0.000 244 140 620 999 999 988 878 156 886 862 311 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 244 140 620 999 999 988 878 156 886 862 311 5(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
0.000 244 140 620 999 999 988 878 156 886 862 311 5(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: 0.000 244 140 620 999 999 988 878 156 886 862 311 5.

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) 0.000 244 140 620 999 999 988 878 156 886 862 311 5 × 2 = 0 + 0.000 488 281 241 999 999 977 756 313 773 724 623;
  • 2) 0.000 488 281 241 999 999 977 756 313 773 724 623 × 2 = 0 + 0.000 976 562 483 999 999 955 512 627 547 449 246;
  • 3) 0.000 976 562 483 999 999 955 512 627 547 449 246 × 2 = 0 + 0.001 953 124 967 999 999 911 025 255 094 898 492;
  • 4) 0.001 953 124 967 999 999 911 025 255 094 898 492 × 2 = 0 + 0.003 906 249 935 999 999 822 050 510 189 796 984;
  • 5) 0.003 906 249 935 999 999 822 050 510 189 796 984 × 2 = 0 + 0.007 812 499 871 999 999 644 101 020 379 593 968;
  • 6) 0.007 812 499 871 999 999 644 101 020 379 593 968 × 2 = 0 + 0.015 624 999 743 999 999 288 202 040 759 187 936;
  • 7) 0.015 624 999 743 999 999 288 202 040 759 187 936 × 2 = 0 + 0.031 249 999 487 999 998 576 404 081 518 375 872;
  • 8) 0.031 249 999 487 999 998 576 404 081 518 375 872 × 2 = 0 + 0.062 499 998 975 999 997 152 808 163 036 751 744;
  • 9) 0.062 499 998 975 999 997 152 808 163 036 751 744 × 2 = 0 + 0.124 999 997 951 999 994 305 616 326 073 503 488;
  • 10) 0.124 999 997 951 999 994 305 616 326 073 503 488 × 2 = 0 + 0.249 999 995 903 999 988 611 232 652 147 006 976;
  • 11) 0.249 999 995 903 999 988 611 232 652 147 006 976 × 2 = 0 + 0.499 999 991 807 999 977 222 465 304 294 013 952;
  • 12) 0.499 999 991 807 999 977 222 465 304 294 013 952 × 2 = 0 + 0.999 999 983 615 999 954 444 930 608 588 027 904;
  • 13) 0.999 999 983 615 999 954 444 930 608 588 027 904 × 2 = 1 + 0.999 999 967 231 999 908 889 861 217 176 055 808;
  • 14) 0.999 999 967 231 999 908 889 861 217 176 055 808 × 2 = 1 + 0.999 999 934 463 999 817 779 722 434 352 111 616;
  • 15) 0.999 999 934 463 999 817 779 722 434 352 111 616 × 2 = 1 + 0.999 999 868 927 999 635 559 444 868 704 223 232;
  • 16) 0.999 999 868 927 999 635 559 444 868 704 223 232 × 2 = 1 + 0.999 999 737 855 999 271 118 889 737 408 446 464;
  • 17) 0.999 999 737 855 999 271 118 889 737 408 446 464 × 2 = 1 + 0.999 999 475 711 998 542 237 779 474 816 892 928;
  • 18) 0.999 999 475 711 998 542 237 779 474 816 892 928 × 2 = 1 + 0.999 998 951 423 997 084 475 558 949 633 785 856;
  • 19) 0.999 998 951 423 997 084 475 558 949 633 785 856 × 2 = 1 + 0.999 997 902 847 994 168 951 117 899 267 571 712;
  • 20) 0.999 997 902 847 994 168 951 117 899 267 571 712 × 2 = 1 + 0.999 995 805 695 988 337 902 235 798 535 143 424;
  • 21) 0.999 995 805 695 988 337 902 235 798 535 143 424 × 2 = 1 + 0.999 991 611 391 976 675 804 471 597 070 286 848;
  • 22) 0.999 991 611 391 976 675 804 471 597 070 286 848 × 2 = 1 + 0.999 983 222 783 953 351 608 943 194 140 573 696;
  • 23) 0.999 983 222 783 953 351 608 943 194 140 573 696 × 2 = 1 + 0.999 966 445 567 906 703 217 886 388 281 147 392;
