0.000 020 830 729 321 671 205 134 999 151 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 151(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
0.000 020 830 729 321 671 205 134 999 151(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: 0.000 020 830 729 321 671 205 134 999 151.

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) 0.000 020 830 729 321 671 205 134 999 151 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 302;
  • 2) 0.000 041 661 458 643 342 410 269 998 302 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 604;
  • 3) 0.000 083 322 917 286 684 820 539 996 604 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 208;
  • 4) 0.000 166 645 834 573 369 641 079 993 208 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 416;
  • 5) 0.000 333 291 669 146 739 282 159 986 416 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 832;
  • 6) 0.000 666 583 338 293 478 564 319 972 832 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 664;
  • 7) 0.001 333 166 676 586 957 128 639 945 664 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 328;
  • 8) 0.002 666 333 353 173 914 257 279 891 328 × 2 = 0 + 0.005 332 666 706 347 828 514 559 782 656;
  • 9) 0.005 332 666 706 347 828 514 559 782 656 × 2 = 0 + 0.010 665 333 412 695 657 029 119 565 312;
  • 10) 0.010 665 333 412 695 657 029 119 565 312 × 2 = 0 + 0.021 330 666 825 391 314 058 239 130 624;
  • 11) 0.021 330 666 825 391 314 058 239 130 624 × 2 = 0 + 0.042 661 333 650 782 628 116 478 261 248;
  • 12) 0.042 661 333 650 782 628 116 478 261 248 × 2 = 0 + 0.085 322 667 301 565 256 232 956 522 496;
  • 13) 0.085 322 667 301 565 256 232 956 522 496 × 2 = 0 + 0.170 645 334 603 130 512 465 913 044 992;
  • 14) 0.170 645 334 603 130 512 465 913 044 992 × 2 = 0 + 0.341 290 669 206 261 024 931 826 089 984;
  • 15) 0.341 290 669 206 261 024 931 826 089 984 × 2 = 0 + 0.682 581 338 412 522 049 863 652 179 968;
  • 16) 0.682 581 338 412 522 049 863 652 179 968 × 2 = 1 + 0.365 162 676 825 044 099 727 304 359 936;
  • 17) 0.365 162 676 825 044 099 727 304 359 936 × 2 = 0 + 0.730 325 353 650 088 199 454 608 719 872;
  • 18) 0.730 325 353 650 088 199 454 608 719 872 × 2 = 1 + 0.460 650 707 300 176 398 909 217 439 744;
  • 19) 0.460 650 707 300 176 398 909 217 439 744 × 2 = 0 + 0.921 301 414 600 352 797 818 434 879 488;
  • 20) 0.921 301 414 600 352 797 818 434 879 488 × 2 = 1 + 0.842 602 829 200 705 595 636 869 758 976;
  • 21) 0.842 602 829 200 705 595 636 869 758 976 × 2 = 1 + 0.685 205 658 401 411 191 273 739 517 952;
  • 22) 0.685 205 658 401 411 191 273 739 517 952 × 2 = 1 + 0.370 411 316 802 822 382 547 479 035 904;
  • 23) 0.370 411 316 802 822 382 547 479 035 904 × 2 = 0 + 0.740 822 633 605 644 765 094 958 071 808;
  • 24) 0.740 822 633 605 644 765 094 958 071 808 × 2 = 1 + 0.481 645 267 211 289 530 189 916 143 616;
  • 25) 0.481 645 267 211 289 530 189 916 143 616 × 2 = 0 + 0.963 290 534 422 579 060 379 832 287 232;
  • 26) 0.963 290 534 422 579 060 379 832 287 232 × 2 = 1 + 0.926 581 068 845 158 120 759 664 574 464;
  • 27) 0.926 581 068 845 158 120 759 664 574 464 × 2 = 1 + 0.853 162 137 690 316 241 519 329 148 928;
  • 28) 0.853 162 137 690 316 241 519 329 148 928 × 2 = 1 + 0.706 324 275 380 632 483 038 658 297 856;
  • 29) 0.706 324 275 380 632 483 038 658 297 856 × 2 = 1 + 0.412 648 550 761 264 966 077 316 595 712;
  • 30) 0.412 648 550 761 264 966 077 316 595 712 × 2 = 0 + 0.825 297 101 522 529 932 154 633 191 424;
  • 31) 0.825 297 101 522 529 932 154 633 191 424 × 2 = 1 + 0.650 594 203 045 059 864 309 266 382 848;
  • 32) 0.650 594 203 045 059 864 309 266 382 848 × 2 = 1 + 0.301 188 406 090 119 728 618 532 765 696;
  • 33) 0.301 188 406 090 119 728 618 532 765 696 × 2 = 0 + 0.602 376 812 180 239 457 237 065 531 392;
  • 34) 0.602 376 812 180 239 457 237 065 531 392 × 2 = 1 + 0.204 753 624 360 478 914 474 131 062 784;
