 0.000 000 131 072;
  • 21) 0.000 000 131 072 × 2 = 0 + 0.000 000 262 144;
  • 22) 0.000 000 262 144 × 2 = 0 + 0.000 000 524 288;
  • 23) 0.000 000 524 288 × 2 = 0 + 0.000 001 048 576;
  • 24) 0.000 001 048 576 × 2 = 0 + 0.000 002 097 152;

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:


0.149 993 896 484 5(10) =


0.0010 0110 0110 0110 0000 0000(2)


5. Positive number before normalization:

66.149 993 896 484 5(10) =


100 0010.0010 0110 0110 0110 0000 0000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


66.149 993 896 484 5(10) =


100 0010.0010 0110 0110 0110 0000 0000(2) =


100 0010.0010 0110 0110 0110 0000 0000(2) × 20 =


1.0000 1000 1001 1001 1001 1000 0000 00(2) × 26


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): 6


Mantissa (not normalized):
1.0000 1000 1001 1001 1001 1000 0000 00


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


6 + 2(8-1) - 1 =


(6 + 127)(10) =


133(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;
  • 133 ÷ 2 = 66 + 1;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 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) =


133(10) =


1000 0101(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. 000 0100 0100 1100 1100 1100 000 0000 =


000 0100 0100 1100 1100 1100


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 0101


Mantissa (23 bits) =
000 0100 0100 1100 1100 1100


The base ten decimal number 66.149 993 896 484 5 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0101 - 000 0100 0100 1100 1100 1100

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