Convert 0.000 569 761 78 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.000 569 761 78(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 569 761 78.

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 78 × 2 = 0 + 0.001 139 523 56;
  • 2) 0.001 139 523 56 × 2 = 0 + 0.002 279 047 12;
  • 3) 0.002 279 047 12 × 2 = 0 + 0.004 558 094 24;
  • 4) 0.004 558 094 24 × 2 = 0 + 0.009 116 188 48;
  • 5) 0.009 116 188 48 × 2 = 0 + 0.018 232 376 96;
  • 6) 0.018 232 376 96 × 2 = 0 + 0.036 464 753 92;
  • 7) 0.036 464 753 92 × 2 = 0 + 0.072 929 507 84;
  • 8) 0.072 929 507 84 × 2 = 0 + 0.145 859 015 68;
  • 9) 0.145 859 015 68 × 2 = 0 + 0.291 718 031 36;
  • 10) 0.291 718 031 36 × 2 = 0 + 0.583 436 062 72;
  • 11) 0.583 436 062 72 × 2 = 1 + 0.166 872 125 44;
  • 12) 0.166 872 125 44 × 2 = 0 + 0.333 744 250 88;
  • 13) 0.333 744 250 88 × 2 = 0 + 0.667 488 501 76;
  • 14) 0.667 488 501 76 × 2 = 1 + 0.334 977 003 52;
  • 15) 0.334 977 003 52 × 2 = 0 + 0.669 954 007 04;
  • 16) 0.669 954 007 04 × 2 = 1 + 0.339 908 014 08;
  • 17) 0.339 908 014 08 × 2 = 0 + 0.679 816 028 16;
  • 18) 0.679 816 028 16 × 2 = 1 + 0.359 632 056 32;
  • 19) 0.359 632 056 32 × 2 = 0 + 0.719 264 112 64;
  • 20) 0.719 264 112 64 × 2 = 1 + 0.438 528 225 28;
  • 21) 0.438 528 225 28 × 2 = 0 + 0.877 056 450 56;
  • 22) 0.877 056 450 56 × 2 = 1 + 0.754 112 901 12;
  • 23) 0.754 112 901 12 × 2 = 1 + 0.508 225 802 24;
  • 24) 0.508 225 802 24 × 2 = 1 + 0.016 451 604 48;
  • 25) 0.016 451 604 48 × 2 = 0 + 0.032 903 208 96;
  • 26) 0.032 903 208 96 × 2 = 0 + 0.065 806 417 92;
  • 27) 0.065 806 417 92 × 2 = 0 + 0.131 612 835 84;
  • 28) 0.131 612 835 84 × 2 = 0 + 0.263 225 671 68;
  • 29) 0.263 225 671 68 × 2 = 0 + 0.526 451 343 36;
  • 30) 0.526 451 343 36 × 2 = 1 + 0.052 902 686 72;
  • 31) 0.052 902 686 72 × 2 = 0 + 0.105 805 373 44;
  • 32) 0.105 805 373 44 × 2 = 0 + 0.211 610 746 88;
  • 33) 0.211 610 746 88 × 2 = 0 + 0.423 221 493 76;
  • 34) 0.423 221 493 76 × 2 = 0 + 0.846 442 987 52;
  • 35) 0.846 442 987 52 × 2 = 1 + 0.692 885 975 04;
  • 36) 0.692 885 975 04 × 2 = 1 + 0.385 771 950 08;
  • 37) 0.385 771 950 08 × 2 = 0 + 0.771 543 900 16;
  • 38) 0.771 543 900 16 × 2 = 1 + 0.543 087 800 32;
  • 39) 0.543 087 800 32 × 2 = 1 + 0.086 175 600 64;
  • 40) 0.086 175 600 64 × 2 = 0 + 0.172 351 201 28;
  • 41) 0.172 351 201 28 × 2 = 0 + 0.344 702 402 56;
  • 42) 0.344 702 402 56 × 2 = 0 + 0.689 404 805 12;
  • 43) 0.689 404 805 12 × 2 = 1 + 0.378 809 610 24;
  • 44) 0.378 809 610 24 × 2 = 0 + 0.757 619 220 48;
  • 45) 0.757 619 220 48 × 2 = 1 + 0.515 238 440 96;
  • 46) 0.515 238 440 96 × 2 = 1 + 0.030 476 881 92;
  • 47) 0.030 476 881 92 × 2 = 0 + 0.060 953 763 84;
  • 48) 0.060 953 763 84 × 2 = 0 + 0.121 907 527 68;
  • 49) 0.121 907 527 68 × 2 = 0 + 0.243 815 055 36;
  • 50) 0.243 815 055 36 × 2 = 0 + 0.487 630 110 72;
  • 51) 0.487 630 110 72 × 2 = 0 + 0.975 260 221 44;
  • 52) 0.975 260 221 44 × 2 = 1 + 0.950 520 442 88;
  • 53) 0.950 520 442 88 × 2 = 1 + 0.901 040 885 76;
  • 54) 0.901 040 885 76 × 2 = 1 + 0.802 081 771 52;
  • 55) 0.802 081 771 52 × 2 = 1 + 0.604 163 543 04;
  • 56) 0.604 163 543 04 × 2 = 1 + 0.208 327 086 08;
  • 57) 0.208 327 086 08 × 2 = 0 + 0.416 654 172 16;
  • 58) 0.416 654 172 16 × 2 = 0 + 0.833 308 344 32;
  • 59) 0.833 308 344 32 × 2 = 1 + 0.666 616 688 64;
  • 60) 0.666 616 688 64 × 2 = 1 + 0.333 233 377 28;
  • 61) 0.333 233 377 28 × 2 = 0 + 0.666 466 754 56;
  • 62) 0.666 466 754 56 × 2 = 1 + 0.332 933 509 12;
  • 63) 0.332 933 509 12 × 2 = 0 + 0.665 867 018 24;

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 78(10) =


0.0000 0000 0010 0101 0101 0111 0000 0100 0011 0110 0010 1100 0001 1111 0011 010(2)


5. Positive number before normalization:

0.000 569 761 78(10) =


0.0000 0000 0010 0101 0101 0111 0000 0100 0011 0110 0010 1100 0001 1111 0011 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 78(10) =


0.0000 0000 0010 0101 0101 0111 0000 0100 0011 0110 0010 1100 0001 1111 0011 010(2) =


0.0000 0000 0010 0101 0101 0111 0000 0100 0011 0110 0010 1100 0001 1111 0011 010(2) × 20 =


1.0010 1010 1011 1000 0010 0001 1011 0001 0110 0000 1111 1001 1010(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 1011 0001 0110 0000 1111 1001 1010


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 1011 0001 0110 0000 1111 1001 1010 =


0010 1010 1011 1000 0010 0001 1011 0001 0110 0000 1111 1001 1010


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 1011 0001 0110 0000 1111 1001 1010


Number 0.000 569 761 78 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 0100 - 0010 1010 1011 1000 0010 0001 1011 0001 0110 0000 1111 1001 1010

(64 bits IEEE 754)

More operations of this kind:

0.000 569 761 77 = ? ... 0.000 569 761 79 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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