0.000 020 830 729 321 671 205 134 999 154 508 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 154 508 5(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
0.000 020 830 729 321 671 205 134 999 154 508 5(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: 0.000 020 830 729 321 671 205 134 999 154 508 5.

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) 0.000 020 830 729 321 671 205 134 999 154 508 5 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 017;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 017 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 034;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 034 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 236 068;
  • 4) 0.000 166 645 834 573 369 641 079 993 236 068 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 472 136;
  • 5) 0.000 333 291 669 146 739 282 159 986 472 136 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 944 272;
  • 6) 0.000 666 583 338 293 478 564 319 972 944 272 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 888 544;
  • 7) 0.001 333 166 676 586 957 128 639 945 888 544 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 777 088;
  • 8) 0.002 666 333 353 173 914 257 279 891 777 088 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 554 176;
  • 9) 0.005 332 666 706 347 828 514 559 783 554 176 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 108 352;
  • 10) 0.010 665 333 412 695 657 029 119 567 108 352 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 216 704;
  • 11) 0.021 330 666 825 391 314 058 239 134 216 704 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 433 408;
  • 12) 0.042 661 333 650 782 628 116 478 268 433 408 × 2 = 0 + 0.085 322 667 301 565 256 232 956 536 866 816;
  • 13) 0.085 322 667 301 565 256 232 956 536 866 816 × 2 = 0 + 0.170 645 334 603 130 512 465 913 073 733 632;
  • 14) 0.170 645 334 603 130 512 465 913 073 733 632 × 2 = 0 + 0.341 290 669 206 261 024 931 826 147 467 264;
  • 15) 0.341 290 669 206 261 024 931 826 147 467 264 × 2 = 0 + 0.682 581 338 412 522 049 863 652 294 934 528;
  • 16) 0.682 581 338 412 522 049 863 652 294 934 528 × 2 = 1 + 0.365 162 676 825 044 099 727 304 589 869 056;
  • 17) 0.365 162 676 825 044 099 727 304 589 869 056 × 2 = 0 + 0.730 325 353 650 088 199 454 609 179 738 112;
  • 18) 0.730 325 353 650 088 199 454 609 179 738 112 × 2 = 1 + 0.460 650 707 300 176 398 909 218 359 476 224;
  • 19) 0.460 650 707 300 176 398 909 218 359 476 224 × 2 = 0 + 0.921 301 414 600 352 797 818 436 718 952 448;
  • 20) 0.921 301 414 600 352 797 818 436 718 952 448 × 2 = 1 + 0.842 602 829 200 705 595 636 873 437 904 896;
  • 21) 0.842 602 829 200 705 595 636 873 437 904 896 × 2 = 1 + 0.685 205 658 401 411 191 273 746 875 809 792;
  • 22) 0.685 205 658 401 411 191 273 746 875 809 792 × 2 = 1 + 0.370 411 316 802 822 382 547 493 751 619 584;
  • 23) 0.370 411 316 802 822 382 547 493 751 619 584 × 2 = 0 + 0.740 822 633 605 644 765 094 987 503 239 168;
  • 24) 0.740 822 633 605 644 765 094 987 503 239 168 × 2 = 1 + 0.481 645 267 211 289 530 189 975 006 478 336;
  • 25) 0.481 645 267 211 289 530 189 975 006 478 336 × 2 = 0 + 0.963 290 534 422 579 060 379 950 012 956 672;
  • 26) 0.963 290 534 422 579 060 379 950 012 956 672 × 2 = 1 + 0.926 581 068 845 158 120 759 900 025 913 344;
  • 27) 0.926 581 068 845 158 120 759 900 025 913 344 × 2 = 1 + 0.853 162 137 690 316 241 519 800 051 826 688;
  • 28) 0.853 162 137 690 316 241 519 800 051 826 688 × 2 = 1 + 0.706 324 275 380 632 483 039 600 103 653 376;
  • 29) 0.706 324 275 380 632 483 039 600 103 653 376 × 2 = 1 + 0.412 648 550 761 264 966 079 200 207 306 752;
  • 30) 0.412 648 550 761 264 966 079 200 207 306 752 × 2 = 0 + 0.825 297 101 522 529 932 158 400 414 613 504;
  • 31) 0.825 297 101 522 529 932 158 400 414 613 504 × 2 = 1 + 0.650 594 203 045 059 864 316 800 829 227 008;
  • 32) 0.650 594 203 045 059 864 316 800 829 227 008 × 2 = 1 + 0.301 188 406 090 119 728 633 601 658 454 016;
  • 33) 0.301 188 406 090 119 728 633 601 658 454 016 × 2 = 0 + 0.602 376 812 180 239 457 267 203 316 908 032;
  • 34) 0.602 376 812 180 239 457 267 203 316 908 032 × 2 = 1 + 0.204 753 624 360 478 914 534 406 633 816 064;
