0.000 020 830 729 321 671 205 134 999 154 496 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 496(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
0.000 020 830 729 321 671 205 134 999 154 496(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: 0.000 020 830 729 321 671 205 134 999 154 496.

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) 0.000 020 830 729 321 671 205 134 999 154 496 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 308 992;
  • 2) 0.000 041 661 458 643 342 410 269 998 308 992 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 617 984;
  • 3) 0.000 083 322 917 286 684 820 539 996 617 984 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 235 968;
  • 4) 0.000 166 645 834 573 369 641 079 993 235 968 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 471 936;
  • 5) 0.000 333 291 669 146 739 282 159 986 471 936 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 943 872;
  • 6) 0.000 666 583 338 293 478 564 319 972 943 872 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 887 744;
  • 7) 0.001 333 166 676 586 957 128 639 945 887 744 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 775 488;
  • 8) 0.002 666 333 353 173 914 257 279 891 775 488 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 550 976;
  • 9) 0.005 332 666 706 347 828 514 559 783 550 976 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 101 952;
  • 10) 0.010 665 333 412 695 657 029 119 567 101 952 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 203 904;
  • 11) 0.021 330 666 825 391 314 058 239 134 203 904 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 407 808;
  • 12) 0.042 661 333 650 782 628 116 478 268 407 808 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 815 616;
  • 13) 0.085 322 667 301 565 256 232 956 536 815 616 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 631 232;
  • 14) 0.170 645 334 603 130 512 465 913 073 631 232 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 262 464;
  • 15) 0.341 290 669 206 261 024 931 826 147 262 464 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 524 928;
  • 16) 0.682 581 338 412 522 049 863 652 294 524 928 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 049 856;
  • 17) 0.365 162 676 825 044 099 727 304 589 049 856 × 2 = 0 + 0.730 325 353 650 088 199 454 609 178 099 712;
  • 18) 0.730 325 353 650 088 199 454 609 178 099 712 × 2 = 1 + 0.460 650 707 300 176 398 909 218 356 199 424;
  • 19) 0.460 650 707 300 176 398 909 218 356 199 424 × 2 = 0 + 0.921 301 414 600 352 797 818 436 712 398 848;
  • 20) 0.921 301 414 600 352 797 818 436 712 398 848 × 2 = 1 + 0.842 602 829 200 705 595 636 873 424 797 696;
  • 21) 0.842 602 829 200 705 595 636 873 424 797 696 × 2 = 1 + 0.685 205 658 401 411 191 273 746 849 595 392;
  • 22) 0.685 205 658 401 411 191 273 746 849 595 392 × 2 = 1 + 0.370 411 316 802 822 382 547 493 699 190 784;
  • 23) 0.370 411 316 802 822 382 547 493 699 190 784 × 2 = 0 + 0.740 822 633 605 644 765 094 987 398 381 568;
  • 24) 0.740 822 633 605 644 765 094 987 398 381 568 × 2 = 1 + 0.481 645 267 211 289 530 189 974 796 763 136;
  • 25) 0.481 645 267 211 289 530 189 974 796 763 136 × 2 = 0 + 0.963 290 534 422 579 060 379 949 593 526 272;
  • 26) 0.963 290 534 422 579 060 379 949 593 526 272 × 2 = 1 + 0.926 581 068 845 158 120 759 899 187 052 544;
  • 27) 0.926 581 068 845 158 120 759 899 187 052 544 × 2 = 1 + 0.853 162 137 690 316 241 519 798 374 105 088;
  • 28) 0.853 162 137 690 316 241 519 798 374 105 088 × 2 = 1 + 0.706 324 275 380 632 483 039 596 748 210 176;
  • 29) 0.706 324 275 380 632 483 039 596 748 210 176 × 2 = 1 + 0.412 648 550 761 264 966 079 193 496 420 352;
  • 30) 0.412 648 550 761 264 966 079 193 496 420 352 × 2 = 0 + 0.825 297 101 522 529 932 158 386 992 840 704;
  • 31) 0.825 297 101 522 529 932 158 386 992 840 704 × 2 = 1 + 0.650 594 203 045 059 864 316 773 985 681 408;
  • 32) 0.650 594 203 045 059 864 316 773 985 681 408 × 2 = 1 + 0.301 188 406 090 119 728 633 547 971 362 816;
  • 33) 0.301 188 406 090 119 728 633 547 971 362 816 × 2 = 0 + 0.602 376 812 180 239 457 267 095 942 725 632;
  • 34) 0.602 376 812 180 239 457 267 095 942 725 632 × 2 = 1 + 0.204 753 624 360 478 914 534 191 885 451 264;
