0.000 020 830 729 321 671 205 134 999 140 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 140 2(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
0.000 020 830 729 321 671 205 134 999 140 2(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: 0.000 020 830 729 321 671 205 134 999 140 2.

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) 0.000 020 830 729 321 671 205 134 999 140 2 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 280 4;
  • 2) 0.000 041 661 458 643 342 410 269 998 280 4 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 560 8;
  • 3) 0.000 083 322 917 286 684 820 539 996 560 8 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 121 6;
  • 4) 0.000 166 645 834 573 369 641 079 993 121 6 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 243 2;
  • 5) 0.000 333 291 669 146 739 282 159 986 243 2 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 486 4;
  • 6) 0.000 666 583 338 293 478 564 319 972 486 4 × 2 = 0 + 0.001 333 166 676 586 957 128 639 944 972 8;
  • 7) 0.001 333 166 676 586 957 128 639 944 972 8 × 2 = 0 + 0.002 666 333 353 173 914 257 279 889 945 6;
  • 8) 0.002 666 333 353 173 914 257 279 889 945 6 × 2 = 0 + 0.005 332 666 706 347 828 514 559 779 891 2;
  • 9) 0.005 332 666 706 347 828 514 559 779 891 2 × 2 = 0 + 0.010 665 333 412 695 657 029 119 559 782 4;
  • 10) 0.010 665 333 412 695 657 029 119 559 782 4 × 2 = 0 + 0.021 330 666 825 391 314 058 239 119 564 8;
  • 11) 0.021 330 666 825 391 314 058 239 119 564 8 × 2 = 0 + 0.042 661 333 650 782 628 116 478 239 129 6;
  • 12) 0.042 661 333 650 782 628 116 478 239 129 6 × 2 = 0 + 0.085 322 667 301 565 256 232 956 478 259 2;
  • 13) 0.085 322 667 301 565 256 232 956 478 259 2 × 2 = 0 + 0.170 645 334 603 130 512 465 912 956 518 4;
  • 14) 0.170 645 334 603 130 512 465 912 956 518 4 × 2 = 0 + 0.341 290 669 206 261 024 931 825 913 036 8;
  • 15) 0.341 290 669 206 261 024 931 825 913 036 8 × 2 = 0 + 0.682 581 338 412 522 049 863 651 826 073 6;
  • 16) 0.682 581 338 412 522 049 863 651 826 073 6 × 2 = 1 + 0.365 162 676 825 044 099 727 303 652 147 2;
  • 17) 0.365 162 676 825 044 099 727 303 652 147 2 × 2 = 0 + 0.730 325 353 650 088 199 454 607 304 294 4;
  • 18) 0.730 325 353 650 088 199 454 607 304 294 4 × 2 = 1 + 0.460 650 707 300 176 398 909 214 608 588 8;
  • 19) 0.460 650 707 300 176 398 909 214 608 588 8 × 2 = 0 + 0.921 301 414 600 352 797 818 429 217 177 6;
  • 20) 0.921 301 414 600 352 797 818 429 217 177 6 × 2 = 1 + 0.842 602 829 200 705 595 636 858 434 355 2;
  • 21) 0.842 602 829 200 705 595 636 858 434 355 2 × 2 = 1 + 0.685 205 658 401 411 191 273 716 868 710 4;
  • 22) 0.685 205 658 401 411 191 273 716 868 710 4 × 2 = 1 + 0.370 411 316 802 822 382 547 433 737 420 8;
  • 23) 0.370 411 316 802 822 382 547 433 737 420 8 × 2 = 0 + 0.740 822 633 605 644 765 094 867 474 841 6;
  • 24) 0.740 822 633 605 644 765 094 867 474 841 6 × 2 = 1 + 0.481 645 267 211 289 530 189 734 949 683 2;
  • 25) 0.481 645 267 211 289 530 189 734 949 683 2 × 2 = 0 + 0.963 290 534 422 579 060 379 469 899 366 4;
  • 26) 0.963 290 534 422 579 060 379 469 899 366 4 × 2 = 1 + 0.926 581 068 845 158 120 758 939 798 732 8;
  • 27) 0.926 581 068 845 158 120 758 939 798 732 8 × 2 = 1 + 0.853 162 137 690 316 241 517 879 597 465 6;
  • 28) 0.853 162 137 690 316 241 517 879 597 465 6 × 2 = 1 + 0.706 324 275 380 632 483 035 759 194 931 2;
  • 29) 0.706 324 275 380 632 483 035 759 194 931 2 × 2 = 1 + 0.412 648 550 761 264 966 071 518 389 862 4;
  • 30) 0.412 648 550 761 264 966 071 518 389 862 4 × 2 = 0 + 0.825 297 101 522 529 932 143 036 779 724 8;
  • 31) 0.825 297 101 522 529 932 143 036 779 724 8 × 2 = 1 + 0.650 594 203 045 059 864 286 073 559 449 6;
  • 32) 0.650 594 203 045 059 864 286 073 559 449 6 × 2 = 1 + 0.301 188 406 090 119 728 572 147 118 899 2;
  • 33) 0.301 188 406 090 119 728 572 147 118 899 2 × 2 = 0 + 0.602 376 812 180 239 457 144 294 237 798 4;
  • 34) 0.602 376 812 180 239 457 144 294 237 798 4 × 2 = 1 + 0.204 753 624 360 478 914 288 588 475 596 8;
