-0.000 806 264 623 585 362 514 063 654 156 855 97 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 855 97(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
-0.000 806 264 623 585 362 514 063 654 156 855 97(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. Start with the positive version of the number:

|-0.000 806 264 623 585 362 514 063 654 156 855 97| = 0.000 806 264 623 585 362 514 063 654 156 855 97


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 806 264 623 585 362 514 063 654 156 855 97.

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) 0.000 806 264 623 585 362 514 063 654 156 855 97 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 711 94;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 711 94 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 423 88;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 423 88 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 847 76;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 847 76 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 695 52;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 695 52 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 391 04;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 391 04 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 782 08;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 782 08 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 077 564 16;
  • 8) 0.103 201 871 818 926 401 800 147 732 077 564 16 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 155 128 32;
  • 9) 0.206 403 743 637 852 803 600 295 464 155 128 32 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 310 256 64;
  • 10) 0.412 807 487 275 705 607 200 590 928 310 256 64 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 620 513 28;
  • 11) 0.825 614 974 551 411 214 401 181 856 620 513 28 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 241 026 56;
  • 12) 0.651 229 949 102 822 428 802 363 713 241 026 56 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 482 053 12;
  • 13) 0.302 459 898 205 644 857 604 727 426 482 053 12 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 964 106 24;
  • 14) 0.604 919 796 411 289 715 209 454 852 964 106 24 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 928 212 48;
  • 15) 0.209 839 592 822 579 430 418 909 705 928 212 48 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 856 424 96;
  • 16) 0.419 679 185 645 158 860 837 819 411 856 424 96 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 712 849 92;
  • 17) 0.839 358 371 290 317 721 675 638 823 712 849 92 × 2 = 1 + 0.678 716 742 580 635 443 351 277 647 425 699 84;
  • 18) 0.678 716 742 580 635 443 351 277 647 425 699 84 × 2 = 1 + 0.357 433 485 161 270 886 702 555 294 851 399 68;
  • 19) 0.357 433 485 161 270 886 702 555 294 851 399 68 × 2 = 0 + 0.714 866 970 322 541 773 405 110 589 702 799 36;
  • 20) 0.714 866 970 322 541 773 405 110 589 702 799 36 × 2 = 1 + 0.429 733 940 645 083 546 810 221 179 405 598 72;
  • 21) 0.429 733 940 645 083 546 810 221 179 405 598 72 × 2 = 0 + 0.859 467 881 290 167 093 620 442 358 811 197 44;
  • 22) 0.859 467 881 290 167 093 620 442 358 811 197 44 × 2 = 1 + 0.718 935 762 580 334 187 240 884 717 622 394 88;
  • 23) 0.718 935 762 580 334 187 240 884 717 622 394 88 × 2 = 1 + 0.437 871 525 160 668 374 481 769 435 244 789 76;
  • 24) 0.437 871 525 160 668 374 481 769 435 244 789 76 × 2 = 0 + 0.875 743 050 321 336 748 963 538 870 489 579 52;
  • 25) 0.875 743 050 321 336 748 963 538 870 489 579 52 × 2 = 1 + 0.751 486 100 642 673 497 927 077 740 979 159 04;
  • 26) 0.751 486 100 642 673 497 927 077 740 979 159 04 × 2 = 1 + 0.502 972 201 285 346 995 854 155 481 958 318 08;
  • 27) 0.502 972 201 285 346 995 854 155 481 958 318 08 × 2 = 1 + 0.005 944 402 570 693 991 708 310 963 916 636 16;
  • 28) 0.005 944 402 570 693 991 708 310 963 916 636 16 × 2 = 0 + 0.011 888 805 141 387 983 416 621 927 833 272 32;
  • 29) 0.011 888 805 141 387 983 416 621 927 833 272 32 × 2 = 0 + 0.023 777 610 282 775 966 833 243 855 666 544 64;
  • 30) 0.023 777 610 282 775 966 833 243 855 666 544 64 × 2 = 0 + 0.047 555 220 565 551 933 666 487 711 333 089 28;
  • 31) 0.047 555 220 565 551 933 666 487 711 333 089 28 × 2 = 0 + 0.095 110 441 131 103 867 332 975 422 666 178 56;
  • 32) 0.095 110 441 131 103 867 332 975 422 666 178 56 × 2 = 0 + 0.190 220 882 262 207 734 665 950 845 332 357 12;
  • 33) 0.190 220 882 262 207 734 665 950 845 332 357 12 × 2 = 0 + 0.380 441 764 524 415 469 331 901 690 664 714 24;
  • 34) 0.380 441 764 524 415 469 331 901 690 664 714 24 × 2 = 0 + 0.760 883 529 048 830 938 663 803 381 329 428 48;
  • 35) 0.760 883 529 048 830 938 663 803 381 329 428 48 × 2 = 1 + 0.521 767 058 097 661 877 327 606 762 658 856 96;
