0.000 020 830 729 321 671 205 134 999 124 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 124(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
0.000 020 830 729 321 671 205 134 999 124(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: 0.000 020 830 729 321 671 205 134 999 124.

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) 0.000 020 830 729 321 671 205 134 999 124 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 248;
  • 2) 0.000 041 661 458 643 342 410 269 998 248 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 496;
  • 3) 0.000 083 322 917 286 684 820 539 996 496 × 2 = 0 + 0.000 166 645 834 573 369 641 079 992 992;
  • 4) 0.000 166 645 834 573 369 641 079 992 992 × 2 = 0 + 0.000 333 291 669 146 739 282 159 985 984;
  • 5) 0.000 333 291 669 146 739 282 159 985 984 × 2 = 0 + 0.000 666 583 338 293 478 564 319 971 968;
  • 6) 0.000 666 583 338 293 478 564 319 971 968 × 2 = 0 + 0.001 333 166 676 586 957 128 639 943 936;
  • 7) 0.001 333 166 676 586 957 128 639 943 936 × 2 = 0 + 0.002 666 333 353 173 914 257 279 887 872;
  • 8) 0.002 666 333 353 173 914 257 279 887 872 × 2 = 0 + 0.005 332 666 706 347 828 514 559 775 744;
  • 9) 0.005 332 666 706 347 828 514 559 775 744 × 2 = 0 + 0.010 665 333 412 695 657 029 119 551 488;
  • 10) 0.010 665 333 412 695 657 029 119 551 488 × 2 = 0 + 0.021 330 666 825 391 314 058 239 102 976;
  • 11) 0.021 330 666 825 391 314 058 239 102 976 × 2 = 0 + 0.042 661 333 650 782 628 116 478 205 952;
  • 12) 0.042 661 333 650 782 628 116 478 205 952 × 2 = 0 + 0.085 322 667 301 565 256 232 956 411 904;
  • 13) 0.085 322 667 301 565 256 232 956 411 904 × 2 = 0 + 0.170 645 334 603 130 512 465 912 823 808;
  • 14) 0.170 645 334 603 130 512 465 912 823 808 × 2 = 0 + 0.341 290 669 206 261 024 931 825 647 616;
  • 15) 0.341 290 669 206 261 024 931 825 647 616 × 2 = 0 + 0.682 581 338 412 522 049 863 651 295 232;
  • 16) 0.682 581 338 412 522 049 863 651 295 232 × 2 = 1 + 0.365 162 676 825 044 099 727 302 590 464;
  • 17) 0.365 162 676 825 044 099 727 302 590 464 × 2 = 0 + 0.730 325 353 650 088 199 454 605 180 928;
  • 18) 0.730 325 353 650 088 199 454 605 180 928 × 2 = 1 + 0.460 650 707 300 176 398 909 210 361 856;
  • 19) 0.460 650 707 300 176 398 909 210 361 856 × 2 = 0 + 0.921 301 414 600 352 797 818 420 723 712;
  • 20) 0.921 301 414 600 352 797 818 420 723 712 × 2 = 1 + 0.842 602 829 200 705 595 636 841 447 424;
  • 21) 0.842 602 829 200 705 595 636 841 447 424 × 2 = 1 + 0.685 205 658 401 411 191 273 682 894 848;
  • 22) 0.685 205 658 401 411 191 273 682 894 848 × 2 = 1 + 0.370 411 316 802 822 382 547 365 789 696;
  • 23) 0.370 411 316 802 822 382 547 365 789 696 × 2 = 0 + 0.740 822 633 605 644 765 094 731 579 392;
  • 24) 0.740 822 633 605 644 765 094 731 579 392 × 2 = 1 + 0.481 645 267 211 289 530 189 463 158 784;
  • 25) 0.481 645 267 211 289 530 189 463 158 784 × 2 = 0 + 0.963 290 534 422 579 060 378 926 317 568;
  • 26) 0.963 290 534 422 579 060 378 926 317 568 × 2 = 1 + 0.926 581 068 845 158 120 757 852 635 136;
  • 27) 0.926 581 068 845 158 120 757 852 635 136 × 2 = 1 + 0.853 162 137 690 316 241 515 705 270 272;
  • 28) 0.853 162 137 690 316 241 515 705 270 272 × 2 = 1 + 0.706 324 275 380 632 483 031 410 540 544;
  • 29) 0.706 324 275 380 632 483 031 410 540 544 × 2 = 1 + 0.412 648 550 761 264 966 062 821 081 088;
  • 30) 0.412 648 550 761 264 966 062 821 081 088 × 2 = 0 + 0.825 297 101 522 529 932 125 642 162 176;
  • 31) 0.825 297 101 522 529 932 125 642 162 176 × 2 = 1 + 0.650 594 203 045 059 864 251 284 324 352;
  • 32) 0.650 594 203 045 059 864 251 284 324 352 × 2 = 1 + 0.301 188 406 090 119 728 502 568 648 704;
  • 33) 0.301 188 406 090 119 728 502 568 648 704 × 2 = 0 + 0.602 376 812 180 239 457 005 137 297 408;
  • 34) 0.602 376 812 180 239 457 005 137 297 408 × 2 = 1 + 0.204 753 624 360 478 914 010 274 594 816;