  • 35) 0.204 753 624 360 478 914 288 588 475 596 8 × 2 = 0 + 0.409 507 248 720 957 828 577 176 951 193 6;
  • 36) 0.409 507 248 720 957 828 577 176 951 193 6 × 2 = 0 + 0.819 014 497 441 915 657 154 353 902 387 2;
  • 37) 0.819 014 497 441 915 657 154 353 902 387 2 × 2 = 1 + 0.638 028 994 883 831 314 308 707 804 774 4;
  • 38) 0.638 028 994 883 831 314 308 707 804 774 4 × 2 = 1 + 0.276 057 989 767 662 628 617 415 609 548 8;
  • 39) 0.276 057 989 767 662 628 617 415 609 548 8 × 2 = 0 + 0.552 115 979 535 325 257 234 831 219 097 6;
  • 40) 0.552 115 979 535 325 257 234 831 219 097 6 × 2 = 1 + 0.104 231 959 070 650 514 469 662 438 195 2;
  • 41) 0.104 231 959 070 650 514 469 662 438 195 2 × 2 = 0 + 0.208 463 918 141 301 028 939 324 876 390 4;
  • 42) 0.208 463 918 141 301 028 939 324 876 390 4 × 2 = 0 + 0.416 927 836 282 602 057 878 649 752 780 8;
  • 43) 0.416 927 836 282 602 057 878 649 752 780 8 × 2 = 0 + 0.833 855 672 565 204 115 757 299 505 561 6;
  • 44) 0.833 855 672 565 204 115 757 299 505 561 6 × 2 = 1 + 0.667 711 345 130 408 231 514 599 011 123 2;
  • 45) 0.667 711 345 130 408 231 514 599 011 123 2 × 2 = 1 + 0.335 422 690 260 816 463 029 198 022 246 4;
  • 46) 0.335 422 690 260 816 463 029 198 022 246 4 × 2 = 0 + 0.670 845 380 521 632 926 058 396 044 492 8;
  • 47) 0.670 845 380 521 632 926 058 396 044 492 8 × 2 = 1 + 0.341 690 761 043 265 852 116 792 088 985 6;
  • 48) 0.341 690 761 043 265 852 116 792 088 985 6 × 2 = 0 + 0.683 381 522 086 531 704 233 584 177 971 2;
  • 49) 0.683 381 522 086 531 704 233 584 177 971 2 × 2 = 1 + 0.366 763 044 173 063 408 467 168 355 942 4;
  • 50) 0.366 763 044 173 063 408 467 168 355 942 4 × 2 = 0 + 0.733 526 088 346 126 816 934 336 711 884 8;
  • 51) 0.733 526 088 346 126 816 934 336 711 884 8 × 2 = 1 + 0.467 052 176 692 253 633 868 673 423 769 6;
  • 52) 0.467 052 176 692 253 633 868 673 423 769 6 × 2 = 0 + 0.934 104 353 384 507 267 737 346 847 539 2;
  • 53) 0.934 104 353 384 507 267 737 346 847 539 2 × 2 = 1 + 0.868 208 706 769 014 535 474 693 695 078 4;
  • 54) 0.868 208 706 769 014 535 474 693 695 078 4 × 2 = 1 + 0.736 417 413 538 029 070 949 387 390 156 8;
  • 55) 0.736 417 413 538 029 070 949 387 390 156 8 × 2 = 1 + 0.472 834 827 076 058 141 898 774 780 313 6;
  • 56) 0.472 834 827 076 058 141 898 774 780 313 6 × 2 = 0 + 0.945 669 654 152 116 283 797 549 560 627 2;
  • 57) 0.945 669 654 152 116 283 797 549 560 627 2 × 2 = 1 + 0.891 339 308 304 232 567 595 099 121 254 4;
  • 58) 0.891 339 308 304 232 567 595 099 121 254 4 × 2 = 1 + 0.782 678 616 608 465 135 190 198 242 508 8;
  • 59) 0.782 678 616 608 465 135 190 198 242 508 8 × 2 = 1 + 0.565 357 233 216 930 270 380 396 485 017 6;
  • 60) 0.565 357 233 216 930 270 380 396 485 017 6 × 2 = 1 + 0.130 714 466 433 860 540 760 792 970 035 2;
  • 61) 0.130 714 466 433 860 540 760 792 970 035 2 × 2 = 0 + 0.261 428 932 867 721 081 521 585 940 070 4;
  • 62) 0.261 428 932 867 721 081 521 585 940 070 4 × 2 = 0 + 0.522 857 865 735 442 163 043 171 880 140 8;
  • 63) 0.522 857 865 735 442 163 043 171 880 140 8 × 2 = 1 + 0.045 715 731 470 884 326 086 343 760 281 6;
  • 64) 0.045 715 731 470 884 326 086 343 760 281 6 × 2 = 0 + 0.091 431 462 941 768 652 172 687 520 563 2;
  • 65) 0.091 431 462 941 768 652 172 687 520 563 2 × 2 = 0 + 0.182 862 925 883 537 304 345 375 041 126 4;
  • 66) 0.182 862 925 883 537 304 345 375 041 126 4 × 2 = 0 + 0.365 725 851 767 074 608 690 750 082 252 8;
  • 67) 0.365 725 851 767 074 608 690 750 082 252 8 × 2 = 0 + 0.731 451 703 534 149 217 381 500 164 505 6;
  • 68) 0.731 451 703 534 149 217 381 500 164 505 6 × 2 = 1 + 0.462 903 407 068 298 434 763 000 329 011 2;

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:


0.000 020 830 729 321 671 205 134 999 140 2(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:

0.000 020 830 729 321 671 205 134 999 140 2(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:


0.000 020 830 729 321 671 205 134 999 140 2(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 0.000 020 830 729 321 671 205 134 999 140 2 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