-0.000 000 000 000 176 557 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 63(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 63(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 63| = 0.000 000 000 000 176 557 63


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 63.

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 63 × 2 = 0 + 0.000 000 000 000 353 115 26;
  • 2) 0.000 000 000 000 353 115 26 × 2 = 0 + 0.000 000 000 000 706 230 52;
  • 3) 0.000 000 000 000 706 230 52 × 2 = 0 + 0.000 000 000 001 412 461 04;
  • 4) 0.000 000 000 001 412 461 04 × 2 = 0 + 0.000 000 000 002 824 922 08;
  • 5) 0.000 000 000 002 824 922 08 × 2 = 0 + 0.000 000 000 005 649 844 16;
  • 6) 0.000 000 000 005 649 844 16 × 2 = 0 + 0.000 000 000 011 299 688 32;
  • 7) 0.000 000 000 011 299 688 32 × 2 = 0 + 0.000 000 000 022 599 376 64;
  • 8) 0.000 000 000 022 599 376 64 × 2 = 0 + 0.000 000 000 045 198 753 28;
  • 9) 0.000 000 000 045 198 753 28 × 2 = 0 + 0.000 000 000 090 397 506 56;
  • 10) 0.000 000 000 090 397 506 56 × 2 = 0 + 0.000 000 000 180 795 013 12;
  • 11) 0.000 000 000 180 795 013 12 × 2 = 0 + 0.000 000 000 361 590 026 24;
  • 12) 0.000 000 000 361 590 026 24 × 2 = 0 + 0.000 000 000 723 180 052 48;
  • 13) 0.000 000 000 723 180 052 48 × 2 = 0 + 0.000 000 001 446 360 104 96;
  • 14) 0.000 000 001 446 360 104 96 × 2 = 0 + 0.000 000 002 892 720 209 92;
  • 15) 0.000 000 002 892 720 209 92 × 2 = 0 + 0.000 000 005 785 440 419 84;
  • 16) 0.000 000 005 785 440 419 84 × 2 = 0 + 0.000 000 011 570 880 839 68;
  • 17) 0.000 000 011 570 880 839 68 × 2 = 0 + 0.000 000 023 141 761 679 36;
  • 18) 0.000 000 023 141 761 679 36 × 2 = 0 + 0.000 000 046 283 523 358 72;
  • 19) 0.000 000 046 283 523 358 72 × 2 = 0 + 0.000 000 092 567 046 717 44;
  • 20) 0.000 000 092 567 046 717 44 × 2 = 0 + 0.000 000 185 134 093 434 88;
  • 21) 0.000 000 185 134 093 434 88 × 2 = 0 + 0.000 000 370 268 186 869 76;
  • 22) 0.000 000 370 268 186 869 76 × 2 = 0 + 0.000 000 740 536 373 739 52;
  • 23) 0.000 000 740 536 373 739 52 × 2 = 0 + 0.000 001 481 072 747 479 04;
  • 24) 0.000 001 481 072 747 479 04 × 2 = 0 + 0.000 002 962 145 494 958 08;
  • 25) 0.000 002 962 145 494 958 08 × 2 = 0 + 0.000 005 924 290 989 916 16;
  • 26) 0.000 005 924 290 989 916 16 × 2 = 0 + 0.000 011 848 581 979 832 32;
  • 27) 0.000 011 848 581 979 832 32 × 2 = 0 + 0.000 023 697 163 959 664 64;
  • 28) 0.000 023 697 163 959 664 64 × 2 = 0 + 0.000 047 394 327 919 329 28;
  • 29) 0.000 047 394 327 919 329 28 × 2 = 0 + 0.000 094 788 655 838 658 56;
  • 30) 0.000 094 788 655 838 658 56 × 2 = 0 + 0.000 189 577 311 677 317 12;
  • 31) 0.000 189 577 311 677 317 12 × 2 = 0 + 0.000 379 154 623 354 634 24;
  • 32) 0.000 379 154 623 354 634 24 × 2 = 0 + 0.000 758 309 246 709 268 48;
  • 33) 0.000 758 309 246 709 268 48 × 2 = 0 + 0.001 516 618 493 418 536 96;
  • 34) 0.001 516 618 493 418 536 96 × 2 = 0 + 0.003 033 236 986 837 073 92;
