0.000 020 830 729 321 671 205 134 999 161 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 161 9(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
0.000 020 830 729 321 671 205 134 999 161 9(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: 0.000 020 830 729 321 671 205 134 999 161 9.

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) 0.000 020 830 729 321 671 205 134 999 161 9 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 323 8;
  • 2) 0.000 041 661 458 643 342 410 269 998 323 8 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 647 6;
  • 3) 0.000 083 322 917 286 684 820 539 996 647 6 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 295 2;
  • 4) 0.000 166 645 834 573 369 641 079 993 295 2 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 590 4;
  • 5) 0.000 333 291 669 146 739 282 159 986 590 4 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 180 8;
  • 6) 0.000 666 583 338 293 478 564 319 973 180 8 × 2 = 0 + 0.001 333 166 676 586 957 128 639 946 361 6;
  • 7) 0.001 333 166 676 586 957 128 639 946 361 6 × 2 = 0 + 0.002 666 333 353 173 914 257 279 892 723 2;
  • 8) 0.002 666 333 353 173 914 257 279 892 723 2 × 2 = 0 + 0.005 332 666 706 347 828 514 559 785 446 4;
  • 9) 0.005 332 666 706 347 828 514 559 785 446 4 × 2 = 0 + 0.010 665 333 412 695 657 029 119 570 892 8;
  • 10) 0.010 665 333 412 695 657 029 119 570 892 8 × 2 = 0 + 0.021 330 666 825 391 314 058 239 141 785 6;
  • 11) 0.021 330 666 825 391 314 058 239 141 785 6 × 2 = 0 + 0.042 661 333 650 782 628 116 478 283 571 2;
  • 12) 0.042 661 333 650 782 628 116 478 283 571 2 × 2 = 0 + 0.085 322 667 301 565 256 232 956 567 142 4;
  • 13) 0.085 322 667 301 565 256 232 956 567 142 4 × 2 = 0 + 0.170 645 334 603 130 512 465 913 134 284 8;
  • 14) 0.170 645 334 603 130 512 465 913 134 284 8 × 2 = 0 + 0.341 290 669 206 261 024 931 826 268 569 6;
  • 15) 0.341 290 669 206 261 024 931 826 268 569 6 × 2 = 0 + 0.682 581 338 412 522 049 863 652 537 139 2;
  • 16) 0.682 581 338 412 522 049 863 652 537 139 2 × 2 = 1 + 0.365 162 676 825 044 099 727 305 074 278 4;
  • 17) 0.365 162 676 825 044 099 727 305 074 278 4 × 2 = 0 + 0.730 325 353 650 088 199 454 610 148 556 8;
  • 18) 0.730 325 353 650 088 199 454 610 148 556 8 × 2 = 1 + 0.460 650 707 300 176 398 909 220 297 113 6;
  • 19) 0.460 650 707 300 176 398 909 220 297 113 6 × 2 = 0 + 0.921 301 414 600 352 797 818 440 594 227 2;
  • 20) 0.921 301 414 600 352 797 818 440 594 227 2 × 2 = 1 + 0.842 602 829 200 705 595 636 881 188 454 4;
  • 21) 0.842 602 829 200 705 595 636 881 188 454 4 × 2 = 1 + 0.685 205 658 401 411 191 273 762 376 908 8;
  • 22) 0.685 205 658 401 411 191 273 762 376 908 8 × 2 = 1 + 0.370 411 316 802 822 382 547 524 753 817 6;
  • 23) 0.370 411 316 802 822 382 547 524 753 817 6 × 2 = 0 + 0.740 822 633 605 644 765 095 049 507 635 2;
  • 24) 0.740 822 633 605 644 765 095 049 507 635 2 × 2 = 1 + 0.481 645 267 211 289 530 190 099 015 270 4;
  • 25) 0.481 645 267 211 289 530 190 099 015 270 4 × 2 = 0 + 0.963 290 534 422 579 060 380 198 030 540 8;
  • 26) 0.963 290 534 422 579 060 380 198 030 540 8 × 2 = 1 + 0.926 581 068 845 158 120 760 396 061 081 6;
  • 27) 0.926 581 068 845 158 120 760 396 061 081 6 × 2 = 1 + 0.853 162 137 690 316 241 520 792 122 163 2;
  • 28) 0.853 162 137 690 316 241 520 792 122 163 2 × 2 = 1 + 0.706 324 275 380 632 483 041 584 244 326 4;
  • 29) 0.706 324 275 380 632 483 041 584 244 326 4 × 2 = 1 + 0.412 648 550 761 264 966 083 168 488 652 8;
  • 30) 0.412 648 550 761 264 966 083 168 488 652 8 × 2 = 0 + 0.825 297 101 522 529 932 166 336 977 305 6;
  • 31) 0.825 297 101 522 529 932 166 336 977 305 6 × 2 = 1 + 0.650 594 203 045 059 864 332 673 954 611 2;
  • 32) 0.650 594 203 045 059 864 332 673 954 611 2 × 2 = 1 + 0.301 188 406 090 119 728 665 347 909 222 4;
  • 33) 0.301 188 406 090 119 728 665 347 909 222 4 × 2 = 0 + 0.602 376 812 180 239 457 330 695 818 444 8;
  • 34) 0.602 376 812 180 239 457 330 695 818 444 8 × 2 = 1 + 0.204 753 624 360 478 914 661 391 636 889 6;
