0.000 244 140 620 999 999 988 878 156 886 862 323 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 244 140 620 999 999 988 878 156 886 862 323(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
0.000 244 140 620 999 999 988 878 156 886 862 323(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: 0.000 244 140 620 999 999 988 878 156 886 862 323.

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) 0.000 244 140 620 999 999 988 878 156 886 862 323 × 2 = 0 + 0.000 488 281 241 999 999 977 756 313 773 724 646;
  • 2) 0.000 488 281 241 999 999 977 756 313 773 724 646 × 2 = 0 + 0.000 976 562 483 999 999 955 512 627 547 449 292;
  • 3) 0.000 976 562 483 999 999 955 512 627 547 449 292 × 2 = 0 + 0.001 953 124 967 999 999 911 025 255 094 898 584;
  • 4) 0.001 953 124 967 999 999 911 025 255 094 898 584 × 2 = 0 + 0.003 906 249 935 999 999 822 050 510 189 797 168;
  • 5) 0.003 906 249 935 999 999 822 050 510 189 797 168 × 2 = 0 + 0.007 812 499 871 999 999 644 101 020 379 594 336;
  • 6) 0.007 812 499 871 999 999 644 101 020 379 594 336 × 2 = 0 + 0.015 624 999 743 999 999 288 202 040 759 188 672;
  • 7) 0.015 624 999 743 999 999 288 202 040 759 188 672 × 2 = 0 + 0.031 249 999 487 999 998 576 404 081 518 377 344;
  • 8) 0.031 249 999 487 999 998 576 404 081 518 377 344 × 2 = 0 + 0.062 499 998 975 999 997 152 808 163 036 754 688;
  • 9) 0.062 499 998 975 999 997 152 808 163 036 754 688 × 2 = 0 + 0.124 999 997 951 999 994 305 616 326 073 509 376;
  • 10) 0.124 999 997 951 999 994 305 616 326 073 509 376 × 2 = 0 + 0.249 999 995 903 999 988 611 232 652 147 018 752;
  • 11) 0.249 999 995 903 999 988 611 232 652 147 018 752 × 2 = 0 + 0.499 999 991 807 999 977 222 465 304 294 037 504;
  • 12) 0.499 999 991 807 999 977 222 465 304 294 037 504 × 2 = 0 + 0.999 999 983 615 999 954 444 930 608 588 075 008;
  • 13) 0.999 999 983 615 999 954 444 930 608 588 075 008 × 2 = 1 + 0.999 999 967 231 999 908 889 861 217 176 150 016;
  • 14) 0.999 999 967 231 999 908 889 861 217 176 150 016 × 2 = 1 + 0.999 999 934 463 999 817 779 722 434 352 300 032;
  • 15) 0.999 999 934 463 999 817 779 722 434 352 300 032 × 2 = 1 + 0.999 999 868 927 999 635 559 444 868 704 600 064;
  • 16) 0.999 999 868 927 999 635 559 444 868 704 600 064 × 2 = 1 + 0.999 999 737 855 999 271 118 889 737 409 200 128;
  • 17) 0.999 999 737 855 999 271 118 889 737 409 200 128 × 2 = 1 + 0.999 999 475 711 998 542 237 779 474 818 400 256;
  • 18) 0.999 999 475 711 998 542 237 779 474 818 400 256 × 2 = 1 + 0.999 998 951 423 997 084 475 558 949 636 800 512;
  • 19) 0.999 998 951 423 997 084 475 558 949 636 800 512 × 2 = 1 + 0.999 997 902 847 994 168 951 117 899 273 601 024;
  • 20) 0.999 997 902 847 994 168 951 117 899 273 601 024 × 2 = 1 + 0.999 995 805 695 988 337 902 235 798 547 202 048;
  • 21) 0.999 995 805 695 988 337 902 235 798 547 202 048 × 2 = 1 + 0.999 991 611 391 976 675 804 471 597 094 404 096;
  • 22) 0.999 991 611 391 976 675 804 471 597 094 404 096 × 2 = 1 + 0.999 983 222 783 953 351 608 943 194 188 808 192;
  • 23) 0.999 983 222 783 953 351 608 943 194 188 808 192 × 2 = 1 + 0.999 966 445 567 906 703 217 886 388 377 616 384;
