0.000 020 830 729 321 671 205 134 999 150 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 150 6(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
0.000 020 830 729 321 671 205 134 999 150 6(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: 0.000 020 830 729 321 671 205 134 999 150 6.

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) 0.000 020 830 729 321 671 205 134 999 150 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 301 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 301 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 602 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 602 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 204 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 204 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 409 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 409 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 819 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 819 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 638 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 638 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 276 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 276 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 782 553 6;
  • 9) 0.005 332 666 706 347 828 514 559 782 553 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 565 107 2;
  • 10) 0.010 665 333 412 695 657 029 119 565 107 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 130 214 4;
  • 11) 0.021 330 666 825 391 314 058 239 130 214 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 260 428 8;
  • 12) 0.042 661 333 650 782 628 116 478 260 428 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 520 857 6;
  • 13) 0.085 322 667 301 565 256 232 956 520 857 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 041 715 2;
  • 14) 0.170 645 334 603 130 512 465 913 041 715 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 083 430 4;
  • 15) 0.341 290 669 206 261 024 931 826 083 430 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 166 860 8;
  • 16) 0.682 581 338 412 522 049 863 652 166 860 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 333 721 6;
  • 17) 0.365 162 676 825 044 099 727 304 333 721 6 × 2 = 0 + 0.730 325 353 650 088 199 454 608 667 443 2;
  • 18) 0.730 325 353 650 088 199 454 608 667 443 2 × 2 = 1 + 0.460 650 707 300 176 398 909 217 334 886 4;
  • 19) 0.460 650 707 300 176 398 909 217 334 886 4 × 2 = 0 + 0.921 301 414 600 352 797 818 434 669 772 8;
  • 20) 0.921 301 414 600 352 797 818 434 669 772 8 × 2 = 1 + 0.842 602 829 200 705 595 636 869 339 545 6;
  • 21) 0.842 602 829 200 705 595 636 869 339 545 6 × 2 = 1 + 0.685 205 658 401 411 191 273 738 679 091 2;
  • 22) 0.685 205 658 401 411 191 273 738 679 091 2 × 2 = 1 + 0.370 411 316 802 822 382 547 477 358 182 4;
  • 23) 0.370 411 316 802 822 382 547 477 358 182 4 × 2 = 0 + 0.740 822 633 605 644 765 094 954 716 364 8;
  • 24) 0.740 822 633 605 644 765 094 954 716 364 8 × 2 = 1 + 0.481 645 267 211 289 530 189 909 432 729 6;
  • 25) 0.481 645 267 211 289 530 189 909 432 729 6 × 2 = 0 + 0.963 290 534 422 579 060 379 818 865 459 2;
  • 26) 0.963 290 534 422 579 060 379 818 865 459 2 × 2 = 1 + 0.926 581 068 845 158 120 759 637 730 918 4;
  • 27) 0.926 581 068 845 158 120 759 637 730 918 4 × 2 = 1 + 0.853 162 137 690 316 241 519 275 461 836 8;
  • 28) 0.853 162 137 690 316 241 519 275 461 836 8 × 2 = 1 + 0.706 324 275 380 632 483 038 550 923 673 6;
  • 29) 0.706 324 275 380 632 483 038 550 923 673 6 × 2 = 1 + 0.412 648 550 761 264 966 077 101 847 347 2;
  • 30) 0.412 648 550 761 264 966 077 101 847 347 2 × 2 = 0 + 0.825 297 101 522 529 932 154 203 694 694 4;
  • 31) 0.825 297 101 522 529 932 154 203 694 694 4 × 2 = 1 + 0.650 594 203 045 059 864 308 407 389 388 8;
  • 32) 0.650 594 203 045 059 864 308 407 389 388 8 × 2 = 1 + 0.301 188 406 090 119 728 616 814 778 777 6;
  • 33) 0.301 188 406 090 119 728 616 814 778 777 6 × 2 = 0 + 0.602 376 812 180 239 457 233 629 557 555 2;
  • 34) 0.602 376 812 180 239 457 233 629 557 555 2 × 2 = 1 + 0.204 753 624 360 478 914 467 259 115 110 4;
