0.000 020 830 729 321 671 205 134 999 157 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 020 830 729 321 671 205 134 999 157 2(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
0.000 020 830 729 321 671 205 134 999 157 2(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: 0.000 020 830 729 321 671 205 134 999 157 2.

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) 0.000 020 830 729 321 671 205 134 999 157 2 × 2 = 0 + 0.000 041 661 458 643 342 410 269 998 314 4;
  • 2) 0.000 041 661 458 643 342 410 269 998 314 4 × 2 = 0 + 0.000 083 322 917 286 684 820 539 996 628 8;
  • 3) 0.000 083 322 917 286 684 820 539 996 628 8 × 2 = 0 + 0.000 166 645 834 573 369 641 079 993 257 6;
  • 4) 0.000 166 645 834 573 369 641 079 993 257 6 × 2 = 0 + 0.000 333 291 669 146 739 282 159 986 515 2;
  • 5) 0.000 333 291 669 146 739 282 159 986 515 2 × 2 = 0 + 0.000 666 583 338 293 478 564 319 973 030 4;
  • 6) 0.000 666 583 338 293 478 564 319 973 030 4 × 2 = 0 + 0.001 333 166 676 586 957 128 639 946 060 8;
  • 7) 0.001 333 166 676 586 957 128 639 946 060 8 × 2 = 0 + 0.002 666 333 353 173 914 257 279 892 121 6;
  • 8) 0.002 666 333 353 173 914 257 279 892 121 6 × 2 = 0 + 0.005 332 666 706 347 828 514 559 784 243 2;
  • 9) 0.005 332 666 706 347 828 514 559 784 243 2 × 2 = 0 + 0.010 665 333 412 695 657 029 119 568 486 4;
  • 10) 0.010 665 333 412 695 657 029 119 568 486 4 × 2 = 0 + 0.021 330 666 825 391 314 058 239 136 972 8;
  • 11) 0.021 330 666 825 391 314 058 239 136 972 8 × 2 = 0 + 0.042 661 333 650 782 628 116 478 273 945 6;
  • 12) 0.042 661 333 650 782 628 116 478 273 945 6 × 2 = 0 + 0.085 322 667 301 565 256 232 956 547 891 2;
  • 13) 0.085 322 667 301 565 256 232 956 547 891 2 × 2 = 0 + 0.170 645 334 603 130 512 465 913 095 782 4;
  • 14) 0.170 645 334 603 130 512 465 913 095 782 4 × 2 = 0 + 0.341 290 669 206 261 024 931 826 191 564 8;
  • 15) 0.341 290 669 206 261 024 931 826 191 564 8 × 2 = 0 + 0.682 581 338 412 522 049 863 652 383 129 6;
  • 16) 0.682 581 338 412 522 049 863 652 383 129 6 × 2 = 1 + 0.365 162 676 825 044 099 727 304 766 259 2;
  • 17) 0.365 162 676 825 044 099 727 304 766 259 2 × 2 = 0 + 0.730 325 353 650 088 199 454 609 532 518 4;
  • 18) 0.730 325 353 650 088 199 454 609 532 518 4 × 2 = 1 + 0.460 650 707 300 176 398 909 219 065 036 8;
  • 19) 0.460 650 707 300 176 398 909 219 065 036 8 × 2 = 0 + 0.921 301 414 600 352 797 818 438 130 073 6;
  • 20) 0.921 301 414 600 352 797 818 438 130 073 6 × 2 = 1 + 0.842 602 829 200 705 595 636 876 260 147 2;
  • 21) 0.842 602 829 200 705 595 636 876 260 147 2 × 2 = 1 + 0.685 205 658 401 411 191 273 752 520 294 4;
  • 22) 0.685 205 658 401 411 191 273 752 520 294 4 × 2 = 1 + 0.370 411 316 802 822 382 547 505 040 588 8;