  • 24) 0.999 966 445 567 906 703 217 886 388 281 147 392 × 2 = 1 + 0.999 932 891 135 813 406 435 772 776 562 294 784;
  • 25) 0.999 932 891 135 813 406 435 772 776 562 294 784 × 2 = 1 + 0.999 865 782 271 626 812 871 545 553 124 589 568;
  • 26) 0.999 865 782 271 626 812 871 545 553 124 589 568 × 2 = 1 + 0.999 731 564 543 253 625 743 091 106 249 179 136;
  • 27) 0.999 731 564 543 253 625 743 091 106 249 179 136 × 2 = 1 + 0.999 463 129 086 507 251 486 182 212 498 358 272;
  • 28) 0.999 463 129 086 507 251 486 182 212 498 358 272 × 2 = 1 + 0.998 926 258 173 014 502 972 364 424 996 716 544;
  • 29) 0.998 926 258 173 014 502 972 364 424 996 716 544 × 2 = 1 + 0.997 852 516 346 029 005 944 728 849 993 433 088;
  • 30) 0.997 852 516 346 029 005 944 728 849 993 433 088 × 2 = 1 + 0.995 705 032 692 058 011 889 457 699 986 866 176;
  • 31) 0.995 705 032 692 058 011 889 457 699 986 866 176 × 2 = 1 + 0.991 410 065 384 116 023 778 915 399 973 732 352;
  • 32) 0.991 410 065 384 116 023 778 915 399 973 732 352 × 2 = 1 + 0.982 820 130 768 232 047 557 830 799 947 464 704;
  • 33) 0.982 820 130 768 232 047 557 830 799 947 464 704 × 2 = 1 + 0.965 640 261 536 464 095 115 661 599 894 929 408;
  • 34) 0.965 640 261 536 464 095 115 661 599 894 929 408 × 2 = 1 + 0.931 280 523 072 928 190 231 323 199 789 858 816;
  • 35) 0.931 280 523 072 928 190 231 323 199 789 858 816 × 2 = 1 + 0.862 561 046 145 856 380 462 646 399 579 717 632;
  • 36) 0.862 561 046 145 856 380 462 646 399 579 717 632 × 2 = 1 + 0.725 122 092 291 712 760 925 292 799 159 435 264;
  • 37) 0.725 122 092 291 712 760 925 292 799 159 435 264 × 2 = 1 + 0.450 244 184 583 425 521 850 585 598 318 870 528;
  • 38) 0.450 244 184 583 425 521 850 585 598 318 870 528 × 2 = 0 + 0.900 488 369 166 851 043 701 171 196 637 741 056;
  • 39) 0.900 488 369 166 851 043 701 171 196 637 741 056 × 2 = 1 + 0.800 976 738 333 702 087 402 342 393 275 482 112;
  • 40) 0.800 976 738 333 702 087 402 342 393 275 482 112 × 2 = 1 + 0.601 953 476 667 404 174 804 684 786 550 964 224;
  • 41) 0.601 953 476 667 404 174 804 684 786 550 964 224 × 2 = 1 + 0.203 906 953 334 808 349 609 369 573 101 928 448;
  • 42) 0.203 906 953 334 808 349 609 369 573 101 928 448 × 2 = 0 + 0.407 813 906 669 616 699 218 739 146 203 856 896;
  • 43) 0.407 813 906 669 616 699 218 739 146 203 856 896 × 2 = 0 + 0.815 627 813 339 233 398 437 478 292 407 713 792;
  • 44) 0.815 627 813 339 233 398 437 478 292 407 713 792 × 2 = 1 + 0.631 255 626 678 466 796 874 956 584 815 427 584;
  • 45) 0.631 255 626 678 466 796 874 956 584 815 427 584 × 2 = 1 + 0.262 511 253 356 933 593 749 913 169 630 855 168;
  • 46) 0.262 511 253 356 933 593 749 913 169 630 855 168 × 2 = 0 + 0.525 022 506 713 867 187 499 826 339 261 710 336;
  • 47) 0.525 022 506 713 867 187 499 826 339 261 710 336 × 2 = 1 + 0.050 045 013 427 734 374 999 652 678 523 420 672;
  • 48) 0.050 045 013 427 734 374 999 652 678 523 420 672 × 2 = 0 + 0.100 090 026 855 468 749 999 305 357 046 841 344;
  • 49) 0.100 090 026 855 468 749 999 305 357 046 841 344 × 2 = 0 + 0.200 180 053 710 937 499 998 610 714 093 682 688;
