0.000 020 830 729 321 671 205 134 999 154 515 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 515 5(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
0.000 020 830 729 321 671 205 134 999 154 515 5(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: 0.000 020 830 729 321 671 205 134 999 154 515 5.

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) 0.000 020 830 729 321 671 205 134 999 154 515 5 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 031;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 031 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 062;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 062 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 124;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 124 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 248;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 248 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 496;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 496 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 992;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 992 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 984;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 984 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 555 968;
  • 9) 0.005 332 666 706 347 828 514 559 783 555 968 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 111 936;
  • 10) 0.010 665 333 412 695 657 029 119 567 111 936 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 223 872;
  • 11) 0.021 330 666 825 391 314 058 239 134 223 872 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 447 744;
  • 12) 0.042 661 333 650 782 628 116 478 268 447 744 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 895 488;
  • 13) 0.085 322 667 301 565 256 232 956 536 895 488 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 790 976;
  • 14) 0.170 645 334 603 130 512 465 913 073 790 976 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 581 952;
  • 15) 0.341 290 669 206 261 024 931 826 147 581 952 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 163 904;
  • 16) 0.682 581 338 412 522 049 863 652 295 163 904 × 2 = 1 + 0.365 162 676 825 044 099 727 304 590 327 808;
  • 17) 0.365 162 676 825 044 099 727 304 590 327 808 × 2 = 0 + 0.730 325 353 650 088 199 454 609 180 655 616;
  • 18) 0.730 325 353 650 088 199 454 609 180 655 616 × 2 = 1 + 0.460 650 707 300 176 398 909 218 361 311 232;
  • 19) 0.460 650 707 300 176 398 909 218 361 311 232 × 2 = 0 + 0.921 301 414 600 352 797 818 436 722 622 464;
  • 20) 0.921 301 414 600 352 797 818 436 722 622 464 × 2 = 1 + 0.842 602 829 200 705 595 636 873 445 244 928;
  • 21) 0.842 602 829 200 705 595 636 873 445 244 928 × 2 = 1 + 0.685 205 658 401 411 191 273 746 890 489 856;
  • 22) 0.685 205 658 401 411 191 273 746 890 489 856 × 2 = 1 + 0.370 411 316 802 822 382 547 493 780 979 712;
  • 23) 0.370 411 316 802 822 382 547 493 780 979 712 × 2 = 0 + 0.740 822 633 605 644 765 094 987 561 959 424;
  • 24) 0.740 822 633 605 644 765 094 987 561 959 424 × 2 = 1 + 0.481 645 267 211 289 530 189 975 123 918 848;
  • 25) 0.481 645 267 211 289 530 189 975 123 918 848 × 2 = 0 + 0.963 290 534 422 579 060 379 950 247 837 696;
  • 26) 0.963 290 534 422 579 060 379 950 247 837 696 × 2 = 1 + 0.926 581 068 845 158 120 759 900 495 675 392;
  • 27) 0.926 581 068 845 158 120 759 900 495 675 392 × 2 = 1 + 0.853 162 137 690 316 241 519 800 991 350 784;
  • 28) 0.853 162 137 690 316 241 519 800 991 350 784 × 2 = 1 + 0.706 324 275 380 632 483 039 601 982 701 568;
  • 29) 0.706 324 275 380 632 483 039 601 982 701 568 × 2 = 1 + 0.412 648 550 761 264 966 079 203 965 403 136;
  • 30) 0.412 648 550 761 264 966 079 203 965 403 136 × 2 = 0 + 0.825 297 101 522 529 932 158 407 930 806 272;
  • 31) 0.825 297 101 522 529 932 158 407 930 806 272 × 2 = 1 + 0.650 594 203 045 059 864 316 815 861 612 544;
  • 32) 0.650 594 203 045 059 864 316 815 861 612 544 × 2 = 1 + 0.301 188 406 090 119 728 633 631 723 225 088;
  • 33) 0.301 188 406 090 119 728 633 631 723 225 088 × 2 = 0 + 0.602 376 812 180 239 457 267 263 446 450 176;
  • 34) 0.602 376 812 180 239 457 267 263 446 450 176 × 2 = 1 + 0.204 753 624 360 478 914 534 526 892 900 352;
  • 35) 0.204 753 624 360 478 914 534 526 892 900 352 × 2 = 0 + 0.409 507 248 720 957 829 069 053 785 800 704;
