-0.000 000 000 000 176 558 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 558 08(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 08(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 08| = 0.000 000 000 000 176 558 08


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 08.

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 08 × 2 = 0 + 0.000 000 000 000 353 116 16;
  • 2) 0.000 000 000 000 353 116 16 × 2 = 0 + 0.000 000 000 000 706 232 32;
  • 3) 0.000 000 000 000 706 232 32 × 2 = 0 + 0.000 000 000 001 412 464 64;
  • 4) 0.000 000 000 001 412 464 64 × 2 = 0 + 0.000 000 000 002 824 929 28;
  • 5) 0.000 000 000 002 824 929 28 × 2 = 0 + 0.000 000 000 005 649 858 56;
  • 6) 0.000 000 000 005 649 858 56 × 2 = 0 + 0.000 000 000 011 299 717 12;
  • 7) 0.000 000 000 011 299 717 12 × 2 = 0 + 0.000 000 000 022 599 434 24;
  • 8) 0.000 000 000 022 599 434 24 × 2 = 0 + 0.000 000 000 045 198 868 48;
  • 9) 0.000 000 000 045 198 868 48 × 2 = 0 + 0.000 000 000 090 397 736 96;
  • 10) 0.000 000 000 090 397 736 96 × 2 = 0 + 0.000 000 000 180 795 473 92;
  • 11) 0.000 000 000 180 795 473 92 × 2 = 0 + 0.000 000 000 361 590 947 84;
  • 12) 0.000 000 000 361 590 947 84 × 2 = 0 + 0.000 000 000 723 181 895 68;
  • 13) 0.000 000 000 723 181 895 68 × 2 = 0 + 0.000 000 001 446 363 791 36;
  • 14) 0.000 000 001 446 363 791 36 × 2 = 0 + 0.000 000 002 892 727 582 72;
  • 15) 0.000 000 002 892 727 582 72 × 2 = 0 + 0.000 000 005 785 455 165 44;
  • 16) 0.000 000 005 785 455 165 44 × 2 = 0 + 0.000 000 011 570 910 330 88;
  • 17) 0.000 000 011 570 910 330 88 × 2 = 0 + 0.000 000 023 141 820 661 76;
  • 18) 0.000 000 023 141 820 661 76 × 2 = 0 + 0.000 000 046 283 641 323 52;
  • 19) 0.000 000 046 283 641 323 52 × 2 = 0 + 0.000 000 092 567 282 647 04;
  • 20) 0.000 000 092 567 282 647 04 × 2 = 0 + 0.000 000 185 134 565 294 08;
  • 21) 0.000 000 185 134 565 294 08 × 2 = 0 + 0.000 000 370 269 130 588 16;
  • 22) 0.000 000 370 269 130 588 16 × 2 = 0 + 0.000 000 740 538 261 176 32;
  • 23) 0.000 000 740 538 261 176 32 × 2 = 0 + 0.000 001 481 076 522 352 64;
  • 24) 0.000 001 481 076 522 352 64 × 2 = 0 + 0.000 002 962 153 044 705 28;
  • 25) 0.000 002 962 153 044 705 28 × 2 = 0 + 0.000 005 924 306 089 410 56;
  • 26) 0.000 005 924 306 089 410 56 × 2 = 0 + 0.000 011 848 612 178 821 12;
  • 27) 0.000 011 848 612 178 821 12 × 2 = 0 + 0.000 023 697 224 357 642 24;
  • 28) 0.000 023 697 224 357 642 24 × 2 = 0 + 0.000 047 394 448 715 284 48;
  • 29) 0.000 047 394 448 715 284 48 × 2 = 0 + 0.000 094 788 897 430 568 96;
  • 30) 0.000 094 788 897 430 568 96 × 2 = 0 + 0.000 189 577 794 861 137 92;
  • 31) 0.000 189 577 794 861 137 92 × 2 = 0 + 0.000 379 155 589 722 275 84;
  • 32) 0.000 379 155 589 722 275 84 × 2 = 0 + 0.000 758 311 179 444 551 68;
  • 33) 0.000 758 311 179 444 551 68 × 2 = 0 + 0.001 516 622 358 889 103 36;
  • 34) 0.001 516 622 358 889 103 36 × 2 = 0 + 0.003 033 244 717 778 206 72;
  • 35) 0.003 033 244 717 778 206 72 × 2 = 0 + 0.006 066 489 435 556 413 44;
