64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 27.566 408 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 27.566 408(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 27.
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;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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.


27(10) =


1 1011(2)


3. Convert to binary (base 2) the fractional part: 0.566 408.

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) 0.566 408 × 2 = 1 + 0.132 816;
  • 2) 0.132 816 × 2 = 0 + 0.265 632;
  • 3) 0.265 632 × 2 = 0 + 0.531 264;
  • 4) 0.531 264 × 2 = 1 + 0.062 528;
  • 5) 0.062 528 × 2 = 0 + 0.125 056;
  • 6) 0.125 056 × 2 = 0 + 0.250 112;
  • 7) 0.250 112 × 2 = 0 + 0.500 224;
  • 8) 0.500 224 × 2 = 1 + 0.000 448;
  • 9) 0.000 448 × 2 = 0 + 0.000 896;
  • 10) 0.000 896 × 2 = 0 + 0.001 792;
  • 11) 0.001 792 × 2 = 0 + 0.003 584;
  • 12) 0.003 584 × 2 = 0 + 0.007 168;
  • 13) 0.007 168 × 2 = 0 + 0.014 336;
  • 14) 0.014 336 × 2 = 0 + 0.028 672;
  • 15) 0.028 672 × 2 = 0 + 0.057 344;
  • 16) 0.057 344 × 2 = 0 + 0.114 688;
  • 17) 0.114 688 × 2 = 0 + 0.229 376;
  • 18) 0.229 376 × 2 = 0 + 0.458 752;
  • 19) 0.458 752 × 2 = 0 + 0.917 504;
  • 20) 0.917 504 × 2 = 1 + 0.835 008;
  • 21) 0.835 008 × 2 = 1 + 0.670 016;
  • 22) 0.670 016 × 2 = 1 + 0.340 032;
  • 23) 0.340 032 × 2 = 0 + 0.680 064;
  • 24) 0.680 064 × 2 = 1 + 0.360 128;
  • 25) 0.360 128 × 2 = 0 + 0.720 256;
  • 26) 0.720 256 × 2 = 1 + 0.440 512;
  • 27) 0.440 512 × 2 = 0 + 0.881 024;
  • 28) 0.881 024 × 2 = 1 + 0.762 048;
  • 29) 0.762 048 × 2 = 1 + 0.524 096;
  • 30) 0.524 096 × 2 = 1 + 0.048 192;
  • 31) 0.048 192 × 2 = 0 + 0.096 384;
  • 32) 0.096 384 × 2 = 0 + 0.192 768;
  • 33) 0.192 768 × 2 = 0 + 0.385 536;
  • 34) 0.385 536 × 2 = 0 + 0.771 072;
  • 35) 0.771 072 × 2 = 1 + 0.542 144;
  • 36) 0.542 144 × 2 = 1 + 0.084 288;
  • 37) 0.084 288 × 2 = 0 + 0.168 576;
  • 38) 0.168 576 × 2 = 0 + 0.337 152;
  • 39) 0.337 152 × 2 = 0 + 0.674 304;
  • 40) 0.674 304 × 2 = 1 + 0.348 608;
  • 41) 0.348 608 × 2 = 0 + 0.697 216;
  • 42) 0.697 216 × 2 = 1 + 0.394 432;
  • 43) 0.394 432 × 2 = 0 + 0.788 864;
  • 44) 0.788 864 × 2 = 1 + 0.577 728;
  • 45) 0.577 728 × 2 = 1 + 0.155 456;
  • 46) 0.155 456 × 2 = 0 + 0.310 912;
  • 47) 0.310 912 × 2 = 0 + 0.621 824;
  • 48) 0.621 824 × 2 = 1 + 0.243 648;
  • 49) 0.243 648 × 2 = 0 + 0.487 296;
  • 50) 0.487 296 × 2 = 0 + 0.974 592;
  • 51) 0.974 592 × 2 = 1 + 0.949 184;
  • 52) 0.949 184 × 2 = 1 + 0.898 368;
  • 53) 0.898 368 × 2 = 1 + 0.796 736;

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:


0.566 408(10) =


0.1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1(2)


5. Positive number before normalization:

27.566 408(10) =


1 1011.1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


27.566 408(10) =


1 1011.1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1(2) =


1 1011.1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1(2) × 20 =


1.1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1(2) × 24


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 4


Mantissa (not normalized):
1.1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0011 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 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) =


1027(10) =


100 0000 0011(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 52 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. 1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001 0 0111 =


1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

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


Exponent (11 bits) =
100 0000 0011


Mantissa (52 bits) =
1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001


The base ten decimal number 27.566 408 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0011 - 1011 1001 0001 0000 0000 0001 1101 0101 1100 0011 0001 0101 1001

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation