0.000 020 830 729 321 671 205 134 999 154 484 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 484(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
0.000 020 830 729 321 671 205 134 999 154 484(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: 0.000 020 830 729 321 671 205 134 999 154 484.

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) 0.000 020 830 729 321 671 205 134 999 154 484 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 308 968;
  • 2) 0.000 041 661 458 643 342 410 269 998 308 968 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 617 936;
  • 3) 0.000 083 322 917 286 684 820 539 996 617 936 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 235 872;
  • 4) 0.000 166 645 834 573 369 641 079 993 235 872 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 471 744;
  • 5) 0.000 333 291 669 146 739 282 159 986 471 744 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 943 488;
  • 6) 0.000 666 583 338 293 478 564 319 972 943 488 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 886 976;
  • 7) 0.001 333 166 676 586 957 128 639 945 886 976 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 773 952;
  • 8) 0.002 666 333 353 173 914 257 279 891 773 952 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 547 904;
  • 9) 0.005 332 666 706 347 828 514 559 783 547 904 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 095 808;
  • 10) 0.010 665 333 412 695 657 029 119 567 095 808 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 191 616;
  • 11) 0.021 330 666 825 391 314 058 239 134 191 616 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 383 232;
  • 12) 0.042 661 333 650 782 628 116 478 268 383 232 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 766 464;
  • 13) 0.085 322 667 301 565 256 232 956 536 766 464 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 532 928;
  • 14) 0.170 645 334 603 130 512 465 913 073 532 928 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 065 856;
  • 15) 0.341 290 669 206 261 024 931 826 147 065 856 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 131 712;
  • 16) 0.682 581 338 412 522 049 863 652 294 131 712 × 2 = 1 + 0.365 162 676 825 044 099 727 304 588 263 424;
  • 17) 0.365 162 676 825 044 099 727 304 588 263 424 × 2 = 0 + 0.730 325 353 650 088 199 454 609 176 526 848;
  • 18) 0.730 325 353 650 088 199 454 609 176 526 848 × 2 = 1 + 0.460 650 707 300 176 398 909 218 353 053 696;
  • 19) 0.460 650 707 300 176 398 909 218 353 053 696 × 2 = 0 + 0.921 301 414 600 352 797 818 436 706 107 392;
  • 20) 0.921 301 414 600 352 797 818 436 706 107 392 × 2 = 1 + 0.842 602 829 200 705 595 636 873 412 214 784;
  • 21) 0.842 602 829 200 705 595 636 873 412 214 784 × 2 = 1 + 0.685 205 658 401 411 191 273 746 824 429 568;
  • 22) 0.685 205 658 401 411 191 273 746 824 429 568 × 2 = 1 + 0.370 411 316 802 822 382 547 493 648 859 136;
  • 23) 0.370 411 316 802 822 382 547 493 648 859 136 × 2 = 0 + 0.740 822 633 605 644 765 094 987 297 718 272;
  • 24) 0.740 822 633 605 644 765 094 987 297 718 272 × 2 = 1 + 0.481 645 267 211 289 530 189 974 595 436 544;
  • 25) 0.481 645 267 211 289 530 189 974 595 436 544 × 2 = 0 + 0.963 290 534 422 579 060 379 949 190 873 088;
  • 26) 0.963 290 534 422 579 060 379 949 190 873 088 × 2 = 1 + 0.926 581 068 845 158 120 759 898 381 746 176;
  • 27) 0.926 581 068 845 158 120 759 898 381 746 176 × 2 = 1 + 0.853 162 137 690 316 241 519 796 763 492 352;
  • 28) 0.853 162 137 690 316 241 519 796 763 492 352 × 2 = 1 + 0.706 324 275 380 632 483 039 593 526 984 704;
  • 29) 0.706 324 275 380 632 483 039 593 526 984 704 × 2 = 1 + 0.412 648 550 761 264 966 079 187 053 969 408;
  • 30) 0.412 648 550 761 264 966 079 187 053 969 408 × 2 = 0 + 0.825 297 101 522 529 932 158 374 107 938 816;
  • 31) 0.825 297 101 522 529 932 158 374 107 938 816 × 2 = 1 + 0.650 594 203 045 059 864 316 748 215 877 632;
  • 32) 0.650 594 203 045 059 864 316 748 215 877 632 × 2 = 1 + 0.301 188 406 090 119 728 633 496 431 755 264;
  • 33) 0.301 188 406 090 119 728 633 496 431 755 264 × 2 = 0 + 0.602 376 812 180 239 457 266 992 863 510 528;
  • 34) 0.602 376 812 180 239 457 266 992 863 510 528 × 2 = 1 + 0.204 753 624 360 478 914 533 985 727 021 056;
