0.000 020 830 729 321 671 205 134 999 154 522 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 522 3(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
0.000 020 830 729 321 671 205 134 999 154 522 3(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: 0.000 020 830 729 321 671 205 134 999 154 522 3.

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) 0.000 020 830 729 321 671 205 134 999 154 522 3 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 044 6;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 044 6 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 089 2;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 089 2 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 178 4;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 178 4 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 356 8;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 356 8 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 713 6;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 713 6 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 889 427 2;
  • 7) 0.001 333 166 676 586 957 128 639 945 889 427 2 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 778 854 4;
  • 8) 0.002 666 333 353 173 914 257 279 891 778 854 4 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 557 708 8;
  • 9) 0.005 332 666 706 347 828 514 559 783 557 708 8 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 115 417 6;
  • 10) 0.010 665 333 412 695 657 029 119 567 115 417 6 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 230 835 2;
  • 11) 0.021 330 666 825 391 314 058 239 134 230 835 2 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 461 670 4;
  • 12) 0.042 661 333 650 782 628 116 478 268 461 670 4 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 923 340 8;
  • 13) 0.085 322 667 301 565 256 232 956 536 923 340 8 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 846 681 6;
  • 14) 0.170 645 334 603 130 512 465 913 073 846 681 6 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 693 363 2;
  • 15) 0.341 290 669 206 261 024 931 826 147 693 363 2 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 386 726 4;
  • 16) 0.682 581 338 412 522 049 863 652 295 386 726 4 × 2 = 1 + 0.365 162 676 825 044 099 727 304 590 773 452 8;
  • 17) 0.365 162 676 825 044 099 727 304 590 773 452 8 × 2 = 0 + 0.730 325 353 650 088 199 454 609 181 546 905 6;
  • 18) 0.730 325 353 650 088 199 454 609 181 546 905 6 × 2 = 1 + 0.460 650 707 300 176 398 909 218 363 093 811 2;
  • 19) 0.460 650 707 300 176 398 909 218 363 093 811 2 × 2 = 0 + 0.921 301 414 600 352 797 818 436 726 187 622 4;
  • 20) 0.921 301 414 600 352 797 818 436 726 187 622 4 × 2 = 1 + 0.842 602 829 200 705 595 636 873 452 375 244 8;
  • 21) 0.842 602 829 200 705 595 636 873 452 375 244 8 × 2 = 1 + 0.685 205 658 401 411 191 273 746 904 750 489 6;
  • 22) 0.685 205 658 401 411 191 273 746 904 750 489 6 × 2 = 1 + 0.370 411 316 802 822 382 547 493 809 500 979 2;
  • 23) 0.370 411 316 802 822 382 547 493 809 500 979 2 × 2 = 0 + 0.740 822 633 605 644 765 094 987 619 001 958 4;
  • 24) 0.740 822 633 605 644 765 094 987 619 001 958 4 × 2 = 1 + 0.481 645 267 211 289 530 189 975 238 003 916 8;
  • 25) 0.481 645 267 211 289 530 189 975 238 003 916 8 × 2 = 0 + 0.963 290 534 422 579 060 379 950 476 007 833 6;
  • 26) 0.963 290 534 422 579 060 379 950 476 007 833 6 × 2 = 1 + 0.926 581 068 845 158 120 759 900 952 015 667 2;
  • 27) 0.926 581 068 845 158 120 759 900 952 015 667 2 × 2 = 1 + 0.853 162 137 690 316 241 519 801 904 031 334 4;
  • 28) 0.853 162 137 690 316 241 519 801 904 031 334 4 × 2 = 1 + 0.706 324 275 380 632 483 039 603 808 062 668 8;
  • 29) 0.706 324 275 380 632 483 039 603 808 062 668 8 × 2 = 1 + 0.412 648 550 761 264 966 079 207 616 125 337 6;
  • 30) 0.412 648 550 761 264 966 079 207 616 125 337 6 × 2 = 0 + 0.825 297 101 522 529 932 158 415 232 250 675 2;
  • 31) 0.825 297 101 522 529 932 158 415 232 250 675 2 × 2 = 1 + 0.650 594 203 045 059 864 316 830 464 501 350 4;
  • 32) 0.650 594 203 045 059 864 316 830 464 501 350 4 × 2 = 1 + 0.301 188 406 090 119 728 633 660 929 002 700 8;
  • 33) 0.301 188 406 090 119 728 633 660 929 002 700 8 × 2 = 0 + 0.602 376 812 180 239 457 267 321 858 005 401 6;
  • 34) 0.602 376 812 180 239 457 267 321 858 005 401 6 × 2 = 1 + 0.204 753 624 360 478 914 534 643 716 010 803 2;