  • 24) 0.999 966 445 567 906 703 217 886 388 377 616 384 × 2 = 1 + 0.999 932 891 135 813 406 435 772 776 755 232 768;
  • 25) 0.999 932 891 135 813 406 435 772 776 755 232 768 × 2 = 1 + 0.999 865 782 271 626 812 871 545 553 510 465 536;
  • 26) 0.999 865 782 271 626 812 871 545 553 510 465 536 × 2 = 1 + 0.999 731 564 543 253 625 743 091 107 020 931 072;
  • 27) 0.999 731 564 543 253 625 743 091 107 020 931 072 × 2 = 1 + 0.999 463 129 086 507 251 486 182 214 041 862 144;
  • 28) 0.999 463 129 086 507 251 486 182 214 041 862 144 × 2 = 1 + 0.998 926 258 173 014 502 972 364 428 083 724 288;
  • 29) 0.998 926 258 173 014 502 972 364 428 083 724 288 × 2 = 1 + 0.997 852 516 346 029 005 944 728 856 167 448 576;
  • 30) 0.997 852 516 346 029 005 944 728 856 167 448 576 × 2 = 1 + 0.995 705 032 692 058 011 889 457 712 334 897 152;
  • 31) 0.995 705 032 692 058 011 889 457 712 334 897 152 × 2 = 1 + 0.991 410 065 384 116 023 778 915 424 669 794 304;
  • 32) 0.991 410 065 384 116 023 778 915 424 669 794 304 × 2 = 1 + 0.982 820 130 768 232 047 557 830 849 339 588 608;
  • 33) 0.982 820 130 768 232 047 557 830 849 339 588 608 × 2 = 1 + 0.965 640 261 536 464 095 115 661 698 679 177 216;
  • 34) 0.965 640 261 536 464 095 115 661 698 679 177 216 × 2 = 1 + 0.931 280 523 072 928 190 231 323 397 358 354 432;
  • 35) 0.931 280 523 072 928 190 231 323 397 358 354 432 × 2 = 1 + 0.862 561 046 145 856 380 462 646 794 716 708 864;
  • 36) 0.862 561 046 145 856 380 462 646 794 716 708 864 × 2 = 1 + 0.725 122 092 291 712 760 925 293 589 433 417 728;
  • 37) 0.725 122 092 291 712 760 925 293 589 433 417 728 × 2 = 1 + 0.450 244 184 583 425 521 850 587 178 866 835 456;
  • 38) 0.450 244 184 583 425 521 850 587 178 866 835 456 × 2 = 0 + 0.900 488 369 166 851 043 701 174 357 733 670 912;
  • 39) 0.900 488 369 166 851 043 701 174 357 733 670 912 × 2 = 1 + 0.800 976 738 333 702 087 402 348 715 467 341 824;
  • 40) 0.800 976 738 333 702 087 402 348 715 467 341 824 × 2 = 1 + 0.601 953 476 667 404 174 804 697 430 934 683 648;
  • 41) 0.601 953 476 667 404 174 804 697 430 934 683 648 × 2 = 1 + 0.203 906 953 334 808 349 609 394 861 869 367 296;
  • 42) 0.203 906 953 334 808 349 609 394 861 869 367 296 × 2 = 0 + 0.407 813 906 669 616 699 218 789 723 738 734 592;
  • 43) 0.407 813 906 669 616 699 218 789 723 738 734 592 × 2 = 0 + 0.815 627 813 339 233 398 437 579 447 477 469 184;
  • 44) 0.815 627 813 339 233 398 437 579 447 477 469 184 × 2 = 1 + 0.631 255 626 678 466 796 875 158 894 954 938 368;
  • 45) 0.631 255 626 678 466 796 875 158 894 954 938 368 × 2 = 1 + 0.262 511 253 356 933 593 750 317 789 909 876 736;
  • 46) 0.262 511 253 356 933 593 750 317 789 909 876 736 × 2 = 0 + 0.525 022 506 713 867 187 500 635 579 819 753 472;
  • 47) 0.525 022 506 713 867 187 500 635 579 819 753 472 × 2 = 1 + 0.050 045 013 427 734 375 001 271 159 639 506 944;
  • 48) 0.050 045 013 427 734 375 001 271 159 639 506 944 × 2 = 0 + 0.100 090 026 855 468 750 002 542 319 279 013 888;
  • 49) 0.100 090 026 855 468 750 002 542 319 279 013 888 × 2 = 0 + 0.200 180 053 710 937 500 005 084 638 558 027 776;