  • 36) 0.409 507 248 720 957 829 069 053 785 800 704 × 2 = 0 + 0.819 014 497 441 915 658 138 107 571 601 408;
  • 37) 0.819 014 497 441 915 658 138 107 571 601 408 × 2 = 1 + 0.638 028 994 883 831 316 276 215 143 202 816;
  • 38) 0.638 028 994 883 831 316 276 215 143 202 816 × 2 = 1 + 0.276 057 989 767 662 632 552 430 286 405 632;
  • 39) 0.276 057 989 767 662 632 552 430 286 405 632 × 2 = 0 + 0.552 115 979 535 325 265 104 860 572 811 264;
  • 40) 0.552 115 979 535 325 265 104 860 572 811 264 × 2 = 1 + 0.104 231 959 070 650 530 209 721 145 622 528;
  • 41) 0.104 231 959 070 650 530 209 721 145 622 528 × 2 = 0 + 0.208 463 918 141 301 060 419 442 291 245 056;
  • 42) 0.208 463 918 141 301 060 419 442 291 245 056 × 2 = 0 + 0.416 927 836 282 602 120 838 884 582 490 112;
  • 43) 0.416 927 836 282 602 120 838 884 582 490 112 × 2 = 0 + 0.833 855 672 565 204 241 677 769 164 980 224;
  • 44) 0.833 855 672 565 204 241 677 769 164 980 224 × 2 = 1 + 0.667 711 345 130 408 483 355 538 329 960 448;
  • 45) 0.667 711 345 130 408 483 355 538 329 960 448 × 2 = 1 + 0.335 422 690 260 816 966 711 076 659 920 896;
  • 46) 0.335 422 690 260 816 966 711 076 659 920 896 × 2 = 0 + 0.670 845 380 521 633 933 422 153 319 841 792;
  • 47) 0.670 845 380 521 633 933 422 153 319 841 792 × 2 = 1 + 0.341 690 761 043 267 866 844 306 639 683 584;
  • 48) 0.341 690 761 043 267 866 844 306 639 683 584 × 2 = 0 + 0.683 381 522 086 535 733 688 613 279 367 168;
  • 49) 0.683 381 522 086 535 733 688 613 279 367 168 × 2 = 1 + 0.366 763 044 173 071 467 377 226 558 734 336;
  • 50) 0.366 763 044 173 071 467 377 226 558 734 336 × 2 = 0 + 0.733 526 088 346 142 934 754 453 117 468 672;
  • 51) 0.733 526 088 346 142 934 754 453 117 468 672 × 2 = 1 + 0.467 052 176 692 285 869 508 906 234 937 344;
  • 52) 0.467 052 176 692 285 869 508 906 234 937 344 × 2 = 0 + 0.934 104 353 384 571 739 017 812 469 874 688;
  • 53) 0.934 104 353 384 571 739 017 812 469 874 688 × 2 = 1 + 0.868 208 706 769 143 478 035 624 939 749 376;
  • 54) 0.868 208 706 769 143 478 035 624 939 749 376 × 2 = 1 + 0.736 417 413 538 286 956 071 249 879 498 752;
  • 55) 0.736 417 413 538 286 956 071 249 879 498 752 × 2 = 1 + 0.472 834 827 076 573 912 142 499 758 997 504;
  • 56) 0.472 834 827 076 573 912 142 499 758 997 504 × 2 = 0 + 0.945 669 654 153 147 824 284 999 517 995 008;
  • 57) 0.945 669 654 153 147 824 284 999 517 995 008 × 2 = 1 + 0.891 339 308 306 295 648 569 999 035 990 016;
  • 58) 0.891 339 308 306 295 648 569 999 035 990 016 × 2 = 1 + 0.782 678 616 612 591 297 139 998 071 980 032;
  • 59) 0.782 678 616 612 591 297 139 998 071 980 032 × 2 = 1 + 0.565 357 233 225 182 594 279 996 143 960 064;
  • 60) 0.565 357 233 225 182 594 279 996 143 960 064 × 2 = 1 + 0.130 714 466 450 365 188 559 992 287 920 128;
  • 61) 0.130 714 466 450 365 188 559 992 287 920 128 × 2 = 0 + 0.261 428 932 900 730 377 119 984 575 840 256;
  • 62) 0.261 428 932 900 730 377 119 984 575 840 256 × 2 = 0 + 0.522 857 865 801 460 754 239 969 151 680 512;
  • 63) 0.522 857 865 801 460 754 239 969 151 680 512 × 2 = 1 + 0.045 715 731 602 921 508 479 938 303 361 024;
  • 64) 0.045 715 731 602 921 508 479 938 303 361 024 × 2 = 0 + 0.091 431 463 205 843 016 959 876 606 722 048;
  • 65) 0.091 431 463 205 843 016 959 876 606 722 048 × 2 = 0 + 0.182 862 926 411 686 033 919 753 213 444 096;
  • 66) 0.182 862 926 411 686 033 919 753 213 444 096 × 2 = 0 + 0.365 725 852 823 372 067 839 506 426 888 192;
  • 67) 0.365 725 852 823 372 067 839 506 426 888 192 × 2 = 0 + 0.731 451 705 646 744 135 679 012 853 776 384;
  • 68) 0.731 451 705 646 744 135 679 012 853 776 384 × 2 = 1 + 0.462 903 411 293 488 271 358 025 707 552 768;

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:


0.000 020 830 729 321 671 205 134 999 154 515 5(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:

0.000 020 830 729 321 671 205 134 999 154 515 5(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:


0.000 020 830 729 321 671 205 134 999 154 515 5(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 0.000 020 830 729 321 671 205 134 999 154 515 5 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

Share

» Convert decimal 0.000 020 830 729 321 671 205 134 999 154 510 5 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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