0.000 020 830 729 321 671 205 134 999 154 506 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 506 1(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
0.000 020 830 729 321 671 205 134 999 154 506 1(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: 0.000 020 830 729 321 671 205 134 999 154 506 1.

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) 0.000 020 830 729 321 671 205 134 999 154 506 1 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 012 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 012 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 024 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 024 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 048 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 048 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 097 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 097 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 195 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 195 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 390 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 390 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 776 780 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 776 780 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 553 561 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 553 561 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 107 123 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 107 123 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 214 246 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 214 246 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 428 492 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 428 492 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 856 985 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 856 985 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 713 971 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 713 971 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 427 942 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 427 942 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 855 884 8;
  • 16) 0.682 581 338 412 522 049 863 652 294 855 884 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 711 769 6;
  • 17) 0.365 162 676 825 044 099 727 304 589 711 769 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 423 539 2;
  • 18) 0.730 325 353 650 088 199 454 609 179 423 539 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 358 847 078 4;
  • 19) 0.460 650 707 300 176 398 909 218 358 847 078 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 717 694 156 8;
  • 20) 0.921 301 414 600 352 797 818 436 717 694 156 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 435 388 313 6;
  • 21) 0.842 602 829 200 705 595 636 873 435 388 313 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 870 776 627 2;
  • 22) 0.685 205 658 401 411 191 273 746 870 776 627 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 741 553 254 4;
  • 23) 0.370 411 316 802 822 382 547 493 741 553 254 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 483 106 508 8;
  • 24) 0.740 822 633 605 644 765 094 987 483 106 508 8 × 2 = 1 + 0.481 645 267 211 289 530 189 974 966 213 017 6;
  • 25) 0.481 645 267 211 289 530 189 974 966 213 017 6 × 2 = 0 + 0.963 290 534 422 579 060 379 949 932 426 035 2;
  • 26) 0.963 290 534 422 579 060 379 949 932 426 035 2 × 2 = 1 + 0.926 581 068 845 158 120 759 899 864 852 070 4;
  • 27) 0.926 581 068 845 158 120 759 899 864 852 070 4 × 2 = 1 + 0.853 162 137 690 316 241 519 799 729 704 140 8;
  • 28) 0.853 162 137 690 316 241 519 799 729 704 140 8 × 2 = 1 + 0.706 324 275 380 632 483 039 599 459 408 281 6;
  • 29) 0.706 324 275 380 632 483 039 599 459 408 281 6 × 2 = 1 + 0.412 648 550 761 264 966 079 198 918 816 563 2;
  • 30) 0.412 648 550 761 264 966 079 198 918 816 563 2 × 2 = 0 + 0.825 297 101 522 529 932 158 397 837 633 126 4;
  • 31) 0.825 297 101 522 529 932 158 397 837 633 126 4 × 2 = 1 + 0.650 594 203 045 059 864 316 795 675 266 252 8;
  • 32) 0.650 594 203 045 059 864 316 795 675 266 252 8 × 2 = 1 + 0.301 188 406 090 119 728 633 591 350 532 505 6;
  • 33) 0.301 188 406 090 119 728 633 591 350 532 505 6 × 2 = 0 + 0.602 376 812 180 239 457 267 182 701 065 011 2;
  • 34) 0.602 376 812 180 239 457 267 182 701 065 011 2 × 2 = 1 + 0.204 753 624 360 478 914 534 365 402 130 022 4;