  • 35) 0.003 033 236 986 837 073 92 × 2 = 0 + 0.006 066 473 973 674 147 84;
  • 36) 0.006 066 473 973 674 147 84 × 2 = 0 + 0.012 132 947 947 348 295 68;
  • 37) 0.012 132 947 947 348 295 68 × 2 = 0 + 0.024 265 895 894 696 591 36;
  • 38) 0.024 265 895 894 696 591 36 × 2 = 0 + 0.048 531 791 789 393 182 72;
  • 39) 0.048 531 791 789 393 182 72 × 2 = 0 + 0.097 063 583 578 786 365 44;
  • 40) 0.097 063 583 578 786 365 44 × 2 = 0 + 0.194 127 167 157 572 730 88;
  • 41) 0.194 127 167 157 572 730 88 × 2 = 0 + 0.388 254 334 315 145 461 76;
  • 42) 0.388 254 334 315 145 461 76 × 2 = 0 + 0.776 508 668 630 290 923 52;
  • 43) 0.776 508 668 630 290 923 52 × 2 = 1 + 0.553 017 337 260 581 847 04;
  • 44) 0.553 017 337 260 581 847 04 × 2 = 1 + 0.106 034 674 521 163 694 08;
  • 45) 0.106 034 674 521 163 694 08 × 2 = 0 + 0.212 069 349 042 327 388 16;
  • 46) 0.212 069 349 042 327 388 16 × 2 = 0 + 0.424 138 698 084 654 776 32;
  • 47) 0.424 138 698 084 654 776 32 × 2 = 0 + 0.848 277 396 169 309 552 64;
  • 48) 0.848 277 396 169 309 552 64 × 2 = 1 + 0.696 554 792 338 619 105 28;
  • 49) 0.696 554 792 338 619 105 28 × 2 = 1 + 0.393 109 584 677 238 210 56;
  • 50) 0.393 109 584 677 238 210 56 × 2 = 0 + 0.786 219 169 354 476 421 12;
  • 51) 0.786 219 169 354 476 421 12 × 2 = 1 + 0.572 438 338 708 952 842 24;
  • 52) 0.572 438 338 708 952 842 24 × 2 = 1 + 0.144 876 677 417 905 684 48;
  • 53) 0.144 876 677 417 905 684 48 × 2 = 0 + 0.289 753 354 835 811 368 96;
  • 54) 0.289 753 354 835 811 368 96 × 2 = 0 + 0.579 506 709 671 622 737 92;
  • 55) 0.579 506 709 671 622 737 92 × 2 = 1 + 0.159 013 419 343 245 475 84;
  • 56) 0.159 013 419 343 245 475 84 × 2 = 0 + 0.318 026 838 686 490 951 68;
  • 57) 0.318 026 838 686 490 951 68 × 2 = 0 + 0.636 053 677 372 981 903 36;
  • 58) 0.636 053 677 372 981 903 36 × 2 = 1 + 0.272 107 354 745 963 806 72;
  • 59) 0.272 107 354 745 963 806 72 × 2 = 0 + 0.544 214 709 491 927 613 44;
  • 60) 0.544 214 709 491 927 613 44 × 2 = 1 + 0.088 429 418 983 855 226 88;
  • 61) 0.088 429 418 983 855 226 88 × 2 = 0 + 0.176 858 837 967 710 453 76;
  • 62) 0.176 858 837 967 710 453 76 × 2 = 0 + 0.353 717 675 935 420 907 52;
  • 63) 0.353 717 675 935 420 907 52 × 2 = 0 + 0.707 435 351 870 841 815 04;
  • 64) 0.707 435 351 870 841 815 04 × 2 = 1 + 0.414 870 703 741 683 630 08;
  • 65) 0.414 870 703 741 683 630 08 × 2 = 0 + 0.829 741 407 483 367 260 16;
  • 66) 0.829 741 407 483 367 260 16 × 2 = 1 + 0.659 482 814 966 734 520 32;
  • 67) 0.659 482 814 966 734 520 32 × 2 = 1 + 0.318 965 629 933 469 040 64;
  • 68) 0.318 965 629 933 469 040 64 × 2 = 0 + 0.637 931 259 866 938 081 28;
  • 69) 0.637 931 259 866 938 081 28 × 2 = 1 + 0.275 862 519 733 876 162 56;
  • 70) 0.275 862 519 733 876 162 56 × 2 = 0 + 0.551 725 039 467 752 325 12;
  • 71) 0.551 725 039 467 752 325 12 × 2 = 1 + 0.103 450 078 935 504 650 24;
  • 72) 0.103 450 078 935 504 650 24 × 2 = 0 + 0.206 900 157 871 009 300 48;
