0.000 020 830 729 321 671 205 134 999 154 531 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 531 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
0.000 020 830 729 321 671 205 134 999 154 531 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: 0.000 020 830 729 321 671 205 134 999 154 531 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) 0.000 020 830 729 321 671 205 134 999 154 531 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 063 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 063 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 126 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 126 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 252 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 252 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 505 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 505 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 945 011 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 945 011 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 890 022 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 890 022 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 780 044 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 780 044 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 560 089 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 560 089 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 120 179 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 120 179 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 240 358 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 240 358 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 480 716 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 480 716 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 961 433 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 961 433 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 922 867 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 922 867 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 845 734 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 845 734 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 691 468 8;
  • 16) 0.682 581 338 412 522 049 863 652 295 691 468 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 591 382 937 6;
  • 17) 0.365 162 676 825 044 099 727 304 591 382 937 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 182 765 875 2;
  • 18) 0.730 325 353 650 088 199 454 609 182 765 875 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 365 531 750 4;
  • 19) 0.460 650 707 300 176 398 909 218 365 531 750 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 731 063 500 8;
  • 20) 0.921 301 414 600 352 797 818 436 731 063 500 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 462 127 001 6;
  • 21) 0.842 602 829 200 705 595 636 873 462 127 001 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 924 254 003 2;
  • 22) 0.685 205 658 401 411 191 273 746 924 254 003 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 848 508 006 4;
  • 23) 0.370 411 316 802 822 382 547 493 848 508 006 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 697 016 012 8;
  • 24) 0.740 822 633 605 644 765 094 987 697 016 012 8 × 2 = 1 + 0.481 645 267 211 289 530 189 975 394 032 025 6;
  • 25) 0.481 645 267 211 289 530 189 975 394 032 025 6 × 2 = 0 + 0.963 290 534 422 579 060 379 950 788 064 051 2;
  • 26) 0.963 290 534 422 579 060 379 950 788 064 051 2 × 2 = 1 + 0.926 581 068 845 158 120 759 901 576 128 102 4;
  • 27) 0.926 581 068 845 158 120 759 901 576 128 102 4 × 2 = 1 + 0.853 162 137 690 316 241 519 803 152 256 204 8;
  • 28) 0.853 162 137 690 316 241 519 803 152 256 204 8 × 2 = 1 + 0.706 324 275 380 632 483 039 606 304 512 409 6;
  • 29) 0.706 324 275 380 632 483 039 606 304 512 409 6 × 2 = 1 + 0.412 648 550 761 264 966 079 212 609 024 819 2;
  • 30) 0.412 648 550 761 264 966 079 212 609 024 819 2 × 2 = 0 + 0.825 297 101 522 529 932 158 425 218 049 638 4;
  • 31) 0.825 297 101 522 529 932 158 425 218 049 638 4 × 2 = 1 + 0.650 594 203 045 059 864 316 850 436 099 276 8;
  • 32) 0.650 594 203 045 059 864 316 850 436 099 276 8 × 2 = 1 + 0.301 188 406 090 119 728 633 700 872 198 553 6;
  • 33) 0.301 188 406 090 119 728 633 700 872 198 553 6 × 2 = 0 + 0.602 376 812 180 239 457 267 401 744 397 107 2;
  • 34) 0.602 376 812 180 239 457 267 401 744 397 107 2 × 2 = 1 + 0.204 753 624 360 478 914 534 803 488 794 214 4;
