-0.000 000 000 000 176 557 423 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 423(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
-0.000 000 000 000 176 557 423(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:

|-0.000 000 000 000 176 557 423| = 0.000 000 000 000 176 557 423


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: 0.000 000 000 000 176 557 423.

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) 0.000 000 000 000 176 557 423 × 2 = 0 + 0.000 000 000 000 353 114 846;
  • 2) 0.000 000 000 000 353 114 846 × 2 = 0 + 0.000 000 000 000 706 229 692;
  • 3) 0.000 000 000 000 706 229 692 × 2 = 0 + 0.000 000 000 001 412 459 384;
  • 4) 0.000 000 000 001 412 459 384 × 2 = 0 + 0.000 000 000 002 824 918 768;
  • 5) 0.000 000 000 002 824 918 768 × 2 = 0 + 0.000 000 000 005 649 837 536;
  • 6) 0.000 000 000 005 649 837 536 × 2 = 0 + 0.000 000 000 011 299 675 072;
  • 7) 0.000 000 000 011 299 675 072 × 2 = 0 + 0.000 000 000 022 599 350 144;
  • 8) 0.000 000 000 022 599 350 144 × 2 = 0 + 0.000 000 000 045 198 700 288;
  • 9) 0.000 000 000 045 198 700 288 × 2 = 0 + 0.000 000 000 090 397 400 576;
  • 10) 0.000 000 000 090 397 400 576 × 2 = 0 + 0.000 000 000 180 794 801 152;
  • 11) 0.000 000 000 180 794 801 152 × 2 = 0 + 0.000 000 000 361 589 602 304;
  • 12) 0.000 000 000 361 589 602 304 × 2 = 0 + 0.000 000 000 723 179 204 608;
  • 13) 0.000 000 000 723 179 204 608 × 2 = 0 + 0.000 000 001 446 358 409 216;
  • 14) 0.000 000 001 446 358 409 216 × 2 = 0 + 0.000 000 002 892 716 818 432;
  • 15) 0.000 000 002 892 716 818 432 × 2 = 0 + 0.000 000 005 785 433 636 864;
  • 16) 0.000 000 005 785 433 636 864 × 2 = 0 + 0.000 000 011 570 867 273 728;
  • 17) 0.000 000 011 570 867 273 728 × 2 = 0 + 0.000 000 023 141 734 547 456;
  • 18) 0.000 000 023 141 734 547 456 × 2 = 0 + 0.000 000 046 283 469 094 912;
  • 19) 0.000 000 046 283 469 094 912 × 2 = 0 + 0.000 000 092 566 938 189 824;
  • 20) 0.000 000 092 566 938 189 824 × 2 = 0 + 0.000 000 185 133 876 379 648;
  • 21) 0.000 000 185 133 876 379 648 × 2 = 0 + 0.000 000 370 267 752 759 296;
  • 22) 0.000 000 370 267 752 759 296 × 2 = 0 + 0.000 000 740 535 505 518 592;
  • 23) 0.000 000 740 535 505 518 592 × 2 = 0 + 0.000 001 481 071 011 037 184;
  • 24) 0.000 001 481 071 011 037 184 × 2 = 0 + 0.000 002 962 142 022 074 368;
  • 25) 0.000 002 962 142 022 074 368 × 2 = 0 + 0.000 005 924 284 044 148 736;
  • 26) 0.000 005 924 284 044 148 736 × 2 = 0 + 0.000 011 848 568 088 297 472;
  • 27) 0.000 011 848 568 088 297 472 × 2 = 0 + 0.000 023 697 136 176 594 944;
  • 28) 0.000 023 697 136 176 594 944 × 2 = 0 + 0.000 047 394 272 353 189 888;
  • 29) 0.000 047 394 272 353 189 888 × 2 = 0 + 0.000 094 788 544 706 379 776;
  • 30) 0.000 094 788 544 706 379 776 × 2 = 0 + 0.000 189 577 089 412 759 552;
  • 31) 0.000 189 577 089 412 759 552 × 2 = 0 + 0.000 379 154 178 825 519 104;
  • 32) 0.000 379 154 178 825 519 104 × 2 = 0 + 0.000 758 308 357 651 038 208;
  • 33) 0.000 758 308 357 651 038 208 × 2 = 0 + 0.001 516 616 715 302 076 416;
