0.000 020 830 729 321 671 205 134 999 154 532 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 532 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
0.000 020 830 729 321 671 205 134 999 154 532 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: 0.000 020 830 729 321 671 205 134 999 154 532 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) 0.000 020 830 729 321 671 205 134 999 154 532 2 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 064 4;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 064 4 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 128 8;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 128 8 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 257 6;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 257 6 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 515 2;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 515 2 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 945 030 4;
  • 6) 0.000 666 583 338 293 478 564 319 972 945 030 4 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 890 060 8;
  • 7) 0.001 333 166 676 586 957 128 639 945 890 060 8 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 780 121 6;
  • 8) 0.002 666 333 353 173 914 257 279 891 780 121 6 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 560 243 2;
  • 9) 0.005 332 666 706 347 828 514 559 783 560 243 2 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 120 486 4;
  • 10) 0.010 665 333 412 695 657 029 119 567 120 486 4 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 240 972 8;
  • 11) 0.021 330 666 825 391 314 058 239 134 240 972 8 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 481 945 6;
  • 12) 0.042 661 333 650 782 628 116 478 268 481 945 6 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 963 891 2;
  • 13) 0.085 322 667 301 565 256 232 956 536 963 891 2 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 927 782 4;
  • 14) 0.170 645 334 603 130 512 465 913 073 927 782 4 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 855 564 8;
  • 15) 0.341 290 669 206 261 024 931 826 147 855 564 8 × 2 = 0 + 0.682 581 338 412 522 049 863 652 295 711 129 6;
  • 16) 0.682 581 338 412 522 049 863 652 295 711 129 6 × 2 = 1 + 0.365 162 676 825 044 099 727 304 591 422 259 2;
  • 17) 0.365 162 676 825 044 099 727 304 591 422 259 2 × 2 = 0 + 0.730 325 353 650 088 199 454 609 182 844 518 4;
  • 18) 0.730 325 353 650 088 199 454 609 182 844 518 4 × 2 = 1 + 0.460 650 707 300 176 398 909 218 365 689 036 8;
  • 19) 0.460 650 707 300 176 398 909 218 365 689 036 8 × 2 = 0 + 0.921 301 414 600 352 797 818 436 731 378 073 6;
  • 20) 0.921 301 414 600 352 797 818 436 731 378 073 6 × 2 = 1 + 0.842 602 829 200 705 595 636 873 462 756 147 2;
  • 21) 0.842 602 829 200 705 595 636 873 462 756 147 2 × 2 = 1 + 0.685 205 658 401 411 191 273 746 925 512 294 4;
  • 22) 0.685 205 658 401 411 191 273 746 925 512 294 4 × 2 = 1 + 0.370 411 316 802 822 382 547 493 851 024 588 8;
  • 23) 0.370 411 316 802 822 382 547 493 851 024 588 8 × 2 = 0 + 0.740 822 633 605 644 765 094 987 702 049 177 6;
  • 24) 0.740 822 633 605 644 765 094 987 702 049 177 6 × 2 = 1 + 0.481 645 267 211 289 530 189 975 404 098 355 2;
  • 25) 0.481 645 267 211 289 530 189 975 404 098 355 2 × 2 = 0 + 0.963 290 534 422 579 060 379 950 808 196 710 4;
  • 26) 0.963 290 534 422 579 060 379 950 808 196 710 4 × 2 = 1 + 0.926 581 068 845 158 120 759 901 616 393 420 8;
  • 27) 0.926 581 068 845 158 120 759 901 616 393 420 8 × 2 = 1 + 0.853 162 137 690 316 241 519 803 232 786 841 6;
  • 28) 0.853 162 137 690 316 241 519 803 232 786 841 6 × 2 = 1 + 0.706 324 275 380 632 483 039 606 465 573 683 2;
  • 29) 0.706 324 275 380 632 483 039 606 465 573 683 2 × 2 = 1 + 0.412 648 550 761 264 966 079 212 931 147 366 4;
  • 30) 0.412 648 550 761 264 966 079 212 931 147 366 4 × 2 = 0 + 0.825 297 101 522 529 932 158 425 862 294 732 8;
  • 31) 0.825 297 101 522 529 932 158 425 862 294 732 8 × 2 = 1 + 0.650 594 203 045 059 864 316 851 724 589 465 6;
  • 32) 0.650 594 203 045 059 864 316 851 724 589 465 6 × 2 = 1 + 0.301 188 406 090 119 728 633 703 449 178 931 2;
  • 33) 0.301 188 406 090 119 728 633 703 449 178 931 2 × 2 = 0 + 0.602 376 812 180 239 457 267 406 898 357 862 4;
  • 34) 0.602 376 812 180 239 457 267 406 898 357 862 4 × 2 = 1 + 0.204 753 624 360 478 914 534 813 796 715 724 8;
