0.000 020 830 729 321 671 205 134 999 142 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 142 9(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
0.000 020 830 729 321 671 205 134 999 142 9(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: 0.000 020 830 729 321 671 205 134 999 142 9.

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) 0.000 020 830 729 321 671 205 134 999 142 9 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 285 8;
  • 2) 0.000 041 661 458 643 342 410 269 998 285 8 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 571 6;
  • 3) 0.000 083 322 917 286 684 820 539 996 571 6 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 143 2;
  • 4) 0.000 166 645 834 573 369 641 079 993 143 2 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 286 4;
  • 5) 0.000 333 291 669 146 739 282 159 986 286 4 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 572 8;
  • 6) 0.000 666 583 338 293 478 564 319 972 572 8 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 145 6;
  • 7) 0.001 333 166 676 586 957 128 639 945 145 6 × 2 = 0 + 0.002 666 333 353 173 914 257 279 890 291 2;
  • 8) 0.002 666 333 353 173 914 257 279 890 291 2 × 2 = 0 + 0.005 332 666 706 347 828 514 559 780 582 4;
  • 9) 0.005 332 666 706 347 828 514 559 780 582 4 × 2 = 0 + 0.010 665 333 412 695 657 029 119 561 164 8;
  • 10) 0.010 665 333 412 695 657 029 119 561 164 8 × 2 = 0 + 0.021 330 666 825 391 314 058 239 122 329 6;
  • 11) 0.021 330 666 825 391 314 058 239 122 329 6 × 2 = 0 + 0.042 661 333 650 782 628 116 478 244 659 2;
  • 12) 0.042 661 333 650 782 628 116 478 244 659 2 × 2 = 0 + 0.085 322 667 301 565 256 232 956 489 318 4;
  • 13) 0.085 322 667 301 565 256 232 956 489 318 4 × 2 = 0 + 0.170 645 334 603 130 512 465 912 978 636 8;
  • 14) 0.170 645 334 603 130 512 465 912 978 636 8 × 2 = 0 + 0.341 290 669 206 261 024 931 825 957 273 6;
  • 15) 0.341 290 669 206 261 024 931 825 957 273 6 × 2 = 0 + 0.682 581 338 412 522 049 863 651 914 547 2;
  • 16) 0.682 581 338 412 522 049 863 651 914 547 2 × 2 = 1 + 0.365 162 676 825 044 099 727 303 829 094 4;
  • 17) 0.365 162 676 825 044 099 727 303 829 094 4 × 2 = 0 + 0.730 325 353 650 088 199 454 607 658 188 8;
  • 18) 0.730 325 353 650 088 199 454 607 658 188 8 × 2 = 1 + 0.460 650 707 300 176 398 909 215 316 377 6;
  • 19) 0.460 650 707 300 176 398 909 215 316 377 6 × 2 = 0 + 0.921 301 414 600 352 797 818 430 632 755 2;
  • 20) 0.921 301 414 600 352 797 818 430 632 755 2 × 2 = 1 + 0.842 602 829 200 705 595 636 861 265 510 4;
  • 21) 0.842 602 829 200 705 595 636 861 265 510 4 × 2 = 1 + 0.685 205 658 401 411 191 273 722 531 020 8;
  • 22) 0.685 205 658 401 411 191 273 722 531 020 8 × 2 = 1 + 0.370 411 316 802 822 382 547 445 062 041 6;
  • 23) 0.370 411 316 802 822 382 547 445 062 041 6 × 2 = 0 + 0.740 822 633 605 644 765 094 890 124 083 2;
  • 24) 0.740 822 633 605 644 765 094 890 124 083 2 × 2 = 1 + 0.481 645 267 211 289 530 189 780 248 166 4;
  • 25) 0.481 645 267 211 289 530 189 780 248 166 4 × 2 = 0 + 0.963 290 534 422 579 060 379 560 496 332 8;
  • 26) 0.963 290 534 422 579 060 379 560 496 332 8 × 2 = 1 + 0.926 581 068 845 158 120 759 120 992 665 6;
  • 27) 0.926 581 068 845 158 120 759 120 992 665 6 × 2 = 1 + 0.853 162 137 690 316 241 518 241 985 331 2;
  • 28) 0.853 162 137 690 316 241 518 241 985 331 2 × 2 = 1 + 0.706 324 275 380 632 483 036 483 970 662 4;
  • 29) 0.706 324 275 380 632 483 036 483 970 662 4 × 2 = 1 + 0.412 648 550 761 264 966 072 967 941 324 8;
  • 30) 0.412 648 550 761 264 966 072 967 941 324 8 × 2 = 0 + 0.825 297 101 522 529 932 145 935 882 649 6;
  • 31) 0.825 297 101 522 529 932 145 935 882 649 6 × 2 = 1 + 0.650 594 203 045 059 864 291 871 765 299 2;
  • 32) 0.650 594 203 045 059 864 291 871 765 299 2 × 2 = 1 + 0.301 188 406 090 119 728 583 743 530 598 4;
  • 33) 0.301 188 406 090 119 728 583 743 530 598 4 × 2 = 0 + 0.602 376 812 180 239 457 167 487 061 196 8;
  • 34) 0.602 376 812 180 239 457 167 487 061 196 8 × 2 = 1 + 0.204 753 624 360 478 914 334 974 122 393 6;
