-0.000 806 264 623 585 362 514 063 654 156 854 74 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 806 264 623 585 362 514 063 654 156 854 74(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
-0.000 806 264 623 585 362 514 063 654 156 854 74(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:

|-0.000 806 264 623 585 362 514 063 654 156 854 74| = 0.000 806 264 623 585 362 514 063 654 156 854 74


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: 0.000 806 264 623 585 362 514 063 654 156 854 74.

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) 0.000 806 264 623 585 362 514 063 654 156 854 74 × 2 = 0 + 0.001 612 529 247 170 725 028 127 308 313 709 48;
  • 2) 0.001 612 529 247 170 725 028 127 308 313 709 48 × 2 = 0 + 0.003 225 058 494 341 450 056 254 616 627 418 96;
  • 3) 0.003 225 058 494 341 450 056 254 616 627 418 96 × 2 = 0 + 0.006 450 116 988 682 900 112 509 233 254 837 92;
  • 4) 0.006 450 116 988 682 900 112 509 233 254 837 92 × 2 = 0 + 0.012 900 233 977 365 800 225 018 466 509 675 84;
  • 5) 0.012 900 233 977 365 800 225 018 466 509 675 84 × 2 = 0 + 0.025 800 467 954 731 600 450 036 933 019 351 68;
  • 6) 0.025 800 467 954 731 600 450 036 933 019 351 68 × 2 = 0 + 0.051 600 935 909 463 200 900 073 866 038 703 36;
  • 7) 0.051 600 935 909 463 200 900 073 866 038 703 36 × 2 = 0 + 0.103 201 871 818 926 401 800 147 732 077 406 72;
  • 8) 0.103 201 871 818 926 401 800 147 732 077 406 72 × 2 = 0 + 0.206 403 743 637 852 803 600 295 464 154 813 44;
  • 9) 0.206 403 743 637 852 803 600 295 464 154 813 44 × 2 = 0 + 0.412 807 487 275 705 607 200 590 928 309 626 88;
  • 10) 0.412 807 487 275 705 607 200 590 928 309 626 88 × 2 = 0 + 0.825 614 974 551 411 214 401 181 856 619 253 76;
  • 11) 0.825 614 974 551 411 214 401 181 856 619 253 76 × 2 = 1 + 0.651 229 949 102 822 428 802 363 713 238 507 52;
  • 12) 0.651 229 949 102 822 428 802 363 713 238 507 52 × 2 = 1 + 0.302 459 898 205 644 857 604 727 426 477 015 04;
  • 13) 0.302 459 898 205 644 857 604 727 426 477 015 04 × 2 = 0 + 0.604 919 796 411 289 715 209 454 852 954 030 08;
  • 14) 0.604 919 796 411 289 715 209 454 852 954 030 08 × 2 = 1 + 0.209 839 592 822 579 430 418 909 705 908 060 16;
  • 15) 0.209 839 592 822 579 430 418 909 705 908 060 16 × 2 = 0 + 0.419 679 185 645 158 860 837 819 411 816 120 32;
  • 16) 0.419 679 185 645 158 860 837 819 411 816 120 32 × 2 = 0 + 0.839 358 371 290 317 721 675 638 823 632 240 64;
  • 17) 0.839 358 371 290 317 721 675 638 823 632 240 64 × 2 = 1 + 0.678 716 742 580 635 443 351 277 647 264 481 28;
  • 18) 0.678 716 742 580 635 443 351 277 647 264 481 28 × 2 = 1 + 0.357 433 485 161 270 886 702 555 294 528 962 56;
  • 19) 0.357 433 485 161 270 886 702 555 294 528 962 56 × 2 = 0 + 0.714 866 970 322 541 773 405 110 589 057 925 12;
