64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 8.533 333 33 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 8.533 333 33(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: 8.
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;
  • 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.


8(10) =


1000(2)


3. Convert to binary (base 2) the fractional part: 0.533 333 33.

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) 0.533 333 33 × 2 = 1 + 0.066 666 66;
  • 2) 0.066 666 66 × 2 = 0 + 0.133 333 32;
  • 3) 0.133 333 32 × 2 = 0 + 0.266 666 64;
  • 4) 0.266 666 64 × 2 = 0 + 0.533 333 28;
  • 5) 0.533 333 28 × 2 = 1 + 0.066 666 56;
  • 6) 0.066 666 56 × 2 = 0 + 0.133 333 12;
  • 7) 0.133 333 12 × 2 = 0 + 0.266 666 24;
  • 8) 0.266 666 24 × 2 = 0 + 0.533 332 48;
  • 9) 0.533 332 48 × 2 = 1 + 0.066 664 96;
  • 10) 0.066 664 96 × 2 = 0 + 0.133 329 92;
  • 11) 0.133 329 92 × 2 = 0 + 0.266 659 84;
  • 12) 0.266 659 84 × 2 = 0 + 0.533 319 68;
  • 13) 0.533 319 68 × 2 = 1 + 0.066 639 36;
  • 14) 0.066 639 36 × 2 = 0 + 0.133 278 72;
  • 15) 0.133 278 72 × 2 = 0 + 0.266 557 44;
  • 16) 0.266 557 44 × 2 = 0 + 0.533 114 88;
  • 17) 0.533 114 88 × 2 = 1 + 0.066 229 76;
  • 18) 0.066 229 76 × 2 = 0 + 0.132 459 52;
  • 19) 0.132 459 52 × 2 = 0 + 0.264 919 04;
  • 20) 0.264 919 04 × 2 = 0 + 0.529 838 08;
  • 21) 0.529 838 08 × 2 = 1 + 0.059 676 16;
  • 22) 0.059 676 16 × 2 = 0 + 0.119 352 32;
  • 23) 0.119 352 32 × 2 = 0 + 0.238 704 64;
  • 24) 0.238 704 64 × 2 = 0 + 0.477 409 28;
  • 25) 0.477 409 28 × 2 = 0 + 0.954 818 56;
  • 26) 0.954 818 56 × 2 = 1 + 0.909 637 12;
  • 27) 0.909 637 12 × 2 = 1 + 0.819 274 24;
  • 28) 0.819 274 24 × 2 = 1 + 0.638 548 48;
  • 29) 0.638 548 48 × 2 = 1 + 0.277 096 96;
  • 30) 0.277 096 96 × 2 = 0 + 0.554 193 92;
  • 31) 0.554 193 92 × 2 = 1 + 0.108 387 84;
  • 32) 0.108 387 84 × 2 = 0 + 0.216 775 68;
  • 33) 0.216 775 68 × 2 = 0 + 0.433 551 36;
  • 34) 0.433 551 36 × 2 = 0 + 0.867 102 72;
  • 35) 0.867 102 72 × 2 = 1 + 0.734 205 44;
  • 36) 0.734 205 44 × 2 = 1 + 0.468 410 88;
  • 37) 0.468 410 88 × 2 = 0 + 0.936 821 76;
  • 38) 0.936 821 76 × 2 = 1 + 0.873 643 52;
  • 39) 0.873 643 52 × 2 = 1 + 0.747 287 04;
  • 40) 0.747 287 04 × 2 = 1 + 0.494 574 08;
  • 41) 0.494 574 08 × 2 = 0 + 0.989 148 16;
  • 42) 0.989 148 16 × 2 = 1 + 0.978 296 32;
  • 43) 0.978 296 32 × 2 = 1 + 0.956 592 64;
  • 44) 0.956 592 64 × 2 = 1 + 0.913 185 28;
  • 45) 0.913 185 28 × 2 = 1 + 0.826 370 56;
  • 46) 0.826 370 56 × 2 = 1 + 0.652 741 12;
  • 47) 0.652 741 12 × 2 = 1 + 0.305 482 24;
  • 48) 0.305 482 24 × 2 = 0 + 0.610 964 48;
  • 49) 0.610 964 48 × 2 = 1 + 0.221 928 96;
  • 50) 0.221 928 96 × 2 = 0 + 0.443 857 92;
  • 51) 0.443 857 92 × 2 = 0 + 0.887 715 84;
  • 52) 0.887 715 84 × 2 = 1 + 0.775 431 68;
  • 53) 0.775 431 68 × 2 = 1 + 0.550 863 36;

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:


0.533 333 33(10) =


0.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2)


5. Positive number before normalization:

8.533 333 33(10) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(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:


8.533 333 33(10) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2) =


1000.1000 1000 1000 1000 1000 1000 0111 1010 0011 0111 0111 1110 1001 1(2) × 20 =


1.0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011(2) × 23


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): 3


Mantissa (not normalized):
1.0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 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) =


1026(10) =


100 0000 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 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. 0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101 0011 =


0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101


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 0010


Mantissa (52 bits) =
0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101


The base ten decimal number 8.533 333 33 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0010 - 0001 0001 0001 0001 0001 0001 0000 1111 0100 0110 1110 1111 1101

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