  • 23) 0.370 411 316 802 822 382 547 505 040 588 8 × 2 = 0 + 0.740 822 633 605 644 765 095 010 081 177 6;
  • 24) 0.740 822 633 605 644 765 095 010 081 177 6 × 2 = 1 + 0.481 645 267 211 289 530 190 020 162 355 2;
  • 25) 0.481 645 267 211 289 530 190 020 162 355 2 × 2 = 0 + 0.963 290 534 422 579 060 380 040 324 710 4;
  • 26) 0.963 290 534 422 579 060 380 040 324 710 4 × 2 = 1 + 0.926 581 068 845 158 120 760 080 649 420 8;
  • 27) 0.926 581 068 845 158 120 760 080 649 420 8 × 2 = 1 + 0.853 162 137 690 316 241 520 161 298 841 6;
  • 28) 0.853 162 137 690 316 241 520 161 298 841 6 × 2 = 1 + 0.706 324 275 380 632 483 040 322 597 683 2;
  • 29) 0.706 324 275 380 632 483 040 322 597 683 2 × 2 = 1 + 0.412 648 550 761 264 966 080 645 195 366 4;
  • 30) 0.412 648 550 761 264 966 080 645 195 366 4 × 2 = 0 + 0.825 297 101 522 529 932 161 290 390 732 8;
  • 31) 0.825 297 101 522 529 932 161 290 390 732 8 × 2 = 1 + 0.650 594 203 045 059 864 322 580 781 465 6;
  • 32) 0.650 594 203 045 059 864 322 580 781 465 6 × 2 = 1 + 0.301 188 406 090 119 728 645 161 562 931 2;
  • 33) 0.301 188 406 090 119 728 645 161 562 931 2 × 2 = 0 + 0.602 376 812 180 239 457 290 323 125 862 4;
  • 34) 0.602 376 812 180 239 457 290 323 125 862 4 × 2 = 1 + 0.204 753 624 360 478 914 580 646 251 724 8;
  • 35) 0.204 753 624 360 478 914 580 646 251 724 8 × 2 = 0 + 0.409 507 248 720 957 829 161 292 503 449 6;
  • 36) 0.409 507 248 720 957 829 161 292 503 449 6 × 2 = 0 + 0.819 014 497 441 915 658 322 585 006 899 2;
  • 37) 0.819 014 497 441 915 658 322 585 006 899 2 × 2 = 1 + 0.638 028 994 883 831 316 645 170 013 798 4;
  • 38) 0.638 028 994 883 831 316 645 170 013 798 4 × 2 = 1 + 0.276 057 989 767 662 633 290 340 027 596 8;
  • 39) 0.276 057 989 767 662 633 290 340 027 596 8 × 2 = 0 + 0.552 115 979 535 325 266 580 680 055 193 6;
  • 40) 0.552 115 979 535 325 266 580 680 055 193 6 × 2 = 1 + 0.104 231 959 070 650 533 161 360 110 387 2;
  • 41) 0.104 231 959 070 650 533 161 360 110 387 2 × 2 = 0 + 0.208 463 918 141 301 066 322 720 220 774 4;
  • 42) 0.208 463 918 141 301 066 322 720 220 774 4 × 2 = 0 + 0.416 927 836 282 602 132 645 440 441 548 8;
  • 43) 0.416 927 836 282 602 132 645 440 441 548 8 × 2 = 0 + 0.833 855 672 565 204 265 290 880 883 097 6;
  • 44) 0.833 855 672 565 204 265 290 880 883 097 6 × 2 = 1 + 0.667 711 345 130 408 530 581 761 766 195 2;
  • 45) 0.667 711 345 130 408 530 581 761 766 195 2 × 2 = 1 + 0.335 422 690 260 817 061 163 523 532 390 4;
  • 46) 0.335 422 690 260 817 061 163 523 532 390 4 × 2 = 0 + 0.670 845 380 521 634 122 327 047 064 780 8;
  • 47) 0.670 845 380 521 634 122 327 047 064 780 8 × 2 = 1 + 0.341 690 761 043 268 244 654 094 129 561 6;
