0.000 020 830 729 321 671 205 134 999 159 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 159(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
0.000 020 830 729 321 671 205 134 999 159(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: 0.000 020 830 729 321 671 205 134 999 159.

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) 0.000 020 830 729 321 671 205 134 999 159 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 318;
  • 2) 0.000 041 661 458 643 342 410 269 998 318 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 636;
  • 3) 0.000 083 322 917 286 684 820 539 996 636 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 272;
  • 4) 0.000 166 645 834 573 369 641 079 993 272 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 544;
  • 5) 0.000 333 291 669 146 739 282 159 986 544 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 088;
  • 6) 0.000 666 583 338 293 478 564 319 973 088 × 2 = 0 + 0.001 333 166 676 586 957 128 639 946 176;
  • 7) 0.001 333 166 676 586 957 128 639 946 176 × 2 = 0 + 0.002 666 333 353 173 914 257 279 892 352;
  • 8) 0.002 666 333 353 173 914 257 279 892 352 × 2 = 0 + 0.005 332 666 706 347 828 514 559 784 704;
  • 9) 0.005 332 666 706 347 828 514 559 784 704 × 2 = 0 + 0.010 665 333 412 695 657 029 119 569 408;
  • 10) 0.010 665 333 412 695 657 029 119 569 408 × 2 = 0 + 0.021 330 666 825 391 314 058 239 138 816;
  • 11) 0.021 330 666 825 391 314 058 239 138 816 × 2 = 0 + 0.042 661 333 650 782 628 116 478 277 632;
  • 12) 0.042 661 333 650 782 628 116 478 277 632 × 2 = 0 + 0.085 322 667 301 565 256 232 956 555 264;
  • 13) 0.085 322 667 301 565 256 232 956 555 264 × 2 = 0 + 0.170 645 334 603 130 512 465 913 110 528;
  • 14) 0.170 645 334 603 130 512 465 913 110 528 × 2 = 0 + 0.341 290 669 206 261 024 931 826 221 056;
  • 15) 0.341 290 669 206 261 024 931 826 221 056 × 2 = 0 + 0.682 581 338 412 522 049 863 652 442 112;
  • 16) 0.682 581 338 412 522 049 863 652 442 112 × 2 = 1 + 0.365 162 676 825 044 099 727 304 884 224;
  • 17) 0.365 162 676 825 044 099 727 304 884 224 × 2 = 0 + 0.730 325 353 650 088 199 454 609 768 448;
  • 18) 0.730 325 353 650 088 199 454 609 768 448 × 2 = 1 + 0.460 650 707 300 176 398 909 219 536 896;
  • 19) 0.460 650 707 300 176 398 909 219 536 896 × 2 = 0 + 0.921 301 414 600 352 797 818 439 073 792;
  • 20) 0.921 301 414 600 352 797 818 439 073 792 × 2 = 1 + 0.842 602 829 200 705 595 636 878 147 584;
  • 21) 0.842 602 829 200 705 595 636 878 147 584 × 2 = 1 + 0.685 205 658 401 411 191 273 756 295 168;
  • 22) 0.685 205 658 401 411 191 273 756 295 168 × 2 = 1 + 0.370 411 316 802 822 382 547 512 590 336;
  • 23) 0.370 411 316 802 822 382 547 512 590 336 × 2 = 0 + 0.740 822 633 605 644 765 095 025 180 672;
  • 24) 0.740 822 633 605 644 765 095 025 180 672 × 2 = 1 + 0.481 645 267 211 289 530 190 050 361 344;
  • 25) 0.481 645 267 211 289 530 190 050 361 344 × 2 = 0 + 0.963 290 534 422 579 060 380 100 722 688;
  • 26) 0.963 290 534 422 579 060 380 100 722 688 × 2 = 1 + 0.926 581 068 845 158 120 760 201 445 376;
  • 27) 0.926 581 068 845 158 120 760 201 445 376 × 2 = 1 + 0.853 162 137 690 316 241 520 402 890 752;
  • 28) 0.853 162 137 690 316 241 520 402 890 752 × 2 = 1 + 0.706 324 275 380 632 483 040 805 781 504;
  • 29) 0.706 324 275 380 632 483 040 805 781 504 × 2 = 1 + 0.412 648 550 761 264 966 081 611 563 008;
  • 30) 0.412 648 550 761 264 966 081 611 563 008 × 2 = 0 + 0.825 297 101 522 529 932 163 223 126 016;
  • 31) 0.825 297 101 522 529 932 163 223 126 016 × 2 = 1 + 0.650 594 203 045 059 864 326 446 252 032;
  • 32) 0.650 594 203 045 059 864 326 446 252 032 × 2 = 1 + 0.301 188 406 090 119 728 652 892 504 064;
  • 33) 0.301 188 406 090 119 728 652 892 504 064 × 2 = 0 + 0.602 376 812 180 239 457 305 785 008 128;
  • 34) 0.602 376 812 180 239 457 305 785 008 128 × 2 = 1 + 0.204 753 624 360 478 914 611 570 016 256;