  • 35) 0.204 753 624 360 478 914 334 974 122 393 6 × 2 = 0 + 0.409 507 248 720 957 828 669 948 244 787 2;
  • 36) 0.409 507 248 720 957 828 669 948 244 787 2 × 2 = 0 + 0.819 014 497 441 915 657 339 896 489 574 4;
  • 37) 0.819 014 497 441 915 657 339 896 489 574 4 × 2 = 1 + 0.638 028 994 883 831 314 679 792 979 148 8;
  • 38) 0.638 028 994 883 831 314 679 792 979 148 8 × 2 = 1 + 0.276 057 989 767 662 629 359 585 958 297 6;
  • 39) 0.276 057 989 767 662 629 359 585 958 297 6 × 2 = 0 + 0.552 115 979 535 325 258 719 171 916 595 2;
  • 40) 0.552 115 979 535 325 258 719 171 916 595 2 × 2 = 1 + 0.104 231 959 070 650 517 438 343 833 190 4;
  • 41) 0.104 231 959 070 650 517 438 343 833 190 4 × 2 = 0 + 0.208 463 918 141 301 034 876 687 666 380 8;
  • 42) 0.208 463 918 141 301 034 876 687 666 380 8 × 2 = 0 + 0.416 927 836 282 602 069 753 375 332 761 6;
  • 43) 0.416 927 836 282 602 069 753 375 332 761 6 × 2 = 0 + 0.833 855 672 565 204 139 506 750 665 523 2;
  • 44) 0.833 855 672 565 204 139 506 750 665 523 2 × 2 = 1 + 0.667 711 345 130 408 279 013 501 331 046 4;
  • 45) 0.667 711 345 130 408 279 013 501 331 046 4 × 2 = 1 + 0.335 422 690 260 816 558 027 002 662 092 8;
  • 46) 0.335 422 690 260 816 558 027 002 662 092 8 × 2 = 0 + 0.670 845 380 521 633 116 054 005 324 185 6;
  • 47) 0.670 845 380 521 633 116 054 005 324 185 6 × 2 = 1 + 0.341 690 761 043 266 232 108 010 648 371 2;
  • 48) 0.341 690 761 043 266 232 108 010 648 371 2 × 2 = 0 + 0.683 381 522 086 532 464 216 021 296 742 4;
  • 49) 0.683 381 522 086 532 464 216 021 296 742 4 × 2 = 1 + 0.366 763 044 173 064 928 432 042 593 484 8;
  • 50) 0.366 763 044 173 064 928 432 042 593 484 8 × 2 = 0 + 0.733 526 088 346 129 856 864 085 186 969 6;
  • 51) 0.733 526 088 346 129 856 864 085 186 969 6 × 2 = 1 + 0.467 052 176 692 259 713 728 170 373 939 2;
  • 52) 0.467 052 176 692 259 713 728 170 373 939 2 × 2 = 0 + 0.934 104 353 384 519 427 456 340 747 878 4;
  • 53) 0.934 104 353 384 519 427 456 340 747 878 4 × 2 = 1 + 0.868 208 706 769 038 854 912 681 495 756 8;
  • 54) 0.868 208 706 769 038 854 912 681 495 756 8 × 2 = 1 + 0.736 417 413 538 077 709 825 362 991 513 6;
  • 55) 0.736 417 413 538 077 709 825 362 991 513 6 × 2 = 1 + 0.472 834 827 076 155 419 650 725 983 027 2;
  • 56) 0.472 834 827 076 155 419 650 725 983 027 2 × 2 = 0 + 0.945 669 654 152 310 839 301 451 966 054 4;
  • 57) 0.945 669 654 152 310 839 301 451 966 054 4 × 2 = 1 + 0.891 339 308 304 621 678 602 903 932 108 8;
  • 58) 0.891 339 308 304 621 678 602 903 932 108 8 × 2 = 1 + 0.782 678 616 609 243 357 205 807 864 217 6;
  • 59) 0.782 678 616 609 243 357 205 807 864 217 6 × 2 = 1 + 0.565 357 233 218 486 714 411 615 728 435 2;
  • 60) 0.565 357 233 218 486 714 411 615 728 435 2 × 2 = 1 + 0.130 714 466 436 973 428 823 231 456 870 4;
  • 61) 0.130 714 466 436 973 428 823 231 456 870 4 × 2 = 0 + 0.261 428 932 873 946 857 646 462 913 740 8;
  • 62) 0.261 428 932 873 946 857 646 462 913 740 8 × 2 = 0 + 0.522 857 865 747 893 715 292 925 827 481 6;
  • 63) 0.522 857 865 747 893 715 292 925 827 481 6 × 2 = 1 + 0.045 715 731 495 787 430 585 851 654 963 2;
  • 64) 0.045 715 731 495 787 430 585 851 654 963 2 × 2 = 0 + 0.091 431 462 991 574 861 171 703 309 926 4;
  • 65) 0.091 431 462 991 574 861 171 703 309 926 4 × 2 = 0 + 0.182 862 925 983 149 722 343 406 619 852 8;
  • 66) 0.182 862 925 983 149 722 343 406 619 852 8 × 2 = 0 + 0.365 725 851 966 299 444 686 813 239 705 6;
  • 67) 0.365 725 851 966 299 444 686 813 239 705 6 × 2 = 0 + 0.731 451 703 932 598 889 373 626 479 411 2;
  • 68) 0.731 451 703 932 598 889 373 626 479 411 2 × 2 = 1 + 0.462 903 407 865 197 778 747 252 958 822 4;

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:


0.000 020 830 729 321 671 205 134 999 142 9(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:

0.000 020 830 729 321 671 205 134 999 142 9(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:


0.000 020 830 729 321 671 205 134 999 142 9(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 0.000 020 830 729 321 671 205 134 999 142 9 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