0.000 020 830 729 321 671 205 134 999 154 7 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 7(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 7(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 7.

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 7 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 309 4;
  • 2) 0.000 041 661 458 643 342 410 269 998 309 4 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 618 8;
  • 3) 0.000 083 322 917 286 684 820 539 996 618 8 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 237 6;
  • 4) 0.000 166 645 834 573 369 641 079 993 237 6 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 475 2;
  • 5) 0.000 333 291 669 146 739 282 159 986 475 2 × 2 = 0 + 0.000 666 583 338 293 478 564 319 972 950 4;
  • 6) 0.000 666 583 338 293 478 564 319 972 950 4 × 2 = 0 + 0.001 333 166 676 586 957 128 639 945 900 8;
  • 7) 0.001 333 166 676 586 957 128 639 945 900 8 × 2 = 0 + 0.002 666 333 353 173 914 257 279 891 801 6;
  • 8) 0.002 666 333 353 173 914 257 279 891 801 6 × 2 = 0 + 0.005 332 666 706 347 828 514 559 783 603 2;
  • 9) 0.005 332 666 706 347 828 514 559 783 603 2 × 2 = 0 + 0.010 665 333 412 695 657 029 119 567 206 4;
  • 10) 0.010 665 333 412 695 657 029 119 567 206 4 × 2 = 0 + 0.021 330 666 825 391 314 058 239 134 412 8;
  • 11) 0.021 330 666 825 391 314 058 239 134 412 8 × 2 = 0 + 0.042 661 333 650 782 628 116 478 268 825 6;
  • 12) 0.042 661 333 650 782 628 116 478 268 825 6 × 2 = 0 + 0.085 322 667 301 565 256 232 956 537 651 2;
  • 13) 0.085 322 667 301 565 256 232 956 537 651 2 × 2 = 0 + 0.170 645 334 603 130 512 465 913 075 302 4;
  • 14) 0.170 645 334 603 130 512 465 913 075 302 4 × 2 = 0 + 0.341 290 669 206 261 024 931 826 150 604 8;
  • 15) 0.341 290 669 206 261 024 931 826 150 604 8 × 2 = 0 + 0.682 581 338 412 522 049 863 652 301 209 6;
  • 16) 0.682 581 338 412 522 049 863 652 301 209 6 × 2 = 1 + 0.365 162 676 825 044 099 727 304 602 419 2;
  • 17) 0.365 162 676 825 044 099 727 304 602 419 2 × 2 = 0 + 0.730 325 353 650 088 199 454 609 204 838 4;
  • 18) 0.730 325 353 650 088 199 454 609 204 838 4 × 2 = 1 + 0.460 650 707 300 176 398 909 218 409 676 8;
  • 19) 0.460 650 707 300 176 398 909 218 409 676 8 × 2 = 0 + 0.921 301 414 600 352 797 818 436 819 353 6;
  • 20) 0.921 301 414 600 352 797 818 436 819 353 6 × 2 = 1 + 0.842 602 829 200 705 595 636 873 638 707 2;
  • 21) 0.842 602 829 200 705 595 636 873 638 707 2 × 2 = 1 + 0.685 205 658 401 411 191 273 747 277 414 4;
  • 22) 0.685 205 658 401 411 191 273 747 277 414 4 × 2 = 1 + 0.370 411 316 802 822 382 547 494 554 828 8;
  • 23) 0.370 411 316 802 822 382 547 494 554 828 8 × 2 = 0 + 0.740 822 633 605 644 765 094 989 109 657 6;
  • 24) 0.740 822 633 605 644 765 094 989 109 657 6 × 2 = 1 + 0.481 645 267 211 289 530 189 978 219 315 2;
  • 25) 0.481 645 267 211 289 530 189 978 219 315 2 × 2 = 0 + 0.963 290 534 422 579 060 379 956 438 630 4;
  • 26) 0.963 290 534 422 579 060 379 956 438 630 4 × 2 = 1 + 0.926 581 068 845 158 120 759 912 877 260 8;
  • 27) 0.926 581 068 845 158 120 759 912 877 260 8 × 2 = 1 + 0.853 162 137 690 316 241 519 825 754 521 6;
  • 28) 0.853 162 137 690 316 241 519 825 754 521 6 × 2 = 1 + 0.706 324 275 380 632 483 039 651 509 043 2;
  • 29) 0.706 324 275 380 632 483 039 651 509 043 2 × 2 = 1 + 0.412 648 550 761 264 966 079 303 018 086 4;
  • 30) 0.412 648 550 761 264 966 079 303 018 086 4 × 2 = 0 + 0.825 297 101 522 529 932 158 606 036 172 8;
  • 31) 0.825 297 101 522 529 932 158 606 036 172 8 × 2 = 1 + 0.650 594 203 045 059 864 317 212 072 345 6;
  • 32) 0.650 594 203 045 059 864 317 212 072 345 6 × 2 = 1 + 0.301 188 406 090 119 728 634 424 144 691 2;
  • 33) 0.301 188 406 090 119 728 634 424 144 691 2 × 2 = 0 + 0.602 376 812 180 239 457 268 848 289 382 4;
  • 34) 0.602 376 812 180 239 457 268 848 289 382 4 × 2 = 1 + 0.204 753 624 360 478 914 537 696 578 764 8;
