0.000 020 830 729 321 671 205 134 999 155 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 155 5(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
0.000 020 830 729 321 671 205 134 999 155 5(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: 0.000 020 830 729 321 671 205 134 999 155 5.

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) 0.000 020 830 729 321 671 205 134 999 155 5 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 311;
  • 2) 0.000 041 661 458 643 342 410 269 998 311 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 622;
  • 3) 0.000 083 322 917 286 684 820 539 996 622 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 244;
  • 4) 0.000 166 645 834 573 369 641 079 993 244 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 488;
  • 5) 0.000 333 291 669 146 739 282 159 986 488 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 976;
  • 6) 0.000 666 583 338 293 478 564 319 972 976 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 952;
  • 7) 0.001 333 166 676 586 957 128 639 945 952 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 904;
  • 8) 0.002 666 333 353 173 914 257 279 891 904 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 808;
  • 9) 0.005 332 666 706 347 828 514 559 783 808 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 616;
  • 10) 0.010 665 333 412 695 657 029 119 567 616 × 2 = 0 + 0.021 330 666 825 391 314 058 239 135 232;
  • 11) 0.021 330 666 825 391 314 058 239 135 232 × 2 = 0 + 0.042 661 333 650 782 628 116 478 270 464;
  • 12) 0.042 661 333 650 782 628 116 478 270 464 × 2 = 0 + 0.085 322 667 301 565 256 232 956 540 928;
  • 13) 0.085 322 667 301 565 256 232 956 540 928 × 2 = 0 + 0.170 645 334 603 130 512 465 913 081 856;
  • 14) 0.170 645 334 603 130 512 465 913 081 856 × 2 = 0 + 0.341 290 669 206 261 024 931 826 163 712;
  • 15) 0.341 290 669 206 261 024 931 826 163 712 × 2 = 0 + 0.682 581 338 412 522 049 863 652 327 424;
  • 16) 0.682 581 338 412 522 049 863 652 327 424 × 2 = 1 + 0.365 162 676 825 044 099 727 304 654 848;
  • 17) 0.365 162 676 825 044 099 727 304 654 848 × 2 = 0 + 0.730 325 353 650 088 199 454 609 309 696;
  • 18) 0.730 325 353 650 088 199 454 609 309 696 × 2 = 1 + 0.460 650 707 300 176 398 909 218 619 392;
  • 19) 0.460 650 707 300 176 398 909 218 619 392 × 2 = 0 + 0.921 301 414 600 352 797 818 437 238 784;
  • 20) 0.921 301 414 600 352 797 818 437 238 784 × 2 = 1 + 0.842 602 829 200 705 595 636 874 477 568;
  • 21) 0.842 602 829 200 705 595 636 874 477 568 × 2 = 1 + 0.685 205 658 401 411 191 273 748 955 136;
  • 22) 0.685 205 658 401 411 191 273 748 955 136 × 2 = 1 + 0.370 411 316 802 822 382 547 497 910 272;
  • 23) 0.370 411 316 802 822 382 547 497 910 272 × 2 = 0 + 0.740 822 633 605 644 765 094 995 820 544;
  • 24) 0.740 822 633 605 644 765 094 995 820 544 × 2 = 1 + 0.481 645 267 211 289 530 189 991 641 088;
  • 25) 0.481 645 267 211 289 530 189 991 641 088 × 2 = 0 + 0.963 290 534 422 579 060 379 983 282 176;
  • 26) 0.963 290 534 422 579 060 379 983 282 176 × 2 = 1 + 0.926 581 068 845 158 120 759 966 564 352;
  • 27) 0.926 581 068 845 158 120 759 966 564 352 × 2 = 1 + 0.853 162 137 690 316 241 519 933 128 704;
  • 28) 0.853 162 137 690 316 241 519 933 128 704 × 2 = 1 + 0.706 324 275 380 632 483 039 866 257 408;
  • 29) 0.706 324 275 380 632 483 039 866 257 408 × 2 = 1 + 0.412 648 550 761 264 966 079 732 514 816;
  • 30) 0.412 648 550 761 264 966 079 732 514 816 × 2 = 0 + 0.825 297 101 522 529 932 159 465 029 632;
  • 31) 0.825 297 101 522 529 932 159 465 029 632 × 2 = 1 + 0.650 594 203 045 059 864 318 930 059 264;
  • 32) 0.650 594 203 045 059 864 318 930 059 264 × 2 = 1 + 0.301 188 406 090 119 728 637 860 118 528;
  • 33) 0.301 188 406 090 119 728 637 860 118 528 × 2 = 0 + 0.602 376 812 180 239 457 275 720 237 056;
  • 34) 0.602 376 812 180 239 457 275 720 237 056 × 2 = 1 + 0.204 753 624 360 478 914 551 440 474 112;
