-0.000 806 264 623 585 362 514 063 654 156 845 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 845 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
-0.000 806 264 623 585 362 514 063 654 156 845 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. Start with the positive version of the number:

|-0.000 806 264 623 585 362 514 063 654 156 845 9| = 0.000 806 264 623 585 362 514 063 654 156 845 9


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: 0.000 806 264 623 585 362 514 063 654 156 845 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) 0.000 806 264 623 585 362 514 063 654 156 845 9 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 691 8;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 691 8 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 383 6;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 383 6 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 767 2;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 767 2 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 534 4;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 534 4 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 068 8;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 068 8 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 137 6;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 137 6 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 076 275 2;
  • 8) 0.103 201 871 818 926 401 800 147 732 076 275 2 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 152 550 4;
  • 9) 0.206 403 743 637 852 803 600 295 464 152 550 4 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 305 100 8;
  • 10) 0.412 807 487 275 705 607 200 590 928 305 100 8 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 610 201 6;
  • 11) 0.825 614 974 551 411 214 401 181 856 610 201 6 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 220 403 2;
  • 12) 0.651 229 949 102 822 428 802 363 713 220 403 2 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 440 806 4;
  • 13) 0.302 459 898 205 644 857 604 727 426 440 806 4 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 881 612 8;
  • 14) 0.604 919 796 411 289 715 209 454 852 881 612 8 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 763 225 6;
  • 15) 0.209 839 592 822 579 430 418 909 705 763 225 6 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 526 451 2;
  • 16) 0.419 679 185 645 158 860 837 819 411 526 451 2 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 052 902 4;
  • 17) 0.839 358 371 290 317 721 675 638 823 052 902 4 × 2 = 1 + 0.678 716 742 580 635 443 351 277 646 105 804 8;
  • 18) 0.678 716 742 580 635 443 351 277 646 105 804 8 × 2 = 1 + 0.357 433 485 161 270 886 702 555 292 211 609 6;
  • 19) 0.357 433 485 161 270 886 702 555 292 211 609 6 × 2 = 0 + 0.714 866 970 322 541 773 405 110 584 423 219 2;
  • 20) 0.714 866 970 322 541 773 405 110 584 423 219 2 × 2 = 1 + 0.429 733 940 645 083 546 810 221 168 846 438 4;
  • 21) 0.429 733 940 645 083 546 810 221 168 846 438 4 × 2 = 0 + 0.859 467 881 290 167 093 620 442 337 692 876 8;
  • 22) 0.859 467 881 290 167 093 620 442 337 692 876 8 × 2 = 1 + 0.718 935 762 580 334 187 240 884 675 385 753 6;
  • 23) 0.718 935 762 580 334 187 240 884 675 385 753 6 × 2 = 1 + 0.437 871 525 160 668 374 481 769 350 771 507 2;
  • 24) 0.437 871 525 160 668 374 481 769 350 771 507 2 × 2 = 0 + 0.875 743 050 321 336 748 963 538 701 543 014 4;
  • 25) 0.875 743 050 321 336 748 963 538 701 543 014 4 × 2 = 1 + 0.751 486 100 642 673 497 927 077 403 086 028 8;
  • 26) 0.751 486 100 642 673 497 927 077 403 086 028 8 × 2 = 1 + 0.502 972 201 285 346 995 854 154 806 172 057 6;
  • 27) 0.502 972 201 285 346 995 854 154 806 172 057 6 × 2 = 1 + 0.005 944 402 570 693 991 708 309 612 344 115 2;
  • 28) 0.005 944 402 570 693 991 708 309 612 344 115 2 × 2 = 0 + 0.011 888 805 141 387 983 416 619 224 688 230 4;
  • 29) 0.011 888 805 141 387 983 416 619 224 688 230 4 × 2 = 0 + 0.023 777 610 282 775 966 833 238 449 376 460 8;
  • 30) 0.023 777 610 282 775 966 833 238 449 376 460 8 × 2 = 0 + 0.047 555 220 565 551 933 666 476 898 752 921 6;
  • 31) 0.047 555 220 565 551 933 666 476 898 752 921 6 × 2 = 0 + 0.095 110 441 131 103 867 332 953 797 505 843 2;
