-0.000 000 000 000 014 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 31(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
-0.000 000 000 000 014 31(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. Start with the positive version of the number:

|-0.000 000 000 000 014 31| = 0.000 000 000 000 014 31


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 014 31.

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) 0.000 000 000 000 014 31 × 2 = 0 + 0.000 000 000 000 028 62;
  • 2) 0.000 000 000 000 028 62 × 2 = 0 + 0.000 000 000 000 057 24;
  • 3) 0.000 000 000 000 057 24 × 2 = 0 + 0.000 000 000 000 114 48;
  • 4) 0.000 000 000 000 114 48 × 2 = 0 + 0.000 000 000 000 228 96;
  • 5) 0.000 000 000 000 228 96 × 2 = 0 + 0.000 000 000 000 457 92;
  • 6) 0.000 000 000 000 457 92 × 2 = 0 + 0.000 000 000 000 915 84;
  • 7) 0.000 000 000 000 915 84 × 2 = 0 + 0.000 000 000 001 831 68;
  • 8) 0.000 000 000 001 831 68 × 2 = 0 + 0.000 000 000 003 663 36;
  • 9) 0.000 000 000 003 663 36 × 2 = 0 + 0.000 000 000 007 326 72;
  • 10) 0.000 000 000 007 326 72 × 2 = 0 + 0.000 000 000 014 653 44;
  • 11) 0.000 000 000 014 653 44 × 2 = 0 + 0.000 000 000 029 306 88;
  • 12) 0.000 000 000 029 306 88 × 2 = 0 + 0.000 000 000 058 613 76;
  • 13) 0.000 000 000 058 613 76 × 2 = 0 + 0.000 000 000 117 227 52;
  • 14) 0.000 000 000 117 227 52 × 2 = 0 + 0.000 000 000 234 455 04;
  • 15) 0.000 000 000 234 455 04 × 2 = 0 + 0.000 000 000 468 910 08;
  • 16) 0.000 000 000 468 910 08 × 2 = 0 + 0.000 000 000 937 820 16;
  • 17) 0.000 000 000 937 820 16 × 2 = 0 + 0.000 000 001 875 640 32;
  • 18) 0.000 000 001 875 640 32 × 2 = 0 + 0.000 000 003 751 280 64;
  • 19) 0.000 000 003 751 280 64 × 2 = 0 + 0.000 000 007 502 561 28;
  • 20) 0.000 000 007 502 561 28 × 2 = 0 + 0.000 000 015 005 122 56;
  • 21) 0.000 000 015 005 122 56 × 2 = 0 + 0.000 000 030 010 245 12;
  • 22) 0.000 000 030 010 245 12 × 2 = 0 + 0.000 000 060 020 490 24;
  • 23) 0.000 000 060 020 490 24 × 2 = 0 + 0.000 000 120 040 980 48;
  • 24) 0.000 000 120 040 980 48 × 2 = 0 + 0.000 000 240 081 960 96;
  • 25) 0.000 000 240 081 960 96 × 2 = 0 + 0.000 000 480 163 921 92;
  • 26) 0.000 000 480 163 921 92 × 2 = 0 + 0.000 000 960 327 843 84;
  • 27) 0.000 000 960 327 843 84 × 2 = 0 + 0.000 001 920 655 687 68;
  • 28) 0.000 001 920 655 687 68 × 2 = 0 + 0.000 003 841 311 375 36;
  • 29) 0.000 003 841 311 375 36 × 2 = 0 + 0.000 007 682 622 750 72;
  • 30) 0.000 007 682 622 750 72 × 2 = 0 + 0.000 015 365 245 501 44;
  • 31) 0.000 015 365 245 501 44 × 2 = 0 + 0.000 030 730 491 002 88;
  • 32) 0.000 030 730 491 002 88 × 2 = 0 + 0.000 061 460 982 005 76;
  • 33) 0.000 061 460 982 005 76 × 2 = 0 + 0.000 122 921 964 011 52;
  • 34) 0.000 122 921 964 011 52 × 2 = 0 + 0.000 245 843 928 023 04;
  • 35) 0.000 245 843 928 023 04 × 2 = 0 + 0.000 491 687 856 046 08;
  • 36) 0.000 491 687 856 046 08 × 2 = 0 + 0.000 983 375 712 092 16;
