-0.000 000 000 000 014 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 41(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 41(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 41| = 0.000 000 000 000 014 41


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 41.

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 41 × 2 = 0 + 0.000 000 000 000 028 82;
  • 2) 0.000 000 000 000 028 82 × 2 = 0 + 0.000 000 000 000 057 64;
  • 3) 0.000 000 000 000 057 64 × 2 = 0 + 0.000 000 000 000 115 28;
  • 4) 0.000 000 000 000 115 28 × 2 = 0 + 0.000 000 000 000 230 56;
  • 5) 0.000 000 000 000 230 56 × 2 = 0 + 0.000 000 000 000 461 12;
  • 6) 0.000 000 000 000 461 12 × 2 = 0 + 0.000 000 000 000 922 24;
  • 7) 0.000 000 000 000 922 24 × 2 = 0 + 0.000 000 000 001 844 48;
  • 8) 0.000 000 000 001 844 48 × 2 = 0 + 0.000 000 000 003 688 96;
  • 9) 0.000 000 000 003 688 96 × 2 = 0 + 0.000 000 000 007 377 92;
  • 10) 0.000 000 000 007 377 92 × 2 = 0 + 0.000 000 000 014 755 84;
  • 11) 0.000 000 000 014 755 84 × 2 = 0 + 0.000 000 000 029 511 68;
  • 12) 0.000 000 000 029 511 68 × 2 = 0 + 0.000 000 000 059 023 36;
  • 13) 0.000 000 000 059 023 36 × 2 = 0 + 0.000 000 000 118 046 72;
  • 14) 0.000 000 000 118 046 72 × 2 = 0 + 0.000 000 000 236 093 44;
  • 15) 0.000 000 000 236 093 44 × 2 = 0 + 0.000 000 000 472 186 88;
  • 16) 0.000 000 000 472 186 88 × 2 = 0 + 0.000 000 000 944 373 76;
  • 17) 0.000 000 000 944 373 76 × 2 = 0 + 0.000 000 001 888 747 52;
  • 18) 0.000 000 001 888 747 52 × 2 = 0 + 0.000 000 003 777 495 04;
  • 19) 0.000 000 003 777 495 04 × 2 = 0 + 0.000 000 007 554 990 08;
  • 20) 0.000 000 007 554 990 08 × 2 = 0 + 0.000 000 015 109 980 16;
  • 21) 0.000 000 015 109 980 16 × 2 = 0 + 0.000 000 030 219 960 32;
  • 22) 0.000 000 030 219 960 32 × 2 = 0 + 0.000 000 060 439 920 64;
  • 23) 0.000 000 060 439 920 64 × 2 = 0 + 0.000 000 120 879 841 28;
  • 24) 0.000 000 120 879 841 28 × 2 = 0 + 0.000 000 241 759 682 56;
  • 25) 0.000 000 241 759 682 56 × 2 = 0 + 0.000 000 483 519 365 12;
  • 26) 0.000 000 483 519 365 12 × 2 = 0 + 0.000 000 967 038 730 24;
  • 27) 0.000 000 967 038 730 24 × 2 = 0 + 0.000 001 934 077 460 48;
  • 28) 0.000 001 934 077 460 48 × 2 = 0 + 0.000 003 868 154 920 96;
  • 29) 0.000 003 868 154 920 96 × 2 = 0 + 0.000 007 736 309 841 92;
  • 30) 0.000 007 736 309 841 92 × 2 = 0 + 0.000 015 472 619 683 84;
  • 31) 0.000 015 472 619 683 84 × 2 = 0 + 0.000 030 945 239 367 68;
  • 32) 0.000 030 945 239 367 68 × 2 = 0 + 0.000 061 890 478 735 36;
  • 33) 0.000 061 890 478 735 36 × 2 = 0 + 0.000 123 780 957 470 72;
  • 34) 0.000 123 780 957 470 72 × 2 = 0 + 0.000 247 561 914 941 44;
  • 35) 0.000 247 561 914 941 44 × 2 = 0 + 0.000 495 123 829 882 88;
  • 36) 0.000 495 123 829 882 88 × 2 = 0 + 0.000 990 247 659 765 76;
