0.000 020 830 729 321 671 205 134 999 131 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 131 6(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
0.000 020 830 729 321 671 205 134 999 131 6(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: 0.000 020 830 729 321 671 205 134 999 131 6.

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) 0.000 020 830 729 321 671 205 134 999 131 6 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 263 2;
  • 2) 0.000 041 661 458 643 342 410 269 998 263 2 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 526 4;
  • 3) 0.000 083 322 917 286 684 820 539 996 526 4 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 052 8;
  • 4) 0.000 166 645 834 573 369 641 079 993 052 8 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 105 6;
  • 5) 0.000 333 291 669 146 739 282 159 986 105 6 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 211 2;
  • 6) 0.000 666 583 338 293 478 564 319 972 211 2 × 2 = 0 + 0.001 333 166 676 586 957 128 639 944 422 4;
  • 7) 0.001 333 166 676 586 957 128 639 944 422 4 × 2 = 0 + 0.002 666 333 353 173 914 257 279 888 844 8;
  • 8) 0.002 666 333 353 173 914 257 279 888 844 8 × 2 = 0 + 0.005 332 666 706 347 828 514 559 777 689 6;
  • 9) 0.005 332 666 706 347 828 514 559 777 689 6 × 2 = 0 + 0.010 665 333 412 695 657 029 119 555 379 2;
  • 10) 0.010 665 333 412 695 657 029 119 555 379 2 × 2 = 0 + 0.021 330 666 825 391 314 058 239 110 758 4;
  • 11) 0.021 330 666 825 391 314 058 239 110 758 4 × 2 = 0 + 0.042 661 333 650 782 628 116 478 221 516 8;
  • 12) 0.042 661 333 650 782 628 116 478 221 516 8 × 2 = 0 + 0.085 322 667 301 565 256 232 956 443 033 6;
  • 13) 0.085 322 667 301 565 256 232 956 443 033 6 × 2 = 0 + 0.170 645 334 603 130 512 465 912 886 067 2;
  • 14) 0.170 645 334 603 130 512 465 912 886 067 2 × 2 = 0 + 0.341 290 669 206 261 024 931 825 772 134 4;
  • 15) 0.341 290 669 206 261 024 931 825 772 134 4 × 2 = 0 + 0.682 581 338 412 522 049 863 651 544 268 8;
  • 16) 0.682 581 338 412 522 049 863 651 544 268 8 × 2 = 1 + 0.365 162 676 825 044 099 727 303 088 537 6;
  • 17) 0.365 162 676 825 044 099 727 303 088 537 6 × 2 = 0 + 0.730 325 353 650 088 199 454 606 177 075 2;
  • 18) 0.730 325 353 650 088 199 454 606 177 075 2 × 2 = 1 + 0.460 650 707 300 176 398 909 212 354 150 4;
  • 19) 0.460 650 707 300 176 398 909 212 354 150 4 × 2 = 0 + 0.921 301 414 600 352 797 818 424 708 300 8;
  • 20) 0.921 301 414 600 352 797 818 424 708 300 8 × 2 = 1 + 0.842 602 829 200 705 595 636 849 416 601 6;
  • 21) 0.842 602 829 200 705 595 636 849 416 601 6 × 2 = 1 + 0.685 205 658 401 411 191 273 698 833 203 2;
  • 22) 0.685 205 658 401 411 191 273 698 833 203 2 × 2 = 1 + 0.370 411 316 802 822 382 547 397 666 406 4;
  • 23) 0.370 411 316 802 822 382 547 397 666 406 4 × 2 = 0 + 0.740 822 633 605 644 765 094 795 332 812 8;
  • 24) 0.740 822 633 605 644 765 094 795 332 812 8 × 2 = 1 + 0.481 645 267 211 289 530 189 590 665 625 6;
  • 25) 0.481 645 267 211 289 530 189 590 665 625 6 × 2 = 0 + 0.963 290 534 422 579 060 379 181 331 251 2;
  • 26) 0.963 290 534 422 579 060 379 181 331 251 2 × 2 = 1 + 0.926 581 068 845 158 120 758 362 662 502 4;
  • 27) 0.926 581 068 845 158 120 758 362 662 502 4 × 2 = 1 + 0.853 162 137 690 316 241 516 725 325 004 8;
  • 28) 0.853 162 137 690 316 241 516 725 325 004 8 × 2 = 1 + 0.706 324 275 380 632 483 033 450 650 009 6;
  • 29) 0.706 324 275 380 632 483 033 450 650 009 6 × 2 = 1 + 0.412 648 550 761 264 966 066 901 300 019 2;
  • 30) 0.412 648 550 761 264 966 066 901 300 019 2 × 2 = 0 + 0.825 297 101 522 529 932 133 802 600 038 4;
  • 31) 0.825 297 101 522 529 932 133 802 600 038 4 × 2 = 1 + 0.650 594 203 045 059 864 267 605 200 076 8;
  • 32) 0.650 594 203 045 059 864 267 605 200 076 8 × 2 = 1 + 0.301 188 406 090 119 728 535 210 400 153 6;
  • 33) 0.301 188 406 090 119 728 535 210 400 153 6 × 2 = 0 + 0.602 376 812 180 239 457 070 420 800 307 2;
  • 34) 0.602 376 812 180 239 457 070 420 800 307 2 × 2 = 1 + 0.204 753 624 360 478 914 140 841 600 614 4;
