0.333 333 333 333 333 314 829 616 256 247 390 992 939 446 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.333 333 333 333 333 314 829 616 256 247 390 992 939 446(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
0.333 333 333 333 333 314 829 616 256 247 390 992 939 446(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: 0.333 333 333 333 333 314 829 616 256 247 390 992 939 446.

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) 0.333 333 333 333 333 314 829 616 256 247 390 992 939 446 × 2 = 0 + 0.666 666 666 666 666 629 659 232 512 494 781 985 878 892;
  • 2) 0.666 666 666 666 666 629 659 232 512 494 781 985 878 892 × 2 = 1 + 0.333 333 333 333 333 259 318 465 024 989 563 971 757 784;
  • 3) 0.333 333 333 333 333 259 318 465 024 989 563 971 757 784 × 2 = 0 + 0.666 666 666 666 666 518 636 930 049 979 127 943 515 568;
  • 4) 0.666 666 666 666 666 518 636 930 049 979 127 943 515 568 × 2 = 1 + 0.333 333 333 333 333 037 273 860 099 958 255 887 031 136;
  • 5) 0.333 333 333 333 333 037 273 860 099 958 255 887 031 136 × 2 = 0 + 0.666 666 666 666 666 074 547 720 199 916 511 774 062 272;
  • 6) 0.666 666 666 666 666 074 547 720 199 916 511 774 062 272 × 2 = 1 + 0.333 333 333 333 332 149 095 440 399 833 023 548 124 544;
  • 7) 0.333 333 333 333 332 149 095 440 399 833 023 548 124 544 × 2 = 0 + 0.666 666 666 666 664 298 190 880 799 666 047 096 249 088;
  • 8) 0.666 666 666 666 664 298 190 880 799 666 047 096 249 088 × 2 = 1 + 0.333 333 333 333 328 596 381 761 599 332 094 192 498 176;
  • 9) 0.333 333 333 333 328 596 381 761 599 332 094 192 498 176 × 2 = 0 + 0.666 666 666 666 657 192 763 523 198 664 188 384 996 352;
  • 10) 0.666 666 666 666 657 192 763 523 198 664 188 384 996 352 × 2 = 1 + 0.333 333 333 333 314 385 527 046 397 328 376 769 992 704;
  • 11) 0.333 333 333 333 314 385 527 046 397 328 376 769 992 704 × 2 = 0 + 0.666 666 666 666 628 771 054 092 794 656 753 539 985 408;
  • 12) 0.666 666 666 666 628 771 054 092 794 656 753 539 985 408 × 2 = 1 + 0.333 333 333 333 257 542 108 185 589 313 507 079 970 816;
  • 13) 0.333 333 333 333 257 542 108 185 589 313 507 079 970 816 × 2 = 0 + 0.666 666 666 666 515 084 216 371 178 627 014 159 941 632;
  • 14) 0.666 666 666 666 515 084 216 371 178 627 014 159 941 632 × 2 = 1 + 0.333 333 333 333 030 168 432 742 357 254 028 319 883 264;
  • 15) 0.333 333 333 333 030 168 432 742 357 254 028 319 883 264 × 2 = 0 + 0.666 666 666 666 060 336 865 484 714 508 056 639 766 528;
  • 16) 0.666 666 666 666 060 336 865 484 714 508 056 639 766 528 × 2 = 1 + 0.333 333 333 332 120 673 730 969 429 016 113 279 533 056;
  • 17) 0.333 333 333 332 120 673 730 969 429 016 113 279 533 056 × 2 = 0 + 0.666 666 666 664 241 347 461 938 858 032 226 559 066 112;
  • 18) 0.666 666 666 664 241 347 461 938 858 032 226 559 066 112 × 2 = 1 + 0.333 333 333 328 482 694 923 877 716 064 453 118 132 224;
  • 19) 0.333 333 333 328 482 694 923 877 716 064 453 118 132 224 × 2 = 0 + 0.666 666 666 656 965 389 847 755 432 128 906 236 264 448;
  • 20) 0.666 666 666 656 965 389 847 755 432 128 906 236 264 448 × 2 = 1 + 0.333 333 333 313 930 779 695 510 864 257 812 472 528 896;
  • 21) 0.333 333 333 313 930 779 695 510 864 257 812 472 528 896 × 2 = 0 + 0.666 666 666 627 861 559 391 021 728 515 624 945 057 792;
  • 22) 0.666 666 666 627 861 559 391 021 728 515 624 945 057 792 × 2 = 1 + 0.333 333 333 255 723 118 782 043 457 031 249 890 115 584;
  • 23) 0.333 333 333 255 723 118 782 043 457 031 249 890 115 584 × 2 = 0 + 0.666 666 666 511 446 237 564 086 914 062 499 780 231 168;
  • 24) 0.666 666 666 511 446 237 564 086 914 062 499 780 231 168 × 2 = 1 + 0.333 333 333 022 892 475 128 173 828 124 999 560 462 336;
  • 25) 0.333 333 333 022 892 475 128 173 828 124 999 560 462 336 × 2 = 0 + 0.666 666 666 045 784 950 256 347 656 249 999 120 924 672;
  • 26) 0.666 666 666 045 784 950 256 347 656 249 999 120 924 672 × 2 = 1 + 0.333 333 332 091 569 900 512 695 312 499 998 241 849 344;
  • 27) 0.333 333 332 091 569 900 512 695 312 499 998 241 849 344 × 2 = 0 + 0.666 666 664 183 139 801 025 390 624 999 996 483 698 688;
  • 28) 0.666 666 664 183 139 801 025 390 624 999 996 483 698 688 × 2 = 1 + 0.333 333 328 366 279 602 050 781 249 999 992 967 397 376;
  • 29) 0.333 333 328 366 279 602 050 781 249 999 992 967 397 376 × 2 = 0 + 0.666 666 656 732 559 204 101 562 499 999 985 934 794 752;
  • 30) 0.666 666 656 732 559 204 101 562 499 999 985 934 794 752 × 2 = 1 + 0.333 333 313 465 118 408 203 124 999 999 971 869 589 504;
