0.000 020 830 729 321 671 205 134 999 154 524 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 524 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
0.000 020 830 729 321 671 205 134 999 154 524 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: 0.000 020 830 729 321 671 205 134 999 154 524 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) 0.000 020 830 729 321 671 205 134 999 154 524 3 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 048 6;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 048 6 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 097 2;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 097 2 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 194 4;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 194 4 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 388 8;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 388 8 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 777 6;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 777 6 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 889 555 2;
  • 7) 0.001 333 166 676 586 957 128 639 945 889 555 2 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 779 110 4;
  • 8) 0.002 666 333 353 173 914 257 279 891 779 110 4 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 558 220 8;
  • 9) 0.005 332 666 706 347 828 514 559 783 558 220 8 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 116 441 6;
  • 10) 0.010 665 333 412 695 657 029 119 567 116 441 6 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 232 883 2;
  • 11) 0.021 330 666 825 391 314 058 239 134 232 883 2 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 465 766 4;
  • 12) 0.042 661 333 650 782 628 116 478 268 465 766 4 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 931 532 8;
  • 13) 0.085 322 667 301 565 256 232 956 536 931 532 8 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 863 065 6;
  • 14) 0.170 645 334 603 130 512 465 913 073 863 065 6 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 726 131 2;
  • 15) 0.341 290 669 206 261 024 931 826 147 726 131 2 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 452 262 4;
  • 16) 0.682 581 338 412 522 049 863 652 295 452 262 4 × 2 = 1 + 0.365 162 676 825 044 099 727 304 590 904 524 8;
  • 17) 0.365 162 676 825 044 099 727 304 590 904 524 8 × 2 = 0 + 0.730 325 353 650 088 199 454 609 181 809 049 6;
  • 18) 0.730 325 353 650 088 199 454 609 181 809 049 6 × 2 = 1 + 0.460 650 707 300 176 398 909 218 363 618 099 2;
  • 19) 0.460 650 707 300 176 398 909 218 363 618 099 2 × 2 = 0 + 0.921 301 414 600 352 797 818 436 727 236 198 4;
  • 20) 0.921 301 414 600 352 797 818 436 727 236 198 4 × 2 = 1 + 0.842 602 829 200 705 595 636 873 454 472 396 8;
  • 21) 0.842 602 829 200 705 595 636 873 454 472 396 8 × 2 = 1 + 0.685 205 658 401 411 191 273 746 908 944 793 6;
  • 22) 0.685 205 658 401 411 191 273 746 908 944 793 6 × 2 = 1 + 0.370 411 316 802 822 382 547 493 817 889 587 2;
  • 23) 0.370 411 316 802 822 382 547 493 817 889 587 2 × 2 = 0 + 0.740 822 633 605 644 765 094 987 635 779 174 4;
  • 24) 0.740 822 633 605 644 765 094 987 635 779 174 4 × 2 = 1 + 0.481 645 267 211 289 530 189 975 271 558 348 8;
  • 25) 0.481 645 267 211 289 530 189 975 271 558 348 8 × 2 = 0 + 0.963 290 534 422 579 060 379 950 543 116 697 6;
  • 26) 0.963 290 534 422 579 060 379 950 543 116 697 6 × 2 = 1 + 0.926 581 068 845 158 120 759 901 086 233 395 2;
  • 27) 0.926 581 068 845 158 120 759 901 086 233 395 2 × 2 = 1 + 0.853 162 137 690 316 241 519 802 172 466 790 4;
  • 28) 0.853 162 137 690 316 241 519 802 172 466 790 4 × 2 = 1 + 0.706 324 275 380 632 483 039 604 344 933 580 8;
  • 29) 0.706 324 275 380 632 483 039 604 344 933 580 8 × 2 = 1 + 0.412 648 550 761 264 966 079 208 689 867 161 6;
  • 30) 0.412 648 550 761 264 966 079 208 689 867 161 6 × 2 = 0 + 0.825 297 101 522 529 932 158 417 379 734 323 2;
  • 31) 0.825 297 101 522 529 932 158 417 379 734 323 2 × 2 = 1 + 0.650 594 203 045 059 864 316 834 759 468 646 4;
  • 32) 0.650 594 203 045 059 864 316 834 759 468 646 4 × 2 = 1 + 0.301 188 406 090 119 728 633 669 518 937 292 8;
  • 33) 0.301 188 406 090 119 728 633 669 518 937 292 8 × 2 = 0 + 0.602 376 812 180 239 457 267 339 037 874 585 6;
  • 34) 0.602 376 812 180 239 457 267 339 037 874 585 6 × 2 = 1 + 0.204 753 624 360 478 914 534 678 075 749 171 2;
