0.000 020 830 729 321 671 205 134 999 154 501 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 501 8(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
0.000 020 830 729 321 671 205 134 999 154 501 8(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: 0.000 020 830 729 321 671 205 134 999 154 501 8.

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) 0.000 020 830 729 321 671 205 134 999 154 501 8 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 003 6;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 003 6 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 007 2;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 007 2 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 014 4;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 014 4 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 028 8;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 028 8 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 057 6;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 057 6 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 115 2;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 115 2 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 776 230 4;
  • 8) 0.002 666 333 353 173 914 257 279 891 776 230 4 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 552 460 8;
  • 9) 0.005 332 666 706 347 828 514 559 783 552 460 8 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 104 921 6;
  • 10) 0.010 665 333 412 695 657 029 119 567 104 921 6 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 209 843 2;
  • 11) 0.021 330 666 825 391 314 058 239 134 209 843 2 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 419 686 4;
  • 12) 0.042 661 333 650 782 628 116 478 268 419 686 4 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 839 372 8;
  • 13) 0.085 322 667 301 565 256 232 956 536 839 372 8 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 678 745 6;
  • 14) 0.170 645 334 603 130 512 465 913 073 678 745 6 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 357 491 2;
  • 15) 0.341 290 669 206 261 024 931 826 147 357 491 2 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 714 982 4;
  • 16) 0.682 581 338 412 522 049 863 652 294 714 982 4 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 429 964 8;
  • 17) 0.365 162 676 825 044 099 727 304 589 429 964 8 × 2 = 0 + 0.730 325 353 650 088 199 454 609 178 859 929 6;
  • 18) 0.730 325 353 650 088 199 454 609 178 859 929 6 × 2 = 1 + 0.460 650 707 300 176 398 909 218 357 719 859 2;
  • 19) 0.460 650 707 300 176 398 909 218 357 719 859 2 × 2 = 0 + 0.921 301 414 600 352 797 818 436 715 439 718 4;
  • 20) 0.921 301 414 600 352 797 818 436 715 439 718 4 × 2 = 1 + 0.842 602 829 200 705 595 636 873 430 879 436 8;
  • 21) 0.842 602 829 200 705 595 636 873 430 879 436 8 × 2 = 1 + 0.685 205 658 401 411 191 273 746 861 758 873 6;
  • 22) 0.685 205 658 401 411 191 273 746 861 758 873 6 × 2 = 1 + 0.370 411 316 802 822 382 547 493 723 517 747 2;
  • 23) 0.370 411 316 802 822 382 547 493 723 517 747 2 × 2 = 0 + 0.740 822 633 605 644 765 094 987 447 035 494 4;
  • 24) 0.740 822 633 605 644 765 094 987 447 035 494 4 × 2 = 1 + 0.481 645 267 211 289 530 189 974 894 070 988 8;
  • 25) 0.481 645 267 211 289 530 189 974 894 070 988 8 × 2 = 0 + 0.963 290 534 422 579 060 379 949 788 141 977 6;
  • 26) 0.963 290 534 422 579 060 379 949 788 141 977 6 × 2 = 1 + 0.926 581 068 845 158 120 759 899 576 283 955 2;
  • 27) 0.926 581 068 845 158 120 759 899 576 283 955 2 × 2 = 1 + 0.853 162 137 690 316 241 519 799 152 567 910 4;
  • 28) 0.853 162 137 690 316 241 519 799 152 567 910 4 × 2 = 1 + 0.706 324 275 380 632 483 039 598 305 135 820 8;
  • 29) 0.706 324 275 380 632 483 039 598 305 135 820 8 × 2 = 1 + 0.412 648 550 761 264 966 079 196 610 271 641 6;
  • 30) 0.412 648 550 761 264 966 079 196 610 271 641 6 × 2 = 0 + 0.825 297 101 522 529 932 158 393 220 543 283 2;
  • 31) 0.825 297 101 522 529 932 158 393 220 543 283 2 × 2 = 1 + 0.650 594 203 045 059 864 316 786 441 086 566 4;
  • 32) 0.650 594 203 045 059 864 316 786 441 086 566 4 × 2 = 1 + 0.301 188 406 090 119 728 633 572 882 173 132 8;
  • 33) 0.301 188 406 090 119 728 633 572 882 173 132 8 × 2 = 0 + 0.602 376 812 180 239 457 267 145 764 346 265 6;
  • 34) 0.602 376 812 180 239 457 267 145 764 346 265 6 × 2 = 1 + 0.204 753 624 360 478 914 534 291 528 692 531 2;
