-0.000 000 000 000 176 557 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 92(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 92(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 92| = 0.000 000 000 000 176 557 92


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 92.

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 92 × 2 = 0 + 0.000 000 000 000 353 115 84;
  • 2) 0.000 000 000 000 353 115 84 × 2 = 0 + 0.000 000 000 000 706 231 68;
  • 3) 0.000 000 000 000 706 231 68 × 2 = 0 + 0.000 000 000 001 412 463 36;
  • 4) 0.000 000 000 001 412 463 36 × 2 = 0 + 0.000 000 000 002 824 926 72;
  • 5) 0.000 000 000 002 824 926 72 × 2 = 0 + 0.000 000 000 005 649 853 44;
  • 6) 0.000 000 000 005 649 853 44 × 2 = 0 + 0.000 000 000 011 299 706 88;
  • 7) 0.000 000 000 011 299 706 88 × 2 = 0 + 0.000 000 000 022 599 413 76;
  • 8) 0.000 000 000 022 599 413 76 × 2 = 0 + 0.000 000 000 045 198 827 52;
  • 9) 0.000 000 000 045 198 827 52 × 2 = 0 + 0.000 000 000 090 397 655 04;
  • 10) 0.000 000 000 090 397 655 04 × 2 = 0 + 0.000 000 000 180 795 310 08;
  • 11) 0.000 000 000 180 795 310 08 × 2 = 0 + 0.000 000 000 361 590 620 16;
  • 12) 0.000 000 000 361 590 620 16 × 2 = 0 + 0.000 000 000 723 181 240 32;
  • 13) 0.000 000 000 723 181 240 32 × 2 = 0 + 0.000 000 001 446 362 480 64;
  • 14) 0.000 000 001 446 362 480 64 × 2 = 0 + 0.000 000 002 892 724 961 28;
  • 15) 0.000 000 002 892 724 961 28 × 2 = 0 + 0.000 000 005 785 449 922 56;
  • 16) 0.000 000 005 785 449 922 56 × 2 = 0 + 0.000 000 011 570 899 845 12;
  • 17) 0.000 000 011 570 899 845 12 × 2 = 0 + 0.000 000 023 141 799 690 24;
  • 18) 0.000 000 023 141 799 690 24 × 2 = 0 + 0.000 000 046 283 599 380 48;
  • 19) 0.000 000 046 283 599 380 48 × 2 = 0 + 0.000 000 092 567 198 760 96;
  • 20) 0.000 000 092 567 198 760 96 × 2 = 0 + 0.000 000 185 134 397 521 92;
  • 21) 0.000 000 185 134 397 521 92 × 2 = 0 + 0.000 000 370 268 795 043 84;
  • 22) 0.000 000 370 268 795 043 84 × 2 = 0 + 0.000 000 740 537 590 087 68;
  • 23) 0.000 000 740 537 590 087 68 × 2 = 0 + 0.000 001 481 075 180 175 36;
  • 24) 0.000 001 481 075 180 175 36 × 2 = 0 + 0.000 002 962 150 360 350 72;
  • 25) 0.000 002 962 150 360 350 72 × 2 = 0 + 0.000 005 924 300 720 701 44;
  • 26) 0.000 005 924 300 720 701 44 × 2 = 0 + 0.000 011 848 601 441 402 88;
  • 27) 0.000 011 848 601 441 402 88 × 2 = 0 + 0.000 023 697 202 882 805 76;
  • 28) 0.000 023 697 202 882 805 76 × 2 = 0 + 0.000 047 394 405 765 611 52;
  • 29) 0.000 047 394 405 765 611 52 × 2 = 0 + 0.000 094 788 811 531 223 04;
  • 30) 0.000 094 788 811 531 223 04 × 2 = 0 + 0.000 189 577 623 062 446 08;
  • 31) 0.000 189 577 623 062 446 08 × 2 = 0 + 0.000 379 155 246 124 892 16;
  • 32) 0.000 379 155 246 124 892 16 × 2 = 0 + 0.000 758 310 492 249 784 32;
  • 33) 0.000 758 310 492 249 784 32 × 2 = 0 + 0.001 516 620 984 499 568 64;
  • 34) 0.001 516 620 984 499 568 64 × 2 = 0 + 0.003 033 241 968 999 137 28;
