Decimal to 64 Bit IEEE 754 Binary: Convert Number 0.666 666 666 666 666 666 62 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.666 666 666 666 666 666 62(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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: 0.666 666 666 666 666 666 62.

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) 0.666 666 666 666 666 666 62 × 2 = 1 + 0.333 333 333 333 333 333 24;
  • 2) 0.333 333 333 333 333 333 24 × 2 = 0 + 0.666 666 666 666 666 666 48;
  • 3) 0.666 666 666 666 666 666 48 × 2 = 1 + 0.333 333 333 333 333 332 96;
  • 4) 0.333 333 333 333 333 332 96 × 2 = 0 + 0.666 666 666 666 666 665 92;
  • 5) 0.666 666 666 666 666 665 92 × 2 = 1 + 0.333 333 333 333 333 331 84;
  • 6) 0.333 333 333 333 333 331 84 × 2 = 0 + 0.666 666 666 666 666 663 68;
  • 7) 0.666 666 666 666 666 663 68 × 2 = 1 + 0.333 333 333 333 333 327 36;
  • 8) 0.333 333 333 333 333 327 36 × 2 = 0 + 0.666 666 666 666 666 654 72;
  • 9) 0.666 666 666 666 666 654 72 × 2 = 1 + 0.333 333 333 333 333 309 44;
  • 10) 0.333 333 333 333 333 309 44 × 2 = 0 + 0.666 666 666 666 666 618 88;
  • 11) 0.666 666 666 666 666 618 88 × 2 = 1 + 0.333 333 333 333 333 237 76;
  • 12) 0.333 333 333 333 333 237 76 × 2 = 0 + 0.666 666 666 666 666 475 52;
  • 13) 0.666 666 666 666 666 475 52 × 2 = 1 + 0.333 333 333 333 332 951 04;
  • 14) 0.333 333 333 333 332 951 04 × 2 = 0 + 0.666 666 666 666 665 902 08;
  • 15) 0.666 666 666 666 665 902 08 × 2 = 1 + 0.333 333 333 333 331 804 16;
  • 16) 0.333 333 333 333 331 804 16 × 2 = 0 + 0.666 666 666 666 663 608 32;
  • 17) 0.666 666 666 666 663 608 32 × 2 = 1 + 0.333 333 333 333 327 216 64;
  • 18) 0.333 333 333 333 327 216 64 × 2 = 0 + 0.666 666 666 666 654 433 28;
  • 19) 0.666 666 666 666 654 433 28 × 2 = 1 + 0.333 333 333 333 308 866 56;
  • 20) 0.333 333 333 333 308 866 56 × 2 = 0 + 0.666 666 666 666 617 733 12;
  • 21) 0.666 666 666 666 617 733 12 × 2 = 1 + 0.333 333 333 333 235 466 24;
  • 22) 0.333 333 333 333 235 466 24 × 2 = 0 + 0.666 666 666 666 470 932 48;
  • 23) 0.666 666 666 666 470 932 48 × 2 = 1 + 0.333 333 333 332 941 864 96;
  • 24) 0.333 333 333 332 941 864 96 × 2 = 0 + 0.666 666 666 665 883 729 92;
  • 25) 0.666 666 666 665 883 729 92 × 2 = 1 + 0.333 333 333 331 767 459 84;
  • 26) 0.333 333 333 331 767 459 84 × 2 = 0 + 0.666 666 666 663 534 919 68;
  • 27) 0.666 666 666 663 534 919 68 × 2 = 1 + 0.333 333 333 327 069 839 36;
  • 28) 0.333 333 333 327 069 839 36 × 2 = 0 + 0.666 666 666 654 139 678 72;
  • 29) 0.666 666 666 654 139 678 72 × 2 = 1 + 0.333 333 333 308 279 357 44;
  • 30) 0.333 333 333 308 279 357 44 × 2 = 0 + 0.666 666 666 616 558 714 88;
  • 31) 0.666 666 666 616 558 714 88 × 2 = 1 + 0.333 333 333 233 117 429 76;
  • 32) 0.333 333 333 233 117 429 76 × 2 = 0 + 0.666 666 666 466 234 859 52;
  • 33) 0.666 666 666 466 234 859 52 × 2 = 1 + 0.333 333 332 932 469 719 04;
  • 34) 0.333 333 332 932 469 719 04 × 2 = 0 + 0.666 666 665 864 939 438 08;
  • 35) 0.666 666 665 864 939 438 08 × 2 = 1 + 0.333 333 331 729 878 876 16;
  • 36) 0.333 333 331 729 878 876 16 × 2 = 0 + 0.666 666 663 459 757 752 32;
  • 37) 0.666 666 663 459 757 752 32 × 2 = 1 + 0.333 333 326 919 515 504 64;
  • 38) 0.333 333 326 919 515 504 64 × 2 = 0 + 0.666 666 653 839 031 009 28;
  • 39) 0.666 666 653 839 031 009 28 × 2 = 1 + 0.333 333 307 678 062 018 56;
  • 40) 0.333 333 307 678 062 018 56 × 2 = 0 + 0.666 666 615 356 124 037 12;
  • 41) 0.666 666 615 356 124 037 12 × 2 = 1 + 0.333 333 230 712 248 074 24;
  • 42) 0.333 333 230 712 248 074 24 × 2 = 0 + 0.666 666 461 424 496 148 48;
  • 43) 0.666 666 461 424 496 148 48 × 2 = 1 + 0.333 332 922 848 992 296 96;
  • 44) 0.333 332 922 848 992 296 96 × 2 = 0 + 0.666 665 845 697 984 593 92;
  • 45) 0.666 665 845 697 984 593 92 × 2 = 1 + 0.333 331 691 395 969 187 84;
  • 46) 0.333 331 691 395 969 187 84 × 2 = 0 + 0.666 663 382 791 938 375 68;
  • 47) 0.666 663 382 791 938 375 68 × 2 = 1 + 0.333 326 765 583 876 751 36;
  • 48) 0.333 326 765 583 876 751 36 × 2 = 0 + 0.666 653 531 167 753 502 72;
  • 49) 0.666 653 531 167 753 502 72 × 2 = 1 + 0.333 307 062 335 507 005 44;
  • 50) 0.333 307 062 335 507 005 44 × 2 = 0 + 0.666 614 124 671 014 010 88;
  • 51) 0.666 614 124 671 014 010 88 × 2 = 1 + 0.333 228 249 342 028 021 76;
  • 52) 0.333 228 249 342 028 021 76 × 2 = 0 + 0.666 456 498 684 056 043 52;
  • 53) 0.666 456 498 684 056 043 52 × 2 = 1 + 0.332 912 997 368 112 087 04;

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:


0.666 666 666 666 666 666 62(10) =


0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2)

5. Positive number before normalization:

0.666 666 666 666 666 666 62(10) =


0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 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:


0.666 666 666 666 666 666 62(10) =


0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) =


0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1(2) × 20 =


1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101(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 0101 0101 0101


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 0101 0101 0101 =


0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101


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 0101 0101 0101


The base ten decimal number 0.666 666 666 666 666 666 62 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101

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