-0.000 000 000 000 176 558 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 558 17(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
-0.000 000 000 000 176 558 17(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:

|-0.000 000 000 000 176 558 17| = 0.000 000 000 000 176 558 17


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: 0.000 000 000 000 176 558 17.

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) 0.000 000 000 000 176 558 17 × 2 = 0 + 0.000 000 000 000 353 116 34;
  • 2) 0.000 000 000 000 353 116 34 × 2 = 0 + 0.000 000 000 000 706 232 68;
  • 3) 0.000 000 000 000 706 232 68 × 2 = 0 + 0.000 000 000 001 412 465 36;
  • 4) 0.000 000 000 001 412 465 36 × 2 = 0 + 0.000 000 000 002 824 930 72;
  • 5) 0.000 000 000 002 824 930 72 × 2 = 0 + 0.000 000 000 005 649 861 44;
  • 6) 0.000 000 000 005 649 861 44 × 2 = 0 + 0.000 000 000 011 299 722 88;
  • 7) 0.000 000 000 011 299 722 88 × 2 = 0 + 0.000 000 000 022 599 445 76;
  • 8) 0.000 000 000 022 599 445 76 × 2 = 0 + 0.000 000 000 045 198 891 52;
  • 9) 0.000 000 000 045 198 891 52 × 2 = 0 + 0.000 000 000 090 397 783 04;
  • 10) 0.000 000 000 090 397 783 04 × 2 = 0 + 0.000 000 000 180 795 566 08;
  • 11) 0.000 000 000 180 795 566 08 × 2 = 0 + 0.000 000 000 361 591 132 16;
  • 12) 0.000 000 000 361 591 132 16 × 2 = 0 + 0.000 000 000 723 182 264 32;
  • 13) 0.000 000 000 723 182 264 32 × 2 = 0 + 0.000 000 001 446 364 528 64;
  • 14) 0.000 000 001 446 364 528 64 × 2 = 0 + 0.000 000 002 892 729 057 28;
  • 15) 0.000 000 002 892 729 057 28 × 2 = 0 + 0.000 000 005 785 458 114 56;
  • 16) 0.000 000 005 785 458 114 56 × 2 = 0 + 0.000 000 011 570 916 229 12;
  • 17) 0.000 000 011 570 916 229 12 × 2 = 0 + 0.000 000 023 141 832 458 24;
  • 18) 0.000 000 023 141 832 458 24 × 2 = 0 + 0.000 000 046 283 664 916 48;
  • 19) 0.000 000 046 283 664 916 48 × 2 = 0 + 0.000 000 092 567 329 832 96;
  • 20) 0.000 000 092 567 329 832 96 × 2 = 0 + 0.000 000 185 134 659 665 92;
  • 21) 0.000 000 185 134 659 665 92 × 2 = 0 + 0.000 000 370 269 319 331 84;
  • 22) 0.000 000 370 269 319 331 84 × 2 = 0 + 0.000 000 740 538 638 663 68;
  • 23) 0.000 000 740 538 638 663 68 × 2 = 0 + 0.000 001 481 077 277 327 36;
  • 24) 0.000 001 481 077 277 327 36 × 2 = 0 + 0.000 002 962 154 554 654 72;
  • 25) 0.000 002 962 154 554 654 72 × 2 = 0 + 0.000 005 924 309 109 309 44;
  • 26) 0.000 005 924 309 109 309 44 × 2 = 0 + 0.000 011 848 618 218 618 88;
  • 27) 0.000 011 848 618 218 618 88 × 2 = 0 + 0.000 023 697 236 437 237 76;
  • 28) 0.000 023 697 236 437 237 76 × 2 = 0 + 0.000 047 394 472 874 475 52;
  • 29) 0.000 047 394 472 874 475 52 × 2 = 0 + 0.000 094 788 945 748 951 04;
  • 30) 0.000 094 788 945 748 951 04 × 2 = 0 + 0.000 189 577 891 497 902 08;
  • 31) 0.000 189 577 891 497 902 08 × 2 = 0 + 0.000 379 155 782 995 804 16;
  • 32) 0.000 379 155 782 995 804 16 × 2 = 0 + 0.000 758 311 565 991 608 32;
  • 33) 0.000 758 311 565 991 608 32 × 2 = 0 + 0.001 516 623 131 983 216 64;
  • 34) 0.001 516 623 131 983 216 64 × 2 = 0 + 0.003 033 246 263 966 433 28;
