0.000 020 830 729 321 671 205 134 999 154 508 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 508 1(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
0.000 020 830 729 321 671 205 134 999 154 508 1(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: 0.000 020 830 729 321 671 205 134 999 154 508 1.

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) 0.000 020 830 729 321 671 205 134 999 154 508 1 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 016 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 016 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 032 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 032 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 064 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 064 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 129 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 129 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 259 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 259 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 518 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 518 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 036 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 036 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 554 073 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 554 073 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 108 147 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 108 147 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 216 294 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 216 294 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 432 588 8;
  • 12) 0.042 661 333 650 782 628 116 478 268 432 588 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 865 177 6;
  • 13) 0.085 322 667 301 565 256 232 956 536 865 177 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 730 355 2;
  • 14) 0.170 645 334 603 130 512 465 913 073 730 355 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 460 710 4;
  • 15) 0.341 290 669 206 261 024 931 826 147 460 710 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 921 420 8;
  • 16) 0.682 581 338 412 522 049 863 652 294 921 420 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 842 841 6;
  • 17) 0.365 162 676 825 044 099 727 304 589 842 841 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 685 683 2;
  • 18) 0.730 325 353 650 088 199 454 609 179 685 683 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 359 371 366 4;
  • 19) 0.460 650 707 300 176 398 909 218 359 371 366 4 × 2 = 0 + 0.921 301 414 600 352 797 818 436 718 742 732 8;
  • 20) 0.921 301 414 600 352 797 818 436 718 742 732 8 × 2 = 1 + 0.842 602 829 200 705 595 636 873 437 485 465 6;
  • 21) 0.842 602 829 200 705 595 636 873 437 485 465 6 × 2 = 1 + 0.685 205 658 401 411 191 273 746 874 970 931 2;
  • 22) 0.685 205 658 401 411 191 273 746 874 970 931 2 × 2 = 1 + 0.370 411 316 802 822 382 547 493 749 941 862 4;
  • 23) 0.370 411 316 802 822 382 547 493 749 941 862 4 × 2 = 0 + 0.740 822 633 605 644 765 094 987 499 883 724 8;
  • 24) 0.740 822 633 605 644 765 094 987 499 883 724 8 × 2 = 1 + 0.481 645 267 211 289 530 189 974 999 767 449 6;
  • 25) 0.481 645 267 211 289 530 189 974 999 767 449 6 × 2 = 0 + 0.963 290 534 422 579 060 379 949 999 534 899 2;
  • 26) 0.963 290 534 422 579 060 379 949 999 534 899 2 × 2 = 1 + 0.926 581 068 845 158 120 759 899 999 069 798 4;
  • 27) 0.926 581 068 845 158 120 759 899 999 069 798 4 × 2 = 1 + 0.853 162 137 690 316 241 519 799 998 139 596 8;
  • 28) 0.853 162 137 690 316 241 519 799 998 139 596 8 × 2 = 1 + 0.706 324 275 380 632 483 039 599 996 279 193 6;
  • 29) 0.706 324 275 380 632 483 039 599 996 279 193 6 × 2 = 1 + 0.412 648 550 761 264 966 079 199 992 558 387 2;
  • 30) 0.412 648 550 761 264 966 079 199 992 558 387 2 × 2 = 0 + 0.825 297 101 522 529 932 158 399 985 116 774 4;
  • 31) 0.825 297 101 522 529 932 158 399 985 116 774 4 × 2 = 1 + 0.650 594 203 045 059 864 316 799 970 233 548 8;
  • 32) 0.650 594 203 045 059 864 316 799 970 233 548 8 × 2 = 1 + 0.301 188 406 090 119 728 633 599 940 467 097 6;
  • 33) 0.301 188 406 090 119 728 633 599 940 467 097 6 × 2 = 0 + 0.602 376 812 180 239 457 267 199 880 934 195 2;
  • 34) 0.602 376 812 180 239 457 267 199 880 934 195 2 × 2 = 1 + 0.204 753 624 360 478 914 534 399 761 868 390 4;
