-0.000 000 000 000 176 556 46 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 46(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
-0.000 000 000 000 176 556 46(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:

|-0.000 000 000 000 176 556 46| = 0.000 000 000 000 176 556 46


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: 0.000 000 000 000 176 556 46.

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) 0.000 000 000 000 176 556 46 × 2 = 0 + 0.000 000 000 000 353 112 92;
  • 2) 0.000 000 000 000 353 112 92 × 2 = 0 + 0.000 000 000 000 706 225 84;
  • 3) 0.000 000 000 000 706 225 84 × 2 = 0 + 0.000 000 000 001 412 451 68;
  • 4) 0.000 000 000 001 412 451 68 × 2 = 0 + 0.000 000 000 002 824 903 36;
  • 5) 0.000 000 000 002 824 903 36 × 2 = 0 + 0.000 000 000 005 649 806 72;
  • 6) 0.000 000 000 005 649 806 72 × 2 = 0 + 0.000 000 000 011 299 613 44;
  • 7) 0.000 000 000 011 299 613 44 × 2 = 0 + 0.000 000 000 022 599 226 88;
  • 8) 0.000 000 000 022 599 226 88 × 2 = 0 + 0.000 000 000 045 198 453 76;
  • 9) 0.000 000 000 045 198 453 76 × 2 = 0 + 0.000 000 000 090 396 907 52;
  • 10) 0.000 000 000 090 396 907 52 × 2 = 0 + 0.000 000 000 180 793 815 04;
  • 11) 0.000 000 000 180 793 815 04 × 2 = 0 + 0.000 000 000 361 587 630 08;
  • 12) 0.000 000 000 361 587 630 08 × 2 = 0 + 0.000 000 000 723 175 260 16;
  • 13) 0.000 000 000 723 175 260 16 × 2 = 0 + 0.000 000 001 446 350 520 32;
  • 14) 0.000 000 001 446 350 520 32 × 2 = 0 + 0.000 000 002 892 701 040 64;
  • 15) 0.000 000 002 892 701 040 64 × 2 = 0 + 0.000 000 005 785 402 081 28;
  • 16) 0.000 000 005 785 402 081 28 × 2 = 0 + 0.000 000 011 570 804 162 56;
  • 17) 0.000 000 011 570 804 162 56 × 2 = 0 + 0.000 000 023 141 608 325 12;
  • 18) 0.000 000 023 141 608 325 12 × 2 = 0 + 0.000 000 046 283 216 650 24;
  • 19) 0.000 000 046 283 216 650 24 × 2 = 0 + 0.000 000 092 566 433 300 48;
  • 20) 0.000 000 092 566 433 300 48 × 2 = 0 + 0.000 000 185 132 866 600 96;
  • 21) 0.000 000 185 132 866 600 96 × 2 = 0 + 0.000 000 370 265 733 201 92;
  • 22) 0.000 000 370 265 733 201 92 × 2 = 0 + 0.000 000 740 531 466 403 84;
  • 23) 0.000 000 740 531 466 403 84 × 2 = 0 + 0.000 001 481 062 932 807 68;
  • 24) 0.000 001 481 062 932 807 68 × 2 = 0 + 0.000 002 962 125 865 615 36;
  • 25) 0.000 002 962 125 865 615 36 × 2 = 0 + 0.000 005 924 251 731 230 72;
  • 26) 0.000 005 924 251 731 230 72 × 2 = 0 + 0.000 011 848 503 462 461 44;
  • 27) 0.000 011 848 503 462 461 44 × 2 = 0 + 0.000 023 697 006 924 922 88;
  • 28) 0.000 023 697 006 924 922 88 × 2 = 0 + 0.000 047 394 013 849 845 76;
  • 29) 0.000 047 394 013 849 845 76 × 2 = 0 + 0.000 094 788 027 699 691 52;
  • 30) 0.000 094 788 027 699 691 52 × 2 = 0 + 0.000 189 576 055 399 383 04;
  • 31) 0.000 189 576 055 399 383 04 × 2 = 0 + 0.000 379 152 110 798 766 08;
  • 32) 0.000 379 152 110 798 766 08 × 2 = 0 + 0.000 758 304 221 597 532 16;
  • 33) 0.000 758 304 221 597 532 16 × 2 = 0 + 0.001 516 608 443 195 064 32;
