0.000 020 830 729 321 671 205 134 999 164 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 164(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
0.000 020 830 729 321 671 205 134 999 164(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: 0.000 020 830 729 321 671 205 134 999 164.

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) 0.000 020 830 729 321 671 205 134 999 164 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 328;
  • 2) 0.000 041 661 458 643 342 410 269 998 328 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 656;
  • 3) 0.000 083 322 917 286 684 820 539 996 656 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 312;
  • 4) 0.000 166 645 834 573 369 641 079 993 312 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 624;
  • 5) 0.000 333 291 669 146 739 282 159 986 624 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 248;
  • 6) 0.000 666 583 338 293 478 564 319 973 248 × 2 = 0 + 0.001 333 166 676 586 957 128 639 946 496;
  • 7) 0.001 333 166 676 586 957 128 639 946 496 × 2 = 0 + 0.002 666 333 353 173 914 257 279 892 992;
  • 8) 0.002 666 333 353 173 914 257 279 892 992 × 2 = 0 + 0.005 332 666 706 347 828 514 559 785 984;
  • 9) 0.005 332 666 706 347 828 514 559 785 984 × 2 = 0 + 0.010 665 333 412 695 657 029 119 571 968;
  • 10) 0.010 665 333 412 695 657 029 119 571 968 × 2 = 0 + 0.021 330 666 825 391 314 058 239 143 936;
  • 11) 0.021 330 666 825 391 314 058 239 143 936 × 2 = 0 + 0.042 661 333 650 782 628 116 478 287 872;
  • 12) 0.042 661 333 650 782 628 116 478 287 872 × 2 = 0 + 0.085 322 667 301 565 256 232 956 575 744;
  • 13) 0.085 322 667 301 565 256 232 956 575 744 × 2 = 0 + 0.170 645 334 603 130 512 465 913 151 488;
  • 14) 0.170 645 334 603 130 512 465 913 151 488 × 2 = 0 + 0.341 290 669 206 261 024 931 826 302 976;
  • 15) 0.341 290 669 206 261 024 931 826 302 976 × 2 = 0 + 0.682 581 338 412 522 049 863 652 605 952;
  • 16) 0.682 581 338 412 522 049 863 652 605 952 × 2 = 1 + 0.365 162 676 825 044 099 727 305 211 904;
  • 17) 0.365 162 676 825 044 099 727 305 211 904 × 2 = 0 + 0.730 325 353 650 088 199 454 610 423 808;
  • 18) 0.730 325 353 650 088 199 454 610 423 808 × 2 = 1 + 0.460 650 707 300 176 398 909 220 847 616;
  • 19) 0.460 650 707 300 176 398 909 220 847 616 × 2 = 0 + 0.921 301 414 600 352 797 818 441 695 232;
  • 20) 0.921 301 414 600 352 797 818 441 695 232 × 2 = 1 + 0.842 602 829 200 705 595 636 883 390 464;
  • 21) 0.842 602 829 200 705 595 636 883 390 464 × 2 = 1 + 0.685 205 658 401 411 191 273 766 780 928;
  • 22) 0.685 205 658 401 411 191 273 766 780 928 × 2 = 1 + 0.370 411 316 802 822 382 547 533 561 856;
  • 23) 0.370 411 316 802 822 382 547 533 561 856 × 2 = 0 + 0.740 822 633 605 644 765 095 067 123 712;
  • 24) 0.740 822 633 605 644 765 095 067 123 712 × 2 = 1 + 0.481 645 267 211 289 530 190 134 247 424;
  • 25) 0.481 645 267 211 289 530 190 134 247 424 × 2 = 0 + 0.963 290 534 422 579 060 380 268 494 848;
  • 26) 0.963 290 534 422 579 060 380 268 494 848 × 2 = 1 + 0.926 581 068 845 158 120 760 536 989 696;
  • 27) 0.926 581 068 845 158 120 760 536 989 696 × 2 = 1 + 0.853 162 137 690 316 241 521 073 979 392;
  • 28) 0.853 162 137 690 316 241 521 073 979 392 × 2 = 1 + 0.706 324 275 380 632 483 042 147 958 784;
  • 29) 0.706 324 275 380 632 483 042 147 958 784 × 2 = 1 + 0.412 648 550 761 264 966 084 295 917 568;
  • 30) 0.412 648 550 761 264 966 084 295 917 568 × 2 = 0 + 0.825 297 101 522 529 932 168 591 835 136;
  • 31) 0.825 297 101 522 529 932 168 591 835 136 × 2 = 1 + 0.650 594 203 045 059 864 337 183 670 272;
  • 32) 0.650 594 203 045 059 864 337 183 670 272 × 2 = 1 + 0.301 188 406 090 119 728 674 367 340 544;
  • 33) 0.301 188 406 090 119 728 674 367 340 544 × 2 = 0 + 0.602 376 812 180 239 457 348 734 681 088;
  • 34) 0.602 376 812 180 239 457 348 734 681 088 × 2 = 1 + 0.204 753 624 360 478 914 697 469 362 176;
