0.666 666 666 666 680 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.666 666 666 666 680 9(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
0.666 666 666 666 680 9(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: 0.666 666 666 666 680 9.

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) 0.666 666 666 666 680 9 × 2 = 1 + 0.333 333 333 333 361 8;
  • 2) 0.333 333 333 333 361 8 × 2 = 0 + 0.666 666 666 666 723 6;
  • 3) 0.666 666 666 666 723 6 × 2 = 1 + 0.333 333 333 333 447 2;
  • 4) 0.333 333 333 333 447 2 × 2 = 0 + 0.666 666 666 666 894 4;
  • 5) 0.666 666 666 666 894 4 × 2 = 1 + 0.333 333 333 333 788 8;
  • 6) 0.333 333 333 333 788 8 × 2 = 0 + 0.666 666 666 667 577 6;
  • 7) 0.666 666 666 667 577 6 × 2 = 1 + 0.333 333 333 335 155 2;
  • 8) 0.333 333 333 335 155 2 × 2 = 0 + 0.666 666 666 670 310 4;
  • 9) 0.666 666 666 670 310 4 × 2 = 1 + 0.333 333 333 340 620 8;
  • 10) 0.333 333 333 340 620 8 × 2 = 0 + 0.666 666 666 681 241 6;
  • 11) 0.666 666 666 681 241 6 × 2 = 1 + 0.333 333 333 362 483 2;
  • 12) 0.333 333 333 362 483 2 × 2 = 0 + 0.666 666 666 724 966 4;
  • 13) 0.666 666 666 724 966 4 × 2 = 1 + 0.333 333 333 449 932 8;
  • 14) 0.333 333 333 449 932 8 × 2 = 0 + 0.666 666 666 899 865 6;
  • 15) 0.666 666 666 899 865 6 × 2 = 1 + 0.333 333 333 799 731 2;
  • 16) 0.333 333 333 799 731 2 × 2 = 0 + 0.666 666 667 599 462 4;
  • 17) 0.666 666 667 599 462 4 × 2 = 1 + 0.333 333 335 198 924 8;
  • 18) 0.333 333 335 198 924 8 × 2 = 0 + 0.666 666 670 397 849 6;
  • 19) 0.666 666 670 397 849 6 × 2 = 1 + 0.333 333 340 795 699 2;
  • 20) 0.333 333 340 795 699 2 × 2 = 0 + 0.666 666 681 591 398 4;
  • 21) 0.666 666 681 591 398 4 × 2 = 1 + 0.333 333 363 182 796 8;
  • 22) 0.333 333 363 182 796 8 × 2 = 0 + 0.666 666 726 365 593 6;
  • 23) 0.666 666 726 365 593 6 × 2 = 1 + 0.333 333 452 731 187 2;
  • 24) 0.333 333 452 731 187 2 × 2 = 0 + 0.666 666 905 462 374 4;
  • 25) 0.666 666 905 462 374 4 × 2 = 1 + 0.333 333 810 924 748 8;
  • 26) 0.333 333 810 924 748 8 × 2 = 0 + 0.666 667 621 849 497 6;
  • 27) 0.666 667 621 849 497 6 × 2 = 1 + 0.333 335 243 698 995 2;
  • 28) 0.333 335 243 698 995 2 × 2 = 0 + 0.666 670 487 397 990 4;
  • 29) 0.666 670 487 397 990 4 × 2 = 1 + 0.333 340 974 795 980 8;
  • 30) 0.333 340 974 795 980 8 × 2 = 0 + 0.666 681 949 591 961 6;
  • 31) 0.666 681 949 591 961 6 × 2 = 1 + 0.333 363 899 183 923 2;
  • 32) 0.333 363 899 183 923 2 × 2 = 0 + 0.666 727 798 367 846 4;
  • 33) 0.666 727 798 367 846 4 × 2 = 1 + 0.333 455 596 735 692 8;
  • 34) 0.333 455 596 735 692 8 × 2 = 0 + 0.666 911 193 471 385 6;
  • 35) 0.666 911 193 471 385 6 × 2 = 1 + 0.333 822 386 942 771 2;
  • 36) 0.333 822 386 942 771 2 × 2 = 0 + 0.667 644 773 885 542 4;
  • 37) 0.667 644 773 885 542 4 × 2 = 1 + 0.335 289 547 771 084 8;
  • 38) 0.335 289 547 771 084 8 × 2 = 0 + 0.670 579 095 542 169 6;
  • 39) 0.670 579 095 542 169 6 × 2 = 1 + 0.341 158 191 084 339 2;
  • 40) 0.341 158 191 084 339 2 × 2 = 0 + 0.682 316 382 168 678 4;
  • 41) 0.682 316 382 168 678 4 × 2 = 1 + 0.364 632 764 337 356 8;
  • 42) 0.364 632 764 337 356 8 × 2 = 0 + 0.729 265 528 674 713 6;
  • 43) 0.729 265 528 674 713 6 × 2 = 1 + 0.458 531 057 349 427 2;
  • 44) 0.458 531 057 349 427 2 × 2 = 0 + 0.917 062 114 698 854 4;
  • 45) 0.917 062 114 698 854 4 × 2 = 1 + 0.834 124 229 397 708 8;
  • 46) 0.834 124 229 397 708 8 × 2 = 1 + 0.668 248 458 795 417 6;
  • 47) 0.668 248 458 795 417 6 × 2 = 1 + 0.336 496 917 590 835 2;
  • 48) 0.336 496 917 590 835 2 × 2 = 0 + 0.672 993 835 181 670 4;
  • 49) 0.672 993 835 181 670 4 × 2 = 1 + 0.345 987 670 363 340 8;
  • 50) 0.345 987 670 363 340 8 × 2 = 0 + 0.691 975 340 726 681 6;
  • 51) 0.691 975 340 726 681 6 × 2 = 1 + 0.383 950 681 453 363 2;
  • 52) 0.383 950 681 453 363 2 × 2 = 0 + 0.767 901 362 906 726 4;
  • 53) 0.767 901 362 906 726 4 × 2 = 1 + 0.535 802 725 813 452 8;

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:


0.666 666 666 666 680 9(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1110 1010 1(2)

5. Positive number before normalization:

0.666 666 666 666 680 9(10) =

0.1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1110 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:


0.666 666 666 666 680 9(10) =

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

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

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1101 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 1101 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 1101 0101 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 1101 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 1101 0101


Decimal number 0.666 666 666 666 680 9 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 0101 1101 0101

» Convert decimal 0.666 666 666 666 678 6 to 64 bit double precision IEEE 754 binary floating point representation standard

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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