0.000 020 830 729 321 671 205 134 999 169 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 169(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
0.000 020 830 729 321 671 205 134 999 169(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: 0.000 020 830 729 321 671 205 134 999 169.

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) 0.000 020 830 729 321 671 205 134 999 169 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 338;
  • 2) 0.000 041 661 458 643 342 410 269 998 338 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 676;
  • 3) 0.000 083 322 917 286 684 820 539 996 676 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 352;
  • 4) 0.000 166 645 834 573 369 641 079 993 352 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 704;
  • 5) 0.000 333 291 669 146 739 282 159 986 704 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 408;
  • 6) 0.000 666 583 338 293 478 564 319 973 408 × 2 = 0 + 0.001 333 166 676 586 957 128 639 946 816;
  • 7) 0.001 333 166 676 586 957 128 639 946 816 × 2 = 0 + 0.002 666 333 353 173 914 257 279 893 632;
  • 8) 0.002 666 333 353 173 914 257 279 893 632 × 2 = 0 + 0.005 332 666 706 347 828 514 559 787 264;
  • 9) 0.005 332 666 706 347 828 514 559 787 264 × 2 = 0 + 0.010 665 333 412 695 657 029 119 574 528;
  • 10) 0.010 665 333 412 695 657 029 119 574 528 × 2 = 0 + 0.021 330 666 825 391 314 058 239 149 056;
  • 11) 0.021 330 666 825 391 314 058 239 149 056 × 2 = 0 + 0.042 661 333 650 782 628 116 478 298 112;
  • 12) 0.042 661 333 650 782 628 116 478 298 112 × 2 = 0 + 0.085 322 667 301 565 256 232 956 596 224;
  • 13) 0.085 322 667 301 565 256 232 956 596 224 × 2 = 0 + 0.170 645 334 603 130 512 465 913 192 448;
  • 14) 0.170 645 334 603 130 512 465 913 192 448 × 2 = 0 + 0.341 290 669 206 261 024 931 826 384 896;
  • 15) 0.341 290 669 206 261 024 931 826 384 896 × 2 = 0 + 0.682 581 338 412 522 049 863 652 769 792;
  • 16) 0.682 581 338 412 522 049 863 652 769 792 × 2 = 1 + 0.365 162 676 825 044 099 727 305 539 584;
  • 17) 0.365 162 676 825 044 099 727 305 539 584 × 2 = 0 + 0.730 325 353 650 088 199 454 611 079 168;
  • 18) 0.730 325 353 650 088 199 454 611 079 168 × 2 = 1 + 0.460 650 707 300 176 398 909 222 158 336;
  • 19) 0.460 650 707 300 176 398 909 222 158 336 × 2 = 0 + 0.921 301 414 600 352 797 818 444 316 672;
  • 20) 0.921 301 414 600 352 797 818 444 316 672 × 2 = 1 + 0.842 602 829 200 705 595 636 888 633 344;
  • 21) 0.842 602 829 200 705 595 636 888 633 344 × 2 = 1 + 0.685 205 658 401 411 191 273 777 266 688;
  • 22) 0.685 205 658 401 411 191 273 777 266 688 × 2 = 1 + 0.370 411 316 802 822 382 547 554 533 376;
  • 23) 0.370 411 316 802 822 382 547 554 533 376 × 2 = 0 + 0.740 822 633 605 644 765 095 109 066 752;
  • 24) 0.740 822 633 605 644 765 095 109 066 752 × 2 = 1 + 0.481 645 267 211 289 530 190 218 133 504;
  • 25) 0.481 645 267 211 289 530 190 218 133 504 × 2 = 0 + 0.963 290 534 422 579 060 380 436 267 008;
  • 26) 0.963 290 534 422 579 060 380 436 267 008 × 2 = 1 + 0.926 581 068 845 158 120 760 872 534 016;
  • 27) 0.926 581 068 845 158 120 760 872 534 016 × 2 = 1 + 0.853 162 137 690 316 241 521 745 068 032;
  • 28) 0.853 162 137 690 316 241 521 745 068 032 × 2 = 1 + 0.706 324 275 380 632 483 043 490 136 064;
  • 29) 0.706 324 275 380 632 483 043 490 136 064 × 2 = 1 + 0.412 648 550 761 264 966 086 980 272 128;
  • 30) 0.412 648 550 761 264 966 086 980 272 128 × 2 = 0 + 0.825 297 101 522 529 932 173 960 544 256;
  • 31) 0.825 297 101 522 529 932 173 960 544 256 × 2 = 1 + 0.650 594 203 045 059 864 347 921 088 512;
  • 32) 0.650 594 203 045 059 864 347 921 088 512 × 2 = 1 + 0.301 188 406 090 119 728 695 842 177 024;
  • 33) 0.301 188 406 090 119 728 695 842 177 024 × 2 = 0 + 0.602 376 812 180 239 457 391 684 354 048;
  • 34) 0.602 376 812 180 239 457 391 684 354 048 × 2 = 1 + 0.204 753 624 360 478 914 783 368 708 096;