  • 35) 0.204 753 624 360 478 914 534 803 488 794 214 4 × 2 = 0 + 0.409 507 248 720 957 829 069 606 977 588 428 8;
  • 36) 0.409 507 248 720 957 829 069 606 977 588 428 8 × 2 = 0 + 0.819 014 497 441 915 658 139 213 955 176 857 6;
  • 37) 0.819 014 497 441 915 658 139 213 955 176 857 6 × 2 = 1 + 0.638 028 994 883 831 316 278 427 910 353 715 2;
  • 38) 0.638 028 994 883 831 316 278 427 910 353 715 2 × 2 = 1 + 0.276 057 989 767 662 632 556 855 820 707 430 4;
  • 39) 0.276 057 989 767 662 632 556 855 820 707 430 4 × 2 = 0 + 0.552 115 979 535 325 265 113 711 641 414 860 8;
  • 40) 0.552 115 979 535 325 265 113 711 641 414 860 8 × 2 = 1 + 0.104 231 959 070 650 530 227 423 282 829 721 6;
  • 41) 0.104 231 959 070 650 530 227 423 282 829 721 6 × 2 = 0 + 0.208 463 918 141 301 060 454 846 565 659 443 2;
  • 42) 0.208 463 918 141 301 060 454 846 565 659 443 2 × 2 = 0 + 0.416 927 836 282 602 120 909 693 131 318 886 4;
  • 43) 0.416 927 836 282 602 120 909 693 131 318 886 4 × 2 = 0 + 0.833 855 672 565 204 241 819 386 262 637 772 8;
  • 44) 0.833 855 672 565 204 241 819 386 262 637 772 8 × 2 = 1 + 0.667 711 345 130 408 483 638 772 525 275 545 6;
  • 45) 0.667 711 345 130 408 483 638 772 525 275 545 6 × 2 = 1 + 0.335 422 690 260 816 967 277 545 050 551 091 2;
  • 46) 0.335 422 690 260 816 967 277 545 050 551 091 2 × 2 = 0 + 0.670 845 380 521 633 934 555 090 101 102 182 4;
  • 47) 0.670 845 380 521 633 934 555 090 101 102 182 4 × 2 = 1 + 0.341 690 761 043 267 869 110 180 202 204 364 8;
  • 48) 0.341 690 761 043 267 869 110 180 202 204 364 8 × 2 = 0 + 0.683 381 522 086 535 738 220 360 404 408 729 6;
  • 49) 0.683 381 522 086 535 738 220 360 404 408 729 6 × 2 = 1 + 0.366 763 044 173 071 476 440 720 808 817 459 2;
  • 50) 0.366 763 044 173 071 476 440 720 808 817 459 2 × 2 = 0 + 0.733 526 088 346 142 952 881 441 617 634 918 4;
  • 51) 0.733 526 088 346 142 952 881 441 617 634 918 4 × 2 = 1 + 0.467 052 176 692 285 905 762 883 235 269 836 8;
  • 52) 0.467 052 176 692 285 905 762 883 235 269 836 8 × 2 = 0 + 0.934 104 353 384 571 811 525 766 470 539 673 6;
  • 53) 0.934 104 353 384 571 811 525 766 470 539 673 6 × 2 = 1 + 0.868 208 706 769 143 623 051 532 941 079 347 2;
  • 54) 0.868 208 706 769 143 623 051 532 941 079 347 2 × 2 = 1 + 0.736 417 413 538 287 246 103 065 882 158 694 4;
  • 55) 0.736 417 413 538 287 246 103 065 882 158 694 4 × 2 = 1 + 0.472 834 827 076 574 492 206 131 764 317 388 8;
  • 56) 0.472 834 827 076 574 492 206 131 764 317 388 8 × 2 = 0 + 0.945 669 654 153 148 984 412 263 528 634 777 6;
  • 57) 0.945 669 654 153 148 984 412 263 528 634 777 6 × 2 = 1 + 0.891 339 308 306 297 968 824 527 057 269 555 2;
  • 58) 0.891 339 308 306 297 968 824 527 057 269 555 2 × 2 = 1 + 0.782 678 616 612 595 937 649 054 114 539 110 4;
  • 59) 0.782 678 616 612 595 937 649 054 114 539 110 4 × 2 = 1 + 0.565 357 233 225 191 875 298 108 229 078 220 8;
  • 60) 0.565 357 233 225 191 875 298 108 229 078 220 8 × 2 = 1 + 0.130 714 466 450 383 750 596 216 458 156 441 6;
  • 61) 0.130 714 466 450 383 750 596 216 458 156 441 6 × 2 = 0 + 0.261 428 932 900 767 501 192 432 916 312 883 2;
  • 62) 0.261 428 932 900 767 501 192 432 916 312 883 2 × 2 = 0 + 0.522 857 865 801 535 002 384 865 832 625 766 4;
  • 63) 0.522 857 865 801 535 002 384 865 832 625 766 4 × 2 = 1 + 0.045 715 731 603 070 004 769 731 665 251 532 8;
  • 64) 0.045 715 731 603 070 004 769 731 665 251 532 8 × 2 = 0 + 0.091 431 463 206 140 009 539 463 330 503 065 6;
  • 65) 0.091 431 463 206 140 009 539 463 330 503 065 6 × 2 = 0 + 0.182 862 926 412 280 019 078 926 661 006 131 2;
  • 66) 0.182 862 926 412 280 019 078 926 661 006 131 2 × 2 = 0 + 0.365 725 852 824 560 038 157 853 322 012 262 4;
  • 67) 0.365 725 852 824 560 038 157 853 322 012 262 4 × 2 = 0 + 0.731 451 705 649 120 076 315 706 644 024 524 8;
  • 68) 0.731 451 705 649 120 076 315 706 644 024 524 8 × 2 = 1 + 0.462 903 411 298 240 152 631 413 288 049 049 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:


0.000 020 830 729 321 671 205 134 999 154 531 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:

0.000 020 830 729 321 671 205 134 999 154 531 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:


0.000 020 830 729 321 671 205 134 999 154 531 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 0.000 020 830 729 321 671 205 134 999 154 531 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