  • 35) 0.204 753 624 360 478 914 474 131 062 784 × 2 = 0 + 0.409 507 248 720 957 828 948 262 125 568;
  • 36) 0.409 507 248 720 957 828 948 262 125 568 × 2 = 0 + 0.819 014 497 441 915 657 896 524 251 136;
  • 37) 0.819 014 497 441 915 657 896 524 251 136 × 2 = 1 + 0.638 028 994 883 831 315 793 048 502 272;
  • 38) 0.638 028 994 883 831 315 793 048 502 272 × 2 = 1 + 0.276 057 989 767 662 631 586 097 004 544;
  • 39) 0.276 057 989 767 662 631 586 097 004 544 × 2 = 0 + 0.552 115 979 535 325 263 172 194 009 088;
  • 40) 0.552 115 979 535 325 263 172 194 009 088 × 2 = 1 + 0.104 231 959 070 650 526 344 388 018 176;
  • 41) 0.104 231 959 070 650 526 344 388 018 176 × 2 = 0 + 0.208 463 918 141 301 052 688 776 036 352;
  • 42) 0.208 463 918 141 301 052 688 776 036 352 × 2 = 0 + 0.416 927 836 282 602 105 377 552 072 704;
  • 43) 0.416 927 836 282 602 105 377 552 072 704 × 2 = 0 + 0.833 855 672 565 204 210 755 104 145 408;
  • 44) 0.833 855 672 565 204 210 755 104 145 408 × 2 = 1 + 0.667 711 345 130 408 421 510 208 290 816;
  • 45) 0.667 711 345 130 408 421 510 208 290 816 × 2 = 1 + 0.335 422 690 260 816 843 020 416 581 632;
  • 46) 0.335 422 690 260 816 843 020 416 581 632 × 2 = 0 + 0.670 845 380 521 633 686 040 833 163 264;
  • 47) 0.670 845 380 521 633 686 040 833 163 264 × 2 = 1 + 0.341 690 761 043 267 372 081 666 326 528;
  • 48) 0.341 690 761 043 267 372 081 666 326 528 × 2 = 0 + 0.683 381 522 086 534 744 163 332 653 056;
  • 49) 0.683 381 522 086 534 744 163 332 653 056 × 2 = 1 + 0.366 763 044 173 069 488 326 665 306 112;
  • 50) 0.366 763 044 173 069 488 326 665 306 112 × 2 = 0 + 0.733 526 088 346 138 976 653 330 612 224;
  • 51) 0.733 526 088 346 138 976 653 330 612 224 × 2 = 1 + 0.467 052 176 692 277 953 306 661 224 448;
  • 52) 0.467 052 176 692 277 953 306 661 224 448 × 2 = 0 + 0.934 104 353 384 555 906 613 322 448 896;
  • 53) 0.934 104 353 384 555 906 613 322 448 896 × 2 = 1 + 0.868 208 706 769 111 813 226 644 897 792;
  • 54) 0.868 208 706 769 111 813 226 644 897 792 × 2 = 1 + 0.736 417 413 538 223 626 453 289 795 584;
  • 55) 0.736 417 413 538 223 626 453 289 795 584 × 2 = 1 + 0.472 834 827 076 447 252 906 579 591 168;
  • 56) 0.472 834 827 076 447 252 906 579 591 168 × 2 = 0 + 0.945 669 654 152 894 505 813 159 182 336;
  • 57) 0.945 669 654 152 894 505 813 159 182 336 × 2 = 1 + 0.891 339 308 305 789 011 626 318 364 672;
  • 58) 0.891 339 308 305 789 011 626 318 364 672 × 2 = 1 + 0.782 678 616 611 578 023 252 636 729 344;
  • 59) 0.782 678 616 611 578 023 252 636 729 344 × 2 = 1 + 0.565 357 233 223 156 046 505 273 458 688;
  • 60) 0.565 357 233 223 156 046 505 273 458 688 × 2 = 1 + 0.130 714 466 446 312 093 010 546 917 376;
  • 61) 0.130 714 466 446 312 093 010 546 917 376 × 2 = 0 + 0.261 428 932 892 624 186 021 093 834 752;
  • 62) 0.261 428 932 892 624 186 021 093 834 752 × 2 = 0 + 0.522 857 865 785 248 372 042 187 669 504;
  • 63) 0.522 857 865 785 248 372 042 187 669 504 × 2 = 1 + 0.045 715 731 570 496 744 084 375 339 008;
  • 64) 0.045 715 731 570 496 744 084 375 339 008 × 2 = 0 + 0.091 431 463 140 993 488 168 750 678 016;
  • 65) 0.091 431 463 140 993 488 168 750 678 016 × 2 = 0 + 0.182 862 926 281 986 976 337 501 356 032;
  • 66) 0.182 862 926 281 986 976 337 501 356 032 × 2 = 0 + 0.365 725 852 563 973 952 675 002 712 064;
  • 67) 0.365 725 852 563 973 952 675 002 712 064 × 2 = 0 + 0.731 451 705 127 947 905 350 005 424 128;
  • 68) 0.731 451 705 127 947 905 350 005 424 128 × 2 = 1 + 0.462 903 410 255 895 810 700 010 848 256;

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:


0.000 020 830 729 321 671 205 134 999 151(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:

0.000 020 830 729 321 671 205 134 999 151(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:


0.000 020 830 729 321 671 205 134 999 151(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 0.000 020 830 729 321 671 205 134 999 151 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