0.000 000 564 358 690 715 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 564 358 690 715(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
0.000 000 564 358 690 715(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: 0.000 000 564 358 690 715.

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) 0.000 000 564 358 690 715 × 2 = 0 + 0.000 001 128 717 381 43;
  • 2) 0.000 001 128 717 381 43 × 2 = 0 + 0.000 002 257 434 762 86;
  • 3) 0.000 002 257 434 762 86 × 2 = 0 + 0.000 004 514 869 525 72;
  • 4) 0.000 004 514 869 525 72 × 2 = 0 + 0.000 009 029 739 051 44;
  • 5) 0.000 009 029 739 051 44 × 2 = 0 + 0.000 018 059 478 102 88;
  • 6) 0.000 018 059 478 102 88 × 2 = 0 + 0.000 036 118 956 205 76;
  • 7) 0.000 036 118 956 205 76 × 2 = 0 + 0.000 072 237 912 411 52;
  • 8) 0.000 072 237 912 411 52 × 2 = 0 + 0.000 144 475 824 823 04;
  • 9) 0.000 144 475 824 823 04 × 2 = 0 + 0.000 288 951 649 646 08;
  • 10) 0.000 288 951 649 646 08 × 2 = 0 + 0.000 577 903 299 292 16;
  • 11) 0.000 577 903 299 292 16 × 2 = 0 + 0.001 155 806 598 584 32;
  • 12) 0.001 155 806 598 584 32 × 2 = 0 + 0.002 311 613 197 168 64;
  • 13) 0.002 311 613 197 168 64 × 2 = 0 + 0.004 623 226 394 337 28;
  • 14) 0.004 623 226 394 337 28 × 2 = 0 + 0.009 246 452 788 674 56;
  • 15) 0.009 246 452 788 674 56 × 2 = 0 + 0.018 492 905 577 349 12;
  • 16) 0.018 492 905 577 349 12 × 2 = 0 + 0.036 985 811 154 698 24;
  • 17) 0.036 985 811 154 698 24 × 2 = 0 + 0.073 971 622 309 396 48;
  • 18) 0.073 971 622 309 396 48 × 2 = 0 + 0.147 943 244 618 792 96;
  • 19) 0.147 943 244 618 792 96 × 2 = 0 + 0.295 886 489 237 585 92;
  • 20) 0.295 886 489 237 585 92 × 2 = 0 + 0.591 772 978 475 171 84;
  • 21) 0.591 772 978 475 171 84 × 2 = 1 + 0.183 545 956 950 343 68;
  • 22) 0.183 545 956 950 343 68 × 2 = 0 + 0.367 091 913 900 687 36;
  • 23) 0.367 091 913 900 687 36 × 2 = 0 + 0.734 183 827 801 374 72;
  • 24) 0.734 183 827 801 374 72 × 2 = 1 + 0.468 367 655 602 749 44;
  • 25) 0.468 367 655 602 749 44 × 2 = 0 + 0.936 735 311 205 498 88;
  • 26) 0.936 735 311 205 498 88 × 2 = 1 + 0.873 470 622 410 997 76;
  • 27) 0.873 470 622 410 997 76 × 2 = 1 + 0.746 941 244 821 995 52;
  • 28) 0.746 941 244 821 995 52 × 2 = 1 + 0.493 882 489 643 991 04;
  • 29) 0.493 882 489 643 991 04 × 2 = 0 + 0.987 764 979 287 982 08;
  • 30) 0.987 764 979 287 982 08 × 2 = 1 + 0.975 529 958 575 964 16;
  • 31) 0.975 529 958 575 964 16 × 2 = 1 + 0.951 059 917 151 928 32;
  • 32) 0.951 059 917 151 928 32 × 2 = 1 + 0.902 119 834 303 856 64;
  • 33) 0.902 119 834 303 856 64 × 2 = 1 + 0.804 239 668 607 713 28;
  • 34) 0.804 239 668 607 713 28 × 2 = 1 + 0.608 479 337 215 426 56;
  • 35) 0.608 479 337 215 426 56 × 2 = 1 + 0.216 958 674 430 853 12;
  • 36) 0.216 958 674 430 853 12 × 2 = 0 + 0.433 917 348 861 706 24;
  • 37) 0.433 917 348 861 706 24 × 2 = 0 + 0.867 834 697 723 412 48;
  • 38) 0.867 834 697 723 412 48 × 2 = 1 + 0.735 669 395 446 824 96;
  • 39) 0.735 669 395 446 824 96 × 2 = 1 + 0.471 338 790 893 649 92;
  • 40) 0.471 338 790 893 649 92 × 2 = 0 + 0.942 677 581 787 299 84;
  • 41) 0.942 677 581 787 299 84 × 2 = 1 + 0.885 355 163 574 599 68;
  • 42) 0.885 355 163 574 599 68 × 2 = 1 + 0.770 710 327 149 199 36;
  • 43) 0.770 710 327 149 199 36 × 2 = 1 + 0.541 420 654 298 398 72;
  • 44) 0.541 420 654 298 398 72 × 2 = 1 + 0.082 841 308 596 797 44;
  • 45) 0.082 841 308 596 797 44 × 2 = 0 + 0.165 682 617 193 594 88;
  • 46) 0.165 682 617 193 594 88 × 2 = 0 + 0.331 365 234 387 189 76;
