0.000 020 830 729 321 671 205 134 999 154 518 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 518 6(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
0.000 020 830 729 321 671 205 134 999 154 518 6(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: 0.000 020 830 729 321 671 205 134 999 154 518 6.

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) 0.000 020 830 729 321 671 205 134 999 154 518 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 037 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 037 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 074 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 074 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 148 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 148 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 297 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 297 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 595 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 595 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 889 190 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 889 190 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 778 380 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 778 380 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 556 761 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 556 761 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 113 523 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 113 523 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 227 046 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 227 046 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 454 092 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 454 092 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 908 185 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 908 185 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 816 371 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 816 371 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 632 742 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 632 742 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 265 484 8;
  • 16) 0.682 581 338 412 522 049 863 652 295 265 484 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 590 530 969 6;
  • 17) 0.365 162 676 825 044 099 727 304 590 530 969 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 181 061 939 2;
  • 18) 0.730 325 353 650 088 199 454 609 181 061 939 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 362 123 878 4;
  • 19) 0.460 650 707 300 176 398 909 218 362 123 878 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 724 247 756 8;
  • 20) 0.921 301 414 600 352 797 818 436 724 247 756 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 448 495 513 6;
  • 21) 0.842 602 829 200 705 595 636 873 448 495 513 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 896 991 027 2;
  • 22) 0.685 205 658 401 411 191 273 746 896 991 027 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 793 982 054 4;
  • 23) 0.370 411 316 802 822 382 547 493 793 982 054 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 587 964 108 8;
  • 24) 0.740 822 633 605 644 765 094 987 587 964 108 8 × 2 = 1 + 0.481 645 267 211 289 530 189 975 175 928 217 6;
  • 25) 0.481 645 267 211 289 530 189 975 175 928 217 6 × 2 = 0 + 0.963 290 534 422 579 060 379 950 351 856 435 2;
  • 26) 0.963 290 534 422 579 060 379 950 351 856 435 2 × 2 = 1 + 0.926 581 068 845 158 120 759 900 703 712 870 4;
  • 27) 0.926 581 068 845 158 120 759 900 703 712 870 4 × 2 = 1 + 0.853 162 137 690 316 241 519 801 407 425 740 8;
  • 28) 0.853 162 137 690 316 241 519 801 407 425 740 8 × 2 = 1 + 0.706 324 275 380 632 483 039 602 814 851 481 6;
  • 29) 0.706 324 275 380 632 483 039 602 814 851 481 6 × 2 = 1 + 0.412 648 550 761 264 966 079 205 629 702 963 2;
  • 30) 0.412 648 550 761 264 966 079 205 629 702 963 2 × 2 = 0 + 0.825 297 101 522 529 932 158 411 259 405 926 4;
  • 31) 0.825 297 101 522 529 932 158 411 259 405 926 4 × 2 = 1 + 0.650 594 203 045 059 864 316 822 518 811 852 8;
  • 32) 0.650 594 203 045 059 864 316 822 518 811 852 8 × 2 = 1 + 0.301 188 406 090 119 728 633 645 037 623 705 6;
  • 33) 0.301 188 406 090 119 728 633 645 037 623 705 6 × 2 = 0 + 0.602 376 812 180 239 457 267 290 075 247 411 2;
  • 34) 0.602 376 812 180 239 457 267 290 075 247 411 2 × 2 = 1 + 0.204 753 624 360 478 914 534 580 150 494 822 4;
