0.402 823 466 385 288 598 117 041 96 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.402 823 466 385 288 598 117 041 96(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
0.402 823 466 385 288 598 117 041 96(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: 0.402 823 466 385 288 598 117 041 96.

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) 0.402 823 466 385 288 598 117 041 96 × 2 = 0 + 0.805 646 932 770 577 196 234 083 92;
  • 2) 0.805 646 932 770 577 196 234 083 92 × 2 = 1 + 0.611 293 865 541 154 392 468 167 84;
  • 3) 0.611 293 865 541 154 392 468 167 84 × 2 = 1 + 0.222 587 731 082 308 784 936 335 68;
  • 4) 0.222 587 731 082 308 784 936 335 68 × 2 = 0 + 0.445 175 462 164 617 569 872 671 36;
  • 5) 0.445 175 462 164 617 569 872 671 36 × 2 = 0 + 0.890 350 924 329 235 139 745 342 72;
  • 6) 0.890 350 924 329 235 139 745 342 72 × 2 = 1 + 0.780 701 848 658 470 279 490 685 44;
  • 7) 0.780 701 848 658 470 279 490 685 44 × 2 = 1 + 0.561 403 697 316 940 558 981 370 88;
  • 8) 0.561 403 697 316 940 558 981 370 88 × 2 = 1 + 0.122 807 394 633 881 117 962 741 76;
  • 9) 0.122 807 394 633 881 117 962 741 76 × 2 = 0 + 0.245 614 789 267 762 235 925 483 52;
  • 10) 0.245 614 789 267 762 235 925 483 52 × 2 = 0 + 0.491 229 578 535 524 471 850 967 04;
  • 11) 0.491 229 578 535 524 471 850 967 04 × 2 = 0 + 0.982 459 157 071 048 943 701 934 08;
  • 12) 0.982 459 157 071 048 943 701 934 08 × 2 = 1 + 0.964 918 314 142 097 887 403 868 16;
  • 13) 0.964 918 314 142 097 887 403 868 16 × 2 = 1 + 0.929 836 628 284 195 774 807 736 32;
  • 14) 0.929 836 628 284 195 774 807 736 32 × 2 = 1 + 0.859 673 256 568 391 549 615 472 64;
  • 15) 0.859 673 256 568 391 549 615 472 64 × 2 = 1 + 0.719 346 513 136 783 099 230 945 28;
  • 16) 0.719 346 513 136 783 099 230 945 28 × 2 = 1 + 0.438 693 026 273 566 198 461 890 56;
  • 17) 0.438 693 026 273 566 198 461 890 56 × 2 = 0 + 0.877 386 052 547 132 396 923 781 12;
  • 18) 0.877 386 052 547 132 396 923 781 12 × 2 = 1 + 0.754 772 105 094 264 793 847 562 24;
  • 19) 0.754 772 105 094 264 793 847 562 24 × 2 = 1 + 0.509 544 210 188 529 587 695 124 48;
  • 20) 0.509 544 210 188 529 587 695 124 48 × 2 = 1 + 0.019 088 420 377 059 175 390 248 96;
  • 21) 0.019 088 420 377 059 175 390 248 96 × 2 = 0 + 0.038 176 840 754 118 350 780 497 92;
  • 22) 0.038 176 840 754 118 350 780 497 92 × 2 = 0 + 0.076 353 681 508 236 701 560 995 84;
  • 23) 0.076 353 681 508 236 701 560 995 84 × 2 = 0 + 0.152 707 363 016 473 403 121 991 68;
  • 24) 0.152 707 363 016 473 403 121 991 68 × 2 = 0 + 0.305 414 726 032 946 806 243 983 36;
  • 25) 0.305 414 726 032 946 806 243 983 36 × 2 = 0 + 0.610 829 452 065 893 612 487 966 72;
  • 26) 0.610 829 452 065 893 612 487 966 72 × 2 = 1 + 0.221 658 904 131 787 224 975 933 44;
  • 27) 0.221 658 904 131 787 224 975 933 44 × 2 = 0 + 0.443 317 808 263 574 449 951 866 88;
  • 28) 0.443 317 808 263 574 449 951 866 88 × 2 = 0 + 0.886 635 616 527 148 899 903 733 76;
  • 29) 0.886 635 616 527 148 899 903 733 76 × 2 = 1 + 0.773 271 233 054 297 799 807 467 52;
  • 30) 0.773 271 233 054 297 799 807 467 52 × 2 = 1 + 0.546 542 466 108 595 599 614 935 04;
  • 31) 0.546 542 466 108 595 599 614 935 04 × 2 = 1 + 0.093 084 932 217 191 199 229 870 08;
  • 32) 0.093 084 932 217 191 199 229 870 08 × 2 = 0 + 0.186 169 864 434 382 398 459 740 16;
  • 33) 0.186 169 864 434 382 398 459 740 16 × 2 = 0 + 0.372 339 728 868 764 796 919 480 32;
