64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.000 569 761 77 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 569 761 77(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: 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: 0.000 569 761 77.

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) 0.000 569 761 77 × 2 = 0 + 0.001 139 523 54;
  • 2) 0.001 139 523 54 × 2 = 0 + 0.002 279 047 08;
  • 3) 0.002 279 047 08 × 2 = 0 + 0.004 558 094 16;
  • 4) 0.004 558 094 16 × 2 = 0 + 0.009 116 188 32;
  • 5) 0.009 116 188 32 × 2 = 0 + 0.018 232 376 64;
  • 6) 0.018 232 376 64 × 2 = 0 + 0.036 464 753 28;
  • 7) 0.036 464 753 28 × 2 = 0 + 0.072 929 506 56;
  • 8) 0.072 929 506 56 × 2 = 0 + 0.145 859 013 12;
  • 9) 0.145 859 013 12 × 2 = 0 + 0.291 718 026 24;
  • 10) 0.291 718 026 24 × 2 = 0 + 0.583 436 052 48;
  • 11) 0.583 436 052 48 × 2 = 1 + 0.166 872 104 96;
  • 12) 0.166 872 104 96 × 2 = 0 + 0.333 744 209 92;
  • 13) 0.333 744 209 92 × 2 = 0 + 0.667 488 419 84;
  • 14) 0.667 488 419 84 × 2 = 1 + 0.334 976 839 68;
  • 15) 0.334 976 839 68 × 2 = 0 + 0.669 953 679 36;
  • 16) 0.669 953 679 36 × 2 = 1 + 0.339 907 358 72;
  • 17) 0.339 907 358 72 × 2 = 0 + 0.679 814 717 44;
  • 18) 0.679 814 717 44 × 2 = 1 + 0.359 629 434 88;
  • 19) 0.359 629 434 88 × 2 = 0 + 0.719 258 869 76;
  • 20) 0.719 258 869 76 × 2 = 1 + 0.438 517 739 52;
  • 21) 0.438 517 739 52 × 2 = 0 + 0.877 035 479 04;
  • 22) 0.877 035 479 04 × 2 = 1 + 0.754 070 958 08;
  • 23) 0.754 070 958 08 × 2 = 1 + 0.508 141 916 16;
  • 24) 0.508 141 916 16 × 2 = 1 + 0.016 283 832 32;
  • 25) 0.016 283 832 32 × 2 = 0 + 0.032 567 664 64;
  • 26) 0.032 567 664 64 × 2 = 0 + 0.065 135 329 28;
  • 27) 0.065 135 329 28 × 2 = 0 + 0.130 270 658 56;
  • 28) 0.130 270 658 56 × 2 = 0 + 0.260 541 317 12;
  • 29) 0.260 541 317 12 × 2 = 0 + 0.521 082 634 24;
  • 30) 0.521 082 634 24 × 2 = 1 + 0.042 165 268 48;
  • 31) 0.042 165 268 48 × 2 = 0 + 0.084 330 536 96;
  • 32) 0.084 330 536 96 × 2 = 0 + 0.168 661 073 92;
  • 33) 0.168 661 073 92 × 2 = 0 + 0.337 322 147 84;
  • 34) 0.337 322 147 84 × 2 = 0 + 0.674 644 295 68;
  • 35) 0.674 644 295 68 × 2 = 1 + 0.349 288 591 36;
  • 36) 0.349 288 591 36 × 2 = 0 + 0.698 577 182 72;
  • 37) 0.698 577 182 72 × 2 = 1 + 0.397 154 365 44;
  • 38) 0.397 154 365 44 × 2 = 0 + 0.794 308 730 88;
  • 39) 0.794 308 730 88 × 2 = 1 + 0.588 617 461 76;
  • 40) 0.588 617 461 76 × 2 = 1 + 0.177 234 923 52;
  • 41) 0.177 234 923 52 × 2 = 0 + 0.354 469 847 04;
  • 42) 0.354 469 847 04 × 2 = 0 + 0.708 939 694 08;
  • 43) 0.708 939 694 08 × 2 = 1 + 0.417 879 388 16;
  • 44) 0.417 879 388 16 × 2 = 0 + 0.835 758 776 32;
  • 45) 0.835 758 776 32 × 2 = 1 + 0.671 517 552 64;
  • 46) 0.671 517 552 64 × 2 = 1 + 0.343 035 105 28;
  • 47) 0.343 035 105 28 × 2 = 0 + 0.686 070 210 56;
  • 48) 0.686 070 210 56 × 2 = 1 + 0.372 140 421 12;
  • 49) 0.372 140 421 12 × 2 = 0 + 0.744 280 842 24;
  • 50) 0.744 280 842 24 × 2 = 1 + 0.488 561 684 48;
  • 51) 0.488 561 684 48 × 2 = 0 + 0.977 123 368 96;
  • 52) 0.977 123 368 96 × 2 = 1 + 0.954 246 737 92;
  • 53) 0.954 246 737 92 × 2 = 1 + 0.908 493 475 84;
  • 54) 0.908 493 475 84 × 2 = 1 + 0.816 986 951 68;
  • 55) 0.816 986 951 68 × 2 = 1 + 0.633 973 903 36;
  • 56) 0.633 973 903 36 × 2 = 1 + 0.267 947 806 72;
  • 57) 0.267 947 806 72 × 2 = 0 + 0.535 895 613 44;
  • 58) 0.535 895 613 44 × 2 = 1 + 0.071 791 226 88;
  • 59) 0.071 791 226 88 × 2 = 0 + 0.143 582 453 76;
  • 60) 0.143 582 453 76 × 2 = 0 + 0.287 164 907 52;
  • 61) 0.287 164 907 52 × 2 = 0 + 0.574 329 815 04;
  • 62) 0.574 329 815 04 × 2 = 1 + 0.148 659 630 08;
  • 63) 0.148 659 630 08 × 2 = 0 + 0.297 319 260 16;

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:


0.000 569 761 77(10) =

0.0000 0000 0010 0101 0101 0111 0000 0100 0010 1011 0010 1101 0101 1111 0100 010(2)


5. Positive number before normalization:

0.000 569 761 77(10) =

0.0000 0000 0010 0101 0101 0111 0000 0100 0010 1011 0010 1101 0101 1111 0100 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 569 761 77(10) =

0.0000 0000 0010 0101 0101 0111 0000 0100 0010 1011 0010 1101 0101 1111 0100 010(2) =

0.0000 0000 0010 0101 0101 0111 0000 0100 0010 1011 0010 1101 0101 1111 0100 010(2) × 20 =

1.0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 0010(2) × 2-11


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): -11

Mantissa (not normalized):
1.0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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. 0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 0010 =

0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 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 0100

Mantissa (52 bits) =
0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 0010


The base ten decimal number 0.000 569 761 77 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 0100 - 0010 1010 1011 1000 0010 0001 0101 1001 0110 1010 1111 1010 0010

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.000 569 761 76 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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