  • 36) 0.521 767 058 097 661 877 327 606 762 658 856 96 × 2 = 1 + 0.043 534 116 195 323 754 655 213 525 317 713 92;
  • 37) 0.043 534 116 195 323 754 655 213 525 317 713 92 × 2 = 0 + 0.087 068 232 390 647 509 310 427 050 635 427 84;
  • 38) 0.087 068 232 390 647 509 310 427 050 635 427 84 × 2 = 0 + 0.174 136 464 781 295 018 620 854 101 270 855 68;
  • 39) 0.174 136 464 781 295 018 620 854 101 270 855 68 × 2 = 0 + 0.348 272 929 562 590 037 241 708 202 541 711 36;
  • 40) 0.348 272 929 562 590 037 241 708 202 541 711 36 × 2 = 0 + 0.696 545 859 125 180 074 483 416 405 083 422 72;
  • 41) 0.696 545 859 125 180 074 483 416 405 083 422 72 × 2 = 1 + 0.393 091 718 250 360 148 966 832 810 166 845 44;
  • 42) 0.393 091 718 250 360 148 966 832 810 166 845 44 × 2 = 0 + 0.786 183 436 500 720 297 933 665 620 333 690 88;
  • 43) 0.786 183 436 500 720 297 933 665 620 333 690 88 × 2 = 1 + 0.572 366 873 001 440 595 867 331 240 667 381 76;
  • 44) 0.572 366 873 001 440 595 867 331 240 667 381 76 × 2 = 1 + 0.144 733 746 002 881 191 734 662 481 334 763 52;
  • 45) 0.144 733 746 002 881 191 734 662 481 334 763 52 × 2 = 0 + 0.289 467 492 005 762 383 469 324 962 669 527 04;
  • 46) 0.289 467 492 005 762 383 469 324 962 669 527 04 × 2 = 0 + 0.578 934 984 011 524 766 938 649 925 339 054 08;
  • 47) 0.578 934 984 011 524 766 938 649 925 339 054 08 × 2 = 1 + 0.157 869 968 023 049 533 877 299 850 678 108 16;
  • 48) 0.157 869 968 023 049 533 877 299 850 678 108 16 × 2 = 0 + 0.315 739 936 046 099 067 754 599 701 356 216 32;
  • 49) 0.315 739 936 046 099 067 754 599 701 356 216 32 × 2 = 0 + 0.631 479 872 092 198 135 509 199 402 712 432 64;
  • 50) 0.631 479 872 092 198 135 509 199 402 712 432 64 × 2 = 1 + 0.262 959 744 184 396 271 018 398 805 424 865 28;
  • 51) 0.262 959 744 184 396 271 018 398 805 424 865 28 × 2 = 0 + 0.525 919 488 368 792 542 036 797 610 849 730 56;
  • 52) 0.525 919 488 368 792 542 036 797 610 849 730 56 × 2 = 1 + 0.051 838 976 737 585 084 073 595 221 699 461 12;
  • 53) 0.051 838 976 737 585 084 073 595 221 699 461 12 × 2 = 0 + 0.103 677 953 475 170 168 147 190 443 398 922 24;
  • 54) 0.103 677 953 475 170 168 147 190 443 398 922 24 × 2 = 0 + 0.207 355 906 950 340 336 294 380 886 797 844 48;
  • 55) 0.207 355 906 950 340 336 294 380 886 797 844 48 × 2 = 0 + 0.414 711 813 900 680 672 588 761 773 595 688 96;
  • 56) 0.414 711 813 900 680 672 588 761 773 595 688 96 × 2 = 0 + 0.829 423 627 801 361 345 177 523 547 191 377 92;
  • 57) 0.829 423 627 801 361 345 177 523 547 191 377 92 × 2 = 1 + 0.658 847 255 602 722 690 355 047 094 382 755 84;
  • 58) 0.658 847 255 602 722 690 355 047 094 382 755 84 × 2 = 1 + 0.317 694 511 205 445 380 710 094 188 765 511 68;
  • 59) 0.317 694 511 205 445 380 710 094 188 765 511 68 × 2 = 0 + 0.635 389 022 410 890 761 420 188 377 531 023 36;
  • 60) 0.635 389 022 410 890 761 420 188 377 531 023 36 × 2 = 1 + 0.270 778 044 821 781 522 840 376 755 062 046 72;
  • 61) 0.270 778 044 821 781 522 840 376 755 062 046 72 × 2 = 0 + 0.541 556 089 643 563 045 680 753 510 124 093 44;
  • 62) 0.541 556 089 643 563 045 680 753 510 124 093 44 × 2 = 1 + 0.083 112 179 287 126 091 361 507 020 248 186 88;
  • 63) 0.083 112 179 287 126 091 361 507 020 248 186 88 × 2 = 0 + 0.166 224 358 574 252 182 723 014 040 496 373 76;

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).


5. 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:


0.000 806 264 623 585 362 514 063 654 156 855 97(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 855 97(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 806 264 623 585 362 514 063 654 156 855 97(10) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) =

0.0000 0000 0011 0100 1101 0110 1110 0000 0011 0000 1011 0010 0101 0000 1101 010(2) × 20 =

1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010(2) × 2-11


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): -11

Mantissa (not normalized):
1.1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-11 + 2(11-1) - 1 =

(-11 + 1 023)(10) =

1 012(10)


10. 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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1012(10) =

011 1111 0100(2)


12. 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. 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010 =

1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0100

Mantissa (52 bits) =
1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010


Decimal number -0.000 806 264 623 585 362 514 063 654 156 855 97 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0100 - 1010 0110 1011 0111 0000 0001 1000 0101 1001 0010 1000 0110 1010

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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