  • 35) 0.204 753 624 360 478 914 467 259 115 110 4 × 2 = 0 + 0.409 507 248 720 957 828 934 518 230 220 8;
  • 36) 0.409 507 248 720 957 828 934 518 230 220 8 × 2 = 0 + 0.819 014 497 441 915 657 869 036 460 441 6;
  • 37) 0.819 014 497 441 915 657 869 036 460 441 6 × 2 = 1 + 0.638 028 994 883 831 315 738 072 920 883 2;
  • 38) 0.638 028 994 883 831 315 738 072 920 883 2 × 2 = 1 + 0.276 057 989 767 662 631 476 145 841 766 4;
  • 39) 0.276 057 989 767 662 631 476 145 841 766 4 × 2 = 0 + 0.552 115 979 535 325 262 952 291 683 532 8;
  • 40) 0.552 115 979 535 325 262 952 291 683 532 8 × 2 = 1 + 0.104 231 959 070 650 525 904 583 367 065 6;
  • 41) 0.104 231 959 070 650 525 904 583 367 065 6 × 2 = 0 + 0.208 463 918 141 301 051 809 166 734 131 2;
  • 42) 0.208 463 918 141 301 051 809 166 734 131 2 × 2 = 0 + 0.416 927 836 282 602 103 618 333 468 262 4;
  • 43) 0.416 927 836 282 602 103 618 333 468 262 4 × 2 = 0 + 0.833 855 672 565 204 207 236 666 936 524 8;
  • 44) 0.833 855 672 565 204 207 236 666 936 524 8 × 2 = 1 + 0.667 711 345 130 408 414 473 333 873 049 6;
  • 45) 0.667 711 345 130 408 414 473 333 873 049 6 × 2 = 1 + 0.335 422 690 260 816 828 946 667 746 099 2;
  • 46) 0.335 422 690 260 816 828 946 667 746 099 2 × 2 = 0 + 0.670 845 380 521 633 657 893 335 492 198 4;
  • 47) 0.670 845 380 521 633 657 893 335 492 198 4 × 2 = 1 + 0.341 690 761 043 267 315 786 670 984 396 8;
  • 48) 0.341 690 761 043 267 315 786 670 984 396 8 × 2 = 0 + 0.683 381 522 086 534 631 573 341 968 793 6;
  • 49) 0.683 381 522 086 534 631 573 341 968 793 6 × 2 = 1 + 0.366 763 044 173 069 263 146 683 937 587 2;
  • 50) 0.366 763 044 173 069 263 146 683 937 587 2 × 2 = 0 + 0.733 526 088 346 138 526 293 367 875 174 4;
  • 51) 0.733 526 088 346 138 526 293 367 875 174 4 × 2 = 1 + 0.467 052 176 692 277 052 586 735 750 348 8;
  • 52) 0.467 052 176 692 277 052 586 735 750 348 8 × 2 = 0 + 0.934 104 353 384 554 105 173 471 500 697 6;
  • 53) 0.934 104 353 384 554 105 173 471 500 697 6 × 2 = 1 + 0.868 208 706 769 108 210 346 943 001 395 2;
  • 54) 0.868 208 706 769 108 210 346 943 001 395 2 × 2 = 1 + 0.736 417 413 538 216 420 693 886 002 790 4;
  • 55) 0.736 417 413 538 216 420 693 886 002 790 4 × 2 = 1 + 0.472 834 827 076 432 841 387 772 005 580 8;
  • 56) 0.472 834 827 076 432 841 387 772 005 580 8 × 2 = 0 + 0.945 669 654 152 865 682 775 544 011 161 6;
  • 57) 0.945 669 654 152 865 682 775 544 011 161 6 × 2 = 1 + 0.891 339 308 305 731 365 551 088 022 323 2;
  • 58) 0.891 339 308 305 731 365 551 088 022 323 2 × 2 = 1 + 0.782 678 616 611 462 731 102 176 044 646 4;
  • 59) 0.782 678 616 611 462 731 102 176 044 646 4 × 2 = 1 + 0.565 357 233 222 925 462 204 352 089 292 8;
  • 60) 0.565 357 233 222 925 462 204 352 089 292 8 × 2 = 1 + 0.130 714 466 445 850 924 408 704 178 585 6;
  • 61) 0.130 714 466 445 850 924 408 704 178 585 6 × 2 = 0 + 0.261 428 932 891 701 848 817 408 357 171 2;
  • 62) 0.261 428 932 891 701 848 817 408 357 171 2 × 2 = 0 + 0.522 857 865 783 403 697 634 816 714 342 4;
  • 63) 0.522 857 865 783 403 697 634 816 714 342 4 × 2 = 1 + 0.045 715 731 566 807 395 269 633 428 684 8;
  • 64) 0.045 715 731 566 807 395 269 633 428 684 8 × 2 = 0 + 0.091 431 463 133 614 790 539 266 857 369 6;
  • 65) 0.091 431 463 133 614 790 539 266 857 369 6 × 2 = 0 + 0.182 862 926 267 229 581 078 533 714 739 2;
  • 66) 0.182 862 926 267 229 581 078 533 714 739 2 × 2 = 0 + 0.365 725 852 534 459 162 157 067 429 478 4;
  • 67) 0.365 725 852 534 459 162 157 067 429 478 4 × 2 = 0 + 0.731 451 705 068 918 324 314 134 858 956 8;
  • 68) 0.731 451 705 068 918 324 314 134 858 956 8 × 2 = 1 + 0.462 903 410 137 836 648 628 269 717 913 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:


0.000 020 830 729 321 671 205 134 999 150 6(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:

0.000 020 830 729 321 671 205 134 999 150 6(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:


0.000 020 830 729 321 671 205 134 999 150 6(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 0.000 020 830 729 321 671 205 134 999 150 6 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