  • 35) 0.204 753 624 360 478 914 534 643 716 010 803 2 × 2 = 0 + 0.409 507 248 720 957 829 069 287 432 021 606 4;
  • 36) 0.409 507 248 720 957 829 069 287 432 021 606 4 × 2 = 0 + 0.819 014 497 441 915 658 138 574 864 043 212 8;
  • 37) 0.819 014 497 441 915 658 138 574 864 043 212 8 × 2 = 1 + 0.638 028 994 883 831 316 277 149 728 086 425 6;
  • 38) 0.638 028 994 883 831 316 277 149 728 086 425 6 × 2 = 1 + 0.276 057 989 767 662 632 554 299 456 172 851 2;
  • 39) 0.276 057 989 767 662 632 554 299 456 172 851 2 × 2 = 0 + 0.552 115 979 535 325 265 108 598 912 345 702 4;
  • 40) 0.552 115 979 535 325 265 108 598 912 345 702 4 × 2 = 1 + 0.104 231 959 070 650 530 217 197 824 691 404 8;
  • 41) 0.104 231 959 070 650 530 217 197 824 691 404 8 × 2 = 0 + 0.208 463 918 141 301 060 434 395 649 382 809 6;
  • 42) 0.208 463 918 141 301 060 434 395 649 382 809 6 × 2 = 0 + 0.416 927 836 282 602 120 868 791 298 765 619 2;
  • 43) 0.416 927 836 282 602 120 868 791 298 765 619 2 × 2 = 0 + 0.833 855 672 565 204 241 737 582 597 531 238 4;
  • 44) 0.833 855 672 565 204 241 737 582 597 531 238 4 × 2 = 1 + 0.667 711 345 130 408 483 475 165 195 062 476 8;
  • 45) 0.667 711 345 130 408 483 475 165 195 062 476 8 × 2 = 1 + 0.335 422 690 260 816 966 950 330 390 124 953 6;
  • 46) 0.335 422 690 260 816 966 950 330 390 124 953 6 × 2 = 0 + 0.670 845 380 521 633 933 900 660 780 249 907 2;
  • 47) 0.670 845 380 521 633 933 900 660 780 249 907 2 × 2 = 1 + 0.341 690 761 043 267 867 801 321 560 499 814 4;
  • 48) 0.341 690 761 043 267 867 801 321 560 499 814 4 × 2 = 0 + 0.683 381 522 086 535 735 602 643 120 999 628 8;
  • 49) 0.683 381 522 086 535 735 602 643 120 999 628 8 × 2 = 1 + 0.366 763 044 173 071 471 205 286 241 999 257 6;
  • 50) 0.366 763 044 173 071 471 205 286 241 999 257 6 × 2 = 0 + 0.733 526 088 346 142 942 410 572 483 998 515 2;
  • 51) 0.733 526 088 346 142 942 410 572 483 998 515 2 × 2 = 1 + 0.467 052 176 692 285 884 821 144 967 997 030 4;
  • 52) 0.467 052 176 692 285 884 821 144 967 997 030 4 × 2 = 0 + 0.934 104 353 384 571 769 642 289 935 994 060 8;
  • 53) 0.934 104 353 384 571 769 642 289 935 994 060 8 × 2 = 1 + 0.868 208 706 769 143 539 284 579 871 988 121 6;
  • 54) 0.868 208 706 769 143 539 284 579 871 988 121 6 × 2 = 1 + 0.736 417 413 538 287 078 569 159 743 976 243 2;
  • 55) 0.736 417 413 538 287 078 569 159 743 976 243 2 × 2 = 1 + 0.472 834 827 076 574 157 138 319 487 952 486 4;
  • 56) 0.472 834 827 076 574 157 138 319 487 952 486 4 × 2 = 0 + 0.945 669 654 153 148 314 276 638 975 904 972 8;
  • 57) 0.945 669 654 153 148 314 276 638 975 904 972 8 × 2 = 1 + 0.891 339 308 306 296 628 553 277 951 809 945 6;
  • 58) 0.891 339 308 306 296 628 553 277 951 809 945 6 × 2 = 1 + 0.782 678 616 612 593 257 106 555 903 619 891 2;
  • 59) 0.782 678 616 612 593 257 106 555 903 619 891 2 × 2 = 1 + 0.565 357 233 225 186 514 213 111 807 239 782 4;
  • 60) 0.565 357 233 225 186 514 213 111 807 239 782 4 × 2 = 1 + 0.130 714 466 450 373 028 426 223 614 479 564 8;
  • 61) 0.130 714 466 450 373 028 426 223 614 479 564 8 × 2 = 0 + 0.261 428 932 900 746 056 852 447 228 959 129 6;
  • 62) 0.261 428 932 900 746 056 852 447 228 959 129 6 × 2 = 0 + 0.522 857 865 801 492 113 704 894 457 918 259 2;
  • 63) 0.522 857 865 801 492 113 704 894 457 918 259 2 × 2 = 1 + 0.045 715 731 602 984 227 409 788 915 836 518 4;
  • 64) 0.045 715 731 602 984 227 409 788 915 836 518 4 × 2 = 0 + 0.091 431 463 205 968 454 819 577 831 673 036 8;
  • 65) 0.091 431 463 205 968 454 819 577 831 673 036 8 × 2 = 0 + 0.182 862 926 411 936 909 639 155 663 346 073 6;
  • 66) 0.182 862 926 411 936 909 639 155 663 346 073 6 × 2 = 0 + 0.365 725 852 823 873 819 278 311 326 692 147 2;
  • 67) 0.365 725 852 823 873 819 278 311 326 692 147 2 × 2 = 0 + 0.731 451 705 647 747 638 556 622 653 384 294 4;
  • 68) 0.731 451 705 647 747 638 556 622 653 384 294 4 × 2 = 1 + 0.462 903 411 295 495 277 113 245 306 768 588 8;

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:


0.000 020 830 729 321 671 205 134 999 154 522 3(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:

0.000 020 830 729 321 671 205 134 999 154 522 3(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:


0.000 020 830 729 321 671 205 134 999 154 522 3(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 0.000 020 830 729 321 671 205 134 999 154 522 3 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