0.000 020 830 729 321 671 205 134 999 154 446 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 446(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
0.000 020 830 729 321 671 205 134 999 154 446(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: 0.000 020 830 729 321 671 205 134 999 154 446.

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) 0.000 020 830 729 321 671 205 134 999 154 446 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 308 892;
  • 2) 0.000 041 661 458 643 342 410 269 998 308 892 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 617 784;
  • 3) 0.000 083 322 917 286 684 820 539 996 617 784 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 235 568;
  • 4) 0.000 166 645 834 573 369 641 079 993 235 568 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 471 136;
  • 5) 0.000 333 291 669 146 739 282 159 986 471 136 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 942 272;
  • 6) 0.000 666 583 338 293 478 564 319 972 942 272 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 884 544;
  • 7) 0.001 333 166 676 586 957 128 639 945 884 544 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 769 088;
  • 8) 0.002 666 333 353 173 914 257 279 891 769 088 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 538 176;
  • 9) 0.005 332 666 706 347 828 514 559 783 538 176 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 076 352;
  • 10) 0.010 665 333 412 695 657 029 119 567 076 352 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 152 704;
  • 11) 0.021 330 666 825 391 314 058 239 134 152 704 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 305 408;
  • 12) 0.042 661 333 650 782 628 116 478 268 305 408 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 610 816;
  • 13) 0.085 322 667 301 565 256 232 956 536 610 816 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 221 632;
  • 14) 0.170 645 334 603 130 512 465 913 073 221 632 × 2 = 0 + 0.341 290 669 206 261 024 931 826 146 443 264;
  • 15) 0.341 290 669 206 261 024 931 826 146 443 264 × 2 = 0 + 0.682 581 338 412 522 049 863 652 292 886 528;
  • 16) 0.682 581 338 412 522 049 863 652 292 886 528 × 2 = 1 + 0.365 162 676 825 044 099 727 304 585 773 056;
  • 17) 0.365 162 676 825 044 099 727 304 585 773 056 × 2 = 0 + 0.730 325 353 650 088 199 454 609 171 546 112;
  • 18) 0.730 325 353 650 088 199 454 609 171 546 112 × 2 = 1 + 0.460 650 707 300 176 398 909 218 343 092 224;
  • 19) 0.460 650 707 300 176 398 909 218 343 092 224 × 2 = 0 + 0.921 301 414 600 352 797 818 436 686 184 448;
  • 20) 0.921 301 414 600 352 797 818 436 686 184 448 × 2 = 1 + 0.842 602 829 200 705 595 636 873 372 368 896;
  • 21) 0.842 602 829 200 705 595 636 873 372 368 896 × 2 = 1 + 0.685 205 658 401 411 191 273 746 744 737 792;
  • 22) 0.685 205 658 401 411 191 273 746 744 737 792 × 2 = 1 + 0.370 411 316 802 822 382 547 493 489 475 584;
  • 23) 0.370 411 316 802 822 382 547 493 489 475 584 × 2 = 0 + 0.740 822 633 605 644 765 094 986 978 951 168;
  • 24) 0.740 822 633 605 644 765 094 986 978 951 168 × 2 = 1 + 0.481 645 267 211 289 530 189 973 957 902 336;
  • 25) 0.481 645 267 211 289 530 189 973 957 902 336 × 2 = 0 + 0.963 290 534 422 579 060 379 947 915 804 672;
  • 26) 0.963 290 534 422 579 060 379 947 915 804 672 × 2 = 1 + 0.926 581 068 845 158 120 759 895 831 609 344;
  • 27) 0.926 581 068 845 158 120 759 895 831 609 344 × 2 = 1 + 0.853 162 137 690 316 241 519 791 663 218 688;
  • 28) 0.853 162 137 690 316 241 519 791 663 218 688 × 2 = 1 + 0.706 324 275 380 632 483 039 583 326 437 376;
  • 29) 0.706 324 275 380 632 483 039 583 326 437 376 × 2 = 1 + 0.412 648 550 761 264 966 079 166 652 874 752;
  • 30) 0.412 648 550 761 264 966 079 166 652 874 752 × 2 = 0 + 0.825 297 101 522 529 932 158 333 305 749 504;
  • 31) 0.825 297 101 522 529 932 158 333 305 749 504 × 2 = 1 + 0.650 594 203 045 059 864 316 666 611 499 008;
  • 32) 0.650 594 203 045 059 864 316 666 611 499 008 × 2 = 1 + 0.301 188 406 090 119 728 633 333 222 998 016;
  • 33) 0.301 188 406 090 119 728 633 333 222 998 016 × 2 = 0 + 0.602 376 812 180 239 457 266 666 445 996 032;
  • 34) 0.602 376 812 180 239 457 266 666 445 996 032 × 2 = 1 + 0.204 753 624 360 478 914 533 332 891 992 064;
  • 35) 0.204 753 624 360 478 914 533 332 891 992 064 × 2 = 0 + 0.409 507 248 720 957 829 066 665 783 984 128;