  • 35) 0.204 753 624 360 478 914 697 469 362 176 × 2 = 0 + 0.409 507 248 720 957 829 394 938 724 352;
  • 36) 0.409 507 248 720 957 829 394 938 724 352 × 2 = 0 + 0.819 014 497 441 915 658 789 877 448 704;
  • 37) 0.819 014 497 441 915 658 789 877 448 704 × 2 = 1 + 0.638 028 994 883 831 317 579 754 897 408;
  • 38) 0.638 028 994 883 831 317 579 754 897 408 × 2 = 1 + 0.276 057 989 767 662 635 159 509 794 816;
  • 39) 0.276 057 989 767 662 635 159 509 794 816 × 2 = 0 + 0.552 115 979 535 325 270 319 019 589 632;
  • 40) 0.552 115 979 535 325 270 319 019 589 632 × 2 = 1 + 0.104 231 959 070 650 540 638 039 179 264;
  • 41) 0.104 231 959 070 650 540 638 039 179 264 × 2 = 0 + 0.208 463 918 141 301 081 276 078 358 528;
  • 42) 0.208 463 918 141 301 081 276 078 358 528 × 2 = 0 + 0.416 927 836 282 602 162 552 156 717 056;
  • 43) 0.416 927 836 282 602 162 552 156 717 056 × 2 = 0 + 0.833 855 672 565 204 325 104 313 434 112;
  • 44) 0.833 855 672 565 204 325 104 313 434 112 × 2 = 1 + 0.667 711 345 130 408 650 208 626 868 224;
  • 45) 0.667 711 345 130 408 650 208 626 868 224 × 2 = 1 + 0.335 422 690 260 817 300 417 253 736 448;
  • 46) 0.335 422 690 260 817 300 417 253 736 448 × 2 = 0 + 0.670 845 380 521 634 600 834 507 472 896;
  • 47) 0.670 845 380 521 634 600 834 507 472 896 × 2 = 1 + 0.341 690 761 043 269 201 669 014 945 792;
  • 48) 0.341 690 761 043 269 201 669 014 945 792 × 2 = 0 + 0.683 381 522 086 538 403 338 029 891 584;
  • 49) 0.683 381 522 086 538 403 338 029 891 584 × 2 = 1 + 0.366 763 044 173 076 806 676 059 783 168;
  • 50) 0.366 763 044 173 076 806 676 059 783 168 × 2 = 0 + 0.733 526 088 346 153 613 352 119 566 336;
  • 51) 0.733 526 088 346 153 613 352 119 566 336 × 2 = 1 + 0.467 052 176 692 307 226 704 239 132 672;
  • 52) 0.467 052 176 692 307 226 704 239 132 672 × 2 = 0 + 0.934 104 353 384 614 453 408 478 265 344;
  • 53) 0.934 104 353 384 614 453 408 478 265 344 × 2 = 1 + 0.868 208 706 769 228 906 816 956 530 688;
  • 54) 0.868 208 706 769 228 906 816 956 530 688 × 2 = 1 + 0.736 417 413 538 457 813 633 913 061 376;
  • 55) 0.736 417 413 538 457 813 633 913 061 376 × 2 = 1 + 0.472 834 827 076 915 627 267 826 122 752;
  • 56) 0.472 834 827 076 915 627 267 826 122 752 × 2 = 0 + 0.945 669 654 153 831 254 535 652 245 504;
  • 57) 0.945 669 654 153 831 254 535 652 245 504 × 2 = 1 + 0.891 339 308 307 662 509 071 304 491 008;
  • 58) 0.891 339 308 307 662 509 071 304 491 008 × 2 = 1 + 0.782 678 616 615 325 018 142 608 982 016;
  • 59) 0.782 678 616 615 325 018 142 608 982 016 × 2 = 1 + 0.565 357 233 230 650 036 285 217 964 032;
  • 60) 0.565 357 233 230 650 036 285 217 964 032 × 2 = 1 + 0.130 714 466 461 300 072 570 435 928 064;
  • 61) 0.130 714 466 461 300 072 570 435 928 064 × 2 = 0 + 0.261 428 932 922 600 145 140 871 856 128;
  • 62) 0.261 428 932 922 600 145 140 871 856 128 × 2 = 0 + 0.522 857 865 845 200 290 281 743 712 256;
  • 63) 0.522 857 865 845 200 290 281 743 712 256 × 2 = 1 + 0.045 715 731 690 400 580 563 487 424 512;
  • 64) 0.045 715 731 690 400 580 563 487 424 512 × 2 = 0 + 0.091 431 463 380 801 161 126 974 849 024;
  • 65) 0.091 431 463 380 801 161 126 974 849 024 × 2 = 0 + 0.182 862 926 761 602 322 253 949 698 048;
  • 66) 0.182 862 926 761 602 322 253 949 698 048 × 2 = 0 + 0.365 725 853 523 204 644 507 899 396 096;
  • 67) 0.365 725 853 523 204 644 507 899 396 096 × 2 = 0 + 0.731 451 707 046 409 289 015 798 792 192;
  • 68) 0.731 451 707 046 409 289 015 798 792 192 × 2 = 1 + 0.462 903 414 092 818 578 031 597 584 384;

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:


0.000 020 830 729 321 671 205 134 999 164(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:

0.000 020 830 729 321 671 205 134 999 164(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:


0.000 020 830 729 321 671 205 134 999 164(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 0.000 020 830 729 321 671 205 134 999 164 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