  • 48) 0.341 690 761 043 268 244 654 094 129 561 6 × 2 = 0 + 0.683 381 522 086 536 489 308 188 259 123 2;
  • 49) 0.683 381 522 086 536 489 308 188 259 123 2 × 2 = 1 + 0.366 763 044 173 072 978 616 376 518 246 4;
  • 50) 0.366 763 044 173 072 978 616 376 518 246 4 × 2 = 0 + 0.733 526 088 346 145 957 232 753 036 492 8;
  • 51) 0.733 526 088 346 145 957 232 753 036 492 8 × 2 = 1 + 0.467 052 176 692 291 914 465 506 072 985 6;
  • 52) 0.467 052 176 692 291 914 465 506 072 985 6 × 2 = 0 + 0.934 104 353 384 583 828 931 012 145 971 2;
  • 53) 0.934 104 353 384 583 828 931 012 145 971 2 × 2 = 1 + 0.868 208 706 769 167 657 862 024 291 942 4;
  • 54) 0.868 208 706 769 167 657 862 024 291 942 4 × 2 = 1 + 0.736 417 413 538 335 315 724 048 583 884 8;
  • 55) 0.736 417 413 538 335 315 724 048 583 884 8 × 2 = 1 + 0.472 834 827 076 670 631 448 097 167 769 6;
  • 56) 0.472 834 827 076 670 631 448 097 167 769 6 × 2 = 0 + 0.945 669 654 153 341 262 896 194 335 539 2;
  • 57) 0.945 669 654 153 341 262 896 194 335 539 2 × 2 = 1 + 0.891 339 308 306 682 525 792 388 671 078 4;
  • 58) 0.891 339 308 306 682 525 792 388 671 078 4 × 2 = 1 + 0.782 678 616 613 365 051 584 777 342 156 8;
  • 59) 0.782 678 616 613 365 051 584 777 342 156 8 × 2 = 1 + 0.565 357 233 226 730 103 169 554 684 313 6;
  • 60) 0.565 357 233 226 730 103 169 554 684 313 6 × 2 = 1 + 0.130 714 466 453 460 206 339 109 368 627 2;
  • 61) 0.130 714 466 453 460 206 339 109 368 627 2 × 2 = 0 + 0.261 428 932 906 920 412 678 218 737 254 4;
  • 62) 0.261 428 932 906 920 412 678 218 737 254 4 × 2 = 0 + 0.522 857 865 813 840 825 356 437 474 508 8;
  • 63) 0.522 857 865 813 840 825 356 437 474 508 8 × 2 = 1 + 0.045 715 731 627 681 650 712 874 949 017 6;
  • 64) 0.045 715 731 627 681 650 712 874 949 017 6 × 2 = 0 + 0.091 431 463 255 363 301 425 749 898 035 2;
  • 65) 0.091 431 463 255 363 301 425 749 898 035 2 × 2 = 0 + 0.182 862 926 510 726 602 851 499 796 070 4;
  • 66) 0.182 862 926 510 726 602 851 499 796 070 4 × 2 = 0 + 0.365 725 853 021 453 205 702 999 592 140 8;
  • 67) 0.365 725 853 021 453 205 702 999 592 140 8 × 2 = 0 + 0.731 451 706 042 906 411 405 999 184 281 6;
  • 68) 0.731 451 706 042 906 411 405 999 184 281 6 × 2 = 1 + 0.462 903 412 085 812 822 811 998 368 563 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:


0.000 020 830 729 321 671 205 134 999 157 2(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:

0.000 020 830 729 321 671 205 134 999 157 2(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:


0.000 020 830 729 321 671 205 134 999 157 2(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 0.000 020 830 729 321 671 205 134 999 157 2 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