  • 35) 0.003 033 241 968 999 137 28 × 2 = 0 + 0.006 066 483 937 998 274 56;
  • 36) 0.006 066 483 937 998 274 56 × 2 = 0 + 0.012 132 967 875 996 549 12;
  • 37) 0.012 132 967 875 996 549 12 × 2 = 0 + 0.024 265 935 751 993 098 24;
  • 38) 0.024 265 935 751 993 098 24 × 2 = 0 + 0.048 531 871 503 986 196 48;
  • 39) 0.048 531 871 503 986 196 48 × 2 = 0 + 0.097 063 743 007 972 392 96;
  • 40) 0.097 063 743 007 972 392 96 × 2 = 0 + 0.194 127 486 015 944 785 92;
  • 41) 0.194 127 486 015 944 785 92 × 2 = 0 + 0.388 254 972 031 889 571 84;
  • 42) 0.388 254 972 031 889 571 84 × 2 = 0 + 0.776 509 944 063 779 143 68;
  • 43) 0.776 509 944 063 779 143 68 × 2 = 1 + 0.553 019 888 127 558 287 36;
  • 44) 0.553 019 888 127 558 287 36 × 2 = 1 + 0.106 039 776 255 116 574 72;
  • 45) 0.106 039 776 255 116 574 72 × 2 = 0 + 0.212 079 552 510 233 149 44;
  • 46) 0.212 079 552 510 233 149 44 × 2 = 0 + 0.424 159 105 020 466 298 88;
  • 47) 0.424 159 105 020 466 298 88 × 2 = 0 + 0.848 318 210 040 932 597 76;
  • 48) 0.848 318 210 040 932 597 76 × 2 = 1 + 0.696 636 420 081 865 195 52;
  • 49) 0.696 636 420 081 865 195 52 × 2 = 1 + 0.393 272 840 163 730 391 04;
  • 50) 0.393 272 840 163 730 391 04 × 2 = 0 + 0.786 545 680 327 460 782 08;
  • 51) 0.786 545 680 327 460 782 08 × 2 = 1 + 0.573 091 360 654 921 564 16;
  • 52) 0.573 091 360 654 921 564 16 × 2 = 1 + 0.146 182 721 309 843 128 32;
  • 53) 0.146 182 721 309 843 128 32 × 2 = 0 + 0.292 365 442 619 686 256 64;
  • 54) 0.292 365 442 619 686 256 64 × 2 = 0 + 0.584 730 885 239 372 513 28;
  • 55) 0.584 730 885 239 372 513 28 × 2 = 1 + 0.169 461 770 478 745 026 56;
  • 56) 0.169 461 770 478 745 026 56 × 2 = 0 + 0.338 923 540 957 490 053 12;
  • 57) 0.338 923 540 957 490 053 12 × 2 = 0 + 0.677 847 081 914 980 106 24;
  • 58) 0.677 847 081 914 980 106 24 × 2 = 1 + 0.355 694 163 829 960 212 48;
  • 59) 0.355 694 163 829 960 212 48 × 2 = 0 + 0.711 388 327 659 920 424 96;
  • 60) 0.711 388 327 659 920 424 96 × 2 = 1 + 0.422 776 655 319 840 849 92;
  • 61) 0.422 776 655 319 840 849 92 × 2 = 0 + 0.845 553 310 639 681 699 84;
  • 62) 0.845 553 310 639 681 699 84 × 2 = 1 + 0.691 106 621 279 363 399 68;
  • 63) 0.691 106 621 279 363 399 68 × 2 = 1 + 0.382 213 242 558 726 799 36;
  • 64) 0.382 213 242 558 726 799 36 × 2 = 0 + 0.764 426 485 117 453 598 72;
  • 65) 0.764 426 485 117 453 598 72 × 2 = 1 + 0.528 852 970 234 907 197 44;
  • 66) 0.528 852 970 234 907 197 44 × 2 = 1 + 0.057 705 940 469 814 394 88;
  • 67) 0.057 705 940 469 814 394 88 × 2 = 0 + 0.115 411 880 939 628 789 76;
  • 68) 0.115 411 880 939 628 789 76 × 2 = 0 + 0.230 823 761 879 257 579 52;
  • 69) 0.230 823 761 879 257 579 52 × 2 = 0 + 0.461 647 523 758 515 159 04;
  • 70) 0.461 647 523 758 515 159 04 × 2 = 0 + 0.923 295 047 517 030 318 08;
  • 71) 0.923 295 047 517 030 318 08 × 2 = 1 + 0.846 590 095 034 060 636 16;
  • 72) 0.846 590 095 034 060 636 16 × 2 = 1 + 0.693 180 190 068 121 272 32;
