0.000 020 830 729 321 671 205 134 999 007 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 007(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
0.000 020 830 729 321 671 205 134 999 007(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: 0.000 020 830 729 321 671 205 134 999 007.

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) 0.000 020 830 729 321 671 205 134 999 007 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 014;
  • 2) 0.000 041 661 458 643 342 410 269 998 014 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 028;
  • 3) 0.000 083 322 917 286 684 820 539 996 028 × 2 = 0 + 0.000 166 645 834 573 369 641 079 992 056;
  • 4) 0.000 166 645 834 573 369 641 079 992 056 × 2 = 0 + 0.000 333 291 669 146 739 282 159 984 112;
  • 5) 0.000 333 291 669 146 739 282 159 984 112 × 2 = 0 + 0.000 666 583 338 293 478 564 319 968 224;
  • 6) 0.000 666 583 338 293 478 564 319 968 224 × 2 = 0 + 0.001 333 166 676 586 957 128 639 936 448;
  • 7) 0.001 333 166 676 586 957 128 639 936 448 × 2 = 0 + 0.002 666 333 353 173 914 257 279 872 896;
  • 8) 0.002 666 333 353 173 914 257 279 872 896 × 2 = 0 + 0.005 332 666 706 347 828 514 559 745 792;
  • 9) 0.005 332 666 706 347 828 514 559 745 792 × 2 = 0 + 0.010 665 333 412 695 657 029 119 491 584;
  • 10) 0.010 665 333 412 695 657 029 119 491 584 × 2 = 0 + 0.021 330 666 825 391 314 058 238 983 168;
  • 11) 0.021 330 666 825 391 314 058 238 983 168 × 2 = 0 + 0.042 661 333 650 782 628 116 477 966 336;
  • 12) 0.042 661 333 650 782 628 116 477 966 336 × 2 = 0 + 0.085 322 667 301 565 256 232 955 932 672;
  • 13) 0.085 322 667 301 565 256 232 955 932 672 × 2 = 0 + 0.170 645 334 603 130 512 465 911 865 344;
  • 14) 0.170 645 334 603 130 512 465 911 865 344 × 2 = 0 + 0.341 290 669 206 261 024 931 823 730 688;
  • 15) 0.341 290 669 206 261 024 931 823 730 688 × 2 = 0 + 0.682 581 338 412 522 049 863 647 461 376;
  • 16) 0.682 581 338 412 522 049 863 647 461 376 × 2 = 1 + 0.365 162 676 825 044 099 727 294 922 752;
  • 17) 0.365 162 676 825 044 099 727 294 922 752 × 2 = 0 + 0.730 325 353 650 088 199 454 589 845 504;
  • 18) 0.730 325 353 650 088 199 454 589 845 504 × 2 = 1 + 0.460 650 707 300 176 398 909 179 691 008;
  • 19) 0.460 650 707 300 176 398 909 179 691 008 × 2 = 0 + 0.921 301 414 600 352 797 818 359 382 016;
  • 20) 0.921 301 414 600 352 797 818 359 382 016 × 2 = 1 + 0.842 602 829 200 705 595 636 718 764 032;
  • 21) 0.842 602 829 200 705 595 636 718 764 032 × 2 = 1 + 0.685 205 658 401 411 191 273 437 528 064;
  • 22) 0.685 205 658 401 411 191 273 437 528 064 × 2 = 1 + 0.370 411 316 802 822 382 546 875 056 128;
  • 23) 0.370 411 316 802 822 382 546 875 056 128 × 2 = 0 + 0.740 822 633 605 644 765 093 750 112 256;
  • 24) 0.740 822 633 605 644 765 093 750 112 256 × 2 = 1 + 0.481 645 267 211 289 530 187 500 224 512;
  • 25) 0.481 645 267 211 289 530 187 500 224 512 × 2 = 0 + 0.963 290 534 422 579 060 375 000 449 024;
  • 26) 0.963 290 534 422 579 060 375 000 449 024 × 2 = 1 + 0.926 581 068 845 158 120 750 000 898 048;
  • 27) 0.926 581 068 845 158 120 750 000 898 048 × 2 = 1 + 0.853 162 137 690 316 241 500 001 796 096;
  • 28) 0.853 162 137 690 316 241 500 001 796 096 × 2 = 1 + 0.706 324 275 380 632 483 000 003 592 192;
  • 29) 0.706 324 275 380 632 483 000 003 592 192 × 2 = 1 + 0.412 648 550 761 264 966 000 007 184 384;
  • 30) 0.412 648 550 761 264 966 000 007 184 384 × 2 = 0 + 0.825 297 101 522 529 932 000 014 368 768;
  • 31) 0.825 297 101 522 529 932 000 014 368 768 × 2 = 1 + 0.650 594 203 045 059 864 000 028 737 536;
  • 32) 0.650 594 203 045 059 864 000 028 737 536 × 2 = 1 + 0.301 188 406 090 119 728 000 057 475 072;
  • 33) 0.301 188 406 090 119 728 000 057 475 072 × 2 = 0 + 0.602 376 812 180 239 456 000 114 950 144;
  • 34) 0.602 376 812 180 239 456 000 114 950 144 × 2 = 1 + 0.204 753 624 360 478 912 000 229 900 288;
