0.000 020 830 729 321 671 205 134 999 177 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 177(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
0.000 020 830 729 321 671 205 134 999 177(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: 0.000 020 830 729 321 671 205 134 999 177.

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) 0.000 020 830 729 321 671 205 134 999 177 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 354;
  • 2) 0.000 041 661 458 643 342 410 269 998 354 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 708;
  • 3) 0.000 083 322 917 286 684 820 539 996 708 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 416;
  • 4) 0.000 166 645 834 573 369 641 079 993 416 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 832;
  • 5) 0.000 333 291 669 146 739 282 159 986 832 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 664;
  • 6) 0.000 666 583 338 293 478 564 319 973 664 × 2 = 0 + 0.001 333 166 676 586 957 128 639 947 328;
  • 7) 0.001 333 166 676 586 957 128 639 947 328 × 2 = 0 + 0.002 666 333 353 173 914 257 279 894 656;
  • 8) 0.002 666 333 353 173 914 257 279 894 656 × 2 = 0 + 0.005 332 666 706 347 828 514 559 789 312;
  • 9) 0.005 332 666 706 347 828 514 559 789 312 × 2 = 0 + 0.010 665 333 412 695 657 029 119 578 624;
  • 10) 0.010 665 333 412 695 657 029 119 578 624 × 2 = 0 + 0.021 330 666 825 391 314 058 239 157 248;
  • 11) 0.021 330 666 825 391 314 058 239 157 248 × 2 = 0 + 0.042 661 333 650 782 628 116 478 314 496;
  • 12) 0.042 661 333 650 782 628 116 478 314 496 × 2 = 0 + 0.085 322 667 301 565 256 232 956 628 992;
  • 13) 0.085 322 667 301 565 256 232 956 628 992 × 2 = 0 + 0.170 645 334 603 130 512 465 913 257 984;
  • 14) 0.170 645 334 603 130 512 465 913 257 984 × 2 = 0 + 0.341 290 669 206 261 024 931 826 515 968;
  • 15) 0.341 290 669 206 261 024 931 826 515 968 × 2 = 0 + 0.682 581 338 412 522 049 863 653 031 936;
  • 16) 0.682 581 338 412 522 049 863 653 031 936 × 2 = 1 + 0.365 162 676 825 044 099 727 306 063 872;
  • 17) 0.365 162 676 825 044 099 727 306 063 872 × 2 = 0 + 0.730 325 353 650 088 199 454 612 127 744;
  • 18) 0.730 325 353 650 088 199 454 612 127 744 × 2 = 1 + 0.460 650 707 300 176 398 909 224 255 488;
  • 19) 0.460 650 707 300 176 398 909 224 255 488 × 2 = 0 + 0.921 301 414 600 352 797 818 448 510 976;
  • 20) 0.921 301 414 600 352 797 818 448 510 976 × 2 = 1 + 0.842 602 829 200 705 595 636 897 021 952;
  • 21) 0.842 602 829 200 705 595 636 897 021 952 × 2 = 1 + 0.685 205 658 401 411 191 273 794 043 904;
  • 22) 0.685 205 658 401 411 191 273 794 043 904 × 2 = 1 + 0.370 411 316 802 822 382 547 588 087 808;
  • 23) 0.370 411 316 802 822 382 547 588 087 808 × 2 = 0 + 0.740 822 633 605 644 765 095 176 175 616;
  • 24) 0.740 822 633 605 644 765 095 176 175 616 × 2 = 1 + 0.481 645 267 211 289 530 190 352 351 232;
  • 25) 0.481 645 267 211 289 530 190 352 351 232 × 2 = 0 + 0.963 290 534 422 579 060 380 704 702 464;
  • 26) 0.963 290 534 422 579 060 380 704 702 464 × 2 = 1 + 0.926 581 068 845 158 120 761 409 404 928;
  • 27) 0.926 581 068 845 158 120 761 409 404 928 × 2 = 1 + 0.853 162 137 690 316 241 522 818 809 856;
  • 28) 0.853 162 137 690 316 241 522 818 809 856 × 2 = 1 + 0.706 324 275 380 632 483 045 637 619 712;
  • 29) 0.706 324 275 380 632 483 045 637 619 712 × 2 = 1 + 0.412 648 550 761 264 966 091 275 239 424;
  • 30) 0.412 648 550 761 264 966 091 275 239 424 × 2 = 0 + 0.825 297 101 522 529 932 182 550 478 848;
  • 31) 0.825 297 101 522 529 932 182 550 478 848 × 2 = 1 + 0.650 594 203 045 059 864 365 100 957 696;
  • 32) 0.650 594 203 045 059 864 365 100 957 696 × 2 = 1 + 0.301 188 406 090 119 728 730 201 915 392;
  • 33) 0.301 188 406 090 119 728 730 201 915 392 × 2 = 0 + 0.602 376 812 180 239 457 460 403 830 784;
  • 34) 0.602 376 812 180 239 457 460 403 830 784 × 2 = 1 + 0.204 753 624 360 478 914 920 807 661 568;