  • 36) 0.006 066 489 435 556 413 44 × 2 = 0 + 0.012 132 978 871 112 826 88;
  • 37) 0.012 132 978 871 112 826 88 × 2 = 0 + 0.024 265 957 742 225 653 76;
  • 38) 0.024 265 957 742 225 653 76 × 2 = 0 + 0.048 531 915 484 451 307 52;
  • 39) 0.048 531 915 484 451 307 52 × 2 = 0 + 0.097 063 830 968 902 615 04;
  • 40) 0.097 063 830 968 902 615 04 × 2 = 0 + 0.194 127 661 937 805 230 08;
  • 41) 0.194 127 661 937 805 230 08 × 2 = 0 + 0.388 255 323 875 610 460 16;
  • 42) 0.388 255 323 875 610 460 16 × 2 = 0 + 0.776 510 647 751 220 920 32;
  • 43) 0.776 510 647 751 220 920 32 × 2 = 1 + 0.553 021 295 502 441 840 64;
  • 44) 0.553 021 295 502 441 840 64 × 2 = 1 + 0.106 042 591 004 883 681 28;
  • 45) 0.106 042 591 004 883 681 28 × 2 = 0 + 0.212 085 182 009 767 362 56;
  • 46) 0.212 085 182 009 767 362 56 × 2 = 0 + 0.424 170 364 019 534 725 12;
  • 47) 0.424 170 364 019 534 725 12 × 2 = 0 + 0.848 340 728 039 069 450 24;
  • 48) 0.848 340 728 039 069 450 24 × 2 = 1 + 0.696 681 456 078 138 900 48;
  • 49) 0.696 681 456 078 138 900 48 × 2 = 1 + 0.393 362 912 156 277 800 96;
  • 50) 0.393 362 912 156 277 800 96 × 2 = 0 + 0.786 725 824 312 555 601 92;
  • 51) 0.786 725 824 312 555 601 92 × 2 = 1 + 0.573 451 648 625 111 203 84;
  • 52) 0.573 451 648 625 111 203 84 × 2 = 1 + 0.146 903 297 250 222 407 68;
  • 53) 0.146 903 297 250 222 407 68 × 2 = 0 + 0.293 806 594 500 444 815 36;
  • 54) 0.293 806 594 500 444 815 36 × 2 = 0 + 0.587 613 189 000 889 630 72;
  • 55) 0.587 613 189 000 889 630 72 × 2 = 1 + 0.175 226 378 001 779 261 44;
  • 56) 0.175 226 378 001 779 261 44 × 2 = 0 + 0.350 452 756 003 558 522 88;
  • 57) 0.350 452 756 003 558 522 88 × 2 = 0 + 0.700 905 512 007 117 045 76;
  • 58) 0.700 905 512 007 117 045 76 × 2 = 1 + 0.401 811 024 014 234 091 52;
  • 59) 0.401 811 024 014 234 091 52 × 2 = 0 + 0.803 622 048 028 468 183 04;
  • 60) 0.803 622 048 028 468 183 04 × 2 = 1 + 0.607 244 096 056 936 366 08;
  • 61) 0.607 244 096 056 936 366 08 × 2 = 1 + 0.214 488 192 113 872 732 16;
  • 62) 0.214 488 192 113 872 732 16 × 2 = 0 + 0.428 976 384 227 745 464 32;
  • 63) 0.428 976 384 227 745 464 32 × 2 = 0 + 0.857 952 768 455 490 928 64;
  • 64) 0.857 952 768 455 490 928 64 × 2 = 1 + 0.715 905 536 910 981 857 28;
  • 65) 0.715 905 536 910 981 857 28 × 2 = 1 + 0.431 811 073 821 963 714 56;
  • 66) 0.431 811 073 821 963 714 56 × 2 = 0 + 0.863 622 147 643 927 429 12;
  • 67) 0.863 622 147 643 927 429 12 × 2 = 1 + 0.727 244 295 287 854 858 24;
  • 68) 0.727 244 295 287 854 858 24 × 2 = 1 + 0.454 488 590 575 709 716 48;
  • 69) 0.454 488 590 575 709 716 48 × 2 = 0 + 0.908 977 181 151 419 432 96;
  • 70) 0.908 977 181 151 419 432 96 × 2 = 1 + 0.817 954 362 302 838 865 92;
  • 71) 0.817 954 362 302 838 865 92 × 2 = 1 + 0.635 908 724 605 677 731 84;
  • 72) 0.635 908 724 605 677 731 84 × 2 = 1 + 0.271 817 449 211 355 463 68;
  • 73) 0.271 817 449 211 355 463 68 × 2 = 0 + 0.543 634 898 422 710 927 36;
