-0.000 000 000 000 176 557 665 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 665(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
-0.000 000 000 000 176 557 665(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:

|-0.000 000 000 000 176 557 665| = 0.000 000 000 000 176 557 665


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: 0.000 000 000 000 176 557 665.

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) 0.000 000 000 000 176 557 665 × 2 = 0 + 0.000 000 000 000 353 115 33;
  • 2) 0.000 000 000 000 353 115 33 × 2 = 0 + 0.000 000 000 000 706 230 66;
  • 3) 0.000 000 000 000 706 230 66 × 2 = 0 + 0.000 000 000 001 412 461 32;
  • 4) 0.000 000 000 001 412 461 32 × 2 = 0 + 0.000 000 000 002 824 922 64;
  • 5) 0.000 000 000 002 824 922 64 × 2 = 0 + 0.000 000 000 005 649 845 28;
  • 6) 0.000 000 000 005 649 845 28 × 2 = 0 + 0.000 000 000 011 299 690 56;
  • 7) 0.000 000 000 011 299 690 56 × 2 = 0 + 0.000 000 000 022 599 381 12;
  • 8) 0.000 000 000 022 599 381 12 × 2 = 0 + 0.000 000 000 045 198 762 24;
  • 9) 0.000 000 000 045 198 762 24 × 2 = 0 + 0.000 000 000 090 397 524 48;
  • 10) 0.000 000 000 090 397 524 48 × 2 = 0 + 0.000 000 000 180 795 048 96;
  • 11) 0.000 000 000 180 795 048 96 × 2 = 0 + 0.000 000 000 361 590 097 92;
  • 12) 0.000 000 000 361 590 097 92 × 2 = 0 + 0.000 000 000 723 180 195 84;
  • 13) 0.000 000 000 723 180 195 84 × 2 = 0 + 0.000 000 001 446 360 391 68;
  • 14) 0.000 000 001 446 360 391 68 × 2 = 0 + 0.000 000 002 892 720 783 36;
  • 15) 0.000 000 002 892 720 783 36 × 2 = 0 + 0.000 000 005 785 441 566 72;
  • 16) 0.000 000 005 785 441 566 72 × 2 = 0 + 0.000 000 011 570 883 133 44;
  • 17) 0.000 000 011 570 883 133 44 × 2 = 0 + 0.000 000 023 141 766 266 88;
  • 18) 0.000 000 023 141 766 266 88 × 2 = 0 + 0.000 000 046 283 532 533 76;
  • 19) 0.000 000 046 283 532 533 76 × 2 = 0 + 0.000 000 092 567 065 067 52;
  • 20) 0.000 000 092 567 065 067 52 × 2 = 0 + 0.000 000 185 134 130 135 04;
  • 21) 0.000 000 185 134 130 135 04 × 2 = 0 + 0.000 000 370 268 260 270 08;
  • 22) 0.000 000 370 268 260 270 08 × 2 = 0 + 0.000 000 740 536 520 540 16;
  • 23) 0.000 000 740 536 520 540 16 × 2 = 0 + 0.000 001 481 073 041 080 32;
  • 24) 0.000 001 481 073 041 080 32 × 2 = 0 + 0.000 002 962 146 082 160 64;
  • 25) 0.000 002 962 146 082 160 64 × 2 = 0 + 0.000 005 924 292 164 321 28;
  • 26) 0.000 005 924 292 164 321 28 × 2 = 0 + 0.000 011 848 584 328 642 56;
  • 27) 0.000 011 848 584 328 642 56 × 2 = 0 + 0.000 023 697 168 657 285 12;
  • 28) 0.000 023 697 168 657 285 12 × 2 = 0 + 0.000 047 394 337 314 570 24;
  • 29) 0.000 047 394 337 314 570 24 × 2 = 0 + 0.000 094 788 674 629 140 48;
  • 30) 0.000 094 788 674 629 140 48 × 2 = 0 + 0.000 189 577 349 258 280 96;
  • 31) 0.000 189 577 349 258 280 96 × 2 = 0 + 0.000 379 154 698 516 561 92;
  • 32) 0.000 379 154 698 516 561 92 × 2 = 0 + 0.000 758 309 397 033 123 84;
  • 33) 0.000 758 309 397 033 123 84 × 2 = 0 + 0.001 516 618 794 066 247 68;
