-0.000 000 000 000 176 557 345 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 345(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
-0.000 000 000 000 176 557 345(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:

|-0.000 000 000 000 176 557 345| = 0.000 000 000 000 176 557 345


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: 0.000 000 000 000 176 557 345.

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) 0.000 000 000 000 176 557 345 × 2 = 0 + 0.000 000 000 000 353 114 69;
  • 2) 0.000 000 000 000 353 114 69 × 2 = 0 + 0.000 000 000 000 706 229 38;
  • 3) 0.000 000 000 000 706 229 38 × 2 = 0 + 0.000 000 000 001 412 458 76;
  • 4) 0.000 000 000 001 412 458 76 × 2 = 0 + 0.000 000 000 002 824 917 52;
  • 5) 0.000 000 000 002 824 917 52 × 2 = 0 + 0.000 000 000 005 649 835 04;
  • 6) 0.000 000 000 005 649 835 04 × 2 = 0 + 0.000 000 000 011 299 670 08;
  • 7) 0.000 000 000 011 299 670 08 × 2 = 0 + 0.000 000 000 022 599 340 16;
  • 8) 0.000 000 000 022 599 340 16 × 2 = 0 + 0.000 000 000 045 198 680 32;
  • 9) 0.000 000 000 045 198 680 32 × 2 = 0 + 0.000 000 000 090 397 360 64;
  • 10) 0.000 000 000 090 397 360 64 × 2 = 0 + 0.000 000 000 180 794 721 28;
  • 11) 0.000 000 000 180 794 721 28 × 2 = 0 + 0.000 000 000 361 589 442 56;
  • 12) 0.000 000 000 361 589 442 56 × 2 = 0 + 0.000 000 000 723 178 885 12;
  • 13) 0.000 000 000 723 178 885 12 × 2 = 0 + 0.000 000 001 446 357 770 24;
  • 14) 0.000 000 001 446 357 770 24 × 2 = 0 + 0.000 000 002 892 715 540 48;
  • 15) 0.000 000 002 892 715 540 48 × 2 = 0 + 0.000 000 005 785 431 080 96;
  • 16) 0.000 000 005 785 431 080 96 × 2 = 0 + 0.000 000 011 570 862 161 92;
  • 17) 0.000 000 011 570 862 161 92 × 2 = 0 + 0.000 000 023 141 724 323 84;
  • 18) 0.000 000 023 141 724 323 84 × 2 = 0 + 0.000 000 046 283 448 647 68;
  • 19) 0.000 000 046 283 448 647 68 × 2 = 0 + 0.000 000 092 566 897 295 36;
  • 20) 0.000 000 092 566 897 295 36 × 2 = 0 + 0.000 000 185 133 794 590 72;
  • 21) 0.000 000 185 133 794 590 72 × 2 = 0 + 0.000 000 370 267 589 181 44;
  • 22) 0.000 000 370 267 589 181 44 × 2 = 0 + 0.000 000 740 535 178 362 88;
  • 23) 0.000 000 740 535 178 362 88 × 2 = 0 + 0.000 001 481 070 356 725 76;
  • 24) 0.000 001 481 070 356 725 76 × 2 = 0 + 0.000 002 962 140 713 451 52;
  • 25) 0.000 002 962 140 713 451 52 × 2 = 0 + 0.000 005 924 281 426 903 04;
  • 26) 0.000 005 924 281 426 903 04 × 2 = 0 + 0.000 011 848 562 853 806 08;
  • 27) 0.000 011 848 562 853 806 08 × 2 = 0 + 0.000 023 697 125 707 612 16;
  • 28) 0.000 023 697 125 707 612 16 × 2 = 0 + 0.000 047 394 251 415 224 32;
  • 29) 0.000 047 394 251 415 224 32 × 2 = 0 + 0.000 094 788 502 830 448 64;
  • 30) 0.000 094 788 502 830 448 64 × 2 = 0 + 0.000 189 577 005 660 897 28;
  • 31) 0.000 189 577 005 660 897 28 × 2 = 0 + 0.000 379 154 011 321 794 56;
  • 32) 0.000 379 154 011 321 794 56 × 2 = 0 + 0.000 758 308 022 643 589 12;
  • 33) 0.000 758 308 022 643 589 12 × 2 = 0 + 0.001 516 616 045 287 178 24;
  • 34) 0.001 516 616 045 287 178 24 × 2 = 0 + 0.003 033 232 090 574 356 48;