  • 35) 0.204 753 624 360 478 914 534 580 150 494 822 4 × 2 = 0 + 0.409 507 248 720 957 829 069 160 300 989 644 8;
  • 36) 0.409 507 248 720 957 829 069 160 300 989 644 8 × 2 = 0 + 0.819 014 497 441 915 658 138 320 601 979 289 6;
  • 37) 0.819 014 497 441 915 658 138 320 601 979 289 6 × 2 = 1 + 0.638 028 994 883 831 316 276 641 203 958 579 2;
  • 38) 0.638 028 994 883 831 316 276 641 203 958 579 2 × 2 = 1 + 0.276 057 989 767 662 632 553 282 407 917 158 4;
  • 39) 0.276 057 989 767 662 632 553 282 407 917 158 4 × 2 = 0 + 0.552 115 979 535 325 265 106 564 815 834 316 8;
  • 40) 0.552 115 979 535 325 265 106 564 815 834 316 8 × 2 = 1 + 0.104 231 959 070 650 530 213 129 631 668 633 6;
  • 41) 0.104 231 959 070 650 530 213 129 631 668 633 6 × 2 = 0 + 0.208 463 918 141 301 060 426 259 263 337 267 2;
  • 42) 0.208 463 918 141 301 060 426 259 263 337 267 2 × 2 = 0 + 0.416 927 836 282 602 120 852 518 526 674 534 4;
  • 43) 0.416 927 836 282 602 120 852 518 526 674 534 4 × 2 = 0 + 0.833 855 672 565 204 241 705 037 053 349 068 8;
  • 44) 0.833 855 672 565 204 241 705 037 053 349 068 8 × 2 = 1 + 0.667 711 345 130 408 483 410 074 106 698 137 6;
  • 45) 0.667 711 345 130 408 483 410 074 106 698 137 6 × 2 = 1 + 0.335 422 690 260 816 966 820 148 213 396 275 2;
  • 46) 0.335 422 690 260 816 966 820 148 213 396 275 2 × 2 = 0 + 0.670 845 380 521 633 933 640 296 426 792 550 4;
  • 47) 0.670 845 380 521 633 933 640 296 426 792 550 4 × 2 = 1 + 0.341 690 761 043 267 867 280 592 853 585 100 8;
  • 48) 0.341 690 761 043 267 867 280 592 853 585 100 8 × 2 = 0 + 0.683 381 522 086 535 734 561 185 707 170 201 6;
  • 49) 0.683 381 522 086 535 734 561 185 707 170 201 6 × 2 = 1 + 0.366 763 044 173 071 469 122 371 414 340 403 2;
  • 50) 0.366 763 044 173 071 469 122 371 414 340 403 2 × 2 = 0 + 0.733 526 088 346 142 938 244 742 828 680 806 4;
  • 51) 0.733 526 088 346 142 938 244 742 828 680 806 4 × 2 = 1 + 0.467 052 176 692 285 876 489 485 657 361 612 8;
  • 52) 0.467 052 176 692 285 876 489 485 657 361 612 8 × 2 = 0 + 0.934 104 353 384 571 752 978 971 314 723 225 6;
  • 53) 0.934 104 353 384 571 752 978 971 314 723 225 6 × 2 = 1 + 0.868 208 706 769 143 505 957 942 629 446 451 2;
  • 54) 0.868 208 706 769 143 505 957 942 629 446 451 2 × 2 = 1 + 0.736 417 413 538 287 011 915 885 258 892 902 4;
  • 55) 0.736 417 413 538 287 011 915 885 258 892 902 4 × 2 = 1 + 0.472 834 827 076 574 023 831 770 517 785 804 8;
  • 56) 0.472 834 827 076 574 023 831 770 517 785 804 8 × 2 = 0 + 0.945 669 654 153 148 047 663 541 035 571 609 6;
  • 57) 0.945 669 654 153 148 047 663 541 035 571 609 6 × 2 = 1 + 0.891 339 308 306 296 095 327 082 071 143 219 2;
  • 58) 0.891 339 308 306 296 095 327 082 071 143 219 2 × 2 = 1 + 0.782 678 616 612 592 190 654 164 142 286 438 4;
  • 59) 0.782 678 616 612 592 190 654 164 142 286 438 4 × 2 = 1 + 0.565 357 233 225 184 381 308 328 284 572 876 8;
  • 60) 0.565 357 233 225 184 381 308 328 284 572 876 8 × 2 = 1 + 0.130 714 466 450 368 762 616 656 569 145 753 6;
  • 61) 0.130 714 466 450 368 762 616 656 569 145 753 6 × 2 = 0 + 0.261 428 932 900 737 525 233 313 138 291 507 2;
  • 62) 0.261 428 932 900 737 525 233 313 138 291 507 2 × 2 = 0 + 0.522 857 865 801 475 050 466 626 276 583 014 4;
  • 63) 0.522 857 865 801 475 050 466 626 276 583 014 4 × 2 = 1 + 0.045 715 731 602 950 100 933 252 553 166 028 8;
  • 64) 0.045 715 731 602 950 100 933 252 553 166 028 8 × 2 = 0 + 0.091 431 463 205 900 201 866 505 106 332 057 6;
  • 65) 0.091 431 463 205 900 201 866 505 106 332 057 6 × 2 = 0 + 0.182 862 926 411 800 403 733 010 212 664 115 2;
  • 66) 0.182 862 926 411 800 403 733 010 212 664 115 2 × 2 = 0 + 0.365 725 852 823 600 807 466 020 425 328 230 4;
  • 67) 0.365 725 852 823 600 807 466 020 425 328 230 4 × 2 = 0 + 0.731 451 705 647 201 614 932 040 850 656 460 8;
  • 68) 0.731 451 705 647 201 614 932 040 850 656 460 8 × 2 = 1 + 0.462 903 411 294 403 229 864 081 701 312 921 6;

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:


0.000 020 830 729 321 671 205 134 999 154 518 6(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:

0.000 020 830 729 321 671 205 134 999 154 518 6(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:


0.000 020 830 729 321 671 205 134 999 154 518 6(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 0.000 020 830 729 321 671 205 134 999 154 518 6 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