  • 35) 0.003 033 246 263 966 433 28 × 2 = 0 + 0.006 066 492 527 932 866 56;
  • 36) 0.006 066 492 527 932 866 56 × 2 = 0 + 0.012 132 985 055 865 733 12;
  • 37) 0.012 132 985 055 865 733 12 × 2 = 0 + 0.024 265 970 111 731 466 24;
  • 38) 0.024 265 970 111 731 466 24 × 2 = 0 + 0.048 531 940 223 462 932 48;
  • 39) 0.048 531 940 223 462 932 48 × 2 = 0 + 0.097 063 880 446 925 864 96;
  • 40) 0.097 063 880 446 925 864 96 × 2 = 0 + 0.194 127 760 893 851 729 92;
  • 41) 0.194 127 760 893 851 729 92 × 2 = 0 + 0.388 255 521 787 703 459 84;
  • 42) 0.388 255 521 787 703 459 84 × 2 = 0 + 0.776 511 043 575 406 919 68;
  • 43) 0.776 511 043 575 406 919 68 × 2 = 1 + 0.553 022 087 150 813 839 36;
  • 44) 0.553 022 087 150 813 839 36 × 2 = 1 + 0.106 044 174 301 627 678 72;
  • 45) 0.106 044 174 301 627 678 72 × 2 = 0 + 0.212 088 348 603 255 357 44;
  • 46) 0.212 088 348 603 255 357 44 × 2 = 0 + 0.424 176 697 206 510 714 88;
  • 47) 0.424 176 697 206 510 714 88 × 2 = 0 + 0.848 353 394 413 021 429 76;
  • 48) 0.848 353 394 413 021 429 76 × 2 = 1 + 0.696 706 788 826 042 859 52;
  • 49) 0.696 706 788 826 042 859 52 × 2 = 1 + 0.393 413 577 652 085 719 04;
  • 50) 0.393 413 577 652 085 719 04 × 2 = 0 + 0.786 827 155 304 171 438 08;
  • 51) 0.786 827 155 304 171 438 08 × 2 = 1 + 0.573 654 310 608 342 876 16;
  • 52) 0.573 654 310 608 342 876 16 × 2 = 1 + 0.147 308 621 216 685 752 32;
  • 53) 0.147 308 621 216 685 752 32 × 2 = 0 + 0.294 617 242 433 371 504 64;
  • 54) 0.294 617 242 433 371 504 64 × 2 = 0 + 0.589 234 484 866 743 009 28;
  • 55) 0.589 234 484 866 743 009 28 × 2 = 1 + 0.178 468 969 733 486 018 56;
  • 56) 0.178 468 969 733 486 018 56 × 2 = 0 + 0.356 937 939 466 972 037 12;
  • 57) 0.356 937 939 466 972 037 12 × 2 = 0 + 0.713 875 878 933 944 074 24;
  • 58) 0.713 875 878 933 944 074 24 × 2 = 1 + 0.427 751 757 867 888 148 48;
  • 59) 0.427 751 757 867 888 148 48 × 2 = 0 + 0.855 503 515 735 776 296 96;
  • 60) 0.855 503 515 735 776 296 96 × 2 = 1 + 0.711 007 031 471 552 593 92;
  • 61) 0.711 007 031 471 552 593 92 × 2 = 1 + 0.422 014 062 943 105 187 84;
  • 62) 0.422 014 062 943 105 187 84 × 2 = 0 + 0.844 028 125 886 210 375 68;
  • 63) 0.844 028 125 886 210 375 68 × 2 = 1 + 0.688 056 251 772 420 751 36;
  • 64) 0.688 056 251 772 420 751 36 × 2 = 1 + 0.376 112 503 544 841 502 72;
  • 65) 0.376 112 503 544 841 502 72 × 2 = 0 + 0.752 225 007 089 683 005 44;
  • 66) 0.752 225 007 089 683 005 44 × 2 = 1 + 0.504 450 014 179 366 010 88;
  • 67) 0.504 450 014 179 366 010 88 × 2 = 1 + 0.008 900 028 358 732 021 76;
  • 68) 0.008 900 028 358 732 021 76 × 2 = 0 + 0.017 800 056 717 464 043 52;
  • 69) 0.017 800 056 717 464 043 52 × 2 = 0 + 0.035 600 113 434 928 087 04;
  • 70) 0.035 600 113 434 928 087 04 × 2 = 0 + 0.071 200 226 869 856 174 08;
  • 71) 0.071 200 226 869 856 174 08 × 2 = 0 + 0.142 400 453 739 712 348 16;
  • 72) 0.142 400 453 739 712 348 16 × 2 = 0 + 0.284 800 907 479 424 696 32;
  • 73) 0.284 800 907 479 424 696 32 × 2 = 0 + 0.569 601 814 958 849 392 64;
