0.666 666 666 666 593 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.666 666 666 666 593(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.666 666 666 666 593(10) to 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: 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: 0.666 666 666 666 593.

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) 0.666 666 666 666 593 × 2 = 1 + 0.333 333 333 333 186;
  • 2) 0.333 333 333 333 186 × 2 = 0 + 0.666 666 666 666 372;
  • 3) 0.666 666 666 666 372 × 2 = 1 + 0.333 333 333 332 744;
  • 4) 0.333 333 333 332 744 × 2 = 0 + 0.666 666 666 665 488;
  • 5) 0.666 666 666 665 488 × 2 = 1 + 0.333 333 333 330 976;
  • 6) 0.333 333 333 330 976 × 2 = 0 + 0.666 666 666 661 952;
  • 7) 0.666 666 666 661 952 × 2 = 1 + 0.333 333 333 323 904;
  • 8) 0.333 333 333 323 904 × 2 = 0 + 0.666 666 666 647 808;
  • 9) 0.666 666 666 647 808 × 2 = 1 + 0.333 333 333 295 616;
  • 10) 0.333 333 333 295 616 × 2 = 0 + 0.666 666 666 591 232;
  • 11) 0.666 666 666 591 232 × 2 = 1 + 0.333 333 333 182 464;
  • 12) 0.333 333 333 182 464 × 2 = 0 + 0.666 666 666 364 928;
  • 13) 0.666 666 666 364 928 × 2 = 1 + 0.333 333 332 729 856;
  • 14) 0.333 333 332 729 856 × 2 = 0 + 0.666 666 665 459 712;
  • 15) 0.666 666 665 459 712 × 2 = 1 + 0.333 333 330 919 424;
  • 16) 0.333 333 330 919 424 × 2 = 0 + 0.666 666 661 838 848;
  • 17) 0.666 666 661 838 848 × 2 = 1 + 0.333 333 323 677 696;
  • 18) 0.333 333 323 677 696 × 2 = 0 + 0.666 666 647 355 392;
  • 19) 0.666 666 647 355 392 × 2 = 1 + 0.333 333 294 710 784;
  • 20) 0.333 333 294 710 784 × 2 = 0 + 0.666 666 589 421 568;
  • 21) 0.666 666 589 421 568 × 2 = 1 + 0.333 333 178 843 136;
  • 22) 0.333 333 178 843 136 × 2 = 0 + 0.666 666 357 686 272;
  • 23) 0.666 666 357 686 272 × 2 = 1 + 0.333 332 715 372 544;
  • 24) 0.333 332 715 372 544 × 2 = 0 + 0.666 665 430 745 088;
  • 25) 0.666 665 430 745 088 × 2 = 1 + 0.333 330 861 490 176;
  • 26) 0.333 330 861 490 176 × 2 = 0 + 0.666 661 722 980 352;
  • 27) 0.666 661 722 980 352 × 2 = 1 + 0.333 323 445 960 704;
  • 28) 0.333 323 445 960 704 × 2 = 0 + 0.666 646 891 921 408;
  • 29) 0.666 646 891 921 408 × 2 = 1 + 0.333 293 783 842 816;
  • 30) 0.333 293 783 842 816 × 2 = 0 + 0.666 587 567 685 632;
  • 31) 0.666 587 567 685 632 × 2 = 1 + 0.333 175 135 371 264;
  • 32) 0.333 175 135 371 264 × 2 = 0 + 0.666 350 270 742 528;
  • 33) 0.666 350 270 742 528 × 2 = 1 + 0.332 700 541 485 056;
  • 34) 0.332 700 541 485 056 × 2 = 0 + 0.665 401 082 970 112;
  • 35) 0.665 401 082 970 112 × 2 = 1 + 0.330 802 165 940 224;
  • 36) 0.330 802 165 940 224 × 2 = 0 + 0.661 604 331 880 448;
  • 37) 0.661 604 331 880 448 × 2 = 1 + 0.323 208 663 760 896;
  • 38) 0.323 208 663 760 896 × 2 = 0 + 0.646 417 327 521 792;
  • 39) 0.646 417 327 521 792 × 2 = 1 + 0.292 834 655 043 584;
  • 40) 0.292 834 655 043 584 × 2 = 0 + 0.585 669 310 087 168;
  • 41) 0.585 669 310 087 168 × 2 = 1 + 0.171 338 620 174 336;
  • 42) 0.171 338 620 174 336 × 2 = 0 + 0.342 677 240 348 672;
  • 43) 0.342 677 240 348 672 × 2 = 0 + 0.685 354 480 697 344;
  • 44) 0.685 354 480 697 344 × 2 = 1 + 0.370 708 961 394 688;
  • 45) 0.370 708 961 394 688 × 2 = 0 + 0.741 417 922 789 376;
  • 46) 0.741 417 922 789 376 × 2 = 1 + 0.482 835 845 578 752;
  • 47) 0.482 835 845 578 752 × 2 = 0 + 0.965 671 691 157 504;
  • 48) 0.965 671 691 157 504 × 2 = 1 + 0.931 343 382 315 008;
  • 49) 0.931 343 382 315 008 × 2 = 1 + 0.862 686 764 630 016;
  • 50) 0.862 686 764 630 016 × 2 = 1 + 0.725 373 529 260 032;
  • 51) 0.725 373 529 260 032 × 2 = 1 + 0.450 747 058 520 064;
  • 52) 0.450 747 058 520 064 × 2 = 0 + 0.901 494 117 040 128;
  • 53) 0.901 494 117 040 128 × 2 = 1 + 0.802 988 234 080 256;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.666 666 666 666 593(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1001 0101 1110 1(2)

5. Positive number before normalization:

0.666 666 666 666 593(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1001 0101 1110 1(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:


0.666 666 666 666 593(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1001 0101 1110 1(2) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1001 0101 1110 1(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 1101(2) × 2-1


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): -1

Mantissa (not normalized):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 1101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1022(10) =

011 1111 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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 1101 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 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) =
011 1111 1110

Mantissa (52 bits) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 1101


Decimal number 0.666 666 666 666 593 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0010 1011 1101

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. 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: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100