  • 34) 0.372 339 728 868 764 796 919 480 32 × 2 = 0 + 0.744 679 457 737 529 593 838 960 64;
  • 35) 0.744 679 457 737 529 593 838 960 64 × 2 = 1 + 0.489 358 915 475 059 187 677 921 28;
  • 36) 0.489 358 915 475 059 187 677 921 28 × 2 = 0 + 0.978 717 830 950 118 375 355 842 56;
  • 37) 0.978 717 830 950 118 375 355 842 56 × 2 = 1 + 0.957 435 661 900 236 750 711 685 12;
  • 38) 0.957 435 661 900 236 750 711 685 12 × 2 = 1 + 0.914 871 323 800 473 501 423 370 24;
  • 39) 0.914 871 323 800 473 501 423 370 24 × 2 = 1 + 0.829 742 647 600 947 002 846 740 48;
  • 40) 0.829 742 647 600 947 002 846 740 48 × 2 = 1 + 0.659 485 295 201 894 005 693 480 96;
  • 41) 0.659 485 295 201 894 005 693 480 96 × 2 = 1 + 0.318 970 590 403 788 011 386 961 92;
  • 42) 0.318 970 590 403 788 011 386 961 92 × 2 = 0 + 0.637 941 180 807 576 022 773 923 84;
  • 43) 0.637 941 180 807 576 022 773 923 84 × 2 = 1 + 0.275 882 361 615 152 045 547 847 68;
  • 44) 0.275 882 361 615 152 045 547 847 68 × 2 = 0 + 0.551 764 723 230 304 091 095 695 36;
  • 45) 0.551 764 723 230 304 091 095 695 36 × 2 = 1 + 0.103 529 446 460 608 182 191 390 72;
  • 46) 0.103 529 446 460 608 182 191 390 72 × 2 = 0 + 0.207 058 892 921 216 364 382 781 44;
  • 47) 0.207 058 892 921 216 364 382 781 44 × 2 = 0 + 0.414 117 785 842 432 728 765 562 88;
  • 48) 0.414 117 785 842 432 728 765 562 88 × 2 = 0 + 0.828 235 571 684 865 457 531 125 76;
  • 49) 0.828 235 571 684 865 457 531 125 76 × 2 = 1 + 0.656 471 143 369 730 915 062 251 52;
  • 50) 0.656 471 143 369 730 915 062 251 52 × 2 = 1 + 0.312 942 286 739 461 830 124 503 04;
  • 51) 0.312 942 286 739 461 830 124 503 04 × 2 = 0 + 0.625 884 573 478 923 660 249 006 08;
  • 52) 0.625 884 573 478 923 660 249 006 08 × 2 = 1 + 0.251 769 146 957 847 320 498 012 16;
  • 53) 0.251 769 146 957 847 320 498 012 16 × 2 = 0 + 0.503 538 293 915 694 640 996 024 32;
  • 54) 0.503 538 293 915 694 640 996 024 32 × 2 = 1 + 0.007 076 587 831 389 281 992 048 64;

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:


0.402 823 466 385 288 598 117 041 96(10) =

0.0110 0111 0001 1111 0111 0000 0100 1110 0010 1111 1010 1000 1101 01(2)

5. Positive number before normalization:

0.402 823 466 385 288 598 117 041 96(10) =

0.0110 0111 0001 1111 0111 0000 0100 1110 0010 1111 1010 1000 1101 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.402 823 466 385 288 598 117 041 96(10) =

0.0110 0111 0001 1111 0111 0000 0100 1110 0010 1111 1010 1000 1101 01(2) =

0.0110 0111 0001 1111 0111 0000 0100 1110 0010 1111 1010 1000 1101 01(2) × 20 =

1.1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101(2) × 2-2


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): -2

Mantissa (not normalized):
1.1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1021(10) =

011 1111 1101(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. 1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101 =

1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101


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 1101

Mantissa (52 bits) =
1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101


Decimal number 0.402 823 466 385 288 598 117 041 96 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 1001 1100 0111 1101 1100 0001 0011 1000 1011 1110 1010 0011 0101

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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