  • 50) 0.200 180 053 710 937 499 998 610 714 093 682 688 × 2 = 0 + 0.400 360 107 421 874 999 997 221 428 187 365 376;
  • 51) 0.400 360 107 421 874 999 997 221 428 187 365 376 × 2 = 0 + 0.800 720 214 843 749 999 994 442 856 374 730 752;
  • 52) 0.800 720 214 843 749 999 994 442 856 374 730 752 × 2 = 1 + 0.601 440 429 687 499 999 988 885 712 749 461 504;
  • 53) 0.601 440 429 687 499 999 988 885 712 749 461 504 × 2 = 1 + 0.202 880 859 374 999 999 977 771 425 498 923 008;
  • 54) 0.202 880 859 374 999 999 977 771 425 498 923 008 × 2 = 0 + 0.405 761 718 749 999 999 955 542 850 997 846 016;
  • 55) 0.405 761 718 749 999 999 955 542 850 997 846 016 × 2 = 0 + 0.811 523 437 499 999 999 911 085 701 995 692 032;
  • 56) 0.811 523 437 499 999 999 911 085 701 995 692 032 × 2 = 1 + 0.623 046 874 999 999 999 822 171 403 991 384 064;
  • 57) 0.623 046 874 999 999 999 822 171 403 991 384 064 × 2 = 1 + 0.246 093 749 999 999 999 644 342 807 982 768 128;
  • 58) 0.246 093 749 999 999 999 644 342 807 982 768 128 × 2 = 0 + 0.492 187 499 999 999 999 288 685 615 965 536 256;
  • 59) 0.492 187 499 999 999 999 288 685 615 965 536 256 × 2 = 0 + 0.984 374 999 999 999 998 577 371 231 931 072 512;
  • 60) 0.984 374 999 999 999 998 577 371 231 931 072 512 × 2 = 1 + 0.968 749 999 999 999 997 154 742 463 862 145 024;
  • 61) 0.968 749 999 999 999 997 154 742 463 862 145 024 × 2 = 1 + 0.937 499 999 999 999 994 309 484 927 724 290 048;
  • 62) 0.937 499 999 999 999 994 309 484 927 724 290 048 × 2 = 1 + 0.874 999 999 999 999 988 618 969 855 448 580 096;
  • 63) 0.874 999 999 999 999 988 618 969 855 448 580 096 × 2 = 1 + 0.749 999 999 999 999 977 237 939 710 897 160 192;
  • 64) 0.749 999 999 999 999 977 237 939 710 897 160 192 × 2 = 1 + 0.499 999 999 999 999 954 475 879 421 794 320 384;
  • 65) 0.499 999 999 999 999 954 475 879 421 794 320 384 × 2 = 0 + 0.999 999 999 999 999 908 951 758 843 588 640 768;

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:


0.000 244 140 620 999 999 988 878 156 886 862 311 5(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 0(2)

5. Positive number before normalization:

0.000 244 140 620 999 999 988 878 156 886 862 311 5(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 0(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:


0.000 244 140 620 999 999 988 878 156 886 862 311 5(10) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 0(2) =

0.0000 0000 0000 1111 1111 1111 1111 1111 1111 1011 1001 1010 0001 1001 1001 1111 0(2) × 20 =

1.1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1110(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 1110


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 1110 =

1111 1111 1111 1111 1111 1111 0111 0011 0100 0011 0011 0011 1110


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 1110


Decimal number 0.000 244 140 620 999 999 988 878 156 886 862 311 5 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 1110

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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