  • 34) 0.001 516 618 794 066 247 68 × 2 = 0 + 0.003 033 237 588 132 495 36;
  • 35) 0.003 033 237 588 132 495 36 × 2 = 0 + 0.006 066 475 176 264 990 72;
  • 36) 0.006 066 475 176 264 990 72 × 2 = 0 + 0.012 132 950 352 529 981 44;
  • 37) 0.012 132 950 352 529 981 44 × 2 = 0 + 0.024 265 900 705 059 962 88;
  • 38) 0.024 265 900 705 059 962 88 × 2 = 0 + 0.048 531 801 410 119 925 76;
  • 39) 0.048 531 801 410 119 925 76 × 2 = 0 + 0.097 063 602 820 239 851 52;
  • 40) 0.097 063 602 820 239 851 52 × 2 = 0 + 0.194 127 205 640 479 703 04;
  • 41) 0.194 127 205 640 479 703 04 × 2 = 0 + 0.388 254 411 280 959 406 08;
  • 42) 0.388 254 411 280 959 406 08 × 2 = 0 + 0.776 508 822 561 918 812 16;
  • 43) 0.776 508 822 561 918 812 16 × 2 = 1 + 0.553 017 645 123 837 624 32;
  • 44) 0.553 017 645 123 837 624 32 × 2 = 1 + 0.106 035 290 247 675 248 64;
  • 45) 0.106 035 290 247 675 248 64 × 2 = 0 + 0.212 070 580 495 350 497 28;
  • 46) 0.212 070 580 495 350 497 28 × 2 = 0 + 0.424 141 160 990 700 994 56;
  • 47) 0.424 141 160 990 700 994 56 × 2 = 0 + 0.848 282 321 981 401 989 12;
  • 48) 0.848 282 321 981 401 989 12 × 2 = 1 + 0.696 564 643 962 803 978 24;
  • 49) 0.696 564 643 962 803 978 24 × 2 = 1 + 0.393 129 287 925 607 956 48;
  • 50) 0.393 129 287 925 607 956 48 × 2 = 0 + 0.786 258 575 851 215 912 96;
  • 51) 0.786 258 575 851 215 912 96 × 2 = 1 + 0.572 517 151 702 431 825 92;
  • 52) 0.572 517 151 702 431 825 92 × 2 = 1 + 0.145 034 303 404 863 651 84;
  • 53) 0.145 034 303 404 863 651 84 × 2 = 0 + 0.290 068 606 809 727 303 68;
  • 54) 0.290 068 606 809 727 303 68 × 2 = 0 + 0.580 137 213 619 454 607 36;
  • 55) 0.580 137 213 619 454 607 36 × 2 = 1 + 0.160 274 427 238 909 214 72;
  • 56) 0.160 274 427 238 909 214 72 × 2 = 0 + 0.320 548 854 477 818 429 44;
  • 57) 0.320 548 854 477 818 429 44 × 2 = 0 + 0.641 097 708 955 636 858 88;
  • 58) 0.641 097 708 955 636 858 88 × 2 = 1 + 0.282 195 417 911 273 717 76;
  • 59) 0.282 195 417 911 273 717 76 × 2 = 0 + 0.564 390 835 822 547 435 52;
  • 60) 0.564 390 835 822 547 435 52 × 2 = 1 + 0.128 781 671 645 094 871 04;
  • 61) 0.128 781 671 645 094 871 04 × 2 = 0 + 0.257 563 343 290 189 742 08;
  • 62) 0.257 563 343 290 189 742 08 × 2 = 0 + 0.515 126 686 580 379 484 16;
  • 63) 0.515 126 686 580 379 484 16 × 2 = 1 + 0.030 253 373 160 758 968 32;
  • 64) 0.030 253 373 160 758 968 32 × 2 = 0 + 0.060 506 746 321 517 936 64;
  • 65) 0.060 506 746 321 517 936 64 × 2 = 0 + 0.121 013 492 643 035 873 28;
  • 66) 0.121 013 492 643 035 873 28 × 2 = 0 + 0.242 026 985 286 071 746 56;
  • 67) 0.242 026 985 286 071 746 56 × 2 = 0 + 0.484 053 970 572 143 493 12;
  • 68) 0.484 053 970 572 143 493 12 × 2 = 0 + 0.968 107 941 144 286 986 24;
  • 69) 0.968 107 941 144 286 986 24 × 2 = 1 + 0.936 215 882 288 573 972 48;
  • 70) 0.936 215 882 288 573 972 48 × 2 = 1 + 0.872 431 764 577 147 944 96;
  • 71) 0.872 431 764 577 147 944 96 × 2 = 1 + 0.744 863 529 154 295 889 92;
  • 72) 0.744 863 529 154 295 889 92 × 2 = 1 + 0.489 727 058 308 591 779 84;