  • 35) 0.204 753 624 360 478 914 534 813 796 715 724 8 × 2 = 0 + 0.409 507 248 720 957 829 069 627 593 431 449 6;
  • 36) 0.409 507 248 720 957 829 069 627 593 431 449 6 × 2 = 0 + 0.819 014 497 441 915 658 139 255 186 862 899 2;
  • 37) 0.819 014 497 441 915 658 139 255 186 862 899 2 × 2 = 1 + 0.638 028 994 883 831 316 278 510 373 725 798 4;
  • 38) 0.638 028 994 883 831 316 278 510 373 725 798 4 × 2 = 1 + 0.276 057 989 767 662 632 557 020 747 451 596 8;
  • 39) 0.276 057 989 767 662 632 557 020 747 451 596 8 × 2 = 0 + 0.552 115 979 535 325 265 114 041 494 903 193 6;
  • 40) 0.552 115 979 535 325 265 114 041 494 903 193 6 × 2 = 1 + 0.104 231 959 070 650 530 228 082 989 806 387 2;
  • 41) 0.104 231 959 070 650 530 228 082 989 806 387 2 × 2 = 0 + 0.208 463 918 141 301 060 456 165 979 612 774 4;
  • 42) 0.208 463 918 141 301 060 456 165 979 612 774 4 × 2 = 0 + 0.416 927 836 282 602 120 912 331 959 225 548 8;
  • 43) 0.416 927 836 282 602 120 912 331 959 225 548 8 × 2 = 0 + 0.833 855 672 565 204 241 824 663 918 451 097 6;
  • 44) 0.833 855 672 565 204 241 824 663 918 451 097 6 × 2 = 1 + 0.667 711 345 130 408 483 649 327 836 902 195 2;
  • 45) 0.667 711 345 130 408 483 649 327 836 902 195 2 × 2 = 1 + 0.335 422 690 260 816 967 298 655 673 804 390 4;
  • 46) 0.335 422 690 260 816 967 298 655 673 804 390 4 × 2 = 0 + 0.670 845 380 521 633 934 597 311 347 608 780 8;
  • 47) 0.670 845 380 521 633 934 597 311 347 608 780 8 × 2 = 1 + 0.341 690 761 043 267 869 194 622 695 217 561 6;
  • 48) 0.341 690 761 043 267 869 194 622 695 217 561 6 × 2 = 0 + 0.683 381 522 086 535 738 389 245 390 435 123 2;
  • 49) 0.683 381 522 086 535 738 389 245 390 435 123 2 × 2 = 1 + 0.366 763 044 173 071 476 778 490 780 870 246 4;
  • 50) 0.366 763 044 173 071 476 778 490 780 870 246 4 × 2 = 0 + 0.733 526 088 346 142 953 556 981 561 740 492 8;
  • 51) 0.733 526 088 346 142 953 556 981 561 740 492 8 × 2 = 1 + 0.467 052 176 692 285 907 113 963 123 480 985 6;
  • 52) 0.467 052 176 692 285 907 113 963 123 480 985 6 × 2 = 0 + 0.934 104 353 384 571 814 227 926 246 961 971 2;
  • 53) 0.934 104 353 384 571 814 227 926 246 961 971 2 × 2 = 1 + 0.868 208 706 769 143 628 455 852 493 923 942 4;
  • 54) 0.868 208 706 769 143 628 455 852 493 923 942 4 × 2 = 1 + 0.736 417 413 538 287 256 911 704 987 847 884 8;
  • 55) 0.736 417 413 538 287 256 911 704 987 847 884 8 × 2 = 1 + 0.472 834 827 076 574 513 823 409 975 695 769 6;
  • 56) 0.472 834 827 076 574 513 823 409 975 695 769 6 × 2 = 0 + 0.945 669 654 153 149 027 646 819 951 391 539 2;
  • 57) 0.945 669 654 153 149 027 646 819 951 391 539 2 × 2 = 1 + 0.891 339 308 306 298 055 293 639 902 783 078 4;
  • 58) 0.891 339 308 306 298 055 293 639 902 783 078 4 × 2 = 1 + 0.782 678 616 612 596 110 587 279 805 566 156 8;
  • 59) 0.782 678 616 612 596 110 587 279 805 566 156 8 × 2 = 1 + 0.565 357 233 225 192 221 174 559 611 132 313 6;
  • 60) 0.565 357 233 225 192 221 174 559 611 132 313 6 × 2 = 1 + 0.130 714 466 450 384 442 349 119 222 264 627 2;
  • 61) 0.130 714 466 450 384 442 349 119 222 264 627 2 × 2 = 0 + 0.261 428 932 900 768 884 698 238 444 529 254 4;
  • 62) 0.261 428 932 900 768 884 698 238 444 529 254 4 × 2 = 0 + 0.522 857 865 801 537 769 396 476 889 058 508 8;
  • 63) 0.522 857 865 801 537 769 396 476 889 058 508 8 × 2 = 1 + 0.045 715 731 603 075 538 792 953 778 117 017 6;
  • 64) 0.045 715 731 603 075 538 792 953 778 117 017 6 × 2 = 0 + 0.091 431 463 206 151 077 585 907 556 234 035 2;
  • 65) 0.091 431 463 206 151 077 585 907 556 234 035 2 × 2 = 0 + 0.182 862 926 412 302 155 171 815 112 468 070 4;
  • 66) 0.182 862 926 412 302 155 171 815 112 468 070 4 × 2 = 0 + 0.365 725 852 824 604 310 343 630 224 936 140 8;
  • 67) 0.365 725 852 824 604 310 343 630 224 936 140 8 × 2 = 0 + 0.731 451 705 649 208 620 687 260 449 872 281 6;
  • 68) 0.731 451 705 649 208 620 687 260 449 872 281 6 × 2 = 1 + 0.462 903 411 298 417 241 374 520 899 744 563 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:


0.000 020 830 729 321 671 205 134 999 154 532 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:

0.000 020 830 729 321 671 205 134 999 154 532 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:


0.000 020 830 729 321 671 205 134 999 154 532 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 0.000 020 830 729 321 671 205 134 999 154 532 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