  • 35) 0.204 753 624 360 478 914 534 399 761 868 390 4 × 2 = 0 + 0.409 507 248 720 957 829 068 799 523 736 780 8;
  • 36) 0.409 507 248 720 957 829 068 799 523 736 780 8 × 2 = 0 + 0.819 014 497 441 915 658 137 599 047 473 561 6;
  • 37) 0.819 014 497 441 915 658 137 599 047 473 561 6 × 2 = 1 + 0.638 028 994 883 831 316 275 198 094 947 123 2;
  • 38) 0.638 028 994 883 831 316 275 198 094 947 123 2 × 2 = 1 + 0.276 057 989 767 662 632 550 396 189 894 246 4;
  • 39) 0.276 057 989 767 662 632 550 396 189 894 246 4 × 2 = 0 + 0.552 115 979 535 325 265 100 792 379 788 492 8;
  • 40) 0.552 115 979 535 325 265 100 792 379 788 492 8 × 2 = 1 + 0.104 231 959 070 650 530 201 584 759 576 985 6;
  • 41) 0.104 231 959 070 650 530 201 584 759 576 985 6 × 2 = 0 + 0.208 463 918 141 301 060 403 169 519 153 971 2;
  • 42) 0.208 463 918 141 301 060 403 169 519 153 971 2 × 2 = 0 + 0.416 927 836 282 602 120 806 339 038 307 942 4;
  • 43) 0.416 927 836 282 602 120 806 339 038 307 942 4 × 2 = 0 + 0.833 855 672 565 204 241 612 678 076 615 884 8;
  • 44) 0.833 855 672 565 204 241 612 678 076 615 884 8 × 2 = 1 + 0.667 711 345 130 408 483 225 356 153 231 769 6;
  • 45) 0.667 711 345 130 408 483 225 356 153 231 769 6 × 2 = 1 + 0.335 422 690 260 816 966 450 712 306 463 539 2;
  • 46) 0.335 422 690 260 816 966 450 712 306 463 539 2 × 2 = 0 + 0.670 845 380 521 633 932 901 424 612 927 078 4;
  • 47) 0.670 845 380 521 633 932 901 424 612 927 078 4 × 2 = 1 + 0.341 690 761 043 267 865 802 849 225 854 156 8;
  • 48) 0.341 690 761 043 267 865 802 849 225 854 156 8 × 2 = 0 + 0.683 381 522 086 535 731 605 698 451 708 313 6;
  • 49) 0.683 381 522 086 535 731 605 698 451 708 313 6 × 2 = 1 + 0.366 763 044 173 071 463 211 396 903 416 627 2;
  • 50) 0.366 763 044 173 071 463 211 396 903 416 627 2 × 2 = 0 + 0.733 526 088 346 142 926 422 793 806 833 254 4;
  • 51) 0.733 526 088 346 142 926 422 793 806 833 254 4 × 2 = 1 + 0.467 052 176 692 285 852 845 587 613 666 508 8;
  • 52) 0.467 052 176 692 285 852 845 587 613 666 508 8 × 2 = 0 + 0.934 104 353 384 571 705 691 175 227 333 017 6;
  • 53) 0.934 104 353 384 571 705 691 175 227 333 017 6 × 2 = 1 + 0.868 208 706 769 143 411 382 350 454 666 035 2;
  • 54) 0.868 208 706 769 143 411 382 350 454 666 035 2 × 2 = 1 + 0.736 417 413 538 286 822 764 700 909 332 070 4;
  • 55) 0.736 417 413 538 286 822 764 700 909 332 070 4 × 2 = 1 + 0.472 834 827 076 573 645 529 401 818 664 140 8;
  • 56) 0.472 834 827 076 573 645 529 401 818 664 140 8 × 2 = 0 + 0.945 669 654 153 147 291 058 803 637 328 281 6;
  • 57) 0.945 669 654 153 147 291 058 803 637 328 281 6 × 2 = 1 + 0.891 339 308 306 294 582 117 607 274 656 563 2;
  • 58) 0.891 339 308 306 294 582 117 607 274 656 563 2 × 2 = 1 + 0.782 678 616 612 589 164 235 214 549 313 126 4;
  • 59) 0.782 678 616 612 589 164 235 214 549 313 126 4 × 2 = 1 + 0.565 357 233 225 178 328 470 429 098 626 252 8;
  • 60) 0.565 357 233 225 178 328 470 429 098 626 252 8 × 2 = 1 + 0.130 714 466 450 356 656 940 858 197 252 505 6;
  • 61) 0.130 714 466 450 356 656 940 858 197 252 505 6 × 2 = 0 + 0.261 428 932 900 713 313 881 716 394 505 011 2;
  • 62) 0.261 428 932 900 713 313 881 716 394 505 011 2 × 2 = 0 + 0.522 857 865 801 426 627 763 432 789 010 022 4;
  • 63) 0.522 857 865 801 426 627 763 432 789 010 022 4 × 2 = 1 + 0.045 715 731 602 853 255 526 865 578 020 044 8;
  • 64) 0.045 715 731 602 853 255 526 865 578 020 044 8 × 2 = 0 + 0.091 431 463 205 706 511 053 731 156 040 089 6;
  • 65) 0.091 431 463 205 706 511 053 731 156 040 089 6 × 2 = 0 + 0.182 862 926 411 413 022 107 462 312 080 179 2;
  • 66) 0.182 862 926 411 413 022 107 462 312 080 179 2 × 2 = 0 + 0.365 725 852 822 826 044 214 924 624 160 358 4;
  • 67) 0.365 725 852 822 826 044 214 924 624 160 358 4 × 2 = 0 + 0.731 451 705 645 652 088 429 849 248 320 716 8;
  • 68) 0.731 451 705 645 652 088 429 849 248 320 716 8 × 2 = 1 + 0.462 903 411 291 304 176 859 698 496 641 433 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:


0.000 020 830 729 321 671 205 134 999 154 508 1(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:

0.000 020 830 729 321 671 205 134 999 154 508 1(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:


0.000 020 830 729 321 671 205 134 999 154 508 1(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 0.000 020 830 729 321 671 205 134 999 154 508 1 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