  • 35) 0.204 753 624 360 478 914 534 191 885 451 264 × 2 = 0 + 0.409 507 248 720 957 829 068 383 770 902 528;
  • 36) 0.409 507 248 720 957 829 068 383 770 902 528 × 2 = 0 + 0.819 014 497 441 915 658 136 767 541 805 056;
  • 37) 0.819 014 497 441 915 658 136 767 541 805 056 × 2 = 1 + 0.638 028 994 883 831 316 273 535 083 610 112;
  • 38) 0.638 028 994 883 831 316 273 535 083 610 112 × 2 = 1 + 0.276 057 989 767 662 632 547 070 167 220 224;
  • 39) 0.276 057 989 767 662 632 547 070 167 220 224 × 2 = 0 + 0.552 115 979 535 325 265 094 140 334 440 448;
  • 40) 0.552 115 979 535 325 265 094 140 334 440 448 × 2 = 1 + 0.104 231 959 070 650 530 188 280 668 880 896;
  • 41) 0.104 231 959 070 650 530 188 280 668 880 896 × 2 = 0 + 0.208 463 918 141 301 060 376 561 337 761 792;
  • 42) 0.208 463 918 141 301 060 376 561 337 761 792 × 2 = 0 + 0.416 927 836 282 602 120 753 122 675 523 584;
  • 43) 0.416 927 836 282 602 120 753 122 675 523 584 × 2 = 0 + 0.833 855 672 565 204 241 506 245 351 047 168;
  • 44) 0.833 855 672 565 204 241 506 245 351 047 168 × 2 = 1 + 0.667 711 345 130 408 483 012 490 702 094 336;
  • 45) 0.667 711 345 130 408 483 012 490 702 094 336 × 2 = 1 + 0.335 422 690 260 816 966 024 981 404 188 672;
  • 46) 0.335 422 690 260 816 966 024 981 404 188 672 × 2 = 0 + 0.670 845 380 521 633 932 049 962 808 377 344;
  • 47) 0.670 845 380 521 633 932 049 962 808 377 344 × 2 = 1 + 0.341 690 761 043 267 864 099 925 616 754 688;
  • 48) 0.341 690 761 043 267 864 099 925 616 754 688 × 2 = 0 + 0.683 381 522 086 535 728 199 851 233 509 376;
  • 49) 0.683 381 522 086 535 728 199 851 233 509 376 × 2 = 1 + 0.366 763 044 173 071 456 399 702 467 018 752;
  • 50) 0.366 763 044 173 071 456 399 702 467 018 752 × 2 = 0 + 0.733 526 088 346 142 912 799 404 934 037 504;
  • 51) 0.733 526 088 346 142 912 799 404 934 037 504 × 2 = 1 + 0.467 052 176 692 285 825 598 809 868 075 008;
  • 52) 0.467 052 176 692 285 825 598 809 868 075 008 × 2 = 0 + 0.934 104 353 384 571 651 197 619 736 150 016;
  • 53) 0.934 104 353 384 571 651 197 619 736 150 016 × 2 = 1 + 0.868 208 706 769 143 302 395 239 472 300 032;
  • 54) 0.868 208 706 769 143 302 395 239 472 300 032 × 2 = 1 + 0.736 417 413 538 286 604 790 478 944 600 064;
  • 55) 0.736 417 413 538 286 604 790 478 944 600 064 × 2 = 1 + 0.472 834 827 076 573 209 580 957 889 200 128;
  • 56) 0.472 834 827 076 573 209 580 957 889 200 128 × 2 = 0 + 0.945 669 654 153 146 419 161 915 778 400 256;
  • 57) 0.945 669 654 153 146 419 161 915 778 400 256 × 2 = 1 + 0.891 339 308 306 292 838 323 831 556 800 512;
  • 58) 0.891 339 308 306 292 838 323 831 556 800 512 × 2 = 1 + 0.782 678 616 612 585 676 647 663 113 601 024;
  • 59) 0.782 678 616 612 585 676 647 663 113 601 024 × 2 = 1 + 0.565 357 233 225 171 353 295 326 227 202 048;
  • 60) 0.565 357 233 225 171 353 295 326 227 202 048 × 2 = 1 + 0.130 714 466 450 342 706 590 652 454 404 096;
  • 61) 0.130 714 466 450 342 706 590 652 454 404 096 × 2 = 0 + 0.261 428 932 900 685 413 181 304 908 808 192;
  • 62) 0.261 428 932 900 685 413 181 304 908 808 192 × 2 = 0 + 0.522 857 865 801 370 826 362 609 817 616 384;
  • 63) 0.522 857 865 801 370 826 362 609 817 616 384 × 2 = 1 + 0.045 715 731 602 741 652 725 219 635 232 768;
  • 64) 0.045 715 731 602 741 652 725 219 635 232 768 × 2 = 0 + 0.091 431 463 205 483 305 450 439 270 465 536;
  • 65) 0.091 431 463 205 483 305 450 439 270 465 536 × 2 = 0 + 0.182 862 926 410 966 610 900 878 540 931 072;
  • 66) 0.182 862 926 410 966 610 900 878 540 931 072 × 2 = 0 + 0.365 725 852 821 933 221 801 757 081 862 144;
  • 67) 0.365 725 852 821 933 221 801 757 081 862 144 × 2 = 0 + 0.731 451 705 643 866 443 603 514 163 724 288;
  • 68) 0.731 451 705 643 866 443 603 514 163 724 288 × 2 = 1 + 0.462 903 411 287 732 887 207 028 327 448 576;

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:


0.000 020 830 729 321 671 205 134 999 154 496(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:

0.000 020 830 729 321 671 205 134 999 154 496(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:


0.000 020 830 729 321 671 205 134 999 154 496(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 0.000 020 830 729 321 671 205 134 999 154 496 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