  • 36) 0.409 507 248 720 957 829 066 665 783 984 128 × 2 = 0 + 0.819 014 497 441 915 658 133 331 567 968 256;
  • 37) 0.819 014 497 441 915 658 133 331 567 968 256 × 2 = 1 + 0.638 028 994 883 831 316 266 663 135 936 512;
  • 38) 0.638 028 994 883 831 316 266 663 135 936 512 × 2 = 1 + 0.276 057 989 767 662 632 533 326 271 873 024;
  • 39) 0.276 057 989 767 662 632 533 326 271 873 024 × 2 = 0 + 0.552 115 979 535 325 265 066 652 543 746 048;
  • 40) 0.552 115 979 535 325 265 066 652 543 746 048 × 2 = 1 + 0.104 231 959 070 650 530 133 305 087 492 096;
  • 41) 0.104 231 959 070 650 530 133 305 087 492 096 × 2 = 0 + 0.208 463 918 141 301 060 266 610 174 984 192;
  • 42) 0.208 463 918 141 301 060 266 610 174 984 192 × 2 = 0 + 0.416 927 836 282 602 120 533 220 349 968 384;
  • 43) 0.416 927 836 282 602 120 533 220 349 968 384 × 2 = 0 + 0.833 855 672 565 204 241 066 440 699 936 768;
  • 44) 0.833 855 672 565 204 241 066 440 699 936 768 × 2 = 1 + 0.667 711 345 130 408 482 132 881 399 873 536;
  • 45) 0.667 711 345 130 408 482 132 881 399 873 536 × 2 = 1 + 0.335 422 690 260 816 964 265 762 799 747 072;
  • 46) 0.335 422 690 260 816 964 265 762 799 747 072 × 2 = 0 + 0.670 845 380 521 633 928 531 525 599 494 144;
  • 47) 0.670 845 380 521 633 928 531 525 599 494 144 × 2 = 1 + 0.341 690 761 043 267 857 063 051 198 988 288;
  • 48) 0.341 690 761 043 267 857 063 051 198 988 288 × 2 = 0 + 0.683 381 522 086 535 714 126 102 397 976 576;
  • 49) 0.683 381 522 086 535 714 126 102 397 976 576 × 2 = 1 + 0.366 763 044 173 071 428 252 204 795 953 152;
  • 50) 0.366 763 044 173 071 428 252 204 795 953 152 × 2 = 0 + 0.733 526 088 346 142 856 504 409 591 906 304;
  • 51) 0.733 526 088 346 142 856 504 409 591 906 304 × 2 = 1 + 0.467 052 176 692 285 713 008 819 183 812 608;
  • 52) 0.467 052 176 692 285 713 008 819 183 812 608 × 2 = 0 + 0.934 104 353 384 571 426 017 638 367 625 216;
  • 53) 0.934 104 353 384 571 426 017 638 367 625 216 × 2 = 1 + 0.868 208 706 769 142 852 035 276 735 250 432;
  • 54) 0.868 208 706 769 142 852 035 276 735 250 432 × 2 = 1 + 0.736 417 413 538 285 704 070 553 470 500 864;
  • 55) 0.736 417 413 538 285 704 070 553 470 500 864 × 2 = 1 + 0.472 834 827 076 571 408 141 106 941 001 728;
  • 56) 0.472 834 827 076 571 408 141 106 941 001 728 × 2 = 0 + 0.945 669 654 153 142 816 282 213 882 003 456;
  • 57) 0.945 669 654 153 142 816 282 213 882 003 456 × 2 = 1 + 0.891 339 308 306 285 632 564 427 764 006 912;
  • 58) 0.891 339 308 306 285 632 564 427 764 006 912 × 2 = 1 + 0.782 678 616 612 571 265 128 855 528 013 824;
  • 59) 0.782 678 616 612 571 265 128 855 528 013 824 × 2 = 1 + 0.565 357 233 225 142 530 257 711 056 027 648;
  • 60) 0.565 357 233 225 142 530 257 711 056 027 648 × 2 = 1 + 0.130 714 466 450 285 060 515 422 112 055 296;
  • 61) 0.130 714 466 450 285 060 515 422 112 055 296 × 2 = 0 + 0.261 428 932 900 570 121 030 844 224 110 592;
  • 62) 0.261 428 932 900 570 121 030 844 224 110 592 × 2 = 0 + 0.522 857 865 801 140 242 061 688 448 221 184;
  • 63) 0.522 857 865 801 140 242 061 688 448 221 184 × 2 = 1 + 0.045 715 731 602 280 484 123 376 896 442 368;
  • 64) 0.045 715 731 602 280 484 123 376 896 442 368 × 2 = 0 + 0.091 431 463 204 560 968 246 753 792 884 736;
  • 65) 0.091 431 463 204 560 968 246 753 792 884 736 × 2 = 0 + 0.182 862 926 409 121 936 493 507 585 769 472;
  • 66) 0.182 862 926 409 121 936 493 507 585 769 472 × 2 = 0 + 0.365 725 852 818 243 872 987 015 171 538 944;
  • 67) 0.365 725 852 818 243 872 987 015 171 538 944 × 2 = 0 + 0.731 451 705 636 487 745 974 030 343 077 888;
  • 68) 0.731 451 705 636 487 745 974 030 343 077 888 × 2 = 1 + 0.462 903 411 272 975 491 948 060 686 155 776;

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:


0.000 020 830 729 321 671 205 134 999 154 446(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:

0.000 020 830 729 321 671 205 134 999 154 446(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:


0.000 020 830 729 321 671 205 134 999 154 446(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 0.000 020 830 729 321 671 205 134 999 154 446 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

» Convert decimal 0.000 020 830 729 321 671 205 134 999 154 474 to 64 bit double precision IEEE 754 binary floating point representation standard

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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