  • 37) 0.000 990 247 659 765 76 × 2 = 0 + 0.001 980 495 319 531 52;
  • 38) 0.001 980 495 319 531 52 × 2 = 0 + 0.003 960 990 639 063 04;
  • 39) 0.003 960 990 639 063 04 × 2 = 0 + 0.007 921 981 278 126 08;
  • 40) 0.007 921 981 278 126 08 × 2 = 0 + 0.015 843 962 556 252 16;
  • 41) 0.015 843 962 556 252 16 × 2 = 0 + 0.031 687 925 112 504 32;
  • 42) 0.031 687 925 112 504 32 × 2 = 0 + 0.063 375 850 225 008 64;
  • 43) 0.063 375 850 225 008 64 × 2 = 0 + 0.126 751 700 450 017 28;
  • 44) 0.126 751 700 450 017 28 × 2 = 0 + 0.253 503 400 900 034 56;
  • 45) 0.253 503 400 900 034 56 × 2 = 0 + 0.507 006 801 800 069 12;
  • 46) 0.507 006 801 800 069 12 × 2 = 1 + 0.014 013 603 600 138 24;
  • 47) 0.014 013 603 600 138 24 × 2 = 0 + 0.028 027 207 200 276 48;
  • 48) 0.028 027 207 200 276 48 × 2 = 0 + 0.056 054 414 400 552 96;
  • 49) 0.056 054 414 400 552 96 × 2 = 0 + 0.112 108 828 801 105 92;
  • 50) 0.112 108 828 801 105 92 × 2 = 0 + 0.224 217 657 602 211 84;
  • 51) 0.224 217 657 602 211 84 × 2 = 0 + 0.448 435 315 204 423 68;
  • 52) 0.448 435 315 204 423 68 × 2 = 0 + 0.896 870 630 408 847 36;
  • 53) 0.896 870 630 408 847 36 × 2 = 1 + 0.793 741 260 817 694 72;
  • 54) 0.793 741 260 817 694 72 × 2 = 1 + 0.587 482 521 635 389 44;
  • 55) 0.587 482 521 635 389 44 × 2 = 1 + 0.174 965 043 270 778 88;
  • 56) 0.174 965 043 270 778 88 × 2 = 0 + 0.349 930 086 541 557 76;
  • 57) 0.349 930 086 541 557 76 × 2 = 0 + 0.699 860 173 083 115 52;
  • 58) 0.699 860 173 083 115 52 × 2 = 1 + 0.399 720 346 166 231 04;
  • 59) 0.399 720 346 166 231 04 × 2 = 0 + 0.799 440 692 332 462 08;
  • 60) 0.799 440 692 332 462 08 × 2 = 1 + 0.598 881 384 664 924 16;
  • 61) 0.598 881 384 664 924 16 × 2 = 1 + 0.197 762 769 329 848 32;
  • 62) 0.197 762 769 329 848 32 × 2 = 0 + 0.395 525 538 659 696 64;
  • 63) 0.395 525 538 659 696 64 × 2 = 0 + 0.791 051 077 319 393 28;
  • 64) 0.791 051 077 319 393 28 × 2 = 1 + 0.582 102 154 638 786 56;
  • 65) 0.582 102 154 638 786 56 × 2 = 1 + 0.164 204 309 277 573 12;
  • 66) 0.164 204 309 277 573 12 × 2 = 0 + 0.328 408 618 555 146 24;
  • 67) 0.328 408 618 555 146 24 × 2 = 0 + 0.656 817 237 110 292 48;
  • 68) 0.656 817 237 110 292 48 × 2 = 1 + 0.313 634 474 220 584 96;
  • 69) 0.313 634 474 220 584 96 × 2 = 0 + 0.627 268 948 441 169 92;
  • 70) 0.627 268 948 441 169 92 × 2 = 1 + 0.254 537 896 882 339 84;
  • 71) 0.254 537 896 882 339 84 × 2 = 0 + 0.509 075 793 764 679 68;
  • 72) 0.509 075 793 764 679 68 × 2 = 1 + 0.018 151 587 529 359 36;
  • 73) 0.018 151 587 529 359 36 × 2 = 0 + 0.036 303 175 058 718 72;
  • 74) 0.036 303 175 058 718 72 × 2 = 0 + 0.072 606 350 117 437 44;
  • 75) 0.072 606 350 117 437 44 × 2 = 0 + 0.145 212 700 234 874 88;
  • 76) 0.145 212 700 234 874 88 × 2 = 0 + 0.290 425 400 469 749 76;
  • 77) 0.290 425 400 469 749 76 × 2 = 0 + 0.580 850 800 939 499 52;
