0.000 999 999 999 999 999 803 976 247 214 620 798 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 999 999 999 999 999 803 976 247 214 620 798 22(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
0.000 999 999 999 999 999 803 976 247 214 620 798 22(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: 0.000 999 999 999 999 999 803 976 247 214 620 798 22.

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) 0.000 999 999 999 999 999 803 976 247 214 620 798 22 × 2 = 0 + 0.001 999 999 999 999 999 607 952 494 429 241 596 44;
  • 2) 0.001 999 999 999 999 999 607 952 494 429 241 596 44 × 2 = 0 + 0.003 999 999 999 999 999 215 904 988 858 483 192 88;
  • 3) 0.003 999 999 999 999 999 215 904 988 858 483 192 88 × 2 = 0 + 0.007 999 999 999 999 998 431 809 977 716 966 385 76;
  • 4) 0.007 999 999 999 999 998 431 809 977 716 966 385 76 × 2 = 0 + 0.015 999 999 999 999 996 863 619 955 433 932 771 52;
  • 5) 0.015 999 999 999 999 996 863 619 955 433 932 771 52 × 2 = 0 + 0.031 999 999 999 999 993 727 239 910 867 865 543 04;
  • 6) 0.031 999 999 999 999 993 727 239 910 867 865 543 04 × 2 = 0 + 0.063 999 999 999 999 987 454 479 821 735 731 086 08;
  • 7) 0.063 999 999 999 999 987 454 479 821 735 731 086 08 × 2 = 0 + 0.127 999 999 999 999 974 908 959 643 471 462 172 16;
  • 8) 0.127 999 999 999 999 974 908 959 643 471 462 172 16 × 2 = 0 + 0.255 999 999 999 999 949 817 919 286 942 924 344 32;
  • 9) 0.255 999 999 999 999 949 817 919 286 942 924 344 32 × 2 = 0 + 0.511 999 999 999 999 899 635 838 573 885 848 688 64;
  • 10) 0.511 999 999 999 999 899 635 838 573 885 848 688 64 × 2 = 1 + 0.023 999 999 999 999 799 271 677 147 771 697 377 28;
  • 11) 0.023 999 999 999 999 799 271 677 147 771 697 377 28 × 2 = 0 + 0.047 999 999 999 999 598 543 354 295 543 394 754 56;
  • 12) 0.047 999 999 999 999 598 543 354 295 543 394 754 56 × 2 = 0 + 0.095 999 999 999 999 197 086 708 591 086 789 509 12;
  • 13) 0.095 999 999 999 999 197 086 708 591 086 789 509 12 × 2 = 0 + 0.191 999 999 999 998 394 173 417 182 173 579 018 24;
  • 14) 0.191 999 999 999 998 394 173 417 182 173 579 018 24 × 2 = 0 + 0.383 999 999 999 996 788 346 834 364 347 158 036 48;
  • 15) 0.383 999 999 999 996 788 346 834 364 347 158 036 48 × 2 = 0 + 0.767 999 999 999 993 576 693 668 728 694 316 072 96;
  • 16) 0.767 999 999 999 993 576 693 668 728 694 316 072 96 × 2 = 1 + 0.535 999 999 999 987 153 387 337 457 388 632 145 92;
  • 17) 0.535 999 999 999 987 153 387 337 457 388 632 145 92 × 2 = 1 + 0.071 999 999 999 974 306 774 674 914 777 264 291 84;
  • 18) 0.071 999 999 999 974 306 774 674 914 777 264 291 84 × 2 = 0 + 0.143 999 999 999 948 613 549 349 829 554 528 583 68;
  • 19) 0.143 999 999 999 948 613 549 349 829 554 528 583 68 × 2 = 0 + 0.287 999 999 999 897 227 098 699 659 109 057 167 36;
  • 20) 0.287 999 999 999 897 227 098 699 659 109 057 167 36 × 2 = 0 + 0.575 999 999 999 794 454 197 399 318 218 114 334 72;
  • 21) 0.575 999 999 999 794 454 197 399 318 218 114 334 72 × 2 = 1 + 0.151 999 999 999 588 908 394 798 636 436 228 669 44;
  • 22) 0.151 999 999 999 588 908 394 798 636 436 228 669 44 × 2 = 0 + 0.303 999 999 999 177 816 789 597 272 872 457 338 88;
