0.000 020 830 729 321 671 205 134 999 181 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 181(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
0.000 020 830 729 321 671 205 134 999 181(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: 0.000 020 830 729 321 671 205 134 999 181.

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) 0.000 020 830 729 321 671 205 134 999 181 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 362;
  • 2) 0.000 041 661 458 643 342 410 269 998 362 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 724;
  • 3) 0.000 083 322 917 286 684 820 539 996 724 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 448;
  • 4) 0.000 166 645 834 573 369 641 079 993 448 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 896;
  • 5) 0.000 333 291 669 146 739 282 159 986 896 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 792;
  • 6) 0.000 666 583 338 293 478 564 319 973 792 × 2 = 0 + 0.001 333 166 676 586 957 128 639 947 584;
  • 7) 0.001 333 166 676 586 957 128 639 947 584 × 2 = 0 + 0.002 666 333 353 173 914 257 279 895 168;
  • 8) 0.002 666 333 353 173 914 257 279 895 168 × 2 = 0 + 0.005 332 666 706 347 828 514 559 790 336;
  • 9) 0.005 332 666 706 347 828 514 559 790 336 × 2 = 0 + 0.010 665 333 412 695 657 029 119 580 672;
  • 10) 0.010 665 333 412 695 657 029 119 580 672 × 2 = 0 + 0.021 330 666 825 391 314 058 239 161 344;
  • 11) 0.021 330 666 825 391 314 058 239 161 344 × 2 = 0 + 0.042 661 333 650 782 628 116 478 322 688;
  • 12) 0.042 661 333 650 782 628 116 478 322 688 × 2 = 0 + 0.085 322 667 301 565 256 232 956 645 376;
  • 13) 0.085 322 667 301 565 256 232 956 645 376 × 2 = 0 + 0.170 645 334 603 130 512 465 913 290 752;
  • 14) 0.170 645 334 603 130 512 465 913 290 752 × 2 = 0 + 0.341 290 669 206 261 024 931 826 581 504;
  • 15) 0.341 290 669 206 261 024 931 826 581 504 × 2 = 0 + 0.682 581 338 412 522 049 863 653 163 008;
  • 16) 0.682 581 338 412 522 049 863 653 163 008 × 2 = 1 + 0.365 162 676 825 044 099 727 306 326 016;
  • 17) 0.365 162 676 825 044 099 727 306 326 016 × 2 = 0 + 0.730 325 353 650 088 199 454 612 652 032;
  • 18) 0.730 325 353 650 088 199 454 612 652 032 × 2 = 1 + 0.460 650 707 300 176 398 909 225 304 064;
  • 19) 0.460 650 707 300 176 398 909 225 304 064 × 2 = 0 + 0.921 301 414 600 352 797 818 450 608 128;
  • 20) 0.921 301 414 600 352 797 818 450 608 128 × 2 = 1 + 0.842 602 829 200 705 595 636 901 216 256;
  • 21) 0.842 602 829 200 705 595 636 901 216 256 × 2 = 1 + 0.685 205 658 401 411 191 273 802 432 512;
  • 22) 0.685 205 658 401 411 191 273 802 432 512 × 2 = 1 + 0.370 411 316 802 822 382 547 604 865 024;
  • 23) 0.370 411 316 802 822 382 547 604 865 024 × 2 = 0 + 0.740 822 633 605 644 765 095 209 730 048;
  • 24) 0.740 822 633 605 644 765 095 209 730 048 × 2 = 1 + 0.481 645 267 211 289 530 190 419 460 096;
  • 25) 0.481 645 267 211 289 530 190 419 460 096 × 2 = 0 + 0.963 290 534 422 579 060 380 838 920 192;
  • 26) 0.963 290 534 422 579 060 380 838 920 192 × 2 = 1 + 0.926 581 068 845 158 120 761 677 840 384;
  • 27) 0.926 581 068 845 158 120 761 677 840 384 × 2 = 1 + 0.853 162 137 690 316 241 523 355 680 768;
  • 28) 0.853 162 137 690 316 241 523 355 680 768 × 2 = 1 + 0.706 324 275 380 632 483 046 711 361 536;
  • 29) 0.706 324 275 380 632 483 046 711 361 536 × 2 = 1 + 0.412 648 550 761 264 966 093 422 723 072;
  • 30) 0.412 648 550 761 264 966 093 422 723 072 × 2 = 0 + 0.825 297 101 522 529 932 186 845 446 144;
  • 31) 0.825 297 101 522 529 932 186 845 446 144 × 2 = 1 + 0.650 594 203 045 059 864 373 690 892 288;
  • 32) 0.650 594 203 045 059 864 373 690 892 288 × 2 = 1 + 0.301 188 406 090 119 728 747 381 784 576;
  • 33) 0.301 188 406 090 119 728 747 381 784 576 × 2 = 0 + 0.602 376 812 180 239 457 494 763 569 152;
  • 34) 0.602 376 812 180 239 457 494 763 569 152 × 2 = 1 + 0.204 753 624 360 478 914 989 527 138 304;
  • 35) 0.204 753 624 360 478 914 989 527 138 304 × 2 = 0 + 0.409 507 248 720 957 829 979 054 276 608;
  • 36) 0.409 507 248 720 957 829 979 054 276 608 × 2 = 0 + 0.819 014 497 441 915 659 958 108 553 216;
  • 37) 0.819 014 497 441 915 659 958 108 553 216 × 2 = 1 + 0.638 028 994 883 831 319 916 217 106 432;
  • 38) 0.638 028 994 883 831 319 916 217 106 432 × 2 = 1 + 0.276 057 989 767 662 639 832 434 212 864;
  • 39) 0.276 057 989 767 662 639 832 434 212 864 × 2 = 0 + 0.552 115 979 535 325 279 664 868 425 728;
  • 40) 0.552 115 979 535 325 279 664 868 425 728 × 2 = 1 + 0.104 231 959 070 650 559 329 736 851 456;