  • 35) 0.204 753 624 360 478 914 534 678 075 749 171 2 × 2 = 0 + 0.409 507 248 720 957 829 069 356 151 498 342 4;
  • 36) 0.409 507 248 720 957 829 069 356 151 498 342 4 × 2 = 0 + 0.819 014 497 441 915 658 138 712 302 996 684 8;
  • 37) 0.819 014 497 441 915 658 138 712 302 996 684 8 × 2 = 1 + 0.638 028 994 883 831 316 277 424 605 993 369 6;
  • 38) 0.638 028 994 883 831 316 277 424 605 993 369 6 × 2 = 1 + 0.276 057 989 767 662 632 554 849 211 986 739 2;
  • 39) 0.276 057 989 767 662 632 554 849 211 986 739 2 × 2 = 0 + 0.552 115 979 535 325 265 109 698 423 973 478 4;
  • 40) 0.552 115 979 535 325 265 109 698 423 973 478 4 × 2 = 1 + 0.104 231 959 070 650 530 219 396 847 946 956 8;
  • 41) 0.104 231 959 070 650 530 219 396 847 946 956 8 × 2 = 0 + 0.208 463 918 141 301 060 438 793 695 893 913 6;
  • 42) 0.208 463 918 141 301 060 438 793 695 893 913 6 × 2 = 0 + 0.416 927 836 282 602 120 877 587 391 787 827 2;
  • 43) 0.416 927 836 282 602 120 877 587 391 787 827 2 × 2 = 0 + 0.833 855 672 565 204 241 755 174 783 575 654 4;
  • 44) 0.833 855 672 565 204 241 755 174 783 575 654 4 × 2 = 1 + 0.667 711 345 130 408 483 510 349 567 151 308 8;
  • 45) 0.667 711 345 130 408 483 510 349 567 151 308 8 × 2 = 1 + 0.335 422 690 260 816 967 020 699 134 302 617 6;
  • 46) 0.335 422 690 260 816 967 020 699 134 302 617 6 × 2 = 0 + 0.670 845 380 521 633 934 041 398 268 605 235 2;
  • 47) 0.670 845 380 521 633 934 041 398 268 605 235 2 × 2 = 1 + 0.341 690 761 043 267 868 082 796 537 210 470 4;
  • 48) 0.341 690 761 043 267 868 082 796 537 210 470 4 × 2 = 0 + 0.683 381 522 086 535 736 165 593 074 420 940 8;
  • 49) 0.683 381 522 086 535 736 165 593 074 420 940 8 × 2 = 1 + 0.366 763 044 173 071 472 331 186 148 841 881 6;
  • 50) 0.366 763 044 173 071 472 331 186 148 841 881 6 × 2 = 0 + 0.733 526 088 346 142 944 662 372 297 683 763 2;
  • 51) 0.733 526 088 346 142 944 662 372 297 683 763 2 × 2 = 1 + 0.467 052 176 692 285 889 324 744 595 367 526 4;
  • 52) 0.467 052 176 692 285 889 324 744 595 367 526 4 × 2 = 0 + 0.934 104 353 384 571 778 649 489 190 735 052 8;
  • 53) 0.934 104 353 384 571 778 649 489 190 735 052 8 × 2 = 1 + 0.868 208 706 769 143 557 298 978 381 470 105 6;
  • 54) 0.868 208 706 769 143 557 298 978 381 470 105 6 × 2 = 1 + 0.736 417 413 538 287 114 597 956 762 940 211 2;
  • 55) 0.736 417 413 538 287 114 597 956 762 940 211 2 × 2 = 1 + 0.472 834 827 076 574 229 195 913 525 880 422 4;
  • 56) 0.472 834 827 076 574 229 195 913 525 880 422 4 × 2 = 0 + 0.945 669 654 153 148 458 391 827 051 760 844 8;
  • 57) 0.945 669 654 153 148 458 391 827 051 760 844 8 × 2 = 1 + 0.891 339 308 306 296 916 783 654 103 521 689 6;
  • 58) 0.891 339 308 306 296 916 783 654 103 521 689 6 × 2 = 1 + 0.782 678 616 612 593 833 567 308 207 043 379 2;
  • 59) 0.782 678 616 612 593 833 567 308 207 043 379 2 × 2 = 1 + 0.565 357 233 225 187 667 134 616 414 086 758 4;
  • 60) 0.565 357 233 225 187 667 134 616 414 086 758 4 × 2 = 1 + 0.130 714 466 450 375 334 269 232 828 173 516 8;
  • 61) 0.130 714 466 450 375 334 269 232 828 173 516 8 × 2 = 0 + 0.261 428 932 900 750 668 538 465 656 347 033 6;
  • 62) 0.261 428 932 900 750 668 538 465 656 347 033 6 × 2 = 0 + 0.522 857 865 801 501 337 076 931 312 694 067 2;
  • 63) 0.522 857 865 801 501 337 076 931 312 694 067 2 × 2 = 1 + 0.045 715 731 603 002 674 153 862 625 388 134 4;
  • 64) 0.045 715 731 603 002 674 153 862 625 388 134 4 × 2 = 0 + 0.091 431 463 206 005 348 307 725 250 776 268 8;
  • 65) 0.091 431 463 206 005 348 307 725 250 776 268 8 × 2 = 0 + 0.182 862 926 412 010 696 615 450 501 552 537 6;
  • 66) 0.182 862 926 412 010 696 615 450 501 552 537 6 × 2 = 0 + 0.365 725 852 824 021 393 230 901 003 105 075 2;
  • 67) 0.365 725 852 824 021 393 230 901 003 105 075 2 × 2 = 0 + 0.731 451 705 648 042 786 461 802 006 210 150 4;
  • 68) 0.731 451 705 648 042 786 461 802 006 210 150 4 × 2 = 1 + 0.462 903 411 296 085 572 923 604 012 420 300 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:


0.000 020 830 729 321 671 205 134 999 154 524 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:

0.000 020 830 729 321 671 205 134 999 154 524 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:


0.000 020 830 729 321 671 205 134 999 154 524 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 0.000 020 830 729 321 671 205 134 999 154 524 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