  • 23) 0.303 999 999 999 177 816 789 597 272 872 457 338 88 × 2 = 0 + 0.607 999 999 998 355 633 579 194 545 744 914 677 76;
  • 24) 0.607 999 999 998 355 633 579 194 545 744 914 677 76 × 2 = 1 + 0.215 999 999 996 711 267 158 389 091 489 829 355 52;
  • 25) 0.215 999 999 996 711 267 158 389 091 489 829 355 52 × 2 = 0 + 0.431 999 999 993 422 534 316 778 182 979 658 711 04;
  • 26) 0.431 999 999 993 422 534 316 778 182 979 658 711 04 × 2 = 0 + 0.863 999 999 986 845 068 633 556 365 959 317 422 08;
  • 27) 0.863 999 999 986 845 068 633 556 365 959 317 422 08 × 2 = 1 + 0.727 999 999 973 690 137 267 112 731 918 634 844 16;
  • 28) 0.727 999 999 973 690 137 267 112 731 918 634 844 16 × 2 = 1 + 0.455 999 999 947 380 274 534 225 463 837 269 688 32;
  • 29) 0.455 999 999 947 380 274 534 225 463 837 269 688 32 × 2 = 0 + 0.911 999 999 894 760 549 068 450 927 674 539 376 64;
  • 30) 0.911 999 999 894 760 549 068 450 927 674 539 376 64 × 2 = 1 + 0.823 999 999 789 521 098 136 901 855 349 078 753 28;
  • 31) 0.823 999 999 789 521 098 136 901 855 349 078 753 28 × 2 = 1 + 0.647 999 999 579 042 196 273 803 710 698 157 506 56;
  • 32) 0.647 999 999 579 042 196 273 803 710 698 157 506 56 × 2 = 1 + 0.295 999 999 158 084 392 547 607 421 396 315 013 12;
  • 33) 0.295 999 999 158 084 392 547 607 421 396 315 013 12 × 2 = 0 + 0.591 999 998 316 168 785 095 214 842 792 630 026 24;
  • 34) 0.591 999 998 316 168 785 095 214 842 792 630 026 24 × 2 = 1 + 0.183 999 996 632 337 570 190 429 685 585 260 052 48;
  • 35) 0.183 999 996 632 337 570 190 429 685 585 260 052 48 × 2 = 0 + 0.367 999 993 264 675 140 380 859 371 170 520 104 96;
  • 36) 0.367 999 993 264 675 140 380 859 371 170 520 104 96 × 2 = 0 + 0.735 999 986 529 350 280 761 718 742 341 040 209 92;
  • 37) 0.735 999 986 529 350 280 761 718 742 341 040 209 92 × 2 = 1 + 0.471 999 973 058 700 561 523 437 484 682 080 419 84;
  • 38) 0.471 999 973 058 700 561 523 437 484 682 080 419 84 × 2 = 0 + 0.943 999 946 117 401 123 046 874 969 364 160 839 68;
  • 39) 0.943 999 946 117 401 123 046 874 969 364 160 839 68 × 2 = 1 + 0.887 999 892 234 802 246 093 749 938 728 321 679 36;
  • 40) 0.887 999 892 234 802 246 093 749 938 728 321 679 36 × 2 = 1 + 0.775 999 784 469 604 492 187 499 877 456 643 358 72;
  • 41) 0.775 999 784 469 604 492 187 499 877 456 643 358 72 × 2 = 1 + 0.551 999 568 939 208 984 374 999 754 913 286 717 44;
  • 42) 0.551 999 568 939 208 984 374 999 754 913 286 717 44 × 2 = 1 + 0.103 999 137 878 417 968 749 999 509 826 573 434 88;
  • 43) 0.103 999 137 878 417 968 749 999 509 826 573 434 88 × 2 = 0 + 0.207 998 275 756 835 937 499 999 019 653 146 869 76;
  • 44) 0.207 998 275 756 835 937 499 999 019 653 146 869 76 × 2 = 0 + 0.415 996 551 513 671 874 999 998 039 306 293 739 52;
  • 45) 0.415 996 551 513 671 874 999 998 039 306 293 739 52 × 2 = 0 + 0.831 993 103 027 343 749 999 996 078 612 587 479 04;
  • 46) 0.831 993 103 027 343 749 999 996 078 612 587 479 04 × 2 = 1 + 0.663 986 206 054 687 499 999 992 157 225 174 958 08;
  • 47) 0.663 986 206 054 687 499 999 992 157 225 174 958 08 × 2 = 1 + 0.327 972 412 109 374 999 999 984 314 450 349 916 16;
  • 48) 0.327 972 412 109 374 999 999 984 314 450 349 916 16 × 2 = 0 + 0.655 944 824 218 749 999 999 968 628 900 699 832 32;