  • 74) 0.543 634 898 422 710 927 36 × 2 = 1 + 0.087 269 796 845 421 854 72;
  • 75) 0.087 269 796 845 421 854 72 × 2 = 0 + 0.174 539 593 690 843 709 44;
  • 76) 0.174 539 593 690 843 709 44 × 2 = 0 + 0.349 079 187 381 687 418 88;
  • 77) 0.349 079 187 381 687 418 88 × 2 = 0 + 0.698 158 374 763 374 837 76;
  • 78) 0.698 158 374 763 374 837 76 × 2 = 1 + 0.396 316 749 526 749 675 52;
  • 79) 0.396 316 749 526 749 675 52 × 2 = 0 + 0.792 633 499 053 499 351 04;
  • 80) 0.792 633 499 053 499 351 04 × 2 = 1 + 0.585 266 998 106 998 702 08;
  • 81) 0.585 266 998 106 998 702 08 × 2 = 1 + 0.170 533 996 213 997 404 16;
  • 82) 0.170 533 996 213 997 404 16 × 2 = 0 + 0.341 067 992 427 994 808 32;
  • 83) 0.341 067 992 427 994 808 32 × 2 = 0 + 0.682 135 984 855 989 616 64;
  • 84) 0.682 135 984 855 989 616 64 × 2 = 1 + 0.364 271 969 711 979 233 28;
  • 85) 0.364 271 969 711 979 233 28 × 2 = 0 + 0.728 543 939 423 958 466 56;
  • 86) 0.728 543 939 423 958 466 56 × 2 = 1 + 0.457 087 878 847 916 933 12;
  • 87) 0.457 087 878 847 916 933 12 × 2 = 0 + 0.914 175 757 695 833 866 24;
  • 88) 0.914 175 757 695 833 866 24 × 2 = 1 + 0.828 351 515 391 667 732 48;
  • 89) 0.828 351 515 391 667 732 48 × 2 = 1 + 0.656 703 030 783 335 464 96;
  • 90) 0.656 703 030 783 335 464 96 × 2 = 1 + 0.313 406 061 566 670 929 92;
  • 91) 0.313 406 061 566 670 929 92 × 2 = 0 + 0.626 812 123 133 341 859 84;
  • 92) 0.626 812 123 133 341 859 84 × 2 = 1 + 0.253 624 246 266 683 719 68;
  • 93) 0.253 624 246 266 683 719 68 × 2 = 0 + 0.507 248 492 533 367 439 36;
  • 94) 0.507 248 492 533 367 439 36 × 2 = 1 + 0.014 496 985 066 734 878 72;
  • 95) 0.014 496 985 066 734 878 72 × 2 = 0 + 0.028 993 970 133 469 757 44;

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 08(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1001 1011 0111 0100 0101 1001 0101 1101 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 558 08(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1001 1011 0111 0100 0101 1001 0101 1101 010(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 08(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1001 1011 0111 0100 0101 1001 0101 1101 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1001 1011 0111 0100 0101 1001 0101 1101 010(2) × 20 =

1.1000 1101 1001 0010 1100 1101 1011 1010 0010 1100 1010 1110 1010(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 1100 1101 1011 1010 0010 1100 1010 1110 1010


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 1100 1101 1011 1010 0010 1100 1010 1110 1010 =

1000 1101 1001 0010 1100 1101 1011 1010 0010 1100 1010 1110 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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 1100 1101 1011 1010 0010 1100 1010 1110 1010


Decimal number -0.000 000 000 000 176 558 08 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1100 1101 1011 1010 0010 1100 1010 1110 1010

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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