  • 35) 0.204 753 624 360 478 914 010 274 594 816 × 2 = 0 + 0.409 507 248 720 957 828 020 549 189 632;
  • 36) 0.409 507 248 720 957 828 020 549 189 632 × 2 = 0 + 0.819 014 497 441 915 656 041 098 379 264;
  • 37) 0.819 014 497 441 915 656 041 098 379 264 × 2 = 1 + 0.638 028 994 883 831 312 082 196 758 528;
  • 38) 0.638 028 994 883 831 312 082 196 758 528 × 2 = 1 + 0.276 057 989 767 662 624 164 393 517 056;
  • 39) 0.276 057 989 767 662 624 164 393 517 056 × 2 = 0 + 0.552 115 979 535 325 248 328 787 034 112;
  • 40) 0.552 115 979 535 325 248 328 787 034 112 × 2 = 1 + 0.104 231 959 070 650 496 657 574 068 224;
  • 41) 0.104 231 959 070 650 496 657 574 068 224 × 2 = 0 + 0.208 463 918 141 300 993 315 148 136 448;
  • 42) 0.208 463 918 141 300 993 315 148 136 448 × 2 = 0 + 0.416 927 836 282 601 986 630 296 272 896;
  • 43) 0.416 927 836 282 601 986 630 296 272 896 × 2 = 0 + 0.833 855 672 565 203 973 260 592 545 792;
  • 44) 0.833 855 672 565 203 973 260 592 545 792 × 2 = 1 + 0.667 711 345 130 407 946 521 185 091 584;
  • 45) 0.667 711 345 130 407 946 521 185 091 584 × 2 = 1 + 0.335 422 690 260 815 893 042 370 183 168;
  • 46) 0.335 422 690 260 815 893 042 370 183 168 × 2 = 0 + 0.670 845 380 521 631 786 084 740 366 336;
  • 47) 0.670 845 380 521 631 786 084 740 366 336 × 2 = 1 + 0.341 690 761 043 263 572 169 480 732 672;
  • 48) 0.341 690 761 043 263 572 169 480 732 672 × 2 = 0 + 0.683 381 522 086 527 144 338 961 465 344;
  • 49) 0.683 381 522 086 527 144 338 961 465 344 × 2 = 1 + 0.366 763 044 173 054 288 677 922 930 688;
  • 50) 0.366 763 044 173 054 288 677 922 930 688 × 2 = 0 + 0.733 526 088 346 108 577 355 845 861 376;
  • 51) 0.733 526 088 346 108 577 355 845 861 376 × 2 = 1 + 0.467 052 176 692 217 154 711 691 722 752;
  • 52) 0.467 052 176 692 217 154 711 691 722 752 × 2 = 0 + 0.934 104 353 384 434 309 423 383 445 504;
  • 53) 0.934 104 353 384 434 309 423 383 445 504 × 2 = 1 + 0.868 208 706 768 868 618 846 766 891 008;
  • 54) 0.868 208 706 768 868 618 846 766 891 008 × 2 = 1 + 0.736 417 413 537 737 237 693 533 782 016;
  • 55) 0.736 417 413 537 737 237 693 533 782 016 × 2 = 1 + 0.472 834 827 075 474 475 387 067 564 032;
  • 56) 0.472 834 827 075 474 475 387 067 564 032 × 2 = 0 + 0.945 669 654 150 948 950 774 135 128 064;
  • 57) 0.945 669 654 150 948 950 774 135 128 064 × 2 = 1 + 0.891 339 308 301 897 901 548 270 256 128;
  • 58) 0.891 339 308 301 897 901 548 270 256 128 × 2 = 1 + 0.782 678 616 603 795 803 096 540 512 256;
  • 59) 0.782 678 616 603 795 803 096 540 512 256 × 2 = 1 + 0.565 357 233 207 591 606 193 081 024 512;
  • 60) 0.565 357 233 207 591 606 193 081 024 512 × 2 = 1 + 0.130 714 466 415 183 212 386 162 049 024;
  • 61) 0.130 714 466 415 183 212 386 162 049 024 × 2 = 0 + 0.261 428 932 830 366 424 772 324 098 048;
  • 62) 0.261 428 932 830 366 424 772 324 098 048 × 2 = 0 + 0.522 857 865 660 732 849 544 648 196 096;
  • 63) 0.522 857 865 660 732 849 544 648 196 096 × 2 = 1 + 0.045 715 731 321 465 699 089 296 392 192;
  • 64) 0.045 715 731 321 465 699 089 296 392 192 × 2 = 0 + 0.091 431 462 642 931 398 178 592 784 384;
  • 65) 0.091 431 462 642 931 398 178 592 784 384 × 2 = 0 + 0.182 862 925 285 862 796 357 185 568 768;
  • 66) 0.182 862 925 285 862 796 357 185 568 768 × 2 = 0 + 0.365 725 850 571 725 592 714 371 137 536;
  • 67) 0.365 725 850 571 725 592 714 371 137 536 × 2 = 0 + 0.731 451 701 143 451 185 428 742 275 072;
  • 68) 0.731 451 701 143 451 185 428 742 275 072 × 2 = 1 + 0.462 903 402 286 902 370 857 484 550 144;

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:


0.000 020 830 729 321 671 205 134 999 124(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:

0.000 020 830 729 321 671 205 134 999 124(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:


0.000 020 830 729 321 671 205 134 999 124(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 0.000 020 830 729 321 671 205 134 999 124 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