0.000 020 830 729 321 671 205 134 999 155 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 155 1(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
0.000 020 830 729 321 671 205 134 999 155 1(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: 0.000 020 830 729 321 671 205 134 999 155 1.

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) 0.000 020 830 729 321 671 205 134 999 155 1 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 310 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 310 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 620 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 620 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 240 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 240 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 481 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 481 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 963 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 963 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 926 4;
  • 7) 0.001 333 166 676 586 957 128 639 945 926 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 852 8;
  • 8) 0.002 666 333 353 173 914 257 279 891 852 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 705 6;
  • 9) 0.005 332 666 706 347 828 514 559 783 705 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 411 2;
  • 10) 0.010 665 333 412 695 657 029 119 567 411 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 822 4;
  • 11) 0.021 330 666 825 391 314 058 239 134 822 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 269 644 8;
  • 12) 0.042 661 333 650 782 628 116 478 269 644 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 539 289 6;
  • 13) 0.085 322 667 301 565 256 232 956 539 289 6 × 2 = 0 + 0.170 645 334 603 130 512 465 913 078 579 2;
  • 14) 0.170 645 334 603 130 512 465 913 078 579 2 × 2 = 0 + 0.341 290 669 206 261 024 931 826 157 158 4;
  • 15) 0.341 290 669 206 261 024 931 826 157 158 4 × 2 = 0 + 0.682 581 338 412 522 049 863 652 314 316 8;
  • 16) 0.682 581 338 412 522 049 863 652 314 316 8 × 2 = 1 + 0.365 162 676 825 044 099 727 304 628 633 6;
  • 17) 0.365 162 676 825 044 099 727 304 628 633 6 × 2 = 0 + 0.730 325 353 650 088 199 454 609 257 267 2;
  • 18) 0.730 325 353 650 088 199 454 609 257 267 2 × 2 = 1 + 0.460 650 707 300 176 398 909 218 514 534 4;
  • 19) 0.460 650 707 300 176 398 909 218 514 534 4 × 2 = 0 + 0.921 301 414 600 352 797 818 437 029 068 8;
  • 20) 0.921 301 414 600 352 797 818 437 029 068 8 × 2 = 1 + 0.842 602 829 200 705 595 636 874 058 137 6;
  • 21) 0.842 602 829 200 705 595 636 874 058 137 6 × 2 = 1 + 0.685 205 658 401 411 191 273 748 116 275 2;
  • 22) 0.685 205 658 401 411 191 273 748 116 275 2 × 2 = 1 + 0.370 411 316 802 822 382 547 496 232 550 4;
  • 23) 0.370 411 316 802 822 382 547 496 232 550 4 × 2 = 0 + 0.740 822 633 605 644 765 094 992 465 100 8;
  • 24) 0.740 822 633 605 644 765 094 992 465 100 8 × 2 = 1 + 0.481 645 267 211 289 530 189 984 930 201 6;
  • 25) 0.481 645 267 211 289 530 189 984 930 201 6 × 2 = 0 + 0.963 290 534 422 579 060 379 969 860 403 2;
  • 26) 0.963 290 534 422 579 060 379 969 860 403 2 × 2 = 1 + 0.926 581 068 845 158 120 759 939 720 806 4;
  • 27) 0.926 581 068 845 158 120 759 939 720 806 4 × 2 = 1 + 0.853 162 137 690 316 241 519 879 441 612 8;
  • 28) 0.853 162 137 690 316 241 519 879 441 612 8 × 2 = 1 + 0.706 324 275 380 632 483 039 758 883 225 6;
  • 29) 0.706 324 275 380 632 483 039 758 883 225 6 × 2 = 1 + 0.412 648 550 761 264 966 079 517 766 451 2;
  • 30) 0.412 648 550 761 264 966 079 517 766 451 2 × 2 = 0 + 0.825 297 101 522 529 932 159 035 532 902 4;
  • 31) 0.825 297 101 522 529 932 159 035 532 902 4 × 2 = 1 + 0.650 594 203 045 059 864 318 071 065 804 8;
  • 32) 0.650 594 203 045 059 864 318 071 065 804 8 × 2 = 1 + 0.301 188 406 090 119 728 636 142 131 609 6;
  • 33) 0.301 188 406 090 119 728 636 142 131 609 6 × 2 = 0 + 0.602 376 812 180 239 457 272 284 263 219 2;
  • 34) 0.602 376 812 180 239 457 272 284 263 219 2 × 2 = 1 + 0.204 753 624 360 478 914 544 568 526 438 4;
