0.000 020 830 729 321 671 205 134 999 173 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 173(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
0.000 020 830 729 321 671 205 134 999 173(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: 0.000 020 830 729 321 671 205 134 999 173.

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) 0.000 020 830 729 321 671 205 134 999 173 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 346;
  • 2) 0.000 041 661 458 643 342 410 269 998 346 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 692;
  • 3) 0.000 083 322 917 286 684 820 539 996 692 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 384;
  • 4) 0.000 166 645 834 573 369 641 079 993 384 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 768;
  • 5) 0.000 333 291 669 146 739 282 159 986 768 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 536;
  • 6) 0.000 666 583 338 293 478 564 319 973 536 × 2 = 0 + 0.001 333 166 676 586 957 128 639 947 072;
  • 7) 0.001 333 166 676 586 957 128 639 947 072 × 2 = 0 + 0.002 666 333 353 173 914 257 279 894 144;
  • 8) 0.002 666 333 353 173 914 257 279 894 144 × 2 = 0 + 0.005 332 666 706 347 828 514 559 788 288;
  • 9) 0.005 332 666 706 347 828 514 559 788 288 × 2 = 0 + 0.010 665 333 412 695 657 029 119 576 576;
  • 10) 0.010 665 333 412 695 657 029 119 576 576 × 2 = 0 + 0.021 330 666 825 391 314 058 239 153 152;
  • 11) 0.021 330 666 825 391 314 058 239 153 152 × 2 = 0 + 0.042 661 333 650 782 628 116 478 306 304;
  • 12) 0.042 661 333 650 782 628 116 478 306 304 × 2 = 0 + 0.085 322 667 301 565 256 232 956 612 608;
  • 13) 0.085 322 667 301 565 256 232 956 612 608 × 2 = 0 + 0.170 645 334 603 130 512 465 913 225 216;
  • 14) 0.170 645 334 603 130 512 465 913 225 216 × 2 = 0 + 0.341 290 669 206 261 024 931 826 450 432;
  • 15) 0.341 290 669 206 261 024 931 826 450 432 × 2 = 0 + 0.682 581 338 412 522 049 863 652 900 864;
  • 16) 0.682 581 338 412 522 049 863 652 900 864 × 2 = 1 + 0.365 162 676 825 044 099 727 305 801 728;
  • 17) 0.365 162 676 825 044 099 727 305 801 728 × 2 = 0 + 0.730 325 353 650 088 199 454 611 603 456;
  • 18) 0.730 325 353 650 088 199 454 611 603 456 × 2 = 1 + 0.460 650 707 300 176 398 909 223 206 912;
  • 19) 0.460 650 707 300 176 398 909 223 206 912 × 2 = 0 + 0.921 301 414 600 352 797 818 446 413 824;
  • 20) 0.921 301 414 600 352 797 818 446 413 824 × 2 = 1 + 0.842 602 829 200 705 595 636 892 827 648;
  • 21) 0.842 602 829 200 705 595 636 892 827 648 × 2 = 1 + 0.685 205 658 401 411 191 273 785 655 296;
  • 22) 0.685 205 658 401 411 191 273 785 655 296 × 2 = 1 + 0.370 411 316 802 822 382 547 571 310 592;
  • 23) 0.370 411 316 802 822 382 547 571 310 592 × 2 = 0 + 0.740 822 633 605 644 765 095 142 621 184;
  • 24) 0.740 822 633 605 644 765 095 142 621 184 × 2 = 1 + 0.481 645 267 211 289 530 190 285 242 368;
  • 25) 0.481 645 267 211 289 530 190 285 242 368 × 2 = 0 + 0.963 290 534 422 579 060 380 570 484 736;
  • 26) 0.963 290 534 422 579 060 380 570 484 736 × 2 = 1 + 0.926 581 068 845 158 120 761 140 969 472;
  • 27) 0.926 581 068 845 158 120 761 140 969 472 × 2 = 1 + 0.853 162 137 690 316 241 522 281 938 944;
  • 28) 0.853 162 137 690 316 241 522 281 938 944 × 2 = 1 + 0.706 324 275 380 632 483 044 563 877 888;
  • 29) 0.706 324 275 380 632 483 044 563 877 888 × 2 = 1 + 0.412 648 550 761 264 966 089 127 755 776;
  • 30) 0.412 648 550 761 264 966 089 127 755 776 × 2 = 0 + 0.825 297 101 522 529 932 178 255 511 552;
  • 31) 0.825 297 101 522 529 932 178 255 511 552 × 2 = 1 + 0.650 594 203 045 059 864 356 511 023 104;
  • 32) 0.650 594 203 045 059 864 356 511 023 104 × 2 = 1 + 0.301 188 406 090 119 728 713 022 046 208;
  • 33) 0.301 188 406 090 119 728 713 022 046 208 × 2 = 0 + 0.602 376 812 180 239 457 426 044 092 416;
  • 34) 0.602 376 812 180 239 457 426 044 092 416 × 2 = 1 + 0.204 753 624 360 478 914 852 088 184 832;