  • 34) 0.001 516 616 715 302 076 416 × 2 = 0 + 0.003 033 233 430 604 152 832;
  • 35) 0.003 033 233 430 604 152 832 × 2 = 0 + 0.006 066 466 861 208 305 664;
  • 36) 0.006 066 466 861 208 305 664 × 2 = 0 + 0.012 132 933 722 416 611 328;
  • 37) 0.012 132 933 722 416 611 328 × 2 = 0 + 0.024 265 867 444 833 222 656;
  • 38) 0.024 265 867 444 833 222 656 × 2 = 0 + 0.048 531 734 889 666 445 312;
  • 39) 0.048 531 734 889 666 445 312 × 2 = 0 + 0.097 063 469 779 332 890 624;
  • 40) 0.097 063 469 779 332 890 624 × 2 = 0 + 0.194 126 939 558 665 781 248;
  • 41) 0.194 126 939 558 665 781 248 × 2 = 0 + 0.388 253 879 117 331 562 496;
  • 42) 0.388 253 879 117 331 562 496 × 2 = 0 + 0.776 507 758 234 663 124 992;
  • 43) 0.776 507 758 234 663 124 992 × 2 = 1 + 0.553 015 516 469 326 249 984;
  • 44) 0.553 015 516 469 326 249 984 × 2 = 1 + 0.106 031 032 938 652 499 968;
  • 45) 0.106 031 032 938 652 499 968 × 2 = 0 + 0.212 062 065 877 304 999 936;
  • 46) 0.212 062 065 877 304 999 936 × 2 = 0 + 0.424 124 131 754 609 999 872;
  • 47) 0.424 124 131 754 609 999 872 × 2 = 0 + 0.848 248 263 509 219 999 744;
  • 48) 0.848 248 263 509 219 999 744 × 2 = 1 + 0.696 496 527 018 439 999 488;
  • 49) 0.696 496 527 018 439 999 488 × 2 = 1 + 0.392 993 054 036 879 998 976;
  • 50) 0.392 993 054 036 879 998 976 × 2 = 0 + 0.785 986 108 073 759 997 952;
  • 51) 0.785 986 108 073 759 997 952 × 2 = 1 + 0.571 972 216 147 519 995 904;
  • 52) 0.571 972 216 147 519 995 904 × 2 = 1 + 0.143 944 432 295 039 991 808;
  • 53) 0.143 944 432 295 039 991 808 × 2 = 0 + 0.287 888 864 590 079 983 616;
  • 54) 0.287 888 864 590 079 983 616 × 2 = 0 + 0.575 777 729 180 159 967 232;
  • 55) 0.575 777 729 180 159 967 232 × 2 = 1 + 0.151 555 458 360 319 934 464;
  • 56) 0.151 555 458 360 319 934 464 × 2 = 0 + 0.303 110 916 720 639 868 928;
  • 57) 0.303 110 916 720 639 868 928 × 2 = 0 + 0.606 221 833 441 279 737 856;
  • 58) 0.606 221 833 441 279 737 856 × 2 = 1 + 0.212 443 666 882 559 475 712;
  • 59) 0.212 443 666 882 559 475 712 × 2 = 0 + 0.424 887 333 765 118 951 424;
  • 60) 0.424 887 333 765 118 951 424 × 2 = 0 + 0.849 774 667 530 237 902 848;
  • 61) 0.849 774 667 530 237 902 848 × 2 = 1 + 0.699 549 335 060 475 805 696;
  • 62) 0.699 549 335 060 475 805 696 × 2 = 1 + 0.399 098 670 120 951 611 392;
  • 63) 0.399 098 670 120 951 611 392 × 2 = 0 + 0.798 197 340 241 903 222 784;
  • 64) 0.798 197 340 241 903 222 784 × 2 = 1 + 0.596 394 680 483 806 445 568;
  • 65) 0.596 394 680 483 806 445 568 × 2 = 1 + 0.192 789 360 967 612 891 136;
  • 66) 0.192 789 360 967 612 891 136 × 2 = 0 + 0.385 578 721 935 225 782 272;
  • 67) 0.385 578 721 935 225 782 272 × 2 = 0 + 0.771 157 443 870 451 564 544;
  • 68) 0.771 157 443 870 451 564 544 × 2 = 1 + 0.542 314 887 740 903 129 088;
  • 69) 0.542 314 887 740 903 129 088 × 2 = 1 + 0.084 629 775 481 806 258 176;
  • 70) 0.084 629 775 481 806 258 176 × 2 = 0 + 0.169 259 550 963 612 516 352;
  • 71) 0.169 259 550 963 612 516 352 × 2 = 0 + 0.338 519 101 927 225 032 704;
  • 72) 0.338 519 101 927 225 032 704 × 2 = 0 + 0.677 038 203 854 450 065 408;