  • 35) 0.204 753 624 360 478 914 551 440 474 112 × 2 = 0 + 0.409 507 248 720 957 829 102 880 948 224;
  • 36) 0.409 507 248 720 957 829 102 880 948 224 × 2 = 0 + 0.819 014 497 441 915 658 205 761 896 448;
  • 37) 0.819 014 497 441 915 658 205 761 896 448 × 2 = 1 + 0.638 028 994 883 831 316 411 523 792 896;
  • 38) 0.638 028 994 883 831 316 411 523 792 896 × 2 = 1 + 0.276 057 989 767 662 632 823 047 585 792;
  • 39) 0.276 057 989 767 662 632 823 047 585 792 × 2 = 0 + 0.552 115 979 535 325 265 646 095 171 584;
  • 40) 0.552 115 979 535 325 265 646 095 171 584 × 2 = 1 + 0.104 231 959 070 650 531 292 190 343 168;
  • 41) 0.104 231 959 070 650 531 292 190 343 168 × 2 = 0 + 0.208 463 918 141 301 062 584 380 686 336;
  • 42) 0.208 463 918 141 301 062 584 380 686 336 × 2 = 0 + 0.416 927 836 282 602 125 168 761 372 672;
  • 43) 0.416 927 836 282 602 125 168 761 372 672 × 2 = 0 + 0.833 855 672 565 204 250 337 522 745 344;
  • 44) 0.833 855 672 565 204 250 337 522 745 344 × 2 = 1 + 0.667 711 345 130 408 500 675 045 490 688;
  • 45) 0.667 711 345 130 408 500 675 045 490 688 × 2 = 1 + 0.335 422 690 260 817 001 350 090 981 376;
  • 46) 0.335 422 690 260 817 001 350 090 981 376 × 2 = 0 + 0.670 845 380 521 634 002 700 181 962 752;
  • 47) 0.670 845 380 521 634 002 700 181 962 752 × 2 = 1 + 0.341 690 761 043 268 005 400 363 925 504;
  • 48) 0.341 690 761 043 268 005 400 363 925 504 × 2 = 0 + 0.683 381 522 086 536 010 800 727 851 008;
  • 49) 0.683 381 522 086 536 010 800 727 851 008 × 2 = 1 + 0.366 763 044 173 072 021 601 455 702 016;
  • 50) 0.366 763 044 173 072 021 601 455 702 016 × 2 = 0 + 0.733 526 088 346 144 043 202 911 404 032;
  • 51) 0.733 526 088 346 144 043 202 911 404 032 × 2 = 1 + 0.467 052 176 692 288 086 405 822 808 064;
  • 52) 0.467 052 176 692 288 086 405 822 808 064 × 2 = 0 + 0.934 104 353 384 576 172 811 645 616 128;
  • 53) 0.934 104 353 384 576 172 811 645 616 128 × 2 = 1 + 0.868 208 706 769 152 345 623 291 232 256;
  • 54) 0.868 208 706 769 152 345 623 291 232 256 × 2 = 1 + 0.736 417 413 538 304 691 246 582 464 512;
  • 55) 0.736 417 413 538 304 691 246 582 464 512 × 2 = 1 + 0.472 834 827 076 609 382 493 164 929 024;
  • 56) 0.472 834 827 076 609 382 493 164 929 024 × 2 = 0 + 0.945 669 654 153 218 764 986 329 858 048;
  • 57) 0.945 669 654 153 218 764 986 329 858 048 × 2 = 1 + 0.891 339 308 306 437 529 972 659 716 096;
  • 58) 0.891 339 308 306 437 529 972 659 716 096 × 2 = 1 + 0.782 678 616 612 875 059 945 319 432 192;
  • 59) 0.782 678 616 612 875 059 945 319 432 192 × 2 = 1 + 0.565 357 233 225 750 119 890 638 864 384;
  • 60) 0.565 357 233 225 750 119 890 638 864 384 × 2 = 1 + 0.130 714 466 451 500 239 781 277 728 768;
  • 61) 0.130 714 466 451 500 239 781 277 728 768 × 2 = 0 + 0.261 428 932 903 000 479 562 555 457 536;
  • 62) 0.261 428 932 903 000 479 562 555 457 536 × 2 = 0 + 0.522 857 865 806 000 959 125 110 915 072;
  • 63) 0.522 857 865 806 000 959 125 110 915 072 × 2 = 1 + 0.045 715 731 612 001 918 250 221 830 144;
  • 64) 0.045 715 731 612 001 918 250 221 830 144 × 2 = 0 + 0.091 431 463 224 003 836 500 443 660 288;
  • 65) 0.091 431 463 224 003 836 500 443 660 288 × 2 = 0 + 0.182 862 926 448 007 673 000 887 320 576;
  • 66) 0.182 862 926 448 007 673 000 887 320 576 × 2 = 0 + 0.365 725 852 896 015 346 001 774 641 152;
  • 67) 0.365 725 852 896 015 346 001 774 641 152 × 2 = 0 + 0.731 451 705 792 030 692 003 549 282 304;
  • 68) 0.731 451 705 792 030 692 003 549 282 304 × 2 = 1 + 0.462 903 411 584 061 384 007 098 564 608;

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:


0.000 020 830 729 321 671 205 134 999 155 5(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:

0.000 020 830 729 321 671 205 134 999 155 5(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:


0.000 020 830 729 321 671 205 134 999 155 5(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 0.000 020 830 729 321 671 205 134 999 155 5 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