  • 73) 0.489 727 058 308 591 779 84 × 2 = 0 + 0.979 454 116 617 183 559 68;
  • 74) 0.979 454 116 617 183 559 68 × 2 = 1 + 0.958 908 233 234 367 119 36;
  • 75) 0.958 908 233 234 367 119 36 × 2 = 1 + 0.917 816 466 468 734 238 72;
  • 76) 0.917 816 466 468 734 238 72 × 2 = 1 + 0.835 632 932 937 468 477 44;
  • 77) 0.835 632 932 937 468 477 44 × 2 = 1 + 0.671 265 865 874 936 954 88;
  • 78) 0.671 265 865 874 936 954 88 × 2 = 1 + 0.342 531 731 749 873 909 76;
  • 79) 0.342 531 731 749 873 909 76 × 2 = 0 + 0.685 063 463 499 747 819 52;
  • 80) 0.685 063 463 499 747 819 52 × 2 = 1 + 0.370 126 926 999 495 639 04;
  • 81) 0.370 126 926 999 495 639 04 × 2 = 0 + 0.740 253 853 998 991 278 08;
  • 82) 0.740 253 853 998 991 278 08 × 2 = 1 + 0.480 507 707 997 982 556 16;
  • 83) 0.480 507 707 997 982 556 16 × 2 = 0 + 0.961 015 415 995 965 112 32;
  • 84) 0.961 015 415 995 965 112 32 × 2 = 1 + 0.922 030 831 991 930 224 64;
  • 85) 0.922 030 831 991 930 224 64 × 2 = 1 + 0.844 061 663 983 860 449 28;
  • 86) 0.844 061 663 983 860 449 28 × 2 = 1 + 0.688 123 327 967 720 898 56;
  • 87) 0.688 123 327 967 720 898 56 × 2 = 1 + 0.376 246 655 935 441 797 12;
  • 88) 0.376 246 655 935 441 797 12 × 2 = 0 + 0.752 493 311 870 883 594 24;
  • 89) 0.752 493 311 870 883 594 24 × 2 = 1 + 0.504 986 623 741 767 188 48;
  • 90) 0.504 986 623 741 767 188 48 × 2 = 1 + 0.009 973 247 483 534 376 96;
  • 91) 0.009 973 247 483 534 376 96 × 2 = 0 + 0.019 946 494 967 068 753 92;
  • 92) 0.019 946 494 967 068 753 92 × 2 = 0 + 0.039 892 989 934 137 507 84;
  • 93) 0.039 892 989 934 137 507 84 × 2 = 0 + 0.079 785 979 868 275 015 68;
  • 94) 0.079 785 979 868 275 015 68 × 2 = 0 + 0.159 571 959 736 550 031 36;
  • 95) 0.159 571 959 736 550 031 36 × 2 = 0 + 0.319 143 919 473 100 062 72;

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:


0.000 000 000 000 176 557 665(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0000 1111 0111 1101 0101 1110 1100 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 665(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0000 1111 0111 1101 0101 1110 1100 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:


0.000 000 000 000 176 557 665(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0000 1111 0111 1101 0101 1110 1100 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0000 1111 0111 1101 0101 1110 1100 000(2) × 20 =

1.1000 1101 1001 0010 1001 0000 0111 1011 1110 1010 1111 0110 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 1001 0000 0111 1011 1110 1010 1111 0110 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 1001 0000 0111 1011 1110 1010 1111 0110 0000 =

1000 1101 1001 0010 1001 0000 0111 1011 1110 1010 1111 0110 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 1001 0000 0111 1011 1110 1010 1111 0110 0000


Decimal number -0.000 000 000 000 176 557 665 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1001 0000 0111 1011 1110 1010 1111 0110 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