  • 73) 0.677 038 203 854 450 065 408 × 2 = 1 + 0.354 076 407 708 900 130 816;
  • 74) 0.354 076 407 708 900 130 816 × 2 = 0 + 0.708 152 815 417 800 261 632;
  • 75) 0.708 152 815 417 800 261 632 × 2 = 1 + 0.416 305 630 835 600 523 264;
  • 76) 0.416 305 630 835 600 523 264 × 2 = 0 + 0.832 611 261 671 201 046 528;
  • 77) 0.832 611 261 671 201 046 528 × 2 = 1 + 0.665 222 523 342 402 093 056;
  • 78) 0.665 222 523 342 402 093 056 × 2 = 1 + 0.330 445 046 684 804 186 112;
  • 79) 0.330 445 046 684 804 186 112 × 2 = 0 + 0.660 890 093 369 608 372 224;
  • 80) 0.660 890 093 369 608 372 224 × 2 = 1 + 0.321 780 186 739 216 744 448;
  • 81) 0.321 780 186 739 216 744 448 × 2 = 0 + 0.643 560 373 478 433 488 896;
  • 82) 0.643 560 373 478 433 488 896 × 2 = 1 + 0.287 120 746 956 866 977 792;
  • 83) 0.287 120 746 956 866 977 792 × 2 = 0 + 0.574 241 493 913 733 955 584;
  • 84) 0.574 241 493 913 733 955 584 × 2 = 1 + 0.148 482 987 827 467 911 168;
  • 85) 0.148 482 987 827 467 911 168 × 2 = 0 + 0.296 965 975 654 935 822 336;
  • 86) 0.296 965 975 654 935 822 336 × 2 = 0 + 0.593 931 951 309 871 644 672;
  • 87) 0.593 931 951 309 871 644 672 × 2 = 1 + 0.187 863 902 619 743 289 344;
  • 88) 0.187 863 902 619 743 289 344 × 2 = 0 + 0.375 727 805 239 486 578 688;
  • 89) 0.375 727 805 239 486 578 688 × 2 = 0 + 0.751 455 610 478 973 157 376;
  • 90) 0.751 455 610 478 973 157 376 × 2 = 1 + 0.502 911 220 957 946 314 752;
  • 91) 0.502 911 220 957 946 314 752 × 2 = 1 + 0.005 822 441 915 892 629 504;
  • 92) 0.005 822 441 915 892 629 504 × 2 = 0 + 0.011 644 883 831 785 259 008;
  • 93) 0.011 644 883 831 785 259 008 × 2 = 0 + 0.023 289 767 663 570 518 016;
  • 94) 0.023 289 767 663 570 518 016 × 2 = 0 + 0.046 579 535 327 141 036 032;
  • 95) 0.046 579 535 327 141 036 032 × 2 = 0 + 0.093 159 070 654 282 072 064;

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:


0.000 000 000 000 176 557 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1001 1000 1010 1101 0101 0010 0110 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1001 1000 1010 1101 0101 0010 0110 000(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:


0.000 000 000 000 176 557 423(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1001 1000 1010 1101 0101 0010 0110 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1101 1001 1000 1010 1101 0101 0010 0110 000(2) × 20 =

1.1000 1101 1001 0010 0110 1100 1100 0101 0110 1010 1001 0011 0000(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 0110 1100 1100 0101 0110 1010 1001 0011 0000


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 0110 1100 1100 0101 0110 1010 1001 0011 0000 =

1000 1101 1001 0010 0110 1100 1100 0101 0110 1010 1001 0011 0000


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 0110 1100 1100 0101 0110 1010 1001 0011 0000


Decimal number -0.000 000 000 000 176 557 423 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 1100 1100 0101 0110 1010 1001 0011 0000

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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