  • 35) 0.003 033 232 090 574 356 48 × 2 = 0 + 0.006 066 464 181 148 712 96;
  • 36) 0.006 066 464 181 148 712 96 × 2 = 0 + 0.012 132 928 362 297 425 92;
  • 37) 0.012 132 928 362 297 425 92 × 2 = 0 + 0.024 265 856 724 594 851 84;
  • 38) 0.024 265 856 724 594 851 84 × 2 = 0 + 0.048 531 713 449 189 703 68;
  • 39) 0.048 531 713 449 189 703 68 × 2 = 0 + 0.097 063 426 898 379 407 36;
  • 40) 0.097 063 426 898 379 407 36 × 2 = 0 + 0.194 126 853 796 758 814 72;
  • 41) 0.194 126 853 796 758 814 72 × 2 = 0 + 0.388 253 707 593 517 629 44;
  • 42) 0.388 253 707 593 517 629 44 × 2 = 0 + 0.776 507 415 187 035 258 88;
  • 43) 0.776 507 415 187 035 258 88 × 2 = 1 + 0.553 014 830 374 070 517 76;
  • 44) 0.553 014 830 374 070 517 76 × 2 = 1 + 0.106 029 660 748 141 035 52;
  • 45) 0.106 029 660 748 141 035 52 × 2 = 0 + 0.212 059 321 496 282 071 04;
  • 46) 0.212 059 321 496 282 071 04 × 2 = 0 + 0.424 118 642 992 564 142 08;
  • 47) 0.424 118 642 992 564 142 08 × 2 = 0 + 0.848 237 285 985 128 284 16;
  • 48) 0.848 237 285 985 128 284 16 × 2 = 1 + 0.696 474 571 970 256 568 32;
  • 49) 0.696 474 571 970 256 568 32 × 2 = 1 + 0.392 949 143 940 513 136 64;
  • 50) 0.392 949 143 940 513 136 64 × 2 = 0 + 0.785 898 287 881 026 273 28;
  • 51) 0.785 898 287 881 026 273 28 × 2 = 1 + 0.571 796 575 762 052 546 56;
  • 52) 0.571 796 575 762 052 546 56 × 2 = 1 + 0.143 593 151 524 105 093 12;
  • 53) 0.143 593 151 524 105 093 12 × 2 = 0 + 0.287 186 303 048 210 186 24;
  • 54) 0.287 186 303 048 210 186 24 × 2 = 0 + 0.574 372 606 096 420 372 48;
  • 55) 0.574 372 606 096 420 372 48 × 2 = 1 + 0.148 745 212 192 840 744 96;
  • 56) 0.148 745 212 192 840 744 96 × 2 = 0 + 0.297 490 424 385 681 489 92;
  • 57) 0.297 490 424 385 681 489 92 × 2 = 0 + 0.594 980 848 771 362 979 84;
  • 58) 0.594 980 848 771 362 979 84 × 2 = 1 + 0.189 961 697 542 725 959 68;
  • 59) 0.189 961 697 542 725 959 68 × 2 = 0 + 0.379 923 395 085 451 919 36;
  • 60) 0.379 923 395 085 451 919 36 × 2 = 0 + 0.759 846 790 170 903 838 72;
  • 61) 0.759 846 790 170 903 838 72 × 2 = 1 + 0.519 693 580 341 807 677 44;
  • 62) 0.519 693 580 341 807 677 44 × 2 = 1 + 0.039 387 160 683 615 354 88;
  • 63) 0.039 387 160 683 615 354 88 × 2 = 0 + 0.078 774 321 367 230 709 76;
  • 64) 0.078 774 321 367 230 709 76 × 2 = 0 + 0.157 548 642 734 461 419 52;
  • 65) 0.157 548 642 734 461 419 52 × 2 = 0 + 0.315 097 285 468 922 839 04;
  • 66) 0.315 097 285 468 922 839 04 × 2 = 0 + 0.630 194 570 937 845 678 08;
  • 67) 0.630 194 570 937 845 678 08 × 2 = 1 + 0.260 389 141 875 691 356 16;
  • 68) 0.260 389 141 875 691 356 16 × 2 = 0 + 0.520 778 283 751 382 712 32;
  • 69) 0.520 778 283 751 382 712 32 × 2 = 1 + 0.041 556 567 502 765 424 64;
  • 70) 0.041 556 567 502 765 424 64 × 2 = 0 + 0.083 113 135 005 530 849 28;
  • 71) 0.083 113 135 005 530 849 28 × 2 = 0 + 0.166 226 270 011 061 698 56;
  • 72) 0.166 226 270 011 061 698 56 × 2 = 0 + 0.332 452 540 022 123 397 12;
  • 73) 0.332 452 540 022 123 397 12 × 2 = 0 + 0.664 905 080 044 246 794 24;
