0.000 020 830 729 321 671 205 134 999 154 512 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 512 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
0.000 020 830 729 321 671 205 134 999 154 512 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: 0.000 020 830 729 321 671 205 134 999 154 512 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) 0.000 020 830 729 321 671 205 134 999 154 512 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 025 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 025 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 050 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 050 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 100 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 100 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 201 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 201 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 403 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 403 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 806 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 806 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 612 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 612 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 555 225 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 555 225 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 110 451 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 110 451 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 220 902 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 220 902 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 441 804 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 441 804 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 883 609 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 883 609 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 767 219 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 767 219 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 534 438 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 534 438 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 068 876 8;
  • 16) 0.682 581 338 412 522 049 863 652 295 068 876 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 590 137 753 6;
  • 17) 0.365 162 676 825 044 099 727 304 590 137 753 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 180 275 507 2;
  • 18) 0.730 325 353 650 088 199 454 609 180 275 507 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 360 551 014 4;
  • 19) 0.460 650 707 300 176 398 909 218 360 551 014 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 721 102 028 8;
  • 20) 0.921 301 414 600 352 797 818 436 721 102 028 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 442 204 057 6;
  • 21) 0.842 602 829 200 705 595 636 873 442 204 057 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 884 408 115 2;
  • 22) 0.685 205 658 401 411 191 273 746 884 408 115 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 768 816 230 4;
  • 23) 0.370 411 316 802 822 382 547 493 768 816 230 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 537 632 460 8;
  • 24) 0.740 822 633 605 644 765 094 987 537 632 460 8 × 2 = 1 + 0.481 645 267 211 289 530 189 975 075 264 921 6;
  • 25) 0.481 645 267 211 289 530 189 975 075 264 921 6 × 2 = 0 + 0.963 290 534 422 579 060 379 950 150 529 843 2;
  • 26) 0.963 290 534 422 579 060 379 950 150 529 843 2 × 2 = 1 + 0.926 581 068 845 158 120 759 900 301 059 686 4;
  • 27) 0.926 581 068 845 158 120 759 900 301 059 686 4 × 2 = 1 + 0.853 162 137 690 316 241 519 800 602 119 372 8;
  • 28) 0.853 162 137 690 316 241 519 800 602 119 372 8 × 2 = 1 + 0.706 324 275 380 632 483 039 601 204 238 745 6;
  • 29) 0.706 324 275 380 632 483 039 601 204 238 745 6 × 2 = 1 + 0.412 648 550 761 264 966 079 202 408 477 491 2;
  • 30) 0.412 648 550 761 264 966 079 202 408 477 491 2 × 2 = 0 + 0.825 297 101 522 529 932 158 404 816 954 982 4;
  • 31) 0.825 297 101 522 529 932 158 404 816 954 982 4 × 2 = 1 + 0.650 594 203 045 059 864 316 809 633 909 964 8;
  • 32) 0.650 594 203 045 059 864 316 809 633 909 964 8 × 2 = 1 + 0.301 188 406 090 119 728 633 619 267 819 929 6;
  • 33) 0.301 188 406 090 119 728 633 619 267 819 929 6 × 2 = 0 + 0.602 376 812 180 239 457 267 238 535 639 859 2;
  • 34) 0.602 376 812 180 239 457 267 238 535 639 859 2 × 2 = 1 + 0.204 753 624 360 478 914 534 477 071 279 718 4;
