0.000 003 669 410 54 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 003 669 410 54(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
0.000 003 669 410 54(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: 0.000 003 669 410 54.

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) 0.000 003 669 410 54 × 2 = 0 + 0.000 007 338 821 08;
  • 2) 0.000 007 338 821 08 × 2 = 0 + 0.000 014 677 642 16;
  • 3) 0.000 014 677 642 16 × 2 = 0 + 0.000 029 355 284 32;
  • 4) 0.000 029 355 284 32 × 2 = 0 + 0.000 058 710 568 64;
  • 5) 0.000 058 710 568 64 × 2 = 0 + 0.000 117 421 137 28;
  • 6) 0.000 117 421 137 28 × 2 = 0 + 0.000 234 842 274 56;
  • 7) 0.000 234 842 274 56 × 2 = 0 + 0.000 469 684 549 12;
  • 8) 0.000 469 684 549 12 × 2 = 0 + 0.000 939 369 098 24;
  • 9) 0.000 939 369 098 24 × 2 = 0 + 0.001 878 738 196 48;
  • 10) 0.001 878 738 196 48 × 2 = 0 + 0.003 757 476 392 96;
  • 11) 0.003 757 476 392 96 × 2 = 0 + 0.007 514 952 785 92;
  • 12) 0.007 514 952 785 92 × 2 = 0 + 0.015 029 905 571 84;
  • 13) 0.015 029 905 571 84 × 2 = 0 + 0.030 059 811 143 68;
  • 14) 0.030 059 811 143 68 × 2 = 0 + 0.060 119 622 287 36;
  • 15) 0.060 119 622 287 36 × 2 = 0 + 0.120 239 244 574 72;
  • 16) 0.120 239 244 574 72 × 2 = 0 + 0.240 478 489 149 44;
  • 17) 0.240 478 489 149 44 × 2 = 0 + 0.480 956 978 298 88;
  • 18) 0.480 956 978 298 88 × 2 = 0 + 0.961 913 956 597 76;
  • 19) 0.961 913 956 597 76 × 2 = 1 + 0.923 827 913 195 52;
  • 20) 0.923 827 913 195 52 × 2 = 1 + 0.847 655 826 391 04;
  • 21) 0.847 655 826 391 04 × 2 = 1 + 0.695 311 652 782 08;
  • 22) 0.695 311 652 782 08 × 2 = 1 + 0.390 623 305 564 16;
  • 23) 0.390 623 305 564 16 × 2 = 0 + 0.781 246 611 128 32;
  • 24) 0.781 246 611 128 32 × 2 = 1 + 0.562 493 222 256 64;
  • 25) 0.562 493 222 256 64 × 2 = 1 + 0.124 986 444 513 28;
  • 26) 0.124 986 444 513 28 × 2 = 0 + 0.249 972 889 026 56;
  • 27) 0.249 972 889 026 56 × 2 = 0 + 0.499 945 778 053 12;
  • 28) 0.499 945 778 053 12 × 2 = 0 + 0.999 891 556 106 24;
  • 29) 0.999 891 556 106 24 × 2 = 1 + 0.999 783 112 212 48;
  • 30) 0.999 783 112 212 48 × 2 = 1 + 0.999 566 224 424 96;
  • 31) 0.999 566 224 424 96 × 2 = 1 + 0.999 132 448 849 92;
  • 32) 0.999 132 448 849 92 × 2 = 1 + 0.998 264 897 699 84;
  • 33) 0.998 264 897 699 84 × 2 = 1 + 0.996 529 795 399 68;
  • 34) 0.996 529 795 399 68 × 2 = 1 + 0.993 059 590 799 36;
  • 35) 0.993 059 590 799 36 × 2 = 1 + 0.986 119 181 598 72;
  • 36) 0.986 119 181 598 72 × 2 = 1 + 0.972 238 363 197 44;
  • 37) 0.972 238 363 197 44 × 2 = 1 + 0.944 476 726 394 88;
  • 38) 0.944 476 726 394 88 × 2 = 1 + 0.888 953 452 789 76;
  • 39) 0.888 953 452 789 76 × 2 = 1 + 0.777 906 905 579 52;
  • 40) 0.777 906 905 579 52 × 2 = 1 + 0.555 813 811 159 04;
  • 41) 0.555 813 811 159 04 × 2 = 1 + 0.111 627 622 318 08;
  • 42) 0.111 627 622 318 08 × 2 = 0 + 0.223 255 244 636 16;
  • 43) 0.223 255 244 636 16 × 2 = 0 + 0.446 510 489 272 32;
  • 44) 0.446 510 489 272 32 × 2 = 0 + 0.893 020 978 544 64;
  • 45) 0.893 020 978 544 64 × 2 = 1 + 0.786 041 957 089 28;
  • 46) 0.786 041 957 089 28 × 2 = 1 + 0.572 083 914 178 56;