  • 20) 0.714 866 970 322 541 773 405 110 589 057 925 12 × 2 = 1 + 0.429 733 940 645 083 546 810 221 178 115 850 24;
  • 21) 0.429 733 940 645 083 546 810 221 178 115 850 24 × 2 = 0 + 0.859 467 881 290 167 093 620 442 356 231 700 48;
  • 22) 0.859 467 881 290 167 093 620 442 356 231 700 48 × 2 = 1 + 0.718 935 762 580 334 187 240 884 712 463 400 96;
  • 23) 0.718 935 762 580 334 187 240 884 712 463 400 96 × 2 = 1 + 0.437 871 525 160 668 374 481 769 424 926 801 92;
  • 24) 0.437 871 525 160 668 374 481 769 424 926 801 92 × 2 = 0 + 0.875 743 050 321 336 748 963 538 849 853 603 84;
  • 25) 0.875 743 050 321 336 748 963 538 849 853 603 84 × 2 = 1 + 0.751 486 100 642 673 497 927 077 699 707 207 68;
  • 26) 0.751 486 100 642 673 497 927 077 699 707 207 68 × 2 = 1 + 0.502 972 201 285 346 995 854 155 399 414 415 36;
  • 27) 0.502 972 201 285 346 995 854 155 399 414 415 36 × 2 = 1 + 0.005 944 402 570 693 991 708 310 798 828 830 72;
  • 28) 0.005 944 402 570 693 991 708 310 798 828 830 72 × 2 = 0 + 0.011 888 805 141 387 983 416 621 597 657 661 44;
  • 29) 0.011 888 805 141 387 983 416 621 597 657 661 44 × 2 = 0 + 0.023 777 610 282 775 966 833 243 195 315 322 88;
  • 30) 0.023 777 610 282 775 966 833 243 195 315 322 88 × 2 = 0 + 0.047 555 220 565 551 933 666 486 390 630 645 76;
  • 31) 0.047 555 220 565 551 933 666 486 390 630 645 76 × 2 = 0 + 0.095 110 441 131 103 867 332 972 781 261 291 52;
  • 32) 0.095 110 441 131 103 867 332 972 781 261 291 52 × 2 = 0 + 0.190 220 882 262 207 734 665 945 562 522 583 04;
  • 33) 0.190 220 882 262 207 734 665 945 562 522 583 04 × 2 = 0 + 0.380 441 764 524 415 469 331 891 125 045 166 08;
  • 34) 0.380 441 764 524 415 469 331 891 125 045 166 08 × 2 = 0 + 0.760 883 529 048 830 938 663 782 250 090 332 16;
  • 35) 0.760 883 529 048 830 938 663 782 250 090 332 16 × 2 = 1 + 0.521 767 058 097 661 877 327 564 500 180 664 32;
  • 36) 0.521 767 058 097 661 877 327 564 500 180 664 32 × 2 = 1 + 0.043 534 116 195 323 754 655 129 000 361 328 64;
  • 37) 0.043 534 116 195 323 754 655 129 000 361 328 64 × 2 = 0 + 0.087 068 232 390 647 509 310 258 000 722 657 28;
  • 38) 0.087 068 232 390 647 509 310 258 000 722 657 28 × 2 = 0 + 0.174 136 464 781 295 018 620 516 001 445 314 56;
  • 39) 0.174 136 464 781 295 018 620 516 001 445 314 56 × 2 = 0 + 0.348 272 929 562 590 037 241 032 002 890 629 12;
  • 40) 0.348 272 929 562 590 037 241 032 002 890 629 12 × 2 = 0 + 0.696 545 859 125 180 074 482 064 005 781 258 24;
  • 41) 0.696 545 859 125 180 074 482 064 005 781 258 24 × 2 = 1 + 0.393 091 718 250 360 148 964 128 011 562 516 48;
  • 42) 0.393 091 718 250 360 148 964 128 011 562 516 48 × 2 = 0 + 0.786 183 436 500 720 297 928 256 023 125 032 96;
  • 43) 0.786 183 436 500 720 297 928 256 023 125 032 96 × 2 = 1 + 0.572 366 873 001 440 595 856 512 046 250 065 92;
  • 44) 0.572 366 873 001 440 595 856 512 046 250 065 92 × 2 = 1 + 0.144 733 746 002 881 191 713 024 092 500 131 84;