  • 35) 0.204 753 624 360 478 914 534 291 528 692 531 2 × 2 = 0 + 0.409 507 248 720 957 829 068 583 057 385 062 4;
  • 36) 0.409 507 248 720 957 829 068 583 057 385 062 4 × 2 = 0 + 0.819 014 497 441 915 658 137 166 114 770 124 8;
  • 37) 0.819 014 497 441 915 658 137 166 114 770 124 8 × 2 = 1 + 0.638 028 994 883 831 316 274 332 229 540 249 6;
  • 38) 0.638 028 994 883 831 316 274 332 229 540 249 6 × 2 = 1 + 0.276 057 989 767 662 632 548 664 459 080 499 2;
  • 39) 0.276 057 989 767 662 632 548 664 459 080 499 2 × 2 = 0 + 0.552 115 979 535 325 265 097 328 918 160 998 4;
  • 40) 0.552 115 979 535 325 265 097 328 918 160 998 4 × 2 = 1 + 0.104 231 959 070 650 530 194 657 836 321 996 8;
  • 41) 0.104 231 959 070 650 530 194 657 836 321 996 8 × 2 = 0 + 0.208 463 918 141 301 060 389 315 672 643 993 6;
  • 42) 0.208 463 918 141 301 060 389 315 672 643 993 6 × 2 = 0 + 0.416 927 836 282 602 120 778 631 345 287 987 2;
  • 43) 0.416 927 836 282 602 120 778 631 345 287 987 2 × 2 = 0 + 0.833 855 672 565 204 241 557 262 690 575 974 4;
  • 44) 0.833 855 672 565 204 241 557 262 690 575 974 4 × 2 = 1 + 0.667 711 345 130 408 483 114 525 381 151 948 8;
  • 45) 0.667 711 345 130 408 483 114 525 381 151 948 8 × 2 = 1 + 0.335 422 690 260 816 966 229 050 762 303 897 6;
  • 46) 0.335 422 690 260 816 966 229 050 762 303 897 6 × 2 = 0 + 0.670 845 380 521 633 932 458 101 524 607 795 2;
  • 47) 0.670 845 380 521 633 932 458 101 524 607 795 2 × 2 = 1 + 0.341 690 761 043 267 864 916 203 049 215 590 4;
  • 48) 0.341 690 761 043 267 864 916 203 049 215 590 4 × 2 = 0 + 0.683 381 522 086 535 729 832 406 098 431 180 8;
  • 49) 0.683 381 522 086 535 729 832 406 098 431 180 8 × 2 = 1 + 0.366 763 044 173 071 459 664 812 196 862 361 6;
  • 50) 0.366 763 044 173 071 459 664 812 196 862 361 6 × 2 = 0 + 0.733 526 088 346 142 919 329 624 393 724 723 2;
  • 51) 0.733 526 088 346 142 919 329 624 393 724 723 2 × 2 = 1 + 0.467 052 176 692 285 838 659 248 787 449 446 4;
  • 52) 0.467 052 176 692 285 838 659 248 787 449 446 4 × 2 = 0 + 0.934 104 353 384 571 677 318 497 574 898 892 8;
  • 53) 0.934 104 353 384 571 677 318 497 574 898 892 8 × 2 = 1 + 0.868 208 706 769 143 354 636 995 149 797 785 6;
  • 54) 0.868 208 706 769 143 354 636 995 149 797 785 6 × 2 = 1 + 0.736 417 413 538 286 709 273 990 299 595 571 2;
  • 55) 0.736 417 413 538 286 709 273 990 299 595 571 2 × 2 = 1 + 0.472 834 827 076 573 418 547 980 599 191 142 4;
  • 56) 0.472 834 827 076 573 418 547 980 599 191 142 4 × 2 = 0 + 0.945 669 654 153 146 837 095 961 198 382 284 8;
  • 57) 0.945 669 654 153 146 837 095 961 198 382 284 8 × 2 = 1 + 0.891 339 308 306 293 674 191 922 396 764 569 6;
  • 58) 0.891 339 308 306 293 674 191 922 396 764 569 6 × 2 = 1 + 0.782 678 616 612 587 348 383 844 793 529 139 2;
  • 59) 0.782 678 616 612 587 348 383 844 793 529 139 2 × 2 = 1 + 0.565 357 233 225 174 696 767 689 587 058 278 4;
  • 60) 0.565 357 233 225 174 696 767 689 587 058 278 4 × 2 = 1 + 0.130 714 466 450 349 393 535 379 174 116 556 8;
  • 61) 0.130 714 466 450 349 393 535 379 174 116 556 8 × 2 = 0 + 0.261 428 932 900 698 787 070 758 348 233 113 6;
  • 62) 0.261 428 932 900 698 787 070 758 348 233 113 6 × 2 = 0 + 0.522 857 865 801 397 574 141 516 696 466 227 2;
  • 63) 0.522 857 865 801 397 574 141 516 696 466 227 2 × 2 = 1 + 0.045 715 731 602 795 148 283 033 392 932 454 4;
  • 64) 0.045 715 731 602 795 148 283 033 392 932 454 4 × 2 = 0 + 0.091 431 463 205 590 296 566 066 785 864 908 8;
  • 65) 0.091 431 463 205 590 296 566 066 785 864 908 8 × 2 = 0 + 0.182 862 926 411 180 593 132 133 571 729 817 6;
  • 66) 0.182 862 926 411 180 593 132 133 571 729 817 6 × 2 = 0 + 0.365 725 852 822 361 186 264 267 143 459 635 2;
  • 67) 0.365 725 852 822 361 186 264 267 143 459 635 2 × 2 = 0 + 0.731 451 705 644 722 372 528 534 286 919 270 4;
  • 68) 0.731 451 705 644 722 372 528 534 286 919 270 4 × 2 = 1 + 0.462 903 411 289 444 745 057 068 573 838 540 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:


0.000 020 830 729 321 671 205 134 999 154 501 8(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:

0.000 020 830 729 321 671 205 134 999 154 501 8(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:


0.000 020 830 729 321 671 205 134 999 154 501 8(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 0.000 020 830 729 321 671 205 134 999 154 501 8 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