  • 47) 0.572 083 914 178 56 × 2 = 1 + 0.144 167 828 357 12;
  • 48) 0.144 167 828 357 12 × 2 = 0 + 0.288 335 656 714 24;
  • 49) 0.288 335 656 714 24 × 2 = 0 + 0.576 671 313 428 48;
  • 50) 0.576 671 313 428 48 × 2 = 1 + 0.153 342 626 856 96;
  • 51) 0.153 342 626 856 96 × 2 = 0 + 0.306 685 253 713 92;
  • 52) 0.306 685 253 713 92 × 2 = 0 + 0.613 370 507 427 84;
  • 53) 0.613 370 507 427 84 × 2 = 1 + 0.226 741 014 855 68;
  • 54) 0.226 741 014 855 68 × 2 = 0 + 0.453 482 029 711 36;
  • 55) 0.453 482 029 711 36 × 2 = 0 + 0.906 964 059 422 72;
  • 56) 0.906 964 059 422 72 × 2 = 1 + 0.813 928 118 845 44;
  • 57) 0.813 928 118 845 44 × 2 = 1 + 0.627 856 237 690 88;
  • 58) 0.627 856 237 690 88 × 2 = 1 + 0.255 712 475 381 76;
  • 59) 0.255 712 475 381 76 × 2 = 0 + 0.511 424 950 763 52;
  • 60) 0.511 424 950 763 52 × 2 = 1 + 0.022 849 901 527 04;
  • 61) 0.022 849 901 527 04 × 2 = 0 + 0.045 699 803 054 08;
  • 62) 0.045 699 803 054 08 × 2 = 0 + 0.091 399 606 108 16;
  • 63) 0.091 399 606 108 16 × 2 = 0 + 0.182 799 212 216 32;
  • 64) 0.182 799 212 216 32 × 2 = 0 + 0.365 598 424 432 64;
  • 65) 0.365 598 424 432 64 × 2 = 0 + 0.731 196 848 865 28;
  • 66) 0.731 196 848 865 28 × 2 = 1 + 0.462 393 697 730 56;
  • 67) 0.462 393 697 730 56 × 2 = 0 + 0.924 787 395 461 12;
  • 68) 0.924 787 395 461 12 × 2 = 1 + 0.849 574 790 922 24;
  • 69) 0.849 574 790 922 24 × 2 = 1 + 0.699 149 581 844 48;
  • 70) 0.699 149 581 844 48 × 2 = 1 + 0.398 299 163 688 96;
  • 71) 0.398 299 163 688 96 × 2 = 0 + 0.796 598 327 377 92;

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:


0.000 003 669 410 54(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1111 1000 1110 0100 1001 1101 0000 0101 110(2)

5. Positive number before normalization:

0.000 003 669 410 54(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1111 1000 1110 0100 1001 1101 0000 0101 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right, so that only one non zero digit remains to the left of it:


0.000 003 669 410 54(10) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1111 1000 1110 0100 1001 1101 0000 0101 110(2) =

0.0000 0000 0000 0000 0011 1101 1000 1111 1111 1111 1000 1110 0100 1001 1101 0000 0101 110(2) × 20 =

1.1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110(2) × 2-19


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): -19

Mantissa (not normalized):
1.1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-19 + 2(11-1) - 1 =

(-19 + 1 023)(10) =

1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1004(10) =

011 1110 1100(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. 1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110 =

1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110


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 1110 1100

Mantissa (52 bits) =
1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110


Decimal number 0.000 003 669 410 54 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1100 - 1110 1100 0111 1111 1111 1100 0111 0010 0100 1110 1000 0010 1110

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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