  • 41) 0.104 231 959 070 650 559 329 736 851 456 × 2 = 0 + 0.208 463 918 141 301 118 659 473 702 912;
  • 42) 0.208 463 918 141 301 118 659 473 702 912 × 2 = 0 + 0.416 927 836 282 602 237 318 947 405 824;
  • 43) 0.416 927 836 282 602 237 318 947 405 824 × 2 = 0 + 0.833 855 672 565 204 474 637 894 811 648;
  • 44) 0.833 855 672 565 204 474 637 894 811 648 × 2 = 1 + 0.667 711 345 130 408 949 275 789 623 296;
  • 45) 0.667 711 345 130 408 949 275 789 623 296 × 2 = 1 + 0.335 422 690 260 817 898 551 579 246 592;
  • 46) 0.335 422 690 260 817 898 551 579 246 592 × 2 = 0 + 0.670 845 380 521 635 797 103 158 493 184;
  • 47) 0.670 845 380 521 635 797 103 158 493 184 × 2 = 1 + 0.341 690 761 043 271 594 206 316 986 368;
  • 48) 0.341 690 761 043 271 594 206 316 986 368 × 2 = 0 + 0.683 381 522 086 543 188 412 633 972 736;
  • 49) 0.683 381 522 086 543 188 412 633 972 736 × 2 = 1 + 0.366 763 044 173 086 376 825 267 945 472;
  • 50) 0.366 763 044 173 086 376 825 267 945 472 × 2 = 0 + 0.733 526 088 346 172 753 650 535 890 944;
  • 51) 0.733 526 088 346 172 753 650 535 890 944 × 2 = 1 + 0.467 052 176 692 345 507 301 071 781 888;
  • 52) 0.467 052 176 692 345 507 301 071 781 888 × 2 = 0 + 0.934 104 353 384 691 014 602 143 563 776;
  • 53) 0.934 104 353 384 691 014 602 143 563 776 × 2 = 1 + 0.868 208 706 769 382 029 204 287 127 552;
  • 54) 0.868 208 706 769 382 029 204 287 127 552 × 2 = 1 + 0.736 417 413 538 764 058 408 574 255 104;
  • 55) 0.736 417 413 538 764 058 408 574 255 104 × 2 = 1 + 0.472 834 827 077 528 116 817 148 510 208;
  • 56) 0.472 834 827 077 528 116 817 148 510 208 × 2 = 0 + 0.945 669 654 155 056 233 634 297 020 416;
  • 57) 0.945 669 654 155 056 233 634 297 020 416 × 2 = 1 + 0.891 339 308 310 112 467 268 594 040 832;
  • 58) 0.891 339 308 310 112 467 268 594 040 832 × 2 = 1 + 0.782 678 616 620 224 934 537 188 081 664;
  • 59) 0.782 678 616 620 224 934 537 188 081 664 × 2 = 1 + 0.565 357 233 240 449 869 074 376 163 328;
  • 60) 0.565 357 233 240 449 869 074 376 163 328 × 2 = 1 + 0.130 714 466 480 899 738 148 752 326 656;
  • 61) 0.130 714 466 480 899 738 148 752 326 656 × 2 = 0 + 0.261 428 932 961 799 476 297 504 653 312;
  • 62) 0.261 428 932 961 799 476 297 504 653 312 × 2 = 0 + 0.522 857 865 923 598 952 595 009 306 624;
  • 63) 0.522 857 865 923 598 952 595 009 306 624 × 2 = 1 + 0.045 715 731 847 197 905 190 018 613 248;
  • 64) 0.045 715 731 847 197 905 190 018 613 248 × 2 = 0 + 0.091 431 463 694 395 810 380 037 226 496;
  • 65) 0.091 431 463 694 395 810 380 037 226 496 × 2 = 0 + 0.182 862 927 388 791 620 760 074 452 992;
  • 66) 0.182 862 927 388 791 620 760 074 452 992 × 2 = 0 + 0.365 725 854 777 583 241 520 148 905 984;
  • 67) 0.365 725 854 777 583 241 520 148 905 984 × 2 = 0 + 0.731 451 709 555 166 483 040 297 811 968;
  • 68) 0.731 451 709 555 166 483 040 297 811 968 × 2 = 1 + 0.462 903 419 110 332 966 080 595 623 936;

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:


0.000 020 830 729 321 671 205 134 999 181(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)

5. Positive number before normalization:

0.000 020 830 729 321 671 205 134 999 181(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right, so that only one non zero digit remains to the left of it:


0.000 020 830 729 321 671 205 134 999 181(10) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) =

0.0000 0000 0000 0001 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 20 =

1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001(2) × 2-16


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): -16

Mantissa (not normalized):
1.0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-16 + 2(11-1) - 1 =

(-16 + 1 023)(10) =

1 007(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;
  • 1 007 ÷ 2 = 503 + 1;
  • 503 ÷ 2 = 251 + 1;
  • 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) =

1007(10) =

011 1110 1111(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 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001 =

0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


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 1111

Mantissa (52 bits) =
0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001


Decimal number 0.000 020 830 729 321 671 205 134 999 181 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 1111 - 0101 1101 0111 1011 0100 1101 0001 1010 1010 1110 1111 0010 0001

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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