  • 73) 0.693 180 190 068 121 272 32 × 2 = 1 + 0.386 360 380 136 242 544 64;
  • 74) 0.386 360 380 136 242 544 64 × 2 = 0 + 0.772 720 760 272 485 089 28;
  • 75) 0.772 720 760 272 485 089 28 × 2 = 1 + 0.545 441 520 544 970 178 56;
  • 76) 0.545 441 520 544 970 178 56 × 2 = 1 + 0.090 883 041 089 940 357 12;
  • 77) 0.090 883 041 089 940 357 12 × 2 = 0 + 0.181 766 082 179 880 714 24;
  • 78) 0.181 766 082 179 880 714 24 × 2 = 0 + 0.363 532 164 359 761 428 48;
  • 79) 0.363 532 164 359 761 428 48 × 2 = 0 + 0.727 064 328 719 522 856 96;
  • 80) 0.727 064 328 719 522 856 96 × 2 = 1 + 0.454 128 657 439 045 713 92;
  • 81) 0.454 128 657 439 045 713 92 × 2 = 0 + 0.908 257 314 878 091 427 84;
  • 82) 0.908 257 314 878 091 427 84 × 2 = 1 + 0.816 514 629 756 182 855 68;
  • 83) 0.816 514 629 756 182 855 68 × 2 = 1 + 0.633 029 259 512 365 711 36;
  • 84) 0.633 029 259 512 365 711 36 × 2 = 1 + 0.266 058 519 024 731 422 72;
  • 85) 0.266 058 519 024 731 422 72 × 2 = 0 + 0.532 117 038 049 462 845 44;
  • 86) 0.532 117 038 049 462 845 44 × 2 = 1 + 0.064 234 076 098 925 690 88;
  • 87) 0.064 234 076 098 925 690 88 × 2 = 0 + 0.128 468 152 197 851 381 76;
  • 88) 0.128 468 152 197 851 381 76 × 2 = 0 + 0.256 936 304 395 702 763 52;
  • 89) 0.256 936 304 395 702 763 52 × 2 = 0 + 0.513 872 608 791 405 527 04;
  • 90) 0.513 872 608 791 405 527 04 × 2 = 1 + 0.027 745 217 582 811 054 08;
  • 91) 0.027 745 217 582 811 054 08 × 2 = 0 + 0.055 490 435 165 622 108 16;
  • 92) 0.055 490 435 165 622 108 16 × 2 = 0 + 0.110 980 870 331 244 216 32;
  • 93) 0.110 980 870 331 244 216 32 × 2 = 0 + 0.221 961 740 662 488 432 64;
  • 94) 0.221 961 740 662 488 432 64 × 2 = 0 + 0.443 923 481 324 976 865 28;
  • 95) 0.443 923 481 324 976 865 28 × 2 = 0 + 0.887 846 962 649 953 730 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:


0.000 000 000 000 176 557 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 1100 0011 1011 0001 0111 0100 0100 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 1100 0011 1011 0001 0111 0100 0100 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 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 1100 0011 1011 0001 0111 0100 0100 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0110 1100 0011 1011 0001 0111 0100 0100 000(2) × 20 =

1.1000 1101 1001 0010 1011 0110 0001 1101 1000 1011 1010 0010 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 1011 0110 0001 1101 1000 1011 1010 0010 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 1011 0110 0001 1101 1000 1011 1010 0010 0000 =

1000 1101 1001 0010 1011 0110 0001 1101 1000 1011 1010 0010 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 1011 0110 0001 1101 1000 1011 1010 0010 0000


Decimal number -0.000 000 000 000 176 557 92 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1011 0110 0001 1101 1000 1011 1010 0010 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