  • 35) 0.204 753 624 360 478 914 852 088 184 832 × 2 = 0 + 0.409 507 248 720 957 829 704 176 369 664;
  • 36) 0.409 507 248 720 957 829 704 176 369 664 × 2 = 0 + 0.819 014 497 441 915 659 408 352 739 328;
  • 37) 0.819 014 497 441 915 659 408 352 739 328 × 2 = 1 + 0.638 028 994 883 831 318 816 705 478 656;
  • 38) 0.638 028 994 883 831 318 816 705 478 656 × 2 = 1 + 0.276 057 989 767 662 637 633 410 957 312;
  • 39) 0.276 057 989 767 662 637 633 410 957 312 × 2 = 0 + 0.552 115 979 535 325 275 266 821 914 624;
  • 40) 0.552 115 979 535 325 275 266 821 914 624 × 2 = 1 + 0.104 231 959 070 650 550 533 643 829 248;
  • 41) 0.104 231 959 070 650 550 533 643 829 248 × 2 = 0 + 0.208 463 918 141 301 101 067 287 658 496;
  • 42) 0.208 463 918 141 301 101 067 287 658 496 × 2 = 0 + 0.416 927 836 282 602 202 134 575 316 992;
  • 43) 0.416 927 836 282 602 202 134 575 316 992 × 2 = 0 + 0.833 855 672 565 204 404 269 150 633 984;
  • 44) 0.833 855 672 565 204 404 269 150 633 984 × 2 = 1 + 0.667 711 345 130 408 808 538 301 267 968;
  • 45) 0.667 711 345 130 408 808 538 301 267 968 × 2 = 1 + 0.335 422 690 260 817 617 076 602 535 936;
  • 46) 0.335 422 690 260 817 617 076 602 535 936 × 2 = 0 + 0.670 845 380 521 635 234 153 205 071 872;
  • 47) 0.670 845 380 521 635 234 153 205 071 872 × 2 = 1 + 0.341 690 761 043 270 468 306 410 143 744;
  • 48) 0.341 690 761 043 270 468 306 410 143 744 × 2 = 0 + 0.683 381 522 086 540 936 612 820 287 488;
  • 49) 0.683 381 522 086 540 936 612 820 287 488 × 2 = 1 + 0.366 763 044 173 081 873 225 640 574 976;
  • 50) 0.366 763 044 173 081 873 225 640 574 976 × 2 = 0 + 0.733 526 088 346 163 746 451 281 149 952;
  • 51) 0.733 526 088 346 163 746 451 281 149 952 × 2 = 1 + 0.467 052 176 692 327 492 902 562 299 904;
  • 52) 0.467 052 176 692 327 492 902 562 299 904 × 2 = 0 + 0.934 104 353 384 654 985 805 124 599 808;
  • 53) 0.934 104 353 384 654 985 805 124 599 808 × 2 = 1 + 0.868 208 706 769 309 971 610 249 199 616;
  • 54) 0.868 208 706 769 309 971 610 249 199 616 × 2 = 1 + 0.736 417 413 538 619 943 220 498 399 232;
  • 55) 0.736 417 413 538 619 943 220 498 399 232 × 2 = 1 + 0.472 834 827 077 239 886 440 996 798 464;
  • 56) 0.472 834 827 077 239 886 440 996 798 464 × 2 = 0 + 0.945 669 654 154 479 772 881 993 596 928;
  • 57) 0.945 669 654 154 479 772 881 993 596 928 × 2 = 1 + 0.891 339 308 308 959 545 763 987 193 856;
  • 58) 0.891 339 308 308 959 545 763 987 193 856 × 2 = 1 + 0.782 678 616 617 919 091 527 974 387 712;
  • 59) 0.782 678 616 617 919 091 527 974 387 712 × 2 = 1 + 0.565 357 233 235 838 183 055 948 775 424;
  • 60) 0.565 357 233 235 838 183 055 948 775 424 × 2 = 1 + 0.130 714 466 471 676 366 111 897 550 848;
  • 61) 0.130 714 466 471 676 366 111 897 550 848 × 2 = 0 + 0.261 428 932 943 352 732 223 795 101 696;
  • 62) 0.261 428 932 943 352 732 223 795 101 696 × 2 = 0 + 0.522 857 865 886 705 464 447 590 203 392;
  • 63) 0.522 857 865 886 705 464 447 590 203 392 × 2 = 1 + 0.045 715 731 773 410 928 895 180 406 784;
  • 64) 0.045 715 731 773 410 928 895 180 406 784 × 2 = 0 + 0.091 431 463 546 821 857 790 360 813 568;
  • 65) 0.091 431 463 546 821 857 790 360 813 568 × 2 = 0 + 0.182 862 927 093 643 715 580 721 627 136;
  • 66) 0.182 862 927 093 643 715 580 721 627 136 × 2 = 0 + 0.365 725 854 187 287 431 161 443 254 272;
  • 67) 0.365 725 854 187 287 431 161 443 254 272 × 2 = 0 + 0.731 451 708 374 574 862 322 886 508 544;
  • 68) 0.731 451 708 374 574 862 322 886 508 544 × 2 = 1 + 0.462 903 416 749 149 724 645 773 017 088;

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:


0.000 020 830 729 321 671 205 134 999 173(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:

0.000 020 830 729 321 671 205 134 999 173(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:


0.000 020 830 729 321 671 205 134 999 173(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 0.000 020 830 729 321 671 205 134 999 173 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