  • 74) 0.569 601 814 958 849 392 64 × 2 = 1 + 0.139 203 629 917 698 785 28;
  • 75) 0.139 203 629 917 698 785 28 × 2 = 0 + 0.278 407 259 835 397 570 56;
  • 76) 0.278 407 259 835 397 570 56 × 2 = 0 + 0.556 814 519 670 795 141 12;
  • 77) 0.556 814 519 670 795 141 12 × 2 = 1 + 0.113 629 039 341 590 282 24;
  • 78) 0.113 629 039 341 590 282 24 × 2 = 0 + 0.227 258 078 683 180 564 48;
  • 79) 0.227 258 078 683 180 564 48 × 2 = 0 + 0.454 516 157 366 361 128 96;
  • 80) 0.454 516 157 366 361 128 96 × 2 = 0 + 0.909 032 314 732 722 257 92;
  • 81) 0.909 032 314 732 722 257 92 × 2 = 1 + 0.818 064 629 465 444 515 84;
  • 82) 0.818 064 629 465 444 515 84 × 2 = 1 + 0.636 129 258 930 889 031 68;
  • 83) 0.636 129 258 930 889 031 68 × 2 = 1 + 0.272 258 517 861 778 063 36;
  • 84) 0.272 258 517 861 778 063 36 × 2 = 0 + 0.544 517 035 723 556 126 72;
  • 85) 0.544 517 035 723 556 126 72 × 2 = 1 + 0.089 034 071 447 112 253 44;
  • 86) 0.089 034 071 447 112 253 44 × 2 = 0 + 0.178 068 142 894 224 506 88;
  • 87) 0.178 068 142 894 224 506 88 × 2 = 0 + 0.356 136 285 788 449 013 76;
  • 88) 0.356 136 285 788 449 013 76 × 2 = 0 + 0.712 272 571 576 898 027 52;
  • 89) 0.712 272 571 576 898 027 52 × 2 = 1 + 0.424 545 143 153 796 055 04;
  • 90) 0.424 545 143 153 796 055 04 × 2 = 0 + 0.849 090 286 307 592 110 08;
  • 91) 0.849 090 286 307 592 110 08 × 2 = 1 + 0.698 180 572 615 184 220 16;
  • 92) 0.698 180 572 615 184 220 16 × 2 = 1 + 0.396 361 145 230 368 440 32;
  • 93) 0.396 361 145 230 368 440 32 × 2 = 0 + 0.792 722 290 460 736 880 64;
  • 94) 0.792 722 290 460 736 880 64 × 2 = 1 + 0.585 444 580 921 473 761 28;
  • 95) 0.585 444 580 921 473 761 28 × 2 = 1 + 0.170 889 161 842 947 522 56;

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:


0.000 000 000 000 176 558 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1011 0110 0000 0100 1000 1110 1000 1011 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 558 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1011 0110 0000 0100 1000 1110 1000 1011 011(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:


0.000 000 000 000 176 558 17(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1011 0110 0000 0100 1000 1110 1000 1011 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1011 0110 0000 0100 1000 1110 1000 1011 011(2) × 20 =

1.1000 1101 1001 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011(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 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011


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 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011 =

1000 1101 1001 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011


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 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011


Decimal number -0.000 000 000 000 176 558 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1101 1011 0000 0010 0100 0111 0100 0101 1011

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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