  • 35) 0.204 753 624 360 478 914 534 477 071 279 718 4 × 2 = 0 + 0.409 507 248 720 957 829 068 954 142 559 436 8;
  • 36) 0.409 507 248 720 957 829 068 954 142 559 436 8 × 2 = 0 + 0.819 014 497 441 915 658 137 908 285 118 873 6;
  • 37) 0.819 014 497 441 915 658 137 908 285 118 873 6 × 2 = 1 + 0.638 028 994 883 831 316 275 816 570 237 747 2;
  • 38) 0.638 028 994 883 831 316 275 816 570 237 747 2 × 2 = 1 + 0.276 057 989 767 662 632 551 633 140 475 494 4;
  • 39) 0.276 057 989 767 662 632 551 633 140 475 494 4 × 2 = 0 + 0.552 115 979 535 325 265 103 266 280 950 988 8;
  • 40) 0.552 115 979 535 325 265 103 266 280 950 988 8 × 2 = 1 + 0.104 231 959 070 650 530 206 532 561 901 977 6;
  • 41) 0.104 231 959 070 650 530 206 532 561 901 977 6 × 2 = 0 + 0.208 463 918 141 301 060 413 065 123 803 955 2;
  • 42) 0.208 463 918 141 301 060 413 065 123 803 955 2 × 2 = 0 + 0.416 927 836 282 602 120 826 130 247 607 910 4;
  • 43) 0.416 927 836 282 602 120 826 130 247 607 910 4 × 2 = 0 + 0.833 855 672 565 204 241 652 260 495 215 820 8;
  • 44) 0.833 855 672 565 204 241 652 260 495 215 820 8 × 2 = 1 + 0.667 711 345 130 408 483 304 520 990 431 641 6;
  • 45) 0.667 711 345 130 408 483 304 520 990 431 641 6 × 2 = 1 + 0.335 422 690 260 816 966 609 041 980 863 283 2;
  • 46) 0.335 422 690 260 816 966 609 041 980 863 283 2 × 2 = 0 + 0.670 845 380 521 633 933 218 083 961 726 566 4;
  • 47) 0.670 845 380 521 633 933 218 083 961 726 566 4 × 2 = 1 + 0.341 690 761 043 267 866 436 167 923 453 132 8;
  • 48) 0.341 690 761 043 267 866 436 167 923 453 132 8 × 2 = 0 + 0.683 381 522 086 535 732 872 335 846 906 265 6;
  • 49) 0.683 381 522 086 535 732 872 335 846 906 265 6 × 2 = 1 + 0.366 763 044 173 071 465 744 671 693 812 531 2;
  • 50) 0.366 763 044 173 071 465 744 671 693 812 531 2 × 2 = 0 + 0.733 526 088 346 142 931 489 343 387 625 062 4;
  • 51) 0.733 526 088 346 142 931 489 343 387 625 062 4 × 2 = 1 + 0.467 052 176 692 285 862 978 686 775 250 124 8;
  • 52) 0.467 052 176 692 285 862 978 686 775 250 124 8 × 2 = 0 + 0.934 104 353 384 571 725 957 373 550 500 249 6;
  • 53) 0.934 104 353 384 571 725 957 373 550 500 249 6 × 2 = 1 + 0.868 208 706 769 143 451 914 747 101 000 499 2;
  • 54) 0.868 208 706 769 143 451 914 747 101 000 499 2 × 2 = 1 + 0.736 417 413 538 286 903 829 494 202 000 998 4;
  • 55) 0.736 417 413 538 286 903 829 494 202 000 998 4 × 2 = 1 + 0.472 834 827 076 573 807 658 988 404 001 996 8;
  • 56) 0.472 834 827 076 573 807 658 988 404 001 996 8 × 2 = 0 + 0.945 669 654 153 147 615 317 976 808 003 993 6;
  • 57) 0.945 669 654 153 147 615 317 976 808 003 993 6 × 2 = 1 + 0.891 339 308 306 295 230 635 953 616 007 987 2;
  • 58) 0.891 339 308 306 295 230 635 953 616 007 987 2 × 2 = 1 + 0.782 678 616 612 590 461 271 907 232 015 974 4;
  • 59) 0.782 678 616 612 590 461 271 907 232 015 974 4 × 2 = 1 + 0.565 357 233 225 180 922 543 814 464 031 948 8;
  • 60) 0.565 357 233 225 180 922 543 814 464 031 948 8 × 2 = 1 + 0.130 714 466 450 361 845 087 628 928 063 897 6;
  • 61) 0.130 714 466 450 361 845 087 628 928 063 897 6 × 2 = 0 + 0.261 428 932 900 723 690 175 257 856 127 795 2;
  • 62) 0.261 428 932 900 723 690 175 257 856 127 795 2 × 2 = 0 + 0.522 857 865 801 447 380 350 515 712 255 590 4;
  • 63) 0.522 857 865 801 447 380 350 515 712 255 590 4 × 2 = 1 + 0.045 715 731 602 894 760 701 031 424 511 180 8;
  • 64) 0.045 715 731 602 894 760 701 031 424 511 180 8 × 2 = 0 + 0.091 431 463 205 789 521 402 062 849 022 361 6;
  • 65) 0.091 431 463 205 789 521 402 062 849 022 361 6 × 2 = 0 + 0.182 862 926 411 579 042 804 125 698 044 723 2;
  • 66) 0.182 862 926 411 579 042 804 125 698 044 723 2 × 2 = 0 + 0.365 725 852 823 158 085 608 251 396 089 446 4;
  • 67) 0.365 725 852 823 158 085 608 251 396 089 446 4 × 2 = 0 + 0.731 451 705 646 316 171 216 502 792 178 892 8;
  • 68) 0.731 451 705 646 316 171 216 502 792 178 892 8 × 2 = 1 + 0.462 903 411 292 632 342 433 005 584 357 785 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:


0.000 020 830 729 321 671 205 134 999 154 512 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:

0.000 020 830 729 321 671 205 134 999 154 512 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:


0.000 020 830 729 321 671 205 134 999 154 512 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 0.000 020 830 729 321 671 205 134 999 154 512 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