0.000 020 830 729 321 671 205 134 999 241 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 241(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
0.000 020 830 729 321 671 205 134 999 241(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: 0.000 020 830 729 321 671 205 134 999 241.

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) 0.000 020 830 729 321 671 205 134 999 241 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 482;
  • 2) 0.000 041 661 458 643 342 410 269 998 482 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 964;
  • 3) 0.000 083 322 917 286 684 820 539 996 964 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 928;
  • 4) 0.000 166 645 834 573 369 641 079 993 928 × 2 = 0 + 0.000 333 291 669 146 739 282 159 987 856;
  • 5) 0.000 333 291 669 146 739 282 159 987 856 × 2 = 0 + 0.000 666 583 338 293 478 564 319 975 712;
  • 6) 0.000 666 583 338 293 478 564 319 975 712 × 2 = 0 + 0.001 333 166 676 586 957 128 639 951 424;
  • 7) 0.001 333 166 676 586 957 128 639 951 424 × 2 = 0 + 0.002 666 333 353 173 914 257 279 902 848;
  • 8) 0.002 666 333 353 173 914 257 279 902 848 × 2 = 0 + 0.005 332 666 706 347 828 514 559 805 696;
  • 9) 0.005 332 666 706 347 828 514 559 805 696 × 2 = 0 + 0.010 665 333 412 695 657 029 119 611 392;
  • 10) 0.010 665 333 412 695 657 029 119 611 392 × 2 = 0 + 0.021 330 666 825 391 314 058 239 222 784;
  • 11) 0.021 330 666 825 391 314 058 239 222 784 × 2 = 0 + 0.042 661 333 650 782 628 116 478 445 568;
  • 12) 0.042 661 333 650 782 628 116 478 445 568 × 2 = 0 + 0.085 322 667 301 565 256 232 956 891 136;
  • 13) 0.085 322 667 301 565 256 232 956 891 136 × 2 = 0 + 0.170 645 334 603 130 512 465 913 782 272;
  • 14) 0.170 645 334 603 130 512 465 913 782 272 × 2 = 0 + 0.341 290 669 206 261 024 931 827 564 544;
  • 15) 0.341 290 669 206 261 024 931 827 564 544 × 2 = 0 + 0.682 581 338 412 522 049 863 655 129 088;
  • 16) 0.682 581 338 412 522 049 863 655 129 088 × 2 = 1 + 0.365 162 676 825 044 099 727 310 258 176;
  • 17) 0.365 162 676 825 044 099 727 310 258 176 × 2 = 0 + 0.730 325 353 650 088 199 454 620 516 352;
  • 18) 0.730 325 353 650 088 199 454 620 516 352 × 2 = 1 + 0.460 650 707 300 176 398 909 241 032 704;
  • 19) 0.460 650 707 300 176 398 909 241 032 704 × 2 = 0 + 0.921 301 414 600 352 797 818 482 065 408;
  • 20) 0.921 301 414 600 352 797 818 482 065 408 × 2 = 1 + 0.842 602 829 200 705 595 636 964 130 816;
  • 21) 0.842 602 829 200 705 595 636 964 130 816 × 2 = 1 + 0.685 205 658 401 411 191 273 928 261 632;
  • 22) 0.685 205 658 401 411 191 273 928 261 632 × 2 = 1 + 0.370 411 316 802 822 382 547 856 523 264;
  • 23) 0.370 411 316 802 822 382 547 856 523 264 × 2 = 0 + 0.740 822 633 605 644 765 095 713 046 528;
  • 24) 0.740 822 633 605 644 765 095 713 046 528 × 2 = 1 + 0.481 645 267 211 289 530 191 426 093 056;
  • 25) 0.481 645 267 211 289 530 191 426 093 056 × 2 = 0 + 0.963 290 534 422 579 060 382 852 186 112;
  • 26) 0.963 290 534 422 579 060 382 852 186 112 × 2 = 1 + 0.926 581 068 845 158 120 765 704 372 224;
  • 27) 0.926 581 068 845 158 120 765 704 372 224 × 2 = 1 + 0.853 162 137 690 316 241 531 408 744 448;
  • 28) 0.853 162 137 690 316 241 531 408 744 448 × 2 = 1 + 0.706 324 275 380 632 483 062 817 488 896;
  • 29) 0.706 324 275 380 632 483 062 817 488 896 × 2 = 1 + 0.412 648 550 761 264 966 125 634 977 792;
  • 30) 0.412 648 550 761 264 966 125 634 977 792 × 2 = 0 + 0.825 297 101 522 529 932 251 269 955 584;
  • 31) 0.825 297 101 522 529 932 251 269 955 584 × 2 = 1 + 0.650 594 203 045 059 864 502 539 911 168;
  • 32) 0.650 594 203 045 059 864 502 539 911 168 × 2 = 1 + 0.301 188 406 090 119 729 005 079 822 336;
  • 33) 0.301 188 406 090 119 729 005 079 822 336 × 2 = 0 + 0.602 376 812 180 239 458 010 159 644 672;
  • 34) 0.602 376 812 180 239 458 010 159 644 672 × 2 = 1 + 0.204 753 624 360 478 916 020 319 289 344;