  • 35) 0.204 753 624 360 478 914 544 568 526 438 4 × 2 = 0 + 0.409 507 248 720 957 829 089 137 052 876 8;
  • 36) 0.409 507 248 720 957 829 089 137 052 876 8 × 2 = 0 + 0.819 014 497 441 915 658 178 274 105 753 6;
  • 37) 0.819 014 497 441 915 658 178 274 105 753 6 × 2 = 1 + 0.638 028 994 883 831 316 356 548 211 507 2;
  • 38) 0.638 028 994 883 831 316 356 548 211 507 2 × 2 = 1 + 0.276 057 989 767 662 632 713 096 423 014 4;
  • 39) 0.276 057 989 767 662 632 713 096 423 014 4 × 2 = 0 + 0.552 115 979 535 325 265 426 192 846 028 8;
  • 40) 0.552 115 979 535 325 265 426 192 846 028 8 × 2 = 1 + 0.104 231 959 070 650 530 852 385 692 057 6;
  • 41) 0.104 231 959 070 650 530 852 385 692 057 6 × 2 = 0 + 0.208 463 918 141 301 061 704 771 384 115 2;
  • 42) 0.208 463 918 141 301 061 704 771 384 115 2 × 2 = 0 + 0.416 927 836 282 602 123 409 542 768 230 4;
  • 43) 0.416 927 836 282 602 123 409 542 768 230 4 × 2 = 0 + 0.833 855 672 565 204 246 819 085 536 460 8;
  • 44) 0.833 855 672 565 204 246 819 085 536 460 8 × 2 = 1 + 0.667 711 345 130 408 493 638 171 072 921 6;
  • 45) 0.667 711 345 130 408 493 638 171 072 921 6 × 2 = 1 + 0.335 422 690 260 816 987 276 342 145 843 2;
  • 46) 0.335 422 690 260 816 987 276 342 145 843 2 × 2 = 0 + 0.670 845 380 521 633 974 552 684 291 686 4;
  • 47) 0.670 845 380 521 633 974 552 684 291 686 4 × 2 = 1 + 0.341 690 761 043 267 949 105 368 583 372 8;
  • 48) 0.341 690 761 043 267 949 105 368 583 372 8 × 2 = 0 + 0.683 381 522 086 535 898 210 737 166 745 6;
  • 49) 0.683 381 522 086 535 898 210 737 166 745 6 × 2 = 1 + 0.366 763 044 173 071 796 421 474 333 491 2;
  • 50) 0.366 763 044 173 071 796 421 474 333 491 2 × 2 = 0 + 0.733 526 088 346 143 592 842 948 666 982 4;
  • 51) 0.733 526 088 346 143 592 842 948 666 982 4 × 2 = 1 + 0.467 052 176 692 287 185 685 897 333 964 8;
  • 52) 0.467 052 176 692 287 185 685 897 333 964 8 × 2 = 0 + 0.934 104 353 384 574 371 371 794 667 929 6;
  • 53) 0.934 104 353 384 574 371 371 794 667 929 6 × 2 = 1 + 0.868 208 706 769 148 742 743 589 335 859 2;
  • 54) 0.868 208 706 769 148 742 743 589 335 859 2 × 2 = 1 + 0.736 417 413 538 297 485 487 178 671 718 4;
  • 55) 0.736 417 413 538 297 485 487 178 671 718 4 × 2 = 1 + 0.472 834 827 076 594 970 974 357 343 436 8;
  • 56) 0.472 834 827 076 594 970 974 357 343 436 8 × 2 = 0 + 0.945 669 654 153 189 941 948 714 686 873 6;
  • 57) 0.945 669 654 153 189 941 948 714 686 873 6 × 2 = 1 + 0.891 339 308 306 379 883 897 429 373 747 2;
  • 58) 0.891 339 308 306 379 883 897 429 373 747 2 × 2 = 1 + 0.782 678 616 612 759 767 794 858 747 494 4;
  • 59) 0.782 678 616 612 759 767 794 858 747 494 4 × 2 = 1 + 0.565 357 233 225 519 535 589 717 494 988 8;
  • 60) 0.565 357 233 225 519 535 589 717 494 988 8 × 2 = 1 + 0.130 714 466 451 039 071 179 434 989 977 6;
  • 61) 0.130 714 466 451 039 071 179 434 989 977 6 × 2 = 0 + 0.261 428 932 902 078 142 358 869 979 955 2;
  • 62) 0.261 428 932 902 078 142 358 869 979 955 2 × 2 = 0 + 0.522 857 865 804 156 284 717 739 959 910 4;
  • 63) 0.522 857 865 804 156 284 717 739 959 910 4 × 2 = 1 + 0.045 715 731 608 312 569 435 479 919 820 8;
  • 64) 0.045 715 731 608 312 569 435 479 919 820 8 × 2 = 0 + 0.091 431 463 216 625 138 870 959 839 641 6;
  • 65) 0.091 431 463 216 625 138 870 959 839 641 6 × 2 = 0 + 0.182 862 926 433 250 277 741 919 679 283 2;
  • 66) 0.182 862 926 433 250 277 741 919 679 283 2 × 2 = 0 + 0.365 725 852 866 500 555 483 839 358 566 4;
  • 67) 0.365 725 852 866 500 555 483 839 358 566 4 × 2 = 0 + 0.731 451 705 733 001 110 967 678 717 132 8;
  • 68) 0.731 451 705 733 001 110 967 678 717 132 8 × 2 = 1 + 0.462 903 411 466 002 221 935 357 434 265 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:


0.000 020 830 729 321 671 205 134 999 155 1(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:

0.000 020 830 729 321 671 205 134 999 155 1(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:


0.000 020 830 729 321 671 205 134 999 155 1(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 0.000 020 830 729 321 671 205 134 999 155 1 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