  • 35) 0.204 753 624 360 478 914 140 841 600 614 4 × 2 = 0 + 0.409 507 248 720 957 828 281 683 201 228 8;
  • 36) 0.409 507 248 720 957 828 281 683 201 228 8 × 2 = 0 + 0.819 014 497 441 915 656 563 366 402 457 6;
  • 37) 0.819 014 497 441 915 656 563 366 402 457 6 × 2 = 1 + 0.638 028 994 883 831 313 126 732 804 915 2;
  • 38) 0.638 028 994 883 831 313 126 732 804 915 2 × 2 = 1 + 0.276 057 989 767 662 626 253 465 609 830 4;
  • 39) 0.276 057 989 767 662 626 253 465 609 830 4 × 2 = 0 + 0.552 115 979 535 325 252 506 931 219 660 8;
  • 40) 0.552 115 979 535 325 252 506 931 219 660 8 × 2 = 1 + 0.104 231 959 070 650 505 013 862 439 321 6;
  • 41) 0.104 231 959 070 650 505 013 862 439 321 6 × 2 = 0 + 0.208 463 918 141 301 010 027 724 878 643 2;
  • 42) 0.208 463 918 141 301 010 027 724 878 643 2 × 2 = 0 + 0.416 927 836 282 602 020 055 449 757 286 4;
  • 43) 0.416 927 836 282 602 020 055 449 757 286 4 × 2 = 0 + 0.833 855 672 565 204 040 110 899 514 572 8;
  • 44) 0.833 855 672 565 204 040 110 899 514 572 8 × 2 = 1 + 0.667 711 345 130 408 080 221 799 029 145 6;
  • 45) 0.667 711 345 130 408 080 221 799 029 145 6 × 2 = 1 + 0.335 422 690 260 816 160 443 598 058 291 2;
  • 46) 0.335 422 690 260 816 160 443 598 058 291 2 × 2 = 0 + 0.670 845 380 521 632 320 887 196 116 582 4;
  • 47) 0.670 845 380 521 632 320 887 196 116 582 4 × 2 = 1 + 0.341 690 761 043 264 641 774 392 233 164 8;
  • 48) 0.341 690 761 043 264 641 774 392 233 164 8 × 2 = 0 + 0.683 381 522 086 529 283 548 784 466 329 6;
  • 49) 0.683 381 522 086 529 283 548 784 466 329 6 × 2 = 1 + 0.366 763 044 173 058 567 097 568 932 659 2;
  • 50) 0.366 763 044 173 058 567 097 568 932 659 2 × 2 = 0 + 0.733 526 088 346 117 134 195 137 865 318 4;
  • 51) 0.733 526 088 346 117 134 195 137 865 318 4 × 2 = 1 + 0.467 052 176 692 234 268 390 275 730 636 8;
  • 52) 0.467 052 176 692 234 268 390 275 730 636 8 × 2 = 0 + 0.934 104 353 384 468 536 780 551 461 273 6;
  • 53) 0.934 104 353 384 468 536 780 551 461 273 6 × 2 = 1 + 0.868 208 706 768 937 073 561 102 922 547 2;
  • 54) 0.868 208 706 768 937 073 561 102 922 547 2 × 2 = 1 + 0.736 417 413 537 874 147 122 205 845 094 4;
  • 55) 0.736 417 413 537 874 147 122 205 845 094 4 × 2 = 1 + 0.472 834 827 075 748 294 244 411 690 188 8;
  • 56) 0.472 834 827 075 748 294 244 411 690 188 8 × 2 = 0 + 0.945 669 654 151 496 588 488 823 380 377 6;
  • 57) 0.945 669 654 151 496 588 488 823 380 377 6 × 2 = 1 + 0.891 339 308 302 993 176 977 646 760 755 2;
  • 58) 0.891 339 308 302 993 176 977 646 760 755 2 × 2 = 1 + 0.782 678 616 605 986 353 955 293 521 510 4;
  • 59) 0.782 678 616 605 986 353 955 293 521 510 4 × 2 = 1 + 0.565 357 233 211 972 707 910 587 043 020 8;
  • 60) 0.565 357 233 211 972 707 910 587 043 020 8 × 2 = 1 + 0.130 714 466 423 945 415 821 174 086 041 6;
  • 61) 0.130 714 466 423 945 415 821 174 086 041 6 × 2 = 0 + 0.261 428 932 847 890 831 642 348 172 083 2;
  • 62) 0.261 428 932 847 890 831 642 348 172 083 2 × 2 = 0 + 0.522 857 865 695 781 663 284 696 344 166 4;
  • 63) 0.522 857 865 695 781 663 284 696 344 166 4 × 2 = 1 + 0.045 715 731 391 563 326 569 392 688 332 8;
  • 64) 0.045 715 731 391 563 326 569 392 688 332 8 × 2 = 0 + 0.091 431 462 783 126 653 138 785 376 665 6;
  • 65) 0.091 431 462 783 126 653 138 785 376 665 6 × 2 = 0 + 0.182 862 925 566 253 306 277 570 753 331 2;
  • 66) 0.182 862 925 566 253 306 277 570 753 331 2 × 2 = 0 + 0.365 725 851 132 506 612 555 141 506 662 4;
  • 67) 0.365 725 851 132 506 612 555 141 506 662 4 × 2 = 0 + 0.731 451 702 265 013 225 110 283 013 324 8;
  • 68) 0.731 451 702 265 013 225 110 283 013 324 8 × 2 = 1 + 0.462 903 404 530 026 450 220 566 026 649 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:


0.000 020 830 729 321 671 205 134 999 131 6(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:

0.000 020 830 729 321 671 205 134 999 131 6(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:


0.000 020 830 729 321 671 205 134 999 131 6(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 0.000 020 830 729 321 671 205 134 999 131 6 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