  • 35) 0.204 753 624 360 478 914 661 391 636 889 6 × 2 = 0 + 0.409 507 248 720 957 829 322 783 273 779 2;
  • 36) 0.409 507 248 720 957 829 322 783 273 779 2 × 2 = 0 + 0.819 014 497 441 915 658 645 566 547 558 4;
  • 37) 0.819 014 497 441 915 658 645 566 547 558 4 × 2 = 1 + 0.638 028 994 883 831 317 291 133 095 116 8;
  • 38) 0.638 028 994 883 831 317 291 133 095 116 8 × 2 = 1 + 0.276 057 989 767 662 634 582 266 190 233 6;
  • 39) 0.276 057 989 767 662 634 582 266 190 233 6 × 2 = 0 + 0.552 115 979 535 325 269 164 532 380 467 2;
  • 40) 0.552 115 979 535 325 269 164 532 380 467 2 × 2 = 1 + 0.104 231 959 070 650 538 329 064 760 934 4;
  • 41) 0.104 231 959 070 650 538 329 064 760 934 4 × 2 = 0 + 0.208 463 918 141 301 076 658 129 521 868 8;
  • 42) 0.208 463 918 141 301 076 658 129 521 868 8 × 2 = 0 + 0.416 927 836 282 602 153 316 259 043 737 6;
  • 43) 0.416 927 836 282 602 153 316 259 043 737 6 × 2 = 0 + 0.833 855 672 565 204 306 632 518 087 475 2;
  • 44) 0.833 855 672 565 204 306 632 518 087 475 2 × 2 = 1 + 0.667 711 345 130 408 613 265 036 174 950 4;
  • 45) 0.667 711 345 130 408 613 265 036 174 950 4 × 2 = 1 + 0.335 422 690 260 817 226 530 072 349 900 8;
  • 46) 0.335 422 690 260 817 226 530 072 349 900 8 × 2 = 0 + 0.670 845 380 521 634 453 060 144 699 801 6;
  • 47) 0.670 845 380 521 634 453 060 144 699 801 6 × 2 = 1 + 0.341 690 761 043 268 906 120 289 399 603 2;
  • 48) 0.341 690 761 043 268 906 120 289 399 603 2 × 2 = 0 + 0.683 381 522 086 537 812 240 578 799 206 4;
  • 49) 0.683 381 522 086 537 812 240 578 799 206 4 × 2 = 1 + 0.366 763 044 173 075 624 481 157 598 412 8;
  • 50) 0.366 763 044 173 075 624 481 157 598 412 8 × 2 = 0 + 0.733 526 088 346 151 248 962 315 196 825 6;
  • 51) 0.733 526 088 346 151 248 962 315 196 825 6 × 2 = 1 + 0.467 052 176 692 302 497 924 630 393 651 2;
  • 52) 0.467 052 176 692 302 497 924 630 393 651 2 × 2 = 0 + 0.934 104 353 384 604 995 849 260 787 302 4;
  • 53) 0.934 104 353 384 604 995 849 260 787 302 4 × 2 = 1 + 0.868 208 706 769 209 991 698 521 574 604 8;
  • 54) 0.868 208 706 769 209 991 698 521 574 604 8 × 2 = 1 + 0.736 417 413 538 419 983 397 043 149 209 6;
  • 55) 0.736 417 413 538 419 983 397 043 149 209 6 × 2 = 1 + 0.472 834 827 076 839 966 794 086 298 419 2;
  • 56) 0.472 834 827 076 839 966 794 086 298 419 2 × 2 = 0 + 0.945 669 654 153 679 933 588 172 596 838 4;
  • 57) 0.945 669 654 153 679 933 588 172 596 838 4 × 2 = 1 + 0.891 339 308 307 359 867 176 345 193 676 8;
  • 58) 0.891 339 308 307 359 867 176 345 193 676 8 × 2 = 1 + 0.782 678 616 614 719 734 352 690 387 353 6;
  • 59) 0.782 678 616 614 719 734 352 690 387 353 6 × 2 = 1 + 0.565 357 233 229 439 468 705 380 774 707 2;
  • 60) 0.565 357 233 229 439 468 705 380 774 707 2 × 2 = 1 + 0.130 714 466 458 878 937 410 761 549 414 4;
  • 61) 0.130 714 466 458 878 937 410 761 549 414 4 × 2 = 0 + 0.261 428 932 917 757 874 821 523 098 828 8;
  • 62) 0.261 428 932 917 757 874 821 523 098 828 8 × 2 = 0 + 0.522 857 865 835 515 749 643 046 197 657 6;
  • 63) 0.522 857 865 835 515 749 643 046 197 657 6 × 2 = 1 + 0.045 715 731 671 031 499 286 092 395 315 2;
  • 64) 0.045 715 731 671 031 499 286 092 395 315 2 × 2 = 0 + 0.091 431 463 342 062 998 572 184 790 630 4;
  • 65) 0.091 431 463 342 062 998 572 184 790 630 4 × 2 = 0 + 0.182 862 926 684 125 997 144 369 581 260 8;
  • 66) 0.182 862 926 684 125 997 144 369 581 260 8 × 2 = 0 + 0.365 725 853 368 251 994 288 739 162 521 6;
  • 67) 0.365 725 853 368 251 994 288 739 162 521 6 × 2 = 0 + 0.731 451 706 736 503 988 577 478 325 043 2;
  • 68) 0.731 451 706 736 503 988 577 478 325 043 2 × 2 = 1 + 0.462 903 413 473 007 977 154 956 650 086 4;

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:


0.000 020 830 729 321 671 205 134 999 161 9(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:

0.000 020 830 729 321 671 205 134 999 161 9(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:


0.000 020 830 729 321 671 205 134 999 161 9(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 0.000 020 830 729 321 671 205 134 999 161 9 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