0.000 020 830 729 321 671 205 134 999 154 803 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 803(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
0.000 020 830 729 321 671 205 134 999 154 803(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: 0.000 020 830 729 321 671 205 134 999 154 803.

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) 0.000 020 830 729 321 671 205 134 999 154 803 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 606;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 606 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 619 212;
  • 3) 0.000 083 322 917 286 684 820 539 996 619 212 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 238 424;
  • 4) 0.000 166 645 834 573 369 641 079 993 238 424 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 476 848;
  • 5) 0.000 333 291 669 146 739 282 159 986 476 848 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 953 696;
  • 6) 0.000 666 583 338 293 478 564 319 972 953 696 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 907 392;
  • 7) 0.001 333 166 676 586 957 128 639 945 907 392 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 814 784;
  • 8) 0.002 666 333 353 173 914 257 279 891 814 784 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 629 568;
  • 9) 0.005 332 666 706 347 828 514 559 783 629 568 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 259 136;
  • 10) 0.010 665 333 412 695 657 029 119 567 259 136 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 518 272;
  • 11) 0.021 330 666 825 391 314 058 239 134 518 272 × 2 = 0 + 0.042 661 333 650 782 628 116 478 269 036 544;
  • 12) 0.042 661 333 650 782 628 116 478 269 036 544 × 2 = 0 + 0.085 322 667 301 565 256 232 956 538 073 088;
  • 13) 0.085 322 667 301 565 256 232 956 538 073 088 × 2 = 0 + 0.170 645 334 603 130 512 465 913 076 146 176;
  • 14) 0.170 645 334 603 130 512 465 913 076 146 176 × 2 = 0 + 0.341 290 669 206 261 024 931 826 152 292 352;
  • 15) 0.341 290 669 206 261 024 931 826 152 292 352 × 2 = 0 + 0.682 581 338 412 522 049 863 652 304 584 704;
  • 16) 0.682 581 338 412 522 049 863 652 304 584 704 × 2 = 1 + 0.365 162 676 825 044 099 727 304 609 169 408;
  • 17) 0.365 162 676 825 044 099 727 304 609 169 408 × 2 = 0 + 0.730 325 353 650 088 199 454 609 218 338 816;
  • 18) 0.730 325 353 650 088 199 454 609 218 338 816 × 2 = 1 + 0.460 650 707 300 176 398 909 218 436 677 632;
  • 19) 0.460 650 707 300 176 398 909 218 436 677 632 × 2 = 0 + 0.921 301 414 600 352 797 818 436 873 355 264;
  • 20) 0.921 301 414 600 352 797 818 436 873 355 264 × 2 = 1 + 0.842 602 829 200 705 595 636 873 746 710 528;
  • 21) 0.842 602 829 200 705 595 636 873 746 710 528 × 2 = 1 + 0.685 205 658 401 411 191 273 747 493 421 056;
  • 22) 0.685 205 658 401 411 191 273 747 493 421 056 × 2 = 1 + 0.370 411 316 802 822 382 547 494 986 842 112;
  • 23) 0.370 411 316 802 822 382 547 494 986 842 112 × 2 = 0 + 0.740 822 633 605 644 765 094 989 973 684 224;
  • 24) 0.740 822 633 605 644 765 094 989 973 684 224 × 2 = 1 + 0.481 645 267 211 289 530 189 979 947 368 448;
  • 25) 0.481 645 267 211 289 530 189 979 947 368 448 × 2 = 0 + 0.963 290 534 422 579 060 379 959 894 736 896;
  • 26) 0.963 290 534 422 579 060 379 959 894 736 896 × 2 = 1 + 0.926 581 068 845 158 120 759 919 789 473 792;
  • 27) 0.926 581 068 845 158 120 759 919 789 473 792 × 2 = 1 + 0.853 162 137 690 316 241 519 839 578 947 584;
  • 28) 0.853 162 137 690 316 241 519 839 578 947 584 × 2 = 1 + 0.706 324 275 380 632 483 039 679 157 895 168;
  • 29) 0.706 324 275 380 632 483 039 679 157 895 168 × 2 = 1 + 0.412 648 550 761 264 966 079 358 315 790 336;
  • 30) 0.412 648 550 761 264 966 079 358 315 790 336 × 2 = 0 + 0.825 297 101 522 529 932 158 716 631 580 672;
  • 31) 0.825 297 101 522 529 932 158 716 631 580 672 × 2 = 1 + 0.650 594 203 045 059 864 317 433 263 161 344;
  • 32) 0.650 594 203 045 059 864 317 433 263 161 344 × 2 = 1 + 0.301 188 406 090 119 728 634 866 526 322 688;
  • 33) 0.301 188 406 090 119 728 634 866 526 322 688 × 2 = 0 + 0.602 376 812 180 239 457 269 733 052 645 376;
  • 34) 0.602 376 812 180 239 457 269 733 052 645 376 × 2 = 1 + 0.204 753 624 360 478 914 539 466 105 290 752;