  • 31) 0.333 333 313 465 118 408 203 124 999 999 971 869 589 504 × 2 = 0 + 0.666 666 626 930 236 816 406 249 999 999 943 739 179 008;
  • 32) 0.666 666 626 930 236 816 406 249 999 999 943 739 179 008 × 2 = 1 + 0.333 333 253 860 473 632 812 499 999 999 887 478 358 016;
  • 33) 0.333 333 253 860 473 632 812 499 999 999 887 478 358 016 × 2 = 0 + 0.666 666 507 720 947 265 624 999 999 999 774 956 716 032;
  • 34) 0.666 666 507 720 947 265 624 999 999 999 774 956 716 032 × 2 = 1 + 0.333 333 015 441 894 531 249 999 999 999 549 913 432 064;
  • 35) 0.333 333 015 441 894 531 249 999 999 999 549 913 432 064 × 2 = 0 + 0.666 666 030 883 789 062 499 999 999 999 099 826 864 128;
  • 36) 0.666 666 030 883 789 062 499 999 999 999 099 826 864 128 × 2 = 1 + 0.333 332 061 767 578 124 999 999 999 998 199 653 728 256;
  • 37) 0.333 332 061 767 578 124 999 999 999 998 199 653 728 256 × 2 = 0 + 0.666 664 123 535 156 249 999 999 999 996 399 307 456 512;
  • 38) 0.666 664 123 535 156 249 999 999 999 996 399 307 456 512 × 2 = 1 + 0.333 328 247 070 312 499 999 999 999 992 798 614 913 024;
  • 39) 0.333 328 247 070 312 499 999 999 999 992 798 614 913 024 × 2 = 0 + 0.666 656 494 140 624 999 999 999 999 985 597 229 826 048;
  • 40) 0.666 656 494 140 624 999 999 999 999 985 597 229 826 048 × 2 = 1 + 0.333 312 988 281 249 999 999 999 999 971 194 459 652 096;
  • 41) 0.333 312 988 281 249 999 999 999 999 971 194 459 652 096 × 2 = 0 + 0.666 625 976 562 499 999 999 999 999 942 388 919 304 192;
  • 42) 0.666 625 976 562 499 999 999 999 999 942 388 919 304 192 × 2 = 1 + 0.333 251 953 124 999 999 999 999 999 884 777 838 608 384;
  • 43) 0.333 251 953 124 999 999 999 999 999 884 777 838 608 384 × 2 = 0 + 0.666 503 906 249 999 999 999 999 999 769 555 677 216 768;
  • 44) 0.666 503 906 249 999 999 999 999 999 769 555 677 216 768 × 2 = 1 + 0.333 007 812 499 999 999 999 999 999 539 111 354 433 536;
  • 45) 0.333 007 812 499 999 999 999 999 999 539 111 354 433 536 × 2 = 0 + 0.666 015 624 999 999 999 999 999 999 078 222 708 867 072;
  • 46) 0.666 015 624 999 999 999 999 999 999 078 222 708 867 072 × 2 = 1 + 0.332 031 249 999 999 999 999 999 998 156 445 417 734 144;
  • 47) 0.332 031 249 999 999 999 999 999 998 156 445 417 734 144 × 2 = 0 + 0.664 062 499 999 999 999 999 999 996 312 890 835 468 288;
  • 48) 0.664 062 499 999 999 999 999 999 996 312 890 835 468 288 × 2 = 1 + 0.328 124 999 999 999 999 999 999 992 625 781 670 936 576;
  • 49) 0.328 124 999 999 999 999 999 999 992 625 781 670 936 576 × 2 = 0 + 0.656 249 999 999 999 999 999 999 985 251 563 341 873 152;
  • 50) 0.656 249 999 999 999 999 999 999 985 251 563 341 873 152 × 2 = 1 + 0.312 499 999 999 999 999 999 999 970 503 126 683 746 304;
  • 51) 0.312 499 999 999 999 999 999 999 970 503 126 683 746 304 × 2 = 0 + 0.624 999 999 999 999 999 999 999 941 006 253 367 492 608;
  • 52) 0.624 999 999 999 999 999 999 999 941 006 253 367 492 608 × 2 = 1 + 0.249 999 999 999 999 999 999 999 882 012 506 734 985 216;
  • 53) 0.249 999 999 999 999 999 999 999 882 012 506 734 985 216 × 2 = 0 + 0.499 999 999 999 999 999 999 999 764 025 013 469 970 432;
  • 54) 0.499 999 999 999 999 999 999 999 764 025 013 469 970 432 × 2 = 0 + 0.999 999 999 999 999 999 999 999 528 050 026 939 940 864;

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:


0.333 333 333 333 333 314 829 616 256 247 390 992 939 446(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2)

5. Positive number before normalization:

0.333 333 333 333 333 314 829 616 256 247 390 992 939 446(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(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:


0.333 333 333 333 333 314 829 616 256 247 390 992 939 446(10) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2) =

0.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 00(2) × 20 =

1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100(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):
1.0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


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. 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100 =

0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


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) =
0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100


Decimal number 0.333 333 333 333 333 314 829 616 256 247 390 992 939 446 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1101 - 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0101 0100

» Convert decimal 0.333 333 333 333 333 314 829 616 256 247 390 992 939 528 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