0.000 020 830 729 321 671 205 134 999 154 471 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 471(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
0.000 020 830 729 321 671 205 134 999 154 471(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: 0.000 020 830 729 321 671 205 134 999 154 471.

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) 0.000 020 830 729 321 671 205 134 999 154 471 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 308 942;
  • 2) 0.000 041 661 458 643 342 410 269 998 308 942 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 617 884;
  • 3) 0.000 083 322 917 286 684 820 539 996 617 884 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 235 768;
  • 4) 0.000 166 645 834 573 369 641 079 993 235 768 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 471 536;
  • 5) 0.000 333 291 669 146 739 282 159 986 471 536 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 943 072;
  • 6) 0.000 666 583 338 293 478 564 319 972 943 072 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 886 144;
  • 7) 0.001 333 166 676 586 957 128 639 945 886 144 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 772 288;
  • 8) 0.002 666 333 353 173 914 257 279 891 772 288 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 544 576;
  • 9) 0.005 332 666 706 347 828 514 559 783 544 576 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 089 152;
  • 10) 0.010 665 333 412 695 657 029 119 567 089 152 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 178 304;
  • 11) 0.021 330 666 825 391 314 058 239 134 178 304 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 356 608;
  • 12) 0.042 661 333 650 782 628 116 478 268 356 608 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 713 216;
  • 13) 0.085 322 667 301 565 256 232 956 536 713 216 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 426 432;
  • 14) 0.170 645 334 603 130 512 465 913 073 426 432 × 2 = 0 + 0.341 290 669 206 261 024 931 826 146 852 864;
  • 15) 0.341 290 669 206 261 024 931 826 146 852 864 × 2 = 0 + 0.682 581 338 412 522 049 863 652 293 705 728;
  • 16) 0.682 581 338 412 522 049 863 652 293 705 728 × 2 = 1 + 0.365 162 676 825 044 099 727 304 587 411 456;
  • 17) 0.365 162 676 825 044 099 727 304 587 411 456 × 2 = 0 + 0.730 325 353 650 088 199 454 609 174 822 912;
  • 18) 0.730 325 353 650 088 199 454 609 174 822 912 × 2 = 1 + 0.460 650 707 300 176 398 909 218 349 645 824;
  • 19) 0.460 650 707 300 176 398 909 218 349 645 824 × 2 = 0 + 0.921 301 414 600 352 797 818 436 699 291 648;
  • 20) 0.921 301 414 600 352 797 818 436 699 291 648 × 2 = 1 + 0.842 602 829 200 705 595 636 873 398 583 296;
  • 21) 0.842 602 829 200 705 595 636 873 398 583 296 × 2 = 1 + 0.685 205 658 401 411 191 273 746 797 166 592;
  • 22) 0.685 205 658 401 411 191 273 746 797 166 592 × 2 = 1 + 0.370 411 316 802 822 382 547 493 594 333 184;
  • 23) 0.370 411 316 802 822 382 547 493 594 333 184 × 2 = 0 + 0.740 822 633 605 644 765 094 987 188 666 368;
  • 24) 0.740 822 633 605 644 765 094 987 188 666 368 × 2 = 1 + 0.481 645 267 211 289 530 189 974 377 332 736;
  • 25) 0.481 645 267 211 289 530 189 974 377 332 736 × 2 = 0 + 0.963 290 534 422 579 060 379 948 754 665 472;
  • 26) 0.963 290 534 422 579 060 379 948 754 665 472 × 2 = 1 + 0.926 581 068 845 158 120 759 897 509 330 944;
  • 27) 0.926 581 068 845 158 120 759 897 509 330 944 × 2 = 1 + 0.853 162 137 690 316 241 519 795 018 661 888;
  • 28) 0.853 162 137 690 316 241 519 795 018 661 888 × 2 = 1 + 0.706 324 275 380 632 483 039 590 037 323 776;
  • 29) 0.706 324 275 380 632 483 039 590 037 323 776 × 2 = 1 + 0.412 648 550 761 264 966 079 180 074 647 552;
  • 30) 0.412 648 550 761 264 966 079 180 074 647 552 × 2 = 0 + 0.825 297 101 522 529 932 158 360 149 295 104;
  • 31) 0.825 297 101 522 529 932 158 360 149 295 104 × 2 = 1 + 0.650 594 203 045 059 864 316 720 298 590 208;
  • 32) 0.650 594 203 045 059 864 316 720 298 590 208 × 2 = 1 + 0.301 188 406 090 119 728 633 440 597 180 416;
  • 33) 0.301 188 406 090 119 728 633 440 597 180 416 × 2 = 0 + 0.602 376 812 180 239 457 266 881 194 360 832;
  • 34) 0.602 376 812 180 239 457 266 881 194 360 832 × 2 = 1 + 0.204 753 624 360 478 914 533 762 388 721 664;