  • 37) 0.000 983 375 712 092 16 × 2 = 0 + 0.001 966 751 424 184 32;
  • 38) 0.001 966 751 424 184 32 × 2 = 0 + 0.003 933 502 848 368 64;
  • 39) 0.003 933 502 848 368 64 × 2 = 0 + 0.007 867 005 696 737 28;
  • 40) 0.007 867 005 696 737 28 × 2 = 0 + 0.015 734 011 393 474 56;
  • 41) 0.015 734 011 393 474 56 × 2 = 0 + 0.031 468 022 786 949 12;
  • 42) 0.031 468 022 786 949 12 × 2 = 0 + 0.062 936 045 573 898 24;
  • 43) 0.062 936 045 573 898 24 × 2 = 0 + 0.125 872 091 147 796 48;
  • 44) 0.125 872 091 147 796 48 × 2 = 0 + 0.251 744 182 295 592 96;
  • 45) 0.251 744 182 295 592 96 × 2 = 0 + 0.503 488 364 591 185 92;
  • 46) 0.503 488 364 591 185 92 × 2 = 1 + 0.006 976 729 182 371 84;
  • 47) 0.006 976 729 182 371 84 × 2 = 0 + 0.013 953 458 364 743 68;
  • 48) 0.013 953 458 364 743 68 × 2 = 0 + 0.027 906 916 729 487 36;
  • 49) 0.027 906 916 729 487 36 × 2 = 0 + 0.055 813 833 458 974 72;
  • 50) 0.055 813 833 458 974 72 × 2 = 0 + 0.111 627 666 917 949 44;
  • 51) 0.111 627 666 917 949 44 × 2 = 0 + 0.223 255 333 835 898 88;
  • 52) 0.223 255 333 835 898 88 × 2 = 0 + 0.446 510 667 671 797 76;
  • 53) 0.446 510 667 671 797 76 × 2 = 0 + 0.893 021 335 343 595 52;
  • 54) 0.893 021 335 343 595 52 × 2 = 1 + 0.786 042 670 687 191 04;
  • 55) 0.786 042 670 687 191 04 × 2 = 1 + 0.572 085 341 374 382 08;
  • 56) 0.572 085 341 374 382 08 × 2 = 1 + 0.144 170 682 748 764 16;
  • 57) 0.144 170 682 748 764 16 × 2 = 0 + 0.288 341 365 497 528 32;
  • 58) 0.288 341 365 497 528 32 × 2 = 0 + 0.576 682 730 995 056 64;
  • 59) 0.576 682 730 995 056 64 × 2 = 1 + 0.153 365 461 990 113 28;
  • 60) 0.153 365 461 990 113 28 × 2 = 0 + 0.306 730 923 980 226 56;
  • 61) 0.306 730 923 980 226 56 × 2 = 0 + 0.613 461 847 960 453 12;
  • 62) 0.613 461 847 960 453 12 × 2 = 1 + 0.226 923 695 920 906 24;
  • 63) 0.226 923 695 920 906 24 × 2 = 0 + 0.453 847 391 841 812 48;
  • 64) 0.453 847 391 841 812 48 × 2 = 0 + 0.907 694 783 683 624 96;
  • 65) 0.907 694 783 683 624 96 × 2 = 1 + 0.815 389 567 367 249 92;
  • 66) 0.815 389 567 367 249 92 × 2 = 1 + 0.630 779 134 734 499 84;
  • 67) 0.630 779 134 734 499 84 × 2 = 1 + 0.261 558 269 468 999 68;
  • 68) 0.261 558 269 468 999 68 × 2 = 0 + 0.523 116 538 937 999 36;
  • 69) 0.523 116 538 937 999 36 × 2 = 1 + 0.046 233 077 875 998 72;
  • 70) 0.046 233 077 875 998 72 × 2 = 0 + 0.092 466 155 751 997 44;
  • 71) 0.092 466 155 751 997 44 × 2 = 0 + 0.184 932 311 503 994 88;
  • 72) 0.184 932 311 503 994 88 × 2 = 0 + 0.369 864 623 007 989 76;
  • 73) 0.369 864 623 007 989 76 × 2 = 0 + 0.739 729 246 015 979 52;
  • 74) 0.739 729 246 015 979 52 × 2 = 1 + 0.479 458 492 031 959 04;
  • 75) 0.479 458 492 031 959 04 × 2 = 0 + 0.958 916 984 063 918 08;
  • 76) 0.958 916 984 063 918 08 × 2 = 1 + 0.917 833 968 127 836 16;
  • 77) 0.917 833 968 127 836 16 × 2 = 1 + 0.835 667 936 255 672 32;
  • 78) 0.835 667 936 255 672 32 × 2 = 1 + 0.671 335 872 511 344 64;