  • 47) 0.331 365 234 387 189 76 × 2 = 0 + 0.662 730 468 774 379 52;
  • 48) 0.662 730 468 774 379 52 × 2 = 1 + 0.325 460 937 548 759 04;
  • 49) 0.325 460 937 548 759 04 × 2 = 0 + 0.650 921 875 097 518 08;
  • 50) 0.650 921 875 097 518 08 × 2 = 1 + 0.301 843 750 195 036 16;
  • 51) 0.301 843 750 195 036 16 × 2 = 0 + 0.603 687 500 390 072 32;
  • 52) 0.603 687 500 390 072 32 × 2 = 1 + 0.207 375 000 780 144 64;
  • 53) 0.207 375 000 780 144 64 × 2 = 0 + 0.414 750 001 560 289 28;
  • 54) 0.414 750 001 560 289 28 × 2 = 0 + 0.829 500 003 120 578 56;
  • 55) 0.829 500 003 120 578 56 × 2 = 1 + 0.659 000 006 241 157 12;
  • 56) 0.659 000 006 241 157 12 × 2 = 1 + 0.318 000 012 482 314 24;
  • 57) 0.318 000 012 482 314 24 × 2 = 0 + 0.636 000 024 964 628 48;
  • 58) 0.636 000 024 964 628 48 × 2 = 1 + 0.272 000 049 929 256 96;
  • 59) 0.272 000 049 929 256 96 × 2 = 0 + 0.544 000 099 858 513 92;
  • 60) 0.544 000 099 858 513 92 × 2 = 1 + 0.088 000 199 717 027 84;
  • 61) 0.088 000 199 717 027 84 × 2 = 0 + 0.176 000 399 434 055 68;
  • 62) 0.176 000 399 434 055 68 × 2 = 0 + 0.352 000 798 868 111 36;
  • 63) 0.352 000 798 868 111 36 × 2 = 0 + 0.704 001 597 736 222 72;
  • 64) 0.704 001 597 736 222 72 × 2 = 1 + 0.408 003 195 472 445 44;
  • 65) 0.408 003 195 472 445 44 × 2 = 0 + 0.816 006 390 944 890 88;
  • 66) 0.816 006 390 944 890 88 × 2 = 1 + 0.632 012 781 889 781 76;
  • 67) 0.632 012 781 889 781 76 × 2 = 1 + 0.264 025 563 779 563 52;
  • 68) 0.264 025 563 779 563 52 × 2 = 0 + 0.528 051 127 559 127 04;
  • 69) 0.528 051 127 559 127 04 × 2 = 1 + 0.056 102 255 118 254 08;
  • 70) 0.056 102 255 118 254 08 × 2 = 0 + 0.112 204 510 236 508 16;
  • 71) 0.112 204 510 236 508 16 × 2 = 0 + 0.224 409 020 473 016 32;
  • 72) 0.224 409 020 473 016 32 × 2 = 0 + 0.448 818 040 946 032 64;
  • 73) 0.448 818 040 946 032 64 × 2 = 0 + 0.897 636 081 892 065 28;

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:


0.000 000 564 358 690 715(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 0101 0001 0110 1000 0(2)

5. Positive number before normalization:

0.000 000 564 358 690 715(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 0101 0001 0110 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 564 358 690 715(10) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 0101 0001 0110 1000 0(2) =

0.0000 0000 0000 0000 0000 1001 0111 0111 1110 0110 1111 0001 0101 0011 0101 0001 0110 1000 0(2) × 20 =

1.0010 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000(2) × 2-21


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): -21

Mantissa (not normalized):
1.0010 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(11-1) - 1 =

(-21 + 1 023)(10) =

1 002(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;
  • 1 002 ÷ 2 = 501 + 0;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 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) =

1002(10) =

011 1110 1010(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 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000 =

0010 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000


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 1010

Mantissa (52 bits) =
0010 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000


Decimal number 0.000 000 564 358 690 715 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1010 - 0010 1110 1111 1100 1101 1110 0010 1010 0110 1010 0010 1101 0000

» Convert decimal 0.000 000 564 358 690 782 to 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