0.000 020 830 729 321 671 205 134 999 154 504 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 504 2(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
0.000 020 830 729 321 671 205 134 999 154 504 2(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: 0.000 020 830 729 321 671 205 134 999 154 504 2.

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) 0.000 020 830 729 321 671 205 134 999 154 504 2 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 008 4;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 008 4 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 016 8;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 016 8 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 033 6;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 033 6 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 067 2;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 067 2 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 134 4;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 134 4 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 268 8;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 268 8 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 776 537 6;
  • 8) 0.002 666 333 353 173 914 257 279 891 776 537 6 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 553 075 2;
  • 9) 0.005 332 666 706 347 828 514 559 783 553 075 2 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 106 150 4;
  • 10) 0.010 665 333 412 695 657 029 119 567 106 150 4 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 212 300 8;
  • 11) 0.021 330 666 825 391 314 058 239 134 212 300 8 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 424 601 6;
  • 12) 0.042 661 333 650 782 628 116 478 268 424 601 6 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 849 203 2;
  • 13) 0.085 322 667 301 565 256 232 956 536 849 203 2 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 698 406 4;
  • 14) 0.170 645 334 603 130 512 465 913 073 698 406 4 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 396 812 8;
  • 15) 0.341 290 669 206 261 024 931 826 147 396 812 8 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 793 625 6;
  • 16) 0.682 581 338 412 522 049 863 652 294 793 625 6 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 587 251 2;
  • 17) 0.365 162 676 825 044 099 727 304 589 587 251 2 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 174 502 4;
  • 18) 0.730 325 353 650 088 199 454 609 179 174 502 4 × 2 = 1 + 0.460 650 707 300 176 398 909 218 358 349 004 8;
  • 19) 0.460 650 707 300 176 398 909 218 358 349 004 8 × 2 = 0 + 0.921 301 414 600 352 797 818 436 716 698 009 6;
  • 20) 0.921 301 414 600 352 797 818 436 716 698 009 6 × 2 = 1 + 0.842 602 829 200 705 595 636 873 433 396 019 2;
  • 21) 0.842 602 829 200 705 595 636 873 433 396 019 2 × 2 = 1 + 0.685 205 658 401 411 191 273 746 866 792 038 4;
  • 22) 0.685 205 658 401 411 191 273 746 866 792 038 4 × 2 = 1 + 0.370 411 316 802 822 382 547 493 733 584 076 8;
  • 23) 0.370 411 316 802 822 382 547 493 733 584 076 8 × 2 = 0 + 0.740 822 633 605 644 765 094 987 467 168 153 6;
  • 24) 0.740 822 633 605 644 765 094 987 467 168 153 6 × 2 = 1 + 0.481 645 267 211 289 530 189 974 934 336 307 2;
  • 25) 0.481 645 267 211 289 530 189 974 934 336 307 2 × 2 = 0 + 0.963 290 534 422 579 060 379 949 868 672 614 4;
  • 26) 0.963 290 534 422 579 060 379 949 868 672 614 4 × 2 = 1 + 0.926 581 068 845 158 120 759 899 737 345 228 8;
  • 27) 0.926 581 068 845 158 120 759 899 737 345 228 8 × 2 = 1 + 0.853 162 137 690 316 241 519 799 474 690 457 6;
  • 28) 0.853 162 137 690 316 241 519 799 474 690 457 6 × 2 = 1 + 0.706 324 275 380 632 483 039 598 949 380 915 2;
  • 29) 0.706 324 275 380 632 483 039 598 949 380 915 2 × 2 = 1 + 0.412 648 550 761 264 966 079 197 898 761 830 4;
  • 30) 0.412 648 550 761 264 966 079 197 898 761 830 4 × 2 = 0 + 0.825 297 101 522 529 932 158 395 797 523 660 8;
  • 31) 0.825 297 101 522 529 932 158 395 797 523 660 8 × 2 = 1 + 0.650 594 203 045 059 864 316 791 595 047 321 6;
  • 32) 0.650 594 203 045 059 864 316 791 595 047 321 6 × 2 = 1 + 0.301 188 406 090 119 728 633 583 190 094 643 2;
  • 33) 0.301 188 406 090 119 728 633 583 190 094 643 2 × 2 = 0 + 0.602 376 812 180 239 457 267 166 380 189 286 4;
  • 34) 0.602 376 812 180 239 457 267 166 380 189 286 4 × 2 = 1 + 0.204 753 624 360 478 914 534 332 760 378 572 8;