  • 34) 0.001 516 608 443 195 064 32 × 2 = 0 + 0.003 033 216 886 390 128 64;
  • 35) 0.003 033 216 886 390 128 64 × 2 = 0 + 0.006 066 433 772 780 257 28;
  • 36) 0.006 066 433 772 780 257 28 × 2 = 0 + 0.012 132 867 545 560 514 56;
  • 37) 0.012 132 867 545 560 514 56 × 2 = 0 + 0.024 265 735 091 121 029 12;
  • 38) 0.024 265 735 091 121 029 12 × 2 = 0 + 0.048 531 470 182 242 058 24;
  • 39) 0.048 531 470 182 242 058 24 × 2 = 0 + 0.097 062 940 364 484 116 48;
  • 40) 0.097 062 940 364 484 116 48 × 2 = 0 + 0.194 125 880 728 968 232 96;
  • 41) 0.194 125 880 728 968 232 96 × 2 = 0 + 0.388 251 761 457 936 465 92;
  • 42) 0.388 251 761 457 936 465 92 × 2 = 0 + 0.776 503 522 915 872 931 84;
  • 43) 0.776 503 522 915 872 931 84 × 2 = 1 + 0.553 007 045 831 745 863 68;
  • 44) 0.553 007 045 831 745 863 68 × 2 = 1 + 0.106 014 091 663 491 727 36;
  • 45) 0.106 014 091 663 491 727 36 × 2 = 0 + 0.212 028 183 326 983 454 72;
  • 46) 0.212 028 183 326 983 454 72 × 2 = 0 + 0.424 056 366 653 966 909 44;
  • 47) 0.424 056 366 653 966 909 44 × 2 = 0 + 0.848 112 733 307 933 818 88;
  • 48) 0.848 112 733 307 933 818 88 × 2 = 1 + 0.696 225 466 615 867 637 76;
  • 49) 0.696 225 466 615 867 637 76 × 2 = 1 + 0.392 450 933 231 735 275 52;
  • 50) 0.392 450 933 231 735 275 52 × 2 = 0 + 0.784 901 866 463 470 551 04;
  • 51) 0.784 901 866 463 470 551 04 × 2 = 1 + 0.569 803 732 926 941 102 08;
  • 52) 0.569 803 732 926 941 102 08 × 2 = 1 + 0.139 607 465 853 882 204 16;
  • 53) 0.139 607 465 853 882 204 16 × 2 = 0 + 0.279 214 931 707 764 408 32;
  • 54) 0.279 214 931 707 764 408 32 × 2 = 0 + 0.558 429 863 415 528 816 64;
  • 55) 0.558 429 863 415 528 816 64 × 2 = 1 + 0.116 859 726 831 057 633 28;
  • 56) 0.116 859 726 831 057 633 28 × 2 = 0 + 0.233 719 453 662 115 266 56;
  • 57) 0.233 719 453 662 115 266 56 × 2 = 0 + 0.467 438 907 324 230 533 12;
  • 58) 0.467 438 907 324 230 533 12 × 2 = 0 + 0.934 877 814 648 461 066 24;
  • 59) 0.934 877 814 648 461 066 24 × 2 = 1 + 0.869 755 629 296 922 132 48;
  • 60) 0.869 755 629 296 922 132 48 × 2 = 1 + 0.739 511 258 593 844 264 96;
  • 61) 0.739 511 258 593 844 264 96 × 2 = 1 + 0.479 022 517 187 688 529 92;
  • 62) 0.479 022 517 187 688 529 92 × 2 = 0 + 0.958 045 034 375 377 059 84;
  • 63) 0.958 045 034 375 377 059 84 × 2 = 1 + 0.916 090 068 750 754 119 68;
  • 64) 0.916 090 068 750 754 119 68 × 2 = 1 + 0.832 180 137 501 508 239 36;
  • 65) 0.832 180 137 501 508 239 36 × 2 = 1 + 0.664 360 275 003 016 478 72;
  • 66) 0.664 360 275 003 016 478 72 × 2 = 1 + 0.328 720 550 006 032 957 44;
  • 67) 0.328 720 550 006 032 957 44 × 2 = 0 + 0.657 441 100 012 065 914 88;
  • 68) 0.657 441 100 012 065 914 88 × 2 = 1 + 0.314 882 200 024 131 829 76;
  • 69) 0.314 882 200 024 131 829 76 × 2 = 0 + 0.629 764 400 048 263 659 52;
  • 70) 0.629 764 400 048 263 659 52 × 2 = 1 + 0.259 528 800 096 527 319 04;
  • 71) 0.259 528 800 096 527 319 04 × 2 = 0 + 0.519 057 600 193 054 638 08;