  • 73) 0.206 900 157 871 009 300 48 × 2 = 0 + 0.413 800 315 742 018 600 96;
  • 74) 0.413 800 315 742 018 600 96 × 2 = 0 + 0.827 600 631 484 037 201 92;
  • 75) 0.827 600 631 484 037 201 92 × 2 = 1 + 0.655 201 262 968 074 403 84;
  • 76) 0.655 201 262 968 074 403 84 × 2 = 1 + 0.310 402 525 936 148 807 68;
  • 77) 0.310 402 525 936 148 807 68 × 2 = 0 + 0.620 805 051 872 297 615 36;
  • 78) 0.620 805 051 872 297 615 36 × 2 = 1 + 0.241 610 103 744 595 230 72;
  • 79) 0.241 610 103 744 595 230 72 × 2 = 0 + 0.483 220 207 489 190 461 44;
  • 80) 0.483 220 207 489 190 461 44 × 2 = 0 + 0.966 440 414 978 380 922 88;
  • 81) 0.966 440 414 978 380 922 88 × 2 = 1 + 0.932 880 829 956 761 845 76;
  • 82) 0.932 880 829 956 761 845 76 × 2 = 1 + 0.865 761 659 913 523 691 52;
  • 83) 0.865 761 659 913 523 691 52 × 2 = 1 + 0.731 523 319 827 047 383 04;
  • 84) 0.731 523 319 827 047 383 04 × 2 = 1 + 0.463 046 639 654 094 766 08;
  • 85) 0.463 046 639 654 094 766 08 × 2 = 0 + 0.926 093 279 308 189 532 16;
  • 86) 0.926 093 279 308 189 532 16 × 2 = 1 + 0.852 186 558 616 379 064 32;
  • 87) 0.852 186 558 616 379 064 32 × 2 = 1 + 0.704 373 117 232 758 128 64;
  • 88) 0.704 373 117 232 758 128 64 × 2 = 1 + 0.408 746 234 465 516 257 28;
  • 89) 0.408 746 234 465 516 257 28 × 2 = 0 + 0.817 492 468 931 032 514 56;
  • 90) 0.817 492 468 931 032 514 56 × 2 = 1 + 0.634 984 937 862 065 029 12;
  • 91) 0.634 984 937 862 065 029 12 × 2 = 1 + 0.269 969 875 724 130 058 24;
  • 92) 0.269 969 875 724 130 058 24 × 2 = 0 + 0.539 939 751 448 260 116 48;
  • 93) 0.539 939 751 448 260 116 48 × 2 = 1 + 0.079 879 502 896 520 232 96;
  • 94) 0.079 879 502 896 520 232 96 × 2 = 0 + 0.159 759 005 793 040 465 92;
  • 95) 0.159 759 005 793 040 465 92 × 2 = 0 + 0.319 518 011 586 080 931 84;

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 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0110 1010 0011 0100 1111 0111 0110 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0110 1010 0011 0100 1111 0111 0110 100(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 63(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0110 1010 0011 0100 1111 0111 0110 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0001 0110 1010 0011 0100 1111 0111 0110 100(2) × 20 =

1.1000 1101 1001 0010 1000 1011 0101 0001 1010 0111 1011 1011 0100(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 1000 1011 0101 0001 1010 0111 1011 1011 0100


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 1000 1011 0101 0001 1010 0111 1011 1011 0100 =

1000 1101 1001 0010 1000 1011 0101 0001 1010 0111 1011 1011 0100


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 1000 1011 0101 0001 1010 0111 1011 1011 0100


Decimal number -0.000 000 000 000 176 557 63 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1000 1011 0101 0001 1010 0111 1011 1011 0100

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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