  • 35) 0.204 753 624 360 478 914 534 332 760 378 572 8 × 2 = 0 + 0.409 507 248 720 957 829 068 665 520 757 145 6;
  • 36) 0.409 507 248 720 957 829 068 665 520 757 145 6 × 2 = 0 + 0.819 014 497 441 915 658 137 331 041 514 291 2;
  • 37) 0.819 014 497 441 915 658 137 331 041 514 291 2 × 2 = 1 + 0.638 028 994 883 831 316 274 662 083 028 582 4;
  • 38) 0.638 028 994 883 831 316 274 662 083 028 582 4 × 2 = 1 + 0.276 057 989 767 662 632 549 324 166 057 164 8;
  • 39) 0.276 057 989 767 662 632 549 324 166 057 164 8 × 2 = 0 + 0.552 115 979 535 325 265 098 648 332 114 329 6;
  • 40) 0.552 115 979 535 325 265 098 648 332 114 329 6 × 2 = 1 + 0.104 231 959 070 650 530 197 296 664 228 659 2;
  • 41) 0.104 231 959 070 650 530 197 296 664 228 659 2 × 2 = 0 + 0.208 463 918 141 301 060 394 593 328 457 318 4;
  • 42) 0.208 463 918 141 301 060 394 593 328 457 318 4 × 2 = 0 + 0.416 927 836 282 602 120 789 186 656 914 636 8;
  • 43) 0.416 927 836 282 602 120 789 186 656 914 636 8 × 2 = 0 + 0.833 855 672 565 204 241 578 373 313 829 273 6;
  • 44) 0.833 855 672 565 204 241 578 373 313 829 273 6 × 2 = 1 + 0.667 711 345 130 408 483 156 746 627 658 547 2;
  • 45) 0.667 711 345 130 408 483 156 746 627 658 547 2 × 2 = 1 + 0.335 422 690 260 816 966 313 493 255 317 094 4;
  • 46) 0.335 422 690 260 816 966 313 493 255 317 094 4 × 2 = 0 + 0.670 845 380 521 633 932 626 986 510 634 188 8;
  • 47) 0.670 845 380 521 633 932 626 986 510 634 188 8 × 2 = 1 + 0.341 690 761 043 267 865 253 973 021 268 377 6;
  • 48) 0.341 690 761 043 267 865 253 973 021 268 377 6 × 2 = 0 + 0.683 381 522 086 535 730 507 946 042 536 755 2;
  • 49) 0.683 381 522 086 535 730 507 946 042 536 755 2 × 2 = 1 + 0.366 763 044 173 071 461 015 892 085 073 510 4;
  • 50) 0.366 763 044 173 071 461 015 892 085 073 510 4 × 2 = 0 + 0.733 526 088 346 142 922 031 784 170 147 020 8;
  • 51) 0.733 526 088 346 142 922 031 784 170 147 020 8 × 2 = 1 + 0.467 052 176 692 285 844 063 568 340 294 041 6;
  • 52) 0.467 052 176 692 285 844 063 568 340 294 041 6 × 2 = 0 + 0.934 104 353 384 571 688 127 136 680 588 083 2;
  • 53) 0.934 104 353 384 571 688 127 136 680 588 083 2 × 2 = 1 + 0.868 208 706 769 143 376 254 273 361 176 166 4;
  • 54) 0.868 208 706 769 143 376 254 273 361 176 166 4 × 2 = 1 + 0.736 417 413 538 286 752 508 546 722 352 332 8;
  • 55) 0.736 417 413 538 286 752 508 546 722 352 332 8 × 2 = 1 + 0.472 834 827 076 573 505 017 093 444 704 665 6;
  • 56) 0.472 834 827 076 573 505 017 093 444 704 665 6 × 2 = 0 + 0.945 669 654 153 147 010 034 186 889 409 331 2;
  • 57) 0.945 669 654 153 147 010 034 186 889 409 331 2 × 2 = 1 + 0.891 339 308 306 294 020 068 373 778 818 662 4;
  • 58) 0.891 339 308 306 294 020 068 373 778 818 662 4 × 2 = 1 + 0.782 678 616 612 588 040 136 747 557 637 324 8;
  • 59) 0.782 678 616 612 588 040 136 747 557 637 324 8 × 2 = 1 + 0.565 357 233 225 176 080 273 495 115 274 649 6;
  • 60) 0.565 357 233 225 176 080 273 495 115 274 649 6 × 2 = 1 + 0.130 714 466 450 352 160 546 990 230 549 299 2;
  • 61) 0.130 714 466 450 352 160 546 990 230 549 299 2 × 2 = 0 + 0.261 428 932 900 704 321 093 980 461 098 598 4;
  • 62) 0.261 428 932 900 704 321 093 980 461 098 598 4 × 2 = 0 + 0.522 857 865 801 408 642 187 960 922 197 196 8;
  • 63) 0.522 857 865 801 408 642 187 960 922 197 196 8 × 2 = 1 + 0.045 715 731 602 817 284 375 921 844 394 393 6;
  • 64) 0.045 715 731 602 817 284 375 921 844 394 393 6 × 2 = 0 + 0.091 431 463 205 634 568 751 843 688 788 787 2;
  • 65) 0.091 431 463 205 634 568 751 843 688 788 787 2 × 2 = 0 + 0.182 862 926 411 269 137 503 687 377 577 574 4;
  • 66) 0.182 862 926 411 269 137 503 687 377 577 574 4 × 2 = 0 + 0.365 725 852 822 538 275 007 374 755 155 148 8;
  • 67) 0.365 725 852 822 538 275 007 374 755 155 148 8 × 2 = 0 + 0.731 451 705 645 076 550 014 749 510 310 297 6;
  • 68) 0.731 451 705 645 076 550 014 749 510 310 297 6 × 2 = 1 + 0.462 903 411 290 153 100 029 499 020 620 595 2;

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:


0.000 020 830 729 321 671 205 134 999 154 504 2(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:

0.000 020 830 729 321 671 205 134 999 154 504 2(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:


0.000 020 830 729 321 671 205 134 999 154 504 2(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 0.000 020 830 729 321 671 205 134 999 154 504 2 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