  • 35) 0.204 753 624 360 478 914 533 985 727 021 056 × 2 = 0 + 0.409 507 248 720 957 829 067 971 454 042 112;
  • 36) 0.409 507 248 720 957 829 067 971 454 042 112 × 2 = 0 + 0.819 014 497 441 915 658 135 942 908 084 224;
  • 37) 0.819 014 497 441 915 658 135 942 908 084 224 × 2 = 1 + 0.638 028 994 883 831 316 271 885 816 168 448;
  • 38) 0.638 028 994 883 831 316 271 885 816 168 448 × 2 = 1 + 0.276 057 989 767 662 632 543 771 632 336 896;
  • 39) 0.276 057 989 767 662 632 543 771 632 336 896 × 2 = 0 + 0.552 115 979 535 325 265 087 543 264 673 792;
  • 40) 0.552 115 979 535 325 265 087 543 264 673 792 × 2 = 1 + 0.104 231 959 070 650 530 175 086 529 347 584;
  • 41) 0.104 231 959 070 650 530 175 086 529 347 584 × 2 = 0 + 0.208 463 918 141 301 060 350 173 058 695 168;
  • 42) 0.208 463 918 141 301 060 350 173 058 695 168 × 2 = 0 + 0.416 927 836 282 602 120 700 346 117 390 336;
  • 43) 0.416 927 836 282 602 120 700 346 117 390 336 × 2 = 0 + 0.833 855 672 565 204 241 400 692 234 780 672;
  • 44) 0.833 855 672 565 204 241 400 692 234 780 672 × 2 = 1 + 0.667 711 345 130 408 482 801 384 469 561 344;
  • 45) 0.667 711 345 130 408 482 801 384 469 561 344 × 2 = 1 + 0.335 422 690 260 816 965 602 768 939 122 688;
  • 46) 0.335 422 690 260 816 965 602 768 939 122 688 × 2 = 0 + 0.670 845 380 521 633 931 205 537 878 245 376;
  • 47) 0.670 845 380 521 633 931 205 537 878 245 376 × 2 = 1 + 0.341 690 761 043 267 862 411 075 756 490 752;
  • 48) 0.341 690 761 043 267 862 411 075 756 490 752 × 2 = 0 + 0.683 381 522 086 535 724 822 151 512 981 504;
  • 49) 0.683 381 522 086 535 724 822 151 512 981 504 × 2 = 1 + 0.366 763 044 173 071 449 644 303 025 963 008;
  • 50) 0.366 763 044 173 071 449 644 303 025 963 008 × 2 = 0 + 0.733 526 088 346 142 899 288 606 051 926 016;
  • 51) 0.733 526 088 346 142 899 288 606 051 926 016 × 2 = 1 + 0.467 052 176 692 285 798 577 212 103 852 032;
  • 52) 0.467 052 176 692 285 798 577 212 103 852 032 × 2 = 0 + 0.934 104 353 384 571 597 154 424 207 704 064;
  • 53) 0.934 104 353 384 571 597 154 424 207 704 064 × 2 = 1 + 0.868 208 706 769 143 194 308 848 415 408 128;
  • 54) 0.868 208 706 769 143 194 308 848 415 408 128 × 2 = 1 + 0.736 417 413 538 286 388 617 696 830 816 256;
  • 55) 0.736 417 413 538 286 388 617 696 830 816 256 × 2 = 1 + 0.472 834 827 076 572 777 235 393 661 632 512;
  • 56) 0.472 834 827 076 572 777 235 393 661 632 512 × 2 = 0 + 0.945 669 654 153 145 554 470 787 323 265 024;
  • 57) 0.945 669 654 153 145 554 470 787 323 265 024 × 2 = 1 + 0.891 339 308 306 291 108 941 574 646 530 048;
  • 58) 0.891 339 308 306 291 108 941 574 646 530 048 × 2 = 1 + 0.782 678 616 612 582 217 883 149 293 060 096;
  • 59) 0.782 678 616 612 582 217 883 149 293 060 096 × 2 = 1 + 0.565 357 233 225 164 435 766 298 586 120 192;
  • 60) 0.565 357 233 225 164 435 766 298 586 120 192 × 2 = 1 + 0.130 714 466 450 328 871 532 597 172 240 384;
  • 61) 0.130 714 466 450 328 871 532 597 172 240 384 × 2 = 0 + 0.261 428 932 900 657 743 065 194 344 480 768;
  • 62) 0.261 428 932 900 657 743 065 194 344 480 768 × 2 = 0 + 0.522 857 865 801 315 486 130 388 688 961 536;
  • 63) 0.522 857 865 801 315 486 130 388 688 961 536 × 2 = 1 + 0.045 715 731 602 630 972 260 777 377 923 072;
  • 64) 0.045 715 731 602 630 972 260 777 377 923 072 × 2 = 0 + 0.091 431 463 205 261 944 521 554 755 846 144;
  • 65) 0.091 431 463 205 261 944 521 554 755 846 144 × 2 = 0 + 0.182 862 926 410 523 889 043 109 511 692 288;
  • 66) 0.182 862 926 410 523 889 043 109 511 692 288 × 2 = 0 + 0.365 725 852 821 047 778 086 219 023 384 576;
  • 67) 0.365 725 852 821 047 778 086 219 023 384 576 × 2 = 0 + 0.731 451 705 642 095 556 172 438 046 769 152;
  • 68) 0.731 451 705 642 095 556 172 438 046 769 152 × 2 = 1 + 0.462 903 411 284 191 112 344 876 093 538 304;

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:


0.000 020 830 729 321 671 205 134 999 154 484(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:

0.000 020 830 729 321 671 205 134 999 154 484(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:


0.000 020 830 729 321 671 205 134 999 154 484(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 0.000 020 830 729 321 671 205 134 999 154 484 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