  • 35) 0.204 753 624 360 478 914 534 406 633 816 064 × 2 = 0 + 0.409 507 248 720 957 829 068 813 267 632 128;
  • 36) 0.409 507 248 720 957 829 068 813 267 632 128 × 2 = 0 + 0.819 014 497 441 915 658 137 626 535 264 256;
  • 37) 0.819 014 497 441 915 658 137 626 535 264 256 × 2 = 1 + 0.638 028 994 883 831 316 275 253 070 528 512;
  • 38) 0.638 028 994 883 831 316 275 253 070 528 512 × 2 = 1 + 0.276 057 989 767 662 632 550 506 141 057 024;
  • 39) 0.276 057 989 767 662 632 550 506 141 057 024 × 2 = 0 + 0.552 115 979 535 325 265 101 012 282 114 048;
  • 40) 0.552 115 979 535 325 265 101 012 282 114 048 × 2 = 1 + 0.104 231 959 070 650 530 202 024 564 228 096;
  • 41) 0.104 231 959 070 650 530 202 024 564 228 096 × 2 = 0 + 0.208 463 918 141 301 060 404 049 128 456 192;
  • 42) 0.208 463 918 141 301 060 404 049 128 456 192 × 2 = 0 + 0.416 927 836 282 602 120 808 098 256 912 384;
  • 43) 0.416 927 836 282 602 120 808 098 256 912 384 × 2 = 0 + 0.833 855 672 565 204 241 616 196 513 824 768;
  • 44) 0.833 855 672 565 204 241 616 196 513 824 768 × 2 = 1 + 0.667 711 345 130 408 483 232 393 027 649 536;
  • 45) 0.667 711 345 130 408 483 232 393 027 649 536 × 2 = 1 + 0.335 422 690 260 816 966 464 786 055 299 072;
  • 46) 0.335 422 690 260 816 966 464 786 055 299 072 × 2 = 0 + 0.670 845 380 521 633 932 929 572 110 598 144;
  • 47) 0.670 845 380 521 633 932 929 572 110 598 144 × 2 = 1 + 0.341 690 761 043 267 865 859 144 221 196 288;
  • 48) 0.341 690 761 043 267 865 859 144 221 196 288 × 2 = 0 + 0.683 381 522 086 535 731 718 288 442 392 576;
  • 49) 0.683 381 522 086 535 731 718 288 442 392 576 × 2 = 1 + 0.366 763 044 173 071 463 436 576 884 785 152;
  • 50) 0.366 763 044 173 071 463 436 576 884 785 152 × 2 = 0 + 0.733 526 088 346 142 926 873 153 769 570 304;
  • 51) 0.733 526 088 346 142 926 873 153 769 570 304 × 2 = 1 + 0.467 052 176 692 285 853 746 307 539 140 608;
  • 52) 0.467 052 176 692 285 853 746 307 539 140 608 × 2 = 0 + 0.934 104 353 384 571 707 492 615 078 281 216;
  • 53) 0.934 104 353 384 571 707 492 615 078 281 216 × 2 = 1 + 0.868 208 706 769 143 414 985 230 156 562 432;
  • 54) 0.868 208 706 769 143 414 985 230 156 562 432 × 2 = 1 + 0.736 417 413 538 286 829 970 460 313 124 864;
  • 55) 0.736 417 413 538 286 829 970 460 313 124 864 × 2 = 1 + 0.472 834 827 076 573 659 940 920 626 249 728;
  • 56) 0.472 834 827 076 573 659 940 920 626 249 728 × 2 = 0 + 0.945 669 654 153 147 319 881 841 252 499 456;
  • 57) 0.945 669 654 153 147 319 881 841 252 499 456 × 2 = 1 + 0.891 339 308 306 294 639 763 682 504 998 912;
  • 58) 0.891 339 308 306 294 639 763 682 504 998 912 × 2 = 1 + 0.782 678 616 612 589 279 527 365 009 997 824;
  • 59) 0.782 678 616 612 589 279 527 365 009 997 824 × 2 = 1 + 0.565 357 233 225 178 559 054 730 019 995 648;
  • 60) 0.565 357 233 225 178 559 054 730 019 995 648 × 2 = 1 + 0.130 714 466 450 357 118 109 460 039 991 296;
  • 61) 0.130 714 466 450 357 118 109 460 039 991 296 × 2 = 0 + 0.261 428 932 900 714 236 218 920 079 982 592;
  • 62) 0.261 428 932 900 714 236 218 920 079 982 592 × 2 = 0 + 0.522 857 865 801 428 472 437 840 159 965 184;
  • 63) 0.522 857 865 801 428 472 437 840 159 965 184 × 2 = 1 + 0.045 715 731 602 856 944 875 680 319 930 368;
  • 64) 0.045 715 731 602 856 944 875 680 319 930 368 × 2 = 0 + 0.091 431 463 205 713 889 751 360 639 860 736;
  • 65) 0.091 431 463 205 713 889 751 360 639 860 736 × 2 = 0 + 0.182 862 926 411 427 779 502 721 279 721 472;
  • 66) 0.182 862 926 411 427 779 502 721 279 721 472 × 2 = 0 + 0.365 725 852 822 855 559 005 442 559 442 944;
  • 67) 0.365 725 852 822 855 559 005 442 559 442 944 × 2 = 0 + 0.731 451 705 645 711 118 010 885 118 885 888;
  • 68) 0.731 451 705 645 711 118 010 885 118 885 888 × 2 = 1 + 0.462 903 411 291 422 236 021 770 237 771 776;

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:


0.000 020 830 729 321 671 205 134 999 154 508 5(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:

0.000 020 830 729 321 671 205 134 999 154 508 5(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:


0.000 020 830 729 321 671 205 134 999 154 508 5(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 0.000 020 830 729 321 671 205 134 999 154 508 5 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