  • 35) 0.204 753 624 360 478 914 920 807 661 568 × 2 = 0 + 0.409 507 248 720 957 829 841 615 323 136;
  • 36) 0.409 507 248 720 957 829 841 615 323 136 × 2 = 0 + 0.819 014 497 441 915 659 683 230 646 272;
  • 37) 0.819 014 497 441 915 659 683 230 646 272 × 2 = 1 + 0.638 028 994 883 831 319 366 461 292 544;
  • 38) 0.638 028 994 883 831 319 366 461 292 544 × 2 = 1 + 0.276 057 989 767 662 638 732 922 585 088;
  • 39) 0.276 057 989 767 662 638 732 922 585 088 × 2 = 0 + 0.552 115 979 535 325 277 465 845 170 176;
  • 40) 0.552 115 979 535 325 277 465 845 170 176 × 2 = 1 + 0.104 231 959 070 650 554 931 690 340 352;
  • 41) 0.104 231 959 070 650 554 931 690 340 352 × 2 = 0 + 0.208 463 918 141 301 109 863 380 680 704;
  • 42) 0.208 463 918 141 301 109 863 380 680 704 × 2 = 0 + 0.416 927 836 282 602 219 726 761 361 408;
  • 43) 0.416 927 836 282 602 219 726 761 361 408 × 2 = 0 + 0.833 855 672 565 204 439 453 522 722 816;
  • 44) 0.833 855 672 565 204 439 453 522 722 816 × 2 = 1 + 0.667 711 345 130 408 878 907 045 445 632;
  • 45) 0.667 711 345 130 408 878 907 045 445 632 × 2 = 1 + 0.335 422 690 260 817 757 814 090 891 264;
  • 46) 0.335 422 690 260 817 757 814 090 891 264 × 2 = 0 + 0.670 845 380 521 635 515 628 181 782 528;
  • 47) 0.670 845 380 521 635 515 628 181 782 528 × 2 = 1 + 0.341 690 761 043 271 031 256 363 565 056;
  • 48) 0.341 690 761 043 271 031 256 363 565 056 × 2 = 0 + 0.683 381 522 086 542 062 512 727 130 112;
  • 49) 0.683 381 522 086 542 062 512 727 130 112 × 2 = 1 + 0.366 763 044 173 084 125 025 454 260 224;
  • 50) 0.366 763 044 173 084 125 025 454 260 224 × 2 = 0 + 0.733 526 088 346 168 250 050 908 520 448;
  • 51) 0.733 526 088 346 168 250 050 908 520 448 × 2 = 1 + 0.467 052 176 692 336 500 101 817 040 896;
  • 52) 0.467 052 176 692 336 500 101 817 040 896 × 2 = 0 + 0.934 104 353 384 673 000 203 634 081 792;
  • 53) 0.934 104 353 384 673 000 203 634 081 792 × 2 = 1 + 0.868 208 706 769 346 000 407 268 163 584;
  • 54) 0.868 208 706 769 346 000 407 268 163 584 × 2 = 1 + 0.736 417 413 538 692 000 814 536 327 168;
  • 55) 0.736 417 413 538 692 000 814 536 327 168 × 2 = 1 + 0.472 834 827 077 384 001 629 072 654 336;
  • 56) 0.472 834 827 077 384 001 629 072 654 336 × 2 = 0 + 0.945 669 654 154 768 003 258 145 308 672;
  • 57) 0.945 669 654 154 768 003 258 145 308 672 × 2 = 1 + 0.891 339 308 309 536 006 516 290 617 344;
  • 58) 0.891 339 308 309 536 006 516 290 617 344 × 2 = 1 + 0.782 678 616 619 072 013 032 581 234 688;
  • 59) 0.782 678 616 619 072 013 032 581 234 688 × 2 = 1 + 0.565 357 233 238 144 026 065 162 469 376;
  • 60) 0.565 357 233 238 144 026 065 162 469 376 × 2 = 1 + 0.130 714 466 476 288 052 130 324 938 752;
  • 61) 0.130 714 466 476 288 052 130 324 938 752 × 2 = 0 + 0.261 428 932 952 576 104 260 649 877 504;
  • 62) 0.261 428 932 952 576 104 260 649 877 504 × 2 = 0 + 0.522 857 865 905 152 208 521 299 755 008;
  • 63) 0.522 857 865 905 152 208 521 299 755 008 × 2 = 1 + 0.045 715 731 810 304 417 042 599 510 016;
  • 64) 0.045 715 731 810 304 417 042 599 510 016 × 2 = 0 + 0.091 431 463 620 608 834 085 199 020 032;
  • 65) 0.091 431 463 620 608 834 085 199 020 032 × 2 = 0 + 0.182 862 927 241 217 668 170 398 040 064;
  • 66) 0.182 862 927 241 217 668 170 398 040 064 × 2 = 0 + 0.365 725 854 482 435 336 340 796 080 128;
  • 67) 0.365 725 854 482 435 336 340 796 080 128 × 2 = 0 + 0.731 451 708 964 870 672 681 592 160 256;
  • 68) 0.731 451 708 964 870 672 681 592 160 256 × 2 = 1 + 0.462 903 417 929 741 345 363 184 320 512;

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:


0.000 020 830 729 321 671 205 134 999 177(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:

0.000 020 830 729 321 671 205 134 999 177(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:


0.000 020 830 729 321 671 205 134 999 177(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 0.000 020 830 729 321 671 205 134 999 177 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