  • 45) 0.144 733 746 002 881 191 713 024 092 500 131 84 × 2 = 0 + 0.289 467 492 005 762 383 426 048 185 000 263 68;
  • 46) 0.289 467 492 005 762 383 426 048 185 000 263 68 × 2 = 0 + 0.578 934 984 011 524 766 852 096 370 000 527 36;
  • 47) 0.578 934 984 011 524 766 852 096 370 000 527 36 × 2 = 1 + 0.157 869 968 023 049 533 704 192 740 001 054 72;
  • 48) 0.157 869 968 023 049 533 704 192 740 001 054 72 × 2 = 0 + 0.315 739 936 046 099 067 408 385 480 002 109 44;
  • 49) 0.315 739 936 046 099 067 408 385 480 002 109 44 × 2 = 0 + 0.631 479 872 092 198 134 816 770 960 004 218 88;
  • 50) 0.631 479 872 092 198 134 816 770 960 004 218 88 × 2 = 1 + 0.262 959 744 184 396 269 633 541 920 008 437 76;
  • 51) 0.262 959 744 184 396 269 633 541 920 008 437 76 × 2 = 0 + 0.525 919 488 368 792 539 267 083 840 016 875 52;
  • 52) 0.525 919 488 368 792 539 267 083 840 016 875 52 × 2 = 1 + 0.051 838 976 737 585 078 534 167 680 033 751 04;
  • 53) 0.051 838 976 737 585 078 534 167 680 033 751 04 × 2 = 0 + 0.103 677 953 475 170 157 068 335 360 067 502 08;
  • 54) 0.103 677 953 475 170 157 068 335 360 067 502 08 × 2 = 0 + 0.207 355 906 950 340 314 136 670 720 135 004 16;
  • 55) 0.207 355 906 950 340 314 136 670 720 135 004 16 × 2 = 0 + 0.414 711 813 900 680 628 273 341 440 270 008 32;
  • 56) 0.414 711 813 900 680 628 273 341 440 270 008 32 × 2 = 0 + 0.829 423 627 801 361 256 546 682 880 540 016 64;
  • 57) 0.829 423 627 801 361 256 546 682 880 540 016 64 × 2 = 1 + 0.658 847 255 602 722 513 093 365 761 080 033 28;
  • 58) 0.658 847 255 602 722 513 093 365 761 080 033 28 × 2 = 1 + 0.317 694 511 205 445 026 186 731 522 160 066 56;
  • 59) 0.317 694 511 205 445 026 186 731 522 160 066 56 × 2 = 0 + 0.635 389 022 410 890 052 373 463 044 320 133 12;
  • 60) 0.635 389 022 410 890 052 373 463 044 320 133 12 × 2 = 1 + 0.270 778 044 821 780 104 746 926 088 640 266 24;
  • 61) 0.270 778 044 821 780 104 746 926 088 640 266 24 × 2 = 0 + 0.541 556 089 643 560 209 493 852 177 280 532 48;
  • 62) 0.541 556 089 643 560 209 493 852 177 280 532 48 × 2 = 1 + 0.083 112 179 287 120 418 987 704 354 561 064 96;
  • 63) 0.083 112 179 287 120 418 987 704 354 561 064 96 × 2 = 0 + 0.166 224 358 574 240 837 975 408 709 122 129 92;

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:


0.000 806 264 623 585 362 514 063 654 156 854 74(10) =


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

6. Positive number before normalization:

0.000 806 264 623 585 362 514 063 654 156 854 74(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:


0.000 806 264 623 585 362 514 063 654 156 854 74(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 -0.000 806 264 623 585 362 514 063 654 156 854 74 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


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