  • 79) 0.671 335 872 511 344 64 × 2 = 1 + 0.342 671 745 022 689 28;
  • 80) 0.342 671 745 022 689 28 × 2 = 0 + 0.685 343 490 045 378 56;
  • 81) 0.685 343 490 045 378 56 × 2 = 1 + 0.370 686 980 090 757 12;
  • 82) 0.370 686 980 090 757 12 × 2 = 0 + 0.741 373 960 181 514 24;
  • 83) 0.741 373 960 181 514 24 × 2 = 1 + 0.482 747 920 363 028 48;
  • 84) 0.482 747 920 363 028 48 × 2 = 0 + 0.965 495 840 726 056 96;
  • 85) 0.965 495 840 726 056 96 × 2 = 1 + 0.930 991 681 452 113 92;
  • 86) 0.930 991 681 452 113 92 × 2 = 1 + 0.861 983 362 904 227 84;
  • 87) 0.861 983 362 904 227 84 × 2 = 1 + 0.723 966 725 808 455 68;
  • 88) 0.723 966 725 808 455 68 × 2 = 1 + 0.447 933 451 616 911 36;
  • 89) 0.447 933 451 616 911 36 × 2 = 0 + 0.895 866 903 233 822 72;
  • 90) 0.895 866 903 233 822 72 × 2 = 1 + 0.791 733 806 467 645 44;
  • 91) 0.791 733 806 467 645 44 × 2 = 1 + 0.583 467 612 935 290 88;
  • 92) 0.583 467 612 935 290 88 × 2 = 1 + 0.166 935 225 870 581 76;
  • 93) 0.166 935 225 870 581 76 × 2 = 0 + 0.333 870 451 741 163 52;
  • 94) 0.333 870 451 741 163 52 × 2 = 0 + 0.667 740 903 482 327 04;
  • 95) 0.667 740 903 482 327 04 × 2 = 1 + 0.335 481 806 964 654 08;
  • 96) 0.335 481 806 964 654 08 × 2 = 0 + 0.670 963 613 929 308 16;
  • 97) 0.670 963 613 929 308 16 × 2 = 1 + 0.341 927 227 858 616 32;
  • 98) 0.341 927 227 858 616 32 × 2 = 0 + 0.683 854 455 717 232 64;

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).


5. 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:


0.000 000 000 000 014 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 0111 0010 0100 1110 1000 0101 1110 1010 1111 0111 0010 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 0111 0010 0100 1110 1000 0101 1110 1010 1111 0111 0010 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 31(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 0111 0010 0100 1110 1000 0101 1110 1010 1111 0111 0010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 0111 0010 0100 1110 1000 0101 1110 1010 1111 0111 0010 10(2) × 20 =

1.0000 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(10)


10. 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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

977(10) =

011 1101 0001(2)


12. 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 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010 =

0000 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1101 0001

Mantissa (52 bits) =
0000 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010


Decimal number -0.000 000 000 000 014 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0001 1100 1001 0011 1010 0001 0111 1010 1011 1101 1100 1010

» Convert decimal -0.000 000 000 000 014 64 to 64 bit double precision IEEE 754 binary floating point representation standard

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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