  • 35) 0.204 753 624 360 478 914 537 696 578 764 8 × 2 = 0 + 0.409 507 248 720 957 829 075 393 157 529 6;
  • 36) 0.409 507 248 720 957 829 075 393 157 529 6 × 2 = 0 + 0.819 014 497 441 915 658 150 786 315 059 2;
  • 37) 0.819 014 497 441 915 658 150 786 315 059 2 × 2 = 1 + 0.638 028 994 883 831 316 301 572 630 118 4;
  • 38) 0.638 028 994 883 831 316 301 572 630 118 4 × 2 = 1 + 0.276 057 989 767 662 632 603 145 260 236 8;
  • 39) 0.276 057 989 767 662 632 603 145 260 236 8 × 2 = 0 + 0.552 115 979 535 325 265 206 290 520 473 6;
  • 40) 0.552 115 979 535 325 265 206 290 520 473 6 × 2 = 1 + 0.104 231 959 070 650 530 412 581 040 947 2;
  • 41) 0.104 231 959 070 650 530 412 581 040 947 2 × 2 = 0 + 0.208 463 918 141 301 060 825 162 081 894 4;
  • 42) 0.208 463 918 141 301 060 825 162 081 894 4 × 2 = 0 + 0.416 927 836 282 602 121 650 324 163 788 8;
  • 43) 0.416 927 836 282 602 121 650 324 163 788 8 × 2 = 0 + 0.833 855 672 565 204 243 300 648 327 577 6;
  • 44) 0.833 855 672 565 204 243 300 648 327 577 6 × 2 = 1 + 0.667 711 345 130 408 486 601 296 655 155 2;
  • 45) 0.667 711 345 130 408 486 601 296 655 155 2 × 2 = 1 + 0.335 422 690 260 816 973 202 593 310 310 4;
  • 46) 0.335 422 690 260 816 973 202 593 310 310 4 × 2 = 0 + 0.670 845 380 521 633 946 405 186 620 620 8;
  • 47) 0.670 845 380 521 633 946 405 186 620 620 8 × 2 = 1 + 0.341 690 761 043 267 892 810 373 241 241 6;
  • 48) 0.341 690 761 043 267 892 810 373 241 241 6 × 2 = 0 + 0.683 381 522 086 535 785 620 746 482 483 2;
  • 49) 0.683 381 522 086 535 785 620 746 482 483 2 × 2 = 1 + 0.366 763 044 173 071 571 241 492 964 966 4;
  • 50) 0.366 763 044 173 071 571 241 492 964 966 4 × 2 = 0 + 0.733 526 088 346 143 142 482 985 929 932 8;
  • 51) 0.733 526 088 346 143 142 482 985 929 932 8 × 2 = 1 + 0.467 052 176 692 286 284 965 971 859 865 6;
  • 52) 0.467 052 176 692 286 284 965 971 859 865 6 × 2 = 0 + 0.934 104 353 384 572 569 931 943 719 731 2;
  • 53) 0.934 104 353 384 572 569 931 943 719 731 2 × 2 = 1 + 0.868 208 706 769 145 139 863 887 439 462 4;
  • 54) 0.868 208 706 769 145 139 863 887 439 462 4 × 2 = 1 + 0.736 417 413 538 290 279 727 774 878 924 8;
  • 55) 0.736 417 413 538 290 279 727 774 878 924 8 × 2 = 1 + 0.472 834 827 076 580 559 455 549 757 849 6;
  • 56) 0.472 834 827 076 580 559 455 549 757 849 6 × 2 = 0 + 0.945 669 654 153 161 118 911 099 515 699 2;
  • 57) 0.945 669 654 153 161 118 911 099 515 699 2 × 2 = 1 + 0.891 339 308 306 322 237 822 199 031 398 4;
  • 58) 0.891 339 308 306 322 237 822 199 031 398 4 × 2 = 1 + 0.782 678 616 612 644 475 644 398 062 796 8;
  • 59) 0.782 678 616 612 644 475 644 398 062 796 8 × 2 = 1 + 0.565 357 233 225 288 951 288 796 125 593 6;
  • 60) 0.565 357 233 225 288 951 288 796 125 593 6 × 2 = 1 + 0.130 714 466 450 577 902 577 592 251 187 2;
  • 61) 0.130 714 466 450 577 902 577 592 251 187 2 × 2 = 0 + 0.261 428 932 901 155 805 155 184 502 374 4;
  • 62) 0.261 428 932 901 155 805 155 184 502 374 4 × 2 = 0 + 0.522 857 865 802 311 610 310 369 004 748 8;
  • 63) 0.522 857 865 802 311 610 310 369 004 748 8 × 2 = 1 + 0.045 715 731 604 623 220 620 738 009 497 6;
  • 64) 0.045 715 731 604 623 220 620 738 009 497 6 × 2 = 0 + 0.091 431 463 209 246 441 241 476 018 995 2;
  • 65) 0.091 431 463 209 246 441 241 476 018 995 2 × 2 = 0 + 0.182 862 926 418 492 882 482 952 037 990 4;
  • 66) 0.182 862 926 418 492 882 482 952 037 990 4 × 2 = 0 + 0.365 725 852 836 985 764 965 904 075 980 8;
  • 67) 0.365 725 852 836 985 764 965 904 075 980 8 × 2 = 0 + 0.731 451 705 673 971 529 931 808 151 961 6;
  • 68) 0.731 451 705 673 971 529 931 808 151 961 6 × 2 = 1 + 0.462 903 411 347 943 059 863 616 303 923 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).


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 7(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 7(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 7(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 7 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