  • 35) 0.204 753 624 360 478 914 533 762 388 721 664 × 2 = 0 + 0.409 507 248 720 957 829 067 524 777 443 328;
  • 36) 0.409 507 248 720 957 829 067 524 777 443 328 × 2 = 0 + 0.819 014 497 441 915 658 135 049 554 886 656;
  • 37) 0.819 014 497 441 915 658 135 049 554 886 656 × 2 = 1 + 0.638 028 994 883 831 316 270 099 109 773 312;
  • 38) 0.638 028 994 883 831 316 270 099 109 773 312 × 2 = 1 + 0.276 057 989 767 662 632 540 198 219 546 624;
  • 39) 0.276 057 989 767 662 632 540 198 219 546 624 × 2 = 0 + 0.552 115 979 535 325 265 080 396 439 093 248;
  • 40) 0.552 115 979 535 325 265 080 396 439 093 248 × 2 = 1 + 0.104 231 959 070 650 530 160 792 878 186 496;
  • 41) 0.104 231 959 070 650 530 160 792 878 186 496 × 2 = 0 + 0.208 463 918 141 301 060 321 585 756 372 992;
  • 42) 0.208 463 918 141 301 060 321 585 756 372 992 × 2 = 0 + 0.416 927 836 282 602 120 643 171 512 745 984;
  • 43) 0.416 927 836 282 602 120 643 171 512 745 984 × 2 = 0 + 0.833 855 672 565 204 241 286 343 025 491 968;
  • 44) 0.833 855 672 565 204 241 286 343 025 491 968 × 2 = 1 + 0.667 711 345 130 408 482 572 686 050 983 936;
  • 45) 0.667 711 345 130 408 482 572 686 050 983 936 × 2 = 1 + 0.335 422 690 260 816 965 145 372 101 967 872;
  • 46) 0.335 422 690 260 816 965 145 372 101 967 872 × 2 = 0 + 0.670 845 380 521 633 930 290 744 203 935 744;
  • 47) 0.670 845 380 521 633 930 290 744 203 935 744 × 2 = 1 + 0.341 690 761 043 267 860 581 488 407 871 488;
  • 48) 0.341 690 761 043 267 860 581 488 407 871 488 × 2 = 0 + 0.683 381 522 086 535 721 162 976 815 742 976;
  • 49) 0.683 381 522 086 535 721 162 976 815 742 976 × 2 = 1 + 0.366 763 044 173 071 442 325 953 631 485 952;
  • 50) 0.366 763 044 173 071 442 325 953 631 485 952 × 2 = 0 + 0.733 526 088 346 142 884 651 907 262 971 904;
  • 51) 0.733 526 088 346 142 884 651 907 262 971 904 × 2 = 1 + 0.467 052 176 692 285 769 303 814 525 943 808;
  • 52) 0.467 052 176 692 285 769 303 814 525 943 808 × 2 = 0 + 0.934 104 353 384 571 538 607 629 051 887 616;
  • 53) 0.934 104 353 384 571 538 607 629 051 887 616 × 2 = 1 + 0.868 208 706 769 143 077 215 258 103 775 232;
  • 54) 0.868 208 706 769 143 077 215 258 103 775 232 × 2 = 1 + 0.736 417 413 538 286 154 430 516 207 550 464;
  • 55) 0.736 417 413 538 286 154 430 516 207 550 464 × 2 = 1 + 0.472 834 827 076 572 308 861 032 415 100 928;
  • 56) 0.472 834 827 076 572 308 861 032 415 100 928 × 2 = 0 + 0.945 669 654 153 144 617 722 064 830 201 856;
  • 57) 0.945 669 654 153 144 617 722 064 830 201 856 × 2 = 1 + 0.891 339 308 306 289 235 444 129 660 403 712;
  • 58) 0.891 339 308 306 289 235 444 129 660 403 712 × 2 = 1 + 0.782 678 616 612 578 470 888 259 320 807 424;
  • 59) 0.782 678 616 612 578 470 888 259 320 807 424 × 2 = 1 + 0.565 357 233 225 156 941 776 518 641 614 848;
  • 60) 0.565 357 233 225 156 941 776 518 641 614 848 × 2 = 1 + 0.130 714 466 450 313 883 553 037 283 229 696;
  • 61) 0.130 714 466 450 313 883 553 037 283 229 696 × 2 = 0 + 0.261 428 932 900 627 767 106 074 566 459 392;
  • 62) 0.261 428 932 900 627 767 106 074 566 459 392 × 2 = 0 + 0.522 857 865 801 255 534 212 149 132 918 784;
  • 63) 0.522 857 865 801 255 534 212 149 132 918 784 × 2 = 1 + 0.045 715 731 602 511 068 424 298 265 837 568;
  • 64) 0.045 715 731 602 511 068 424 298 265 837 568 × 2 = 0 + 0.091 431 463 205 022 136 848 596 531 675 136;
  • 65) 0.091 431 463 205 022 136 848 596 531 675 136 × 2 = 0 + 0.182 862 926 410 044 273 697 193 063 350 272;
  • 66) 0.182 862 926 410 044 273 697 193 063 350 272 × 2 = 0 + 0.365 725 852 820 088 547 394 386 126 700 544;
  • 67) 0.365 725 852 820 088 547 394 386 126 700 544 × 2 = 0 + 0.731 451 705 640 177 094 788 772 253 401 088;
  • 68) 0.731 451 705 640 177 094 788 772 253 401 088 × 2 = 1 + 0.462 903 411 280 354 189 577 544 506 802 176;

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:


0.000 020 830 729 321 671 205 134 999 154 471(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:

0.000 020 830 729 321 671 205 134 999 154 471(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:


0.000 020 830 729 321 671 205 134 999 154 471(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 0.000 020 830 729 321 671 205 134 999 154 471 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