  • 72) 0.519 057 600 193 054 638 08 × 2 = 1 + 0.038 115 200 386 109 276 16;
  • 73) 0.038 115 200 386 109 276 16 × 2 = 0 + 0.076 230 400 772 218 552 32;
  • 74) 0.076 230 400 772 218 552 32 × 2 = 0 + 0.152 460 801 544 437 104 64;
  • 75) 0.152 460 801 544 437 104 64 × 2 = 0 + 0.304 921 603 088 874 209 28;
  • 76) 0.304 921 603 088 874 209 28 × 2 = 0 + 0.609 843 206 177 748 418 56;
  • 77) 0.609 843 206 177 748 418 56 × 2 = 1 + 0.219 686 412 355 496 837 12;
  • 78) 0.219 686 412 355 496 837 12 × 2 = 0 + 0.439 372 824 710 993 674 24;
  • 79) 0.439 372 824 710 993 674 24 × 2 = 0 + 0.878 745 649 421 987 348 48;
  • 80) 0.878 745 649 421 987 348 48 × 2 = 1 + 0.757 491 298 843 974 696 96;
  • 81) 0.757 491 298 843 974 696 96 × 2 = 1 + 0.514 982 597 687 949 393 92;
  • 82) 0.514 982 597 687 949 393 92 × 2 = 1 + 0.029 965 195 375 898 787 84;
  • 83) 0.029 965 195 375 898 787 84 × 2 = 0 + 0.059 930 390 751 797 575 68;
  • 84) 0.059 930 390 751 797 575 68 × 2 = 0 + 0.119 860 781 503 595 151 36;
  • 85) 0.119 860 781 503 595 151 36 × 2 = 0 + 0.239 721 563 007 190 302 72;
  • 86) 0.239 721 563 007 190 302 72 × 2 = 0 + 0.479 443 126 014 380 605 44;
  • 87) 0.479 443 126 014 380 605 44 × 2 = 0 + 0.958 886 252 028 761 210 88;
  • 88) 0.958 886 252 028 761 210 88 × 2 = 1 + 0.917 772 504 057 522 421 76;
  • 89) 0.917 772 504 057 522 421 76 × 2 = 1 + 0.835 545 008 115 044 843 52;
  • 90) 0.835 545 008 115 044 843 52 × 2 = 1 + 0.671 090 016 230 089 687 04;
  • 91) 0.671 090 016 230 089 687 04 × 2 = 1 + 0.342 180 032 460 179 374 08;
  • 92) 0.342 180 032 460 179 374 08 × 2 = 0 + 0.684 360 064 920 358 748 16;
  • 93) 0.684 360 064 920 358 748 16 × 2 = 1 + 0.368 720 129 840 717 496 32;
  • 94) 0.368 720 129 840 717 496 32 × 2 = 0 + 0.737 440 259 681 434 992 64;
  • 95) 0.737 440 259 681 434 992 64 × 2 = 1 + 0.474 880 519 362 869 985 28;

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:


0.000 000 000 000 176 556 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1011 1101 0101 0000 1001 1100 0001 1110 101(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1011 1101 0101 0000 1001 1100 0001 1110 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 556 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1011 1101 0101 0000 1001 1100 0001 1110 101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1011 1101 0101 0000 1001 1100 0001 1110 101(2) × 20 =

1.1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101 =

1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101


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 0100

Mantissa (52 bits) =
1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101


Decimal number -0.000 000 000 000 176 556 46 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1101 1110 1010 1000 0100 1110 0000 1111 0101

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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