  • 32) 0.095 110 441 131 103 867 332 953 797 505 843 2 × 2 = 0 + 0.190 220 882 262 207 734 665 907 595 011 686 4;
  • 33) 0.190 220 882 262 207 734 665 907 595 011 686 4 × 2 = 0 + 0.380 441 764 524 415 469 331 815 190 023 372 8;
  • 34) 0.380 441 764 524 415 469 331 815 190 023 372 8 × 2 = 0 + 0.760 883 529 048 830 938 663 630 380 046 745 6;
  • 35) 0.760 883 529 048 830 938 663 630 380 046 745 6 × 2 = 1 + 0.521 767 058 097 661 877 327 260 760 093 491 2;
  • 36) 0.521 767 058 097 661 877 327 260 760 093 491 2 × 2 = 1 + 0.043 534 116 195 323 754 654 521 520 186 982 4;
  • 37) 0.043 534 116 195 323 754 654 521 520 186 982 4 × 2 = 0 + 0.087 068 232 390 647 509 309 043 040 373 964 8;
  • 38) 0.087 068 232 390 647 509 309 043 040 373 964 8 × 2 = 0 + 0.174 136 464 781 295 018 618 086 080 747 929 6;
  • 39) 0.174 136 464 781 295 018 618 086 080 747 929 6 × 2 = 0 + 0.348 272 929 562 590 037 236 172 161 495 859 2;
  • 40) 0.348 272 929 562 590 037 236 172 161 495 859 2 × 2 = 0 + 0.696 545 859 125 180 074 472 344 322 991 718 4;
  • 41) 0.696 545 859 125 180 074 472 344 322 991 718 4 × 2 = 1 + 0.393 091 718 250 360 148 944 688 645 983 436 8;
  • 42) 0.393 091 718 250 360 148 944 688 645 983 436 8 × 2 = 0 + 0.786 183 436 500 720 297 889 377 291 966 873 6;
  • 43) 0.786 183 436 500 720 297 889 377 291 966 873 6 × 2 = 1 + 0.572 366 873 001 440 595 778 754 583 933 747 2;
  • 44) 0.572 366 873 001 440 595 778 754 583 933 747 2 × 2 = 1 + 0.144 733 746 002 881 191 557 509 167 867 494 4;
  • 45) 0.144 733 746 002 881 191 557 509 167 867 494 4 × 2 = 0 + 0.289 467 492 005 762 383 115 018 335 734 988 8;
  • 46) 0.289 467 492 005 762 383 115 018 335 734 988 8 × 2 = 0 + 0.578 934 984 011 524 766 230 036 671 469 977 6;
  • 47) 0.578 934 984 011 524 766 230 036 671 469 977 6 × 2 = 1 + 0.157 869 968 023 049 532 460 073 342 939 955 2;
  • 48) 0.157 869 968 023 049 532 460 073 342 939 955 2 × 2 = 0 + 0.315 739 936 046 099 064 920 146 685 879 910 4;
  • 49) 0.315 739 936 046 099 064 920 146 685 879 910 4 × 2 = 0 + 0.631 479 872 092 198 129 840 293 371 759 820 8;
  • 50) 0.631 479 872 092 198 129 840 293 371 759 820 8 × 2 = 1 + 0.262 959 744 184 396 259 680 586 743 519 641 6;
  • 51) 0.262 959 744 184 396 259 680 586 743 519 641 6 × 2 = 0 + 0.525 919 488 368 792 519 361 173 487 039 283 2;
  • 52) 0.525 919 488 368 792 519 361 173 487 039 283 2 × 2 = 1 + 0.051 838 976 737 585 038 722 346 974 078 566 4;
  • 53) 0.051 838 976 737 585 038 722 346 974 078 566 4 × 2 = 0 + 0.103 677 953 475 170 077 444 693 948 157 132 8;
  • 54) 0.103 677 953 475 170 077 444 693 948 157 132 8 × 2 = 0 + 0.207 355 906 950 340 154 889 387 896 314 265 6;
  • 55) 0.207 355 906 950 340 154 889 387 896 314 265 6 × 2 = 0 + 0.414 711 813 900 680 309 778 775 792 628 531 2;
  • 56) 0.414 711 813 900 680 309 778 775 792 628 531 2 × 2 = 0 + 0.829 423 627 801 360 619 557 551 585 257 062 4;
  • 57) 0.829 423 627 801 360 619 557 551 585 257 062 4 × 2 = 1 + 0.658 847 255 602 721 239 115 103 170 514 124 8;
  • 58) 0.658 847 255 602 721 239 115 103 170 514 124 8 × 2 = 1 + 0.317 694 511 205 442 478 230 206 341 028 249 6;
  • 59) 0.317 694 511 205 442 478 230 206 341 028 249 6 × 2 = 0 + 0.635 389 022 410 884 956 460 412 682 056 499 2;
  • 60) 0.635 389 022 410 884 956 460 412 682 056 499 2 × 2 = 1 + 0.270 778 044 821 769 912 920 825 364 112 998 4;
  • 61) 0.270 778 044 821 769 912 920 825 364 112 998 4 × 2 = 0 + 0.541 556 089 643 539 825 841 650 728 225 996 8;
  • 62) 0.541 556 089 643 539 825 841 650 728 225 996 8 × 2 = 1 + 0.083 112 179 287 079 651 683 301 456 451 993 6;
  • 63) 0.083 112 179 287 079 651 683 301 456 451 993 6 × 2 = 0 + 0.166 224 358 574 159 303 366 602 912 903 987 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).


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:


0.000 806 264 623 585 362 514 063 654 156 845 9(10) =

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

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 845 9(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:


0.000 806 264 623 585 362 514 063 654 156 845 9(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 -0.000 806 264 623 585 362 514 063 654 156 845 9 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