  • 35) 0.204 753 624 360 478 914 611 570 016 256 × 2 = 0 + 0.409 507 248 720 957 829 223 140 032 512;
  • 36) 0.409 507 248 720 957 829 223 140 032 512 × 2 = 0 + 0.819 014 497 441 915 658 446 280 065 024;
  • 37) 0.819 014 497 441 915 658 446 280 065 024 × 2 = 1 + 0.638 028 994 883 831 316 892 560 130 048;
  • 38) 0.638 028 994 883 831 316 892 560 130 048 × 2 = 1 + 0.276 057 989 767 662 633 785 120 260 096;
  • 39) 0.276 057 989 767 662 633 785 120 260 096 × 2 = 0 + 0.552 115 979 535 325 267 570 240 520 192;
  • 40) 0.552 115 979 535 325 267 570 240 520 192 × 2 = 1 + 0.104 231 959 070 650 535 140 481 040 384;
  • 41) 0.104 231 959 070 650 535 140 481 040 384 × 2 = 0 + 0.208 463 918 141 301 070 280 962 080 768;
  • 42) 0.208 463 918 141 301 070 280 962 080 768 × 2 = 0 + 0.416 927 836 282 602 140 561 924 161 536;
  • 43) 0.416 927 836 282 602 140 561 924 161 536 × 2 = 0 + 0.833 855 672 565 204 281 123 848 323 072;
  • 44) 0.833 855 672 565 204 281 123 848 323 072 × 2 = 1 + 0.667 711 345 130 408 562 247 696 646 144;
  • 45) 0.667 711 345 130 408 562 247 696 646 144 × 2 = 1 + 0.335 422 690 260 817 124 495 393 292 288;
  • 46) 0.335 422 690 260 817 124 495 393 292 288 × 2 = 0 + 0.670 845 380 521 634 248 990 786 584 576;
  • 47) 0.670 845 380 521 634 248 990 786 584 576 × 2 = 1 + 0.341 690 761 043 268 497 981 573 169 152;
  • 48) 0.341 690 761 043 268 497 981 573 169 152 × 2 = 0 + 0.683 381 522 086 536 995 963 146 338 304;
  • 49) 0.683 381 522 086 536 995 963 146 338 304 × 2 = 1 + 0.366 763 044 173 073 991 926 292 676 608;
  • 50) 0.366 763 044 173 073 991 926 292 676 608 × 2 = 0 + 0.733 526 088 346 147 983 852 585 353 216;
  • 51) 0.733 526 088 346 147 983 852 585 353 216 × 2 = 1 + 0.467 052 176 692 295 967 705 170 706 432;
  • 52) 0.467 052 176 692 295 967 705 170 706 432 × 2 = 0 + 0.934 104 353 384 591 935 410 341 412 864;
  • 53) 0.934 104 353 384 591 935 410 341 412 864 × 2 = 1 + 0.868 208 706 769 183 870 820 682 825 728;
  • 54) 0.868 208 706 769 183 870 820 682 825 728 × 2 = 1 + 0.736 417 413 538 367 741 641 365 651 456;
  • 55) 0.736 417 413 538 367 741 641 365 651 456 × 2 = 1 + 0.472 834 827 076 735 483 282 731 302 912;
  • 56) 0.472 834 827 076 735 483 282 731 302 912 × 2 = 0 + 0.945 669 654 153 470 966 565 462 605 824;
  • 57) 0.945 669 654 153 470 966 565 462 605 824 × 2 = 1 + 0.891 339 308 306 941 933 130 925 211 648;
  • 58) 0.891 339 308 306 941 933 130 925 211 648 × 2 = 1 + 0.782 678 616 613 883 866 261 850 423 296;
  • 59) 0.782 678 616 613 883 866 261 850 423 296 × 2 = 1 + 0.565 357 233 227 767 732 523 700 846 592;
  • 60) 0.565 357 233 227 767 732 523 700 846 592 × 2 = 1 + 0.130 714 466 455 535 465 047 401 693 184;
  • 61) 0.130 714 466 455 535 465 047 401 693 184 × 2 = 0 + 0.261 428 932 911 070 930 094 803 386 368;
  • 62) 0.261 428 932 911 070 930 094 803 386 368 × 2 = 0 + 0.522 857 865 822 141 860 189 606 772 736;
  • 63) 0.522 857 865 822 141 860 189 606 772 736 × 2 = 1 + 0.045 715 731 644 283 720 379 213 545 472;
  • 64) 0.045 715 731 644 283 720 379 213 545 472 × 2 = 0 + 0.091 431 463 288 567 440 758 427 090 944;
  • 65) 0.091 431 463 288 567 440 758 427 090 944 × 2 = 0 + 0.182 862 926 577 134 881 516 854 181 888;
  • 66) 0.182 862 926 577 134 881 516 854 181 888 × 2 = 0 + 0.365 725 853 154 269 763 033 708 363 776;
  • 67) 0.365 725 853 154 269 763 033 708 363 776 × 2 = 0 + 0.731 451 706 308 539 526 067 416 727 552;
  • 68) 0.731 451 706 308 539 526 067 416 727 552 × 2 = 1 + 0.462 903 412 617 079 052 134 833 455 104;

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:


0.000 020 830 729 321 671 205 134 999 159(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:

0.000 020 830 729 321 671 205 134 999 159(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:


0.000 020 830 729 321 671 205 134 999 159(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 0.000 020 830 729 321 671 205 134 999 159 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