Decimal to 64 Bit IEEE 754 Binary: Convert Number 0.936 294 003 9 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.936 294 003 9(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: 0.936 294 003 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) 0.936 294 003 9 × 2 = 1 + 0.872 588 007 8;
  • 2) 0.872 588 007 8 × 2 = 1 + 0.745 176 015 6;
  • 3) 0.745 176 015 6 × 2 = 1 + 0.490 352 031 2;
  • 4) 0.490 352 031 2 × 2 = 0 + 0.980 704 062 4;
  • 5) 0.980 704 062 4 × 2 = 1 + 0.961 408 124 8;
  • 6) 0.961 408 124 8 × 2 = 1 + 0.922 816 249 6;
  • 7) 0.922 816 249 6 × 2 = 1 + 0.845 632 499 2;
  • 8) 0.845 632 499 2 × 2 = 1 + 0.691 264 998 4;
  • 9) 0.691 264 998 4 × 2 = 1 + 0.382 529 996 8;
  • 10) 0.382 529 996 8 × 2 = 0 + 0.765 059 993 6;
  • 11) 0.765 059 993 6 × 2 = 1 + 0.530 119 987 2;
  • 12) 0.530 119 987 2 × 2 = 1 + 0.060 239 974 4;
  • 13) 0.060 239 974 4 × 2 = 0 + 0.120 479 948 8;
  • 14) 0.120 479 948 8 × 2 = 0 + 0.240 959 897 6;
  • 15) 0.240 959 897 6 × 2 = 0 + 0.481 919 795 2;
  • 16) 0.481 919 795 2 × 2 = 0 + 0.963 839 590 4;
  • 17) 0.963 839 590 4 × 2 = 1 + 0.927 679 180 8;
  • 18) 0.927 679 180 8 × 2 = 1 + 0.855 358 361 6;
  • 19) 0.855 358 361 6 × 2 = 1 + 0.710 716 723 2;
  • 20) 0.710 716 723 2 × 2 = 1 + 0.421 433 446 4;
  • 21) 0.421 433 446 4 × 2 = 0 + 0.842 866 892 8;
  • 22) 0.842 866 892 8 × 2 = 1 + 0.685 733 785 6;
  • 23) 0.685 733 785 6 × 2 = 1 + 0.371 467 571 2;
  • 24) 0.371 467 571 2 × 2 = 0 + 0.742 935 142 4;
  • 25) 0.742 935 142 4 × 2 = 1 + 0.485 870 284 8;
  • 26) 0.485 870 284 8 × 2 = 0 + 0.971 740 569 6;
  • 27) 0.971 740 569 6 × 2 = 1 + 0.943 481 139 2;
  • 28) 0.943 481 139 2 × 2 = 1 + 0.886 962 278 4;
  • 29) 0.886 962 278 4 × 2 = 1 + 0.773 924 556 8;
  • 30) 0.773 924 556 8 × 2 = 1 + 0.547 849 113 6;
  • 31) 0.547 849 113 6 × 2 = 1 + 0.095 698 227 2;
  • 32) 0.095 698 227 2 × 2 = 0 + 0.191 396 454 4;
  • 33) 0.191 396 454 4 × 2 = 0 + 0.382 792 908 8;
  • 34) 0.382 792 908 8 × 2 = 0 + 0.765 585 817 6;
  • 35) 0.765 585 817 6 × 2 = 1 + 0.531 171 635 2;
  • 36) 0.531 171 635 2 × 2 = 1 + 0.062 343 270 4;
  • 37) 0.062 343 270 4 × 2 = 0 + 0.124 686 540 8;
  • 38) 0.124 686 540 8 × 2 = 0 + 0.249 373 081 6;
  • 39) 0.249 373 081 6 × 2 = 0 + 0.498 746 163 2;
  • 40) 0.498 746 163 2 × 2 = 0 + 0.997 492 326 4;
  • 41) 0.997 492 326 4 × 2 = 1 + 0.994 984 652 8;
  • 42) 0.994 984 652 8 × 2 = 1 + 0.989 969 305 6;
  • 43) 0.989 969 305 6 × 2 = 1 + 0.979 938 611 2;
  • 44) 0.979 938 611 2 × 2 = 1 + 0.959 877 222 4;
  • 45) 0.959 877 222 4 × 2 = 1 + 0.919 754 444 8;
  • 46) 0.919 754 444 8 × 2 = 1 + 0.839 508 889 6;
  • 47) 0.839 508 889 6 × 2 = 1 + 0.679 017 779 2;
  • 48) 0.679 017 779 2 × 2 = 1 + 0.358 035 558 4;
  • 49) 0.358 035 558 4 × 2 = 0 + 0.716 071 116 8;
  • 50) 0.716 071 116 8 × 2 = 1 + 0.432 142 233 6;
  • 51) 0.432 142 233 6 × 2 = 0 + 0.864 284 467 2;
  • 52) 0.864 284 467 2 × 2 = 1 + 0.728 568 934 4;
  • 53) 0.728 568 934 4 × 2 = 1 + 0.457 137 868 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:


0.936 294 003 9(10) =

0.1110 1111 1011 0000 1111 0110 1011 1110 0011 0000 1111 1111 0101 1(2)


5. Positive number before normalization:

0.936 294 003 9(10) =

0.1110 1111 1011 0000 1111 0110 1011 1110 0011 0000 1111 1111 0101 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:


0.936 294 003 9(10) =

0.1110 1111 1011 0000 1111 0110 1011 1110 0011 0000 1111 1111 0101 1(2) =

0.1110 1111 1011 0000 1111 0110 1011 1110 0011 0000 1111 1111 0101 1(2) × 20 =

1.1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011(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):
1.1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011


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. 1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011 =

1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011


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) =
1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011


The base ten decimal number 0.936 294 003 9 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1110 - 1101 1111 0110 0001 1110 1101 0111 1100 0110 0001 1111 1110 1011

» Decimal number in base ten 0.936 293 997 1 converted and written as 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