  • 35) 0.204 753 624 360 478 914 539 466 105 290 752 × 2 = 0 + 0.409 507 248 720 957 829 078 932 210 581 504;
  • 36) 0.409 507 248 720 957 829 078 932 210 581 504 × 2 = 0 + 0.819 014 497 441 915 658 157 864 421 163 008;
  • 37) 0.819 014 497 441 915 658 157 864 421 163 008 × 2 = 1 + 0.638 028 994 883 831 316 315 728 842 326 016;
  • 38) 0.638 028 994 883 831 316 315 728 842 326 016 × 2 = 1 + 0.276 057 989 767 662 632 631 457 684 652 032;
  • 39) 0.276 057 989 767 662 632 631 457 684 652 032 × 2 = 0 + 0.552 115 979 535 325 265 262 915 369 304 064;
  • 40) 0.552 115 979 535 325 265 262 915 369 304 064 × 2 = 1 + 0.104 231 959 070 650 530 525 830 738 608 128;
  • 41) 0.104 231 959 070 650 530 525 830 738 608 128 × 2 = 0 + 0.208 463 918 141 301 061 051 661 477 216 256;
  • 42) 0.208 463 918 141 301 061 051 661 477 216 256 × 2 = 0 + 0.416 927 836 282 602 122 103 322 954 432 512;
  • 43) 0.416 927 836 282 602 122 103 322 954 432 512 × 2 = 0 + 0.833 855 672 565 204 244 206 645 908 865 024;
  • 44) 0.833 855 672 565 204 244 206 645 908 865 024 × 2 = 1 + 0.667 711 345 130 408 488 413 291 817 730 048;
  • 45) 0.667 711 345 130 408 488 413 291 817 730 048 × 2 = 1 + 0.335 422 690 260 816 976 826 583 635 460 096;
  • 46) 0.335 422 690 260 816 976 826 583 635 460 096 × 2 = 0 + 0.670 845 380 521 633 953 653 167 270 920 192;
  • 47) 0.670 845 380 521 633 953 653 167 270 920 192 × 2 = 1 + 0.341 690 761 043 267 907 306 334 541 840 384;
  • 48) 0.341 690 761 043 267 907 306 334 541 840 384 × 2 = 0 + 0.683 381 522 086 535 814 612 669 083 680 768;
  • 49) 0.683 381 522 086 535 814 612 669 083 680 768 × 2 = 1 + 0.366 763 044 173 071 629 225 338 167 361 536;
  • 50) 0.366 763 044 173 071 629 225 338 167 361 536 × 2 = 0 + 0.733 526 088 346 143 258 450 676 334 723 072;
  • 51) 0.733 526 088 346 143 258 450 676 334 723 072 × 2 = 1 + 0.467 052 176 692 286 516 901 352 669 446 144;
  • 52) 0.467 052 176 692 286 516 901 352 669 446 144 × 2 = 0 + 0.934 104 353 384 573 033 802 705 338 892 288;
  • 53) 0.934 104 353 384 573 033 802 705 338 892 288 × 2 = 1 + 0.868 208 706 769 146 067 605 410 677 784 576;
  • 54) 0.868 208 706 769 146 067 605 410 677 784 576 × 2 = 1 + 0.736 417 413 538 292 135 210 821 355 569 152;
  • 55) 0.736 417 413 538 292 135 210 821 355 569 152 × 2 = 1 + 0.472 834 827 076 584 270 421 642 711 138 304;
  • 56) 0.472 834 827 076 584 270 421 642 711 138 304 × 2 = 0 + 0.945 669 654 153 168 540 843 285 422 276 608;
  • 57) 0.945 669 654 153 168 540 843 285 422 276 608 × 2 = 1 + 0.891 339 308 306 337 081 686 570 844 553 216;
  • 58) 0.891 339 308 306 337 081 686 570 844 553 216 × 2 = 1 + 0.782 678 616 612 674 163 373 141 689 106 432;
  • 59) 0.782 678 616 612 674 163 373 141 689 106 432 × 2 = 1 + 0.565 357 233 225 348 326 746 283 378 212 864;
  • 60) 0.565 357 233 225 348 326 746 283 378 212 864 × 2 = 1 + 0.130 714 466 450 696 653 492 566 756 425 728;
  • 61) 0.130 714 466 450 696 653 492 566 756 425 728 × 2 = 0 + 0.261 428 932 901 393 306 985 133 512 851 456;
  • 62) 0.261 428 932 901 393 306 985 133 512 851 456 × 2 = 0 + 0.522 857 865 802 786 613 970 267 025 702 912;
  • 63) 0.522 857 865 802 786 613 970 267 025 702 912 × 2 = 1 + 0.045 715 731 605 573 227 940 534 051 405 824;
  • 64) 0.045 715 731 605 573 227 940 534 051 405 824 × 2 = 0 + 0.091 431 463 211 146 455 881 068 102 811 648;
  • 65) 0.091 431 463 211 146 455 881 068 102 811 648 × 2 = 0 + 0.182 862 926 422 292 911 762 136 205 623 296;
  • 66) 0.182 862 926 422 292 911 762 136 205 623 296 × 2 = 0 + 0.365 725 852 844 585 823 524 272 411 246 592;
  • 67) 0.365 725 852 844 585 823 524 272 411 246 592 × 2 = 0 + 0.731 451 705 689 171 647 048 544 822 493 184;
  • 68) 0.731 451 705 689 171 647 048 544 822 493 184 × 2 = 1 + 0.462 903 411 378 343 294 097 089 644 986 368;

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:


0.000 020 830 729 321 671 205 134 999 154 803(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:

0.000 020 830 729 321 671 205 134 999 154 803(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:


0.000 020 830 729 321 671 205 134 999 154 803(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 0.000 020 830 729 321 671 205 134 999 154 803 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