  • 49) 0.655 944 824 218 749 999 999 968 628 900 699 832 32 × 2 = 1 + 0.311 889 648 437 499 999 999 937 257 801 399 664 64;
  • 50) 0.311 889 648 437 499 999 999 937 257 801 399 664 64 × 2 = 0 + 0.623 779 296 874 999 999 999 874 515 602 799 329 28;
  • 51) 0.623 779 296 874 999 999 999 874 515 602 799 329 28 × 2 = 1 + 0.247 558 593 749 999 999 999 749 031 205 598 658 56;
  • 52) 0.247 558 593 749 999 999 999 749 031 205 598 658 56 × 2 = 0 + 0.495 117 187 499 999 999 999 498 062 411 197 317 12;
  • 53) 0.495 117 187 499 999 999 999 498 062 411 197 317 12 × 2 = 0 + 0.990 234 374 999 999 999 998 996 124 822 394 634 24;
  • 54) 0.990 234 374 999 999 999 998 996 124 822 394 634 24 × 2 = 1 + 0.980 468 749 999 999 999 997 992 249 644 789 268 48;
  • 55) 0.980 468 749 999 999 999 997 992 249 644 789 268 48 × 2 = 1 + 0.960 937 499 999 999 999 995 984 499 289 578 536 96;
  • 56) 0.960 937 499 999 999 999 995 984 499 289 578 536 96 × 2 = 1 + 0.921 874 999 999 999 999 991 968 998 579 157 073 92;
  • 57) 0.921 874 999 999 999 999 991 968 998 579 157 073 92 × 2 = 1 + 0.843 749 999 999 999 999 983 937 997 158 314 147 84;
  • 58) 0.843 749 999 999 999 999 983 937 997 158 314 147 84 × 2 = 1 + 0.687 499 999 999 999 999 967 875 994 316 628 295 68;
  • 59) 0.687 499 999 999 999 999 967 875 994 316 628 295 68 × 2 = 1 + 0.374 999 999 999 999 999 935 751 988 633 256 591 36;
  • 60) 0.374 999 999 999 999 999 935 751 988 633 256 591 36 × 2 = 0 + 0.749 999 999 999 999 999 871 503 977 266 513 182 72;
  • 61) 0.749 999 999 999 999 999 871 503 977 266 513 182 72 × 2 = 1 + 0.499 999 999 999 999 999 743 007 954 533 026 365 44;
  • 62) 0.499 999 999 999 999 999 743 007 954 533 026 365 44 × 2 = 0 + 0.999 999 999 999 999 999 486 015 909 066 052 730 88;

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:


0.000 999 999 999 999 999 803 976 247 214 620 798 22(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2)

5. Positive number before normalization:

0.000 999 999 999 999 999 803 976 247 214 620 798 22(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.000 999 999 999 999 999 803 976 247 214 620 798 22(10) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2) =

0.0000 0000 0100 0001 1000 1001 0011 0111 0100 1011 1100 0110 1010 0111 1110 10(2) × 20 =

1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010(2) × 2-10


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): -10

Mantissa (not normalized):
1.0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 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) =

1013(10) =

011 1111 0101(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. 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010 =

0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


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 1111 0101

Mantissa (52 bits) =
0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010


Decimal number 0.000 999 999 999 999 999 803 976 247 214 620 798 22 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0000 0110 0010 0100 1101 1101 0010 1111 0001 1010 1001 1111 1010

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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