0.555 555 555 555 555 582 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.555 555 555 555 555 582 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
0.555 555 555 555 555 582 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: 0.555 555 555 555 555 582 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) 0.555 555 555 555 555 582 9 × 2 = 1 + 0.111 111 111 111 111 165 8;
  • 2) 0.111 111 111 111 111 165 8 × 2 = 0 + 0.222 222 222 222 222 331 6;
  • 3) 0.222 222 222 222 222 331 6 × 2 = 0 + 0.444 444 444 444 444 663 2;
  • 4) 0.444 444 444 444 444 663 2 × 2 = 0 + 0.888 888 888 888 889 326 4;
  • 5) 0.888 888 888 888 889 326 4 × 2 = 1 + 0.777 777 777 777 778 652 8;
  • 6) 0.777 777 777 777 778 652 8 × 2 = 1 + 0.555 555 555 555 557 305 6;
  • 7) 0.555 555 555 555 557 305 6 × 2 = 1 + 0.111 111 111 111 114 611 2;
  • 8) 0.111 111 111 111 114 611 2 × 2 = 0 + 0.222 222 222 222 229 222 4;
  • 9) 0.222 222 222 222 229 222 4 × 2 = 0 + 0.444 444 444 444 458 444 8;
  • 10) 0.444 444 444 444 458 444 8 × 2 = 0 + 0.888 888 888 888 916 889 6;
  • 11) 0.888 888 888 888 916 889 6 × 2 = 1 + 0.777 777 777 777 833 779 2;
  • 12) 0.777 777 777 777 833 779 2 × 2 = 1 + 0.555 555 555 555 667 558 4;
  • 13) 0.555 555 555 555 667 558 4 × 2 = 1 + 0.111 111 111 111 335 116 8;
  • 14) 0.111 111 111 111 335 116 8 × 2 = 0 + 0.222 222 222 222 670 233 6;
  • 15) 0.222 222 222 222 670 233 6 × 2 = 0 + 0.444 444 444 445 340 467 2;
  • 16) 0.444 444 444 445 340 467 2 × 2 = 0 + 0.888 888 888 890 680 934 4;
  • 17) 0.888 888 888 890 680 934 4 × 2 = 1 + 0.777 777 777 781 361 868 8;
  • 18) 0.777 777 777 781 361 868 8 × 2 = 1 + 0.555 555 555 562 723 737 6;
  • 19) 0.555 555 555 562 723 737 6 × 2 = 1 + 0.111 111 111 125 447 475 2;
  • 20) 0.111 111 111 125 447 475 2 × 2 = 0 + 0.222 222 222 250 894 950 4;
  • 21) 0.222 222 222 250 894 950 4 × 2 = 0 + 0.444 444 444 501 789 900 8;
  • 22) 0.444 444 444 501 789 900 8 × 2 = 0 + 0.888 888 889 003 579 801 6;
  • 23) 0.888 888 889 003 579 801 6 × 2 = 1 + 0.777 777 778 007 159 603 2;
  • 24) 0.777 777 778 007 159 603 2 × 2 = 1 + 0.555 555 556 014 319 206 4;
  • 25) 0.555 555 556 014 319 206 4 × 2 = 1 + 0.111 111 112 028 638 412 8;
  • 26) 0.111 111 112 028 638 412 8 × 2 = 0 + 0.222 222 224 057 276 825 6;
  • 27) 0.222 222 224 057 276 825 6 × 2 = 0 + 0.444 444 448 114 553 651 2;
  • 28) 0.444 444 448 114 553 651 2 × 2 = 0 + 0.888 888 896 229 107 302 4;
  • 29) 0.888 888 896 229 107 302 4 × 2 = 1 + 0.777 777 792 458 214 604 8;
  • 30) 0.777 777 792 458 214 604 8 × 2 = 1 + 0.555 555 584 916 429 209 6;
  • 31) 0.555 555 584 916 429 209 6 × 2 = 1 + 0.111 111 169 832 858 419 2;
  • 32) 0.111 111 169 832 858 419 2 × 2 = 0 + 0.222 222 339 665 716 838 4;
  • 33) 0.222 222 339 665 716 838 4 × 2 = 0 + 0.444 444 679 331 433 676 8;
  • 34) 0.444 444 679 331 433 676 8 × 2 = 0 + 0.888 889 358 662 867 353 6;
  • 35) 0.888 889 358 662 867 353 6 × 2 = 1 + 0.777 778 717 325 734 707 2;
  • 36) 0.777 778 717 325 734 707 2 × 2 = 1 + 0.555 557 434 651 469 414 4;
  • 37) 0.555 557 434 651 469 414 4 × 2 = 1 + 0.111 114 869 302 938 828 8;
  • 38) 0.111 114 869 302 938 828 8 × 2 = 0 + 0.222 229 738 605 877 657 6;
  • 39) 0.222 229 738 605 877 657 6 × 2 = 0 + 0.444 459 477 211 755 315 2;
  • 40) 0.444 459 477 211 755 315 2 × 2 = 0 + 0.888 918 954 423 510 630 4;
  • 41) 0.888 918 954 423 510 630 4 × 2 = 1 + 0.777 837 908 847 021 260 8;
  • 42) 0.777 837 908 847 021 260 8 × 2 = 1 + 0.555 675 817 694 042 521 6;
  • 43) 0.555 675 817 694 042 521 6 × 2 = 1 + 0.111 351 635 388 085 043 2;
  • 44) 0.111 351 635 388 085 043 2 × 2 = 0 + 0.222 703 270 776 170 086 4;
  • 45) 0.222 703 270 776 170 086 4 × 2 = 0 + 0.445 406 541 552 340 172 8;
  • 46) 0.445 406 541 552 340 172 8 × 2 = 0 + 0.890 813 083 104 680 345 6;
  • 47) 0.890 813 083 104 680 345 6 × 2 = 1 + 0.781 626 166 209 360 691 2;
  • 48) 0.781 626 166 209 360 691 2 × 2 = 1 + 0.563 252 332 418 721 382 4;
  • 49) 0.563 252 332 418 721 382 4 × 2 = 1 + 0.126 504 664 837 442 764 8;
  • 50) 0.126 504 664 837 442 764 8 × 2 = 0 + 0.253 009 329 674 885 529 6;
  • 51) 0.253 009 329 674 885 529 6 × 2 = 0 + 0.506 018 659 349 771 059 2;
  • 52) 0.506 018 659 349 771 059 2 × 2 = 1 + 0.012 037 318 699 542 118 4;
  • 53) 0.012 037 318 699 542 118 4 × 2 = 0 + 0.024 074 637 399 084 236 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:


0.555 555 555 555 555 582 9(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 0(2)

5. Positive number before normalization:

0.555 555 555 555 555 582 9(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 0(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:


0.555 555 555 555 555 582 9(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 0(2) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1001 0(2) × 20 =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010(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):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010


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. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010


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) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010


Decimal number 0.555 555 555 555 555 582 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010

» Convert decimal 0.555 555 555 555 555 592 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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