  • 35) 0.204 753 624 360 478 914 783 368 708 096 × 2 = 0 + 0.409 507 248 720 957 829 566 737 416 192;
  • 36) 0.409 507 248 720 957 829 566 737 416 192 × 2 = 0 + 0.819 014 497 441 915 659 133 474 832 384;
  • 37) 0.819 014 497 441 915 659 133 474 832 384 × 2 = 1 + 0.638 028 994 883 831 318 266 949 664 768;
  • 38) 0.638 028 994 883 831 318 266 949 664 768 × 2 = 1 + 0.276 057 989 767 662 636 533 899 329 536;
  • 39) 0.276 057 989 767 662 636 533 899 329 536 × 2 = 0 + 0.552 115 979 535 325 273 067 798 659 072;
  • 40) 0.552 115 979 535 325 273 067 798 659 072 × 2 = 1 + 0.104 231 959 070 650 546 135 597 318 144;
  • 41) 0.104 231 959 070 650 546 135 597 318 144 × 2 = 0 + 0.208 463 918 141 301 092 271 194 636 288;
  • 42) 0.208 463 918 141 301 092 271 194 636 288 × 2 = 0 + 0.416 927 836 282 602 184 542 389 272 576;
  • 43) 0.416 927 836 282 602 184 542 389 272 576 × 2 = 0 + 0.833 855 672 565 204 369 084 778 545 152;
  • 44) 0.833 855 672 565 204 369 084 778 545 152 × 2 = 1 + 0.667 711 345 130 408 738 169 557 090 304;
  • 45) 0.667 711 345 130 408 738 169 557 090 304 × 2 = 1 + 0.335 422 690 260 817 476 339 114 180 608;
  • 46) 0.335 422 690 260 817 476 339 114 180 608 × 2 = 0 + 0.670 845 380 521 634 952 678 228 361 216;
  • 47) 0.670 845 380 521 634 952 678 228 361 216 × 2 = 1 + 0.341 690 761 043 269 905 356 456 722 432;
  • 48) 0.341 690 761 043 269 905 356 456 722 432 × 2 = 0 + 0.683 381 522 086 539 810 712 913 444 864;
  • 49) 0.683 381 522 086 539 810 712 913 444 864 × 2 = 1 + 0.366 763 044 173 079 621 425 826 889 728;
  • 50) 0.366 763 044 173 079 621 425 826 889 728 × 2 = 0 + 0.733 526 088 346 159 242 851 653 779 456;
  • 51) 0.733 526 088 346 159 242 851 653 779 456 × 2 = 1 + 0.467 052 176 692 318 485 703 307 558 912;
  • 52) 0.467 052 176 692 318 485 703 307 558 912 × 2 = 0 + 0.934 104 353 384 636 971 406 615 117 824;
  • 53) 0.934 104 353 384 636 971 406 615 117 824 × 2 = 1 + 0.868 208 706 769 273 942 813 230 235 648;
  • 54) 0.868 208 706 769 273 942 813 230 235 648 × 2 = 1 + 0.736 417 413 538 547 885 626 460 471 296;
  • 55) 0.736 417 413 538 547 885 626 460 471 296 × 2 = 1 + 0.472 834 827 077 095 771 252 920 942 592;
  • 56) 0.472 834 827 077 095 771 252 920 942 592 × 2 = 0 + 0.945 669 654 154 191 542 505 841 885 184;
  • 57) 0.945 669 654 154 191 542 505 841 885 184 × 2 = 1 + 0.891 339 308 308 383 085 011 683 770 368;
  • 58) 0.891 339 308 308 383 085 011 683 770 368 × 2 = 1 + 0.782 678 616 616 766 170 023 367 540 736;
  • 59) 0.782 678 616 616 766 170 023 367 540 736 × 2 = 1 + 0.565 357 233 233 532 340 046 735 081 472;
  • 60) 0.565 357 233 233 532 340 046 735 081 472 × 2 = 1 + 0.130 714 466 467 064 680 093 470 162 944;
  • 61) 0.130 714 466 467 064 680 093 470 162 944 × 2 = 0 + 0.261 428 932 934 129 360 186 940 325 888;
  • 62) 0.261 428 932 934 129 360 186 940 325 888 × 2 = 0 + 0.522 857 865 868 258 720 373 880 651 776;
  • 63) 0.522 857 865 868 258 720 373 880 651 776 × 2 = 1 + 0.045 715 731 736 517 440 747 761 303 552;
  • 64) 0.045 715 731 736 517 440 747 761 303 552 × 2 = 0 + 0.091 431 463 473 034 881 495 522 607 104;
  • 65) 0.091 431 463 473 034 881 495 522 607 104 × 2 = 0 + 0.182 862 926 946 069 762 991 045 214 208;
  • 66) 0.182 862 926 946 069 762 991 045 214 208 × 2 = 0 + 0.365 725 853 892 139 525 982 090 428 416;
  • 67) 0.365 725 853 892 139 525 982 090 428 416 × 2 = 0 + 0.731 451 707 784 279 051 964 180 856 832;
  • 68) 0.731 451 707 784 279 051 964 180 856 832 × 2 = 1 + 0.462 903 415 568 558 103 928 361 713 664;

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:


0.000 020 830 729 321 671 205 134 999 169(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:

0.000 020 830 729 321 671 205 134 999 169(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:


0.000 020 830 729 321 671 205 134 999 169(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 0.000 020 830 729 321 671 205 134 999 169 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