  • 78) 0.580 850 800 939 499 52 × 2 = 1 + 0.161 701 601 878 999 04;
  • 79) 0.161 701 601 878 999 04 × 2 = 0 + 0.323 403 203 757 998 08;
  • 80) 0.323 403 203 757 998 08 × 2 = 0 + 0.646 806 407 515 996 16;
  • 81) 0.646 806 407 515 996 16 × 2 = 1 + 0.293 612 815 031 992 32;
  • 82) 0.293 612 815 031 992 32 × 2 = 0 + 0.587 225 630 063 984 64;
  • 83) 0.587 225 630 063 984 64 × 2 = 1 + 0.174 451 260 127 969 28;
  • 84) 0.174 451 260 127 969 28 × 2 = 0 + 0.348 902 520 255 938 56;
  • 85) 0.348 902 520 255 938 56 × 2 = 0 + 0.697 805 040 511 877 12;
  • 86) 0.697 805 040 511 877 12 × 2 = 1 + 0.395 610 081 023 754 24;
  • 87) 0.395 610 081 023 754 24 × 2 = 0 + 0.791 220 162 047 508 48;
  • 88) 0.791 220 162 047 508 48 × 2 = 1 + 0.582 440 324 095 016 96;
  • 89) 0.582 440 324 095 016 96 × 2 = 1 + 0.164 880 648 190 033 92;
  • 90) 0.164 880 648 190 033 92 × 2 = 0 + 0.329 761 296 380 067 84;
  • 91) 0.329 761 296 380 067 84 × 2 = 0 + 0.659 522 592 760 135 68;
  • 92) 0.659 522 592 760 135 68 × 2 = 1 + 0.319 045 185 520 271 36;
  • 93) 0.319 045 185 520 271 36 × 2 = 0 + 0.638 090 371 040 542 72;
  • 94) 0.638 090 371 040 542 72 × 2 = 1 + 0.276 180 742 081 085 44;
  • 95) 0.276 180 742 081 085 44 × 2 = 0 + 0.552 361 484 162 170 88;
  • 96) 0.552 361 484 162 170 88 × 2 = 1 + 0.104 722 968 324 341 76;
  • 97) 0.104 722 968 324 341 76 × 2 = 0 + 0.209 445 936 648 683 52;
  • 98) 0.209 445 936 648 683 52 × 2 = 0 + 0.418 891 873 297 367 04;

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 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1110 0101 1001 1001 0101 0000 0100 1010 0101 1001 0101 00(2)

6. Positive number before normalization:

0.000 000 000 000 014 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1110 0101 1001 1001 0101 0000 0100 1010 0101 1001 0101 00(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 41(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1110 0101 1001 1001 0101 0000 0100 1010 0101 1001 0101 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1110 0101 1001 1001 0101 0000 0100 1010 0101 1001 0101 00(2) × 20 =

1.0000 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100(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 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100


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 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100 =

0000 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100


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 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100


Decimal number -0.000 000 000 000 014 41 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 1001 0110 0110 0101 0100 0001 0010 1001 0110 0101 0100

» Convert decimal -0.000 000 000 000 013 46 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