  • 35) 0.204 753 624 360 478 914 534 365 402 130 022 4 × 2 = 0 + 0.409 507 248 720 957 829 068 730 804 260 044 8;
  • 36) 0.409 507 248 720 957 829 068 730 804 260 044 8 × 2 = 0 + 0.819 014 497 441 915 658 137 461 608 520 089 6;
  • 37) 0.819 014 497 441 915 658 137 461 608 520 089 6 × 2 = 1 + 0.638 028 994 883 831 316 274 923 217 040 179 2;
  • 38) 0.638 028 994 883 831 316 274 923 217 040 179 2 × 2 = 1 + 0.276 057 989 767 662 632 549 846 434 080 358 4;
  • 39) 0.276 057 989 767 662 632 549 846 434 080 358 4 × 2 = 0 + 0.552 115 979 535 325 265 099 692 868 160 716 8;
  • 40) 0.552 115 979 535 325 265 099 692 868 160 716 8 × 2 = 1 + 0.104 231 959 070 650 530 199 385 736 321 433 6;
  • 41) 0.104 231 959 070 650 530 199 385 736 321 433 6 × 2 = 0 + 0.208 463 918 141 301 060 398 771 472 642 867 2;
  • 42) 0.208 463 918 141 301 060 398 771 472 642 867 2 × 2 = 0 + 0.416 927 836 282 602 120 797 542 945 285 734 4;
  • 43) 0.416 927 836 282 602 120 797 542 945 285 734 4 × 2 = 0 + 0.833 855 672 565 204 241 595 085 890 571 468 8;
  • 44) 0.833 855 672 565 204 241 595 085 890 571 468 8 × 2 = 1 + 0.667 711 345 130 408 483 190 171 781 142 937 6;
  • 45) 0.667 711 345 130 408 483 190 171 781 142 937 6 × 2 = 1 + 0.335 422 690 260 816 966 380 343 562 285 875 2;
  • 46) 0.335 422 690 260 816 966 380 343 562 285 875 2 × 2 = 0 + 0.670 845 380 521 633 932 760 687 124 571 750 4;
  • 47) 0.670 845 380 521 633 932 760 687 124 571 750 4 × 2 = 1 + 0.341 690 761 043 267 865 521 374 249 143 500 8;
  • 48) 0.341 690 761 043 267 865 521 374 249 143 500 8 × 2 = 0 + 0.683 381 522 086 535 731 042 748 498 287 001 6;
  • 49) 0.683 381 522 086 535 731 042 748 498 287 001 6 × 2 = 1 + 0.366 763 044 173 071 462 085 496 996 574 003 2;
  • 50) 0.366 763 044 173 071 462 085 496 996 574 003 2 × 2 = 0 + 0.733 526 088 346 142 924 170 993 993 148 006 4;
  • 51) 0.733 526 088 346 142 924 170 993 993 148 006 4 × 2 = 1 + 0.467 052 176 692 285 848 341 987 986 296 012 8;
  • 52) 0.467 052 176 692 285 848 341 987 986 296 012 8 × 2 = 0 + 0.934 104 353 384 571 696 683 975 972 592 025 6;
  • 53) 0.934 104 353 384 571 696 683 975 972 592 025 6 × 2 = 1 + 0.868 208 706 769 143 393 367 951 945 184 051 2;
  • 54) 0.868 208 706 769 143 393 367 951 945 184 051 2 × 2 = 1 + 0.736 417 413 538 286 786 735 903 890 368 102 4;
  • 55) 0.736 417 413 538 286 786 735 903 890 368 102 4 × 2 = 1 + 0.472 834 827 076 573 573 471 807 780 736 204 8;
  • 56) 0.472 834 827 076 573 573 471 807 780 736 204 8 × 2 = 0 + 0.945 669 654 153 147 146 943 615 561 472 409 6;
  • 57) 0.945 669 654 153 147 146 943 615 561 472 409 6 × 2 = 1 + 0.891 339 308 306 294 293 887 231 122 944 819 2;
  • 58) 0.891 339 308 306 294 293 887 231 122 944 819 2 × 2 = 1 + 0.782 678 616 612 588 587 774 462 245 889 638 4;
  • 59) 0.782 678 616 612 588 587 774 462 245 889 638 4 × 2 = 1 + 0.565 357 233 225 177 175 548 924 491 779 276 8;
  • 60) 0.565 357 233 225 177 175 548 924 491 779 276 8 × 2 = 1 + 0.130 714 466 450 354 351 097 848 983 558 553 6;
  • 61) 0.130 714 466 450 354 351 097 848 983 558 553 6 × 2 = 0 + 0.261 428 932 900 708 702 195 697 967 117 107 2;
  • 62) 0.261 428 932 900 708 702 195 697 967 117 107 2 × 2 = 0 + 0.522 857 865 801 417 404 391 395 934 234 214 4;
  • 63) 0.522 857 865 801 417 404 391 395 934 234 214 4 × 2 = 1 + 0.045 715 731 602 834 808 782 791 868 468 428 8;
  • 64) 0.045 715 731 602 834 808 782 791 868 468 428 8 × 2 = 0 + 0.091 431 463 205 669 617 565 583 736 936 857 6;
  • 65) 0.091 431 463 205 669 617 565 583 736 936 857 6 × 2 = 0 + 0.182 862 926 411 339 235 131 167 473 873 715 2;
  • 66) 0.182 862 926 411 339 235 131 167 473 873 715 2 × 2 = 0 + 0.365 725 852 822 678 470 262 334 947 747 430 4;
  • 67) 0.365 725 852 822 678 470 262 334 947 747 430 4 × 2 = 0 + 0.731 451 705 645 356 940 524 669 895 494 860 8;
  • 68) 0.731 451 705 645 356 940 524 669 895 494 860 8 × 2 = 1 + 0.462 903 411 290 713 881 049 339 790 989 721 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:


0.000 020 830 729 321 671 205 134 999 154 506 1(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:

0.000 020 830 729 321 671 205 134 999 154 506 1(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:


0.000 020 830 729 321 671 205 134 999 154 506 1(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 0.000 020 830 729 321 671 205 134 999 154 506 1 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