  • 50) 0.200 180 053 710 937 500 005 084 638 558 027 776 × 2 = 0 + 0.400 360 107 421 875 000 010 169 277 116 055 552;
  • 51) 0.400 360 107 421 875 000 010 169 277 116 055 552 × 2 = 0 + 0.800 720 214 843 750 000 020 338 554 232 111 104;
  • 52) 0.800 720 214 843 750 000 020 338 554 232 111 104 × 2 = 1 + 0.601 440 429 687 500 000 040 677 108 464 222 208;
  • 53) 0.601 440 429 687 500 000 040 677 108 464 222 208 × 2 = 1 + 0.202 880 859 375 000 000 081 354 216 928 444 416;
  • 54) 0.202 880 859 375 000 000 081 354 216 928 444 416 × 2 = 0 + 0.405 761 718 750 000 000 162 708 433 856 888 832;
  • 55) 0.405 761 718 750 000 000 162 708 433 856 888 832 × 2 = 0 + 0.811 523 437 500 000 000 325 416 867 713 777 664;
  • 56) 0.811 523 437 500 000 000 325 416 867 713 777 664 × 2 = 1 + 0.623 046 875 000 000 000 650 833 735 427 555 328;
  • 57) 0.623 046 875 000 000 000 650 833 735 427 555 328 × 2 = 1 + 0.246 093 750 000 000 001 301 667 470 855 110 656;
  • 58) 0.246 093 750 000 000 001 301 667 470 855 110 656 × 2 = 0 + 0.492 187 500 000 000 002 603 334 941 710 221 312;
  • 59) 0.492 187 500 000 000 002 603 334 941 710 221 312 × 2 = 0 + 0.984 375 000 000 000 005 206 669 883 420 442 624;
  • 60) 0.984 375 000 000 000 005 206 669 883 420 442 624 × 2 = 1 + 0.968 750 000 000 000 010 413 339 766 840 885 248;
  • 61) 0.968 750 000 000 000 010 413 339 766 840 885 248 × 2 = 1 + 0.937 500 000 000 000 020 826 679 533 681 770 496;
  • 62) 0.937 500 000 000 000 020 826 679 533 681 770 496 × 2 = 1 + 0.875 000 000 000 000 041 653 359 067 363 540 992;
  • 63) 0.875 000 000 000 000 041 653 359 067 363 540 992 × 2 = 1 + 0.750 000 000 000 000 083 306 718 134 727 081 984;
  • 64) 0.750 000 000 000 000 083 306 718 134 727 081 984 × 2 = 1 + 0.500 000 000 000 000 166 613 436 269 454 163 968;
  • 65) 0.500 000 000 000 000 166 613 436 269 454 163 968 × 2 = 1 + 0.000 000 000 000 000 333 226 872 538 908 327 936;

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:


0.000 244 140 620 999 999 988 878 156 886 862 323(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 1(2)

5. Positive number before normalization:

0.000 244 140 620 999 999 988 878 156 886 862 323(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 244 140 620 999 999 988 878 156 886 862 323(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 1(2) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 1(2) × 20 =

1.1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111(2) × 2-13


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): -13

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1010(10) =

011 1111 0010(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. 1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111 =

1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111


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 1111 0010

Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111


Decimal number 0.000 244 140 620 999 999 988 878 156 886 862 323 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0010 - 1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1111

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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