  • 35) 0.204 753 624 360 478 916 020 319 289 344 × 2 = 0 + 0.409 507 248 720 957 832 040 638 578 688;
  • 36) 0.409 507 248 720 957 832 040 638 578 688 × 2 = 0 + 0.819 014 497 441 915 664 081 277 157 376;
  • 37) 0.819 014 497 441 915 664 081 277 157 376 × 2 = 1 + 0.638 028 994 883 831 328 162 554 314 752;
  • 38) 0.638 028 994 883 831 328 162 554 314 752 × 2 = 1 + 0.276 057 989 767 662 656 325 108 629 504;
  • 39) 0.276 057 989 767 662 656 325 108 629 504 × 2 = 0 + 0.552 115 979 535 325 312 650 217 259 008;
  • 40) 0.552 115 979 535 325 312 650 217 259 008 × 2 = 1 + 0.104 231 959 070 650 625 300 434 518 016;
  • 41) 0.104 231 959 070 650 625 300 434 518 016 × 2 = 0 + 0.208 463 918 141 301 250 600 869 036 032;
  • 42) 0.208 463 918 141 301 250 600 869 036 032 × 2 = 0 + 0.416 927 836 282 602 501 201 738 072 064;
  • 43) 0.416 927 836 282 602 501 201 738 072 064 × 2 = 0 + 0.833 855 672 565 205 002 403 476 144 128;
  • 44) 0.833 855 672 565 205 002 403 476 144 128 × 2 = 1 + 0.667 711 345 130 410 004 806 952 288 256;
  • 45) 0.667 711 345 130 410 004 806 952 288 256 × 2 = 1 + 0.335 422 690 260 820 009 613 904 576 512;
  • 46) 0.335 422 690 260 820 009 613 904 576 512 × 2 = 0 + 0.670 845 380 521 640 019 227 809 153 024;
  • 47) 0.670 845 380 521 640 019 227 809 153 024 × 2 = 1 + 0.341 690 761 043 280 038 455 618 306 048;
  • 48) 0.341 690 761 043 280 038 455 618 306 048 × 2 = 0 + 0.683 381 522 086 560 076 911 236 612 096;
  • 49) 0.683 381 522 086 560 076 911 236 612 096 × 2 = 1 + 0.366 763 044 173 120 153 822 473 224 192;
  • 50) 0.366 763 044 173 120 153 822 473 224 192 × 2 = 0 + 0.733 526 088 346 240 307 644 946 448 384;
  • 51) 0.733 526 088 346 240 307 644 946 448 384 × 2 = 1 + 0.467 052 176 692 480 615 289 892 896 768;
  • 52) 0.467 052 176 692 480 615 289 892 896 768 × 2 = 0 + 0.934 104 353 384 961 230 579 785 793 536;
  • 53) 0.934 104 353 384 961 230 579 785 793 536 × 2 = 1 + 0.868 208 706 769 922 461 159 571 587 072;
  • 54) 0.868 208 706 769 922 461 159 571 587 072 × 2 = 1 + 0.736 417 413 539 844 922 319 143 174 144;
  • 55) 0.736 417 413 539 844 922 319 143 174 144 × 2 = 1 + 0.472 834 827 079 689 844 638 286 348 288;
  • 56) 0.472 834 827 079 689 844 638 286 348 288 × 2 = 0 + 0.945 669 654 159 379 689 276 572 696 576;
  • 57) 0.945 669 654 159 379 689 276 572 696 576 × 2 = 1 + 0.891 339 308 318 759 378 553 145 393 152;
  • 58) 0.891 339 308 318 759 378 553 145 393 152 × 2 = 1 + 0.782 678 616 637 518 757 106 290 786 304;
  • 59) 0.782 678 616 637 518 757 106 290 786 304 × 2 = 1 + 0.565 357 233 275 037 514 212 581 572 608;
  • 60) 0.565 357 233 275 037 514 212 581 572 608 × 2 = 1 + 0.130 714 466 550 075 028 425 163 145 216;
  • 61) 0.130 714 466 550 075 028 425 163 145 216 × 2 = 0 + 0.261 428 933 100 150 056 850 326 290 432;
  • 62) 0.261 428 933 100 150 056 850 326 290 432 × 2 = 0 + 0.522 857 866 200 300 113 700 652 580 864;
  • 63) 0.522 857 866 200 300 113 700 652 580 864 × 2 = 1 + 0.045 715 732 400 600 227 401 305 161 728;
  • 64) 0.045 715 732 400 600 227 401 305 161 728 × 2 = 0 + 0.091 431 464 801 200 454 802 610 323 456;
  • 65) 0.091 431 464 801 200 454 802 610 323 456 × 2 = 0 + 0.182 862 929 602 400 909 605 220 646 912;
  • 66) 0.182 862 929 602 400 909 605 220 646 912 × 2 = 0 + 0.365 725 859 204 801 819 210 441 293 824;
  • 67) 0.365 725 859 204 801 819 210 441 293 824 × 2 = 0 + 0.731 451 718 409 603 638 420 882 587 648;
  • 68) 0.731 451 718 409 603 638 420 882 587 648 × 2 = 1 + 0.462 903 436 819 207 276 841 765 175 296;

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:


0.000 020 830 729 321 671 205 134 999 241(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:

0.000 020 830 729 321 671 205 134 999 241(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:


0.000 020 830 729 321 671 205 134 999 241(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 0.000 020 830 729 321 671 205 134 999 241 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