  • 35) 0.204 753 624 360 478 912 000 229 900 288 × 2 = 0 + 0.409 507 248 720 957 824 000 459 800 576;
  • 36) 0.409 507 248 720 957 824 000 459 800 576 × 2 = 0 + 0.819 014 497 441 915 648 000 919 601 152;
  • 37) 0.819 014 497 441 915 648 000 919 601 152 × 2 = 1 + 0.638 028 994 883 831 296 001 839 202 304;
  • 38) 0.638 028 994 883 831 296 001 839 202 304 × 2 = 1 + 0.276 057 989 767 662 592 003 678 404 608;
  • 39) 0.276 057 989 767 662 592 003 678 404 608 × 2 = 0 + 0.552 115 979 535 325 184 007 356 809 216;
  • 40) 0.552 115 979 535 325 184 007 356 809 216 × 2 = 1 + 0.104 231 959 070 650 368 014 713 618 432;
  • 41) 0.104 231 959 070 650 368 014 713 618 432 × 2 = 0 + 0.208 463 918 141 300 736 029 427 236 864;
  • 42) 0.208 463 918 141 300 736 029 427 236 864 × 2 = 0 + 0.416 927 836 282 601 472 058 854 473 728;
  • 43) 0.416 927 836 282 601 472 058 854 473 728 × 2 = 0 + 0.833 855 672 565 202 944 117 708 947 456;
  • 44) 0.833 855 672 565 202 944 117 708 947 456 × 2 = 1 + 0.667 711 345 130 405 888 235 417 894 912;
  • 45) 0.667 711 345 130 405 888 235 417 894 912 × 2 = 1 + 0.335 422 690 260 811 776 470 835 789 824;
  • 46) 0.335 422 690 260 811 776 470 835 789 824 × 2 = 0 + 0.670 845 380 521 623 552 941 671 579 648;
  • 47) 0.670 845 380 521 623 552 941 671 579 648 × 2 = 1 + 0.341 690 761 043 247 105 883 343 159 296;
  • 48) 0.341 690 761 043 247 105 883 343 159 296 × 2 = 0 + 0.683 381 522 086 494 211 766 686 318 592;
  • 49) 0.683 381 522 086 494 211 766 686 318 592 × 2 = 1 + 0.366 763 044 172 988 423 533 372 637 184;
  • 50) 0.366 763 044 172 988 423 533 372 637 184 × 2 = 0 + 0.733 526 088 345 976 847 066 745 274 368;
  • 51) 0.733 526 088 345 976 847 066 745 274 368 × 2 = 1 + 0.467 052 176 691 953 694 133 490 548 736;
  • 52) 0.467 052 176 691 953 694 133 490 548 736 × 2 = 0 + 0.934 104 353 383 907 388 266 981 097 472;
  • 53) 0.934 104 353 383 907 388 266 981 097 472 × 2 = 1 + 0.868 208 706 767 814 776 533 962 194 944;
  • 54) 0.868 208 706 767 814 776 533 962 194 944 × 2 = 1 + 0.736 417 413 535 629 553 067 924 389 888;
  • 55) 0.736 417 413 535 629 553 067 924 389 888 × 2 = 1 + 0.472 834 827 071 259 106 135 848 779 776;
  • 56) 0.472 834 827 071 259 106 135 848 779 776 × 2 = 0 + 0.945 669 654 142 518 212 271 697 559 552;
  • 57) 0.945 669 654 142 518 212 271 697 559 552 × 2 = 1 + 0.891 339 308 285 036 424 543 395 119 104;
  • 58) 0.891 339 308 285 036 424 543 395 119 104 × 2 = 1 + 0.782 678 616 570 072 849 086 790 238 208;
  • 59) 0.782 678 616 570 072 849 086 790 238 208 × 2 = 1 + 0.565 357 233 140 145 698 173 580 476 416;
  • 60) 0.565 357 233 140 145 698 173 580 476 416 × 2 = 1 + 0.130 714 466 280 291 396 347 160 952 832;
  • 61) 0.130 714 466 280 291 396 347 160 952 832 × 2 = 0 + 0.261 428 932 560 582 792 694 321 905 664;
  • 62) 0.261 428 932 560 582 792 694 321 905 664 × 2 = 0 + 0.522 857 865 121 165 585 388 643 811 328;
  • 63) 0.522 857 865 121 165 585 388 643 811 328 × 2 = 1 + 0.045 715 730 242 331 170 777 287 622 656;
  • 64) 0.045 715 730 242 331 170 777 287 622 656 × 2 = 0 + 0.091 431 460 484 662 341 554 575 245 312;
  • 65) 0.091 431 460 484 662 341 554 575 245 312 × 2 = 0 + 0.182 862 920 969 324 683 109 150 490 624;
  • 66) 0.182 862 920 969 324 683 109 150 490 624 × 2 = 0 + 0.365 725 841 938 649 366 218 300 981 248;
  • 67) 0.365 725 841 938 649 366 218 300 981 248 × 2 = 0 + 0.731 451 683 877 298 732 436 601 962 496;
  • 68) 0.731 451 683 877 298 732 436 601 962 496 × 2 = 1 + 0.462 903 367 754 597 464 873 203 924 992;

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:


0.000 020 830 729 321 671 205 134 999 007(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:

0.000 020 830 729 321 671 205 134 999 007(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:


0.000 020 830 729 321 671 205 134 999 007(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 0.000 020 830 729 321 671 205 134 999 007 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