  • 74) 0.664 905 080 044 246 794 24 × 2 = 1 + 0.329 810 160 088 493 588 48;
  • 75) 0.329 810 160 088 493 588 48 × 2 = 0 + 0.659 620 320 176 987 176 96;
  • 76) 0.659 620 320 176 987 176 96 × 2 = 1 + 0.319 240 640 353 974 353 92;
  • 77) 0.319 240 640 353 974 353 92 × 2 = 0 + 0.638 481 280 707 948 707 84;
  • 78) 0.638 481 280 707 948 707 84 × 2 = 1 + 0.276 962 561 415 897 415 68;
  • 79) 0.276 962 561 415 897 415 68 × 2 = 0 + 0.553 925 122 831 794 831 36;
  • 80) 0.553 925 122 831 794 831 36 × 2 = 1 + 0.107 850 245 663 589 662 72;
  • 81) 0.107 850 245 663 589 662 72 × 2 = 0 + 0.215 700 491 327 179 325 44;
  • 82) 0.215 700 491 327 179 325 44 × 2 = 0 + 0.431 400 982 654 358 650 88;
  • 83) 0.431 400 982 654 358 650 88 × 2 = 0 + 0.862 801 965 308 717 301 76;
  • 84) 0.862 801 965 308 717 301 76 × 2 = 1 + 0.725 603 930 617 434 603 52;
  • 85) 0.725 603 930 617 434 603 52 × 2 = 1 + 0.451 207 861 234 869 207 04;
  • 86) 0.451 207 861 234 869 207 04 × 2 = 0 + 0.902 415 722 469 738 414 08;
  • 87) 0.902 415 722 469 738 414 08 × 2 = 1 + 0.804 831 444 939 476 828 16;
  • 88) 0.804 831 444 939 476 828 16 × 2 = 1 + 0.609 662 889 878 953 656 32;
  • 89) 0.609 662 889 878 953 656 32 × 2 = 1 + 0.219 325 779 757 907 312 64;
  • 90) 0.219 325 779 757 907 312 64 × 2 = 0 + 0.438 651 559 515 814 625 28;
  • 91) 0.438 651 559 515 814 625 28 × 2 = 0 + 0.877 303 119 031 629 250 56;
  • 92) 0.877 303 119 031 629 250 56 × 2 = 1 + 0.754 606 238 063 258 501 12;
  • 93) 0.754 606 238 063 258 501 12 × 2 = 1 + 0.509 212 476 126 517 002 24;
  • 94) 0.509 212 476 126 517 002 24 × 2 = 1 + 0.018 424 952 253 034 004 48;
  • 95) 0.018 424 952 253 034 004 48 × 2 = 0 + 0.036 849 904 506 068 008 96;

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:


0.000 000 000 000 176 557 345(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 0010 1000 0101 0101 0001 1011 1001 110(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 345(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 0010 1000 0101 0101 0001 1011 1001 110(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:


0.000 000 000 000 176 557 345(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 0010 1000 0101 0101 0001 1011 1001 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 0010 1000 0101 0101 0001 1011 1001 110(2) × 20 =

1.1000 1101 1001 0010 0110 0001 0100 0010 1010 1000 1101 1100 1110(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 0001 0100 0010 1010 1000 1101 1100 1110


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 0001 0100 0010 1010 1000 1101 1100 1110 =

1000 1101 1001 0010 0110 0001 0100 0010 1010 1000 1101 1100 1110


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 0001 0100 0010 1010 1000 1101 1100 1110


Decimal number -0.000 000 000 000 176 557 345 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 0001 0100 0010 1010 1000 1101 1100 1110

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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